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Preface, We feel proud to present the revised edition of this well-acclaimed book,, which is in accordance with the current Board examination pattern and, includes the valuable feedback received from our esteemed readers., Though revised, the book retains its qualities which have made it, so popular among the teachers and students. It now has a large number, of questions from the NCERT textbook and previous years CBSE board, papers with full explanatory solutions. A large number of multiple-choice, questions (MCQs) on all topics have also been included., A separate exercise for Very-Short-Answer and Short-Answer, Questions has also been given at the end of each chapter, to provide extra, questions for ample practice., We hope that the present edition will be of immense help to students, who wish to sit for the CBSE class X board examination. We hope that we, shall continue to receive invaluable feedback from teachers and students, for the improvement of the book., —Authors, , (iii)
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Mathematics Syllabus, For Class 10, , Unit I: Number Systems, 1. Real Numbers, , (15 Periods), , Euclid’s division lemma, Fundamental Theorem of Arithmetic—, statements after reviewing work done earlier and after illustrating and, motivating through examples, proofs of irrationality of 2 , 3 , 5 ., Decimal representation of rational numbers in terms of terminating/, nonterminating recurring decimals., , Unit II: Algebra, 1. Polynomials, , (7 Periods), , Zeros of a polynomial. Relationship between zeros and coefficients of, quadratic polynomials. Statement and simple problems on division, algorithm for polynomials with real coefficients., 2. Pair of Linear Equations in Two Variables, , (15 Periods), , Pair of linear equations in two variables and graphical method of their, solution, consistency/inconsistency., Algebraic conditions for number of solutions. Solution of a pair of, linear equations in two variables algebraically—by substitution, by, elimination and by cross multiplication method. Simple situational, problems. Simple problems on equations reducible to linear equations., 3. Quadratic Equations, , (15 Periods), , Standard form of a quadratic equation ax bx c 0, (a ! 0) . Solutions, of quadratic equations (only real roots) by factorisation, by completing, the square and by using quadratic formula. Relationship between, discriminant and nature of roots., 2, , Situational problems based on quadratic equations related to day-to-day, activities to be incorporated., 4. Arithmetic Progressions, , (8 Periods), , Motivation for studying Arithmetic Progression. Derivation of the nth, term and sum of the first n terms of AP and their application in solving, daily life problems., (v)
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Unit III: Coordinate Geometry, 1. Lines (In two dimensions), , (14 Periods), , Review: Concepts of coordinate geometry, graphs of linear equations., Distance between two points. Section formula (internal division). Area, of a triangle., , Unit IV: Geometry, 1. Triangles, , (15 Periods), , Definitions, examples, counterexamples of similar triangles., 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect, the other two sides in distinct points, the other two sides are divided, in the same ratio., 2. (Motivate) If a line divides two sides of a triangle in the same ratio,, the line is parallel to the third side., 3. (Motivate) If in two triangles, the corresponding angles are equal,, their corresponding sides are proportional and the triangles, are similar., 4. (Motivate) If the corresponding sides of two triangles are, proportional, their corresponding angles are equal and the two, triangles are similar., 5. (Motivate) If one angle of a triangle is equal to one angle of another, triangle and the sides including these angles are proportional, the, two triangles are similar., 6. (Motivate) If a perpendicular is drawn from the vertex of the right, angle of a right triangle to the hypotenuse, the triangles on each, side of the perpendicular are similar to the whole triangle and to, each other., 7. (Prove) The ratio of the areas of two similar triangles is equal to the, ratio of the squares on their corresponding sides., 8. (Prove) In a right triangle, the square on the hypotenuse is equal to, the sum of the squares on the other two sides., 9. (Prove) In a triangle, if the square on one side is equal to the sum of, the squares on the other two sides, the angles opposite to the first, side is a right angle., 2. Circles, , (8 Periods), , Tangents to a circle at a point of contact., 1. (Prove) The tangent at any point of a circle is perpendicular to the, radius through the point of contact., (vi)
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2. (Prove) The lengths of tangents drawn from an external point to, circle are equal., 3. Constructions, , (8 Periods), , 1. Division of a line segment in a given ratio (internally)., 2. Tangent to a circle from a point outside it., 3. Construction of a triangle similar to a given triangle., , Unit V: Trigonometry, 1. Introduction to Trigonometry, , (10 Periods), , Trigonometric ratios of an acute angle of a right-angled triangle. Proof of, their existence (well defined): motivate the ratios, whichever are defined, at 0 and 90. Values (with proofs) of the trigonometric ratios of 30, 45, and 60. Relationships between the ratios., 2. Trigonometric Identities, , (15 Periods), , Proof and applications of the identity sin A cos A 1. Only simple, identities to be given. Trigonometric ratios of complementary angles., 2, , 3. Heights and Distances, , 2, , (8 Periods), , Simple problems on heights and distances. Problems should not involve, more than two right triangles. Angles of elevation/depression should, be only 30, 45 and 60., , Unit VI: Mensuration, 1. Areas Related to Circles, , (12 Periods), , Motivate the area of a circle; area of sectors and segments of a circle., Problems based on areas and perimeter/circumference of the abovesaid plane figures. (In calculating area of segment of a circle, problems, should be restricted to central angle of 60, 90 and 120 only. Plane, figures involving triangles, simple quadrilaterals and circle should, be taken.), 2. Surface Areas and Volumes, , (12 Periods), , (i) Surface areas and volumes of combinations of any two of the, following: cubes, cuboids, spheres, hemispheres and right-circular, cylinders/cones. Frustum of a cone., (ii) Problems involving converting one type of metallic solid into, another and other mixed problems. (Problems with combination of, not more than two different solids to be taken.), (vii)
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Unit VII: Statistics and Probability, 1. Statistics, , (18 Periods), , Mean, median and mode of grouped data (bimodal situations to be, avoided). Cumulative frequency graph., 2. Probability, , (10 Periods), , Classical definition of probability. Simple problems on single events, (not using set notation)., , , , (viii)
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Weightage, MATHEMATICS, CLASS 10, Time: 3 Hours, , Max. Marks: 80, , The weightage or the distribution of marks over different dimensions of, the question paper shall be as follows:, Weightage to Content/Subject Units, S. No., , Unit, , 1. Number Systems, , , 6, , Real Numbers, 20, , 2. Algebra, , , , Polynomials, Linear Equations in Two Variables, , , , Quadratic Equations, , , , Arithmetic Progression, , 3. Coordinate Geometry, , , 6, , Coordinate Geometry, 15, , 4. Geometry, , , , , Triangles, Circles, Constructions, 12, , 5. Trigonometry, , , , , , , Trigonometric Ratios, T-Ratios of Some Particular Angles, Trigonometric Identities, Trigonometric Ratios of Complementary Angles, Heights and Distances, , 6. Mensuration, , , , , Marks, , 10, , Perimeter and Area of Plane Figures, Area of Circle, Sector and Segment, Volume and Surface Areas of Solids, (ix)
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11, , 7. Statistics and Probability, , , , , Mean, Median, Mode of Grouped, Cumulative Frequency Graph and Ogive, , Data,, , Probability, Total, , 80, , , , (x)
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Contents, Number Systems, , 1. Real Numbers, , 1, , Algebra, , 2. Polynomials, , 43, , 3. Linear Equations in Two Variables, , 71, , 4. Quadratic Equations, , 165, , 5. Arithmetic Progression, , 244, , Coordinate Geometry, , 6. Coordinate Geometry, , 297, , Geometry, , 7. Triangles, , 351, , 8. Circles, , 461, , 9. Constructions, , 512, , Trigonometry, , 10. Trigonometric Ratios, , 528, , 11. T-Ratios of Some Particular Angles, , 546, , 12. Trigonometric Ratios of Complementary Angles, , 556, , 13. Trigonometric Identities, , 567, , 14. Heights and Distances, , 612, , Mensuration, , 15. Perimeter and Area of Plane Figures, , 654, , 16. Area of Circle, Sector and Segment, , 678, , 17. Volume and Surface Areas of Solids, , 742, , Statistics and Probability, , 18. Mean, Median, Mode of Grouped Data, Cumulative, Frequency Graph and Ogive, (xi), , 826
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19. Probability, , 887, , Sample Paper I, , 929, , Sample Paper II, , 934, , , (xii)
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Real Numbers, , 1, , In class IX we studied about real numbers, especially about irrational, numbers. In this chapter, we shall continue our discussion on real numbers., We begin with two important results, namely Euclid’s division lemma and, the fundamental theorem of arithmetic., LEMMA, , A lemma is a proven statement used for proving another statement., , EUCLID’S DIVISION LEMMA, , For any two given positive integers a and b, there exist unique whole numbers q, and r such that, [CBSE 2009C], a = bq + r, where 0 # r < b., Here, we call a as dividend, b as divisor, q as quotient and r as remainder., Dividend = (divisor × quotient) + remainder., Example, , Suppose we divide 117 by 14. Then, we get, 8 as quotient and 5 as remainder., Here dividend = 117, divisor = 14, quotient = 8, and remainder = 5., , 14) 117 (8, –112, 5, , Clearly, 117 = (14 # 8) + 5., EXAMPLE 1, , A number when divided by 73 gives 34 as quotient and 23 as, remainder. Find the number., , SOLUTION, , Here divisor = 73, quotient = 34 and remainder = 23., By Euclid's division lemma, we have, dividend = (divisor × quotient) + remainder, = (73 # 34) + 23, = (2482 + 23) = 2505., Hence, the required number is 2505., , An algorithm is a series of well-defined steps which gives a method, for solving a certain type of problem., ALGORITHM, , It is a technique to compute the HCF of two given, positive integers, say a and b with a > b, in the following steps., EUCLID’S DIVISION ALGORITHM, , 1
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2, , Secondary School Mathematics for Class 10, , Step 1., , On dividing a by b, we get the quotient q and remainder r such that, a bq r, where 0 # r b., , Step 2., , If r 0 then HCF (a, b) b., If r ! 0 then apply the division lemma to b and r., , Step 3., , Continue the process till the remainder is 0., The last divisor will be the required HCF., , EXAMPLE 2, , Use Euclid‘s algorithm to find the HCF of 272 and 1032., , SOLUTION, , We find HCF(272, 1032) using the following steps., Since 1032 > 272, we 272) 1032 (3, divide 1032 by 272 to, 816, get 3 as quotient and, 216) 272 (1, 216 as remainder., 216, So, by Euclid‘s division, 56) 216 (3, lemma, we get, 168, 1032 = 272 # 3 + 216., 48) 56 (1, Step 2. Since the remainder, 48, 216 ! 0, we divide, 8) 48 (6, 272 by 216 to get 1 as, 48, quotient and 56 as, 0, remainder., , by Euclid‘s division lemma, we get, Step 1., , 272 = 216 # 1 + 56., Step 3., , Since the remainder 56 ! 0, we divide 216 by 56 to get, 3 as quotient and 48 as remainder., , , , , , Step 4., , Since the remainder 48 ! 0, we divide 56 by 48 to get, 1 as quotient and 8 as remainder., , , Step 5., , by Euclid‘s division lemma, we get, 216 = 56 # 3 + 48., , by Euclid‘s division lemma, we get, 56 = 48 # 1 + 8., , Since the remainder 8 ! 0, we divide 48 by 8 to get 6, as quotient and 0 as remainder., , , by Euclid‘s division lemma, we get, 48 = 8 # 6 + 0.
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Real Numbers, , 3, , The remainder has now become 0, so our procedure stops., Hence, HCF(272, 1032) = 8., Note, , 8 = HCF(48, 8) = HCF(56, 48) = HCF(216, 56) = HCF(272, 216) = HCF(1032, 272)., , REMARK, , This method is also known as successive division method., , EXAMPLE 3, , Use Euclid‘s algorithm to find HCF(196, 38220)., , SOLUTION, , We find HCF(196, 38220), using the following steps., Since 38220 > 196, we divide 38220 by 196, to get 195 as quotient and 0 as remainder., , , by Euclid‘s division lemma, we get, 38220 = 196 # 195 + 0., , Since the remainder is 0, so our procedure, stops., , , HCF(196, 38220) = 196., , 196) 38220 (195, 196, 1862, 1764, 980, 980, 0, , EXAMPLE 4, , Use Euclid‘s algorithm to find HCF of 1651 and 2032. Express the, HCF in the form 1651m 2032n., , SOLUTION, , We find HCF(1651, 2032) using the following steps., Since 2032 > 1651, we divide, 1651) 2032 (1, 2032 by 1651 to get 1 as quotient, 1651, and 381 as remainder., 381) 1651 (4, by Euclid‘s division lemma,, 1524, we get, 127) 381 (3, … (i), 2032 = 1651 # 1 + 381., 381, Step 2. Since the remainder 381 ! 0, we, 0, divide 1651 by 381 to get 4 as, quotient and 127 as remainder., Step 1., , , , by Euclid‘s division lemma, we get, 1651 = 381 # 4 + 127., , Step 3., , … (ii), , Since the remainder 127 ! 0, we divide 381 by 127 to, get 3 as quotient and 0 as remainder., , , by Euclid‘s division lemma, we get, 381 127 # 3 0., , … (iii), , The remainder is now 0, so our procedure stops., , , HCF(1651, 2032) = 127.
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4, , Secondary School Mathematics for Class 10, , Now, from (ii), we get, 1651 381# 4 127, &, &, &, , 127 1651 381# 4, 127 1651 (2032 1651#1)# 4 [from (i)], 127 1651 2032 # 4 1651# 4, , 127 1651# 5 2032 #( 4)., Hence, m 5, n 4., , &, , SOME APPLICATIONS OF EUCLID’S DIVISION LEMMA, EXAMPLE 1, , Show that every positive even integer is of the form 2m and that every, positive odd integer is of the form (2m + 1), where m is some integer., , SOLUTION, , Let n be an arbitrary positive integer., On dividing n by 2, let m be the quotient and r be the remainder., Then, by Euclid‘s division lemma, we have, n = 2m + r, where 0 # r < 2., , , n = 2m or (2m + 1), for some integer m., , Case I, , When n = 2m., In this case, n is clearly even., , Case II When n = 2m + 1., , In this case, n is clearly odd., Hence, every positive even integer is of the form 2m and every, positive odd integer is of the form (2m + 1) for some integer m., EXAMPLE 2, , Show that any positive integer is of the form 3m or (3m + 1) or, (3m + 2) for some integer m., , SOLUTION, , Let n be an arbitrary positive integer., On dividing n by 3, let m be the quotient and r be the remainder., Then, by Euclid‘s division lemma, we have, n = 3m + r, where 0 # r < 3., , , n = 3m or (3m + 1) or (3m + 2), for some int eger m., , Thus, any positive integer is of the form 3m or (3m 1) or, (3m + 2) for some integer m., EXAMPLE 3, , Show that any positive odd integer is of the form (4m + 1) or (4m + 3), for some integer m., , SOLUTION, , Let n be an arbitrary odd positive integer.
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Real Numbers, , 5, , On dividing n by 4, let m be the quotient and r be the remainder., So, by Euclid‘s division lemma, we have, n = 4m + r, where 0 # r < 4., , , n = 4m or (4m + 1) or (4m + 2) or (4m + 3) ., , Clearly, 4m and (4m + 2) are even and since n is odd, so n ! 4m, and n ! (4m 2) ., , , n (4m 1) or (4m 3), for some integer m., , Hence, any positive odd integer is of the form (4m 1) or, (4m 3) for some integer m., EXAMPLE 4, , SOLUTION, , Show that every positive odd integer is of the form (6m 1) or, (6m 3) or (6m 5) for some integer m., Let n be a given positive odd integer., On dividing n by 6, let m be the quotient and r be the remainder., Then, by Euclid‘s division lemma, we have, n 6m r, where 0 # r 6, n 6m r, where r 0, 1, 2, 3, 4, 5, n 6m or (6m 1) or (6m 2) or (6m 3) or (6m 4), or (6m + 5) ., But, n = 6m, (6m + 2), (6m + 4) give even values of n., Thus, when n is odd, it is of the form (6m + 1) or (6m + 3) or, (6m 5) for some integer m., , EXAMPLE 5, , Using Euclid‘s division lemma, show that the square of any positive, integer is either of the form 3m or (3m 1) for some integer m., [CBSE 2008], , SOLUTION, , Let n be an arbitrary positive integer., On dividing n by 3, let q be the quotient and r be the remainder., Then, by Euclid‘s division lemma, we have, n = 3q + r, where 0 # r < 3., , , n 2 = 9q 2 + r 2 + 6qr, , Case I, , … (i), where 0 # r < 3., , When r 0., Putting r = 0 in (i), we get, n 2 = 9q 2 = 3 (3q 2) = 3m, where m = 3q 2 is an integer.
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6, , Secondary School Mathematics for Class 10, Case II When r 1., , Putting r = 1 in (i), we get, n 2 = (9q 2 + 1 + 6q) = 3 (3q 2 + 2q) + 1 = 3 (3q 2 + 2q) + 1, = 3m + 1, where m = (3q 2 + 2q) is an integer., Case III When r 2., , Putting r = 2 in (i), we get, n 2 = (9q 2 + 4 + 12q) = 3 (3q 2 + 4q + 1) + 1, = 3m + 1, where m = (3q 2 + 4q + 1) is an int eger., Hence, the square of any positive integer is of the form 3m or, (3m 1) for some integer m., EXAMPLE 6, , Using Euclid‘s division lemma, show that the cube of any positive, integer is of the form 9m or (9m 1) or (9m 8) for some integer m., [CBSE 2009C], , SOLUTION, , Let n be an arbitrary positive integer., On dividing n by 3, let q be the quotient and r be the remainder., So, by Euclid‘s division lemma, we have, n = 3q + r, where 0 # r < 3., , , n 3 = (3q + r) 3 = 27q 3 + r 3 + 9qr (3q + r), = (27q 3 + 27q 2 r + 9qr 2) + r 3, , Case I, , … (i), where 0 # r 3., , When r 0., Putting r 0 in (i), we get, n 3 = 27q 3 = 9 (3q 3) = 9m, where m = 3q 3 is an integer., , Case II When r 1., , Putting r 1 in (i), we get, n 3 = (27q 3 + 27q 2 + 9q) + 1 = 9q (3q 2 + 3q + 1) + 1, = 9m + 1, where m = q (3q 2 + 3q + 1) is an int eger., Case III When r 2., , Putting r 2 in (i), we get, n 3 = (27q 3 + 54q 2 + 36q) + 8 = 9q (3q 2 + 6q + 4) + 8, = 9m + 8, where m = q (3q 2 + 6q + 4) is an integer., Hence, the cube of any positive integer is of the form 9m or, (9m + 1) or (9m + 8) for some integer m.
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Real Numbers, EXAMPLE 7, , SOLUTION, , 7, , Show that one and only one out of n, (n 1) and (n 2) is divisible, by 3, where n is any positive integer., On dividing n by 3, let q be the quotient and r be the remainder., Then, n 3q r, where 0 # r < 3, , , n = 3q + r, where r = 0, 1 or 2, , , , n = 3q or n = (3q + 1) or n = (3q + 2) ., , Case I, , If n 3q then n is clearly divisible by 3., , Case II If n (3q 1) then (n 2) (3q 3) 3 (q 1), which is, , clearly divisible by 3., In this case, (n 2) is divisible by 3., Case III If n (3q 2) then (n 1) (3q 3) 3 (q 1), which is, , clearly divisible by 3., In this case, (n 1) is divisible by 3., Hence, one and only one out of n, (n 1) and (n 2) is divisible, by 3., EXAMPLE 8, , SOLUTION, , Show that one and only one out of n, n 2, n 4 is divisible by 3,, where n is any positive integer., [CBSE 2008C], On dividing n by 3, let q be the quotient and r be the remainder., Then, n = 3q + r, where 0 # r < 3, , , n = 3q + r, where r = 0, 1, 2, , , , n = 3q or n = 3q + 1 or n = 3q + 2., , Case I, , If n = 3q then n is divisible by 3., , Case II, , If n 3q 1 then (n 2) 3q 3 3 (q 1), which is, divisible by 3., So, in this case, (n 2) is divisible by 3., , Case III When n 3q 2 then (n 4) 3q 6 3 (q 2), which, , is divisible by 3., So, in this case, (n 4) is divisible by 3., Hence, one and only one out of n, n 2, n 4 is divisible by 3., EXAMPLE 9, SOLUTION, , If n is an odd integer then show that n 2 1 is divisible by 8., We know that every odd integer is of the form 4q 1 or 4q 3, for some integer q.
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8, , Secondary School Mathematics for Class 10, Case I, , When n 4q 1., Then, n 2 1 (4q 1) 2 1, 16q 2 8q 1 1, 16q 2 8q, 8 (q 2q 2), which is divisible by 8., , Case II When n 4q 3., , Then, n 2 1 (4q 3) 2 1, 16q 2 24q 9 1, 16q 2 24q 8, 8(2q 2 3q 1), which is divisible by 8., Hence, if n is an odd integer then n 2 1 is divisible by 8., f, , EXERCISE 1A, , 1. What do you mean by Euclid‘s division lemma?, 2. A number when divided by 61 gives 27 as quotient and 32 as remainder., Find the number., 3. By what number should 1365 be divided to get 31 as quotient and 32 as, remainder?, 4. Using Euclid’s division algorithm, find the HCF of, (i) 405 and 2520, , (ii) 504 and 1188, , (iii) 960 and 1575., , 5. Show that every positive integer is either even or odd., 6. Show that any positive odd integer is of the form (6m + 1) or (6m + 3), or (6m + 5), where m is some integer., 7. Show that any positive odd integer is of the form (4m + 1) or (4m + 3),, where m is some integer., 8. For any positive integer n, prove that n 3 n is divisible by 6., 9. Prove that if x and y are both odd positive integers then x 2 y 2 is even, but not divisible by 4., 10. Use Euclid’s algorithm to find HCF of 1190 and 1445. Express the HCF, in the form 1190m 1445n., ANSWERS (EXERCISE 1A), , 2. 1679, , 3. 43, , 10. 85; m 6, n 5, , 4. (i) 45, , (ii) 36, , (iii) 15
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Real Numbers, , 9, , HINTS TO SOME SELECTED QUESTIONS, 3. Let the required divisor be x., Then, dividend = (divisor × quotient) + remainder, , , 1365 = (x # 31) + 32. Find x., , 5. Let n be an arbitrary positive integer., On dividing n by 2, let m be the quotient and r be the remainder., Then, n = 2m + r, where 0 # r < 2 [by Euclid’s division lemma]., , , n = 2m or n = 2m + 1 for some integer m., , So, n is either even or odd., , Every composite number can be uniquely, expressed as a product of primes, except for the order in which these prime, factors occurs., , FUNDAMENTAL THEOREM OF ARITHMETIC, , Examples, , (i) 12 = 2 # 2 # 3, , (ii) 69 = 3 # 23, , (iii) 105 = 3 # 5 # 7, , (iv) 234 = 2 # 3 # 3 # 13 (v) 462 = 2 # 3 # 7 # 11, (vi) 651 = 3 # 7 # 31, The above factorisation can easily be verified by actual division., , SOLVED EXAMPLES, EXAMPLE 1, , Show that each of the following is a composite number:, (i) 5 # 11 # 13 + 13, , SOLUTION, , (ii) 6 # 5 # 4 # 3 # 2 # 1 + 5, , We have, (i) 5 # 11 # 13 + 13 = 13 # (5 # 11 + 1) = (13 # 56) ., Clearly, it shows that the given number has more than, two factors. Hence, it is a composite number., (ii) 6 # 5 # 4 # 3 # 2 # 1 + 5 = 5 # (6 # 4 # 3 # 2 # 1 + 1), = (5 # 145) ., Clearly, it shows that the given number has more than, two factors. Hence, it is a composite number., , EXAMPLE 2, , Show that any number of the form 4 n, nd N can never end with the, digit 0.
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10, SOLUTION, , Secondary School Mathematics for Class 10, , If 4 n ends with 0 then it must have 5 as a factor., But, 4 n (2 2) n 2 2n shows that 2 is the only prime factor, of 4 n ., Also, we know from the fundamental theorem of arithmetic, that the prime factorisation of each number is unique., So, 5 is not a factor of 4 n ., Hence, 4 n can never end with the digit 0., , EXAMPLE 3, , Show that any number of the form 6 n, where nd N can never end, with the digit 0., , SOLUTION, , If 6 n ends with 0 then it must have 5 as a factor., But, 6 n (2 # 3) n (2 n # 3 n) shows that 2 and 3 are the only, prime factors of 6 n ., Also, we know from the fundamental theorem of arithmetic, that the prime factorisation of each number is unique., So, 5 is not a factor of 6 n ., Hence, 6 n can never end with the digit 0., , EXAMPLE 4, , Find the HCF and LCM of 126 and 156 using prime factorisation, method., , SOLUTION, , We have, , , , 126 = (2 # 3 # 3 # 7) = (2 # 3 2 # 7), , and 156 = (2 # 2 # 3 # 13) = (2 2 # 3 # 13) ., , , HCF(126, 156) = product of common terms with lowest, , power, = (21 # 31) = (2 # 3) = 6
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Real Numbers, , 11, , and LCM(126, 156) = product of prime factors with highest, power, = (2 2 # 3 2 # 7 # 13) = (4 # 9 # 7 # 13), 3276., , , HCF = 6 and LCM = 3276., , EXAMPLE 5, , Find the HCF and LCM of 612 and 1314 using prime factorisation, method., , SOLUTION, , We have, , , , 612 = (2 # 2 # 3 # 3 # 17) = (2 2 # 3 2 # 17), , and 1314 = (2 # 3 # 3 # 73) = (2 # 3 2 # 73) ., , , HCF(612, 1314) = product of common terms with lowest, , power, = (2 # 3 2) = (2 # 9) = 18, and LCM(612, 1314) = product of prime factors with highest, power, = (2 2 # 3 2 # 17 # 73) = (4 # 9 # 17 # 73), = 44676., Hence, HCF = 18 and LCM = 44676., AN IMPORTANT PROPERTY, , Product of two given numbers = product of their HCF and LCM., , Thus, (a # b) = HCF (a, b) # LCM (a, b) ., CAUTION, , The above result is true for two numbers only.
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12, , Secondary School Mathematics for Class 10, , EXAMPLE 6, , The HCF of two numbers is 23 and their LCM is 1449. If one of the, numbers is 161, find the other., , SOLUTION, , For two numbers a and b, we know that, (a # b) = {HCF(a, b)} # {LCM(a, b)} ., Here a = 161, HCF = 23 and LCM = 1449., And, we have to find b., , , (161 # b) (23 # 1449) & b , , (23 # 1449), 207., 161, , Hence, the other number is 207., EXAMPLE 7, , Given that HCF(252, 594) = 18, find LCM(252, 594)., , SOLUTION, , We have, product of two given numbers, their HCF, (252 # 594), =, = 8316., 18, Hence, LCM(252, 594) = 8316., LCM =, , EXAMPLE 8, , 148, Find the simplest form of 185 ·, , SOLUTION, , We find HCF(148, 185), which is 37., So, we divide the numerator and denominator, of the given fraction by 37., , , 148) 185 (1, 148, , 148 = 148 ' 37 = 4 ·, 185 185 ' 37 5, , 4, Hence, the simplest form of the given fraction is 5 ·, , 37) 148 (4, 148, 0, , EXAMPLE 9, , Find the HCF and LCM of 108, 120 and 252 using prime factorisation, method., , SOLUTION, , By prime factorisation, we get, 108 = (2 2 # 3 3), 120 = (2 3 # 3 # 5), 2, , 2, , 252 = (2 # 3 # 7), , 2 108, , 2 120, , 2 252, , 2, , 54, , 2, , 60, , 2 126, , 3, , 27, , 2, , 30, , 3, , 63, , 3, , 9, , 3, , 15, , 3, , 21, , 3, , 5, , 7, , HCF(108, 120, 252) = product of common terms with lowest, , power, = (2 2 # 3) = (4 # 3) = 12.
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Real Numbers, , 13, , LCM(108, 120, 252) = product of prime factors with highest, , power, = (2 3 # 3 3 # 5 # 7) = 7560., , EXAMPLE 10, , SOLUTION, , HCF = 12 and LCM = 7560., , Find the largest number which divides 245 and 1037, leaving, remainder 5 in each case., Clearly, the required number divides (245 5) = 240 and, (1037 5) = 1032 exactly., So, the required number is HCF(240, 1032). 2 240, 2 120, Now, 240 = (2 4 # 3 # 5), and 1032 = (2 3 # 3 # 43) ., , , HCF(240, 1032) = (2 3 # 3) = 24., , 2, 2, 3, , Hence, the required number is 24., EXAMPLE 11, , SOLUTION, , 60, 30, 15, 5, , 2, 2, 2, 3, , 1032, 516, 258, 129, 43, , Find the largest number which divides 129 and 545, leaving, remainders 3 and 5 respectively., Clearly, the required number divides (129 3) = 126 and, (545 – 5) = 540 exactly., 2 126, 2 540, required number = HCF(126, 540)., 3 63, 2 270, Now, 126 = (2 # 3 # 3 # 7) = (2 # 3 2 # 7), and 540 = (2 # 2 # 3 # 3 # 3 # 5), , 3, , 21, 7, , = (2 2 # 3 3 # 5) ., , , 3 135, 3 45, 3 15, 5, , HCF(126, 540) = product of common terms with lowest, , power, = (2 # 3 2) = (2 # 9) = 18., Hence, the required number is 18., EXAMPLE 12, , SOLUTION, , Find the largest number that will divide 398, 436 and 542, leaving, remainders 7, 11 and 15 respectively., Clearly, the required number divides (398 7) = 391,, (436 11) = 425 and (542 15) = 527 exactly., , , required number = HCF(391, 425, 527). 17 391, , Now, 391 = (17 # 23),, 425 = (5 2 # 17),, 527 = (17 # 31) ., , 23, , 5 425, 5 85, 17
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14, , Secondary School Mathematics for Class 10, , , , HCF(391, 425, 527) = 17., , Hence, the required number is 17., EXAMPLE 13, , Two tanks contain 504 and 735 litres of milk respectively. Find the, maximum capacity of a container which can measure the milk of, either tank an exact number of times., , SOLUTION, , Resolving 504 and 735 into prime, factors, we get, 504 = (2 3 # 3 2 # 7), and 735 = (5 # 3 # 7 2) ., , 2 504, 2 252, 2 126, 3 63, 3 21, 7, , , , HCF (504, 735) = (3 # 7) = 21., , , , capacity of the required container = 21 litres., , 5 735, 3 147, 7 49, 7, , EXAMPLE 14, , An army contingent of 612 members is to march behind an army, band of 48 members in a parade. The two groups are to march in the, same number of columns. What is the maximum number of columns, in which they can march?, , SOLUTION, , Clearly, the maximum number of, columns = HCF(612, 48)., , 2 612, 2 306, 3 153, and, 48 = (2 4 # 3) ., 3 51, 2, HCF(612, 48) = (2 # 3) = (4 # 3) = 12., 17, , Now, 612 = (2 2 # 3 2 # 17), , , , required number of columns = 12., , 2 48, 2 24, 2 12, 2 6, 3, , EXAMPLE 15, , A sweetseller has 420 kaju burfis and 150 badam burfis. He wants, to stack them in such a way that each stack has the same number,, and they take up the least area of the tray. How many of these can be, placed in each stack? How many stacks are formed?, , SOLUTION, , Maximum number of burfis in each, stack = HCF(420, 150)., Now, 420 = (2 2 # 3 # 5 # 7), and, , 150 = (2 # 3 # 5 2) ., , 2 420, 2 210, 3 105, 5 35, 7, , , , HCF(420, 150) (2 # 3 # 5) 30., , , , maximum number of burfis in each stack = 30., , , , 420 150, number of stacks = a 30 + 30 k = (14 + 5) = 19., , 2 150, 3 75, 5 25, 5
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Real Numbers, , 15, , EXAMPLE 16, , Ravi and Sikha drive around a circular sports field. Ravi takes, 16 minutes to take one round, while Sikha completes the round in, 20 minutes. If both start at the same point, at the same time and, go in the same direction, after how much time will they meet at the, starting point?, , SOLUTION, , Required number of minutes = LCM(16, 20)., Now, 16 = 2 4 and 20 = (2 2 # 5) ., , , LCM(16, 20) (2 4 # 5) (16 # 5) 80., , Hence, both will meet at the starting point after 80 minutes., EXAMPLE 17, , In a school there are two sections, namely A and B, of class X. There, are 30 students in section A and 28 students in section B. Find the, minimum number of books required for their class library so that, they can be distributed equally among students of section A or, section B., , SOLUTION, , Clearly, the required number of books are to be distributed, equally among the students of section A or B., So, the number of these books must be a multiple of 30 as well, as that of 28., Consequently, the required number is LCM(30, 28)., Now, 30 = 2 # 3 # 5, and, , , 28 = 2 2 # 7., LCM(30, 28) = product of prime factors with highest power, , = (2 2 # 3 # 5 # 7) = (4 # 3 # 5 # 7) = 420., Hence, the required number of books = 420., f, , EXERCISE 1B, , 1. Using prime factorisation, find the HCF and LCM of:, (i) 36, 84, (iv) 144, 198, , (ii) 23, 31, , (iii) 96, 404, , (v) 396, 1080, , (vi) 1152, 1664, , In each case, verify that:, HCF × LCM = product of given numbers., , 2. Using prime factorisation, find the HCF and LCM of:, (i) 8, 9, 25, , (ii) 12, 15, 21, , (iii) 17, 23, 29, , (iv) 24, 36, 40, , (v) 30, 72, 432, , (vi) 21, 28, 36, 45
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16, , Secondary School Mathematics for Class 10, , 3. The HCF of two numbers is 23 and their LCM is 1449. If one of the, numbers is 161, find the other., 4. The HCF of two numbers is 145 and their LCM is 2175. If one of the, numbers is 725, find the other., 5. The HCF of two numbers is 18 and their product is 12960. Find their, LCM., 6. Is it possible to have two numbers whose HCF is 18 and LCM is 760?, Give reason., HINT, , HCF always divides LCM completely., , 7. Find the simplest form of:, 69, 473, (i) 92, (ii) 645, , 1095, (iii) 1168, , 368, (iv) 496, , 8. Find the largest number which divides 438 and 606, leaving remainder, 6 in each case., 9. Find the largest number which divides 320 and 457, leaving remainders, 5 and 7 respectively., 10. Find the least number which when divided by 35, 56 and 91 leaves the, same remainder 7 in each case., 11. Find the smallest number which when divided by 28 and 32 leaves, remainders 8 and 12 respectively., 12. Find the smallest number which when increased by 17 is exactly, divisible by both 468 and 520., 13. Find the greatest number of four digits which is exactly divisible by 15,, 24 and 36., 14. Find the largest four-digits number which when divided by 4, 7 and 13, leaves a remainder of 3 in each case., 15. Find the least number which should be added to 2497 so that the sum, is exactly divisible by 5, 6, 4 and 3., 16. Find the greatest number that will divide 43, 91 and 183 so as to leave, the same remainder in each case., 17. Find the least number which when divided by 20, 25, 35 and 40 leaves, remainders 14, 19, 29 and 34 respectively., 18. In a seminar, the number of participants in Hindi, English and, mathematics are 60, 84 and 108 respectively. Find the minimum number, of rooms required, if in each room, the same number of participants are, to be seated and all of them being in the same subject.
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Real Numbers, , 17, , 19. Three sets of English, mathematics and science books containing 336,, 240 and 96 books respectively have to be stacked in such a way that, all the books are stored subjectwise and the height of each stack is the, same. How many stacks will be there?, 20. Three pieces of timber 42 m, 49 m and 63 m long have to be divided into, planks of the same length. What is the greatest possible length of each, plank? How many planks are formed?, 21. Find the greatest possible length which can be used to measure exactly, the lengths 7 m, 3 m 85 cm and 12 m 95 cm., 22. Find the maximum number of students among whom 1001 pens and, 910 pencils can be distributed in such a way that each student gets the, same number of pens and the same number of pencils., 23. Find the least number of square tiles required to pave the ceiling of a, room 15 m 17 cm long and 9 m 2 cm broad., 24. Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the, least length of cloth that can be measured an exact number of times,, using any of the rods., 25. An electronic device makes a beep after every 60 seconds. Another, device makes a beep after every 62 seconds. They beeped together at, 10 a.m. At what time will they beep together at the earliest?, 26. The traffic lights at three different road crossings change after every, 48 seconds, 72 seconds and 108 seconds respectively. If they all change, simultaneously at 8 a.m. then at what time will they again change, simultaneously?, 27. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8,, 10, 12 minutes respectively. In 30 hours, how many times do they toll, together?, ANSWERS (EXERCISE 1B), , 1. (i) HCF = 12, LCM = 252, , (iii) HCF = 4, LCM = 9696, , (ii) HCF = 1, LCM = 713, (iv) HCF = 18, LCM = 1584, , (v) HCF = 36, LCM = 11880 (vi) HCF = 128, LCM = 14976, 2. (i) HCF = 1, LCM = 1800, , (ii) HCF = 3, LCM = 420, , (iii) HCF = 1, LCM = 11339 (iv) HCF = 4, LCM = 360, (v) HCF = 6, LCM = 2160, 3. 207, , 4. 435, , 5. 720, , (vi) HCF = 1, LCM = 1260
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18, , Secondary School Mathematics for Class 10, , 6. No, since HCF does not divide LCM exactly, , 3, , 7. (i) 4, , 11, (ii) 15, , 15, (iii) 16, , 23, (iv) 31, , 8. 24, , 9. 45, , 10. 3647, , 11. 204, , 12. 4663, , 13. 9720, , 14. 9831, , 15. 23, , 16. 4, , 17. 1394, , 18. 21, , 19. 14, 24. 9.6 m, , 20. 7 m, 22 planks, , 21. 35 cm 22. 91, , 23. 814, , 25. 10 : 31 hrs, , 26. 8 : 7 : 12 hrs, , 27. 16 times, , HINTS TO SOME SELECTED QUESTIONS, 8. Required number = HCF(432, 600) = 24., 9. Required number = HCF(315, 450) = 45., 10. Required number = {LCM(35, 56, 91) + 7} = (3640 + 7) = 3647., 11. Here, (28 8) = 20 and (32 12) = 20., , , required number = {LCM (28, 32) 20} = (224 20) = 204., , 12. Required number = {LCM (468, 520) 17} = (4680 17) = 4663., 13. Greatest number of four digits = 9999., LCM(15, 24, 36) = 360., , On dividing 9999 by 360, remainder = 279., , , required number = (9999 279) = 9720., , 14. Required number = (greatest number of 4-digits divisible by 4, 7 and 13) + 3., 15. LCM of 5, 6, 4 and 3 = 60., On dividing 2497 by 60, remainder = 37., , , number to be added (60 37) 23., , 16. Required number = HCF of (91 43), (183 91) and (183 43), i.e., HCF of 48, 92, and 140., 17. Here, (20 14) 6, (25 19) 6, (35 29) 6, (40 34) 6., , , required number = (LCM of 20, 25, 35, 40) – 6., , 18. Maximum number of participants in each room = HCF(60, 84, 108) = 12., 60 84 108, Minimum number of rooms required = a 12 + 12 + 12 k = 21., 19. HCF(336, 240, 96) = 48. So, we make stacks of 48 books each., 336 240 96, Number of stacks = a 48 + 48 + 48 k = (7 + 5 + 2) = 14., 20. Required length of each plank in metres = HCF(42 m, 49 m, 63 m) = 7 m., 42 49 63, Number of planks = a 7 + 7 + 7 k = (6 + 7 + 9) = 22., 21. Required length = HCF(700 cm, 385 cm, 1295 cm) = 35 cm.
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Real Numbers, , 19, , 22. Maximum number of students = HCF(1001, 910) = 91., 23. Side of each square tile = HCF(1517 cm, 902 cm) = 41 cm., 1517 # 902, Required number of tiles = a 41 # 41 k = 814., 24. Required length = LCM(64 cm, 80 cm, 96 cm) = 960 cm = 9.6 m., 25. Interval of beeping together = LCM (60 seconds, 62 seconds) = 1860 seconds, = 1860 min = 31 min ., 60, So, they will beep together again at 10 : 31 a.m., 26. Interval of change = LCM (48 seconds, 72 seconds, 108 seconds) = 432 seconds, = 7 min 12 seconds., Required time of simultaneous change = 8 : 7 : 12 hours., 27. LCM of 2, 4, 6, 8, 10, 12 = 120., After every 2 hours they toll together., 30, Required number of times = a 2 + 1 k times = 16 times., , RATIONAL NUMBERS, NATURAL NUMBERS, , Counting numbers 1, 2, 3, 4, …, etc., are known as natural, , numbers., WHOLE NUMBERS, , All counting numbers together with 0 form the collection of, , whole numbers., Thus, 0, 1, 2, 3, 4, 5, …, etc., are whole numbers., All counting numbers, negatives of counting numbers and 0 form the, collection of all integers., , INTEGERS, , Thus, …, –4, –3, –2, –1, 0, 1, 2, 3, …, etc., are integers., p, RATIONAL NUMBERS The numbers of the form, q , where p and q are integers, and, q ! 0 are called rational numbers., Every rational number when expressed in, decimal form is expressible either in terminating or in nonterminating repeating, decimal form., , RATIONAL NUMBERS IN DECIMAL FORM, , AN IMPORTANT OBSERVATION, , To Test Whether a Given Rational Number is a Terminating or Repeating Decimal, , p, Let x be a rational number whose simplest form is q , where p and q are integers, and q ! 0. Then,
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20, , Secondary School Mathematics for Class 10, , (i) x is a terminating decimal only when q is of the form (2 m # 5 n) for some, non-negative integers m and n., (ii) x is a nonterminating repeating decimal, if q ! (2 m # 5 n) ., AN IMPORTANT TEST, , p, Let q be the simplest form of a given rational number., p, (i) If q = (2 m # 5 n) for some non-negative integers m and n then q, is a terminating decimal., , p, , (ii) If q ! (2 m # 5 n) then q is a nonterminating repeating decimal., , SOLVED EXAMPLES, EXAMPLE 1, , Without actual division, show that each of the following rational, numbers is a terminating decimal. Express each in decimal form., (i), , SOLUTION, , 31, (2 2 # 5 3), , 33, (ii) 50, , (i) The given number is, , 41, (iii) 1000, , 17, (iv) 625, , 31 ·, (2 2 # 5 3), , Clearly, none of 2 and 5 is a factor of 31., So, the given rational is in its simplest form., Clearly, (2 2 # 5 3) is of the form (2 m # 5 n) ., So, the given number is a terminating decimal., 31, 62, 31 # 2 , 62 62, Now, 2, (2 # 5 3) (2 3 # 5 3) (2 # 5) 3 (10) 3 1000, 0.062., 33, (ii) The given number is 50 ·, Now, 50 = (2 # 5 2) and none of 2 and 5 is a factor of 33., So, the given rational number is in its simplest form., Clearly, 50 = (2 # 5 2) = (2 m # 5 n), where m = 1 and n = 2., So, the given number is a terminating decimal., 33, 33 = 33 # 2 = 66, Now, 50 =, (2 # 5 2) (2 2 # 5 2) (2 # 5) 2, = 66 2 = 66 = 0.66., 100, (10)
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Real Numbers, , 21, , 41, (iii) The given number is 1000 ·, Now, 1000 = (8 # 125) = (2 3 # 5 3) ., Clearly, none of 2 and 5 is a factor of 41., So, the given number is in its simplest form., Now, 1000 = (2 3 # 5 3) which is of the form (2 m # 5 n),, where m = 3 and n = 3., So, the given number is a terminating decimal., 41, And, 1000 = 0.041., 17, (iv) The given number is 625 ·, And, 625 = 5 4 and 5 is not a factor of 17., So, the given number is in its simplest form., Now, 625 = 5 4 is of the form (2 m # 5 n), where m = 0, n = 4., So, the given number is a terminating decimal., 17, 17 17 # 2 4 = 17 # 16 = 272, Now, 625 = 4 = 4, 5, 5 # 2 4 (5 # 2) 4 (10) 4, = 272 = 0.0272., 10000, EXAMPLE 2, , SOLUTION, , Without actual division, show that each of the following rational, numbers is a nonterminating repeating decimal., 66, 17, 121, 53, (i) 3, (ii) 90, (iii) 343, (iv) 180, (2 # 3 2 # 7 5), (i) Given number is, , 121, ·, (2 3 # 3 2 # 7 5), , Clearly, none of 2, 3 and 7 is a factor of 121., So, the given rational number is in its simplest form., And, (2 3 # 3 2 # 7 5) ! (2 m # 5 n) ., , , 121, is a nonterminating repeating decimal., (2 3 # 3 2 # 7 5), , 17, (ii) Given number is 90 ·, And, 90 = (2 # 3 2 # 5) ., Clearly, none of 2, 3 and 5 is a factor of 17.
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22, , Secondary School Mathematics for Class 10, , , , 17, 90 is in its simplest form., , Also, 90 = (2 # 3 2 # 5) ! (2 m # 5 n) ., , , 17, 90 is a nonterminating repeating decimal., , 53, (iii) Given number is 343 ·, Now, 343 = 7 3 and 7 is not a factor of 53., 53, , 343 is in its simplest form., Also, 343 7 3 ! (2 m # 5 n) ., 53, 343 is a nonterminating repeating decimal., 66, (iv) Given number is 180 and HCF(66, 180) = 6., 66 = 66 ' 6 = 11 ·, , 180 180 ' 6 30, Now, 30 = (2 # 3 # 5) and none of 2, 3, 5 is a factor of 11., , , 11, 30 is in its simplest form., Also, 30 = (2 # 3 # 5) ! (2 m # 5 n) ., , , , , , EXAMPLE 3, , 66, 11, 30 and hence 180 is a nonterminating repeating, decimal., , The decimal expansion of the rational number, after how many places of decimals?, , SOLUTION, , 43, , will terminate, 24 $ 53, [CBSE 2009], , We have, 43 = 43 # 5 = 215 = 215 = 215 =, 0.0215., 2 4 $ 5 3 2 4 # 5 4 (2 # 5) 4 10 4 10000, So, it will terminate after 4 places of decimals., , EXAMPLE 4, , Express each of the following as a rational number in simplest form., (a) 0.6, , SOLUTION, , (b) 1.8, , (c) 0.16, , (a) Let, , x = 0.6. Then,, x = 0.666 …, , 10x = 6.666 …, On subtracting (i) from (ii), we get, 6 2, 9x 6 & x 9 3 ·, , … (i), … (ii)
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Real Numbers, , 23, , 2, Hence, 0.6 = 3 ·, (b) Let, , , x = 1.8. Then,, x = 1.888 …, 10x = 18.888 …, , … (i), … (ii), , On subtracting (i) from (ii), we get, 17, 8, 9x 17 & x 9 1 9 ·, 8, Hence, 1.8 = 1 9 ·, (c) Let, , x = 0.16. Then,, x = 0.1666 …, , 10x = 1.6666 …, And, 100x = 16.6666 …, , … (i), … (ii), … (iii), , On subtracting (ii) from (iii), we get, 15 1, 90x 15 & x 90 6 ·, 1, , 0.16 = 6 ·, EXAMPLE 5, , Express 0.32 as a fraction in simplest form., , SOLUTION, , Let x = 0.32. Then,, x = 0.3232 …, , … (i), , , , … (ii), , 100x = 32.3232 …, , On subtracting (i) from (ii), we get, 32, 99x 32 & x 99 ·, 32, Hence, 0.32 = 99 ·, EXAMPLE 6, , Express 0.254 as a fraction in simplest form., , SOLUTION, , Let x = 0.254. Then,, x = 0.2545454 …, , … (i), , , , … (ii), , 10x = 2.545454 …, and 1000x = 254.545454 …, On subtracting (ii) from (iii), we get, 252 126, 42 14, 990x 252 & x 990 495 165 55 ·, 14, 0.254 55 ·, , … (iii)
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24, , Secondary School Mathematics for Class 10, , Show that each of the following numbers is rational. What can you, say about the prime factors of their denominators?, (i) 23.123456789, (ii) 32.123456789, , EXAMPLE 7, , (i) Clearly, the given number 23.123456789 is a terminating, decimal. So, it is rational and the prime factors of its, denominator are 2 or 5 or both., , SOLUTION, , (ii) Clearly, the given number 32.123456789 is a, nonterminating repeating decimal. So, it is rational and the, prime factors of its denominator are other than 2 or 5 also., EXAMPLE 8, , Decide whether the number 0.12012001200012 … is rational or not., Give reason to support your answer., , SOLUTION, , Clearly, the given number 0.12012001200012 … is a, nonterminating and nonrepeating decimal. So, it is not, rational., , f, , EXERCISE 1C, , 1. Without actual division, show that each of the following rational, numbers is a terminating decimal. Express each in decimal form., (i), , 24, (ii) 125, , 23, (2 3 # 5 2), , 15, (iv) 1600, , 171, (iii) 800, , 19, (vi) 3125, , 17, (v) 320, , 2. Without actual division, show that each of the following rational, numbers is a nonterminating repeating decimal., (i), , 11, (2 3 # 3), , 77, (v) 210, , (ii), , 73, 129, (iii), (2 2 # 3 3 # 5), (2 2 # 5 3 # 7 2), , 32, (vi) 147, , 29, (vii) 343, , 9, (iv) 35, 64, (viii) 455, , 3. Express each of the following as a fraction in simplest form., (i) 0.8, , (ii) 2.4, , (v) 2.24, , (vi) 0.365, , (iii) 0.24, , (iv) 0.12, , ANSWERS (EXERCISE 1C), , 1. (i) 0.115, , (ii) 0.192, , (vi) 0.00608, , (iii) 0.21375, , (iv) 0.009375, , (v) 0.053125
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Real Numbers, , 8, , 3. (i) 9, , 22, (ii) 9, , 8, (iii) 33, , 11, (iv) 90, , 101, (v) 45, , 25, , 181, (vi) 495, , HINTS TO SOME SELECTED QUESTIONS, 1. (i), , 23, = 23 # 5 = 115 = 115 = 0.115., (2 3 # 5 2) (2 3 # 5 3) (10) 3 1000, , 24, 24 2 3, 24 # 8 = 192 = 192 =, (ii) 125 = 3 # 3 =, 0.192., 5, 2, (5 # 2) 3 (10) 3 1000, 171 171, 1, 21.375, (iii) 800 = 8 # 100 = 100 = 0.21375., 15, 15, 1, 0.9375, (iv) 1600 = 16 # 100 = 100 = 0.009375., 17, 17 # 5, 85, 85, 1, 5.3125, (v) 320 = 320 # 5 = 1600 = 16 # 100 = 100 = 0.053125., 19, 19 # 8, 152, 152, 1, 6.08, (vi) 3125 = 3125 # 8 = 25000 = 25 # 1000 = 1000 = 0.00608., 9, 9, 2. (iv) 35 = #, (5 7), 29, 29, (vii) 343 = 3, 7, , 77, 11, 11, (v) 210 = 30 = # #, (2 3 5), , 32, 32, (vi) 147 =, (3 # 7 2), , 64, 64, (viii) 455 = # #, (5 7 13), , IRRATIONAL NUMBERS, IRRATIONAL NUMBERS The numbers which when expressed in decimal form are, expressible as nonterminating and nonrepeating decimals are known as irrational, numbers., , Examples, , Type 1., , Note that every nonterminating and nonrepeating decimal is, irrational., (i) Clearly, 0.1010010001… is a nonterminating, nonrepeating decimal. So, it is irrational., , and, , (ii) 0.2020020002 , 0.3030030003…, etc., are all irrational., (iii) 0.12112111211112… is irrational,, 0.13113111311113… is irrational, and so on., (iv) 0.232232223… is irrational,, 0.343343334… is irrational, and so on., Type 2., , If m is a positive integer which is not a perfect square then m, is irrational., Thus, 2 , 3 , 5 , 6 , 7 , 8 , 10 , 11 , etc., are all irrational.
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26, , Secondary School Mathematics for Class 10, , If m is a positive integer which is not a perfect cube then, is irrational., , Type 3., , Thus,, , 3, , 3, , m, , 2 , 3 3 , 3 5 , 3 6 , etc., are all irrational., , 22, is irrational, while 7 is rational., , Type 4., , SOME RESULTS ON IRRATIONALS, THEOREM 1, , PROOF, , Let p be a prime number and a be a positive integer. If p divides a 2, then show that p divides a., , Let p be a prime number and a be a positive integer such that p, divides a 2 ., We know that every positive integer can be expressed as the, product of primes., Let a p1 p2 … pn, where p1, p2, …, pn are primes, not necessarily all, distinct. Then,, a 2 (p1 p2 … pn) (p1 p2 … pn), , , a 2 (p12 p 22 … p n2) ., , Now, p divides a2, , , p is a prime factor of a 2, , , , p is one of p1, p2, …, pn, , , , p divides a, , [a prime factors of a2 are p1, p2, …, pn], [a a p1 p2 … pn] ., , Thus, (p divides a 2) (p divides a)., Using the above result, we can prove the following., THEOREM 2, PROOF, , Prove that 2 is irrational., , [CBSE 2008, ’09], , a, If possible, let 2 be rational and let its simplest form be ·, b, Then, a and b are integers having no common factor other than 1,, and b ! 0., a, a2, Now, 2 & 2 2, b, b, , [on squaring both sides], , & 2b 2 a 2, & 2 divides a 2 [a 2 divides 2b 2], , … (i)
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Real Numbers, , 27, , & 2 divides a, [a 2 is prime and divides b 2 & 2 divides b]., Let a 2c for some integer c., Putting a 2c in (i), we get, 2b 2 4c 2 & b 2 2c 2, , & 2 divides b 2 [a 2 divides 2c 2], & 2 divides b, [a 2 is prime and 2 divides b 2 & 2 divides b]., Thus, 2 is a common factor of a and b., But, this contradicts the fact that a and b have no common factor, other than 1., The contradiction arises by assuming that 2 is rational., Hence, 2 is irrational., THEOREM 3, PROOF, , Prove that 3 is irrational., , [CBSE 2008, ’09C], , a·, b, Then, a and b are integers having no common factor other than 1,, and b ! 0., If possible, let 3 be rational and let its simplest form be, , a, a2, Now, 3 & 3 2, b, b, & 3b 2 a 2, & 3 divides a 2, & 3 divides a, , [on squaring both sides], … (i), [a 3 divides 3b 2], , [a 3 is prime and 3 divides a 2 & 3 divides a]., Let a = 3c for some integer c., Putting a = 3c in (i), we get, 3b 2 9c 2 & b 2 3c 2, & 3 divides b 2 [a 3 divides 3c 2], & 3 divides b, [a 3 is prime and 3 divides b 2 & 3 divides b]., Thus, 3 is a common factor of a and b., But, this contradicts the fact that a and b have no common factor, other than 1., The contradiction arises by assuming that 3 is rational., Hence, 3 is irrational.
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28, , Secondary School Mathematics for Class 10, , THEOREM 4, PROOF, , Prove that, , If possible, let, , 5 is irrational., , [CBSE 2008, ’09], , a, 5 be rational and let its simplest form be b ·, , Then, a and b are integers having no common factor other than 1,, and b ! 0., a, a2, Now, 5 & 5 2, b, b, , & 5b 2 a 2, & 5 divides a 2, & 5 divides a, , [on squaring both sides], … (i), [a 5 divides 5b ], 2, , [a 5 is prime and 5 divides a 2 & 5 divides a]., Let a = 5c for some integer c., Putting a = 5c in (i), we get, 5b 2 25c 2 & b 2 5c 2, , & 5 divides b 2 [a 5 divides 5c2], & 5 divides b, [a 5 is prime and 5 divides b 2 & 5 divides b]., Thus, 5 is a common factor of a and b., But, this contradicts the fact that a and b have no common factor, other than 1., The contradiction arises by assuming that, Hence,, THEOREM 5, PROOF, , 5 is rational., , 5 is irrational., , Prove that, , If possible, let, , 11 is irrational., a, 11 be rational and let its simplest form be b ·, , Then, a and b are integers having no common factor other than 1,, and b ! 0., a, a2, Now, 11 & 11 2, b, b, , & 11b 2 a 2, & 11 divides a 2, & 11 divides a, , [on squaring both sides], … (i), [a 11 divides 11b ], 2, , [a 11 is prime and 11 divides a 2 & 11 divides a]., Let a 11c for some positive integer c.
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Real Numbers, , 29, , Putting a 11c in (i), we get, 11b 2 121c 2 & b 2 11c 2, [a 11 divides 11c 2], , & 11 divides b 2, & 11 divides b, , [a 11 is prime and 11 divides b 2 & 11 divides b]., Thus, 11 is a common factor of a and b., But, this contradicts the fact that a and b have no common factor, other than 1., The contradiction arises by assuming that, Hence,, THEOREM 6, PROOF, , 11 is rational., , 11 is irrational., , If p is a prime number then prove that, , p is irrational., , Let p be a prime number and if possible, let, , p be rational., , m, p = n , where m and n are integers, having no common factor other than 1, and n ! 0., , Let its simplest form be, m, m2, Then, p n & p 2, n, , [on squaring both sides], , & pn 2 m 2, & p divides m 2, & p divides m, , … (i), [a p divides pn 2], , [a p is prime and p divides m 2 & p divides m]., Let m = pq for some integer q., Putting m = pq in (i), we get, pn 2 p 2 q 2 & n 2 pq 2, , & p divides n2, & p divides n, , [a p divides pq2], , [a p is prime and p divides n 2 & p divides n]., Thus, p is a common factor of m and n., But, this contradicts the fact that m and n have no common factor, other than 1., The contradiction arises by assuming that p is rational., Hence, p is irrational.
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30, , Secondary School Mathematics for Class 10, , THEOREM 7, , PROOF, , If a is rational and, irrational., , b is irrational then prove that (a b ) is, , Let a be rational and b be irrational. Then, we have to prove that, (a b ) is irrational., If possible, let (a b ) be rational. Then,, (a b ) is rational and a is rational, , , , {(a b ) a} is rational [a difference of rationals is rational], b is rational., , This contradicts the fact that b is irrational., The contradiction arises by assuming that (a b ) is rational., Hence, (a b ) is irrational., THEOREM 8, , PROOF, , If a is a nonzero rational and b is irrational then show that a b is, irrational., , Let a be a nonzero rational and let b be irrational., Then, we have to show that a b is irrational., If possible, let a b be rational., p, Then, a b q , where p and q are nonzero integers, having no, common factor other than 1., p, p, Now, a b q & b aq ·, But, p and aq are both rational and aq ! 0., p, , aq is rational., Thus, from (i), it follows that b is rational., This contradicts the fact that b is irrational., The contradiction arises by assuming that a b is rational., Hence, a b is irrational., , SOLVED EXAMPLES, EXAMPLE 1, , Show that (2 3 ) is an irrational number., , SOLUTION, , Let us assume, to the contrary, that (2 3 ) is rational., , ... (i)
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Real Numbers, , 31, , Then, there exist co-primes a and b (b ! 0) such that, a, (2 3 ) , b, a, a 2b ·, , 3 2 & 3 , b, b, a 2b, Since a and b are integers, so, is rational., b, Thus, 3 is also rational., But, this contradicts the fact that, assumption is incorrect., , 3 is irrational. So, our, , Hence, (2 + 3 ) is irrational., EXAMPLE 2, , Show that 2 3 is irrational., , SOLUTION, , Let us assume, to the contrary, that 2 3 is rational., Then, there exist co-primes a and b (b ! 0) such that, a, a ·, 2 3 & 3, 2b, b, a, Since a and b are integers, so, is rational., 2b, Thus,, , 3 is also rational., , But, this contradicts the fact that, assumption is incorrect., , 3 is irrational. So, our, , Hence, 2 3 is irrational., 1, is irrational., 2, , EXAMPLE 3, , Show that, , SOLUTION, , Let us assume, to the contrary, that, , 1, is rational., 2, , Then, there exist co-primes a and b (b ! 0) such that, 1 a, &, 2 b, , b, 2a·, , b, Since a and b are integers, so a is rational., Thus, 2 is also rational., But, this contradicts the fact that, assumption is incorrect., Hence,, , 1, is irrational., 2, , 2 is irrational. So, our
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32, , Secondary School Mathematics for Class 10, , EXAMPLE 4, , Prove that (3 5 2 ) is irrational., , SOLUTION, , Let us assume, to the contrary, that (3 5 2 ) is rational., Then, there exist co-primes a and b (b ! 0) such that, a, 35 2 , b, a 3b, a, 5 2 3 , b, b, 3b, a, ·, , 2, 5b, a 3b, Since a and b are integers, so, is rational., 5b, Thus, 2 is also rational., But, this contradicts the fact that, assumption is incorrect., , 2 is irrational. So, our, , Hence, (3 5 2 ) is irrational., EXAMPLE 5, , Prove that ( 2 3 ) is irrational., , SOLUTION, , Let us assume that ( 2 3 ) is irrational., Then, there exist co-primes a and b such that, a, 2 3, b, a, , 3 2, b, 2, a, ( 3 ) 2 b 2l, [squaring both sides], b, , , , , a 2 2a, 22, b2 b, 2a, a2, 2 2 1, b, b, 2 2, a, b ·, 2, 2ab, , 3, , Since a and b are integers, so, , a2 b2, is rational., 2ab, , Thus, 2 is also rational., But, this contradicts the fact that, assumption is incorrect., , 2 is irrational. So, our, , Hence, ( 2 3 ) is irrational., EXAMPLE 6, , Prove that p q is irrational, where p and q are primes., , SOLUTION, , Let us assume that p q is rational.
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Real Numbers, , 33, , Then, there exist co-primes a and b such that, a, p q , b, a, , p q, b, 2, a, ( p) 2 b ql, [squaring both sides], b, a 2 2a, p 2, q q, b, b, , , 2a, a2, q 2 qp, b, b, , a 2 b b 3 (q p), b, ·, q (a 2 b 2 q b 2 p)# , 2a, 2a, a 2 b b 3 (q p), Since a, b, p, q are integers so, is rational., 2a, , , Thus, q is also rational., But, q being prime, q is irrational., Since, a contradiction arises so our assumption is incorrect., Hence, ( p q ) is irrational., , f, , EXERCISE 1D, , 1. Define (i) rational numbers (ii) irrational numbers (iii) real numbers., 2. Classify the following numbers as rational or irrational:, (i), , 22, 7, , (v) 5.636363 …, , (ii) 3.1416, , (iii) , , (iv) 3.142857, , (vi) 2.040040004 … (vii) 1.535335333 …, , (viii) 3.121221222 … (ix), , 21, , (x), , 3, , 3, , 3. Prove that each of the following numbers is irrational., (i), , (iiii) (3 + 2 ), (v) (5 + 3 2 ), (vii), , (ii) (2 3 ), , [CBSE 2008], , [CBSE 2009], , (iv) (2 + 5 ), , [CBSE 2008C], , [CBSE 2008], , (vi) 3 7, , 6, , 3, 5, , (ix) ( 3 + 5 ), , (viii) (2 3 5 ), , [CBSE 2010]
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34, , Secondary School Mathematics for Class 10, , 4. Prove that, , 1, is irrational., 3, , 1 = 1 #, 3, 3, , HINT, , 3, = 1 · 3., 3, 3, , 5. (i) Give an example of two irrationals whose sum is rational., (ii) Give an example of two irrationals whose product is rational., HINT, , (i) Take (2 +, , 3 ) and (2 3 ) ., , (ii) Take (3 +, , 2 ) and (3 2 ) ., , 6. State whether the given statement is true or false., (i) The sum of two rationals is always rational., (ii) The product of two rationals is always rational., (iii) The sum of two irrationals is always an irrational., (iv) The product of two irrationals is always an irrational., (v) The sum of a rational and an irrational is irrational., (vi) The product of a rational and an irrational is irrational., 7. Prove that (2 3 1) is an irrational number., , [CBSE 2010], , 8. Prove that (4 5 2 ) is an irrational number., , [CBSE 2010], , 9. Prove that (5 2 3 ) is an irrational number., , [CBSE 2010], , 10. Prove that 5 2 is irrational., 11. Prove that, HINT, , 2, is irrational., 7, , 2 2, #, e, 7, 7, , 7, 7, , o 72 $ 7 ., ANSWERS (EXERCISE 1D), , 2. (i) rational, , (ii) rational, , (iii) irrational, , (iv) rational, , (v) rational, , (vi) irrational, , (vii) irrational, , (viii) irrational, , (ix) irrational, , (x) irrational, , 6. (i) True, , (ii) True, , (iii) False, , f, , (iv) False, , EXERCISE 1E, , Very-Short-Answer Questions, 1. State Euclid‘s division lemma., , (v) True, , (vi) True
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Real Numbers, , 35, , 2. State fundamental theorem of arithmetic., 3. Express 360 as product of its prime factors., 4. If a and b are two prime numbers then find HCF(a, b)., 5. If a and b are two prime numbers then find LCM(a, b)., 6. If the product of two numbers is 1050 and their HCF is 25, find their LCM., 7. What is a composite number?, 8. If a and b are relatively prime then what is their HCF?, a, has a terminating decimal expansion, what is, b, [CBSE 2008], the condition to be satisfied by b?, , 9. If the rational number, , 10. Simplify:, , (2 45 + 3 20 ), ·, 2 5, , [CBSE 2010], , 11. Write the decimal expansion of, , 73 ·, (24 # 53), , [CBSE 2009], , 12. Show that there is no value of n for which (2n # 5n) ends in 5., 13. Is it possible to have two numbers whose HCF is 25 and LCM is 520?, 14. Give an example of two irrationals whose sum is rational., 15. Give an example of two irrationals whose product is rational., 16. If a and b are relatively prime, what is their LCM?, 17. The LCM of two numbers is 1200. Show that the HCF of these numbers, cannot be 500. Why?, Short-Answer Questions, 18. Express 0.4 as a rational number in simplest form., 19. Express 0.23 as a rational number in simplest form., 20. Explain why 0.15015001500015 … is an irrational number., 2, 21. Show that 3 is irrational., 22. Write a rational number between, , 3 and 2., , 23. Explain why 3.1416 is a rational number., ANSWERS (EXERCISE 1E), , 3. (23 # 32 # 5) 4. 1, , 5. ab, , 6. 42, , 8. 1, , 9. b = (2 # 5 ), where m and n are some non-negative integers, m, , 10. 6, , n, , 11. 0.0365, 13. No 16. ab, 4, 23, 18. 9 19. 99 22. 1.8, , 17. since 500 is not a factor of 1200
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36, , Secondary School Mathematics for Class 10, HINTS TO SOME SELECTED QUESTIONS, , 7. A number having at least 3 factors is called a composite number., 11., , 73, = 73 # 5 = 365 = 365 = 365 = 0.0365., (24 # 53) (24 # 54) (2 # 5) 4 (10) 4 10000, , 12. (2n # 5n) = (2 # 5) n = 10n, which always ends in a zero., 13. HCF always divides the LCM completely., 14. (2 + 3 ) and (2 3 ) ., 15. (3 + 2 ) and (3 2 ) ., 20. Given number is nonterminating and nonrepeating decimal., 22. Clearly,, , 3 = 1.732 … . So, we may take 1.8 as the required rational number between, , 3 and 2., 23. Clearly, it is a nonterminating repeating decimal., , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. Which of the following is a pair of co-primes?, (a) (14, 35), , (b) (18, 25), , (c) (31, 93), , (d) (32, 62), , 2. If a = (2 # 3 # 5 ) and b = (2 # 3 # 5) then HCF(a, b) = ?, 2, , 3, , 4, , (a) 90, , 3, , 2, , (b) 180, , (c) 360, , (d) 540, , 3. HCF of (2 # 3 # 5), (2 # 3 # 5 ) and (2 # 3 # 5 # 7) is, 3, , (a) 30, , 2, , 2, , 3, , 2, , (b) 48, , 4, , 3, , (c) 60, , (d) 105, , (c) 1120, , (d) 1680, , 4. LCM of (23 # 3 # 5) and (24 # 5 # 7) is, (a) 40, , (b) 560, , 5. The HCF of two numbers is 27 and their LCM is 162. If one of the, numbers is 54, what is the other number?, (a) 36, , (b) 45, , (c) 9, , (d) 81, , 6. The product of two numbers is 1600 and their HCF is 5. The LCM of the, numbers is, (a) 8000, , (b) 1600, , (c) 320, , (d) 1605, , 7. What is the largest number that divides each one of 1152 and 1664, exactly?, (a) 32, , (b) 64, , (c) 128, , (d) 256, , 8. What is the largest number that divides 70 and 125, leaving remainders, 5 and 8 respectively?, (a) 13, , (b) 9, , (c) 3, , (d) 585
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Real Numbers, , 37, , 9. What is the largest number that divides 245 and 1029, leaving remainder, 5 in each case?, (a) 15, , (b) 16, , (c) 9, , (d) 5, , 13, (c) 16, , 15, (d) 16, , 1095, 10. The simplest form of 1168 is, 17, (a) 26, , 25, (b) 26, , 11. Euclid’s division lemma states that for any positive integers a and b,, there exist unique integers q and r such that a = bq + r, where r must, satisfy, (a) 1 < r < b, , (b) 0 < r b, , (c) 0 r < b, , (d) 0 < r < b, , 12. A number when divided by 143 leaves 31 as remainder. What will be, the remainder when the same number is divided by 13?, (a) 0, , (b) 1, , (c) 3, , (d) 5, , 13. Which of the following is an irrational number?, (a), , 22, 7, , (b) 3.1416, , (c) 3.1416, , (d) 3.141141114 …, , 14. is, (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , 15. 2.35 is, (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , 16. 2.13113111311113 ... is, (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , 17. The number 3.24636363 … is, (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , 18. Which of the following rational numbers is expressible as a terminating, decimal?, 124, (a) 165, , (b), , 131, 30, , (c), , 2027, 625, , (d), , 1625, 462
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38, , Secondary School Mathematics for Class 10, , 37, 19. The decimal expansion of the rational number 2, will terminate, 2 #5, after, (a) one decimal place, (c) three decimal places, , (b) two decimal places, (d) four decimal places, , 14753, 20. The decimal expansion of the number 1250 will terminate after, (a) one decimal place, (c) three decimal places, , (b) two decimal places, (d) four decimal places, , 21. The number 1.732 is, (a) an irrational number, (c) an integer, , (b) a rational number, (d) a whole number, , 22. a and b are two positive integers such that the least prime factor of a, is 3 and the least prime factor of b is 5. Then, the least prime factor of, (a + b) is, (a) 2, 23., , (c) 5, , (d) 8, , 2 is, (a), (b), (c), (d), , 24., , (b) 3, , a rational number, an irrational number, a terminating decimal, a nonterminating repeating decimal, , 1, is, 2, (a) a fraction, (c) an irrational number, , (b) a rational number, (d) none of these, , 25. (2 + 2 ) is, (a) an integer, (c) an irrational number, , (b) a rational number, (d) none of these, , 26. What is the least number that is divisible by all the natural numbers, from 1 to 10 (both inclusive)?, (a) 100, , (b) 1260, , (c) 2520, , (d) 5040, , ANSWERS (MCQ), , 1. (b) 2. (b) 3. (c), 4. (d) 5. (d) 6. (c), 7. (c), 10. (d) 11. (c) 12. (d) 13. (d) 14. (c) 15. (b) 16. (c), 19. (b) 20. (d) 21. (b) 22. (a) 23. (b) 24. (c) 25. (c), , 8. (a), 9. (b), 17. (b) 18. (c), 26. (c)
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Real Numbers, , 39, , HINTS TO SOME SELECTED QUESTIONS, 2. HCF(a, b) = product of common terms with lowest power, = (22 # 32 # 5) = (4 # 9 # 5) = 180., 3. HCF = product of common terms with lowest power, = (2 2 # 3 # 5) = (4 # 3 # 5) = 60., 4. LCM = product of prime factors with highest power, = (2 4 # 3 # 5 # 7) = (16 # 3 # 5 # 7) = 1680., 5. Other number =, 6. LCM =, , HCF # LCM = 27 # 162 =, 81., 54, given number, , product of two numbers 1600, =, = 320., 5, their HCF, , 7. Required number = HCF(1152, 1664) = 128., 8. Required number = HCF{(70 – 5), (125 – 8)} = HCF(65, 117) = 13., 9. Required number = HCF{(245 – 5), (1029 – 5)} = HCF(240, 1024) = 16., 10. HCF{1095, 1168} = 73., , , 1095 = 1095 ' 73 = 15, 1168 1168 ' 73 16 $, , 11. On dividing a by b, let q be the quotient and r be the remainder., Then, we have, a = bq + r, where 0 # r < b., 12. Let the given number when divided by 143 give q as quotient and 31 as remainder., Then, number = 143q + 31 = {13 # 11q + 13 # 2 + 5} = 13 # (11q + 2) + 5., So, the same number when divided by 13 gives 5 as remainder., 13. 3.141141114 … is a nonterminating, nonrepeating decimal. So, it is irrational., 14. is an irrational number., 15. 2. 35 = 2.353535 …, which is a repeating decimal., , , 2. 35 is rational., , 16. 2.13113111311113 … is a nonterminating, nonrepeating decimal. So, it is irrational., 17. The number 3.24636363 … is a nonterminating repeating decimal., So, it is a rational number., 18., , 2027 = 2027, $, 625, (5 4 # 2 0), So, it is expressible as a terminating decimal., , 19., , 37 = 37 # 5 = 185 = 185 = 185 =, 1.85., 2 2 # 5 2 2 # 5 2 (2 # 5) 2 (10) 2 100, So, it will terminate after 2 decimal places.
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40, , Secondary School Mathematics for Class 10, , 14753 8 14753 # 8 14753 # 8, 20. 1250 # 8 = 10000 =, $, (10) 4, So, it will terminate after 4 decimal places., 1732, 21. 1.732 = 1000 , which is a rational number., 22. Clearly, 2 is neither a factor of a nor that of b., , , a and b are both odd., , Hence, (a + b) is even., , 24., , least prime factor of (a + b) is 2., , 1 = 1 #, 2, 2, , 2 1, = $ 2., 2 2, , 1, Here, 2 is rational and, , 2 is irrational., , And, the product of a rational and an irrational is irrational., 1, 1, 2 $ 2 and hence 2 is irrational., , , , 25. The sum of a rational and an irrational is irrational., Here, 2 is rational and, , , 2 is irrational., , (2 + 2 ) is irrational., , 26. Required number = LCM{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, = LCM{1, 2, 3, 2 2, 5, 2 # 3, 7, 2 3, 3 2, 2 # 5}, = (1 # 2 3 # 3 2 # 5 # 7) = (8 # 9 # 5 # 7) = 2520., , SUMMARY OF RESULTS, 1. Euclid’s Division Lemma, , Given positive integers a and b, there exist whole numbers q and r satisfying, a = bq + r, where 0 # r < b., According to this, we find the HCF of two, positive integer a and b with a > b in following steps., , 2. Euclid’s Division Algorithm, , Step 1., , Apply the division lemma to find q and r such that a = bq + r,, where 0 # r < b., , Step 2. If r = 0 then HCF = b. If r ! 0, apply Euclid’s division lemma to, , b and r., Step 3. Continue the process till the remainder is zero. The divisor at this, , stage is HCF(a, b).
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Real Numbers, , 41, , 3. The Fundamental Theorem of Arithmetic, , Every composite number can be expressed as a product of primes, and this, factorisation is unique, apart from the order in which the prime factors occur., 4. If p is prime and p divides a 2 then p divides a, where a is a positive integer., 5. To prove that, , 2 , 3 , 5 , 6 , 7 , etc., are irrationals., , 6. Let x be a rational number which can be expressed as a terminating decimal., , p, If we put it in the simplest form q then q = (2 m # 5 n) for some non-negative, integers m and n., p, , 7. Let x = q, , be a rational number such that q ! (2 m # 5 n) then x has a, , nonterminating repeating decimal expansion., 8. A number which can be expressed as a nonterminating and nonrepeating, , decimal is an irrational number., 2 , 3 , 5 , 6 , 7 , 8 , 10 , …, , e, etc., are all irrational numbers., , TEST YOURSELF, MCQ, 71, 1. The decimal representation of 150 is, (a) a terminating decimal, (b) a nonterminating, repeating decimal, (c) a nonterminating and nonrepeating decimal, (d) none of these, 2. Which of the following has a terminating decimal expansion?, (a), , 32, 91, , (b), , 19, 80, , (c), , 23, 45, , (d), , 25, 42, , 3. On dividing a positive integer n by 9, we get 7 as remainder. What will, be the remainder if (3n 1) is divided by 9?, (a) 1, , (b) 2, , (c) 3, , (d) 4, , (b) 1.42, , (c) 0.141, , (d) None of these, , 4. 0.68 0.73 ?, (a) 1.41, Short-Answer Questions, 5. Show that any number of the form 4 n, nd N can never end with the, digit 0.
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42, , Secondary School Mathematics for Class 10, , 6. The HCF of two numbers is 27 and their LCM is 162. If one of the number, is 81, find the other., 17, 7. Examine whether 30 is a terminating decimal., 148, 8. Find the simplest form of 185 $, 9. Which of the following numbers are irrational?, (a), , (b), , 2, , (e) , , (f), , 3, , (c) 3.142857, , 6, , 22, 7, , (d) 2.3, , (g) 0.232332333… (h) 5.2741, , 10. Prove that (2 + 3 ) is irrational., 11. Find the HCF and LCM of 12, 15, 18, 27., 12. Give an example of two irrationals whose sum is rational., 13. Give prime factorisation of 4620., 14. Find the HCF of 1008 and 1080 by prime factorisation method., 8 10, 16, 15. Find the HCF and LCM of 9 , 27 and 81 $, 16. Find the largest number which divides 546 and 764, leaving remainders, 6 and 8 respectively., Long-Answer Questions, 17. Prove that 3 is an irrational number., 18. Show that every positive odd integer is of the form (4q + 1) or (4q + 3), for some integer q., 19. Show that one and only one out of n, (n + 2) and (n + 4) is divisible by, 3, where n is any positive integer., 20. Show that (4 3 2 ) is irrational., ANSWERS (TEST YOURSELF), , 1. (b), 9., , 2. (b), , 3. (b), , 4. (b), , 2 , 3 6 , , 0.232332333…, , 12. (2 3 ) and (2 3 ), 14. 72, , 15. HCF =, , 6. 54, , 7. No, , 8., , 4, 5, , 11. HCF = 3, LCM = 540, 13. (2 2 # 3 # 5 # 7 # 11), , 80, 2 ,, LCM , 9, 81, , , 16. 108
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Polynomials, , 43, , An expression of the form p(x) = a0 + a1 x + a2 x 2 + … + an x n, where, an ! 0, is called a polynomial in x of degree n., , POLYNOMIALS, , Here a0, a1, a2, …, an are real numbers and each power of x is a non-negative, integer., EXAMPLES, , (i) 3x 5 is a polynomial in x of degree 1., (ii) 8x 2 5x + 3 is a polynomial in x of degree 2., 4, (iii) 2y 3 + 9 y 2 5y + 3 is a polynomial in y of degree 3., (iv) 3z 4 5z 3 + 2z 2 8z + 1 is a polynomial in z of degree 4., , REMARK, , 6, 1, Note that ( x 3),, ,, ,, (x 5) (x 2 3x 1), polynomials., , LINEAR POLYNOMIAL, , etc.,, , are, , not, , A polynomial of degree 1 is called a linear polynomial., , A linear polynomial is of the form p(x) = ax + b, where a ! 0., 5, Thus, (3x 5), ( 2 x + 3), a x 8 k, etc., are all linear polynomials., QUADRATIC POLYNOMIAL, , A polynomial of degree 2 is called a quadratic polynomial., , A quadratic polynomial is of the form p(x) = ax 2 + bx + c, where a ! 0., Thus, (3x 2 5x 8), (2x 2 2 2 x 6), (y 2 3y 3 ), etc., are all quadratic, polynomials., A polynomial of degree 3 is called a cubic polynomial. A cubic, polynomial is of the form p(x) = ax 3 + bx 2 + cx + d, where a ! 0., , CUBIC POLYNOMIAL, , Thus, (2x 3 3x 2 + 8x + 1), ( 2 y 3 2y 2 + y 8), (z 3 + 2z 2 3 z + 3), etc.,, are all cubic polynomials., BIQUADRATIC POLYNOMIAL, , A polynomial of degree 4 is called a biquadratic, , polynomial., A biquadratic polynomial is of the form p(x) = ax 4 + bx 3 + cx 2 + dx + e, where, a ! 0., Thus, (2x 4 + 3x 3 5x 2 + 9x + 1), (4y 4 5y 3 + 6y 2 8y + 3), etc., are all, biquadratic polynomials., 43
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44, , Secondary School Mathematics for Class 10, , VALUE OF A POLYNOMIAL AT A GIVEN POINT, , If p(x) is a polynomial in x and if is any real number then the value obtained by, putting x = in p(x) is called the value of p(x) at x = ., The value of p(x) at x = is denoted by p()., Let p(x) = 2x 2 3x + 5. Then,, , EXAMPLE, , p(2) = {2 # 2 2 3 # 2 + 5} = (8 6 + 5) = 7,, p(1) {2 # (1) 2 3 # (1) 5} (2 3 5) 10., ZEROS OF A POLYNOMIAL, , A real number is called a zero of the polynomial, , p(x), if p() = 0., Let p(x) x 2 2x 3. Find (i) p(3) and (ii) p(1) ., What do you conclude?, , EXAMPLE, , We have p(x) x 2 2x 3., , SOLUTION, , (i) p(3) (3 2 2 # 3 3) (9 6 3) 0, and (ii) p(1) {(1) 2 2 # (1) 3} (1 2 3) 0., This shows that 3 and –1 are the zeros of the polynomial p(x)., RELATION BETWEEN THE ZEROS AND COEFFICIENTS OF A QUADRATIC, POLYNOMIAL, , Let and be the zeros of a quadratic polynomial, p(x) ax 2 bx c, where a ! 0., Then, (x ) and (x ) are the factors of p(x)., (ax 2 bx c) k(x )(x ), where k is a constant, k $ {x 2 ( ) x }, kx 2 k( ) x k() ., On comparing coefficients of like powers of x on both sides, we get, k a, k( ) b and k() c, a( ) b and a() c, b, c, ( ) a and a $, , , sum of zeros , , product of zeros , , (coefficient of x), ,, (coefficient of x 2), constant term ·, coefficient of x 2, , [a k a]
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Polynomials, , 45, , SUMMARY, , I. If and are the zeros of p(x) ax 2 bx c, a ! 0 then, b, (i) + = a, c, (ii) a ·, II. A quadratic polynomial whose zeros are and is given by, p(x) {x 2 ( )x } ., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Find the zeros of the polynomial 2x 2 + 5x 12 and verify the, relationship between its zeros and coefficients., Let the given polynomial be denoted by f(x). Then,, , , , f (x) = 2x 2 + 5x 12, = 2x 2 + 8x 3x 12, = 2x(x + 4) 3(x + 4), = (x + 4)(2x 3) ., f (x) 0 & (x 4)(2x 3) 0, , & x 4 0 or 2x 3 0, 3, & x 4 or x 2 ·, So, the zeros of f(x) are –4 and, , 3·, 2, , 5 (coefficient of x), 3, Sum of the zeros = a 4 + 2 k = 2 =, ,, (coefficient of x 2), 3 12 constant term ·, product of the zeros (4)# , 2, 2, (coefficient of x 2), EXAMPLE 2, , SOLUTION, , Find the zeros of the polynomial 6x 2 3 7x and verify the, relationship between the zeros and the coefficients., [CBSE 2008], Let the given polynomial be denoted by f(x). Then,, f (x) = 6x 2 3 7x, = 6x 2 7x 3 [in standard form], = 6x 2 9x + 2x 3, = 3x (2x 3) + (2x 3), = (2x 3)(3x + 1) .
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46, , Secondary School Mathematics for Class 10, , f (x) 0 & (2x 3) (3x 1) 0, & 2x 3 0 or 3x 1 0, , 3, & x 2 or x 31 ·, 1, 3, ·, So, the zeros of f(x) are 2 and, 3, 3 1, 3 1, 7 (coefficient of x) ·, Sum of zeros & a k0 a k , 2, 3, 2 3, 6 (coefficient of x 2), 3, 1, 3, constant term 2 ·, Product of zeros # a k , 2, 3, 6, coefficient of x, , , EXAMPLE 3, , SOLUTION, , Find the zeros of the polynomial f (x) = x 2 2 and verify the, relationship between its zeros and coefficients., We have, f (x) = (x 2 2) = {x 2 ( 2 ) 2} = (x + 2 )(x 2 )., , , f (x) 0 & (x 2 )(x 2 ) 0, & x 2 0 or x 2 0, & x 2 or x 2 ., , So, the zeros of f (x) are 2 and 2 ., 0 (coefficient of x), Sum of zeros = ( 2 + 2 ) = 0 = 1 =, ,, (coefficient of x 2), 2, constant term 2 ·, product of zeros ( 2 ) # ( 2 ) , 1, coefficient of x, EXAMPLE 4, , SOLUTION, , Obtain the zeros of the quadratic polynomial 3 x 2 8x + 4 3 and, verify the relation between its zeros and coefficients., [CBSE 2008C], We have, , , , f (x) = 3 x 2 8x + 4 3 = 3 x 2 6x 2x + 4 3, = 3 x (x 2 3 ) 2 (x 2 3 ) = (x 2 3 ) ( 3 x 2) ., f (x) 0 & (x 2 3 )( 3 x 2) 0, , & (x 2 3 ) 0 or ( 3 x 2) 0, & x 2 3 or x 2 ·, , 3, 2 ·, So, the zeros of f(x) are 2 3 and, 3, 2 = 8 = (coefficient of x), Sum of zeros = d2 3 +, ,, n, (coefficient of x 2), 3, 3, product of zeros d2 3 #, , 2 4 3 constant term ·, n, coefficient of x 2, 3, 3
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Polynomials, , 47, , EXAMPLE 5, , Find a quadratic polynomial, the sum and product of whose zeros are, –5 and 6 respectively., , SOLUTION, , Let and be the zeros of the required polynomial f(x)., Then, ( + ) = 5 and = 6., , , f (x) = x 2 ( + ) x + , = x 2 (5) x + 6, = x 2 + 5x + 6., , Hence, the required polynomial is f (x) = x 2 + 5x + 6., EXAMPLE 6, , Find the quadratic polynomial, the sum of whose zeros is 2 and, their product is –12. Hence, find the zeros of the polynomial., , SOLUTION, , Let and be the zeros of the required polynomial f(x)., Then, ( + ) = 2 and = 12., , , f (x) = x 2 ( + ) x + , = x 2 2 x 12., , So, the required polynomial is f (x) = x 2 2 x 12., Now, f (x) = x 2 2 x 12, = x 2 3 2 x + 2 2 x 12 [note it], = x (x 3 2 ) + 2 2 (x 3 2 ), = (x 3 2 )(x + 2 2 ) ., , , f (x) 0 & (x 3 2 )(x 2 2 ) 0, , & x 3 2 0 or x 2 2 0, & x 3 2 or x 2 ., Hence, the required polynomial is f (x) = x 2 2 x whose, zeros are 3 2 and 2 2 ., EXAMPLE 7, , SOLUTION, , If the product of the zeros of the polynomial (ax 2 6x 6 is 4, find, the value of a., [CBSE 2009], Let and be the zeros of the polynomial (ax 2 6x 6) ., constant term 6 ·, Then, , a, coefficient of x 2, But, = 4 (given)., 3, 6, , ·, , 6 & a 6 , , a 4 & 4a, 2, 4, 3, ·, Hence, a , 2
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48, , Secondary School Mathematics for Class 10, , EXAMPLE 8, , If one zero of the polynomial (a 2 + 9) x 2 + 13x + 6a is reciprocal of the, other, find the value of a., [CBSE 2008], , SOLUTION, , Let one zero of the given polynomial be ., 1, Then, the other zero is ·, 1, product of zeros = b # l = 1., constant term 6a ·, But, product of zeros , coefficient of x 2 (a 2 9), 6a , , 1 & a 2 9 6a, (a 2 9), , , Hence, a = 3., EXAMPLE 9, , & a 2 9 6a 0, & (a 3)2 0, & a 3 0 & a 3., , Find a quadratic polynomial whose zeros are 1 and –3. Verify the, relation between the coefficients and zeros of the polynomial., [CBSE 2008C], , SOLUTION, , Let = 1 and = 3., Sum of zeros = ( + ) = 1 + (3) = 2., Product of zeros = = 1 # (3) = 3., So, the required polynomial is, x2 ( + ) x + = x2 (2) x + (3), = x2 + 2x 3., 2 (coefficient of x), Sum of zeros = 2 = 1 =, ,, (coefficient of x2), 3, constant term 2 ·, product of zeros 3 , 1, coefficient of x, , f, , EXERCISE 2A, , Find the zeros of the following quadratic polynomials and verify the relationship, between the zeros and the coefficients:, , 1. x2 + 7x + 12, 3. x2 + 3x 10, , 2. x2 2x 8, 4. 4x2 4x 3, , 5. 5x2 4 8x [CBSE 2008], 7. 2x2 11x + 15, , 6. 2 3 x2 5x + 3, 8. 4x2 4x + 1, , [CBSE 2008C], [CBSE 2011]
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Polynomials, , 9. x2 5, 11. 5y2 + 10y, , 49, , 10. 8x2 4, 12. 3x2 x 4, , 13. Find the quadratic polynomial whose zeros are 2 and –6. Verify the, relation between the coefficients and the zeros of the polynomial., 1, 2, · Verify the, 14. Find the quadratic polynomial whose zeros are 3 and, 4, relation between the coefficients and the zeros of the polynomial., 15. Find the quadratic polynomial, sum of whose zeros is 8 and their, product is 12. Hence, find the zeros of the polynomial., [CBSE 2008], 16. Find the quadratic polynomial, the sum of whose zeros is 0 and their, product is –1. Hence, find the zeros of the polynomial., 5, 17. Find the quadratic polynomial, the sum of whose zeros is b 2 l and their, product is 1. Hence, find the zeros of the polynomial., , 18. Find the quadratic polynomial, the sum of whose roots is 2 and their, 1, product is ·, 3, 2, 19. If x = 3 and x = 3 are the roots of the quadratic equation ax2 + 7x + b = 0, [CBSE 2011], then find the values of a and b., 20. If (x + a) is a factor of the polynomial 2x2 + 2ax + 5x + 10, find the value, of a., [CBSE 2009], 2, 3, 2, 21. One zero of the polynomial 3x 16x 15x 18 is · Find the other, 3, zeros of the polynomial., ANSWERS (EXERCISE 2A), , 1. –4, –3, , 3, , 6. 2 ,, 11. 0, –2, , 1, 3, , 2. 4, –2, , 5, , 3. –5, 2, , 1 1, , 7. 3, 2, , 8. 2 , 2, , 4, , 12. 3 , 1, , 3 1, , 4. 2 , 2, 9., , 5, 5, , 2, , 5. 2, 5, 10., , 1 1, ,, 2, 2, , 13. x2 + 4x 12 14. 12x2 5x 2, , 15. x2 8x + 12, {6, 2}, , 16. (x2 1), {1, 1}, , 17. (2x2 5x + 2), &2, 2 0, , 18. 3x2 3 2 x + 1, , 19. a = 3, b = 6, , 20. a = 2, , 1, , HINTS TO SOME SELECTED QUESTIONS, 5. 5x2 8x 4 = 5x2 10x + 2x 4, = 5x (x 2) + 2 (x 2) = (x 2) (5x + 2) ., , 21. –3, –3
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50, , Secondary School Mathematics for Class 10, , 6. 2 3 x2 5x + 3 = 2 3 x2 3x 2x + 3, = 3 x (2x 3 ) (2x 3 ), = (2x 3 ) ( 3 x 1) ., 8. 4x2 4x + 1 = (2x 1) 2 ., 9. (x2 5) = (x 5 )(x + 5 ) ., 10. 8x2 4 = 4 (2x2 1) = 4 ( 2 x 1)( 2 x + 1) ., 11. 5y2 + 10y = 5y (y + 2) ., 12. 3x2 x 4 = 3x2 4x + 3x 4 = x (3x 4) + (3x 4) = (3x 4)(x + 1) ., 7, 2, ·, 19. Sum of the roots a 3k , 3, 3, 2, Product of roots = 3 # (3) = 2., 7 7, b, 'a , and a 21 & a 3 and b 6., 3, 20. Let f (x) = 2x2 + 2ax + 5x + 10., Since (x + a) is a factor of f(x), we have f ( a) = 0., , , 2(a) 2 + 2a (a) + 5 (a) + 10 = 0, , , , 2a 2 2a 2 5a 10 0 & 5a 10 & a 2., , 2, 21. ax k is a factor of the given polynomial and therefore, (3x 2) is also its factor., 3, On dividing the given polynomial by (3x 2), we get (x 2 6x 9) as quotient, i.e.,, (x 3) 2 0 & x 3., , RELATION BETWEEN THE ZEROS AND COEFFICIENTS OF A CUBIC POLYNOMIAL, , Let , and be the zeros of a cubic polynomial, p(x) = ax3 + bx2 + cx + d, where a ! 0., Then, (x ), (x ) and (x ) are the factors of p(x) ., , , (ax3 + bx2 + cx + d) = k(x )(x )(x ) for some constant k, = k{x3 ( + + )x2 + ( + + )x ()}, = kx3 k( + + ) x2 + k( + + )x k()., , Comparing coefficients of like powers of x on both sides, we get, k = a, k( + + ) = b, k( + + ) = c, k() = d, , , , a( + + ) = b, a( + + ) = c, a() = d [a k a], b, d, c, ( ) a , ( ) a , a ·
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Polynomials, , 51, , SUMMARY, , I. If and are the zeros of p(x) = ax 3 + bx 2 + cx + d then, b, (i) ( + + ) = a, , c, (ii) ( + + ) = a, , d, (iii) a ·, , II. A cubic polynomial whose zeros are and is given by, p(x) {x 3 ( )x 2 ( )x } ., , SOLVED EXAMPLES, EXAMPLE 1, , 1, Verify that 3, –1 and 3 are the zeros of the cubic polynomial, p(x) = 3x3 5x2 11x 3 and verify the relation between its zeros, and coefficients., , SOLUTION, , The given polynomial is p(x) = 3x3 5x2 11x 3., , , p(3) = {3 # 33 5 # 32 11 # 3 3} = (81 45 33 3) = 0;, p(1) = {3 # (1) 3 5 # (1) 2 11 # (1) 3}, = (3 5 + 11 3) = 0;, , 1, 1 3, 1 2, 1, and p b 3 l = (3 # b 3 l 5 # b 3 l 11 # b 3 l 32, , , = '3 # b 1 l 5 # 1 + 11 31 = b 1 5 + 11 3l, 27, 9, 3, 9 9, 3, (1 5 + 33 27), =, = 0., 9, 1, 3, –1 and 3 are the zeros of p(x) ., 1, Let = 3, = 1 and = 3 $ Then,, 2, 1, 5 (coefficient of x ), ( + + ) = b3 1 3 l = 3 =, 3 ;, (coefficient of x ), (coefficient of x), 11, 1, ( + + ) = b3 + 3 1l = 3 =, ;, (coefficient of x3), 1, 3 (constant term) ·, &3 # (1) # a k0 1 , 3, 3 (coefficient of x 3), EXAMPLE 2, , Find a cubic polynomial with the sum of its zeros, sum of the products, of its zeros taken two at a time and the product of its zeros as 2, –7, and –14 respectively.
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52, SOLUTION, , Secondary School Mathematics for Class 10, , Let be the zeros of the required polynomial. Then,, + + = 2, + + = 7 and = 14., So, the required polynomial is, p(x) = x3 ( + + )x2 + ( + + )x , x 3 2x 2 7x (14) x 3 2x 2 7x 14., , EXAMPLE 3, , SOLUTION, , If the zeros of the polynomial x3 3x2 + x + 1 are (a b), a, (a + b),, find a and b., Given polynomial is f (x) = x3 3x2 + x + 1., Let = (a b), = a and = (a + b) . Then,, 3 & (a b) a (a b) 3 & 3a 3 & a 1., 1 & a (a b) a (a b) (a b)(a b) 1, , & 3a 2 b 2 1 & (3 # 1 2) b 2 1, & b2 2 & b ! 2 ., , EXAMPLE 4, SOLUTION, , a = 1 and b = ! 2 ., , 1, Find a cubic polynomial whose zeros are 3, 2 and 1., 1, Let = 3, = 2 and = 1. Then,, 1, 5, ( + + ) = b3 + 2 1l = 2 ,, 4, 3 1, ( + + ) = b 2 2 3l = 2 = 2, 3, 1, and = &3 # 2 # (1)0 = 2 $, Hence, the required polynomial is, 5, 3, x3 ( + + )x2 + ( + + )x = x3 2 x2 2x + 2 $, Thus, 2x3 5x2 4x + 3 is the desired polynomial., , DIVISION ALGORITHM FOR POLYNOMIALS, , If f(x) and g(x) are any two polynomials with g(x) ! 0 then we can find polynomials, q(x) and r(x) such that, f (x) = q(x) # g(x) + r(x),, where r(x) = 0 or {degree of r(x)} < {degree of g(x)} ., We may write it as, Dividend = (Quotient × Divisor) + Remainder.
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Polynomials, SOLUTION, , 55, , Let f (x) = (x3 3x2 + x + 2), q(x) = (x 2) and r(x) = ( 2x + 4)., Then, f (x) = g(x) $ q(x) + r(x), {f (x) r(x)}, ·, g(x) , q(x), Now, {f (x) r(x)} = (x3 3x2 + x + 2) (2x + 4), , … (i), , = (x3 3x2 + 3x 2)., (x3 3x2 + 3x 2), [using (i)], $, (x 2), On dividing (x3 3x2 + 3x 2) by (x 2), we get g(x)., , , g(x) =, , x 2) x3 3x2 + 3x 2 (x2 x + 1, x3 2x2, x2 + 3x 2, x2 + 2x, x2, x2, ×, , , g(x) = (x2 x + 1) ., , AN IMPORTANT NOTE, , If is a zero of the polynomial f(x) then (x ) is a divisor of f(x)., EXAMPLE 10, , SOLUTION, , It being given that 1 is a zero of the polynomial (7x x3 6), find its, other zeros., Let f (x) = x3 + 7x 6., Since 1 is a zero of f (x), so (x 1) is a factor of f (x)., On dividing f (x) by (x 1), we get, + 7x 6 ( x2 x + 6, x 1) x3, 3, 2, x + x, x2 + 7x 6, x2 + x, 6x 6, 6x 6, ×, , , f (x) (x 3 7x 6) (x 1)(x 2 x 6), (x 1)(x 2 x 6) (x 1)(x 2 3x 2x 6), (1 x) [x(x 3) 2(x 3)] (1 x)(x 3)(x 2) .
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56, , Secondary School Mathematics for Class 10, , , , f (x) 0 & (1 x)(x 3)(x 2) 0, , & (1 x) 0 or (x 3) 0 or (x 2) 0, & x 1 or x 3 or x 2., Thus, the other zeros are –3 and 2., EXAMPLE 11, , SOLUTION, , Obtain all zeros of the polynomial (2x3 4x x2 + 2), if two of its, zeros are 2 and 2 ., [CBSE 2008C], The given polynomial is f (x) = 2x3 x2 4x + 2., Since 2 and 2 are the zeros of f (x), it follows that each one, of (x 2 ) and (x + 2 ) is a factor of f (x)., Consequently, (x 2 )(x + 2 ) = (x2 2) is a factor of f (x)., On dividing f (x) = 2x3 x2 4x + 2 by (x2 2), we get, x2 2) 2x3 x2 4x + 2 (2x 1, 4x, 2x3, x2, x2, , , ×, , +2, +2, , f (x) 0 & (x 2)(2x 1) 0, 2, , & (x 2 ) (x 2 )(2x 1) 0, & (x 2 ) 0 or (x 2 ) 0 or (2x 1) 0, & x 2 or x 2 or x 12 ·, Hence, all zeros of f (x) are 2 , 2 and, EXAMPLE 12, , SOLUTION, , 1·, 2, , If two zeros of the polynomial (x4 6x3 26x2 + 138x 35) are, (2 + 3 ) and (2 3 ), find other zeros., Let f (x) = x4 6x3 26x2 + 138x 35., Let = (2 + 3 ) and = (2 3 ). Then,, ( + ) = 4 and = (4 3) = 1., So, the quadratic polynomial whose roots are and is, given by, x2 ( + ) x + = (x2 4x + 1)., , , (x2 4x + 1) is a factor of f(x)., , On dividing f (x) by (x2 4x + 1), we get
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58, , Secondary School Mathematics for Class 10, , , , f (x) = 3x4 15x3 + 13x2 + 25x 30, = (3x2 5) (x2 5x + 6), = ( 3 x + 5 )( 3 x 5 )(x 2)(x 3)., , , , f (x) 0 & ( 3 x 5 ) 0 or ( 3 x 5 ) 0, or (x 2) = 0 or (x 3) = 0, , 5, or x , 3, 5, Hence, all zeros of f(x) are 3 , , , &x, , f, , 5, or x 2 or x 3., 3, 5, 3 , 2 and 3., , EXERCISE 2B, , 1. Verify that 3, –2, 1 are the zeros of the cubic polynomial, p(x) = x3 2x2 5x + 6 and verify the relation between its zeros and, coefficients., 1, 2. Verify that 5, 2 and 3 are the zeros of the cubic polynomial, p(x) = 3x3 10x2 27x + 10 and verify the relation between its zeros and, coefficients., 3. Find a cubic polynomial whose zeros are 2, –3 and 4., 1, 4. Find a cubic polynomial whose zeros are 2 , 1 and 3., 5. Find a cubic polynomial with the sum, sum of the product of its zeros, taken two at a time, and the product of its zeros as 5, –2 and –24, respectively., Find the quotient and the remainder when, , 6. f (x) = x3 3x2 + 5x 3 is divided by g(x) = x2 2., 7. f (x) = x4 3x2 + 4x + 5 is divided by g(x) = x2 + 1 x., 8. f (x) = x4 5x + 6 is divided by g(x) = 2 x2., 9. By actual division, show that x2 3 is a factor of 2x4 + 3x3 2x2 9x 12., 10. On dividing 3x3 + x2 + 2x + 5 by a polynomial g(x), the quotient and, remainder are (3x 5) and (9x + 10) respectively. Find g(x)., HINT, , g(x) =, , (3x3 + x2 + 2x + 5) (9x + 10), ·, (3x 5), , 11. Verify division algorithm for the polynomials f (x) = 8 + 20x + x2 6x3, and g(x) = 2 + 5x 3x2 ., 12. It is given that –1 is one of the zeros of the polynomial x3 + 2x2 11x 12., Find all the zeros of the given polynomial.
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Polynomials, , 59, , 13. If 1 and –2 are two zeros of the polynomial (x3 4x2 7x + 10), find its, third zero., 14. If 3 and –3 are two zeros of the polynomial (x4 + x3 11x2 9x + 18), find, all the zeros of the given polynomial., 15. If 2 and –2 are two zeros of the polynomial (x4 + x3 34x2 4x + 120),, find all the zeros of the given polynomial., [CBSE 2008], 16. Find all the zeros of (x4 + x3 23x2 3x + 60), if it is given that two of its, zeros are 3 and 3 ., [CBSE 2009C], 17. Find all the zeros of (2x4 3x3 5x2 + 9x 3), it being given that two of, its zeros are 3 and 3 ., 18. Obtain all other zeros of (x4 + 4x3 2x2 20x 15) if two of its zeros are, [CBSE 2009C], 5 and 5 ., 19. Find all the zeros of the polynomial (2x4 11x3 + 7x2 + 13x 7), it being, given that two of its zeros are (3 + 2 ) and (3 2 )., ANSWERS (EXERCISE 2B), , 3. x3 3x2 10x + 24, , 4. 2x3 + 3x2 8x + 3, , 6. q(x) = (x 3), r(x) = (7x 9), 8. q(x) = x2 2, r(x) = 5x + 10, 12. – 4, –1, 3 13. 5, 16., , 3 , 3 , 4, 5, , 14. 1, 2, 3, 3, 17., , 1, 19. (3 + 2 ), (3 2 ), 2 , 1, f, , 5. x3 5x2 2x + 24, , 7. q(x) = x2 + x 3, r(x) = 8, 10. g(x) = x2 + 2x + 1, 15. 2, 2, 6, 5, , 1, 3 , 3 , 1, 2, , 18. 1, 3, , EXERCISE 2C, , Very-Short-Answer Questions, 1. If one zero of the polynomial x2 4x + 1 is (2 + 3 ), write the other zero., [CBSE 2010], , 2. Find the zeros of the polynomial x2 + x p(p + 1)., 3. Find the zeros of the polynomial x2 3x m(m + 3)., , [CBSE 2011], [CBSE 2011], , 4. If are the zeros of a polynomial such that + = 6 and = 4 then, write the polynomial., [CBSE 2010], 2, 5. If one zero of the quadratic polynomial kx + 3x + k is 2 then find the, value of k., 6. If 3 is a zero of the polynomial 2x2 + x + k, find the value of k. [CBSE 2010]
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60, , Secondary School Mathematics for Class 10, , 7. If – 4 is a zero of the polynomial x2 x (2k + 2) then find the value of k., [CBSE 2009], , 8. If 1 is a zero of the polynomial ax 3(a 1) x 1 then find the value, of a., 9. If –2 is a zero of the polynomial 3x2 + 4x + 2k then find the value of k., 2, , [CBSE 2010], , 10. Write the zeros of the polynomial x x 6., 2, , [CBSE 2008], , 11. If the sum of the zeros of the quadratic polynomial kx 3x + 5 is 1,, write the value of k., 12. If the product of the zeros of the quadratic polynomial x2 4x + k is 3, then write the value of k., 13. If (x + a) is a factor of (2x2 + 2ax + 5x + 10), find the value of a. [CBSE 2010], 2, , 14. If (a b), a and (a + b) are zeros of the polynomial 2x3 6x2 + 5x 7, write, the value of a., 15. If x3 + x2 ax + b is divisible by (x2 x), write the values of a and b., 16. If and are the zeros of the polynomial 2x2 + 7x + 5, write the value of, [CBSE 2010], + + ., 17. State division algorithm for polynomials., 18. The sum of the zeros and the product of zeros of a quadratic polynomial, 1, are 2 and –3 respectively. Write the polynomial., Short-Answer Questions, 19. Write the zeros of the quadratic polynomial f (x) = 6x2 3., 20. Write the zeros of the quadratic polynomial f (x) = 4 3 x2 + 5x 2 3 ., 21. If and are the zeros of the polynomial f (x) = x2 5x + k such that, = 1, find the value of k., 22. If and are the zeros of the polynomial f (x) = 6x2 + x 2, find the, , value of d n ·, , 23. If and are the zeros of the polynomial f (x) = 5x2 7x + 1, find the, 1 1, value of c m ·, , 24. If and are the zeros of the polynomial f (x) = x2 + x 2, find the value, 1 1, of c m ·,
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Polynomials, , 61, , 25. If the zeros of the polynomial f (x) = x3 3x2 + x + 1 are (a b), a and, (a + b), find a and b., ANSWERS (EXERCISE 2C), , 1. (2 3 ), , 6, , 5. k = 5, 10. 3 and –2, , 2. (p + 1) and p, , 3. (m + 3) and m, , 4. x2 6x + 4, , 6. k = 21, , 7. k = 9, , 9. k = 2, , 11. k = 3, 16. –1, , 15. a = 2 and b = 0, 20. x =, , 12. k = 3, , 8. a = 1, 13. a = 2, , 1, , 18. x2 + 2 x 3 19. x =, , 25, , 3, 2, or x = 4 21. k = 6, 3, , 22. 12, , 23. 7, , 25. a = 1 and b = ! 2, HINTS TO SOME SELECTED QUESTIONS, 1. Let the other zero be ., (coefficient of x) (4), =, = 4., Then, sum of zeros =, 1, (coefficient of x2), , 2+, , 2. x, , (2 , , 3 ) 4 & (2 , , x p (p + 1) = x, , 2+, , 3)., , (p + 1) x px p (p + 1), , = x {x + (p + 1)} p {x + (p + 1)}, = {x + (p + 1)}{x p} ., So, its zeros are (p + 1) and p., 3. x2 3x m(m + 3) = x2 (m + 3) x + mx m(m + 3), = x {x (m + 3)} + m {x (m + 3)}, = {x (m + 3)}{x + m} ., So, its zeros are (m + 3) and m., 4. Required polynomial is x2 ( + ) x + = x2 6x + 4., 5. Since 2 is a zero of kx2 + 3x + k, we have 4k + 6 + k = 0., 6, ·, 5k 6 & k , 5, 6. Since 3 is a zero of 2x2 + x + k, we have 18 + 3 + k = 0 & k = 21., 7. Since –4 is a zero of x2 x (2k + 2), we have 16 + 4 2k 2 = 0., , , 2k = 18 & k = 9., , 8. Since 1 is a zero of ax2 3 (a 1) x 1, we have a 3 (a 1) 1 = 0., a 3a 3 1 0 & 2a 2 & a 1., 9. Since –2 is a zero of 3x2 + 4x + 2k, we have 12 8 + 2k = 0., , , 2k = 4 & k = 2., , 14. a = 1, , 1, 1, or x =, 2, 2, 3, 24. 2
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62, , Secondary School Mathematics for Class 10, , 10. x2 x 6 = x2 3x + 2x 6 = x (x 3) + 2 (x 3) = (x 3) (x + 2) ., So, its zeros are 3 and –2., 11. Sum of the zeros , , , (coefficient of x), (coefficient of x 2), , 3·, k, , 3, 1 & k 3., k, , constant term k ·, So, k = 3., coefficient of x 2 1, 13. Let f (x) = 2x2 + 2ax + 5x + 10. Then, f ( a) = 0., 12. Product of the zero , , , , 2a 2 2a 2 5a 10 0 & 5a 10 & a 2., (coefficient of x2) ( 6), =, = 3., 2, (coefficient of x3), (a b) a (a b) 3 & 3a 3 & a 1., , 14. ( + + ) =, , , 15. On dividing x3 + x2 ax + b by x2 x, we get, x2 x ) x3 + x2 ax + b (x + 2, x3 x2, 2x2 ax, 2x2 2x, (2 a) x + b, , , (2 a 0 and b 0) & a 2 and b 0., 7, 5, and ·, 2, 2, (7 + 5) 2, 7 5, =, = 1., + + = b 2 + 2 l =, 2, 2, , 16. Clearly, , , , 1, 18. + = 2 and = 3., 1, So, the polynomial is x2 ( + ) x + = x2 + 2 x 3., 19. 6x2 3 = 3 (2x2 1) = 3 ( 2 x 1)( 2 x + 1) ., 1, 1, ·, So, its zeros are, and, 2, 2, 20. f (x) = 4 3 x2 + 5x 2 3, = 4 3 x2 + 8x 3x 2 3 = 4x ( 3 x + 2) 3 ( 3 x + 2), = ( 3 x + 2)(4x 3 ) ., 2 3, ·, ,, 3 4, 21. ( + ) = 5 and = k. Also, = 1 (given)., So, its zeros are, , , , ( ) 2 ( ) 2 4 & 5 2 1 2 4, , , , 4 (25 1) 24 & 6., , Hence, k = 6.
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Polynomials, , 63, , 1, 2 1, 22. + = 6 and = 6 = 3 ·, , , 1 +2, (2 + 2) ( + ) 2 2 b 36 3 l, , , d + n =, =, =, = b 25 # 3 l = 25 ·, 36 1, 12, , , , , b 1l, 3, , 7, 1, 23. + = 5 and = 5 $, ( + ), 1 1, 7 5, c + m = = b 5 # 1 l = 7., 24. + = 1 and = 2., ( + ) 2 ( ) 2 = 4, , , ( ) 2 = ( + ) 2 4 = (1) 2 4 # (2) = 9, , , , = 3., , , , 1 1 m = ( ) = 3 = 3 ·, c, 2, 2, , , , 25. + + = 3, + + = 1 and = 1, , , (a b) a (a b) 3 & 3a 3 & a 1, , and (a b)# a #(a b) 1 & a (a 2 b 2) 1, , & 1 b 2 1 & b 2 2 & b ! 2 ., , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. Which of the following is a polynomial?, (a) x2 5x + 4 x + 3, 1, (c) x +, x, , (b) x3/2 x + x1/2 + 1, (d), , 2 x2 3 3 x + 6, , 2. Which of the following is not a polynomial?, 3 x2 2 3 x + 5, 3, 1 , (c) 2 x3 + 6x2 , x 8, 2, , (b) 9x2 4x + 2, 3, (d) x + x, , (a), , 3. The zeros of the polynomial x2 2x 3 are, (a) –3, 1, , (b) –3, –1, , (c) 3, –1, , (d) 3, 1, , 4. The zeros of the polynomial x 2 x 12 are, 2, , (a), , 2, 2, , (b) 3 2 , 2 2, , (c) 3 2 , 2 2, , (d) 3 2 , 2 2, , 5. The zeros of the polynomial 4x2 + 5 2 x 3 are, (a) 3 2 , 2, , 2, (b) 3 2 , 2, , (c), , 3 2, 2, 2 , 4, , (d) none of these
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64, , Secondary School Mathematics for Class 10, , 1, 6. The zeros of the polynomial x2 + 6 x 2 are, 4 3, 3 4, (a) –3, 4, (b) 2 , 3, (c) 3 , 2, (d) none of these, 11x 2, 7. The zeros of the polynomial 7x2 3 3 are, 2 1, 2 1, 2 1, (b) 7 , 3, (c) 3 , 7, (d) none of these, (a) 3 , 7, 8. The sum and the product of the zeros of a quadratic polynomial are, 3 and –10 respectively. The quadratic polynomial is, (a) x2 3x + 10 (b) x2 + 3x 10, (c) x2 3x 10 (d) x2 + 3x + 10, 9. A quadratic polynomial whose zeros are 5 and –3, is, (a) x2 + 2x 15 (b) x2 2x + 15, (c) x2 2x 15 (d) none of these, 1, 3, 10. A quadratic polynomial whose zeros are 5 and 2 , is, (a) 10x2 + x + 3, , (b) 10x2 + x 3, , (c) 10x2 x + 3 (d) 10x2 x 3, , 11. The zeros of the quadratic polynomial x2 + 88x + 125 are, (a) both positive, , (b) both negative, , (c) one positive and one negative (d) both equal, 12. If and are the zeros of x2 + 5x + 8 then the value of ( + ) is, (a) 5, , (b) –5, , (c) 8, , (d) –8, , 13. If and are the zeros of 2x + 5x 9 then the value of is, 5, 9, 5, 9, (a) 2, (b) 2, (c) 2, (d) 2, 2, , 14. If one zero of the quadratic polynomial kx2 + 3x + k is 2 then the value, of k is, 6, 5, 5, 6, (b) 6, (c) 5, (d) 5, (a) 6, 15. If one zero of the quadratic polynomial (k 1) x2 + kx + 1 is – 4 then the, value of k is, 5, 4, 4, 5, (b) 4, (c) 3, (d) 3, (a) 4, 16. If –2 and 3 are the zeros of the quadratic polynomial x2 + (a + 1) x + b, then, (a) a = 2, b = 6, (b) a = 2, b = 6, (c) a = 2, b = 6, (d) a = 2, b = 6, 17. If one zero of 3x2 + 8x + k be the reciprocal of the other then k = ?, 1, 1, (a) 3, (b) –3, (c) 3, (d) 3
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Polynomials, , 65, , 18. If the sum of the zeros of the quadratic polynomial kx2 + 2x + 3k is equal, to the product of its zeros then k = ?, 2, 1, 2, 1, (a) 3, (b) 3, (c) 3, (d) 3, 1 1, 19. If are the zeros of the polynomial x2 + 6x + 2 then c + m = ?, (a) 3, (b) –3, (c) 12, (d) –12, 3, , 20. If are the zeros of the polynomial x 6x2 x + 30 then, ( + + ) = ?, (a) –1, (b) 1, (c) –5, (d) 30, 3, 2, 21. If are the zeros of the polynomial 2x + x 13x + 6 then = ?, 13, 1, (c) 2, (d) 2, 22. If be the zeros of the polynomial p(x) such that ( + + ) = 3,, ( + + ) = 10 and = 24 then p(x) = ?, (a) –3, , (b) 3, , (a) x3 + 3x2 10x + 24, (c) x3 3x2 10x + 24, , (b) x3 + 3x2 + 10x 24, (d) None of these, 23. If two of the zeros of the cubic polynomial ax3 + bx2 + cx + d are 0 then, the third zero is, b, d, b, c, (a) a, (b) a, (c) a, (d) a, 24. If one of the zeros of the cubic polynomial ax3 + bx2 + cx + d is 0 then the, product of the other two zeros is, c, b, c, (a) a, (b) a, (c) 0, (d) a, 25. If one of the zeros of the cubic polynomial x3 + ax2 + bx + c is –1 then the, product of the other two zeros is, (a) a b 1, (b) b a 1, (c) 1 a + b, (d) 1 + a b, 26. If be the zeros of the polynomial 2x2 + 5x + k such that, 21, 2 + 2 + = 4 then k = ?, (a) 3, (b) –3, (c) –2, (d) 2, 27. On dividing a polynomial p(x) by a nonzero polynomial q(x), let g(x), be the quotient and r(x) be the remainder then p(x) = q(x) $ g(x) + r(x),, where, (a) r(x) = 0 always, (b) deg r(x) < deg g(x) always, (c) either r(x) = 0 or deg r(x) < deg g(x), (d) r(x) = g(x)
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66, , Secondary School Mathematics for Class 10, , 28. Which of the following is a true statement?, (a) x2 + 5x 3 is a linear polynomial., (b) x2 + 4x 1 is a binomial., (c) x + 1 is a monomial., (d) 5x3 is a monomial., ANSWERS (MCQ), , 1., 10., 19., 28., , 4. (b) 5. (c), 6. (b) 7. (a), 8. (c), 9. (c), (d) 2. (d) 3. (c), (d) 11. (b) 12. (b) 13. (c) 14. (d) 15. (b) 16. (c) 17. (a) 18. (d), (b) 20. (a) 21. (a) 22. (c) 23. (a) 24. (b) 25. (c) 26. (d) 27. (c), (d), HINTS TO SOME SELECTED QUESTIONS, , 1. Clearly, 2 x2 3 3 x + 6 is a polynomial., 3, 2. Clearly, x + x is not a polynomial., 3. x2 2x 3 = x2 3x + x 3, = x (x 3) + (x 3) = (x 3)(x + 1) ., (x 3)(x 1) 0 & x 3 or x 1., , , , 4. x2 2 x 12 = x2 3 2 x + 2 2 x 12, = x (x 3 2 ) + 2 2 (x 3 2 ) = (x 3 2 )(x + 2 2 ) ., x = 3 2 or x = 2 2 ., , , 2+, , 5. 4x, , 5 2 x 3 = 4x2 + 6 2 x 2 x 3, = 2 2 x ( 2 x + 3) ( 2 x + 3) = ( 2 x + 3)(2 2 x 1) ., , , , x, , 2 3 2, 2, 2, 3, 1, ·, , , #, #, or x , 2, 4, 2, 2, 2 2, 2, , 6x 2 x 12 ·, 1, 6. x 2 x 2 , 6, 6, 6x2 + x 12 = 6x2 + 9x 8x 12 = 3x (2x + 3) 4 (2x + 3), = (2x + 3)(3x 4), , , 3, 4, the zeros are 2 and ·, 3, , 11x 2 21x2 11x 2 ·, 7. 7x2 3 3 =, 3, Now, 21x2 11x 2 = 21x2 14x + 3x 2, , , , = 7x (3x 2) + (3x 2) = (3x 2)(7x + 1) ., 2 1 ·, the zeros are ,, 3 7
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Polynomials, 8. The required polynomial is, x2 ( + ) x + = x2 3x 10., 9. + = 5 + (3) = 2, = 5 # (3) = 15., Required polynomial is x2 2x 15., 3, 1, 3 1, 3, 1, ·, 10. a k , # a k , 5 2, 5, 2, 10, 10, 1, 3, Required polynomial is x2 10 x 10 , i.e., 10x2 x 3., 11. Let and be the zeros of the given polynomial., Then, + = 88 and = 125., This happens when and are both negative., b, 12. For ax2 + bx + c, we have a ·, For x2 + 5x + 8, we have + = 5., c, 13. For ax2 + bx + c, we have a ·, 9, ·, For 2x2 + 5x 9, we have , 2, 14. x = 2 satisfies kx2 + 3x + k = 0., , , 4k 6 k 0 & 5k 6 & k , , 6, ·, 5, , 15. x = 4 satisfies (k 1) x2 + kx + 1 = 0., , , 16 (k 1) 4k 1 0 & 12k 15 & k , , 5·, 4, , 16. + = 3 + (2) = 1 and = 3 # (2) = 6., , , (a 1) 1 & a 1 1 & a 2., , Also, b = 6., k, 1 & k 3., 3, 2 3k, 2, , , 3 & k 2·, 18. &, &, 3, k, k, k, , =, =, +, 19. , 6 and 2., ( + ) 6, 1, 1, c + m = = 2 = 3., 17. 1 &, , For ax3 + bx2 + cx + d, we have, b, d, c, ( ) a , ( ) a and a ·, c 1, 20. ( + + ) = a = 1 = 1., d 6, 21. = a = 2 = 3., Note, , 22. p(x) = x3 ( + + ) x2 + ( + + ) x , = x3 3x2 10x + 24., , 67
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68, , Secondary School Mathematics for Class 10, , 23. Let , 0, 0 be the zeros of ax3 + bx2 + cx + d. Then,, b, b, b, sum of zeros a & 0 0 a & a ·, b, Hence, the third zero is a ·, 24. Let , , 0 be the zeros of ax3 + bx2 + cx + d. Then,, sum of the products of zeros, taken two at a time is given by, c, c, ( # 0 # 0) a & a ·, c, the product of the other two zeros is a ·, 25. Since –1 is a zero of x3 + ax2 + bx + c, we have, (1) 3 a #(1) 2 b #(1) c 0 & a b c 1 0 & c 1 a b., Also, product of all zeros is given by, #(1) c & c & 1 a b., 5, k, 26. , and ·, 2, 2, 21, 21, 2 2 , & ( ) 2 , 4, 4, , , 5, k 21, &a 2 k 2 , 4, 2, , 25 21, k, & 2 a k 1 & k 2., 4, 4, , SUMMARY OF RESULTS, 1., , I. Polynomial: An expression of the form p(x) = a0 + a1 x + a2 x2 + … + an xn,, where an ! 0, is called a polynomial in x of degree n., II. A polynomial is said to be linear, quadratic, cubic and biquadratic, according as its degree is 1, 2, 3 and 4 respectively., General form:, , (i) linear equation: ax + b., (ii) quadratic equation: ax2 + bx + c., (iii) cubic equation: ax3 + bx2 + cx + d., (iv) biquadratic equation: ax4 + bx3 + cx2 + dx + e., III. Value of a polynomial at a point: The value of a polynomial p(x) at, x = is obtained by putting x = and it is denoted by p()., IV. Zeros of a polynomial: A real number is called a zero of p(x), if, p() = 0.
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Polynomials, , 2., , 69, , I. If are the zeros of p(x) ax 2 bx c, a ! 0 then, b, (i) + = a ,, , c, , (ii) = a ·, , II. If and are the zeros of a quadratic polynomial p(x) then, p(x) = {x2 ( + )x + } ., , 3., , I. If are the zeros of p(x) ax 3 bx 2 cx d, a ! 0 then, d, b, c, (i) + + = a ,, (ii) + + = a ,, (iii) = a ·, II. If and be the zeros of a polynomial p(x) then, p(x) = x3 ( + + ) x2 + ( + + ) x ., , TEST YOURSELF, MCQ, 1. Zeros of p (x) = x2 2x 3 are, (a) 1, –3, , (b) 3, –1, , (c) –3, –1, (d) 1, 3, 3, 2, 2. If are the zeros of the polynomial x 6x x + 30 then the value, of ( + + ) is, (a) –1, , (b) 1, , (c) –5, , (d) 30, , 3. If are the zeros of kx 2x + 3k such that + = then k = ?, 2, 1, 2, 1, (a) 3, (b) 3, (c) 3, (d) 3, 4. It is given that the difference between the zeros of 4x2 8kx + 9 is 4 and, k > 0. Then, k = ?, 3, 5, 7, 1, (a) 2, (b) 2, (c) 2, (d) 2, Short-Answer Questions, 5. Find the zeros of the polynomial x2 + 2x 195., 2, , 6. If one zero of the polynomial (a2 + 9) x2 + 13x + 6a is the reciprocal of the, other, find the value of a., 7. Find a quadratic polynomial whose zeros are 2 and –5., 8. If the zeros of the polynomial x3 3x2 + x + 1 are (a b), a and (a + b), find, the values of a and b., 9. Verify that 2 is a zero of the polynomial x3 + 4x2 3x 18., 10. Find the quadratic polynomial, the sum of whose zeros is –5 and their, product is 6.
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70, , Secondary School Mathematics for Class 10, , 11. Find a cubic polynomial whose zeros are 3, 5 and –2., 12. Using, remainder, theorem,, find, the, remainder, 3, 2, , , =, +, +, p(x) x 3x 5x 4 is divided by (x 2) ., 13. Show that (x + 2) is a factor of f (x) = x3 + 4x2 + x 6., , when, , 14. If are the zeros of the polynomial p(x) = 6x3 + 3x2 5x + 1, find the, 1 1 1, value of c m ·, , 15. If are the zeros of the polynomial f (x) = x2 5x + k such that, = 1, find the value of k., 16. Show that the polynomial f (x) = x4 + 4x2 + 6 has no zero., Long-Answer Questions, 17. If one zero of the polynomial p(x) = x3 6x2 + 11x 6 is 3, find the other, two zeros., 18. If two zeros of the polynomial p(x) = 2x4 3x3 3x2 + 6x 2 are 2 and, 2 , find its other two zeros., 19. Find the quotient when p(x) = 3x4 + 5x3 7x2 + 2x + 2 is divided by, (x2 + 3x + 1) ., 20. Use remainder theorem to find the value of k, it being given that when, x3 + 2x2 + kx + 3 is divided by (x 3) then the remainder is 21., ANSWERS (TEST YOURSELF), , 1. (a), , 2. (a), +, 7. x 3x 10, 2, , 12. 14, , 3. (c), , 4. (c), =, 8. a 1, b = ! 2, , 14. 5, , 15. k = 6, , 5. –15, 13, , 6. a = 3, , 10. x + 5x + 6, , 11. x3 6x2 x + 30, , 17. 1, 2, , 18. 1, 2, , 2, , 19. 3x2 4x + 2 20. k = 9, , , , 1
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Linear Equations in Two Variables, , 71, , LINEAR EQUATIONS IN TWO VARIABLES, , An equation of the form ax + by + c = 0, where a, b, c are real numbers (a ! 0, b ! 0),, is called a linear equation in two variables x and y., Examples, , Each of the equations, (i) 3x 4y + 2 = 0, , (ii) 2x + 5y = 9, , (iii) 0.4x + 0.3y = 2.7, , (iv), , 2x 3y = 0, , is a linear equation in x and y., SOLUTION OF A LINEAR EQUATION, , We say that x = and y = is a solution of ax + by + c = 0 if a + b + c = 0., EXAMPLE, , Show that x = 3 and y 2 is a solution of 5x 3y 9., , SOLUTION, , Substituting x = 3 and y = 2 in the given equation, we get, LHS = 5 # 3 3 # 2 = (15 6) = 9 = RHS., , , , x = 3 and y = 2 is a solution of 5x 3y = 9., , SIMULTANEOUS LINEAR EQUATIONS IN TWO VARIABLES, , Two linear equations in two unknowns x and y are said to form a system of, simultaneous linear equations if each of them is satisfied by the same pair of values, of x and y., Example, , Consider the system of linear equations, x + y = 10, x y = 2., By substitution, you will find that each of these equations is, satisfied by the values x = 6 and y = 4., Hence, the given equations form a system of simultaneous, linear equations in x and y., , SOLUTION OF A GIVEN SYSTEM OF TWO SIMULTANEOUS EQUATIONS, , A pair of values of x and y satisfying each of the equations in a given system of two, simultaneous equations in x and y is called a solution of the system., 71
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72, EXAMPLE 1, , Secondary School Mathematics for Class 10, , Show that x = 5, y = 2 is a solution of the system of linear equations, 2x + 3y = 16, x 2y = 1., , SOLUTION, , The given equations are, 2x + 3y = 16, , … (i), , x 2y = 1., , … (ii), , Putting x = 5 and y = 2 in (i), we get, LHS = (2 # 5 + 3 # 2) = 16 = RHS., , Putting x = 5 and y = 2 in (ii), we get, LHS = (5 2 # 2) = 1 = RHS., , Thus, x = 5 and y = 2 satisfy both (i) and (ii)., Hence, x = 5, y = 2 is a solution of the given system of, equations., EXAMPLE 2, , Show that x = 3, y = 2 is not a solution of the system of linear, equations 3x 2y = 5, 2x + y = 7., , SOLUTION, , The given equations are, 3x 2y = 5, , … (i), , 2x + y = 7., , … (ii), , Putting x = 3 and y = 2 in (i), we get, LHS = (3 # 3 2 # 2) = 5 = RHS., , Putting x = 3 and y = 2 in (ii), we get, LHS = (2 # 3 + 2) = 8 ! RHS., , Thus, the values x = 3, y = 2 do not satisfy (ii)., Hence, x = 3, y = 2 is not a solution of the given system of, equations., CONSISTENT AND INCONSISTENT SYSTEMS OF LINEAR EQUATIONS, CONSISTENT SYSTEM OF LINEAR EQUATIONS, , A system of two linear equations in two unknowns is said to be consistent if it has, at least one solution., INCONSISTENT SYSTEM OF LINEAR EQUATIONS, , A system of two linear equations in two unknowns is said to be inconsistent if it, has no solution at all.
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Linear Equations in Two Variables, Example, , 73, , Consider the system of linear equations, x + y = 3, 2x + 2y = 7., Clearly, we cannot find values of x and y which may satisfy, both the given equations simultaneously., Hence, the given system is inconsistent., , SOLVING SIMULTANEOUS LINEAR EQUATIONS, (GRAPHICAL METHOD), METHOD, , Let the given system of linear equations be, a1 x + b1 y + c1 = 0, a2 x + b2 y + c2 = 0., , … (i), … (ii), , On the same graph paper, we draw the graph of each one of the, given linear equations., Each such graph is always a straight line., Let the lines L1 and L2 represent the graphs of (i) and (ii) respectively., Now, the following cases arise., Case I, , When the lines L1 and L2 intersect at a, point, Let the graph lines L1 and L2 intersect, at a point P (, ), as shown in the, adjoining figure., Then, x = , y = is the unique, solution of the given system of, equations., , Case II, , When the lines L1 and L2 are coincident, When the two graph lines L1 and, L2 coincide, the given system, of equations has infinitely many, solutions.
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74, , Secondary School Mathematics for Class 10, , Case III When the lines L1 and L2 are parallel, , In this case, there is no common, solution of the given system of, equations., Thus, in this case, the given, system is inconsistent., , SUMMARY, , A system of two linear equations in x and y has, (i) a unique solution if the graph lines intersect at a point, (ii) infinitely many solutions if the two graph lines coincide, (iii) no solution if the two graph lines are parallel, , SOLVED EXAMPLES, GRAPHS OF EQUATIONS HAVING UNIQUE SOLUTIONS, Note, , Such graph lines intersect at a point., , EXAMPLE 1, , Solve graphically the system of linear equations, x + 2y = 3, 4x + 3y = 2., , SOLUTION, , On a graph paper, draw a horizontal line X lOX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of x + 2y = 3, , x 2y 3 & 2y (3 x), (3 x), & y 2 ·, , … (i), , Putting x = 3 in (i), we get y = 3., Putting x = 1 in (i), we get y = 2., Putting x = 1 in (i), we get, , y = 1., , Thus, we have the following table for the equation x + 2y = 3., x, , –3, , –1, , 1, , y, , 3, , 2, , 1, , Now, plot the points A(3, 3), B (1, 2) and C(1, 1) on the graph, paper.
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Linear Equations in Two Variables, , 75, , Join AB and BC to get the graph line ABC. Extend it on, both ways., Thus, the line ABC is the graph of x + 2y = 3., Graph of 4x + 3y = 2, , 4x 3y 2 & 3y (2 4x), (2 4x), & y 3 ·, Putting x = 4 in (ii), we get y = 6., , … (ii), , Putting x = 1 in (ii), we get y = 2., y = 2., , Putting x = 2 in (ii), we get, , Thus, we have the following table for the equation 4x + 3y = 2., x, , –4, , –1, , 2, , y, , 6, , 2, , –2, , Now, on the same graph paper as above, plot the points P(4, 6), and Q (2, 2). The point B(1, 2) has already been plotted. Join, PB and BQ to get the line PBQ. Extend it on both ways., Thus, the line PBQ is the graph of 4x + 3y = 2., , The two graph lines intersect at the point B(1, 2) ., x = 1 and y = 2 is the solution of the given system of, equations., EXAMPLE 2, , Solve graphically the system of linear equations, 4x 5y + 16 = 0 and 2x + y 6 = 0.
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76, , Secondary School Mathematics for Class 10, , Determine the vertices of the triangle formed by these lines and the, x-axis., [CBSE 2006], SOLUTION, , On a graph paper, draw a horizontal line X l OX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of 4x 5y + 16 = 0, , 4x 5y 16 0 & 5y 4x 16, (4x 16), ·, & y, 5, , … (i), , Putting x = 4 in (i), we get y = 0., Putting x = 1 in (i), we get, , y = 4., , Putting x = 6 in (i), we get, , y = 8., , Thus, we have the following table for 4x 5y + 16 = 0., x, , –4, , 1, , 6, , y, , 0, , 4, , 8, , Now, plot the points A(4, 0), B(1, 4) and C(6, 8) on the graph, paper., Join AB and BC to get the graph line ABC. Extend it on, both ways., Thus, the line ABC is the graph of 4x 5y + 16 = 0., Graph of 2x + y 6 = 0, , 2x y 6 0 & y (6 2x) ., , … (ii), , Putting x = 0 in (ii), we get y = 6., Putting x = 2 in (ii), we get y = 2., Putting x = 3 in (ii), we get y = 0., Thus, we have the following table for 2x + y 6 = 0., x, , 0, , 2, , 3, , y, , 6, , 2, , 0, , On the same graph paper as above, plot the points, P(0, 6), Q(2, 2) and R(3, 0) ., Join PQ and QR to get the graph line PQR. Extend it on, both ways., Thus, the line PQR is the graph of 2x + y 6 = 0.
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Linear Equations in Two Variables, , 77, , The two graph lines ABC and PQR intersect at the point B(1, 4) ., x = 1 and y = 4 is the solution of the given system of, equations., These lines form 3 BAR with the x-axis, whose vertices are, B(1, 4), A(4, 0) and R(3, 0) ., EXAMPLE 3, , Solve the following system of linear equations graphically:, 4x 5y 20 = 0 and 3x + 5y 15 = 0., Determine the vertices of the triangle formed by the lines representing, the above equations and the y-axis., [CBSE 2004], , SOLUTION, , On a graph paper, draw a horizontal line X lOX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of 4x 5y 20 = 0, , 4x 5y 20 0 & 5y (4x 20), (4x 20), ·, & y, 5, Putting x = 0 in (i), we get y = 4., , … (i), , Putting x = 2 in (i), we get y = 2.4., Putting x = 5 in (i), we get y = 0., Thus, we have the following table for 4x 5y 20 = 0., x, , 0, , 2, , 5, , y, , –4, , –2.4, , 0
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78, , Secondary School Mathematics for Class 10, , Now, plot the points A(0, 4), B(2, 2.4) and C (5, 0) on the, graph paper., Join AB and BC to get the graph line ABC. Extend it on both, ways., Thus, the line ABC is the graph of 4x 5y 20 = 0., Graph of 3x + 5y 15 = 0, , 3x 5y 15 0 & 5y (15 3x), (15 3x), ·, 5, Putting x = 5 in (ii), we get y = 6., , & y, , Putting x = 0 in (ii), we get, , y = 3., , Putting x = 5 in (ii), we get, , y = 0., , … (ii), , Thus, we have the following table for 3x + 5y 15 = 0., x, , –5, , 0, , 5, , y, , 6, , 3, , 0, , On the same graph paper as above, plot the points P(5, 6) and, Q(0, 3) . The third point C(5, 0) has already been plotted., Join PQ and QC to get the graph line PQC. Extend it on, both ways.
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Linear Equations in Two Variables, , 79, , Thus, the line PQC is the graph of 3x + 5y 15 = 0., The two graph lines intersect at the point C(5, 0) ., , , x = 5, y = 0 is the solution of the given system of equations., , Clearly, the given equations are represented by the graph, lines ABC and PQC respectively., The vertices of 3 AQC formed by these lines and the y-axis are, A(0, 4), Q(0, 3) and C(5, 0) ., EXAMPLE 4, , Solve the following system of equations graphically:, 3x + 2y 11 = 0 and 2x 3y + 10 = 0., , [CBSE 2006C], , Shade the region bounded by these lines and the x-axis., SOLUTION, , On a graph paper, draw a horizontal line X l OX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of 3x + 2y 11 = 0, , 3x 2y 11 0 & 2y (11 3x), (11 3x), ·, & y, 2, Putting x = 1 in (i), we get y = 7., Putting x = 1 in (i), we get, , y = 4., , Putting x = 3 in (i), we get, , y = 1., , … (i), , Thus, we have the following table for 3x + 2y 11 = 0., x, , –1, , 1, , 3, , y, , 7, , 4, , 1, , Now, plot the points A(1, 7), B(1, 4) and C(3, 1) on the graph, paper., Join AB and BC to obtain the graph line ABC. Extend it on both, ways., Thus, the line ABC is the graph of the equation 3x + 2y 11 = 0., Graph of 2x 3y + 10 = 0, , 2x 3y 10 0 & 3y 2x 10 & y , Putting x = 2 in (ii), we get y = 2., Putting x = 1 in (ii), we get, , y = 4., , Putting x = 4 in (ii), we get y = 6., , (2x 10), ·, 3, , … (ii)
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80, , Secondary School Mathematics for Class 10, , Thus, we have the following table for 2x 3y + 10 = 0., x, , –2, , 1, , 4, , y, , 2, , 4, , 6, , On the same graph paper as above, plot the points P(2, 2), and Q(4, 6) . The third point B(1, 4) has already been plotted., Now, join PB and BQ to obtain the graph line PBQ. Extend it, on both ways., Thus, the line PBQ is the graph of the equation 2x 3y + 10 = 0., , The two graph lines intersect at the point B(1, 4) ., , , x = 1, y = 4 is the solution of the given system of equations., , These graph lines intersect the x-axis at R and S., The region bounded by these lines and the x-axis has been, shaded., The shaded region is the 3 BRS with B(1, 4), R(–5, 0) and, 11, Sa 3 , 0 k $, EXAMPLE 5, , Solve the following system of linear equations graphically:, 3x + y 11 = 0, x y 1 = 0., Shade the region bounded by these lines and the y-axis., Find the coordinates of the points where the graph lines cut the, y-axis., [CBSE 2002C]
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Linear Equations in Two Variables, SOLUTION, , 81, , On a graph paper, draw a horizontal line X l OX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of 3x + y 11 = 0, , 3x y 11 0 & y (11 3x) ., , … (i), , Putting x = 2 in (i), we get y = 5., Putting x = 3 in (i), we get y = 2., Putting x = 5 in (i), we get y = 4., Thus, we have the following table for equation (i)., x, , 2, , 3, , 5, , y, , 5, , 2, , –4, , On the graph paper, plot the points A(2, 5), B(3, 2) and C(5, 4)., Join AB and BC to get the graph line ABC., Thus, the line ABC is the graph of the equation 3x + y 11 = 0., Graph of x y 1 = 0, , x y 1 0 & y (x 1) ., , … (ii), , Putting x = 3 in (ii), we get y = 4., y = 1., , Putting x = 0 in (ii), we get, , Putting x = 3 in (ii), we get y = 2., Thus, we have the following table for equation (ii)., x, , –3, , 0, , 3, , –y, , –4, , –1, , 2, , On the same graph paper as above, plot the points P(3, 4), and Q(0, 1). The third point B(3, 2) is already plotted., Join PQ and QB to get the graph line PQB., Thus, line PQB is the graph of the equation x y 1 = 0., The two graph lines intersect at the point B(3, 2)., , , x = 3, y = 2 is the solution of the given system of equations., , The region bounded by these lines and the y-axis has been, shaded.
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82, , Secondary School Mathematics for Class 10, , On extending the graph lines on both sides, we find that, these graph lines intersect the y-axis at the points, Q(0, 1) and R(0, 11) ., GRAPH OF EQUATIONS HAVING INFINITELY MANY SOLUTIONS, Note, , Such graph lines just coincide., , EXAMPLE 6, , Show graphically that the system of equations, 3x y = 2, 9x 3y = 6, has an infinite number of solutions., , SOLUTION, , On a graph paper, draw a horizontal line X l OX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of 3x y = 2, , 3x y 2 & y (3x 2) ., Putting x = 1 in (i), we get y = 5., , … (i)
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Linear Equations in Two Variables, , 83, , Putting x = 0 in (i), we get y = 2., Putting x = 2 in (i), we get y = 4., Thus, we have the following table for 3x y = 2., x, , –1, , 0, , 2, , y, , –5, , –2, , 4, , Now, plot the points A(1, 5), B (0, 2) and C(2, 4) on the, graph paper., Join AB and BC to get the graph line ABC., Extend the graph line ABC on both sides., Thus, the line ABC is the graph of the equation 3x y = 2., Graph of 9x 3y = 6, , 9x 3y 6 & 3y (9x 6), (9x 6), ·, 3, Putting x = 2 in (ii), we get y = 8., Putting x = 1 in (ii), we get y = 1., Putting x = 2 in (ii), we get y = 4., , & y, , Thus, we have the following table for 9x 3y = 6., x, , –2, , 1, , 2, , y, , –8, , 1, , 4, , Now, plot the points P(2, 8) and, Q(1, 1) on the same graph paper., The point C(2, 4) has already been, plotted., Join PQ and QC to obtain the line, PQC., Thus, the line PQC is the graph of, 9x 3y = 6., Thus, we find that the two graph, lines coincide. Hence, the given, system of equations has an infinite, number of solutions., , … (ii)
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84, , Secondary School Mathematics for Class 10, , GRAPH OF EQUATIONS HAVING NO SOLUTION, Note, , Such graph lines are parallel., , EXAMPLE 7, , Show graphically that the system of linear equations, 2x 3y = 5, 6y 4x = 3, is inconsistent, i.e., has no solution., , SOLUTION, , On a graph paper, draw a horizontal line X l OX and a vertical, line YOYl as the x-axis and the y-axis respectively., Graph of 2x 3y = 5, , 2x 3y 5 & 3y (2x 5), (2x 5), & y 3 ·, Putting x = 2 in (i), we get y = 3., Putting x = 1 in (i), we get, , … (i), , y = 1., , Putting x = 4 in (i), we get y = 1., Thus, we have the following table for the equation 2x 3y = 5., x, , –2, , 1, , 4, , y, , –3, , –1, , 1, , Now, plot the points A(2, 3), B(1, 1) and C(4, 1) on the graph, paper., Join AB and BC to get the graph line ABC. Extend it on both, ways., Thus, the line ABC is the graph of the equation 2x 3y = 5., Graph of 6y 4x = 3, , 6y 4x 3 & 6y (3 4x), (3 4x), & y 6 ·, , … (ii), , 3, Putting x = 3 in (ii), we get y = 2 $, 1, Putting x = 0 in (ii), we get y = 2 $, 5, Putting x = 3 in (ii), we get y = 2 $, Thus, we have the following table for the equation 6y 4x = 3.
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Linear Equations in Two Variables, , x, , –3, , 0, , 3, , y, , 3, 2, , 1, 2, , 5, 2, , 85, , 3, 1, 5, Now, plot the points P a 3, 2 k, Q a0, 2 k and R a 3, 2 k on the, same graph paper as above. Join PQ and QR to get the graph, line PQR. Extend it on both ways., Then, the line PQR is the graph of the equation 6y 4x = 3., , It is clear from the graph that the two graph lines are parallel, and do not intersect even when produced., Hence, the given system of equations has no solution., EXAMPLE 8, , Show graphically that the system of linear equations, x y = 8, 3x 3y = 16, is inconsistent, i.e., has no solution., , SOLUTION, , On a graph paper, draw a horizontal line X l OX and a vertical, line YOYl as the x-axis and y-axis respectively., Graph of x y = 8, , The first equation is x y = 8., Now, x y 8 & y x 8., Putting x = 3 in (i), we get y = 5., Putting x = 4 in (i), we get y = 4., Putting x = 5 in (i), we get y = 3., , … (i)
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86, , Secondary School Mathematics for Class 10, , Thus, we have the following table for the equation x y = 8., x, , 3, , 4, , 5, , y, , –5, , –4, , –3, , Now, we plot the points A (3, 5), B (4, 4) and C(5, 3) on the, graph paper. Join AB and BC to get the graph line ABC., Graph of 3x 3y = 16, , The second equation is 3x 3y = 16., (3x 16), ·, … (ii), 3, 10, 1, Putting x = 2 in (ii), we get y = 3 = 3 3 = 3.3., 7, 1, Putting x = 3 in (ii), we get y = 3 = 2 3 = 2.3., 16, 1, Putting x = 0 in (ii), we get y = 3 = 5 3 = 5.3., Thus, we have the following table for the equation 3x 3y = 16., , 3x 3y 16 & 3y 3x 16 & y , , x, , 3, , 2, , 0, , y, , –2.3, , –3.3, , –5.3, , Now, we plot D(3, 2.3), E(2, 3.3) and F(0, 5.3) on the same, graph paper as above., Join DE and EF to get the graph line DEF.
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88, , Secondary School Mathematics for Class 10, , Show graphically that each of the following given systems of equations has, infinitely many solutions:, , 22. 2x + 3y = 6, 4x + 6y = 12., 23. 3x y = 5, 6x 2y = 10., , [CBSE 2010], , 24. 2x + y = 6, 6x + 3y = 18., 25. x 2y = 5, 3x 6y = 15., Show graphically that each of the following given systems of equations is, inconsistent, i.e., has no solution:, , 26. x 2y = 6, 3x 6y = 0., 27. 2x + 3y = 4, 4x + 6y = 12., 28. 2x + y = 6, 6x + 3y = 20., 29. Draw the graphs of the following equations on the same graph paper:, 2x + y = 2, 2x + y = 6., Find the coordinates of the vertices of the trapezium formed by these, lines. Also, find the area of the trapezium so formed., [HOTS], HINT, , The line 2x + y = 2 cuts the x-axis at A(1, 0) and the y-axis at B(0, 2)., The line 2x + y = 6 cuts the x-axis at C(3, 0) and the y-axis at D(0, 6)., Area of trap. ABDC = ar(3 OCD) ar(3 OAB), = ` 1 # 3 # 6 j ` 1 # 1 # 2 j = 8 sq units., 2, 2, ANSWERS (EXERCISE 3A), , 1. x = 4, y = 2, , 2. x = 2, y = 1, , 3. x = 1, y = 2, , 4. x = 3, y = 2, , 5. x = 2, y = 3, 9. x = 1, y = 2, , 6. x = 1, y = 2, 10. x = 2, y = 2, , 7. x = 2, y = 3, , 8. x = 2, y = 3, , 11. (x = 1, y = 2); A(1, 2), B(3, 0), C(2, 0); ar(3 ABC) = 5 sq units, 12. (x = 1, y = 2); A(1, 2), B(2, 0), C(5, 0); ar(3 ABC) = 7 sq units, 13. (x = 2, y = 4); A(2, 4), B(1, 0), C(5, 0); ar(3 ABC) = 12 sq units, 14. (x = 2, y = 3); A(2, 3), B(1, 0), C(4, 0); ar(3 ABC) = 7.5 sq units, 15. (x = 2, y = 2); A(2, 2), B(2, 0), C(3, 0); ar(3 ABC) = 5 sq units, 16. (x = 3, y = 4); A(3, 4), B(0, 2), C(0, 6); ar(3 ABC) = 6 sq units, 17. (x = 2, y = 4); A(2, 4), B(0, 4), C(0, 7); ar(3 ABC) = 11 sq units, 18. (x = 5, y = 0); A(5, 0), B(0, 5), C(0, 3); ar(3 ABC) = 20 sq units, 19. (x = 3, y = 2); A(3, 2), B(0, 0.8), C(0, 8); ar(3 ABC) = 10.8 sq units, 20. (x = 2, y = 3); A(2, 3), B(0, 7), C(0, 1); ar(3 ABC) = 8 sq units
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Linear Equations in Two Variables, , 89, , 21. (x = 6, y = 0), A(6, 0), B(0, 4), C(0, 2), ar(3 ABC) = 18 sq units., 29. 8 sq units, , SOLVING SIMULTANEOUS LINEAR EQUATIONS, (BY ALGEBRAIC METHODS), (I) SUBSTITUTION METHOD, METHOD, , Suppose we are given two linear equations in x and y. For solving, these equations by the substitution method, we proceed according, to the following steps., , Step 1., , Express y in terms of x in one of the given equations., , Step 2., , Substitute this value of y in terms of x in the other equation. This, gives a linear equation in x., , Step 3., , Solve the linear equation in x obtained in Step 2., , Step 4., , Substitute this value of x in the relation taken in Step 1 to obtain a, linear equation in y., , Step 5., , Solve the above linear equation in y to get the value of y., , REMARK, , We may interchange the role of x and y in the above method., , SOLVED EXAMPLES, EXAMPLE 1, , Solve for x and y, using substitution method:, 2x + y = 7, 4x 3y + 1 = 0., , SOLUTION, , The given system of equations is, 2x + y = 7, , … (i), , 4x 3y = 1., , … (ii), , From (i), we get y = (7 2x) ., Substituting y = (7 2x) in (ii), we get, 4x 3 (7 2x) = 1, , , 4x 21 + 6x = 1, , 10x = 20, , , x = 2.
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90, , Secondary School Mathematics for Class 10, , Substituting x = 2 in (i), we get, 2 # 2 y 7 & y 7 4 3., Hence, the solution is x = 2, y = 3., EXAMPLE 2, , Solve for x and y, using substitution method:, 3x 5y = x + y = 13, 2, 3 2, 2, 3, 6 $, , SOLUTION, , The given system of equations may be written as, 9x 10y + 12 = 0, , … (i), , 2x + 3y 13 = 0., , … (ii), , 13 2x, From (ii), we get y =, $, 3, , 13 2x, Substituting y =, in (i), we get, 3, 10(13 2x), + 12 = 0, 9x , 3, 27x 10(13 2x) + 36 = 0, , , 27x 130 + 20x + 36 = 0, , 94, 47x 94 0 & 47x 94 & x 47 2., Substituting x = 2 in (i), we get, , , 30, 9 # 2 10y 12 0 & 10y 30 & y 10 3., Hence, x = 2 and y = 3 is the required solution., (II) ELIMINATION METHOD, METHOD, , In this method, we eliminate one of the unknown variables and, proceed using the following steps., , Step 1., , Multiply the given equations by suitable numbers so as to make the, coefficients of one of the unknown variables numerically equal., , Step 2., , If the numerically equal coefficients are opposite in sign then add the, new equations., Otherwise, subtract them., , Step 3., , The resulting equation is linear in one unknown variable., Solve it to get the value of one of the unknown quantities., , Step 4., , Substitute this value in any of the given equations., , Step 5., , Solve it to get the value of the other unknown variable.
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Linear Equations in Two Variables, EXAMPLE 3, , 91, , Solve for x and y using elimination method:, 10x 3y 75, 6x 5y 11., , SOLUTION, , The given equations are, 10x 3y 75, , … (i), , 6x 5y 11., , … (ii), , Multiplying (i) by 5 and (ii) by 3, we get, 50x 15y 375, , … (iii), , 18x 15y 33., , … (iv), , Adding (iii) and (iv), we get, 408, 68x 408 & x 68 & x 6., Putting x 6 in (i), we get, (10 # 6) 3y 75 & 60 3y 75, , EXAMPLE 4, , , , 3y 75 60 & 3y 15 & y 5., , , , x 6, y 5., , Solve for x and y:, 11x 15y 23 0, 7x 2y 20 0., , SOLUTION, , The given equations are, 11x 15y 23, , … (i), , 7x 2y 20., , … (ii), , Multiplying (i) by 2 and (ii) by 15 and adding the results,, we get, 22x 105x 46 300, 127x 254, 254, x 127 2., Putting x 2 in (i), we get, 22 15y 23, 15y 23 22, 45, 15y 45 & y 15 3., Hence, x 2 and y 3.
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92, EXAMPLE 5, , Secondary School Mathematics for Class 10, , Solve for x and y:, , [CBSE 2000, ‘04C, ‘05], , 0.4x 1.5y 6.5,, 0.3x 0.2y 0.9., SOLUTION, , Multiplying each of the equations by 10, we get, 4x 15y 65, 3x 2y 9., , … (i), … (ii), , Multiplying (i) by 2 and (ii) by 15 and adding, we get, 8x 45x 130 135, 53x 265, 265, x 53 5., Putting x 5 in (ii), we get, 15 2y 9 & 2y 9 15 & 2y 6 & y 3., Hence, x 5 and y 3., EXAMPLE 6, , SOLUTION, , Solve for x and y:, 4 6 , 5., x 3y 8, x 4y, 1, Putting x u, the given equations become:, 4u 3y 8, , [CBSE 2010], , … (i), , 6u 4y 5., , … (ii), , Multiplying (i) by 4 and (ii) by 3 and adding the results, we get, 16u 18u 32 15, 34u 17, 17 1, u 34 2 ·, 1, Putting u 2 in (i), we get, b 4 # 1 l 3y 8 & 2 3y 8, 2, & 3y 8 2 & 3y 6 & y 2., , , … (iii), , 1, 1 1, Now, u 2 & x 2 & x 2., Hence, x 2 and y 2., EXAMPLE 7, , Solve for x and y:, 23 , 5 4 , 2 (x ! 0 and y ! 0) ., x y 13, x y, , [CBSE 2008C]
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Linear Equations in Two Variables, SOLUTION, , 93, , 1, 1, Putting x u and y v, the given equations become, 2u 3v 13, , … (i), , 5u 4v 2., , … (ii), , Multiplying (i) by 4 and (ii) by 3 and adding the results, we get, 8u 15u 52 6, 23u 46, 46, u 23 2., , , Putting u 2 in (i), we get, (2 # 2) 3v 13 & 3v 13 4 9 & v 3., 1, 1, Now, u 2 & x 2 & 2x 1 & x 2 ·, 1, 1, And, v 3 & y 3 & 3y 1 & y 3 ·, 1, 1, Hence, x 2 and y 3 ·, EXAMPLE 8, , SOLUTION, , Solve for x and y:, 1 1 1 1 , 1, x 2y 8 (x !0, y !0) ., 2x y, , [CBSE 2004C], , 1, 1, Putting x u and y v, the given equations become, u , 1 & u 2v 2, 2 v, v, u 2 8 & 2u v 16., , … (i), … (ii), , Multiplying (ii) by 2 and adding the result with (i), we get, u 4u 2 32, 5u 30, 30, u 5 6., Putting u 6 in (i), we get, , , 6 2v 2 & 2v 8 & v 4., 1, 1, Now, u 6 & x 6 & 6x 1 & x 6 ·, 1, 1, And, v 4 & y 4 & 4y 1 & y 4 ·, 1, 1, Hence, x 6 and y 4 ·
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94, EXAMPLE 9, , SOLUTION, , Secondary School Mathematics for Class 10, , Solve for x and y:, 1 1 , 1 1 , 7x 6y 3, 2x 3y 5 (x ! 0, y ! 0)., 1, 1, Putting x u and y v, the given equations become, uv , , 7 6 3 & 6u 7v 126, uv , , 2 3 5 & 3u 2v 30., , … (i), … (ii), , Multiplying (i) by 2 and (ii) by 7 and adding the results,, we get, 12u 21u 252 210, 33u 462, 462, u 33 14., Putting u 14 in (i), we get, , , (6 # 14) 7v 126, 42, 7v 126 84 42 & v 7 6., 1, 1, Now, u 14 & x 14 & 14x 1 & x 14 ·, 1, 1, And, v 6 & y 6 & 6y 1 & y 6 ·, 1, 1, Hence, x 14 and y 6 ·, EXAMPLE 10, , SOLUTION, , Solve for x and y:, 3a 2b a 3b , x, y 5 0, x y 2 0 (x ! 0, y ! 0) ., , [CBSE 2005C], , 1, 1, Putting x u and y v, the given equations become, 3au 2bv 5, , … (i), , au 3bv 2., , … (ii), , Multiplying (ii) by 3 and subtracting (i) from the result,, we get, 9bv 2bv 6 5, 11 1 ·, 11bv 11 & v , 11b b, 1, Putting v in (ii), we get, b, , 1, au 3 2 & au 1 & u a ·
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Linear Equations in Two Variables, , 95, , 1, 1 1, Now, u a & x a & x a & x a., 1, 1 1, And, v & y & y b., b, b, , , Hence, x, a and y b., EXAMPLE 11, , Solve for x and y:, 6x 3y 7xy, 3x 9y 11xy (x ! 0, y ! 0)., , SOLUTION, , On dividing each of the given equations by xy, we get, 63 , y x 7, 39 , y x 11., , … (i), … (ii), , 1, 1, Putting x u and y v in (i) and (ii), we get, 6v 3u 7, , … (iii), , 3v 9u 11., , … (iv), , Multiplying (iii) by 3 and subtracting (iv) from the result,, we get, 18v 3v 21 11, 15v 10, 10 2, v 15 3 ·, 2, Putting v 3 in (iii), we get, b 6 # 2 l 3u 7 & 4 3u 7 & 3u 3 & u 1., 3, 1, Now, u 1 & x 1 & x 1., 2, 1 2, 3, And, v 3 & y 3 & 2y 3 & y 2 ·, 3, Hence, x 1 and y 2 ·, EXAMPLE 12, , Solve the following system of equations for x and y:, 5 1 , 6, 3, 2, 1., x1 y2, x 1 y 2, , SOLUTION, , Putting, , [CBSE 2008C, ’09], , 1 , 1, y and v, the given equations become, x1, y 2, , 5u v 2, , … (i), , 6u 3v 1., , … (ii)
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96, , Secondary School Mathematics for Class 10, , Multiplying (i) by 3 and adding the result with (ii), we get, 15u 6u 6 1, 21u 7, 7, 1, u 21 3 ·, 1, Putting u 3 in (i), we get, b5 # 1 l v 2 & 5 v 2 & v 2 5 & v 1 ·, 3, 3, 3, 3, 1, 1, 1, Now, u 3 & 3 & x 1 3 & x 4, x 1, [by cross multiplication]., 1, 1, 1, Similarly, v 3 & 3 & y 2 3 & y 5, y 2, [by cross multiplication]., Hence, x 4 and y 5., EXAMPLE 13, , Solve for x and y:, 15 22 , 40, 55, 5, 13, x ! y and x ! y., xy xy, x y x y, [CBSE 2008C], , SOLUTION, , 1, 1, Putting u and v, the given equations become, x y, x y, … (i), 15u 22v 5, … (ii), 40u 55v 13., Multiplying (ii) by 2 and (i) by 5 and subtracting the results,, we get, 80u 75u 26 25, 5u 1, 1, u 5·, 1, Putting u 5 in (i), we get, b15 # 1 l 22v 5, 5, 3 22v 5, 2, 1, 22v 2 & v 22 11 ·, 1, 1, 1, Now, u 5 & 5 & x y 5., x y, 1, 1, 1, And, v 11 & 11 & x y 11., x y, , , On adding (iii) and (iv), we get 2x 16 & x 8., , … (iii), … (iv)
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Linear Equations in Two Variables, , 97, , On subtracting (iii) from (iv), we get 2y 6 & y 3., Hence, x 8 and y 3., EXAMPLE 14, , Solve for x and y:, 1, 12, 7, 4, , , 1,, 2,, 2 (2x 3y) 7 (3x 2y) 2 (2x 3y) (3x 2y), where (2x 3y) ! 0 and (3x 2y) ! 0., , SOLUTION, , Putting, , [CBSE 2004C], , 1, 1, v, the given equations, u and, 3x 2y, (2x 3y), , become, u 12v 1, , , 2, 7, 2 & 7u 24v 7, and 7u 4v 2., , … (i), … (ii), , Subtracting (ii) from (i), we get, 5, 1, 20v 5 & v 20 4 ·, 1, Putting v 4 in (i), we get, 1, 7u b 24 # 4 l 7, 1, 7u 6 7 & 7u 1 & u 7 ·, 1, 1, 1 & 2x 3y 7, Now, u 7 &, (2x 3y) 7, 1, 1, 1 & 3x 2y 4., and v 4 &, (3x 2y) 4, , … (iii), … (iv), , Multiplying (iii) by 2 and (iv) by 3 and adding the results,, we get, 4x 9x 14 12, 13x 26 & x 2., Putting x 2 in (iii), we get, (2 # 2) 3y 7 & 3y (7 4) 3 & y 1., Hence, x 2 and y 1., EXAMPLE 15, , Solve for x and y:, 2 3 , 4 9 , 2,, 1 (x ! 0, y ! 0) ., x, y, x, y, , SOLUTION, , Putting, , 1 , 1 , u and, v, the given equations become, x, y
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98, , Secondary School Mathematics for Class 10, , 2u 3v 2, , … (i), , 4u 9v 1., , … (ii), , Multiplying (i) by 3 and adding the result with (ii), we get, 6u 4u 6 1, 10u 5, 5, 1, u 10 & u 2 ·, 1, Putting u 2 in (i), we get, a 2 # 12 k 3v 2, 1, 1 3v 2 & 3 v 1 & v 3 ·, 1, 1 1, Now, u 2 &, & x 2 & x 4., x 2, 1, 1 1, And, v 3 &, & y 3 & y 9., y 3, Hence, x 4 and x 9., EXAMPLE 16, , Solve the following system of linear equations:, (a b) x (a b) y a 2 2ab b 2,, (a b) (x y) a 2 b 2 ., , SOLUTION, , [CBSE 2004, ‘07C, ‘08], , The given equations may be written as, (a b) x (a b) y a 2 2ab b 2, , … (i), , (a b) x (a b) y a b ., , … (ii), , 2, , 2, , Subtracting (i) from (ii), we get, {(a b) (a b)} x 2ab 2b 2, 2b (a b), 2bx 2b (a b) & x , 2b, x (a b) ., Putting x (a b) in (ii), we get, (a b) 2 (a b) y a 2 b 2, (a b) y (a 2 b 2) (a b) 2, (a b) y (a 2 b 2) (a 2 b 2 2ab), 2ab, (a b) y 2ab & y ·, (a b), 2ab, x (a b) and y is the required solution., (a b)
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Linear Equations in Two Variables, EXAMPLE 17, , Solve for x and y:, ax by a b 0, bx ay a b 0., , SOLUTION, , 99, , [CBSE 2000, ‘04C, ‘05], , The given equations may be written as, ax by a b, , … (i), , bx ay a b., , … (ii), , Multiplying (i) by a and (ii) by b and adding the results,, we get, (a 2 b 2) x (a 2 b 2) & x 1., Putting x 1 in (i), we get, (a # 1) by a b, , , a by a b, , b, 1., by b & y , b, Hence, x 1 and y 1., EXAMPLE 18, , Solve the following system of linear equations:, 2(ax by) (a 4b) 0, 2(bx ay) (b 4a) 0., , SOLUTION, , [CBSE 2004], , The given equations may be written as, 2ax 2by a 4b, , … (i), , 2bx 2ay 4a b., , … (ii), , Multiplying (i) by a and (ii) by b and adding, we get, (2a 2 2b 2) x (a 2 b 2), 1, 2 (a 2 b 2) x (a 2 b 2) & x 2 ·, 1, Putting x 2 in (i), we get, 1, 2a # a 2 k 2by a 4b, a 2by a 4b, , , 4b, 2by 4b & y 2b 2., 1, Hence, x 2 and y 2., , , EXAMPLE 19, , Solve for x and y:, ax by , , a a b, ax by 2ab., b, , [CBSE 2000, ‘04C, ‘05, ‘07C, ‘09]
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100, , Secondary School Mathematics for Class 10, , The given equations may be written as, , SOLUTION, , a 2 x b 2 y a 2 b ab 2, ax by 2ab., , … (i), … (ii), , Multiplying (ii) by b and subtracting the result from (i),, we get, (a 2 ab) x a 2 b ab 2, (a 2 ab) x b (a 2 ab) & x b., Putting x b in (ii), we get, ab, ab by 2ab & by ab & y b a., Hence, x b and y a., EXAMPLE 20, , Solve for x and y:, 2 x 3 y 0, 5 x 2 y 0., , [CBSE 2000, ‘04C, '05], , The given system of equations is, , SOLUTION, , 2x 3y 0, , … (i), , 5 x 2 y 0., , … (ii), , Multiplying (i) by 2 and (ii) by 3 and adding, we get, (2 15 ) x 0 & x 0., Putting x 0 in (ii), we get, 2 y 0 & y 0., Hence, x 0 and y 0., A SPECIAL CASE, , METHOD Step 1., , When the coefficients of x and y in one equation are interchanged, in the other., Add the given equations and from it obtain an equation, of the form x y a., , Subtract the given equations and from it obtain an, equation of the form x y b., Now, x y a and x y b can be solved easily., , Step 2., , EXAMPLE 21, , Solve for x and y:, 37x 43y 123, 43x 37y 117., , SOLUTION, , [CBSE 2008], , The given equations are, 37x 43y 123, , … (i), , 43x 37y 117., , … (ii)
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Linear Equations in Two Variables, , 101, , Clearly, the coefficients of x and y in one equation are, interchanged in the other., Adding (i) and (ii), we get, (37 43) x (43 37) y (123 117), 80x 80y 240, 80(x y) 240, , , 240, x y 80 & x y 3., , … (iii), , Subtracting (i) from (ii), we get, 6x 6y 6, 6(x y) 6 & x y 1, , … (iv), , Adding (iii) and (iv), we get 2x 2 & x 1., Subtracting (iv) from (iii), we get 2y 4 & y 2., Hence, x 1 and y 2., EXAMPLE 22, , Solve for x and y:, 152x 378y 74, 378x 152y 604., , SOLUTION, , The given equations are, 152x 378y 74, , … (i), , 378x 152y 604., , … (ii), , Clearly, the coefficients of x and y in one equation are, interchanged in the other., Adding (i) and (ii), we get, {152 (378)} x {(378) 152} y (74 604), (226) x (226) y 678, (226)(x y) 678, 678, (x y) , & x y 3., 226, , … (iii), , Subtracting (ii) from (i), we get, (152 378) x (378 152) y (74 604), , , 530x 530y 530, , , , 530(x y) 530 & x y 1., , Adding (iii) and (iv), we get 2x 4 & x 2., , … (iv)
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102, , Secondary School Mathematics for Class 10, , Subtracting (iv) from (iii), we get 2y 2 & y 1., Hence, x 2 and y 1., SOME MORE EXAMPLES, EXAMPLE 23, , SOLUTION, , Solve for x and y:, x1 y1 x1 y1 , 8, 3, 9., 2, 3, 2, , [CBSE 2007], , The given equations may be written as, 3(x 1) 2 (y 1) 48 & 3x 2y 47, , … (i), , and 2(x 1) 3(y 1) 54 & 2x 3y 53., , … (ii), , Multiplying (i) by 2 and (ii) by 3 and subtracting, we get, (4 9) y 94 159, 65, 5y 65 & y & y 13., 5, Putting y 13 in (i), we get, 3x (2 # 13) 47 & 3x (47 26) 21, 21 , x, 7., 3, Hence, x 7 and y 13., EXAMPLE 24, , Solve for x and y:, 2 3 9 4 9 21, x y xy , x y xy (x ! 0, y ! 0)., , SOLUTION, , Multiplying each equation throughout by xy, we get, 2y 3x 9, , … (i), , 4y 9x 21., , … (ii), , Multiplying (i) by 3 and subtracting (ii) from the result,, we get, (6 4) y (27 21) & 2y 6 & y 3., Putting y 3 in (i), we get, (2 # 3) 3x 9 & 6 3x 9 & 3x 3 & x 1., Hence, x 1 and y 3., EXAMPLE 25, , Solve for x and y:, xy, xy, 6,, 6 (x ! 0, y ! 0 and x ! y) ., xy 5 yx
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Linear Equations in Two Variables, SOLUTION, , The given equations may be written as, xy 5, &11 5, xy, y x 6, 6, yx 1, 1, & 1 1, xy, x y 6, 6, , 103, , … (i), … (ii), , Adding (i) and (ii), we get, 2 51 6 , , x a 6 6 k 6 1 & x 2., Subtracting (ii) from (i), we get, 2 51 4 2, , y a 6 6 k 6 3 & y 3., Hence, x 2 and y 3., EXAMPLE 26, , Solve for x and y:, 7x 2y, 8x 7y, 5,, 15 (x ! 0, y ! 0) ., xy, xy, , SOLUTION, , We have, 7x 2y, 5& 7 2 5, xy, y x, 8x 7y, 15 & 8 7 15., xy, y x, , … (i), … (ii), , Multiplying (i) by 7 and (ii) by 2 and adding the results,, we get, 16 m (35 30), c 49, y, y, , , 65 , , , y 65 & 65y 65 & y 1., , Putting y 1 in (ii), we get, 87 , 7 , , , , 1 x 15 & x (15 8) 7 & 7x 7 & x 1., Hence, x 1 and y 1., f, , EXERCISE 3B, , Solve for x and y:, , 1. x y 3,, 4x 3y 26., , 2. x y 3,, xy , 3 2 6.
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Linear Equations in Two Variables, , 105, , 28., , 3 2 , 2,, xy xy, 9 4 , 1., xy xy, , 29., , 5 2 1, ,, x1 y1 2, 10 2 5, , x ! 1 and y ! 1., x1 y1 2, , 30., , 44 30 , 10,, xy xy, , 31., , 10 2 , 4,, xy xy, , 55 40 , 13., xy xy, 32. 71x 37y 253,, 37x 71y 287., , 15 9 , 2., xy xy, , [CBSE 2007C], , 34. 23x 29y 98,, 29x 23y 110., , 33. 217x 131y 913,, 131x 217y 827., 35., , 2x 5y, 6,, xy, 4x 5y, 3., xy, , 1, 5, , 3,, 2, 2(x 2y) 3(3x 2y), 5, 3, , 61 ·, 4(x 2y) 5(3x 2y) 60, , 36., , 1, 1, , 3,, (3x y) (3x y) 4, , 1, 1, , 1·, 8, 2(3x y) 2(3x y), , 37., , 38., , 2, 3, , 17 ,, 5, (3x 2y) (3x 2y), 5, 1, , 2., (3x 2y) (3x 2y), , 39. 3(2x y) 7xy,, 3(x 3y) 11xy (x ! 0 and y ! 0) ., x y, 41. a 2,, b, , ax by a 2 b 2 ., x y, 43. a 0,, b, ax by (a 2 b 2) ., , 40. x y a b,, ax by a 2 b 2 ., 42. px qy p q,, qx py p q., 44. 6(ax by) 3a 2b,, 6(bx ay) 3b 2a., bx ay , 46. a , ,, b a b 0, bx ay 2ab 0. [CBSE 2006], 48. x y a b,, ax by a 2 b 2 ., y, x y, x, 50. a a b, 2 2 2., b, a, b, , 45. ax by a 2 b 2,, x y 2a ., bx ay 2 2, 47. a , ,, b a b, x y 2ab., 49. a 2 x b 2 y c 2,, , [CBSE 2005], , [CBSE 2006C], , [CBSE 2010], , b2 x a2 y d2 ., , ANSWERS (EXERCISE 3B), , 1. x 5, y 2, , 2. x 9, y 6, , 3. x 15, y 10, , 4. x 2, y 3
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106, , Secondary School Mathematics for Class 10, , 5. x 2, y 5, , 6. x 1, y 1, 7. x 14, y 9, , 3, 2, 1, 1, 9. x 2 , y 3 10. x 3, y 4, 11. x 2 , y 3, 13. x 2, y 3, 14. x 0.5, y 0.7 15. x 5, y 1, , 12. x 4, y 3, , 1, , 20. x 3, y 1, , 17. x 2, y 6, , 1, , 19. x 3, y 2, , 22. x 4 , y 7, , 1, , 1, , 23. x 2 , y 3, , 24. x 2 , y 3, , 25. x 3, y 4, , 26. x 2 , y 3, , 1, , 1, , 27. x 3, y 2, , 28. x 2 , y 2, , 29. x 4, y 5, , 30. x 8, y 3, , 33. x 3, y 2, , 34. x 3, y 1, , 31. x 8 , y 8, 35. x 1, y 2, , 37. x 2 , y 4, , 38. x 1, y 1, , 39. x 1, y 2, , 40. x a, y b, , 41. x a, y b, , 42. x 1, y 1, , 43. x a, y b, , 44. x 2 , y 3, 47. x ab, y ab, , 5, , 1, , 21, , 49. x , , 9, , 3, , 45. x a b, y a b, 48. x a, y b, , 1, , 16. x 3, y 2, , 21. x 2 , y 3, , 1, , 1, , 18. x 5 , y 2, , 8. x 6, y 36, , 46. x a, y b, , 1, , 1, , 5, , 1, , 32. x 2, y 3, 36. x 1, y 1, , 1, , 1, , (a c b d ), (a 2 d 2 b 2 c 2), 50. x a 2, y b 2, ,y, 4, 4, (a b ), (a 4 b 4), 2 2, , 2, , 2, , HINTS TO SOME SELECTED QUESTIONS, 9. The second equation is 18x 3y 29., 11. Given equations are 6x 15y 8, 18x 12y 5., 7 4x, 12. 3 y & 7 4x 3y & 4x 3y 7., 13. Multiply each of the given equations throughout by 10., 14. Multiply each of the given equations throughout by 10., 15. The given equations are, 2x 7y 3, 3x 4y 19., 16. 6x 5y 7x 3y 1 & x 2y 1., 7x 3y 1 2x 12y 2 & 5x 9y 3., 17., , x y 8 x 2y 14, , & 3 (x y 8) 2 (x 2y 14), 2, 3, & x y 4., x 2y 14 3x y 12, , & 11 (x 2y 14) 3 (3x y 12), 3, 11, & 2x 19y 118., Now, solve (i) and (ii)., , 18. Multiply (i) by 3 and (ii) by 5 and subtract., 19. Multiply (i) by 4 and (ii) by 3 and add., , … (i), , … (ii)
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Linear Equations in Two Variables, , 107, , 1, 1, 23. Put x u and y v to get, 3u 2v, 5u 3v 1 and 2 3 5 or 9u 4v 30., Now, solve 5u 3v 1 and 9u 4v 30., 1, 1, 24. Putting x u and y v, we get, uv , , 2 3 2 & 3u 2v 12, , … (i), , u v 13, , 3 2, 6 & 2u 3v 13., Add (i) and (ii). Subtract (ii) from (i)., , … (ii), , 4 6, 8 9, 25. Divide each equation throughout by xy to get y x 3, y x 5., 27. Put, , 1 , 1, u and v., xy, x y, , 29. Put, , 1 , 1, u and v., x1, y 1, , 32. Add (i) and (ii) to get 108 (x y) 540 & x y 5., Subtract (ii) from (i) to get 34 (x y) 34 & x y 1., 2 5, 4 5, 35. y x 6, y x 3. Now, add., 36. Put, , 1, 1, u and, v., (3x y), (3x y), , 37. Put, , 1, 1, u and, v., (x 2y), (3x 2y), , 6 3, 39. 6x 3y 7xy & y x 7., , … (i), , 3 9, 3x 9y 11xy & y x 11., , … (ii), , 40. Multiply (i) by b and add (ii) to the result., 41. bx ay 2ab, , and ax by a 2 b 2., , … (ii), , and ax by (a 2 b 2)., , … (ii), , and 6bx 6ay 3b 2a., , … (ii), , … (i), , and b 2 x aby 2ab 2 0., , … (ii), , … (i), , and x y 2ab., , … (ii), , … (i), , Multiply (i) by b and (ii) by a and add., 42. Multiply (i) by p and (ii) by q and add., 43. bx ay 0, , … (i), , Multiply (i) by b and (ii) by a and add., 44. 6ax 6by 3a 2b, , … (i), , Multiply (i) by a and (ii) by b and add., 45. Multiply (ii) by b and add the result with (i)., 46. b 2 x a 2 y a 2 b ab 2 0, Now, subtract (ii) from (i)., 47. b 2 x a 2 y a 3 b ab 3, , Multiply (ii) by b 2 and subtract the result from (i).
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Linear Equations in Two Variables, , 109, , Numbers with downward arrows are multiplied first;, and from this product, the product of numbers with, upward arrows is subtracted., , Rule, , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Solve the system of equations 2x 3y 17, 3x 2y 6 by the, method of cross multiplication., The given equations may be written as, 2x 3y 17 0, , … (i), , 3x 2y 6 0., , … (ii), , By cross multiplication, we have, , , , y, x, , {3 # ( 6) (2) # (17)} {(17) # 3 ( 6) # 2}, , , , , , , 1, {2 # (2) 3 # 3}, , y, x, 1, , , (18 34) (51 12) (4 9), x y 1, 52 39 13, 52, 39, x 4, y 3., 13, 13, , Hence, x 4 and y 3 is the required solution., 4x 7y 28 0, 5y 7x 9 0., , EXAMPLE 2, , Solve, , SOLUTION, , The given equations are, 4x 7y 28 0, , … (i), , 7x 5y 9 0., , … (ii), , By cross multiplication, we have, , , , y, x, 1, , , {(7)# 9 5# 28} {28#(7) 9# 4} {4 # 5 (7)#(7)}
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110, , Secondary School Mathematics for Class 10, , y, x, 1, , , (63 140) (196 36) (20 49), x y 1, , 203 232 29, 203, 232, x a k 7 and y a k 8., 29, 29, Hence, x 7 and y 8 is the required solution., , , EXAMPLE 3, SOLUTION, , Solve, , 23 , 5 4 , 2, where x ! 0 and y ! 0., x y 13, x y, , 1, 1, Taking x u and y v, the given equations become, 2u 3v 13 0, 5u 4v 2 0., , … (i), … (ii), , By cross multiplication, we have, , , , u, v, 1, , , [3# 2 (4)#(13)] [(13)# 5 2# 2] [2#(4) 5# 3], , u v 1, 46 69 23, 46, 69, u a k 2, v a k 3, 23, 23, 1 1 , x 2, y 3, , , 1, 1, x 2, y 3 ·, 1, 1, Hence, x 2 and y 3 is the required solution., , , ax by c, bx ay 1 c., , EXAMPLE 4, , Solve, , SOLUTION, , The given equations may be written as, ax by c 0, , … (i), , bx ay (1 c) 0., , … (ii), , By cross multiplication, we have
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Linear Equations in Two Variables, , 111, , y, 1, x, , , b(1 c) ac cb a (1 c) (a 2 b 2), c(a b) b, c(a b) a, ·, x, and y , (a 2 b 2), (a 2 b 2), c(a b) b, c(a b) a, ·, Hence, x , and y , 2, 2, (a b ), (a 2 b 2), , , f, , EXERCISE 3C, , Solve each of the following systems of equations by using the method of cross, multiplication:, , 1. x 2y 1 0,, 2x 3y 12 0., , 2. 3x 2y 3 0,, 4x 3y 47 0., , 3. 6x 5y 16 0,, 7x 13y 10 0., 5. 2x 5y 1,, 2x 3y 3., , 4. 3x 2y 25 0,, 2x y 10 0., 6. 2x y 35,, 3x 4y 65., x y, 8. 6 15 4,, x y 19 ·, 3 12, 4, , 7. 7x 2y 3,, 22x 3y 16., , 1 1, 9. x y 7,, 23 , x y 17 (x ! 0, y ! 0)., 5 2 , 10., 1 0,, (x y) (x y), 15 7 , 10 0 (x ! y, x ! y) ., (x y) (x y), 11., , ax by , a a b,, b, , ax by 2ab. [CBSE 2007C], , 12. 2ax 3by (a 2b),, 3ax 2by (2a b) ., , a b, ab 2 a 2 b, 13. x y 0, x y (a 2 b 2), where x ! 0 and y ! 0., ANSWERS (EXERCISE 3C), , 1. x 3, y 2, 5. x 3, y 1, , 2. x 5, y 9, , 6. x 15, y 5, 1, 1, 9. x 4 , y 3, 10. x 3, y 2, (4a b), (4b a), , 12. x , 5a , y, 5b, , 3. x 6, y 4, , 4. x 5, y 20, , 7. x 1, y 2, , 8. x 18, y 15, , 11. x b, y a, 13. x a, y b
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Linear Equations in Two Variables, , 113, , SOLVED EXAMPLES, EXAMPLE 1, , Show that the system of equations, 2x 5y 17, 5x 3y 14, has a unique solution. Find the solution., , SOLUTION, , The given system of equations is, 2x 5y 17 0, 5x 3y 14 0., These equations are of the form, a1 x b1 y c1 0, a2 x b2 y c2 0,, where a1 2, b1 5, c1 17 and a2 5, b2 3, c2 14., a, 17 17, 2 b, 5 c, a1 5 , 1 3 , c1 14 ·, b2, 14, 2, 2, a, b, Thus, a1 ! 1 ·, b2, 2, Hence, the given system of equations has a unique solution., By cross multiplication, we have, , y, x, 1, , , (70 51) (85 28) (6 25), y, x, 1, , 19, 57, 19, 19, 57, x 1 and y 3., 19, 19, , , Hence, x 1 and y 3 is the required solution., , , EXAMPLE 2, , Find the values of k for which the system of equations, x 2y 3, 3x ky 1, has a unique solution., , SOLUTION, , The given system of equations is, x 2y 3 0, 3x ky 1 0., These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 1, b1 2, c1 3 and a2 3, b2 k, c2 1., a, b, For a unique solution, we must have a1 ! 1 ·, b, 2, 2
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Linear Equations in Two Variables, , , , , 115, , 2 , 3, 7, , (k 1) (2k 1) (5k 4), 2 , 3, 3, 7, , and, (k 1) (2k 1), (2k 1) (5k 4), , , , 4k 2 3k 3 and 15k 12 14k 7, , , , k 5 and k 5., , Hence, k 5., EXAMPLE 5, , Find the value of k for which the given system of equations has, infinitely many solutions:, kx 3y k 3, 12x ky k., , SOLUTION, , The given system of equations is, kx 3y (3 k) 0,, 12x ky k 0., These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 k, b1 3, c1 (3 k) and a2 12, b2 k, c2 k., Let the given system of equations have infinitely many, solutions., a, b, c, Then, a1 1 c1, b2, 2, 2, k 3 (3 k), 12, k, k, k 3 k3, 12, k, k, k 3, 3 k3, 12, k and k, k, 2, 2, , , , k 36 and k 6k 0, (k 6 or k 6) and k(k 6) 0, (k 6 or k 6) and (k 0 or k 6), , , k 6., , Hence, the given system of equations has infinitely many, solutions when k 6., EXAMPLE 6, , Find the value of k for which the given system of equations has, infinitely many solutions:, x (k 1) y 5,, (k 1) x 9y (1 8k) 0., , [CBSE 2003]
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116, SOLUTION, , Secondary School Mathematics for Class 10, , The given system of equations is, x (k 1) y 5 0, , … (i), , (k 1) x 9y (1 8k) 0., , … (ii), , These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 1, b1 (k 1), c1 5, and, , , a2 (k 1), b2 9, c2 (1 8k) ., c1, a1, b1 (k 1), 5, 5, 1, , , , ·, a2 (k 1) , b2, 9 and c2 (1 8k) (8k 1), , Let the given system of equations have infinitely many, solutions., a, b, c, Then, a1 1 c1, b2, 2, 2, , 1 (k 1) , 5, , 9, (k 1), (8k 1), (k 1), 1 (k 1), 5, , , 9 and 9, (k 1), (8k 1), (k 1) 2 9 and (k 1) (8k 1) 45, (k 1 3 or k 1 3) and 8k 2 7k 46 0, (k 2 or k 4) and (k 2) (8k 23) 0, 23, (k 2 or k 4) and ak 2 or k 8 k, k 2., Hence, the given system of equations will have infinitely, many solutions when k 2., EXAMPLE 7, , Find the values of a and b for which the following pair of linear, equations have an infinite number of solutions:, 2x 3y 7, (a b) x (a b) y 3a b 2., , SOLUTION, , [CBSE 2008C], , The given equations are, 2x 3y 7 0,, , … (i), , (a b) x (a b) y (2 3a b) 0., , … (ii), , These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 2, b1 3, c1 7, and, , a2 (a b), b2 (a b), c2 (2 3a b) .
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118, , Secondary School Mathematics for Class 10, , m 5n 0, , … (i), , and m 6n 1 0., , … (ii), , On solving (i) and (ii), we get m 5 and n 1., Hence, the required values are m 5 and n 1., EXAMPLE 9, , Show that the system of equations, 3x 5y 7, 6x 10y 3, has no solution., , SOLUTION, , The given system of equations is, 3x 5y 7 0 and 6x 10y 3 0., These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 3, b1 5, c1 7 and a2 6, b2 10, c2 3., a, c, 5, 7 7, 3 1 b, 1, a1 6 2 , 1 2 and c1 3 ·, b2, 10, 3, 2, 2, a1 b1, c1, Clearly, a ! c ·, b2, 2, 2, Hence, the given system of equations has no solution., , EXAMPLE 10, , Find the value of k for which the system of equations, x 2y 5, 3x ky 15 0, has no solution., , SOLUTION, , The given system of equations is, x 2y 5 0, , … (i), , 3x ky 15 0., , … (ii), , These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 1, b1 2, c1 5 and a2 3, b2 k, c2 15., a, c, 5, 1 b, 2, 1, a1 3 , 1 and c1 3 ·, b2 k, 15, 2, 2, Let the given system of equations have no solution., a, b, c, Then, a1 1 ! c1, b2, 2, 2, , , , , 1 2 1, 3 k!3, 1 2, 2 1, 3 k and k ! 3, k 6 and k ! 6, which is impossible.
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Linear Equations in Two Variables, , 119, , Hence, there is no value of k for which the given system of, equations has no solution., EXAMPLE 11, , Find the values of k for which the pair of linear equations, kx 3y k 2 and 12x ky k, has no solution., , SOLUTION, , [CBSE 2009], , The given equations are, kx 3y (2 k) 0 and 12x ky k 0., These equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 k, b1 3, c1 (2 k) and a2 12, b2 k, c2 k., a, c, (2 k) (k 2), k b, 3, ·, , a1 12 ; 1 and c1 , b, k, k, k, 2, 2, 2, Let the given system of equations have no solution., a, b, c, Then, a1 1 ! c1, b2, 2, 2, (k 2), k 3, 12, k ! k, (k 2), k, 3, 3, 12 and !, k, k, k, k 2 36 and k 2 2k ! 3k, k 2 36 and k 2 5k ! 0, (k 6 or k 6) and k (k 5) ! 0., Case 1. When k 6., In this case, k(k 5) 6(6 5) 6 # 1 6 ! 0., When k 6., In this case, k(k 5) (6)(6 5) ( 6)#(11) 66 ! 0., Thus, in each case, the given system has no solution., Hence, k 6 or k 6., Case 2., , EXAMPLE 12, , Find the value of k for which the pair of linear equations, (3k 1) x 3y 2 0, (k 2 1) x (k 2) y 5 0, has no solution., , SOLUTION, , The given linear equations are of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 (3k 1), b1 3, c1 2, and, , a2 (k 2 1), b2 (k 2), c2 5.
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Linear Equations in Two Variables, , 121, , 1, 1, ·, !, (k 1) (2k 1), 1, 1, Clearly, when k 2, then !, (k 1), (2k 1), , , k 2 and, , ;as, , 1, 1, E·, !, (2 1) (4 1), , Hence, the given system of equations will have no solution,, when k 2., EXAMPLE 14, , Find the values of k for which the system of equations, kx y 2, 6x 2y 3, has (i) a unique solution, (ii) no solution., (iii) Is there a value of k for which the given system has infinitely, many solutions?, , SOLUTION, , The given system of equations is, kx y 2 0, 6x 2y 3 0., This is of the form, a1 x b1 y c1 0 and a2 x b2 y c2 0,, where a1 k, b1 1, c1 2 and a2 6, b2 2, c2 3., a, b, (i) For a unique solution, we must have a1 ! 1 ·, b2, 2, 1, k, k, 1, 6 ! & 6 ! 2 & k ! 3., 2, Hence, the given system of equations will have a unique, solution when k ! 3., a, b, c, (ii) For no solution, we must have a1 1 ! c1 ·, b2, 2, 2, 2, k 1, 6 !, 2, 3, k1, 2, 6 2!3, k 1, k, 2, 6 2 and 6 ! 3 & k 3 and k ! 4., Clearly, k 3 also satisfies the condition k ! 4., Hence, the given system of equations will have no, solution when k 3., (iii) For infinitely many solutions, we must have, a1 b1 c1, , a2 b2 c2 ,
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122, , Secondary School Mathematics for Class 10, , k 1 1, 1 1, i.e., 6 2 3 , which is never possible, as 2 ! 3 ·, Hence, there is no real value of k for which the given, system of equations has infinitely many solutions., HOMOGENEOUS SYSTEM OF EQUATIONS, , The system of equations a1 x b1 y 0 and a2 x b2 y 0 has, a, b, (i) the zero solution, x 0 and y 0 when a1 ! 1 ·, b2, 2, a, b, (ii) an infinite number of nonzero solutions when a1 1 ·, b2, 2, Note, , The homogeneous system of equations is always consistent., , EXAMPLE 15, , Find the value of k for which the system of equations, 3x 5y 0, kx 10y 0, has a nonzero solution., , SOLUTION, , The given equations are of the form, a1 x b1 y 0 and a2 x b2 y 0,, where a1 3, b1 5 and a2 k, b2 10., a, b, 3, 5 1, Then, a1 and 1 10 2 ·, k, b, 2, 2, Let the given system of equations have a nonzero solution., a, b, 3 1, Then, a1 1 & 2 & k 6., b, k, 2, 2, Hence, the required value of k is 6., , f, , EXERCISE 3D, , Show that each of the following systems of equations has a unique solution, and solve it:, , 1. 3x 5y 12, 5x 3y 4., x y, 3. 3 2 3, x 2y 2., , 2. 2x 3y 17, 4x y 13., , Find the value of k for which each of the following systems of equations has a, unique solution:, , 4. 2x 3y 5 0, kx 6y 8 0., 6. 5x 7y 5 0, 2x ky 1 0., , 5. x ky 2, 3x 2y 5 0., 7. 4x ky 8 0, x y 1 0.
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124, , Secondary School Mathematics for Class 10, , 22. (2a 1) x 3y 5,, 3x (b 1) y 2., 23. 2x 3y 7,, (a b) x (a b 3) y 4a b., 24. 2x 3y 7,, (a b 1) x (a 2b 2) y 4(a b) 1., 25. 2x 3y 7,, (a b) x (2a b) y 21., 26. 2x 3y 7,, 2ax (a b) y 28., , [CBSE 2001C], , [CBSE 2002], , [CBSE 2003], , [CBSE 2001], , [CBSE 2001], , Find the value of k for which each of the following systems of equations has, no solution:, , 27. 8x 5y 9, kx 10y 15., 28. kx 3y 3, 12x ky 6., 29. 3x y 5 0, 6x 2y k 0 (k ! 0)., 30. kx 3y k 3, 12x ky k., , [CBSE 2008], [CBSE 2009], , 31. Find the value of k for which the system of equations, 5x 3y 0, 2x ky 0, has a nonzero solution., ANSWERS (EXERCISE 3D), , 1. x 1, y 3, , 2. x 4, y 3, , 2, , 3. x 6, y 2, , 4. k ! 4, , 14, , 5. k ! 3, 6. k ! 5, 7. k ! 4, 8. k is any real number, 9. k is any real number other than 6 and –6, , 3, , 3, , 12. (i) k ! 2 (ii) k 2, 15. k = 7, , 13. (i) k ! 6 (ii) k = 6, , 14. (i) k ! 10 (ii) k = 10, , 16. k = 5, , 17. k = 2, 18. k = 4 19. k = 3, 20. k = 6, 17, 11, 21. a 3, b 4 22. a 4 , b 5 23. a 5, b 1 24. a 3, b 2, 25. a 5, b 1 26. a 4, b 8, 27. k 16, 28. k 6, 6, 29. k ! 10, 30. k 6, 31. k 5, HINTS TO SOME SELECTED QUESTIONS, k, 3, 9. 12 ! & k 2 ! 36 & k ! 6 and k ! 6., k, So, for a unique solution, we may take any real value of k, other than 6 and –6.
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Linear Equations in Two Variables, , 125, , 1, 3, k, 2, k, 12. (i) 3 ! & 3 ! 2 & k ! 2 ·, 4, 3, k, 2, 5, k 1, k, 1, (ii) 3 ! 10 & 3 2 and 3 ! 2 & k 2 ·, 4, 1, 2, 13. (i) 3 ! & k ! 6., k, 5, 1, 12, 1 2, (ii) 3, ! 15 & 3 ! 3 & k 6., k, k, 1, 2, 14. (i) 5 ! & k ! 10., k, 3, 12, (ii) 5, ! 7 & k 10., k, 1 2 3, 1 3, Clearly, 5 7 is never true, as 5 7 is false., k, k 3 3 k, k3 3, 3, k, 20., &, and 12 ·, k, k 12, k, k, k, , , k 2 6k 0 and k 2 36 & k (k 6) 0 and k 2 36, , , , (k 0 or k 6) and (k 6 or k 6), , k 6., a1, 3 2 1, 21. 6 , 1 2b 6 3, a1 1, 3 1, , 6, 3 and 1 2b 3, 3a 3 6 and 1 2b 9, 3a 9 and 2b 8 & a 3 and b 4., 3, k 3, k, 3, 3, 1, 28. 12, ! & 12 and ! 2, k, k, k, 6, , , & k 2 36 and k ! 6 & k 6., , 3, k, k 3, 30. 12 !, & k 2 36 and k (k 6) ! 0 & k 6., k, k, , WORD PROBLEMS, SOLVED EXAMPLES, PROBLEMS ON MONEY MATTERS, EXAMPLE 1, , 8 chairs and 5 tables for a classroom cost ` 10500, while 5 chairs and, 3 tables cost ` 6450. Find the cost of each chair and that of each table., , SOLUTION, , Let the cost of each chair be ` x and that of each table be ` y., Then, 8x 5y 10500, , … (i), , and 5x 3y 6450., , … (ii), , On multiplying (ii) by 5 and (i) by 3 and subtracting the, results, we get
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126, , Secondary School Mathematics for Class 10, , 25x 24x 32250 31500 & x 750., Putting x 750 in (i), we get, 8 # 750 5y 10500 & 6000 5y 10500, , , 5y (10500 6000) 4500, , , , y 900., , , , cost of each chair = ` 750 and cost of each table = ` 900., , EXAMPLE 2, , The coach of a cricket team buys 7 bats and 6 balls for ` 13200. Later,, he buys 3 bats and 5 balls for ` 5900. Find the cost of each bat and, each ball., , SOLUTION, , Let the cost of each bat be ` x and the cost of each ball be ` y., Then, 7x 6y 13200., , … (i), , And, 3x 5y 5900., , … (ii), , On multiplying (i) by 5, (ii) by 6 and subtracting the results,, we get, 35x 18x 66000 35400 & 17x 30600, 30600 , x, 1800., 17, Putting x 1800 in (ii), we get, , , 5400 5y 5900 & 5y 5900 5400, 5y 500, , , , y 100., , , , cost of each bat = ` 1800 and cost of each ball = ` 100., , EXAMPLE 3, , 37 pens and 53 pencils together cost ` 955, while 53 pens and, 37 pencils together cost ` 1115. Find the cost of a pen and that of, a pencil., , SOLUTION, , Let the cost of each pen be ` x and that of a pencil be ` y., Then, 37x 53y 955., , … (i), , And, 53x 37y 1115., , … (ii), , Adding (i) and (ii), we get, 90x 90y 2070 & 90(x y) 2070, , , , 2070, 90, x y 23., , xy , , … (iii)
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Linear Equations in Two Variables, , 127, , On subtracting (i) from (ii), we get, 16x 16y 160 & 16 (x y) 160, (x y) 10., , … (iv), 33 , , , Adding (iii) and (iv), we get 2x 33 & x 2 16.50., 13, Subtracting (iv) from (iii), we get 2y 13 & y 2 6.50., cost of each pen = ` 16.50 and cost of each pencil = ` 6.50., EXAMPLE 4, , Taxi charges in a city consist of fixed charges and the remaining, depending upon the distance travelled in kilometres. If a person, travels 60 km, he pays ` 960, and for travelling 80 km, he pays, ` 1260. Find the fixed charges and the rate per kilometre., , SOLUTION, , Let the fixed charges be ` x and the other charges be ` y per km., Then, x 60y 960., , … (i), , And, x 80y 1260., , … (ii), , On subtracting (i) from (ii), we get, 300, 20y 300 & y 20 & y 15., Putting y 15 in (i), we get, x (60 # 15) 960 & x 960 900 & x 60., , , fixed charges = ` 60 and the rate per km = ` 15 per km., , EXAMPLE 5, , A part of monthly hostel charges in a school is fixed and the remaining, depends on the number of days one has taken food in the mess. When, a student A takes food for 22 days, he has to pay ` 4250 as hostel, charges, whereas a student B, who takes food for 28 days, pays, ` 5150 as hostel charges. Find the fixed charges and the cost of food, per day., , SOLUTION, , Let the fixed charges be ` x per month and the cost of meals, per day be ` y. Then, we have, x 22y 4250, , … (i), , and x 28y 5150., , … (ii), , On subtracting (i) from (ii), we get, 6y 900 & y 150., Putting y 150 in (i), we get, x (22 # 150) 4250
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128, , Secondary School Mathematics for Class 10, , , , , , x 3300 4250, x 4250 3300 950., x 950 and y 150., , Hence, the fixed charges are ` 950 per month and the cost of, food is ` 150 per day., EXAMPLE 6, , The monthly incomes of A and B are in the ratio 8 : 7 and their, expenditures are in the ratio 19 : 16. If each saves ` 5000 per month,, find the monthly income of each., , SOLUTION, , Let the monthly incomes of A and B be ` 8x and ` 7x, respectively, and let their expenditures be ` 19y and ` 16y, respectively., Then, A’s monthly savings = ` (8x 19y) ., And, B’s monthly savings = ` (7x 16y) ., But, the monthly saving of each is ` 5000., 8x 19y 5000, , … (i), , and 7x 16y 5000., , … (ii), , , , Multiplying (ii) by 19, (i) by 16 and subtracting the results,, we get, (19 # 7 16 # 8) x (19 # 5000 16 # 5000), (133 128) x 5000 #(19 16), 5x 15000 & x 3000., , , A’s monthly income = ` (8x) = ` (8 # 3000) ` 24000., , And, B’s monthly income = ` (7x) = ` (7 # 3000) ` 21000., EXAMPLE 7, , On selling a TV at 5% gain and a fridge at 10% gain, a shopkeeper, gains ` 3250. But, if he sells the TV at 10% gain and the fridge at, 5% loss, he gains ` 1500. Find the actual cost price of TV and that of, the fridge., , SOLUTION, , Let the cost price of the TV set be ` x and that of the fridge, be ` y., Then, total CP of TV and fridge = ` (x y) ., 105x 110y m c 21x 11y m ·, SP in first case = ` c, ` 20, 100, 100, 10, , , y, 21x 11y, x, gain in this case = ` 'c 20 10 m (x y)1 ` a 20 10 k ·
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Linear Equations in Two Variables, , 129, , y, x, So, 20 10 3250 & x 2y 65000., … (i), 110x 95y m c11x 19y m ·, SP in second case = ` c, ` 10, 20, 100 100, 19, y, y, 11x, x, gain in this case = ` 'c 10 20 m (x y)1 ` a10 20 k ·, y, x, So, 10 20 1500 & 2x y 30000., … (ii), Multiplying (ii) by 2 and adding the result with (i), we get, 5x 60000 65000 & 5x 125000 & x 25000., Putting x 25000 in (i), we get, 25000 2y 65000 & 2y 65000 25000 40000, , , y 20000., , , , x 25000 and y 20000., , Hence, the CP of the TV set is ` 25000 and that of the fridge is, ` 20000., EXAMPLE 8, , A man invested an amount at 12% per annum simple interest and, another amount at 10% per annum simple interest. He received an, annual interest of ` 2600. But, if he had interchanged the amounts, invested, he would have received ` 140 less. What amounts did he, invest at the different rates?, , SOLUTION, , Let the amount invested at 12% be ` x and that invested at 10%, be ` y. Then,, total annual interest, x # 12 # 1 y # 10 # 1 m c 6x 5y m ·, =`c, `, 50, 100, 100, 6x 5y, 2600 & 6x 5y 130000., , … (i), 50, Again, the amount invested at 12% is ` y and that invested at, 10% is ` x., Total annual interest at the new rates, y # 12 # 1 x # 10 # 1, 6y 5x, m·, m `c, , =`c, 50, 100, 100, But, interest received at the new rates = ` (2600 140) = ` 2460., , , 6y 5x, 2460 & 5x 6y 123000., 50, , … (ii)
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130, , Secondary School Mathematics for Class 10, , Adding (i) and (ii), we get, 11x 11y 253000 & 11 (x y) 253000, , , x y 23000., , … (iii), , Subtracting (ii) from (i), we get, x y 7000., , … (iv), , Adding (iii) and (iv), we get 2x 30000 & x 15000., Putting x 15000 in (i), we get, 15000 y 23000 & y 23000 15000 8000., , , x 15000 and y 8000., , Hence, the amount at 12% is ` 15000 and that at 10% is ` 8000., EXAMPLE 9, , Each one of A and B has some money. If A gives ` 30 to B then B, will have twice the money left with A. But, if B gives ` 10 to A then, A will have thrice as much as is left with B. How much money does, each have?, , SOLUTION, , Suppose A and B have ` x and ` y respectively., When A gives ` 30 to B., , Case I, , Then, money with A = ` (x 30), money with B = ` (y 30) ., , and, , , (y 30) 2(x 30) & 2x y 90., , Case II, , … (i), , When B gives ` 10 to A., , Then, money with A = ` (x 10), and, , , money with B = ` (y 10) ., (x 10) 3(y 10) & x 3y 40., , … (ii), , On multiplying (ii) by 2 and subtracting the result from (i),, we get, 170, 5y 170 & y 5 34., Putting y 34 in (i), we get, , , , 124, 2x 34 90 & 2x (90 34) 124 & x 2 62., x 62 and y 34., , Hence, A has ` 62 and B has ` 34.
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Linear Equations in Two Variables, , 131, , PROBLEMS ON NUMBERS, EXAMPLE 10, , The students of a class are made to stand in rows. If 4 students are, extra in each row, there would be 2 rows less. If 4 students are less in, each row, there would be 4 rows more. Find the number of students, in the class., , SOLUTION, , Let the number of rows be x and the number of students in, each row be y., Then, the total number of students = xy., When there are 4 more students in each row., , Case I, , Then, number of students in each row (y 4)., And, number of rows (x 2)., Total number of students (x 2) (y 4)., , , (x 2) (y 4) xy & 4x 2y 8, , & 2x y 4., Case II, , … (i), , When 4 students are removed from each row., , Then, number of students in each row (y 4)., And, number of rows (x 4)., Total number of students (x 4)(y 4)., , , (x 4)(y 4) xy & 4y 4x 16, , & 4(y x) 16, & (y x) 4., , … (ii), , Adding (i) and (ii), we get x 8., Putting x 8 in (ii), we get, y 8 4 & y 12., Thus, x 8 and y 12., This shows that there are 8 rows and there are 12 students in, each row., Hence, the number of students in the class xy 8 #12 96., EXAMPLE 11, , The sum of two numbers is 1000 and the difference between their, squares is 256000. Find the numbers., , SOLUTION, , Let the larger number be x and the smaller number be y., Then, x y 1000, , … (i), , and x y 256000., , … (ii), , 2, , 2
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132, , Secondary School Mathematics for Class 10, , On dividing (ii) by (i), we get, x 2 y 2 256000, , , 1000 & x y 256., xy, , … (iii), , Adding (i) and (iii), we get, 2x 1256 & x 628., Substituting x = 628 in (i), we get y 372., Hence, the required numbers are 628 and 372., EXAMPLE 12, , If three times the larger of two numbers is divided by the smaller, one, we get 4 as the quotient and 3 as the remainder. Also, if seven, times the smaller number is divided by the larger one, we get 5 as the, quotient and 1 as the remainder. Find the numbers., , SOLUTION, , Let the larger number be x and the smaller one be y., We know that, dividend = (divisor × quotient) + remainder., Using the above result and the given conditions, we have, 3x 4y 3 & 3x 4y 3, and 7y 5x 1 & 5x 7y 1., , … (i), … (ii), , Multiplying (i) by 5, (ii) by 3 and subtracting, we get y 18., Putting y 18 in (i), we get, 3x (4 # 18) 3 & 3x 72 3, , , 3x 75 & x 25., , Hence, the required numbers are 25 and 18., EXAMPLE 13, , SOLUTION, , 8, The sum of two numbers is 8 and the sum of their reciprocals is 15 ·, Find the numbers., [CBSE 2009], Let the required numbers be x and y., Then, x y 8., xy 8, 1 1, 8, And, x y 15 & xy 15, 8 8, & xy, 15, , xy, 15, ., &, , , (x y) (x y) 2 4xy, 8 2 4 # 15 64 60 4 !2., , … (i), , [using (i)]
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Linear Equations in Two Variables, , 133, , Thus, we have, xy 8, xy 2, , … (i), … (ii), , 3, , or, , ), , xy 8, , … (iii), , x y 2, , … (iv), , On solving (i) and (ii), we get x 5 and y 3., On solving (iii) and (iv), we get x 3 and y 5., Hence, the required numbers are 5 and 3., EXAMPLE 14, , The difference of two numbers is 4 and the difference of their, 4, reciprocals is 21 · Find the numbers., [CBSE 2008], , SOLUTION, , Let the larger number be x and the smaller number be y., Then, x y 4., , … (i), ;a x y & 1y 1xE, , 1 1, 4, And, y x 21, , , , , xy, 4, xy, 21, 4 4, , xy 21 & xy 21, , [using (i)]., , (x y) (x y) 2 4xy, , 4 2 4 # 21 16 84 100 !10., Thus, we have, xy 4, , … (i), , x y 10, , … (ii), , 3, , or, , ), , xy 4, , … (iii), , x y 10, , … (iv), , On solving (i) and (ii), we get x 7 and y 3., On solving (iii) and (iv), we get x 3 and y 7., Hence, the required numbers are (7 and 3) or (–3 and –7)., EXAMPLE 15, , The sum of the digits of a two-digit number is 12. The number, obtained by interchanging its digits exceeds the given number by 18., Find the number., [CBSE 2006], , SOLUTION, , Let the ten‘s digit of the required number be x and the unit‘s, digit be y. Then,, … (i), x y 12., Required number (10x y)., Number obtained on reversing the digits (10y x).
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134, , Secondary School Mathematics for Class 10, , , , (10y x) (10x y) 18 & 9y 9x 18, , & y x 2., , … (ii), , On adding (i) and (ii), we get, 2y 14 & y 7., Putting y 7 in (i), we get, x 7 12 & x 12 7 5., , , x 5 and y 7., , Hence, the required number is 57., EXAMPLE 16, , The sum of a two-digit number and the number obtained by reversing, the order of its digits is 99. If the digits differ by 3, find the number., [CBSE 2002], , SOLUTION, , Let the ten‘s and unit‘s digits of the required number be x and, y respectively., Then, the number (10x y)., The number obtained on reversing the digits (10y x) ., , , (10y x) (10x y) 99 & 11(x y) 99 & x y 9., , Also, (x y) !3., Thus, we have, xy 9, , … (i), , xy 3, , … (ii), , 3, , or, , ), , xy 9, , … (iii), , x y 3, , … (iv), , From (i) and (ii), we get x 6, y 3., From (iii) and (iv), we get x 3, y 6., Hence, the required number is 63 or 36., EXAMPLE 17, , Seven times a two-digit number is equal to four times the number, obtained by reversing the order of its digits. If the difference between, the digits is 3, find the number., , SOLUTION, , Let the ten‘s and unit‘s digits of the required number be x and, y respectively., Then, the number (10x y)., The number obtained by reversing the digits (10y x)., , , 7(10x y) 4(10y x) & 33(2x y) 0, , , , 2x y 0 & y 2x, , … (i)
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Linear Equations in Two Variables, , 135, , Thus, unit‘s digit = 2 times the ten‘s digit., , , (unit‘s digit) > (ten‘s digit) and so y x., , y x 3., , … (ii), , Using (i) in (ii), we get (2x x) 3 & x 3., On substituting x 3 in (i), we get y 2 # 3 6., Hence, the required number is 36., EXAMPLE 18, , A two-digit number is such that the product of its digits is 14. If 45, is added to the number, the digits interchange their places. Find the, number., [CBSE 2005], , SOLUTION, , Let the ten‘s and unit‘s digits of the required number be x and, y respectively. Then, xy 14., Required number (10x y)., Number obtained on reversing its digits (10y x)., , , (10x y) 45 (10y x), , , , 9(y x) 45 & y x 5., , … (i), , Now, (y x) (y x) 4xy, 2, , 2, , (y x) (y x) 2 4xy 25 4 # 14 81, , , yx 9, , … (ii), , [£ digits are never negative]., , On adding (i) and (ii), we get, 2y 14 & y 7., Putting y 7 in (ii), we get, 7 x 9 & x 9 7 2., , , x 2 and y 7., , Hence, the required number is 27., EXAMPLE 19, , A two-digit number is four times the sum of its digits and twice the, product of its digits. Find the number., [CBSE 2005], , SOLUTION, , Let the ten‘s digit of the required number be x and its unit‘s, digit be y., Then, 10x y 4(x y) & 6x 3y 0 & 2x y 0., , … (i), , Also, 10x y 2xy., , … (ii)
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136, , Secondary School Mathematics for Class 10, , Putting y 2x from (i) in (ii), we get, 10x 2x 4x 2 & 4x 2 12x 0, , & 4x(x 3) 0, & x 3 0 & x 3 [£ ten‘s digit, x ! 0] ., Putting x 3 in (i), we get y 6., Thus, ten‘s digit = 3 and unit‘s digit = 6., Hence, the required number is 36., EXAMPLE 20, , SOLUTION, , 1, A fraction becomes 3 , if 2 is added to both of its numerator and, denominator. If 3 is added to both of its numerator and denominator, 2, then it becomes 5 · Find the fraction., [CBSE 2009C], x·, Let the required fraction be y Then,, x2 1, & 3x 6 y 2, y2 3, … (i), & 3x y 4., Also,, , x3 2, & 5x 15 2y 6, y3 5, & 5x 2y 9., , … (ii), , Multiplying (i) by 2 and subtracting (ii) from the result,, we get, 6x 5x 8 9 & x 1 ., Putting x 1 in (i), we get, 3 y 4 & y 7., Thus, x 1 and y 7., 1, Hence, the required fraction is 7 ·, EXAMPLE 21, , SOLUTION, , The sum of numerator and denominator of a fraction is 3 less than, twice the denominator. If each of the numerator and denominator is, 1, decreased by 1, the fraction becomes 2 · Find the fraction. [CBSE 2010], x, Let the required fraction be y · Then,, … (i), (x y) 2y 3 & x y 3., And,, , x1 1, & 2x 2 y 1, y1 2, , & 2x y 1., , … (ii)
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Linear Equations in Two Variables, , 137, , On subtracting (i) from (ii), we get x 4., Putting x 4 in (i), we get y x 3 4 3 7., , , x 4 and y 7., , 4, Hence, the required fraction is 7 ·, EXAMPLE 22, , SOLUTION, , In a given fraction, if the numerator is multiplied by 2 and the, 6, denominator is reduced by 5, we get a 5 k · But if the numerator of the, given fraction is increased by 8 and the denominator is doubled, we, 2, get a 5 k · Find the fraction., x, Let the required fraction be y · Then,, 2x 6, & 10x 6(y 5), y5 5, & 10x 6y 30, , & 5x 3y 15, , … (i), , x8 2, and 2y 5 & 5(x 8) 4y, & 5x 4y 40., , … (ii), , On subtracting (ii) from (i), we get y 25., Putting y 25 in (i), we get, 5x (3 # 25) 15 & 5x 75 15 & 5x 60 & x 12., x 12 and y 25., 12, Hence, the required fraction is 25 ·, PROBLEMS ON AGES, EXAMPLE 23, , Five years ago, A was thrice as old as B and ten years later A shall be, twice as old as B. What are the present ages of A and B? [CBSE 2002C], , SOLUTION, , Let the present ages of B and A be x years and y years, respectively. Then,, B’s age 5 years ago (x 5) years, and A’s age 5 years ago (y 5) years., , , (y 5) 3(x 5) & 3x y 10., , … (i), , B’s age 10 years hence (x 10) years., A’s age 10 years hence (y 10) years., , , (y 10) 2(x 10) & 2x y 10., , … (ii)
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138, , Secondary School Mathematics for Class 10, , On subtracting (ii) from (i), we get x 20., Putting x 20 in (i), we get, (3 # 20) y 10 & y 60 10 50., , , x 20 and y 50., , Hence, B’s present age = 20 years, and, , A’s present age = 50 years., , EXAMPLE 24, , A man‘s age is three times the sum of the ages of his two sons. After, 5 years, his age will be twice the sum of the ages of his two sons. Find, the age of the man., [CBSE 2003], , SOLUTION, , Let the present age of the man be x years and the sum of the, present ages of his two sons be y years. Then,, x 3y & x 3y 0., , … (i), , Man‘s age after 5 years (x 5) years., Sum of the ages of his two sons after 5 years (y 5 5) years, (y 10) years., , , (x 5) 2(y 10) & x 2y 15., , … (ii), , On subtracting (i) from (ii), we get y 15., Putting y 15 in (i), we get x 45., , , x 45 and y 15., , Hence, the present age of the man is 45 years., PROBLEMS ON TIME AND DISTANCE, EXAMPLE 25, , A man travels 370 km, partly by train and partly by car. If he covers, 250 km by train and the rest by car, it takes him 4 hours. But, if, he travels 130 km by train and the rest by car, he takes 18 minutes, longer. Find the speed of the train and that of the car., , SOLUTION, , Let the speed of the train be x km/hr and that of the car be, y km/hr., Case I, , Distance covered by train = 250 km., , Distance covered by car (370 250) km 120 km., 250, Time taken to cover 250 km by train x hours., 120, Time taken to cover 120 km by car y hours.
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Linear Equations in Two Variables, , Total time taken = 4 hours., 250 120 , 125 60, , 4 & x y 2., x, y, Case II, , 139, , … (i), , Distance covered by train = 130 km., , Distance covered by car (370 130) km 240 km., 130, Time taken to cover 130 km by train x hours., 240, Time taken to cover 240 km by car y hours., 18, 3, 43, Total time taken 4 60 hours 4 10 hours 10 hours., 130 240 43, 1300 2400 , , … (ii), 43., x, y, y, 10 & x, 1, 1, Putting x u and y v, equations (i) and (ii) become, 125u 60v 2, , … (iii), , and 1300u 2400v 43. … (iv), , On multiplying (iii) by 40 and subtracting (iv) from the result,, we get, 5000u 1300u 80 43 & 3700u 37, 37, 1, 1, 1, u 3700 100 & x 100 & x 100., 1, Putting u 100 in (iv), we get, 1 , a1300 # 100, k 2400v 43 & 2400v 43 13 30, 30, 1, 1, 1, v 2400 80 & y 80 & y 80., , , x 100 and y 80., , Hence, the speed of the train is 100 km/hr and that of the car, is 80 km/hr., EXAMPLE 26, , Places A and B are 100 km apart on a highway. One car starts from, A and another from B at the same time. If the cars travel in the same, direction at different speeds, they meet in 5 hours. If they travel, towards each other, they meet in 1 hour. What are the speeds of the, two cars?, [CBSE 2009], , SOLUTION, , Let the speeds of the cars from A and B be x km/hr and, y km/hr respectively., Case I, , When the two cars move in the same direction:, , In this case, let them meet at M after 5 hours.
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140, , Secondary School Mathematics for Class 10, , Then, AM 5x km., And, BM 5y km., , , AM BM AB & 5x 5y 100, , , , 5(x y) 100 & x y 20., , Case II, , … (i), , When the two cars move in the opposite directions:, , Let one car move from A to B, and let the other move from, B to A. Let them meet at N, after 1 hour., Then, AN x km and BN y km., , , AN BN AB & x y 100., , … (ii), , Adding (i) and (ii), we get 2x 120 & x 60., Putting x 60 in (ii), we get 60 y 100 & y 40., , , speed of the car from A 60 km/hr,, , and speed of the car from B 40 km/hr., EXAMPLE 27, , A train covered a certain distance at a uniform speed. If the train, had been 6 kmph faster, it would have taken 4 hours less than the, scheduled time. And, if the train were slower by 6 kmph, it would, have taken 6 hours more than the scheduled time. Find the length of, the journey., , SOLUTION, , Let the original speed of the train be x kmph and let the time, taken to complete the journey be y hours., Length of whole journey (xy) km., Case I, , When speed (x 6) kmph and time taken (y 4) hours., , Length of total journey (x 6)(y 4) km., , , xy (x 6)(y 4) & xy xy 4x 6y 24, , , , 4x 6y 24 & 2x 3y 12., , Case II, , … (i), , When speed (x 6) kmph and time taken (y 6) hours., , Length of total journey (x 6)(y 6) km.
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Linear Equations in Two Variables, , , , xy (x 6)(y 6) & xy xy 6x 6y 36, , , , 6x 6y 36 & x y 6., , 141, , … (ii), , On multiplying (ii) by 3 and subtracting (i) from the result,, we get, 3x 2x 18 12 & x 30., On putting x 30 in (ii), we get y 30 6 24., , , x 30 and y 24., , So, length of journey = (xy) km (30 # 24) km = 720 km., PROBLEMS ON BOATS AND STREAM, EXAMPLE 28, , A man can row downstream 20 km in 2 hours, and upstream 4 km, in 2 hours. Find his speed of rowing in still water. Also, find the, speed of the current., , SOLUTION, , Let the speed of the man in still water be x km/hr and let the, speed of the current be y km/hr., Speed downstream (x y) km/hr., Speed upstream (x y) km/hr., But, speed downstream , And, speed upstream , , , x y 10, , distance 20, time, 2 km/hr = 10 km/hr., , distance 4, time, 2 km/hr = 2 km/hr., , … (i), , and x y 2, , … (ii)., , Adding (i) and (ii), we get 2x 12 & x 6., Putting x 6 in (i), we get 6 y 10 & y 4., Thus, x 6 and y 4., Hence, the speed of the man in still water is 6 km/hr and the, speed of the current is 4 km/hr., EXAMPLE 29, , A boat goes 16 km upstream and 24 km downstream in 6 hours., Also, it covers 12 km upstream and 36 km downstream in the, same time. Find the speed of the boat in still water and that of, the stream., , SOLUTION, , Let the speed of the boat in still water be x km/hr and the, speed of the stream be y km/hr. Then,
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142, , Secondary School Mathematics for Class 10, , speed upstream (x y) km/hr, and speed downstream (x y) km/hr., Time taken to cover 16 km upstream , , 16, hours., (x y), , Time taken to cover 24 km downstream , , 24, hours., (x y), , Total time taken = 6 hours., , , 16 24 , 6., xy xy, , … (i), , Again, time taken to cover 12 km upstream , Time taken to cover 36 km downstream , , 12, hours., (x y), , 36, hours., (x y), , Total time taken = 6 hours., , , 12 36 , 6., xy xy, , Putting, , … (ii), , 1 , 1, u and v in (i) and (ii), we get, (x y), (x y), , 16u 24v 6 & 8u 12v 3,, , … (iii), , 12u 36v 6 & 2u 6v 1., , … (iv), , On multiplying (iv) by 4 and subtracting (iii) from the result,, we get, 1, 12v 1 & v 12, 1 1, , & x y 12., x y 12, , … (v), , On multiplying (iv) by 2 and subtracting the result from (iii),, we get, , , , 1, 4u 1 & u 4, 1 1, & x y 4., xy 4, , On adding (v) and (vi), we get 2x 16 & x 8., On subtracting (vi) from (v), we get 2y 8 & y 4., , , speed of the boat in still water = 8 km/hr,, , and speed of the stream = 4 km/hr., , … (vi)
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Linear Equations in Two Variables, , 143, , PROBLEMS ON TIME AND WORK, EXAMPLE 30, , 8 men and 12 boys can finish a piece of work in 5 days, while 6 men, and 8 boys can finish it in 7 days. Find the time taken by 1 man alone, and that by 1 boy alone to finish the work., , SOLUTION, , Suppose 1 man alone can finish the work in x days and 1 boy, alone can finish it in y days., 1, Then, 1 man’s 1 day’s work x ·, 1, And, 1 boy’s 1 day’s work y ·, 8 men and 12 boys can finish the work in 5 days, 1, (8 men’s 1 day’s work) + (12 boys’ 1 day’s work) 5, 8 12 1, x y 5, 1, 1, 1, 8u 12v 5, … (i), where x u and y v., , , Again, 6 men and 8 boys can finish the work in 7 days, 1, (6 men’s 1 day’s work) + (8 boys’ 1 day’s work) 7, 6 8 1, xy7, 1, 6u 8v 7 ·, … (ii), On multiplying (i) by 3, (ii) by 4 and subtracting the results,, we get, 3 4, 1, 1, 1, 1, 4v a 5 7 k 35 & v 140 & y 140 & y 140., 1, On putting v 140 in (ii), we get, 1, 8, 12, 12, 1, 1, 6u a7 140 k 140 & u a140 # 6 k 70, 1, 1, x 70 & x 70., one man alone can finish the work in 70 days,, , , , and one boy alone can finish the work in 140 days., PROBLEMS ON AREA, EXAMPLE 31, , If the length of a rectangle is reduced by 5 units and its breadth is, increased by 2 units then the area of the rectangle is reduced by 80 sq, units. However, if we increase its length by 10 units and decrease
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144, , Secondary School Mathematics for Class 10, , the breadth by 5 units, its area is increased by 50 sq units. Find the, length and breadth of the rectangle., SOLUTION, , Let the length and breadth of the rectangle be x units and y, units respectively., Then, area of the rectangle = xy sq units., Case I, , When the length is reduced by 5 units and the breadth is, increased by 2 units., , Then, new length (x 5) units, and new breadth (y 2) units., , , new area (x 5)(y 2) sq units., , , , xy (x 5)(y 2) 80 & 5y 2x 70., , Case II, , … (i), , When the length is increased by 10 units and the breadth is, decreased by 5 units., , Then, new length (x 10) units, and new breadth (y 5) units., new area (x 10)(y 5) sq units., , , (x 10)(y 5) xy 50, , 10y 5x 100 & 2y x 20., , … (ii), , On multiplying (ii) by 2 and subtracting the result from (i), we, get y 30., Putting y 30 in (ii), we get, (2 # 30) x 20 & 60 x 20 & x (60 20) 40., , , x 40 and y 30., , Hence, length = 40 units and breadth = 30 units., PROBLEMS ON GEOMETRY, EXAMPLE 32, , In a 3ABC, +C 3+B 2(+A +B). Find the angles., , SOLUTION, , Let +A xc and +B yc., Then, +C 3+B (3y)c., Now, +C 2(+A +B), , , 3y 2(x y) & 2x y 0., , We know that the sum of the angles of a triangle is 180c., , … (i)
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Linear Equations in Two Variables, , 145, , +A +B +C 180c & x y 3y 180, , & x 4y 180., , … (ii), , On multiplying (i) by 4 and adding the result with (ii), we get, 8x x 180 & 9x 180 & x 20., Putting x 20 in (i), we get y (2 # 20) 40., Thus, x 20 and y 40., +A 20c, +B 40c and +C (3 # 40c) 120c., EXAMPLE 33, , Find the four angles of a cyclic quadrilateral ABCD in which, +A (2x 1)c, +B (y 5)c, +C (2y 15)c and +D (4x 7)c., , SOLUTION, , We know that the sum of the opposite angles of a cyclic, quadrilateral is 180c., +A +C 180c and +B +D 180c., Now, +A +C 180c & (2x 1) (2y 15) 180, , & 2(x y) 166, & x y 83., And, +B +D 180c & (y 5) (4x 7) 180, & 4x y 182., , … (i), … (ii), , On subtracting (i) from (ii), we get, 3x 182 83 99 & x 33., Putting x 33 in (i), we get, 33 y 83 & y (83 33) & y 50., Thus, x 33 and y 50., +A {(2 # 33) 1}c 65c, +B (50 5)c 55c., +C {(2 # 50) 15}c 115c and +D {(4 # 33) 7}c 125c., Hence, +A 65c, +B 55c, +C 115c and +D 125c., , f, , EXERCISE 3E, , 1. 5 chairs and 4 tables together cost ` 5600, while 4 chairs and 3 tables, together cost ` 4340. Find the cost of a chair and that of a table., 2. 23 spoons and 17 forks together cost ` 1770, while 17 spoons and, 23 forks together cost ` 1830. Find the cost of a spoon and that of a fork.
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146, , Secondary School Mathematics for Class 10, , 3. A lady has only 25-paisa and 50-paisa coins in her purse. If she has, 50 coins in all totalling ` 19.50, how many coins of each kind does, she have?, 4. The sum of two numbers is 137 and their difference is 43. Find the, numbers., 5. Find two numbers such that the sum of twice the first and thrice, the second is 92, and four times the first exceeds seven times the second, by 2., 6. Find two numbers such that the sum of thrice the first and the second, is 142, and four times the first exceeds the second by 138., 7. If 45 is subtracted from twice the greater of two numbers, it results in, the other number. If 21 is subtracted from twice the smaller number, it, results in the greater number. Find the numbers., 8. If three times the larger of two numbers is divided by the smaller, we, get 4 as the quotient and 8 as the remainder. If five times the smaller is, divided by the larger, we get 3 as the quotient and 5 as the remainder., Find the numbers., 9. If 2 is added to each of two given numbers, their ratio becomes 1 : 2., However, if 4 is subtracted from each of the given numbers, the ratio, becomes 5 : 11. Find the numbers., 10. The difference between two numbers is 14 and the difference between, their squares is 448. Find the numbers., 11. The sum of the digits of a two-digit number is 12. The number obtained, by interchanging its digits exceeds the given number by 18. Find the, [CBSE 2006], number., 12. A number consisting of two digits is seven times the sum of its digits., When 27 is subtracted from the number, the digits are reversed. Find, the number., 13. The sum of the digits of a two-digit number is 15. The number obtained, by interchanging the digits exceeds the given number by 9. Find the, number., [CBSE 2004], 14. A two-digit number is 3 more than 4 times the sum of its digits. If 18 is, added to the number, the digits are reversed. Find the number., 15. A number consists of two digits. When it is divided by the sum of, its digits, the quotient is 6 with no remainder. When the number is, diminished by 9, the digits are reversed. Find the number. [CBSE 1999C]
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Linear Equations in Two Variables, , 147, , 16. A two-digit number is such that the product of its digits is 35. If 18, is added to the number, the digits interchange their places. Find the, [CBSE 2006], number., 17. A two-digit number is such that the product of its digits is 18. When 63, is subtracted from the number, the digits interchange their places. Find, [CBSE 2006C], the number., 18. The sum of a two-digit number and the number obtained by reversing, the order of its digits is 121, and the two digits differ by 3. Find the, number., 19. The sum of the numerator and denominator of a fraction is 8. If 3 is, added to both of the numerator and the denominator, the fraction, 3, becomes 4 · Find the fraction., [CBSE 2003], 1, 20. If 2 is added to the numerator of a fraction, it reduces to a 2 k and if 1 is, 1, subtracted from the denominator, it reduces to a 3 k · Find the fraction., 21. The denominator of a fraction is greater than its numerator by 11. If 8 is, 3, added to both its numerator and denominator, it becomes 4 · Find the, fraction., 1, 22. Find a fraction which becomes a 2 k when 1 is subtracted from the, numerator and 2 is added to the denominator, and the fraction becomes, a13 k when 7 is subtracted from the numerator and 2 is subtracted from, the denominator., 23. The sum of the numerator and denominator of a fraction is 4 more than, twice the numerator. If the numerator and denominator are increased, by 3, they are in the ratio 2 : 3. Determine the fraction., [CBSE 2010], 1, 24. The sum of two numbers is 16 and the sum of their reciprocals is 3 ·, Find the numbers., [CBSE 2005], 25. There are two classrooms A and B. If 10 students are sent from A to B,, the number of students in each room becomes the same. If 20 students, are sent from B to A, the number of students in A becomes double the, number of students in B. Find the number of students in each room., 26. Taxi charges in a city consist of fixed charges and the remaining, depending upon the distance travelled in kilometres. If a man travels
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148, , Secondary School Mathematics for Class 10, , 80 km, he pays ` 1330, and travelling 90 km, he pays ` 1490. Find the, fixed charges and rate per km., 27. A part of monthly hostel charges in a college hostel are fixed and the, remaining depends on the number of days one has taken food in the, mess. When a student A takes food for 25 days, he has to pay ` 4500,, whereas a student B who takes food for 30 days, has to pay ` 5200. Find, the fixed charges per month and the cost of the food per day., 28. A man invested an amount at 10% per annum and another amount, at 8% per annum simple interest. Thus, he received ` 1350 as annual, interest. Had he interchanged the amounts invested, he would, have received ` 45 less as interest. What amounts did he invest at, different rates?, 29. The monthly incomes of A and B are in the ratio 5 : 4 and their monthly, expenditures are in the ratio 7 : 5. If each saves ` 9000 per month, find, the monthly income of each., 30. A man sold a chair and a table together for ` 1520, thereby making a, profit of 25% on chair and 10% on table. By selling them together for, ` 1535, he would have made a profit of 10% on the chair and 25% on the, table. Find the cost price of each., 31. Points A and B are 70 km apart on a highway. A car starts from A, and another car starts from B simultaneously. If they travel in the same, direction, they meet in 7 hours. But, if they travel towards each other,, they meet in 1 hour. Find the speed of each car., [CBSE 2007C], 32. A train covered a certain distance at a uniform speed. If the train had, been 5 kmph faster, it would have taken 3 hours less than the scheduled, time. And, if the train were slower by 4 kmph, it would have taken, 3 hours more than the scheduled time. Find the length of the journey., 33. Abdul travelled 300 km by train and 200 km by taxi taking 5 hours, 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he, takes 6 minutes longer. Find the speed of the train and that of the taxi., [CBSE 2006C], , 34. Places A and B are 160 km apart on a highway. One car starts from A, and another from B at the same time. If they travel in the same direction,, they meet in 8 hours. But, if they travel towards each other, they meet, in 2 hours. Find the speed of each car., [CBSE 2009C], 35. A sailor goes 8 km downstream in 40 minutes and returns in 1 hour., Find the speed of the sailor in still water and the speed of the current.
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Linear Equations in Two Variables, , 149, , 36. A boat goes 12 km upstream and 40 km downstream in 8 hours. It can, go 16 km upstream and 32 km downstream in the same time. Find the, speed of the boat in still water and the speed of the stream., 37. 2 men and 5 boys can finish a piece of work in 4 days, while 3 men and, 6 boys can finish it in 3 days. Find the time taken by one man alone to, finish the work and that taken by one boy alone to finish the work., 38. The length of a room exceeds its breadth by 3 metres. If the length is, increased by 3 metres and the breadth is decreased by 2 metres, the, area remains the same. Find the length and the breadth of the room., 39. The area of a rectangle gets reduced by 8 m 2, when its length is reduced, by 5 m and its breadth is increased by 3 m. If we increase the length by, 3 m and breadth by 2 m, the area is increased by 74 m 2 . Find the length, and the breadth of the rectangle., 40. The area of a rectangle gets reduced by 67 square metres, when its, length is increased by 3 m and breadth is decreased by 4 m. If the length, is reduced by 1 m and breadth is increased by 4 m, the area is increased, by 89 square metres. Find the dimensions of the rectangle., 41. A railway half ticket costs half the full fare and the reservation charge, is the same on half ticket as on full ticket. One reserved first class ticket, from Mumbai to Delhi costs ` 4150 while one full and one half reserved, first class tickets cost ` 6255. What is the basic first class full fare and, what is the reservation charge?, [HOTS], 42. Five years hence, a man‘s age will be three times the age of his son., Five years ago, the man was seven times as old as his son. Find their, present ages., 43. Two years ago, a man was five times as old as his son. Two years later,, his age will be 8 more than three times the age of his son. Find their, present ages., [CBSE 2008], 44. If twice the son‘s age in years is added to the father‘s age, the sum is, 70. But, if twice the father‘s age is added to the son‘s age, the sum is 95., Find the ages of father and son., 45. The present age of a woman is 3 years more than three times the age, of her daughter. Three years hence, the woman’s age will be 10 years, more than twice the age of her daughter. Find their present ages., 46. On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery, seller gains ` 7. If he sells the tea set at 5% gain and the lemon set at 10%, gain, he gains ` 13. Find the actual price of each of the tea set and the, lemon set., [HOTS]
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150, , Secondary School Mathematics for Class 10, , 47. A lending library has a fixed charge for the first three days and an, additional charge for each day thereafter. Mona paid ` 27 for a book, kept for 7 days, while Tanvy paid ` 21 for the book she kept for 5 days., Find the fixed charge and the charge for each extra day., 48. A chemist has one solution containing 50% acid and a second one, containing 25% acid. How much of each should be used to make, 10 litres of a 40% acid solution?, [HOTS], 49. A jeweller has bars of 18-carat gold and 12-carat gold. How much of, each must be melted together to obtain a bar of 16-carat gold, weighing, 120 g? (Given: Pure gold is 24-carat)., [HOTS], 50. 90% and 97% pure acid solutions are mixed to obtain 21 litres of 95%, pure acid solution. Find the quantity of each type of acids to be mixed, to form the mixture., [HOTS], 51. The larger of the two supplementary angles exceeds the smaller by 18., Find them., 52. In a 3ABC, +A xc, +B (3x 2)c, +C yc and +C +B 9c. Find the, three angles., 53. In a cyclic quadrilateral ABCD, it is given that +A (2x 4)c,, +B (y 3)c, +C (2y 10)c and +D (4x 5)c. Find the four angles., ANSWERS (EXERCISE 3E), , 1. ` 560, ` 700, , 2. ` 40, ` 50, , 3. 22, 28, , 4. 90, 47, , 6. 40, 22, , 7. 37, 29, , 8. 20, 13, , 9. 34, 70, , 11. 57, , 12. 63 13. 78, , 3, 3, 18. 47 or 74, 19. 5 20. 10, 25. 100 in A and 80 in B, , 5. 25, 14, 10. 23, 9, , 14. 35, , 15. 54 16. 57, 17. 92, 25, 5, 15, 21. 36 22. 26 23. 9, 24. 12 and 4, 26. fixed charges = ` 50, rate per km = ` 16, , 27. fixed charges = ` 1000, cost of food per day = ` 140, 28. ` 8500 at 10% p.a. and ` 6250 at 8% p.a., , 29. ` 30000, ` 24000, , 30. ` 600, ` 700 31. 40 km/hr, 30 km/hr, , 32. 1080 km, , 33. 100 kmph, 80 kmph 34. 50 kmph, 30 kmph, , 35. 10 kmph, 2 kmph, , 36. 6 kmph, 2 kmph, , 37. 18 days, 36 days, , 38. length = 15 m, breadth = 12 m, , 39. length = 19 m, breadth = 10 m, , 40. length = 28 m, breadth = 19 m, 41. first class full fare = ` 4090, reservation charges = ` 60, 42. 40 years, 10 years, , 43. 42 years, 10 years, , 44. 40 years, 15 years
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Linear Equations in Two Variables, , 151, , 45. 33 years, 10 years, , 46. tea set = ` 100, lemon set = ` 80, , 47. ` 15, ` 3 per day, , 48. 50% solution = 6 litres, 25% solution = 4 litres, , 49. 80 g, 40 g, , 50. 6 litres, 15 lires, , 51. 99, 81, , 52. +A 25c, +B 73c, +C 82c, 53. +A 70c, +B 53c, +C 110c, +D 127c, HINTS TO SOME SELECTED QUESTIONS, 3. Let the number of 25-p and 50-p coins be x and y respectively., x y 1950, Then, x y 50 and 4 2 100 ·, 8. Let the larger number be x and the smaller number be y., Then, 3x 4y 8 and 5y 3x 5., 10. x y 14 and x 2 y 2 448., (x 2 y 2) 448, , , xy , 14 32., (x y), 15. Let the ten‘s digit be x and the unit‘s digit be y. Then,, 10x y, 6 and 10x y 9 10y x., xy, 16. Let the ten‘s digit be x and the unit‘s digit be y. Then,, xy 35 and 10x y 18 10y x & y x 2., , , (y x) 2 (y x) 2 4xy & (y x) (y x) 2 4xy 144 12., , Now, solve y x 2 and y x 12., 18. Let the ten‘s digit be x and the unit‘s digit be y., Then, (10x y) (10y x) 121 & 11 (x y) 121 & x y 11., (x y 11, x y 3) or (x y 11, y x 3) ., x, 21. Let the required fraction be y ·, Then, x 11 y & x y 11., x8 3, And, 4 & 4x 32 3y 24 & 4x 3y 8., y 8, x, 23. Let the required fraction be y ·, Then, x y 4 2x & y x 4., x3 2, & 3x 9 2y 6 & 2y 3x 3., y3 3, yx 1, 1 1 1, 24. x y 16 and x y 3 & xy 3 & xy 16 # 3 48., (x y) 2 (x y) 2 4xy (16) 2 (4 # 48) 256 192 64, , , x y 8 or x y 8., , , , (x y 16, x y 8) or (x y 16, x y 8) ., , 25. Let the number of students in room A and room B be x and y respectively. Then,, x 10 y 10 & x y 20., , … (i)
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152, , Secondary School Mathematics for Class 10, And, (x 20) 2 (y 20) & x 2y 60., , … (ii), , Now, solve (i) and (ii) to get x and y., 26. Let the fixed charges be ` x and rate per km be ` y. Then,, x 80y 1330 and x 90y 1490., 27. Let the fixed charges be ` x and the charges for food be ` y per day. Then,, x 25y 4500 and x 30y 5200., 28. Suppose he invested ` x at 10% p.a. and ` y at 8% p.a., (x # 10 # 1) (y # 8 # 1), , 1350 & 5x 4y 67500., 100, 100, (x # 8 # 1) (y # 10 # 1), , 1350 45 & 4x 5y 65250., 100, 100, Add (i) and (ii). Then, subtract (ii) from (i)., , … (i), … (ii), , 29. 5x 7y 9000 and 4x 5y 9000. Find x and y., 30. Let the CP of a chair be ` x and that of a table be ` y. Then,, 125x 110y , , , 100, 100 1520 & 25x 22y 30400., 110x 125y , , , 100, 100 1535 & 22x 25y 30700., , On adding (i) and (ii), we get 47 (x y) 61100 & x y 1300., , … (i), … (ii), , On subtracting (i) from (ii), we get 3 (y x) 300 & y x 100., , , x 600 and y 700., , 31. Let the speed of the car from A be x kmph and that of the car from B be y kmph., Then, 7x 7y 70 & x y 10., , … (i), , And, x y 70., , … (ii), , 32. Let the original speed of the train be x kmph and let the time taken to complete the, journey be y hours., Then, length of the journey xy km., Case I, , , , Speed (x 5) kmph and time taken (y 3) hours., , xy (x 5) (y 3) & 5y 3x 15., , Case II, , , , … (i), , Speed (x 4) kmph and time taken (y 3) hours., , xy (x 4) (y 3) & 3x 4y 12., , … (ii), , From (i) and (ii), we get x 40 and y 27., , , length of the journey xy 40 # 27 km = 1080 km., , 33. Let the speed of the train be x kmph and that of the taxi be y kmph., 300 200 11, 1, 1, Then, x y 2 & 600u 400v 11 … (i), where x u and y v., 260 240 28, And, x y 5 & 325u 300v 7. … (ii), 35. Let the speed of the sailor in still water be x kmph and the speed of the current be, y kmph.
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Linear Equations in Two Variables, , 153, , Then, speed downstream (x y) km/hr., And, speed upstream (x y) km/hr., 8 40 2, & x y 12, x y 60 3, 8, and 1 & x y 8., x y, , , … (i), … (ii), , Solve (i) and (ii)., 1, 1, 37. Let 1 man‘s 1 day‘s work be x and 1 boy‘s 1 day‘s work be y ·, 2 5 1, 1, Then, x y 4 & 2u 5v 4, 3 6 1, 1, and x y 3 & 3u 6v 3 ·, 38. Let length x metres and breadth y metres., Then, x y 3 and (x 3)(y 2) xy., , , x y 3 and 3y 2x 6., , 39. Let length = x metres and breadth = y metres. Then,, xy (x 5) (y 3) 8 & 3x 5y 7., And, (x 3) (y 2) xy 74 & 2x 3y 68., 40. Let length = x metres and breadth = y metres. Then,, xy (x 3)(y 4) 67 & 4x 3y 55, and (x 1)(y 4) xy 89 & 4x y 93., 41. Let the basic first class full fare be ` x and the reservation charge be ` y. Then,, … (i), x y 4150., 1, And, (x y) a 2 x yk 6255 & 3x 4y 12510., … (ii), 42. Let the present ages of the man and his son be x years and y years respectively. Then,, x 5 3 (y 5) & x 3y 10, and x 5 7 (y 5) & x 7y 30., , … (i), … (ii), , 43. Let the present ages of the man and his son be x years and y years respectively. Then,, (x 2) 5 (y 2) & x 5y 8, , … (i), , and (x 2) 3 (y 2) 8 & x 3y 12., , … (ii), , 44. Let father‘s age be x years and son‘s age be y years. Then,, x 2y 70 and 2x y 95., 45. Let woman‘s present age be x years and daughter‘s present age be y years. Then,, x 3y 3 & x 3y 3, and, x 3 2 (y 3) 10 & x 2y 13., 46. Let the cost price of the tea set be ` x and that of the lemon set be ` y. Then,, 115y, (19x 23y), c 95x , m (x y) 7 &, (x y) 7, 20, 100 100, & 3y x 140., , … (i)
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154, , Secondary School Mathematics for Class 10, (21x 22y), 105x 110y, (x y) 13, And, c 100 100 m (x y) 13 &, 20, & x 2y 260., , … (ii), , Now, solve (i) and (ii)., 47. Let the fixed charge be ` x and charge for each extra day be ` y., Then, x (7 3) y 27 & x 4y 27, , … (i), , and, x (5 3) y 21 & x 2y 21., , … (ii), , 48. Let x litres of 50% solution be mixed with y litres of 25% solution., Then, x y 10, 50x 25y 40 # 10, and, 100 100 100 & 2x y 16., , … (i), … (ii), , 49. Let x g of 18-carat gold be mixed with y g of 12-carat gold to get 120 g of 16-carat gold., Then, x y 120., 18, Gold % in 18-carat gold a 24 # 100k % 75%, 12, Gold % in 12-carat gold a 24 # 100k % 50%, 16, 200, Gold % in 16-carat gold a 24 # 100k % 3 %, 200, 75% of x 50% of y 3 % of 120, 75x 50y 200 # 120, 100 100 3 # 100 & 3x 2y 320., Now, solve (i) and (ii)., , … (i), , … (ii), , 50. Let x litres of 90% pure solution be mixed with y litres of 97% pure solution to get, 21 litres of 95% pure solution., Then, x y 21, 90x 97y 95 # 21, and 100 100 100 & 90x 97y 1995., Now, solve (i) and (ii)., , … (i), … (ii), , 51. x y 18 and x y 180. Find x and y., 52. +A +B +C 180c & x (3x 2) y 180 & 4x y 182., +C +B 9c & y 3x 2 9 & y 3x 7., , … (i), … (ii), , From (i) and (ii), we get 7x 175 & x 25., Putting x 25 in (i), we get y 82., , , +A 25c, +B (3 # 25 2)c 73c, +C 82c., , 53. We have, +A +C 180c and +B +D 180c., (2x 4) (2y 10) 180 & x y 83, , … (i), , and (y 3) (4x 5) 180 & 4x y 182., , … (ii), , , , From (i) and (ii), we get, , , 3x 99 & x 33. And, 33 y 83 & y 50.
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Linear Equations in Two Variables, , f, , 155, , EXERCISE 3F, , Very-Short and Short-Answer Questions, 1. Write the number of solutions of the following pair of linear equations:, [CBSE 2009], x 2y 8 0, 2x 4y 16., 2. Find the value of k for which the following pair of linear equations have, infinitely many solutions:, [CBSE 2010], 2x 3y 7, (k 1) x (k 2) y 3k., 3. For what value of k does the following pair of linear equations have, infinitely many solutions?, 10x 5y (k 5) 0 and 20x 10y k 0., 4. For what value of k will the following pair of linear equations have no, solution?, [CBSE 2010], 2x 3y 9, 6x (k 2) y (3k 2) ., 5. Write the number of solutions of the following pair of linear equations:, x 3y 4 0 and 2x 6y 7 0., 6. Write the value of k for which the system of equations 3x ky 0,, 2x y 0 has a unique solution., 7. The difference between two numbers is 5 and the difference between, their squares is 65. Find the numbers., 8. The cost of 5 pens and 8 pencils is ` 120, while the cost of 8 pens and, 5 pencils is ` 153. Find the cost of 1 pen and that of 1 pencil., 9. The sum of two numbers is 80. The larger number exceeds four times, the smaller one by 5. Find the numbers., 10. A number consists of two digits whose sum is 10. If 18 is subtracted, from the number, its digits are reversed. Find the number., 11. A man purchased 47 stamps of 20 p and 25 p for ` 10. Find the number, of each type of stamps., 12. A man has some hens and cows. If the number of heads be 48 and, number of feet be 140, how many cows are there?, 2 3, 9, 4 9 21, 13. If x y xy and x y xy , find the values of x and y., x, x y 5, 14. If 4 3 12 and 2 y 1 then find the value of (x y) ., 15. If 12x 17y 53 and 17x 12y 63 then find the value of (x y) .
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156, , Secondary School Mathematics for Class 10, , 16. Find the value of k for which the system 3x 5y 0, kx 10y 0 has a, nonzero solution., 17. Find k for which the system kx y 2 and 6x 2y 3 has a unique, solution., 18. Find k for which the system 2x 3y 5 0, 4x ky 10 0 has an, infinite number of solutions., 19. Show that the system 2x 3y 1 0, 4x 6y 4 0 has no solution., 20. Find k for which the system x 2y 3 and 5x ky 7 0 is inconsistent., 21. Solve:, , 3 2 , 9, 4, 2 and 1., xy xy, x y x y, ANSWERS (EXERCISE 3F), , 1. infinitely many 2. k 7, , 3. k 10, , 4. k 11, , 3, 6. k ! 2, , 8. ` 16, ` 5, , 9. 65 and 15, , 7. 9 and 4, , 11. 35 and 12, , 12. 22, , 16. k = 6 17. k ! 3 18. k = 6, , 3, , 2, , 20. k 5, , 15. 4, , 5, , 1, , 21. x 2 , y 2, , 1. Given equations are x 2y 8 0 and 2x 4y 16 0., , a1 b1 c1, ·, a2 b2 c2, , So, there are infinitely many solutions., 2. 2x 3y 7 0 and (k 1) x (k 2) y 3k 0., a, b, c, For infinitely many solutions, we have a1 1 c1 ·, b2, 2, 2, 2 3, 3 7, , and, (k 1) (k 2), (k 2) 3k, 2k 4 3k 3 and 9k 7k 14 & k 7., a, b, c, 3. For infinitely many solutions, we have a1 1 c1 ·, b2, 2, 2, 10 5 k 5, k5 1, , , 20 10, &, 2 & 2k 10 k & k 10., k, k, a, b, c, 4. For no solution, we have a1 1 ! c1 ·, b2, 2, 2, , , 2 3, 2, 9, 3 1, 9, 1, 6 k 2 and 6 ! 3k 2 & k 2 3 and 3k 2 ! 3, 29, k 2 9 and 3k 2 ! 27 & k 11 and k ! 3 ·, Hence, k 11., , , 10. 64, , 13. x = 1, y = 3 14. 2, , HINTS TO SOME SELECTED QUESTIONS, , , , 5. 0
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Linear Equations in Two Variables, , 157, , a, b, c, 5. a1 1 ! c1 ·, b2, 2, 2, Hence, the given system has no solution., a, b, 6. For unique solution, we have a1 ! 1 ·, b2, 2, 3, 3, k, 2 ! &k! 2 ·, 1, 7. Let the required numbers be x and y. Then,, x 2 y 2 65, , , , 5 13 & x y 13., xy, Solving x y 5 and x y 13, we get x 9 and y 4., x y 5 and x 2 y 2 65 &, , 8. Let the cost of 1 pen be ` x and that of 1 pencil be ` y. Then,, 5x 8y 120, , … (i) and 8x 5y 153., , … (ii), , Adding (i) and (ii), we get 13 (x y) 273 & x y 21., , … (iii), , Subtracting (i) from (ii), we get 3 (x y) 33 & x y 11., , … (iv), , From (iii) and (iv), we get, , x 16 and y 5., , 9. Let the required numbers be x and y such that x y. Then,, x y 80, , … (i) and x 4y 5., , … (ii), , Solve (i) and (ii)., 10. Let the ten‘s digit be x and unit‘s digit be y. Then,, x y 10, , … (i) and (10x y) 18 (10y x) & x y 2. … (ii), , Solve (i) and (ii)., 11. Let x stamps of 20 p and y stamps of 25 p be purchased. Then,, x y 47, , … (i) and 20x 25y 1000 & 4x 5y 200., , … (ii), , Solve (i) and (ii)., 12. Let there be x cows and y hens. Then,, x y 48, , … (i) and 4x 2y 140 & 2x y 70., , … (ii), , Subtracting (i) from (ii), we get x 22., 13. The given equations are, 2y 3x 9, , … (i) and 4y 9x 21., , … (ii), , … (i) and x 2y 2., , … (ii), , Solve (i) and (ii)., 14. Given equations are 3x 4y 5, , Solve (i) and (ii) for x and y and find (x y) ., 15. Adding the given equations, we get, 116, 29 (x y) 116 & x y 29 4., 16. The given system has a nonzero solution when, , 3 5, & k 6., k 10, , 1, k, 17. For a unique solution, we have 6 ! & k ! 3., 2
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158, , Secondary School Mathematics for Class 10, , 18. For an infinite number of solutions, we have, 2 3 5, , 4 k 10 & k 6., a, b, c, 19. Here a1 1 ! c1 · Hence, the given system has no solution., b2, 2, 2, a, b, c, 20. The system will be inconsistent when a1 1 ! c1 ·, b2, 2, 2, 3, 1 2, 2, 5 ! 7 & k 5·, k, 1 , 1, 21. Put , u and v to get 3u 2v 2 and 9u 4v 1., x y, x y, 1, 1, This gives u 3 and v 2 · So, x y 3 and x y 2., Solve for x and y., , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. If 2x 3y 12 and 3x 2y 5 then, (a) x 2, y 3, (b) x 2, y 3, , (c) x 3, y 2 (d) x 3, y 2, , 2 1, then, xy 5, (b) x 5, y 3, (c) x 6, y 4 (d), (a) x 4, y 2, y, y, 2, 2x, 1, x, If 3 2 6 0 and 2 3 3 then, (b) x 2, y 3 (c) x 2, y 3 (d), (a) x 2, y 3, 12, 31 , If x y 4 and y x 11 then, 1, (b) x 2, y 3 (c) x 2 , y 3 (d), (a) x 2, y 3, 2x y 2 3x y 1 3x 2y 1, , , If, then, 5, 3, 6, (b) x 1, y 1 (c) x 1, y 2 (d), (a) x 1, y 1, 3 2 , 9, 4, If , 2 and 1 then, x y xy, x y x y, 1, 3, 5, 1, 3, 1, (a) x 2 , y 2 (b) x 2 , y 2, (c) x 2 , y 2 (d), If 4x 6y 3xy and 8x 9y 5xy then, (a) x 2, y 3, (b) x 1, y 2, (c) x 3, y 4 (d), , 2. If x y 2 and, , 3., 4., , 5., 6., , 7., , x 7, y 5, , x 2, y 3, 1, 1, x 2,y 3, , x 2, y 1, , 1, 5, x 2, y 2, x 1, y 1, , 8. If 29x 37y 103 and 37x 29y 95 then, (a) x 1, y 2, (b) x 2, y 1, (c) x 3, y 2 (d) x 2, y 3, 9. If 2 x y 2 x y 8 then the value of y is, 3, 1, (b) 2, (c) 0, (a) 2, , (d) none of these
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Linear Equations in Two Variables, , 2 3, 1 1, 10. If x y 6 and x 2y 2 then, 2, 2, (a) x 1, y 3, (b) x 3 , y 1, , 159, , 3, 3, (c) x 1, y 2 (d) x 2 , y 1, , 11. The system kx y 2 and 6x 2y 3 has a unique solution only when, (a) k 0, (b) k ! 0, (c) k 3, (d) k ! 3, 12. The system x 2y 3 and 3x ky 1 has a unique solution only when, (a) k 6, (b) k ! 6, (c) k 0, (d) k ! 0, 13. The system x 2y 3 and 5x ky 7 0 has no solution, when, 7, (a) k 10, (b) k ! 10, (c) k 3, (d) k 21, , 14. If the lines given by 3x 2ky 2 and 2x 5y 1 0 are parallel then the, value of k is, 5, 2, 3, 15, (a) 4, (b) 5, (c) 2, (d) 4, 15. For what value of k do the equations kx 2y 3 and 3x y 5 represent, two lines intersecting at a unique point?, (a) k 3, (b) k 3, (c) k 6, , (d) all real values except –6, 16. The pair of equations x 2y 5 0 and 3x 6y 1 0 has, (a) a unique solution, , (b) exactly two solutions, , (c) infinitely many solutions, (d) no solution, 17. The pair of equations 2x 3y 5 and 4x 6y 15 has, (a) a unique solution, , (b) exactly two solutions, , (c) infinitely many solutions, , (d) no solution, , 18. If a pair of linear equations is consistent then their graph lines will be, (a) parallel, , (b) always coincident, , (c) always intersecting, , (d) intersecting or coincident, , 19. If a pair of linear equations is inconsistent then their graph lines will be, (a) parallel, , (b) always coincident, , (c) always intersecting, (d) intersecting or coincident, 20. In a 3ABC, +C 3+B 2 (+A +B), then +B ?, (a) 20, , (b) 40, , (c) 60, , (d) 80, , 21. In a cyclic quadrilateral ABCD, it is being given that +A (x y 10)c,, +B (y 20)c, +C (x y 30)c and +D (x y)c. Then, +B ?, (a) 70, , (b) 80, , (c) 100, , (d) 110
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160, , Secondary School Mathematics for Class 10, , 22. The sum of the digits of a two-digit number is 15. The number obtained, by interchanging the digits exceeds the given number by 9. The number is, (a) 96, , (b) 69, , (c) 87, , (d) 78, , 23. In a given fraction, if 1 is subtracted from the numerator and 2 is added, 1, to the denominator, it becomes 2 $ If 7 is subtracted from the numerator, 1, and 2 is subtracted from the denominator, it becomes 3 $ The fraction is, 16, 15, 16, 13, (b) 26, (c) 27, (d) 21, (a) 24, 24. 5 years hence, the age of a man shall be 3 times the age of his son while, 5 years earlier the age of the man was 7 times the age of his son. The, present age of the man is, (a) 45 years, , (b) 50 years, , (c) 47 years, , (d) 40 years, , 25. The graphs of the equations 6x 2y 9 0 and 3x y 12 0 are two, lines which are, (a) coincident, (b) parallel, (c) intersecting exactly at one point, (d) perpendicular to each other, 26. The graphs of the equations 2x 3y 2 0 and x 2y 8 0 are two, lines which are, (a), (b), (c), (d), , coincident, parallel, intersecting exactly at one point, perpendicular to each other, , 24, 27. The graphs of the equations 5x 15y 8 and 3x 9y 5 are two lines, which are, (a), (b), (c), (d), , coincident, parallel, intersecting exactly at one point, perpendicular to each other, ANSWERS (MCQ), , 1. (c), 2. (c), 3. (a), 4. (d) 5. (a), 6. (b) 7. (c), 8. (a), 9. (c), 10. (b) 11. (d) 12. (b) 13. (a) 14. (d) 15. (d) 16. (d) 17. (d) 18. (d), 19. (a) 20. (b) 21. (b) 22. (d) 23. (b) 24. (d) 25. (b) 26. (c) 27. (a)
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Linear Equations in Two Variables, , 161, , HINTS TO SOME SELECTED QUESTIONS, 3. 4x 3y 1, 3x 4y 18. Solve., 1, 1, 4. Putting x u and y v, we get u 2v 4 and u 3v 11., 5., , 2x y 2 3x y 1, , & 6x 3y 6 15x 5y 5 & 9x 8y 1., 5, 3, 3x y 1 3x 2y 1, , & 18x 6y 6 9x 6y 3 & 9x 12y 3., 3, 6, Solve (i) and (ii) to get x 1 and y 1., , … (i), … (ii), , 1 , 1, u and v to get 3u 2v 2 and 9u 4v 1., xy, x y, 1, 1, Solve for u and v to get u 3 and v 2 ·, x y 3 and x y 2., , 6. Put, , 1, 1, 7. Divide throughout by xy and put x u and y v to get, … (i) and 8v 9u 5., 4v 6u 3, 1, 1, This gives u 3 and v 4 · Hence, x 3 and y 4., 8. Adding (i) and (ii), we get 66 (x y) 198 & x y 3., , … (ii), , Subtracting (ii) from (i), we get 8 (y x) 8 & y x 1., 3, 3, 9. 2 x y 2 x y 2 3/2 & x y 2 and x y 2 · So, y 0., 1, 1, 1, 10. Put x u and y v. Then, 2u 3v 6, … (i) and u 2 v 2 & 2u v 4. … (ii), a, b, 11. For a unique solution, we have a1 ! 1 ·, b2, 2, 1, k, 6 ! & k ! 3., 2, a, b, 2, 1, 12. For a unique solution, we have a1 ! 1 · So, 3 !, & k ! 6., k, b2, 2, a, b, c, 13. For no solution, we have a1 1 ! c1 ·, b2, 2, 2, 3, 14 ·, 12, , 5, ! 7 & k 10 and k ! 3 Hence, k 10., k, a, b, c, 14. For parallel lines, we have a1 1 ! c1 ·, b2, 2, 2, 2, 3 2k, 15, 15, 2 5 ! 1 & k 4 and k ! 5 & k 4 ·, a, b, 15. For a unique intersecting point, we have a1 ! 1 ·, b2, 2, 2, k, 3 ! 1 & k ! 6. Hence, correct answer is (d)., c, a, 1 b, 1, 5, 1, 2, 16. Here, a1 3 , 1 3 and c1 1 ·, b2, 3, 6, 2, 2, , , a1 b1, c1, ·, , a2 b2 ! c2 So, the given system has no solution.
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162, , Secondary School Mathematics for Class 10, , c, a, 5 1, 2 1 b, 3 1, 17. Here, a1 4 2 , 1 6 2 and c1 3 ·, b2, 15, 2, 2, , , a1 b1, c1, ·, , a2 b2 ! c2 So, the given system has no solution., , 18. If a pair of linear equations is consistent then their graph lines will be intersecting or, coincident., 20. Let C 3B 2(A B) xc., xc, xc, Then, C xc, B ` 3 j and (A B) ` 2 j ·, x, (A B) C 180c & 2 x 180 & 3x 360 & x 120., 120 c, +B a, k 40c., 3, 21. +A +C 180c and +B +D 180c., This gives x y 100 and x 2y 160., 22. Let the ten‘s digit be x and the unit‘s digit be y. Then,, x y 15, , … (i), , and (10y x) (10x y) 9 & y x 1., , x, 23. Let the required fraction be y · Then,, x1 1, & 2x 2 y 2 & 2x y 4, y2 2, x7 1, & 3x 21 y 2 & 3x y 19., y2 3, , … (ii), , … (i), … (ii), , 24. Let the present ages of the man and his son be x years and y years respectively. Then,, (x 5) 3(y 5) & x 3y 10, , … (i), , (x 5) 7(y 5) & x 7y 30., , … (ii), , a, c, 2 2, 6 2 b, 9, 3, 25. a1 3 1 , 1 1 and c1 12 4 ·, b2, 1, 2, 2, a1 b1, c1, ·, , a2 b2 ! c2, So, the system is inconsistent and hence the lines are parallel., , , a, b, a, 2 b, 3, 26. a1 1 , 1 · So, a1 ! 1 ·, b2, b2, 2, 2, 2, Thus, the system has a unique solution and therefore the lines intersect exactly at, one point., 27. The equations are 5x 15y 8 0 and 15x 45y 24 0., , , a1, b, c1, , , 5 1 1 15 1, 8 1, a2 15 3 , b2 45 3 and c2 24 3, , , , a1 b1 c1, , , a2 b2 c2 and therefore the system has infinitely many solutions., , Hence, the graph lines are coincident.
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Linear Equations in Two Variables, , 163, , TEST YOURSELF, MCQ, 1. The graphic representation of the equations x 2y 3 and 2x 4y 7 0, gives a pair of, (a) parallel lines, , (b) intersecting lines, , (c) coincident lines, , (d) none of these, , 2. If 2x 3y 7 and (a b)x (a b 3)y 4a b have an infinite number of, solutions then, (a) a 5, b 1, (b) a 5, b 1, (c) a 5, b 1, , (d) a 5, b 1, 3. The pair of equations 2x y 5, 3x 2y 8 has, (a) a unique solution, , (b) two solutions, , (c) no solution, (d) infinitely many solutions, 4. If x y and y 0, which of the following is wrong?, (a) x 2 y 0, , (b) x y 0, , (c) xy 0, , 1 1, (d) x y 0, , Short-Answer Questions, 1, 1, 5. Show that the system of equations x 2y 2 0 and 2 x 2 y 1 0, has a unique solution., 6. For what values of k is the system of equations kx 3y k 2,, 12x ky k inconsistent?, 3x 5y 7, 7. Show that the equations 9x 10y 21, 2 3 2 have infinitely many, solutions., 8. Solve the system of equations x 2y 0, 3x 4y 20., 9. Show that the paths represented by the equations x 3y 2 and, 2x 6y 5 are parallel., 10. The difference between two numbers is 26 and one number is three, times the other. Find the numbers., 11. Solve: 23x 29y 98, 29x 23y 110., 12. Solve: 6x 3y 7xy and 3x 9y 11xy., 13. Find the value of k for which the system of equations 3x y 1 and, kx 2y 5 has (i) a unique solution, (ii) no solution., 14. In a 3ABC, +C 3+B 2(+A +B) . Find the measure of each one of, +A, +B and +C.
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164, , Secondary School Mathematics for Class 10, , 15. 5 pencils and 7 pens together cost ` 195 while 7 pencils and 5 pens, together cost ` 153. Find the cost of each one of the pencil and the pen., 16. Solve the following system of equations graphically:, 2x 3y 1, 4x 3y 1 0., Long-Answer Questions, 17. Find the angles of a cyclic quadrilateral ABCD in which +A (4x 20)c,, +B (3x 5)c, +C (4y)c and +D (7y 5)c ., 18. Solve for x and y:, , 35 14 , 14, 35, 19, 37., xy xy, x y x y, , 19. If 1 is added to both the numerator and the denominator of a fraction,, 4, it becomes 5 · If, however, 5 is subtracted from both the numerator and, 1, the denominator, the fraction becomes 2 · Find the fraction., 20. Solve:, , ax by , , a a b, ax by 2ab., b, ANSWERS (TEST YOURSELF), , 1. (a), , 2. (d), , 3. (a), , 4. (d), , 8. x 4, y 2, , 6. k !6, , 3, , 11. x 3, y 1 12. x 1, y 2 13. (i) k ! 6 (ii) k 6, 14. +A 20c, +B 40c, +C 120c, 15. ` 4 per pencil, ` 25 per pen, 10. 39, 13, , 16. x 1, y 1, , 17. +A 120c, +B 70c, +C 60c, +D 110c, , 18. x 4, y 3, , 19. 9, , 7, , 20. x b, y a, ,
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Quadratic Equations, , 165, , Quadratic Equations, , 4, , HISTORY, , Babylonians were the first to solve the quadratic equations of the form, x 2 px + q = 0., , Brahmagupta (AD 598–665), an Indian mathematician gave an explicit, formula to solve a quadratic equation of the form ax 2 bx = c., An Arab mathematician, Al-Khwarizmi in AD 800 also studied, quadratic equations of various types., An ancient Indian mathematician Shridharacharya derived the, well-known Quadratic Formula for solving a quadratic equation, ax 2 + bx + c = 0 by the method of completing the square. This is being, used as the standard formula for solving such an equation., According to this formula, the roots of ax 2 + bx + c = 0 are given by, b D, b + D, = 2, and =, =, 2a , where D (b 4ac) is called the, 2a, discriminant of this equation., , A quadratic equation in the variable x is an equation of the, form ax + bx + c = 0, where a, b, c are real numbers and a ! 0., , QUADRATIC EQUATION, 2, , A real number is called a root of the quadratic, , , , equation ax bx c 0, a ! 0 if a 2 + b + c = 0., , ROOTS OF A QUADRATIC EQUATION, 2, , NOTE 1., , If is a root of ax 2 + bx + c = 0 then we say that, (i) x = satisfies the equation ax 2 + bx + c = 0, or (ii) x = is a solution of the equation ax 2 + bx + c = 0., , NOTE 2., , The roots of a quadratic equation ax 2 + bx + c = 0 are called the, zeros of the polynomial ax 2 + bx + c., , SOLVING A QUADRATIC EQUATION, , Solving a quadratic equation means finding, , its roots., 165
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166, , Secondary School Mathematics for Class 10, , SOLVED EXAMPLES, EXAMPLE 1, , Which of the following are quadratic equations?, (i) x 2 5x 3 0, (iii) 3x 2 2 x 8 0, 1, (v) x x x 2, , SOLUTION, , (ii) 2x 2 3 2 x 6 0, (iv) 2x 2 3 0, 1, 1, (vi) x 2 2 4, 4, x, , (i) Clearly, (x 2 5x 3) is a quadratic polynomial., , , x 2 5x 3 0 is a quadratic equation., , (ii) Clearly, (2x 2 3 2 x 6) is a quadratic polynomial., , , 2x 2 3 2 x 6 0 is a quadratic equation., , (iii) 3x 2 2 x 8 is not of the form ax 2 bx c 0., , , 3x 2 2 x 8 0 is not a quadratic equation., , (iv) 2x 2 3 0 is of the form ax 2 bx c 0, where a 2, b = 0, and c 3., 2x 2 3 0 is a quadratic equation., 1, (v) x x x 2 & x 2 1 x 3 & x 3 x 2 1 0., , , And, (x 3 x 2 1) being a polynomial of degree 3, it is not, quadratic., 1, Hence, x x x 2 is not a quadratic equation., 1 17, & 4x 4 4 17x 2 & 4x 4 17x 2 4 0., 4, x2, Clearly, 4x 4 17x 2 4 is a polynomial of degree 4., , (vi) x 2 , , , EXAMPLE 2, , x2 , , 1 17, is not a quadratic equation., 4, x2, , Check whether the following are quadratic equations:, (i) (2x 1) (x 3) (x 4) (x 2) (ii) (x 2) 3 2x (x 2 1), (iii) (x 1) 3 x 3 x 6, , SOLUTION, , (iv) x (x 3) 6 (x 2) (x 2), , We have, (i) (2x 1) (x 3) (x 4) (x 2), 2x 2 7x 3 x 2 2x 8 & x 2 9x 11 0., This is of the form ax 2 bx c 0, where a 1, b 9 and, c 11., Hence, the given equation is a quadratic equation.
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Quadratic Equations, , 167, , (ii) (x + 2) 3 = 2x (x 2 1), , , , , , x 3 + 8 + 6x (x + 2) = 2x 3 2x, , , , x 3 + 6x 2 + 12x + 8 = 2x 3 2x, x 3 6x 2 14x 8 = 0., , , , This is not of the form ax 2 + bx + c = 0., Hence, the given equation is not a quadratic equation., (iii) (x + 1) 3 = x 3 + x + 6, , , x 3 + 1 + 3x (x + 1) = x 3 + x + 6, , , , 3x 2 + 2x 5 = 0., , This is of the form ax 2 + bx + c = 0, where a = 3, b = 2 and, c = 5., Hence, the given equation is a quadratic equation., (iv) x (x + 3) + 6 = (x + 2) (x 2), , , , x 2 + 3x + 6 = x 2 4, 3x + 10 = 0., , This is not of the form ax 2 + bx + c = 0., Hence, the given equation is not a quadratic equation., EXAMPLE 3, , For the quadratic equation 2x 2 5x 3 = 0, show that, (i) x = 3 is its solution., (iii) x = 4 is not its solution., , SOLUTION, , 1, (ii) x = 2 is its solution., , The given equation is 2x 2 5x 3 = 0., (i) On substituting x = 3 in the given equation, we get, LHS = 2 # 3 2 5 # 3 3 = (18 15 3) = 0 = RHS., , x = 3 is a solution of 2x 2 5x 3 = 0., 1, (ii) On substituting x = 2 in the given equation, we get, 1 2, , l 5 #b 1 l 3, LHS = 2 # b, 2, 2, , , , , = &2 # 1 + 5 # 1 3 0, 2, 4, 15, ', 31 0 RHS., 2 2, 1, x = 2 is a solution of 2x 2 5x 3 = 0.
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168, , Secondary School Mathematics for Class 10, , (iii) On substituting x = 4 in the given equation, we get, = 0., LHS = 2 # 4 2 5 # 4 3 = (32 20 3) = 9 Y, Thus, LHS ≠ RHS., x = 4 is not a solution of 2x 2 5x 3 = 0., EXAMPLE 4, , SOLUTION, , Show that, 2 and 2 2, x 2 + 2 x 4 = 0., , are the roots of the equation, , The given equation is x 2 + 2 x 4 = 0., Putting x = 2 in the given equation, we get, LHS = ( 2 ) 2 + ( 2 # 2 ) 4 = (2 + 2 4) = 0 = RHS., , , , 2 is a root of the given equation., , Putting x = 2 2 in the given equation, we get, LHS = (2 2 ) 2 + 2 # (2 2 ) 4 = (8 4 4) = 0 = RHS., , , , 2 2 is a root of the given equation., , Hence,, , 2 and 2 2 are the roots of the given equation., , SOLVING A QUADRATIC EQUATION BY FACTORISATION, , Let the given quadratic equation be ax 2 + bx + c = 0, where a ! 0. Let, (ax 2 + bx + c) be expressible as the product of two linear expressions, say, (px + q) and (rx + s), where p, q, r, s are real numbers such that p ! 0 and, r ! 0. Then,, ax 2 bx c 0 & (px q) (rx s) 0, & (px q) 0 or (rx s) 0, q, & x p or x rs ·, (x + 2)(3x 5) = 0., , EXAMPLE 5, , Solve:, , SOLUTION, , We have, (x 2)(3x 5) 0 & x 2 0 or 3x 5 0, 5, & x 2 or x 3 ·, Hence, the roots of the given equation are –2 and 35 ·, 5x 2 8x = 0., , EXAMPLE 6, , Solve:, , SOLUTION, , We have, 5x 2 8x 0 & x (5x 8) 0 & x 0 or 5x 8 0, 8, & x 0 or x 5 ·
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170, , Secondary School Mathematics for Class 10, , 4 3 x 2 + 5x 2 3 = 0., , EXAMPLE 10, , Solve:, , SOLUTION, , Here, 4 3 # (2 3 ) = 24, 8 # (3) = 24 and 8 + (3) = 5., , [CBSE 2013], , , , 4 3 x 2 + 5x 2 3 = 0 4 3 x 2 + 8x 3x 2 3 = 0, , , , 4x ( 3 x + 2) 3 ( 3 x + 2) = 0, , , , ( 3 x + 2) (4x 3 ) = 0 , , , , x=, , , , x =e, , 3 x + 2 = 0 or 4x 3 = 0, , 3, 2, or x = 4, 3, 2, #, 3, , 2 3, 3, 3, o=, or x = 4 ·, 3, 3, , 4x 2 12x + 9 = 0., , EXAMPLE 11, , Solve:, , SOLUTION, , Here, 4 # 9 = 36, (6) # (6) = 36, and (6) + (6) = 12., , , 4x 2 12x + 9 = 0, , , , 4x 2 6x 6x + 9 = 0 2x (2x 3) 3 (2x 3) = 0, , (2x 3) (2x 3) = 0 (2x 3) 2 = 0, 3, 2x 3 0 & x ·, 2, 3, =, Hence, x 2 is the repeated root of the given equation., , , EXAMPLE 12, , SOLUTION, , Solve the following equation by using factorisation method:, [CBSE 2012, ‘15], 4x 2 4ax + (a 2 b 2) = 0., We may write, 4a = {2 (a + b)} + {2 (a b)} ., Also, {2 (a + b)} # {2 (a b)} = 4 (a 2 b 2) ., , , 4x 2 4ax + (a 2 b 2) = 0, , , , 4x 2 2 (a + b) x 2 (a b) x + (a 2 b 2) = 0, , , , 2x {2x (a + b)} (a b) {2x (a + b)} = 0, , {2x (a b)}#{2x (a b)} 0, 2x (a b) 0 or 2x (a b) 0, (a b), (a b), ·, x, or x , 2, 2, (a b), (a b), Hence,, and, are the roots of the given equation., 2, 2, , , EXAMPLE 13, , SOLUTION, , Solve the following equation by using factorisation method:, [CBSE 2015], 9x 2 6b 2 x (a 4 b 4) 0., We may write, 6b 2 = 3 (a 2 b 2) 3 (a 2 + b 2) .
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Quadratic Equations, , 171, , Also, {3 (a 2 b 2)}#{3 (a 2 b 2)} 9 (a 4 b 4), , , 9x 2 6b 2 x (a 4 b 4) = 0, , , , 9x 2 + 3 (a 2 b 2) x 3 (a 2 + b 2) x (a 4 b 4) = 0, , , , 3x {3x + (a 2 b 2)} (a 2 + b 2) {3x + (a 2 b 2)} = 0, , , , {3x + (a 2 b 2)} # {3x (a 2 + b 2)} = 0, , , , 3x + (a 2 b 2) = 0 or 3x (a 2 + b 2) = 0, , (b 2 a 2), (a 2 b 2), ·, or x , 3, 3, (b 2 a 2), (a 2 + b 2), Hence,, and, are the required roots of the, 3, 3, given equation., , , x, , x + 3 = 3x 7 ·, x + 2 2x 3, , EXAMPLE 14, , Solve:, , SOLUTION, , By cross multiplication, we get, (x + 3) (2x 3) = (x + 2) (3x 7), , , 2x 2 + 3x 9 = 3x 2 x 14 x 2 4x 5 = 0, , , , x 2 5x + x 5 = 0 x (x 5) + (x 5) = 0, , , , (x 5) (x + 1) = 0 x 5 = 0 or x + 1 = 0, , , , x = 5 or x = 1., , Hence, 5 and –1 are the roots of the given equation., EXAMPLE 15, SOLUTION, , 14 5, 1, , x ! 3, 1., x3, x1, The given equation may be written as, 14 5 , 1, x3 x1, 14 (x 1) 5 (x 3), 1, , (x 3) (x 1), (9x 1), 1 & x 2 4x 3 9 x 1, , 2, x 4x 3, x 2 5x 4 0 & x 2 4x x 4 0, x (x 4) (x 4) 0 & (x 4)(x 1) 0, Solve:, , , , , x 4 0 or x 1 0, x 4 or x 1., , Hence, 4 and 1 are the roots of the given equation., , [CBSE 2014]
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172, , Secondary School Mathematics for Class 10, , 1 1 11, , x ! 4, 7., (x 4) (x 7) 30, , EXAMPLE 16, , Solve:, , SOLUTION, , We have, 1 1 11, (x 4) (x 7) 30, (x 7) (x 4) 11, 11, , 11, , 2, 30, (x 4) (x 7), (x 3x 28) 30, 1, 1 [on dividing both sides by 11], , 2, (x 3x 28) 30, (x 2 3x 28) 30, , [CBSE 2008, ‘10], , [by cross multiplication], , x 3x 2 0 & x 2x x 2 0, x (x 2) (x 2) 0 & (x 2)(x 1) 0, x 2 0 or x 1 0, x 2 or x 1., Hence, 2 and 1 are the roots of the given equation., 2, , 2, , 1, 1 1 1 , [x ! 0, x ! (a b)] ., (a b x) a b x, , EXAMPLE 17, , Solve:, , SOLUTION, , We have, 1, 111, (a b x) a b x, x (a b x) b a, 1, 1 11 , , , (a b x) x a b, x (a b x), ab, , , , [CBSE 2013], , (a + b), (a + b), =, ab, x (a + b + x), 1, = 1 [on dividing both sides by (a + b)], x (a + b + x) ab, , , , x (a + b + x) = ab [by cross multiplication], , , , x 2 + ax + bx + ab = 0 x (x + a) + b (x + a) = 0, , , , (x + a) (x + b) = 0 x + a = 0 or x + b = 0, , , , x = a or x = b., , Hence, –a and –b are the roots of the given equation., x2 + x4 = 1 Y, 3 3 , x = 3, 5 ., x3 x5, , EXAMPLE 18, , Solve:, , SOLUTION, , We have, x 2 x 4 10, x3 x5 3, (x 2)(x 5) (x 4)(x 3) 10, , , 3, (x 3)(x 5), , [CBSE 2014]
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Quadratic Equations, , , , (x 2 7x 10) (x 2 7x 12) 10, , 3, (x 2 8x 15), , , , (2x 14x 22) 10, , 3, (x 8x 15), , , , 173, , 3 (2x 14x 22) 10 (x 8x 15), [by cross multiplication], , 6x 2 42x 66 10x 2 80x 150, 4x 2 38x 84 0 & 2x 2 19x 42 0, 2x 2 12x 7x 42 0 & 2x (x 6) 7 (x 6) 0, (x 6)(2x 7) 0 & x 6 0 or 2x 7 0, 7, x 6 or x ·, 2, 7, Hence, 6 and 2 are the roots of the given equation., 2x 1 x 3 , m 5, x ! 3, 1 ·, m 3c, x3, 2x 1, 2, , EXAMPLE 19, , Solve: 2 c, , SOLUTION, , On putting, , [CBSE 2014], , 2x 1 =, y, the given equation becomes, x+3, , 3, 2y y 5 & 2y 2 3 5y, & 2y 2 5y 3 0 & 2y 2 6y y 3 0, & 2y (y 3) (y 3) 0 & (y 3) (2y 1) 0, & y 3 0 or 2y 1 0, 1, , & y 3 or y 2 ·, , 2x 1 , 3, x3, & 2x 1 3 (x 3) [by cross multiplication], & 2x 1 3x 9, & x 10., , Case I, , y3&, , Case II, , y, , , 1, , & 2xx 31 21, 2, & 2 (2x 1) (x 3), , & 5x 1, 1, , &x 5 ·, 1, Hence, –10 and 5 are the roots of the given equation.
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174, , Secondary School Mathematics for Class 10, , 3, 2 , 23 , x ! 0, 1, 2., (x 1) 2 (x 2) 5x, , EXAMPLE 20, , Solve:, , SOLUTION, , We have, 3, 2 , 23, (x 1) 2 (x 2) 5x, 4 (x 2) 3 (x 1) 23, 7x 5, , 23, , , 5x, 2 (x 1) (x 2), 2 (x 2 x 2) 5x, 5x (7x 5) 46 (x 2 x 2) [by cross multiplication], , [CBSE 2015], , 35x 2 25x 46x 2 46x 92, 11x 2 21x 92 0 11x 2 44x 23x 92 0, 11x (x 4) 23 (x 4) 0 (x 4)(11x 23) 0, x 4 0 or 11x 23 0, 23, ·, x 4 or x , 11, 23, Hence, 4 and, are the roots of the given equation., 11, , 1 , 1, 11 , x ! 3, 1 , 9 ·, 2 7, (x 3) (2x 1) (7x 9), , EXAMPLE 21, , Solve:, , SOLUTION, , We have, 1 , 1, 11, (x 3) (2x 1) (7x 9), (2x 1) (x 3), (3x 2), 11 2, 11, , (x 3)(2x 1), (7x 9), 2x 5x 3 (7x 9), , [CBSE 2009C[, , (3x 2)(7x 9) 11(2x 2 5x 3) [by cross multiplication], 21x 2 41x 18 22x 2 55x 33, x 2 14x 51 0 & x 2 17x 3x 51 0, x (x 17) 3 (x 17) 0 & (x 17)(x 3) 0, x 17 0 or x 3 0, x 17 or x 3., Hence, –17 and 3 are the roots of the given equation., EXAMPLE 22, , Solve: 5(x + 1) 5(2 x) 5 3 1., , SOLUTION, , We have, , , , 5 ( x 1) 5 ( 2 x) 5 3 1, 25, 5 x · 5 5 2 · 5 x 126 5 x · 5 x 126, 5, 25, x, 5y y 126, where 5 y, , , , 5y 2 126y 25 0 5y 2 125y y 25 0, ,
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Quadratic Equations, , 175, , , , 5y (y 25) (y 25) 0 (y 25)(5y 1) 0, , , , y 25 0 or 5y 1 0, 1, y 25 or y , 5, 5 x 25 5 2 or 5 x 5 1, x 2 or x 1., , , , , , Hence, 2 and –1 are the roots of the given equation., f, , EXERCISE 4A, , 1. Which of the following are quadratic equations in x?, (i) x 2 x + 3 = 0, (iii), , 2 x 2 + 7x + 5 2 = 0, , (v) x 2 3x x + 4 = 0, 2, (vii) x + x = x 2, (ix) (x + 2) 3 = x 3 8, , 5, (ii) 2x 2 + 2 x 3 = 0, 1, 1, (iv) 3 x 2 + 5 x 2 = 0, 6, (vi) x x = 3, 1, (viii) x 2 2 = 5, x, (x) (2x + 3) (3x + 2) = 6 (x 1) (x 2), , 1 2, 1, (xi) b x + x l = 2 b x + x l + 3, 2. Which of the following are the roots of 3x 2 + 2x 1 = 0 ?, 1, (ii) 3, , (i) –1, , 1, (iii) 2, , 3. (i) Find the value of k for which x 1 is a root of the equation, x 2 kx 3 0. Also, find the other root., 3, (ii) Find the values of a and b for which x and x 2 are the roots, 4, of the equation ax 2 bx 6 0., 4. Show that x , ad 2 b, , bc, ad, , is a solution of the quadratic equation, , ax 2c 2 , l x bc 0., b, d, , [CBSE 2017], , Solve each of the following quadratic equations:, , 5. (2x 3)(3x 1) 0, 7. 3x 2 243 0, 9. x 2 6x 5 0, 11. x 2 12x 35 0, , 6. 4x 2 5x 0, 8. 2x 2 x 6 0, 10. 9x 2 3x 2 0, 12. x 2 18x 77
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176, , Secondary School Mathematics for Class 10, , 13. 6x 2 11x 3 0, 15. 3x 2 2x 1 0, , 14. 6x 2 x 12 0, 16. 4x 2 9x 100, , 17. 15x 2 28 x, 19. 48x 2 13x 1 0, , 18. 4 11x 3x 2, 20. x 2 2 2 x 6 0, 22., , 3 x 2 11x 6 3 0, , 23. 3 7 x 2 4x 7 0, , 24., , 7 x 2 6x 13 7 0, , 25. 4 6 x 2 13x 2 6 0, , 26. 3x 2 2 6 x 2 0, , [CBSE 2010, ’12], , [CBSE 2011], , 28. x 3 5 x 10 0, , [CBSE 2011], , [CBSE 2015], , 30. x 3 3 x 30 0, , [CBSE 2015], , [CBSE 2013], , 32. 5x 13x 8 0, , 21., , 27., , 3 x 2 10x 8 3 0, , 3x 2 2x2 3 0, 2, , 29. x ( 3 1) x 3 0, 2, , 31., , 2 x 7x 5 2 0, 2, , [CBSE 2017], , 33. x (1 2 ) x 2 0, , 2, 2, , 2, , 2, , 35. 100x 2 20x 1 0, 1, 37. 10x x 3, 39. 2x 2 ax a 2 0, , [CBSE 2015], , 40. 4x 4bx (a b ) 0, 2, , 2, , [CBSE 2013C], , 34. 9x 6x 1 0, 1, 36. 2x 2 x 0, 8, 2 5 , 38. 2 x 2 0, x, , 2, , 2, , [CBSE 2015, ’17], , 41. 4x 4a x (a b ) 0, , [CBSE 2015], , 42. x 5x (a a 6) 0, , [CBSE 2015], , 43. x 2ax (4b a ) 0, , [CBSE 2015], , 44. x (2b 1) x (b b 20) 0, , [CBSE 2015], , 45. x 6x (a 2a 8) 0, , [CBSE 2015], , 46. abx (b ac) x bc 0, , [CBSE 2014], , 47. x 4ax b 4a 0, , [CBSE 2012], , 2, , 2, , 4, , 2, , 4, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 48. 4x 2 (a b ) x a b 0, 2, , 2, , 2, , 2 2, , 49. 12abx 2 (9a 2 8b 2) x 6ab 0, , [CBSE 2006], , 50. a b x b x a x 1 0, 2 2, , 2, , 2, , 2, , [CBSE 2005], , 51. 9x 9 (a b) x (2a 5ab 2b ) 0, 2, , 2, , 16, 15, 52. x 1 , x ! 0, 1, x 1, 3, 5, 4, 53. x 3 , , x ! 0,, 2x 3, 2, 3, 1, 2, 1, 54. , , x ! 1,, 3, x 1 2 3x 1, , 2, , [CBSE 2009], [CBSE 2014], [CBSE 2014], [CBSE 2014]
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Quadratic Equations, , 1 1 6, , x ! 1, 5, x1 x5 7, 3, 1 1 1,, (ii), 1 x! , 5, 2x 3 x 5, 9, 2, 1, 1, 1, 1, , , 2a b 2x 2a b 2x, x3 1x 1, 4 , x ! 2, 0, x, x2, 4, 3x 4 7 5, 4, , x!, 7, 3, 3x 4 2, x, x1, 1, (i) x 4 , x ! 0, 1, x 1, 4, , , x 1 2x 1 , 1, (ii), 2, x ! , 1, 2, 2x 1 x 1, x x1 4, 2 , x ! 0, 1, x, x1, 15, x4 x6 1, 3 , x ! 5, 7, x5 x7, 3, x1 x3 1, 3 , x ! 2, 4, x2 x4, 3, 1 2 6, , x ! 0, 1, 2, x2 x1 x, 5, 1, 2, (i) , x ! 1, 2, 4, x 1 x 2 x 4, 3, 5, 1, 1, (ii) , x ! 1, , 4, 5, x 1 5x 1 x 4, 3x 1 2x 3 , 1 3, 3 c m 2 c m 5, x ! ,, 2x 3, 3x 1, 3 2, 3 1, 7x 1 5x 3 , 3 c m 4 c m 11, x ! ,, 5x 3, 7x 1, 5 7, 3, 4x 3 , 2x 1 , c, m 10 c, m 3, x ! 1 ,, 4x 3, 2x 1, 2 4, , 177, , 55. (i), , 56., 57., 58., 59., , 60., 61., 62., 63., 64., , 65., 66., 67., , x 2, x, 68. c m 5 c m 6 0, x ! 1, x 1, x 1, a b , 69., 2, x ! b, a, (x b) (x a), 70., , a, b, , (a b), x ! 1 , 1, a b, (ax 1) (bx 1), , 71. 3(x + 2) 3 x 10, 72. 4(x + 1) 4(1 x) 10, 73. 2 2x 3 · 2(x + 2) 32 0, , [CBSE 2017], [CBSE 2013], , [CBSE 2010], , [CBSE 2017], [CBSE 2014], [CBSE 2014], [CBSE 2017], [CBSE 2013], [CBSE 2013C], [CBSE 2017], [CBSE 2014], , [CBSE 2014]
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178, , Secondary School Mathematics for Class 10, ANSWERS (EXERCISE 4A), , 1, 2. 1 and, , 1. (i), (ii), (iii), (iv), (vi), (ix), 3. (i) k 4, other root = 3, 6. x 0 or x , , 5, 4, , 1, 2, or x , 3, 3, 3, 1, 13. x , or x , 2, 3, 25, 16. x , or x 4, 4, 1, 1, 19. x or x , 3, 16, 10. x , , 12. x = 11 or x = 7, , 1, 3, 1, 18. x 4 or x , 3, 15. x 1 or x , , 2, 3, , 13, or x 7, 7, 2, 27. x 6 or x , 3, 24. x , , 22. x 3 3 or x , 25. x , , 2 2, 3, , or x , , 36., 39., 41., 43., , 11. x = –7 or x = –5, , 3, or x , 2, 4, 17. x , or x , 3, , 4, 3, 7, 5, , 14. x , , 20. x 2 or x 3 2, , 2, 3, , 3, 4 2, , 28. x 5 or x 2 5, , 30. x 2 3 or x 5 3 31. x 2 or x , 33., , 1, 3, or x , 2, 3, 3, 8. x 2 or x , 2, 5. x , , (ii) a 4, b 5, , 7. x = 9 or x = –9, , 9. x = –5 or x = –1, , 21. x 4 3 or x , , 3, , 5, , 23. x , 26. x , , 7, 1, or x , 3, 7, 2, 3, , or x , , 2, 3, , 29. x 3 or x 1, 32. x 1 or x , , 8, 5, , 2, 1, 1, 1, 1, 34. x , x , 35. x , x , x 1 or x 2, 3, 3, 10, 10, 1, 1, 1, 1, 1, 37. x or x , 38. x 2 or x , x ,x, 2, 5, 2, 4, 4, , , (, a, b, ), ab, a, 40. x , or x , x a or x , 2, 2, 2, 2, 2, 2, 2, a b, a b, 42. x (a 3) or x (a 2), x, or x , 2, 2, 44. x = (b 5) or x = (b + 4), x = (a 2b) or x = (a + 2b), b, , c, , 45. x (a 4) or x (a 2), , 46. x a or x , b, , 47. x (2a b) or x (2a b), , 48. x , , 2b, 3a, or x , 3a, 4b, (a 2b), (2a b), 51. x , or x , 3, 3, 49. x , , a2, b2, or x , 2, 2, 1, 1, 50. x 2 or x 2, a, b, 52. x 4 or x 4, , 53. x 2 or x 1
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Quadratic Equations, , 54. x 3 or x 1, , 55. (i) x 6 or x 2, , 179, , (ii) x , , 37, or x 6, 20, , b, 2, 1, 57. x 4 or x , 58. x 6 or x 2, 9, 2, 2, 5, 1, 3, 4, 59. (i) x or x , (ii) x 2, 60. x , or x , 3, 3, 2, 2, 5, 4, 1, 61. x 8 or x 5, 62. x 5 or x , 63. x 3 or x , 2, 3, 2, 3, 11, 64. (i) x 2 or x , (ii) x , 65. x 7 or x 0, or x 1, 2, 17, 3, 4, 1, 66. x = 1 or x = 0, 67. x , 68. x , or x , or x 2, 3, 8, 2, (a b), (a b), 2, 69. x (a b) or x , 70. x , or x , 2, ab, (a b), 1, 1, 71. x 2 or x 0, 72. x or x , 73. x 2 or x 3, 2, 2, 56. x a or x , , HINTS TO SOME SELECTED QUESTIONS, 16. 4x, , 2, , 9x 100 0 & 4x 2 25x 16x 100 0., , 17. 15x 2 x 28 0 & 15x 2 21x 20x 28 0., 18. 3x 2 11x 4 0 & 3x 2 12x x 4 0., 19. 48x 2 13x 1 0 & 48x 2 16x 3x 1 0., 20. x 2 2 2 x 6 0 & x 2 3 2 x 2 x 6 0 & x (x 3 2 ) 2 (x 3 2 ) 0., 21., , 3 x 2 10x 8 3 0 & 3 x 2 12x 2x 8 3 0 & 3 x (x 4 3 ) 2 (x 4 3 ) 0., , 22., , 3 x 2 11x 6 3 0 & 3 x 2 9x 2x 6 3 0 & 3 x (x 3 3 ) 2 (x 3 3 ) 0., , 23. 3 7 x 2 4x 7 0 & 3 7 x 2 7x 3x 7 0 & 7 x (3x 7 ) (3x 7 ) 0., 24., , 7 x 2 6x 13 7 0 & 7 x 2 13x 7x 13 7 0, & x ( 7 x 13) 7 ( 7 x 13) 0., , 25. 4 6 x 2 13x 2 6 0 & 4 6 x 2 16x 3x 2 6 0, & 4 2 x ( 3 x 2 2 ) 3 ( 3 x 2 2 ) 0., 26. 3x 2 2 6 x 2 0 & 3x 2 6 x 6 x 2 0 & 3 x ( 3 x 2 ) 2 ( 3 x 2 ) 0., 27., , 3 x2 2 2 x 2 3 0 & 3 x2 3 2 x 2 x 2 3 0, & 3 x (x 6 ) 2 (x 6 ) 0., , 28. x 2 3 5 x 10 0 & x 2 2 5 x 5 x 10 0 & x (x 2 5 ) 5 (x 2 5 ) 0., 29. x 2 ( 3 1) x 3 0 & x 2 3 x x 3 0 & x (x 3 ) (x 3 ) 0., 30. x 2 3 3 x 30 0 & x 2 5 3 x 2 3 x 30 0 & x (x 5 3 ) 2 3 (x 5 3 ) 0., 31., , 2 x 2 7x 5 2 0 & 2 x 2 5x 2x 5 2 0 & x ( 2 x 5) 2 ( 2 x 5) 0., , 32. 5x 2 13x 8 0 & 5x 2 5x 8x 8 0.
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Quadratic Equations, , , 181, , 1, 1, (a x b) · < F 0, (x b) (x a), , 2x (a b), H 0, (a x b) >, (x a) (x b), (a x b) [2x (a b)] 0, (a b), ·, x (a b) or x , 2, 70. The given equation is, , , , , , b, a, ( b2 ( a2 0, (ax 1), (bx 1), (a abx b) (a abx b), , 0, (ax 1), (bx 1), 1, (a abx b) · c ax 1 , , 71. 3 x # 3 2 , 72. 4 x # 4 , , 1, m 0., bx 1, , 1 , 1, 10 & 9y y 10, where y 3 x ., 3x, , 4 , 2, 10 & 2y y 5, where y 4 x ., 4x, , 73. 2 2x 3 # 2 2 # 2 x 32 0 & y 2 12y 32 0, where 2 x y, & y 2 8y 4y 32 0., , SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, METHOD, , In fact, we can convert any quadratic equation to the form, (ax + b) 2 c 2 = 0 and then we can easily find its roots., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , Solve the equation x 2 10x 2 0 by the method of completing, the square., We have, x 2 10x 2 0, x 2 10x = 2, x2 2# x # 5 52 2 52, (x 5) 2 (2 25) 27, , [adding 5 2 on both sides], , , , x 5 = ! 27 = !3 3, , , , x 5 3 3 or x 5 3 3, x (5 3 3 ) or x (5 3 3 ) ., , , , [taking square root on both sides], , Hence, (5 3 3 ) and (5 3 3 ) are the roots of the given, equation.
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182, EXAMPLE 2, , SOLUTION, , Secondary School Mathematics for Class 10, , Solve the equation 3x 2 5x 2 0 by the method of completing, the square., We have, 3x 2 5x 2 0, , , 9x 2 15x 6 0, , , , 9x 15x 6, , [multiplying each term by 3], , 2, , 5 2, 5 5 2, (3x) 2 2 # 3x # c m 6 c m, 2, 2 2, 5 2, [adding c m on both sides], 2, 2, 2, (24 25) 1, 25, 5, 1, c m, c3x m c6 m , 2, 2, 4, 4, 4, 5, 1, c3x m !, [taking square root on both sides], 2, 2, 5 1, 5 1, 3x or 3x , 2 2, 2, 2, 1 5, 1 5, 6, 4, 3x c m 3 or 3x c m 2, 2 2, 2 2, 2, 2, 2, 3x 3 or 3x 2 & x 1 or x ·, 3, 2, Hence, 1 and are the roots of the given equation., 3, EXAMPLE 3, , SOLUTION, , Solve the equation 2x 2 x 4 0 by the method of completing the, square., We have, , , 2x 2 x 4 0, 4x 2 2x 8 0, , , , 4x 2x 8, , [multiplying both sides by 2], , 2, , 1 2, 1 1 2, (2x) 2 2 # 2x # c 2 m 8 c 2 m, 2, 1 2, [adding c 2 m on both sides], 2, , 1, 1 2, 33 33, d, n, c2x m c8 m , 2, 4, 4, 2, , , 2x , , 1 d 33 n, !, 2, 2, , , , 2x , , 1 33, 1 33, or 2x , 2, 2, 2, 2, , [taking square root on both sides]
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Quadratic Equations, , 183, , ( 33 1), 33 1, n, 2, 2, 2, 33 1 ( 33 1), , or 2x , 2, 2, 2, , , ( 33 1), ( 33 1), ·, x, or x , 4, 4, 1 33, 1 33, Hence,, are the roots of the given, and, 4, 4, equation., , , 2x d, , EXAMPLE 4, , By using the method of completing the square, show that the equation, 4x 2 3x 5 0 has no real roots., , SOLUTION, , We have, , , 4x 2 3x 5 0, 4x 2 3x 5, , 3 2, 3 3 2, (2x) 2 2 # 2x # c m 5 c m, 4, 4 4, 3 2, [adding c m on both sides], 4, 2, (80 9) 71, 3, 9, , c2x m c5 m , 0., 4, 16, 16, 16, 3 2, But, c2x m cannot be negative for any real value of x., 4, So, there is no real value of x satisfying the given equation., Hence, the given equation has no real roots., EXAMPLE 5, , SOLUTION, , 1, Solve the equation 10x x = 3 by the method of completing the, square., We have, 1, 10x x 3, 10x 2 1 3x, 10x 2 3x 1, 100x 2 30x 10 [multiplying each side by 10], 3 2, 3 3 2, (10x) 2 2 #10x # c m 10 c m, 2, 2, 2, 3 2, [adding c m on both sides], 2, 9, 3 2, 49 7 2, c m, c10x m c10 m , 2, 2, 4, 4
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184, , Secondary School Mathematics for Class 10, , 3 7, [taking square root on both sides], !, 2, 2, 3 7, 3 7, 10x or 10x , 2 2, 2, 2, 7 3, 4, 7 3, 10 , 2, 10x c m , 5 or 10x c m , 2 2, 2 2, 2, 2, 2 1, 5 1, ·, , 10x 5 or 10x 2 & x , or x , 5, 10, 10 2, 1, 1, Hence, and, are the roots of the given equation., 5, 2, 10x , , EXAMPLE 6, , SOLUTION, , Solve the equation a 2 x 2 3abx 2b 2 0 by the method of completing, the square., We have, a 2 x 2 3abx 2b 2 0, a 2 x 2 3abx 2b 2, 3b 2, 3b 3b 2, (ax) 2 2 #(ax)# c m 2b 2 c m, 2, 2, 2, 3b 2, [adding a k on both sides], 2, 2, 2, 2, 2, 3b, 9b (8b 9b ) b 2 b 2, c m, m, cax m c 2b 2 , 2, 2, 4, 4, 4, cax , , 3b, b, m !, [taking square root on both sides], 2, 2, b, 3b, 3b, b, cax m or cax m , 2, 2, 2, 2, b 3b, b 3b, 4b , 2b , m, ax c, 2b or ax c m , b, 2 2, 2, 2, 2, 2, 2b, b, x a or x a ·, 2b, b, Hence, a and a are the roots of the given equation., EXAMPLE 7, , SOLUTION, , Solve the equation x 2 ( 3 1) x 3 0 by the method of, completing the square., We have, x 2 ( 3 1) x 3 0, , , x 2 ( 3 1) x 3, , , , x 2 2 # x #d, , 31, 31, 31, nd, n 3 d, n, 2, 2, 2, 2, , 2
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Quadratic Equations, , 185, , [adding d, (x , , , ( 3 1), ( 3 1) 2, 2 (, 32, 2, 4, 2, , 31, n on both sides], 2, 2, , 2, ( 3 1) 2 4 3, 31, d, n, 4, 2, , (x , , ( 3 1), ( 3 1), 2 !, 2, 2, [taking square root on both sides], , , , x, , ( 3 1) ( 3 1), ( 3 1) ( 3 1), , , or x , 2, 2, 2, 2, , , , x, , ( 3 1) ( 3 1) 2 3, , , 3, 2, 2, 2, , or x , , , ( 3 1) ( 3 1) 2, , 1, 2, 2, 2, , x 3 or x 1., , Hence, 3 and 1 are the roots of the given equation., f, , EXERCISE 4B, , Solve each of the following equations by using the method of completing the, square:, , 1. x 2 6x 3 0, , 2. x 2 4x 1 0, , 3. x 2 8x 2 0, , 4. 4x 2 4 3 x 3 0, , 5. 2x 2 5x 3 0, , 6. 3x 2 x 2 0, , 7. 8x 2 14x 15 0, , 8. 7x 2 3x 4 0, 2 5, 10. 5x 2 6x 2 0, 11. 2 x 2 0, x, 13. x 2 ( 2 1) x 2 0, 15., , 9. 3x 2 2x 1 0, 12. 4x 2 4bx (a 2 b 2) 0, 14., , 2 x 2 3x 2 2 0, , 3 x 2 10x 7 3 0, , 16. By using the method of completing the square, show that the equation, 2x 2 x 4 0 has no real roots., ANSWERS (EXERCISE 4B), , 1. x (3 6 ) or x (3 6 ), , 2. x (2 3 ) or x (2 3 ), , 3. x (4 3 2 ) or x (4 3 2 ), 5. x , , 1, or x 3, 2, , 6. x 1 or x , , 3, 3, or x , 2, 2, 3, 5, 7. x or x , 2, 4, 4. x , , 2, 3
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186, , Secondary School Mathematics for Class 10, , 8. x 1 or x , , 4, 7, , 9. x 1 or x , , 3 19, 3 19, or x , 5, 5, (a b), (a b), 12. x , or x , 2, 2, 1, 14. x , or x 2 2, 2, 10. x , , 1, 3, 11. x 2 or x , , 1, 2, , 13. x 2 or x 1, 15. x 3 or x , , 7, 3, , QUADRATIC FORMULA [SHRIDHARACHARYA’S RULE], Consider the quadratic equation ax 2 + bx + c = 0, where a, b, c are real, = 0. Then,, numbers and a Y, , , ax 2 + bx + c = 0, ax 2 + bx = c, , , , c, b, x2 + a $ x = a, , , , b 2 c, b 2, b, x 2 + a $ x + b 2a l = a + b 2a l, , , , 2, c, b2, b x + 2ba l = c a + 2 m, 4a, , , , 2, (b 2 4ac), b x + 2ba l =, 4a 2, , [dividing throughout by a], b 2, [adding b 2a l on both sides], , ! b 2 4ac, b, m, , when (b 2 4ac) $ 0, 2a, 2a, b 2 4ac, b, x, !, 2a, 2a, b ! b 2 4ac, ·, x, 2a, This is called the quadratic formula or Shridharacharya’s rule., , , cx , , Thus, ax 2 bx c 0 has two roots and , given by, , , b b 2 4ac, b b 2 4ac, ·, and , 2a, 2a, , For the equation ax 2 bx c 0, the expression D (b 2 4ac), is called the discriminant., , DISCRIMINANT, , AN IMPORTANT NOTE, , (b 4ac) $ 0., 2, , The roots of ax 2 bx c 0 are real only when
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Quadratic Equations, , 187, , Taking (b 2 4ac) D, the roots of ax 2 bx c 0 are given by, , , b D, b D, ·, and , 2a, 2a, , SOLVED EXAMPLES, EXAMPLE 1, , Show that the equation 9x 2 7x 2 0 has real roots and solve it., , SOLUTION, , The given equation is 9x 2 7x 2 0., Comparing it with ax 2 bx c 0, we get, a 9, b 7 and c 2., , , D (b 2 4ac) (7 2 4 # 9 #(2)] 121 0., , So, the given equation has real roots., Now, D 121 11., b D (7 11), , 4 2,, 2a, 2# 9, 18 9, b D (7 11) 18, , , 1., , 2a, 2#9, 18, 2, Hence, the required roots are and –1., 9, , , , , EXAMPLE 2, , Show that the equation x 2 6x 6 0 has real roots and solve it., , SOLUTION, , The given equation is x 2 6x 6 0., Comparing it with ax 2 bx c 0, we get, a 1, b 6 and c 6., , , D (b 2 4ac) (36 4 #1# 6) 12 0., , So, the given equation has real roots., Now,, , , D = 12 = 2 3 ., , b D (6 2 3 ) (6 2 3 ), , , (3 3 ),, 2a, 2, 2 #1, b D (6 2 3 ) (6 2 3 ), , , (3 3 ) ., , 2a, 2, 2 #1, , , Hence, (3 3 ) and (3 3 ) are the roots of the given, equation., EXAMPLE 3, , Show that the equation 2x 2 + 5 3 x + 6 = 0 has real roots and, solve it.
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188, SOLUTION, , Secondary School Mathematics for Class 10, , The given equation is 2x 2 + 5 3 x + 6 = 0., Comparing it with ax 2 + bx + c = 0, we get, a 2, b 5 3 and c 6., , , D (b 2 4ac) [(5 3 ) 2 4 # 2 # 6] (75 48) 27 0., , So, the given equation has real roots., Now,, , D = 27 = 3 3 ., , b D (5 3 3 3 ) 2 3 3, , , , ,, 2a, 2# 2, 2, 4, b D (5 3 3 3 ) 8 3, , , 2 3 ., , 2a, 2# 2, 4, 3, Hence,, and 2 3 are the roots of the given equation., 2, , , EXAMPLE 4, , , , Using quadratic formula, solve for x:, p 2 x 2 + (p 2 q 2) x q 2 = 0., , SOLUTION, , [CBSE 2014], , The given equation is p 2 x 2 + (p 2 q 2) x q 2 = 0., Comparing it with ax 2 + bx + c = 0, we get, a = p 2, b = (p 2 q 2) and c = q 2 ., `, , D = (b 2 4ac) = (p 2 q 2) 2 4 # p 2 # (q 2), = (p 2 q 2) 2 + 4p 2 q 2 = (p 2 + q 2) 2 > 0., , So, the given equation has real roots., Now, D (p 2 q 2) ., , , b D (p 2 q 2) (p 2 q 2) 2q 2 q 2, , 2 2,, 2a, 2p 2, 2p, p, 2, 2, 2, 2, 2, b D (p q ) (p q ) 2p, , , 1., , 2a, 2p 2, 2p 2, , , Hence,, EXAMPLE 5, , SOLUTION, , q2, p2, , and –1 are the roots of the given equation., , Using quadratic formula, solve for x:, 9x 2 9 (a b) x (2a 2 5ab 2b 2) 0., , [CBSE 2009], , The given equation is 9x 2 9 (a b) x (2a 2 5ab 2b 2) 0., This is of the form Ax 2 Bx C 0, where, A 9, B 9 (a b) and C (2a 2 5ab 2b 2) .
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Quadratic Equations, , , , 189, , D (B 2 4AC) 81 (a b) 2 36 (2a 2 5ab 2b 2), = 81 (a 2 + b 2 + 2ab) 36 (2a 2 + 5ab + 2b 2), = 9a 2 + 9b 2 18ab = 9 (a 2 + b 2 2ab), = 9 (a b) 2 $ 0., , So, the given equation has real roots., Now,, , D = 9 (a b) 2 = 3 (a b) ., , B + D 9 (a + b) + 3 (a b) 6 (2a + b) (2a + b), =, =, =, 3 ,, 18, 2#9, 2A, B D 9 (a + b) 3 (a b) 6 (a + 2b) (a + 2b), =, =, =, $, =, 3, 18, 2#9, 2A, (2a + b), (a + 2b), Hence,, and, are the roots of the given, 3, 3, equation., , , EXAMPLE 6, , SOLUTION, , =, , Using quadratic formula, solve for x:, abx 2 + (b 2 ac) x bc = 0., , [CBSE 2014], , The given equation is abx + (b ac) x bc = 0., 2, , 2, , This is of the form Ax 2 + Bx + C = 0, where, A = ab, B = (b 2 ac) and C = bc., , , D = (B 2 4AC) = (b 2 ac) 2 + 4ab 2 c, , = b 4 + a 2 c 2 2ab 2 c + 4ab 2 c, = b 4 + a 2 c 2 + 2ab 2 c = (b 2 + ac) 2 > 0., So, the given equation has real roots., Now,, , D = (b 2 + ac) ., , B + D (b 2 ac) + (b 2 + ac) 2ac c, =, =, =, 2A, 2ab, 2ab b ,, B D (b 2 ac) + (b 2 + ac) 2b 2 b, ·, =, =, =, =, a, 2A, 2ab, 2ab, b, c, Hence,, and a are the roots of the given equation., b, , , EXAMPLE 7, SOLUTION, , =, , 1, 1 , Solve for x: x , 3, x ! 0, 2., (x 2), The given equation may be written as, (x 2) x, 3 & 3x (x 2) 2, x (x 2), & 3x 2 6x 2 0., , [CBSE 2010], , … (i)
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190, , Secondary School Mathematics for Class 10, , This equation is of the form ax 2 + bx + c = 0, where a = 3,, b = – 6 and c = 2., D = (b 2 4ac) = (6) 2 4 # 3 # 2 = 36 24 = 12 > 0., So, the given equation has real roots., Now,, , , D = 12 = 2 3 ., , =, , b + D 6 + 2 3 6 + 2 3 3 + 3, =, =, =, 2a, 6, 3 ,, 2#3, , =, , b D 6 2 3 6 2 3 3 3, ·, =, =, =, 2a, 6, 3, 2#3, , Hence, the required values of x are, EXAMPLE 8, SOLUTION, , (3 + 3 ), (3 3 ), ·, and, 3, 3, , x1 x3 1, 3 , x ! 2, 4., x2 x4, 3, The given equation is, x 1 x 3 10, x2 x4 3, (x 1)(x 4) (x 3)(x 2) 10, , , 3, (x 2)(x 4), Solve for x:, , [CBSE 2014], , , , (x 2 5x 4) (x 2 5x 6) 10, 2x 2 10x 10 10, , &, 2, 3, 3, (x 6x 8), x 2 6x 8, , , , 3 (2x 2 10x 10) 10 (x 2 6x 8), , 6x 2 30x 30 10x 2 60x 80, 4x 2 30x 50 0 & 2x 2 15x 25 0., , … (i), , This equation is of the form ax + bx + c = 0, where a = 2,, b = –15 and c = 25., 2, , D = (b 2 4ac) = {(15) 2 4 # 2 # 25} = (225 200) = 25 > 0., So, the given equation has real roots., Now,, , , D = 25 = 5., , =, , b + D (15 + 5) 20, =, =, = 5,, 2a, 4, 2#2, , =, , b D (15 5) 10 5, =, =, = $, 2a, 2, 4, 2#2, , 5, Hence, the required values of x are 5 and 2 $
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Quadratic Equations, , f, , 191, , EXERCISE 4C, , Find the discriminant of each of the following equations:, , 1., , (i) 2x 2 7x + 6 = 0, (iii) 2x 2 5 2 x + 4 = 0, , (ii) 3x 2 2x + 8 = 0, (iv) 3 x 2 + 2 2 x 2 3 = 0, (vi) 1 x = 2x 2, , (v) (x 1) (2x 1) = 0, , Find the roots of each of the following equations, if they exist, by applying the, quadratic formula:, , 2. x 2 4x 1 = 0, 4. 2x 2 + x 4 = 0, , 3. x 2 6x + 4 = 0, 5. 25x 2 + 30x + 7 = 0, , 6. 16x 2 = 24x + 1, 8. 2x 2 2 2 x + 1 = 0, , 7. 15x 2 28 = x, 9. 2 x 2 + 7x + 5 2 = 0, , [CBSE 2013, ’17], , 10., , 3 x + 10x 8 3 = 0, , [CBSE 2011], , 11., , 3x 2 2x2 3 = 0, , [CBSE 2015], , 2, 2, , 12. 2x + 6 3 x 60 = 0, 2, , [CBSE 2011, ’15], , 13. 4 3 x + 5x 2 3 = 0, , [CBSE 2013], , 14. 3x 2 6 x + 2 = 0, , [CBSE 2012], , 2, , 2, , 15. 2 3 x 5x + 3 = 0, 2, , 16. x + x + 2 = 0, 18. x 2 ( 3 + 1) x + 3 = 0, 2, , HINT, , [CBSE 2011], , 17. 2x + ax a = 0, 2, , 2, , [CBSE 2015], , D = ( 3 + 1) 4 3 = ( 3 1) ., 2, , 2, , 19. 2x 2 + 5 3 x + 6 = 0, , 20. 3x 2 2x + 2 = 0, , 1, 21. x x 3, x ! 0, , 1, 1, 22. x 3, x ! 0, 2, x 2, , 1, 23. x x 3, x ! 0, m, n, 24. n x 2 m 1 2x, HINT, , [CBSE 2010], [CBSE 2010], , m 2 n , 2 2, 2, , n x 2x b m 1l 0 & m x 2mnx (n mn) 0., , 25. 36x 2 12ax + (a 2 b 2) = 0, 27. x 2 2ax (4b 2 a 2) = 0, 28. x 2 + 6x (a 2 + 2a 8) = 0, 29. x + 5x (a 2 + a 6) = 0, 30. x 2 4ax b 2 + 4a 2 = 0, 31. 4x 2 4a 2 x + (a 4 b 4) = 0, 2, , [CBSE 2015], , 26. x 2 2ax + (a 2 b 2) = 0, [CBSE 2015], [CBSE 2015], [CBSE 2015], [CBSE 2012], [CBSE 2015]
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192, , Secondary School Mathematics for Class 10, , 32. 4x 2 + 4bx (a 2 b 2) = 0, 33. x 2 (2b 1) x + (b 2 b 20) = 0, , [CBSE 2015], [CBSE 2015], , 34. 3a x 8abx 4b 0, a ! 0, 35. a 2 b 2 x 2 (4b 4 3a 4) x 12a 2 b 2 0, a ! 0 and b ! 0, 36. 12abx 2 (9a 2 8b 2) x 6ab = 0, where a ! 0 and b ! 0, 2, , 2, , 2, , [CBSE 2006], [CBSE 2009], , ANSWERS (EXERCISE 4C), , 1. (i) 1, , (ii) –92 (iii) 18 (iv) 32 (v) 1 (vi) 9, , 2. x (2 5 ) or x (2 5 ), 3. x = (3 + 5 ) or x = (3 5 ), (1 + 33 ), (1 33 ), or x =, 4, 4, (3 + 10 ), (3 10 ), 6. x =, or x =, 4, 4, 4. x =, , 8. x =, , 1, 1, ,, 2, 2, , 11. x = 6 or x =, , 2, 3,, , 14. x =, , 5. x =, , (3 + 2 ), (3 2 ), or x =, 5, 5, 7, , 4, , 7. x = 5 or x = 3, , 9. x = 2 or x =, , 5, 2, , 10. x =, , 2 3, = 4 3, 3 or x, , 2, 2, 3, 12. x 2 3 or x 5 3 13. x = 4 or x =, 3, 3, 3, , 2, 3, , 15. x = 2 or x =, , a, , 17. x = 2 or x = a, , 18. x = 1 or x = 3, , 20. Do not exist, , 21. x =, , 1, 3, , 16. Do not exist, 19. x =, , 3, = 2 3, 2 or x, , 3+ 5, 3 5, or x = 2, 2, , 22. x =, , 3+ 3, 3 3, or x = 3, 3, , 24. x =, , n + mn, n mn, or x =, m, m, , 3 + 13, 3 13, or x =, 2, 2, (a + b), (a b), 25. x = 6, or x = 6, 23. x =, , 26. x = (a + b) or x = (a b), , 27. x = (a + 2b) or x = (a 2b), , 28. x = (a 2) or x = (4 + a), , 29. x (a 2) or x (a 3), , 30. x = (2a b) or x = (2a + b), , 1, 1, 31. x = 2 (a 2 + b 2) or x = 2 (a 2 b 2), , 1, , 1, , 32. x = 2 (a b) or x = 2 (a + b), , 2b, , 2b, , 34. x = 3a or x = a, , 35. x =, , 33. x = (b + 4) or x = (b 5), , 2b, 3a 2, 3a, 4b 2, =, 36. x =, 2 or x =, 3a, 4b or x, a, b2
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Quadratic Equations, , 193, , NATURE OF THE ROOTS OF A QUADRATIC EQUATION, = 0. Then, the discriminant, Let the given equation be ax 2 + bx + c = 0, where a Y, 2, is given by D = (b 4ac) . And, the roots of the given equation are, =, Case I, , b + D, b D, ·, and =, 2a, 2a, , When D > 0, , In this case, the roots are real and distinct. These roots are given by, =, Case II, , b + D, b D, ·, and =, 2a, 2a, , When D = 0, , In this case, the roots are real and equal., (b), Each root = 2a ·, Case III When D < 0, , In this case, the roots are imaginary, and we say that the given equation, has no real roots., SUMMARY, , Value, , Nature of roots, , D>0, , Real and unequal, , D=0, , Real and equal, , D<0, , No real roots, , Roots, , b ! D, 2a, Each root =, , b, 2a, , None, , SOLVED EXAMPLES, EXAMPLE 1, , Find the nature of the roots of the quadratic equation 4x 2 5x + 3 = 0., , SOLUTION, , The given equation is 4x 2 5x + 3 = 0., This is of the form ax 2 + bx + c = 0, where a = 4, b = 5 and, c = 3., , , D = (b 2 4ac) = {(5) 2 4 # 4 # 3} = (25 48) = 23 < 0., , Hence, the given equation has no real roots.
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194, EXAMPLE 2, , SOLUTION, , Secondary School Mathematics for Class 10, , Show that the equation 2x 2 6x + 3 = 0 has real roots and find, these roots., The given equation is 2x 2 6x + 3 = 0., This is of the form ax 2 bx c 0, where a 2, b 6 and c 3., , , D = (b 2 4ac) = [( 6) 2 4 # 2 # 3] = (36 24) = 12 > 0., , So, the given equation has real unequal roots., Solving 2x 2 6x + 3 = 0 by quadratic formula, we have, x=, , 6 ! 36 4 # 2 # 3 6 ! 36 24 6 ! 12, =, =, 4, 4, (2 # 2), , 6!2 3, 3! 3, &x 2 ·, 4, (3 + 3 ), (3 3 ), So,, are the roots of the given equation., and, 2, 2, , , EXAMPLE 3, , SOLUTION, , x, , Show that the equation x 2 + ax 4 = 0 has real and distinct roots, for all real values of a., The given equation is x 2 + ax 4 = 0., This is of the form Ax 2 + Bx + C = 0, where A = 1, B = a and, C = –4., D = (B 2 4AC) = {a 2 4 # 1 # ( 4)} = (a 2 + 16) > 0 for all, real values of a., Thus, D > 0 for all real values of a., Hence, the given equation has real and distinct roots for all, real values of a., , EXAMPLE 4, , SOLUTION, , Find the nature of the roots of the quadratic equation, 3x 2 4 3 x + 4 = 0 and hence solve it., The given equation is 3x 2 4 3 x + 4 = 0., This is of the form ax 2 bx c 0, where a 3, b 4 3 and, c = 4., D = (b 2 4ac) = {( 4 3 ) 2 4 # 3 # 4} = (48 48) = 0., This shows that the given quadratic equation has real and, equal roots., b 4 3 2 3, Each root = 2a = 6 = 3 $, Hence,, , 2 3, 2 3, 3 and 3 are the roots of the given equation.
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Quadratic Equations, EXAMPLE 5, , SOLUTION, , 195, , Find the values of k for which the quadratic equation 2x 2 kx 3 0, has two real equal roots., The given equation is 2x 2 + kx + 3 = 0., This is of the form ax 2 bx c 0, where a 2, b k and c 3., D = (b 2 4ac) = (k 2 4 # 2 # 3) = (k 2 24) ., For real and equal roots, we much have, D 0 & k 2 24 0 & k 2 24 & k ! 24 !2 6 ., Hence, 2 6 and 2 6 are the required values of k., , EXAMPLE 6, , SOLUTION, , Find the value of k for which the roots of the quadratic equation, [CBSE 2017], kx(x 2) 6 0 are equal., The given equation is kx 2 2kx + 6 = 0., This is of the form ax 2 bx c 0, where a k, b 2k and, c = 6., D = (b 2 4ac) = (4k 2 4 # k # 6) = (4k 2 24k) ., For equal roots, we must have, D 0 & 4k 2 24k 0 & 4k (k 6) 0 & k 0 or k 6., Now, k = 0, we get 6 = 0, which is absurd., = 0 and hence k = 6., kY, , EXAMPLE 7, , Find the value of k for which the quadratic equation, [CBSE 2014], (k + 4) x 2 + (k + 1) x + 1 = 0 has two real equal roots., , SOLUTION, , The given equation is (k + 4) x 2 + (k + 1) x + 1 = 0., This is of the form ax 2 + bx + c = 0, where a = (k + 4), b = (k + 1), and c = 1., , , D = (b 2 4ac) = (k + 1) 2 4 # (k + 4) # 1 = (k + 1) 2 4 (k + 4), = (k 2 + 1 + 2k 4k 16) = (k 2 2k 15) ., , For equal roots, we must have, D 0 & k 2 2k 15 0, & k 2 5k 3k 15 0 & k (k 5) 3 (k 5) 0, & (k 5) (k 3) 0 & k 5 0 or k 3 0, & k 5 or k 3., Hence, the required value of k is 5 or –3., EXAMPLE 8, , Find the nonzero value of k for which the quadratic equation, kx 2 + 1 2 (k 1) x + x 2 = 0 has equal roots. Hence, find the roots, of the equation., [CBSE 2015]
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196, SOLUTION, , Secondary School Mathematics for Class 10, , The given equation is (k + 1) x 2 2 (k 1) x + 1 = 0., This is of the form ax 2 bx c 0, where a (k 1), b 2 (k 1), and c 1., , , D = (b 2 4ac) = {4 (k 1) 2 4 # (k + 1) # 1} = 4 (k 2 3k) ., , For equal roots, we must have, D 0 & 4 (k 2 3k) 0 & 4k (k 3) 0 & k 0 or k 3., , , the required nonzero value of k is 3., , Putting k = 3, the given equation becomes, 4x 2 4x 1 0 & (2x 1) 2 0 & (2x 1) 0 & x , Hence, the required roots are, EXAMPLE 9, , SOLUTION, , 1·, 2, , 1, 1, and ·, 2, 2, , If – 4 is a root of the equation x 2 + px 4 = 0 and the equation, x 2 + px + q = 0 has equal roots, find the values of p and q., Since – 4 is a root of the equation x 2 + px 4 = 0, we have, ( 4) 2 p ( 4) 4 0 & 4p (16 4) 12 & p 3., Now, the roots of x 2 + px + q = 0 being equal, we have, 9, p 2 4q 0 & 3 2 4q 0 & 4q 9 & q ·, 4, 9, Hence, p = 3 and q ·, 4, , EXAMPLE 10, , If –2 is a root of the equation 3x 2 + 7x + p = 0, find the value of k so, that the roots of the equation x 2 + k (4x + k 1) + p = 0 are equal., , SOLUTION, , Since –2 is a root of the equation 3x + 7x + p = 0, we have, , [CBSE 2015], 2, , 3 #(2) 2 7 #(2) p 0 & 12 14 p 0 & p 2., For p = 2, the other given equation becomes, x 2 + 4kx + k (k 1) + 2 = 0., This is of the form ax 2 + bx + c = 0, where, a = 1, b = 4k and c = (k 2 k + 2) ., D = (b 2 4ac) = {16k 2 4 # 1 # (k 2 k + 2)} = (12k 2 + 4k 8) ., For equal roots we must have, D 0 & 12k 2 4k 8 0 & 4 (3k 2 k 2) 0, & 3k 2 k 2 0 & 3k 2 3k 2k 2 0, & 3k (k 1) 2 (k 1) 0 & (k 1)(3k 2) 0
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Quadratic Equations, , 197, , & k 1 0 or 3k 2 0 & k 1 or k 23 ·, 2, Hence, the required value of k is –1 or 3 ·, EXAMPLE 11, , Prove that both the roots of the equation, (x a) (x b) + (x b) (x c) + (x c) (x a) = 0, are real but they are equal only when a = b = c., , SOLUTION, , The given equation may be written as, 3x 2 2x (a + b + c) + (ab + bc + ca) = 0., , , D = 4 (a + b + c) 2 12 (ab + bc + ca), = 4 [(a + b + c) 2 3 (ab + bc + ca)], = 4 (a 2 + b 2 + c 2 ab bc ca), = 2 (2a 2 + 2b 2 + 2c 2 2ab 2bc 2ca), = 2 [(a b) 2 + (b c) 2 + (c a) 2] $ 0, [a (a b) 2 $ 0, (b c) 2 $ 0 and (c a) 2 $ 0] ., , This shows that both the roots of the given equation are real., For equal roots, we must have D = 0., Now, D 0 & (a b) 2 (b c) 2 (c a) 2 0, & (a b) 0, (b c) 0 and (c a) 0, & a b c., Hence, the roots are equal only when a = b = c., EXAMPLE 12, , SOLUTION, , If the roots of the equation (b c) x 2 + (c a) x + (a b) = 0 are, equal, prove that 2b = a + c., [CBSE 2002C, ‘06, ’17], Clearly, x = 1 satisfies the given equation., Since its roots are equal, so 1 and 1 are its roots., , , product of roots of the given equation = (1 # 1) = 1., , ab, But, product of roots = $, b c, , [B, , C, product of roots = A ], , ab , 1 & a b b c & 2b a c., bc, Hence, 2b = a + c., , , EXAMPLE 13, , Show that the equation 3x 2 + 7x + 8 = 0 is not true for any real, value of x., , SOLUTION, , The given equation is 3x 2 + 7x + 8 = 0., , , D = (7 2 4 # 3 # 8) = (49 96) = 47 < 0.
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198, , Secondary School Mathematics for Class 10, , So, the given equation has no real roots., Hence, the given equation is not true for any real value of x., EXAMPLE 14, , Show that the equation 2 (a 2 + b 2) x 2 + 2 (a + b) x + 1 = 0 has no real, = b., roots, when a Y, , SOLUTION, , The given equation is 2 (a 2 + b 2) x 2 + 2 (a + b) x + 1 = 0 ., , , D 4 (a b) 2 8 (a 2 b 2), 4 (a 2 b 2 2ab) 4 (a b) 2 0, when a b ! 0., , So, the given equation has no real roots, when a ! b., EXAMPLE 15, , Find the values of k for which the equation x 2 + 5kx + 16 = 0 has no, real roots., [CBSE 2013C], , SOLUTION, , The given equation is x 2 + 5kx + 16 = 0., This is of the form ax 2 bx c 0, where a 1, b 5k and, c = 16., , , D = (b 2 4ac) = (25k 2 4 # 1 # 16) = (25k 2 64) ., , Since, the given equation has no real root, we have, D 0 & 25k 2 64 0 & 25k 2 64, 8 2, 64, & k 2 25 & k 2 c 5 m, 8, 8, & 5 k 5·, 8, 8, Hence, the required real values of k are such that, k ·, 5, 5, EXAMPLE 16, , SOLUTION, , Find the values of k for which the given equation has real roots:, (i) kx 2 6x 2 = 0 (ii) 3x 2 + 2x + k = 0 (iii) 2x 2 + kx + 2 = 0, (i) The given equation is kx 2 6x 2 = 0., , , D = [( 6) 2 4 # k # ( 2)] = (36 + 8k) ., , The given equation will have real roots if D $ 0., 36, 9, Now, D $ 0 & 36 8k $ 0 & k $, &k$ 2 ·, 8, (ii) The given equation is 3x 2 + 2x + k = 0., , , D = (2 2 4 # 3 # k) = (4 12k) ., , The given equation will have real roots if D $ 0., 1, Now, D $ 0 & 4 12k $ 0 & 12k # 4 & k # ·, 3
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Quadratic Equations, , 199, , (iii) The given equation is 2x 2 + kx + 2 = 0., , , D = (k 2 4 # 2 # 2) = (k 2 16) ., , The given equation will have real roots if D $ 0., Now, D $ 0 & (k 2 16) $ 0 & k 2 $ 16 & k $ 4 or k # 4., EXAMPLE 17, , Determine the positive value of p for which the equations, x 2 + 2px + 64 = 0 and x 2 8x + 2p = 0 will both have real roots., , SOLUTION, , Let D1 and D2 be the discriminants of the first and second, given equations respectively., , [CBSE 2013C], , For real roots, we must have D1 $ 0 and D2 $ 0., Now, D1 $ 0 and D2 $ 0, , , (4p 2 4 # 64) $ 0 and (64 8p) $ 0, , , , p 2 64 $ 0 and 64 8p $ 0, , , , p 2 $ 64 and 8p # 64, , , , p $ 8 and p # 8 [a p is positive], , , , p = 8., , Hence, p = 8., f, , EXERCISE 4D, , 1. Find the nature of the roots of the following quadratic equations:, (i) 2x 2 8x + 5 = 0, (ii) 3x 2 2 6 x + 2 = 0, (iii) 5x 2 4x + 1 = 0, (v) 12x 2 4 15 x + 5 = 0, , (iv) 5x (x 2) + 6 = 0, (vi) x 2 x + 2 = 0, , 2. If a and b are distinct real numbers, show that the quadratic equation, 2 (a 2 + b 2) x 2 + 2 (a + b) x + 1 = 0 has no real roots., 3. Show that the roots of the equation x 2 + px q 2 = 0 are real for all real, values of p and q., 4. For what values of k are the roots of the quadratic equation, [CBSE 2008C], 3x 2 + 2kx + 27 = 0 real and equal?, 5. For what value of k are the roots of the quadratic equation, [CBSE 2013], kx (x 2 5 ) + 10 = 0 real and equal?, 6. For what values of p are the roots of the equation 4x 2 + px + 3 = 0 real, and equal?, [CBSE 2014]
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200, , Secondary School Mathematics for Class 10, , 7. Find the nonzero value of k for which the roots of the quadratic equation, [CBSE 2014], 9x 2 3kx + k = 0 are real and equal., 8. (i) Find the values of k for which the quadratic equation, [CBSE 2014], (3k + 1) x 2 + 2 (k + 1) x + 1 = 0 has real and equal roots., (ii) Find the value of k for which the equation x 2 k (2x k 1) 2 0, has real and equal roots., [CBSE 2017], 9. Find the values of p for which the quadratic, (2p + 1) x 2 (7p + 2) x + (7p 3) = 0 has real and equal roots., , equation, [CBSE 2014], , 10. Find the values of p for which the quadratic equation, = 1 has equal roots. Hence, find, (p + 1) x 2 6 (p + 1) x + 3 (p + 9) = 0, p Y, the roots of the equation., [CBSE 2015], 2, 11. If –5 is a root of the quadratic equation 2x + px 15 = 0 and the, quadratic equation p (x 2 + x) + k = 0 has equal roots, find the value of k., [CBSE 2014], , 12. If 3 is a root of the quadratic equation x x + k = 0, find the value of p, so that the roots of the equation x 2 + k (2x + k + 2) + p = 0 are equal., 2, , [CBSE 2015], , 13. If –4 is a root of the equation x + 2x + 4p = 0, find the value of k for, which the quadratic equation x 2 + px (1 + 3k) + 7 (3 + 2k) = 0 has equal, roots., [CBSE 2015], 2, 2, 2, 2, 14. If the quadratic equation (1 + m ) x + 2mcx + c a = 0 has equal roots,, prove that c 2 = a 2 (1 + m 2) ., [CBSE 2014, ’17], 2, , 15. If the roots of the equation (c 2 ab) x 2 2 (a 2 bc) x + (b 2 ac) = 0 are real, and equal, show that either a = 0 or (a 3 + b 3 + c 3) = 3abc., [CBSE 2017], 16. Find the values of p for which the quadratic equation 2x 2 + px + 8 = 0, has real roots., 17. Find the value of for which the equation ( 12) x 2 + 2 ( 12) x + 2 = 0, has equal roots., [CBSE 2013], 2, +, +, =, 18. Find the value of k for which the roots of 9x 8kx 16 0 are real and, equal., 19. Find the values of k for which the given quadratic equation has real and, distinct roots:, (i) kx 2 + 6x + 1 = 0, (ii) x 2 kx + 9 = 0, (iii) 9x 2 + 3kx + 4 = 0, , (iv) 5x 2 kx + 1 = 0, , 20. If a and b are real and a ! b then show that the roots of the equation, (a b) x 2 + 5 (a + b) x 2 (a b) = 0 are real and unequal.
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Quadratic Equations, , 201, , 21. (i) If the roots of the equation (a 2 + b 2) x 2 2 (ac + bd) x + (c 2 + d 2) = 0 are, a c, equal, prove that = ·, [CBSE 2017], b d, (ii) If ad ! bc then prove that the equation, (a 2 b 2) x 2 2(ac bd) x (c 2 d 2) 0 has no real roots., , [CBSE 2017], , 22. If the roots of the equations ax 2 + 2bx + c = 0 and bx 2 2 ac x + b = 0, are simultaneously real then prove that b 2 = ac., ANSWERS (EXERCISE 4D), , 1. (i) Real and unequal, , (v) Real and equal, 4. k = 9 or k = 9, 7. k = 4, , (ii) Real and equal, (vi) Not real, 5. k = 0 or k = 2, , 8. (i) k = 0 or k = 1, , 49, , 10. p = 3 or p = –1, , 11. k = 28, 16. p ≥ 8 or p ≤ –8 17. = 14, 19. (i) k < 9, , (iii) Not real, , (ii) k > 6 or k < –6, , 6. p = 4 3 or p = 4 3, , 4, 9. p = 4 or p = 7, 10, 12. p = 12 13. k = 2 or k = 9, 18. k = 3 or k = –3, (ii) k 2, , (iii) k > 4 or k < –4, , (iv) k > 2 5 or k < 2 5, , HINTS TO SOME SELECTED QUESTIONS, 2. D 4 (a b) 2 8 (a 2 b 2), 4 [(a b) 2 2 (a 2 b 2)] 4 (a 2 b 2 2ab), 4 (a 2 b 2 2ab) 4 (a b) 2 0., 3. D (p 2 4q 2) $ 0., 4. D 4k 2 324. For real and equal roots, we must have, D 0., 5. Given equation is kx 2 2 5 kx 10 0., , , D 20k 2 40k 20k (k 2) ., , For real and equal roots, we must have D 0., 6. For real and equal roots, we must have D 0., , , D 0 & p 2 48 0 & p 2 48 & p ! 48 & p 4 3 or 4 3 ., , 9. D (7p 2) 2 4 (2p 1) (7p 3) 7p 2 24p 16., , , D 0 & 7p 2 24p 16 0 & 7p 2 28p 4p 16 0, & 7p (p 4) 4 (p 4) 0 & (p 4) (7p 4) 0., , 15. D 4 (a, , 2, , bc) 2 4 (c 2 ab) (b 2 ac), , Now, D 0 & (a 2 bc) 2 (c 2 ab)(b 2 ac) 0, & a 4 3a 2 bc ac 3 ab 3 0, , (iv) Not real
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202, , Secondary School Mathematics for Class 10, & a (a 3 b 3 c 3 3abc) 0, & a 0 or a 3 b 3 c 3 3abc., , 16. D (p, , 2, , 64) . So, D $ 0 & p 2 $ 64 & p $ 8 or p # 8., , 19. (i) 36 4k 0 & 36 4k & 4k 36 & k 9., (ii) k 2 36 0 & k 2 36 & k 6 or k 6., (iii) 9k 2 144 0 & k 2 16 & k 4 or k 4., (iv) k 2 20 0 & k 2 20 & k 2 5 or k 2 5 ., 20. D 25 (a b) 2 8 (a b) 2, 17 (a b) 2 8 [(a b) 2 (a b) 2] 17 (a b) 2 16 (a 2 b 2) 0., 21. (i) 4 (ac bd) 2 4 (a 2 b 2)(c 2 d 2) 0, , , , , (a 2 b 2)(c 2 d 2) (ac bd) 2 0, , , , , , , , , , a 2 d 2 b 2 c 2 2abcd 0 & (ad bc) 2 0 & ad bc 0, a c, ad bc & $, b d, , 22. 4b 2 4ac $ 0 & b 2 ac $ 0 & b 2 $ ac., 4ac 4b 2 $ 0 & ac b 2 $ 0 & b 2 # ac, Hence, b 2 ac., , WORD PROBLEMS ON QUADRATIC EQUATIONS, SOLVED EXAMPLES, PROBLEMS ON NUMBERS, EXAMPLE 1, , If the sum of two natural numbers is 27 and their product is 182,, find the numbers., , SOLUTION, , Let the required numbers be x and (27 – x). Then,, x (27 x) = 182, , , 27x x 2 = 182 x 2 27x + 182 = 0, , , , x 2 13x 14x + 182 = 0, , , , x (x 13) 14 (x 13) = 0, , , , (x 13)(x 14) 0, , , , x 13 = 0 or x 14 = 0, , , , x = 13 or x = 14., , Hence, the required numbers are 13 and 14.
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Quadratic Equations, , 203, , EXAMPLE 2, , The sum of the squares of two consecutive odd numbers is 394. Find, the numbers., [CBSE 2014], , SOLUTION, , Let the required consecutive odd numbers be x and, (x + 2). Then,, x 2 + (x + 2) 2 = 394, , , , 2x 2 + 4x 390 = 0 x 2 + 2x 195 = 0, x 2 + 15x 13x 195 = 0, , , , x (x + 15) 13 (x + 15) = 0, (x 15)(x 13) 0, , , , x + 15 = 0 or x 13 = 0, , , , x = 15 or x = 13, , , , x = 13., , , , [rejecting x = –15], , Hence, the required numbers are 13 and 15., EXAMPLE 3, , The sum of the squares of two consecutive even numbers is 340. Find, the numbers., [CBSE 2014], , SOLUTION, , Let the required consecutive even numbers be x and (x + 2)., Then,, x 2 + (x + 2) 2 = 340, , , , 2x 2 + 4x 336 = 0 x 2 + 2x 168 = 0, x 2 + 14x 12x 168 = 0, , , , x (x + 14) 12 (x + 14) = 0, (x + 14) (x 12) = 0, , , , x + 14 = 0 or x 12 = 0, , , , x = 14 or x = 12, , , , x = 12 [rejecting x = 14], , , , Hence, the required numbers are 12 and 14., EXAMPLE 4, , The sum of the squares of two consecutive multiples of 7 is 637., Find the multiples., [CBSE 2014], , SOLUTION, , Let the required consecutive multiples of 7 be 7x and 7 (x + 1) ., Then, (7x) 2 + {7 (x + 1)} 2 = 637, , , 49x 2 + (7x + 7) 2 = 637 98x 2 + 98x 588 = 0, , , , x 2 + x 6 = 0 x 2 + 3x 2x 6 = 0
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204, , Secondary School Mathematics for Class 10, , , , x (x + 3) 2 (x + 3) = 0 (x + 3) (x 2) = 0, , , , x + 3 = 0 or x 2 = 0, , , , x = 3 or x = 2, , , , x=2, , [neglecting x = –3], , Hence, the required numbers are (7 × 2) and (7 × 3), i.e., 14, and 21., EXAMPLE 5, , The sum of two natural numbers is 9 and the sum of their reciprocals, 1, is 2 · Find the numbers., [CBSE 2012], , SOLUTION, , Let the required natural numbers be x and (9 – x). Then,, 1+ 1 =1, x (9 x) 2, , , , (9 x) + x 1, =, 2, x (9 x), x (9 x) = 18, , , , x 9x + 18 = 0 x 2 6x 3x + 18 = 0, , , , x (x 6) 3 (x 6) = 0 (x 6) (x 3) = 0, , , , x 6 = 0 or x 3 = 0 x = 6 or x = 3., , , , [by cross multiplication], , 2, , Hence, the required natural numbers are 6 and 3., EXAMPLE 6, , SOLUTION, , The difference of two natural numbers is 5 and the difference of their, 1, reciprocals is · Find the numbers., [CBSE 2014], 10, Let the required natural numbers be x and (x 5) . Then,, 1, 1, 1, 1, x x5 & x & x·, x 5, x 5, , , , 1 1 = 1, x 5 x 10, x (x 5) 1, 5, , 1 [by cross multiplication], & , 10, (x 5) x, (x 5) x 10, (x 5) x = 50 x 2 5x 50 = 0, , , , x 2 10x + 5x 50 = 0 x (x 10) + 5 (x 10) = 0, , , , (x 10) (x + 5) = 0, , , , x 10 = 0 or x + 5 = 0, , , , x = 10 or x = 5, , , , x = 10 [a –5 is not a natural number], , , , , Hence, the required natural numbers are 10 and 5.
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Quadratic Equations, , 205, , EXAMPLE 7, , The difference of squares of two numbers is 180. The square of the, smaller number is 8 times the larger number. Find the two numbers., , SOLUTION, , Let the required numbers be x and y such that x > y., , [CBSE 2012], , Then, x 2 y 2 = 180., , … (i), , And, y = 8x., , … (ii), , 2, , From (i) and (ii), we get, x 2 8x 180 = 0, x 2 18x 10x 180 0 & x (x 18) 10 (x 18) 0, (x 18)(x 10) 0 & x 18 0 or x 10 0, x = 18 or x = –10., Now, x 18 & y 2 (8 #18) 144, & y 12 or y 12., Also, x 10 & y 2 [8 #(10)] 80, which is not possible., Hence, the numbers are (18 and 12) or (18 and –12)., EXAMPLE 8, , The numerator of a fraction is 3 less than its denominator. If 2 is, added to both of its numerator and denominator then the sum of the, 29, new fraction and original fraction is 20 · Find the original fraction., [CBSE 2015], , SOLUTION, , Let the denominator of the required fraction be x., Then, its numerator = (x 3) ., (x 3), So, the original fraction is, x $, (x 3) + 2 (x 3) 29, +, =, , x, 20, x+2, (x 1) (x 3) 29, +, =, , x, 20, (x + 2), x (x 1) + (x 3) (x + 2) 29, =, , 20, (x + 2) x, , , , (x 2 x) + (x 2 x 6) 29, =, 20, x 2 + 2x, 2, 2, 20 (2x 2x 6) = 29 (x + 2x), , , , 40x 2 40x 120 = 29x 2 + 58x, 11x 2 98x 120 = 0, , , , 11x 2 110x + 12x 120 = 0, ,
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206, , Secondary School Mathematics for Class 10, , , , , , , , 11x (x 10) + 12 (x 10) = 0, (x 10) (11x + 12) = 0, x 10 = 0 or 11x + 12 = 0, 12, x = 10 or x = 11, x = 10, [a denominator of a fraction is never negative], , denominator = 10 and numerator = (10 3) = 7., 7, Hence, the required fraction is 10 ·, , , , EXAMPLE 9, , The denominator of a fraction is one more than twice the numerator., 16, If the sum of the fraction and its reciprocal is 2 21 , find the fraction., , SOLUTION, , Let the numerator of the required fraction be x., Then, its denominator = (2x + 1) ., (2x + 1), x, fraction =, and its reciprocal =, $, x, (2x + 1), , , (2x + 1) 58, x, +, =, x, 21, (2x + 1), , , , 21 # [x 2 + (2x + 1) 2] = 58x (2x + 1), , , , 21 # [5x 2 + 4x + 1] = 116x 2 + 58x, , , , , , 11x 2 26x 21 = 0, 11x 2 33x 7x 21 0 & 11x (x 3) 7 (x 3) 0, (x 3)(11x 7) 0 & x 3 0 or 11x 7 0, 7, x = 3 or x = 11, x = 3 [a numerator cannot be a negative fraction]., x, = 3·, required fraction =, (2x + 1) 7, , , , , EXAMPLE 10, , A two-digit number is 5 times the sum of its digits and is also equal, to 5 more than twice the product of its digits. Find the number., , SOLUTION, , Let the tens and units digits of the required number be x and y, respectively. Then,, 5x, … (i), 10x y 5 (x y) & 4y 5x & y , 4, 5x , 5x, [using (i)], 10x y 2xy 5 & 10x , 2x # 5, 4, 4, 2, 45x 10x, & 4 4 5 & 10x 2 45x 20 0, & 2x 2 9x 4 0 & 2x 2 8x x 4 0
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Quadratic Equations, , 207, , & 2x (x 4) (x 4) 0, & (x 4)(2x 1) 0, & x 4 0 or 2x 1 0, & x 4 or x 12, & x 4 [a a digit cannot be a fraction] ., Putting x = 4 in (i), we get y = 5., x = 4 and y = 5., Hence, the required number is 45., EXAMPLE 11, , A two-digit number is such that the product of its digits is 18. When, 63 is subtracted from the number, the digits interchange their places., Find the number., [CBSE 2006C], , SOLUTION, , Let the tens and units digits of the required number be x and y, respectively. Then,, 18, … (i), xy 18 & y x, And, (10x + y) 63 = 10y + x, , , 9x 9y 63 & x y 7, , … (ii), , 18, Putting y = x from (i) into (ii), we get, 18, x x = 7, , , x 2 18 7x & x 2 7x 18 0, , , , x 2 9x 2x 18 0 & x (x 9) 2 (x 9) 0, , , , (x 9)(x 2) 0 & x 9 0 or x 2 0, , , , x = 9 or x = 2, , , , x=9, , [a a digit cannot be negative]., , Putting x = 9 in (i), we get y = 2., Thus, the tens digit is 9 and the units digit is 2., Hence, the required number is 92., GENERAL PROBLEMS ON MONEY MATTERS, EXAMPLE 12, , A person on tour has ` 4200 for his expenses. If he extends his tour, for 3 days, he has to cut down his daily expenses by ` 70. Find the, original duration of the tour., [CBSE 2008C]
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208, SOLUTION, , Secondary School Mathematics for Class 10, , Let the original duration of the tour be x days. Then,, 4200 4200 =, 70, x, (x + 3), , , , (x 3) x, 1, 70, F 70 &, (x 3), 4200, x (x 3), 2, x (x 3) 180 & x 3x 180 0, , , , x 2 15x 12x 180 0 & x (x 15) 12 (x 15) 0, , , , (x 15)(x 12) 0 & x 15 0 or x 12 0, , , , x = 15 or x = 12, , , , x = 12, , , , original duration of the tour is 12 days., , , , 1, 4200 #<x , , [a number of days cannot be negative] ., , EXAMPLE 13, , A bookseller buys a number of books for ` 1760. If he had bought 4, more books for the same amount, each book would have cost ` 22 less., How many books did he buy?, , SOLUTION, , Let the bookseller buy x books for ` 1760., 1760, Then, cost of each book = ` x ·, Again, cost of (x + 4) books = ` 1760, 1760 ·, cost of each book, now = `, (x 4), 1760 1760 =, , 22, x, (x + 4), (x 4) x 1, 1, 1 22, , x, &, 80, (x 4) 1760, x (x 4), 4, 1 & x 2 4x 320, , (x 2 4x) 80, , , , , , x 2 4x 320 0 & x 2 20x 16x 320 0, x (x 20) 16 (x 20) 0 & (x 20)(x 16) 0, x 20 0 or x 16 0 & x 20 or x 16, x = 16 [a number of books cannot be negative], , Hence, the bookseller bought 16 books., EXAMPLE 14, , Some students planned a picnic. The total budget for hiring a bus, was ` 1440. Later on, eight of these refused to go and instead paid, their total share of money towards the fee of one economically weaker, student of their class, and thus, the cost for each member who went, for picnic, increased by ` 30.
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Quadratic Equations, , 209, , (i) How many students attended the picnic?, (ii) How much money in total was paid towards the fee? Which, value is reflected in this question?, [CBSE 2013C], SOLUTION, , Let x students planned the picnic., Then, (x 8) students attended the picnic., Total bus charges = ` 1440., , , 1440 1440 =, 30, x, (x 8), , , , x (x 8), 1 1 30, 1, &, x, , 1440, 48, (x 8), (x 8) x, , , , 8, 1 & x 2 8x 384, (x 2 8x) 48, , , , , , , x 2 8x 384 0 & x 2 24x 16x 384 0, x (x 24) 16 (x 24) 0 & (x 24)(x 16) 0, x 24 0 or x 16 0 & x 24 or x 16, x = 24 [a number of students cannot be negative], , [by cross multiplication], , Thus, 24 students planned the picnic., (i) Number of students who attended the picnic = (24 – 8) = 16., (ii) Share of 24 students = ` 1440, 1440, Share of 8 students = ` b 24 # 8 l = ` 480., , , money paid towards the fee = ` 480., , The value reflected in the given question is 'charity'., EXAMPLE 15, , The total cost of a certain length of a piece of wire is ` 200. If the piece, was 5 metres longer and each metre of wire costs ` 2 less, the cost of, the piece would have remained unchanged. How long is the piece and, what is its original rate per metre?, [CBSE 2015], , SOLUTION, , Let the original length of the piece of wire be x metres., Cost of this piece = ` 200., , , its original rate per metre = `, , New length (x 5) metres., New cost per metre = `, , , 200 200 =, 2, x, (x + 5), , 200 ·, (x + 5), , 200 ·, x
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210, , Secondary School Mathematics for Class 10, , , , , , , , , (x 5) x, 1 1 2, 1, x (x 5) 200 & x (x 5), 100, 5, 1, , & x 2 5x 500, (x 2 5x) 100, x 2 5x 500 0 & x 2 25x 20x 500 0, x (x 25) 20 (x 25) 0 & (x 25)(x 20) 0, x 25 0 or x 20 0 & x 25 or x 20, x = 20 [a length cannot be negative], , , , original length of wire = 20 m., 200, Original rate = ` 20 per m = ` 10 per m., EXAMPLE 16, , ` 6500 were divided equally among a certain number of persons., Had there been 15 more persons, each would have got ` 30 less. Find, the original number of persons., , SOLUTION, , Let the original number of persons be x., Total amount to be divided = ` 6500., 6500, Share of each = ` x $, New number of persons = (x + 15)., 6500 ·, Now, share of each = `, (x + 15), 6500 6500 =, , 30, x, (x + 15), , , (x 15) x, 1, 1, 30, 3, x (x 15) 6500 & x (x 15), 650, , , , 15, 3 & 3 (x 2 15x) 9750, (x 2 15x) 650, , , , , x 2 15x 3250 & x 2 15x 3250 0, x 2 65x 50x 3250 0 & x (x 65) 50 (x 65) 0, , , , (x 65)(x 50) 0 & x 65 0 or x 50 0, , , , x = 65 or x = 50, , , , x = 50 [a number of persons cannot be negative], , Hence, the original number of persons is 50., PROBLEMS ON AGES, EXAMPLE 17, , A girl is twice as old as her sister. Four years hence, the product of, their ages (in years) will be 160. Find their present ages. [CBSE 2010]
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Quadratic Equations, SOLUTION, , 211, , Let the present age of her sister be x years., Then, girl’s present age = 2x years., Product of their ages 4 years hence = (x + 4) (2x + 4) ., , , (x + 4) (2x + 4) = 160, , , , 2x 2 12x 144 0 & x 2 6x 72 0, , , , x 2 12x 6x 72 0 & x (x 12) 6 (x 12) 0, , , , (x 12)(x 6) 0 & x 12 0 or x 6 0, , , , x = 12 or x = 6, , , , x=6, , , , sister’s present age = 6 years and girl’s present age = 12 years., , [a the age cannot be negative], , EXAMPLE 18, , The age of a man is twice the square of the age of his son. Eight years, hence, the age of the man will be 4 years more than three times the, age of his son. Find their present ages., [CBSE 2009C], , SOLUTION, , Let the present age of the son be x years., Then, the present age of the man is (2x 2) years., Age of the son 8 years hence = (x + 8) years., Age of the man 8 years hence (2x 2 + 8) years., , , (2x 2 + 8) = 3 (x + 8) + 4, , , , , 2x 2 3x 20 0 & 2x 2 8x 5x 20 0, 2x (x 4) 5 (x 4) 0 & (x 4)(2x 5) 0, x 4 = 0 or 2x + 5 = 0, , , , , , , 5, x = 4 or x = 2, x = 4 [a age cannot be negative]., son's present age = 4 years, and, man‘s present age = (2 # 4 2) years = 32 years., , EXAMPLE 19, , The sum of the reciprocals of Arun‘s ages (in years) 3 years ago and, 1, five years from now is 3 $ Find his present age., , SOLUTION, , Let Arun's present age be x years., Arun‘s age 3 years ago = (x – 3) years., Arun‘s age 5 years hence = (x + 5) years., 1 + 1, =1, , (x 3) (x + 5) 3
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212, , Secondary School Mathematics for Class 10, , , , (x 5) (x 3) 1, & (x 3)(x 5) 6 (x 1), 3, (x 3)(x 5), , , , x 2 2x 15 6x 6 & x 2 4x 21 0, , , , x 2 7x 3x 21 0 & x (x 7) 3 (x 7) 0, , , , (x 7) (x 3) 0 & x 7 0 or x 3 0, , , , x = 7 or x = 3, , x = 7 [a age cannot be negative], Hence, Arun’s present age is 7 years., EXAMPLE 20, , The sum of the ages of a man and his son is 45 years. Five years ago,, the product of their ages (in years) was 124. Find their present ages., , SOLUTION, , Let the present age of the man be x years., Then, the present age of the son = (45 – x) years., Age of the man 5 years ago = (x – 5) years., Age of the son 5 years ago = (45 – x – 5) years = (40 – x) years., , , (x 5) (40 x) = 124, , x 2 45x 200 124 & x 2 45x 324 0, x 2 36x 9x 324 0 & x (x 36) 9 (x 36) 0, (x 36)(x 9) 0 & x 36 0 or x 9 0, x = 36 or x = 9., But, the man‘s age cannot be 9 years. So, we reject x = 9., Hence, the man‘s present age = 36 years., And, the son‘s present age = (45 – 36) years = 9 years., EXAMPLE 21, , 7 years ago Varun’s age was five times the square of Swati’s age., 3 years hence, Swati’s age will be two-fifth of Varun’s age. Find their, present ages., [CBSE 2006C], , SOLUTION, , Let Swati‘s age 7 years ago be x years., Then, Varun‘s age 7 years ago = 5x 2 years., Swati‘s present age = (x + 7) years., Varun‘s present age = (5x 2 + 7) years., Swati‘s age 3 years hence = (x + 7 + 3) years = (x + 10) years., Varun‘s age 3 years hence = (5x 2 + 7 + 3) years, = (5x 2 + 10) years.
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Quadratic Equations, , 213, , , , 2, (x + 10) = 5 (5x 2 + 10), 5x 50 10x 2 20 & 10x 2 5x 30 0, 2x 2 x 6 0 & 2x 2 4x 3x 6 0, 2x (x 2) 3 (x 2) 0 & (x 2)(2x 3) 0, 3, x 2 0 or 2x 3 0 & x 2 or x , 2, x=2, [a age cannot be negative], , , , Swati‘s present age = (2 + 7) years = 9 years., , , , , , , , Varun‘s present age = (5 # 2 2 + 7) years = 27 years., PROBLEMS ON SPEED AND TIME, EXAMPLE 22, , A bus travels at a certain average speed for a distance of 75 km and, then travels a distance of 90 km at an average speed of 10 km/hr, more than the first speed. If it takes 3 hours to complete the total, journey, find its original speed., [CBSE 2015], , SOLUTION, , Let the original speed of the bus be x km/hr., 75, Time taken to cover 75 km = x hours., New speed = (x + 10) km/hr., Time taken to cover 90 km with new speed =, , 90, hours., (x + 10), , Total time taken to cover the whole journey = 3 hours., , , 75 + 90 =, x (x + 10) 3, , , , 75 (x 10) 90x, 3 & 165x 750 3x 2 30x, x (x 10), 3x 2 135x 750 0 & x 2 45x 250 0, , , , x 2 50x 5x 250 0 & x (x 50) 5 (x 50) 0, , , , (x 50)(x 5) 0 & x 50 0 or x 5 0, , , , x 50 or x 5 & x 50 [a speed cannot be negative], , , , Hence, the original speed of the bus was 50 km/hr., EXAMPLE 23, , In a flight of 2800 km, an aircraft was slowed down due to bad, weather. Its average speed is reduced by 100 km/hr and time of, flight increased by 30 minutes. Find the original duration of the flight., , SOLUTION, , Let the original speed of the aircraft be x km/hr., , [CBSE 2012]
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214, , Secondary School Mathematics for Class 10, , 2800, Time taken to cover 2800 km = x hours., Reduced speed = (x 100) km/hr., Time taken to cover 2800 km at this speed =, , , , , , , 2800, hours., (x 100), , 2800 2800 = 30, x, 60, (x 100), x (x 100), 1, 1, 1 , 1, &, x, , 2 # 2800, 5600, (x 100), (x 100) x, 100, 1, , & x 2 100x 560000 0, (x 2 100x) 5600, x 2 800x + 700x 560000 = 0, x (x 800) 700 (x 800) 0, , (x 800)(x 700) 0, x 800 = 0 or x + 700 = 0, , , , , x = 800 or x = 700, x = 800 [a speed cannot be negative], original speed of the aircraft = 800 km/hr., 2800, original duration of the flight = 800 hours, = 3 hours 30 minutes., , EXAMPLE 24, , An aeroplane left 30 minutes later than its scheduled time and, in order to reach its destination 1500 km away in time, it had to, increase its speed by 250 km/hr from its usual speed. Determine its, usual speed., [CBSE 2005C], , SOLUTION, , Let the usual speed be x km/hr., Actual speed = (x + 250) km/hr., 1500, Time taken at usual speed = b x l hr., 1500, Time taken at actual speed = b x + 250 l hr., 1, Difference between the two times taken = 2 hr., 1500 1500 = 1, , x, x + 250 2, (x 250) x, 1, 1, 1 &, 1, x, , 3000, (x 250) 3000, x (x 250), , , 250, 1 & x 2 250x 750000 0, (x 2 250x) 3000
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Quadratic Equations, , 215, , , , x 2 + 1000x 750x 750000 = 0, , , , x (x 1000) 750 (x 1000) 0 & (x 1000)(x 750) 0, , , , x 1000 0 or x 750 0 & x 1000 or x 750, , , , x = 750, , [a speed cannot be negative], , Hence, the usual speed of the aeroplane was 750 km/hr., EXAMPLE 25, , If a man walks 1 km/hr faster than his usual speed then he covers a, distance of 3 km in 15 minutes less time. Find his usual speed., , SOLUTION, , Let the usual speed of the man be x km/hr., , [CBSE 2014], , 3, Time taken to cover 3 km at usual speed = x hours., Actual speed of the man = (x + 1) km/hr., 3, hours., Time taken to cover 3 km at actual speed = +, (x 1), 3, 3 15, 3, 3, x, & x 14, (x 1) 60, (x 1), , , (x 1) x 1, 1 1 1, , x (x 1) 12 & x (x 1), 12, , , , 1, 1 & x 2 x 12 0, , (x x) 12, , , , x 2 4x 3x 12 0 & x (x 4) 3 (x 4) 0, , , , (x 4) (x 3) 0 & x 4 0 or x 3 0, , , , x 4 or x 3 & x 3, , 2, , [a speed cannot be negative], , Hence, the usual speed of the man is 3 km/hr., EXAMPLE 26, , A motorboat whose speed in still water is 24 km/hr, takes 1 hour, more to go 32 km upstream than to return downstream to the same, spot. Find the speed of the stream., [CBSE 2014], , SOLUTION, , Speed of the motorboat in still water = 24 km/hr., Let the speed of the stream be x km/hr., Then, speed upstream = (24 – x) km/hr., Speed downstream = (24 + x) km/hr., 32, hours., (24 x), 32, Time taken to return 32 km downstream =, hours., (24 + x), 32 32 =, , 1, (24 x) (24 + x), Time taken to go 32 km upstream =
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216, , Secondary School Mathematics for Class 10, , , , , (24 x) (24 x) 1, 1, 1, , 1 &, , 32, (24 x) (24 x) 32, (24 x) (24 x), 2x, 1, , & 576 x 2 64x, (576 x 2) 32, , , , , x 2 64x 576 0 & x 2 72x 8x 576 0, x (x 72) 8 (x 72) 0 & (x 72)(x 8) 0, , , , x + 72 = 0 or x 8 = 0, , , , x = 72 or x = 8, , , , x=8, , [a speed of the stream cannot be negative], , Hence, the speed of the stream is 8 km/hr., EXAMPLE 27, , A sailor can row a boat 8 km downstream and return back to the, starting point in 1 hour 40 minutes. If the speed of the stream is, 2 km/hr, find the speed of the boat in still water., , SOLUTION, , Let the speed of the boat in still water be x km/hr., Speed of the stream = 2 km/hr., speed downstream = (x + 2) km/hr,, speed upstream = (x – 2) km/hr., Time taken to cover 8 km downstream and return back to the, 40, 8, 8, starting point = + + $ But, this time is given as 1, (x 2) (x 2), 60, 2, 5, hours = 1 3 hours = 3 hours., , , 8 + 8 = 5, x+2 x2 3, , , , (x 2) (x 2), 1 1 5, 5, &, , , x 2 x 2 24, 24, (x 2)(x 2), , , , 2x 5, & 5x 2 20 48x, (x 2 4) 24, , , , 5x 2 48x 20 0 & 5x 2 50x 2x 20 0, , , , 5x (x 10) 2 (x 10) 0 & (x 10)(5x 2) 0, , , , x 10 = 0 or 5x + 2 = 0, 2, x = 10 or x = 5, x = 10 [a speed of the boat cannot be negative], , , , , Hence, the speed of the boat in still water is 10 km/hr.
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Quadratic Equations, , 217, , PROBLEMS ON TIME AND WORK, EXAMPLE 28, , A takes 6 days less than the time taken by B to finish a piece of work., If both A and B together can finish it in 4 days, find the time taken by, B to finish the work., [CBSE 2017], , SOLUTION, , Suppose B alone takes x days to finish the work and A alone, can finish it in (x – 6) days., 1, B‘s 1 day‘s work = x $, 1, A‘s 1 day‘s work = $, (x 6), 1, (A + B)‘s 1 day‘s work = 4 $, 1, 1, 1, x+ = 4, (x 6), , , , , , , , (x 6) x 1, (2x 6) 1, & 2, , 4, x (x 6), (x 6x) 4, x 2 6x 8x 24 & x 2 14x 24 0, x 2 12x 2x 24 0 & x (x 12) 2 (x 12) 0, (x 12)(x 2) 0 & x 12 0 or x 2 0, x = 12 or x = 2, x 12 [a x 2 & (x 6) 0], , Hence, B alone can finish the work in 12 days., EXAMPLE 29, , 3, Two water taps together can fill a tank in 9 8 hours. The tap of, larger diameter takes 10 hours less than the smaller one to fill the, tank separately. Find the time in which each tap can separately fill, the tank., , SOLUTION, , [CBSE 2008], , Let the smaller tap fill the tank in x hours., Then, the larger tap fills it in (x – 10) hours., 75, Time taken by both together to fill the tank = 8 hours., 1, Part filled by the smaller tap in 1 hr = x $, 1, Part filled by the larger tap in 1 hr = , $, (x 10), 8, Part filled by both the taps in 1 hr = 75 $, 1, 1, = 8, x+ , (x 10) 75
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218, , Secondary School Mathematics for Class 10, , , , (x 10) x 8, (2x 10), , 8, &, 75, x (x 10), x (x 10) 75, 75 (2x 10) = 8x (x 10) [by cross multiplication], , , , , , , 150x 750 = 8x 2 80x, 8x 2 230x 750 0 & 4x 2 115x 375 0, 4x 2 100x 15x 375 0 & 4x (x 25) 15 (x 25) 0, (x 25)(4x 15) 0 & x 25 0 or 4x 15 0, , , , 15, x = 25 or x = 4, 15, & (x 10) 0E·, x 25 ;a x , 4, Hence, the time taken by the smaller tap to fill the tank, = 25 hours., , , And, the time taken by the larger tap to fill the tank, = (25 –10) hours = 15 hours., PROBLEMS ON AREAS, EXAMPLE 30, , The perimeter of a rectangular field is 82 m and its area is 400 m2., Find the dimensions of the field., , SOLUTION, , We know that 2(length + breadth) = perimeter., 1, 1, (length + breadth) = 2 # perimeter = 2 # 82 m = 41 m., Let the length of the field be x metres., Then, its breadth = (41 – x) m., , , area of the field = x (41 x) m 2 = (41x x 2) m 2 ., , But, area = 400 m 2 (given), , , 41x x 2 = 400, , , , , , , x 2 41x 400 0 & x 2 25x 16x 400 0, x (x 25) 16 (x 25) 0 & (x 25) (x 16) 0, x 25 0 or x 16 0 & x 25 or x 16., length of the field = 25 m and breadth of the field = 16 m., , EXAMPLE 31, , The diagonal of a rectangular field is 16 m more than the shorter, side. If the longer side is 14 m more than the shorter side then find, the lengths of the sides of the field., [CBSE 2015], , SOLUTION, , Let the shorter side of the field be x metres.
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Quadratic Equations, , 219, , Then, longer side = (x + 14) m., And, diagonal = (x + 16) m., , , (x + 16) 2 (x + 14) 2 = x 2, , , , (2x 30)# 2 x 2 & x 2 4x 60 0, , , , x 2 10x 6x 60 0 & x (x 10) 6 (x 10) 0, , , , , , , (x 10) (x 6) 0 & x 10 0 or x 6 0, x 10 or x 6, x 10, [a breadth cannot be negative], breadth = 10 m and length = (10 + 14) m = 24 m., , EXAMPLE 32, , A rectangular field is 20 m long and 14 m wide. There is a path of, equal width all around it, having an area of 111 sq m. Find the width, of the path., , SOLUTION, , Let the width of the path be x metres., Length of the field including the path, = (20 + 2x) m., Breadth of the field including the path, = (14 + 2x) m., Area of the field including the path = (20 + 2x) (14 + 2x) m 2 ., Area of the field excluding the path = (20 # 14) m 2 = 280 m 2 ., , , the area of the path = [(20 + 2x) (14 + 2x) 280] m 2 ., , , , (20 + 2x) (14 + 2x) 280 = 111, , , , , 4x 2 + 68x 111 = 0 4x 2 + 74x 6x 111 = 0, 2x (2x + 37) 3 (2x + 37) = 0 (2x + 37) (2x 3) = 0, , 37, 3, x = 2 or x = 2, 3, x = 2 = 1.5, [a width can never be negative]., Hence, the width of the path is 1.5 m., , , EXAMPLE 33, , Sum of the areas of two squares is 260 m 2 . If the difference of their, perimeters is 24 m then find the sides of the two squares., [CBSE 2009C], , SOLUTION, , Let the sides of the two squares be a metres and b metres., Then, their areas are (a 2) m 2 and (b 2) m 2 respectively., And, their perimeters are (4a) m and (4b) m respectively.
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220, , Secondary School Mathematics for Class 10, , , , 4a 4b 24 & 4 (a b) 24, , & a b 6 & b (a 6), , … (i), , Sum of their areas = 260 m ., 2, , , , a 2 + b 2 = 260, , , , a 2 + (a 6) 2 = 260, , , , 2a 12a 224 0 & a 2 6a 112 0, , , , a 2 14a + 8a 112 = 0, , , , a (a 14) 8 (a 14) 0 & (a 14)(a 8) 0, , , , a 14 0 or a 8 0 & a 14 or a 8, , , , a = 14 [a side of a square cannot be negative]., , , , a = 14 and b = (14 6) = 8., , [using (i)], , 2, , Hence, the sides of the square are 14 m and 8 m respectively., EXAMPLE 34, , The area of a right-angled triangle is 600 sq cm. If the base of the, triangle exceeds the altitude by 10 cm, find the dimensions of the, triangle., , SOLUTION, , Let the altitude of the triangle be x cm., Then, its base = (x + 10) cm., 1, Area of the triangle = 2 x (x + 10) cm 2 ., 1, 2 x (x + 10) = 600, x (x + 10) = 1200, , , x 2 + 10x 1200 = 0, , , , x 2 + 40x 30x 1200 = 0, , , , x (x + 40) 30 (x + 40) = 0, , , , (x + 40) (x 30) = 0, , x = 40 or x = 30, x = 30 [a altitude cannot be negative]., Thus, the altitude of the triangle = 30 cm., And, the base of the triangle = (30 + 10) cm = 40 cm., EXAMPLE 35, , The hypotenuse of a right-angled triangle is 6 cm more than twice, the shortest side. If the third side is 2 cm less than the hypotenuse,, find the sides of the triangle., [CBSE 2007]
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Quadratic Equations, SOLUTION, , 221, , Let the shortest side of the triangle be x cm., Then, its hypotenuse = (2x + 6) cm., And, its third side = (2x + 6 – 2) cm = (2x + 4) cm., By Pythagoras’ theorem, we have, (2x + 6) 2 = x 2 + (2x + 4) 2, , , , , , , , 4x 2 + 24x + 36 = 5x 2 + 16x + 16, x 2 8x 20 0, x 2 10x 2x 20 0, x (x 10) 2 (x 10) 0 & (x 10)(x 2) 0, x 10 0 or x 2 0 & x 10 or x 2, x = 10 [a the side of a triangle cannot be negative], , , , shorter side = 10 cm, longer side = (2 # 10 + 4) cm = 24 cm, and hypotenuse = (2 # 10 + 6) cm = 26 cm., , Hence, the sides of the given triangle are 10 cm, 24 cm and 26 cm., f, , EXERCISE 4E, , PROBLEMS ON NUMBERS, , 1. The sum of a natural number and its square is 156. Find the number., 2. The sum of a natural number and its positive square root is 132. Find, the number., 3. The sum of two natural numbers is 28 and their product is 192. Find the, numbers., 4. The sum of the squares of two consecutive positive integers is 365. Find, the integers., 5. The sum of the squares of two consecutive positive odd numbers is 514., Find the numbers., 6. The sum of the squares of two consecutive positive even numbers is, 452. Find the numbers., 7. The product of two consecutive positive integers is 306. Find the, integers., 8. Two natural numbers differ by 3 and their product is 504. Find the, numbers., 9. Find two consecutive multiples of 3 whose product is 648., 10. Find two consecutive positive odd integers whose product is 483., 11. Find two consecutive positive even integers whose product is 288.
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222, , Secondary School Mathematics for Class 10, , 12. The sum of two natural numbers is 9 and the sum of their reciprocals is, 1·, Find the numbers., [CBSE 2012], 2, 13. The sum of two natural numbers is 15 and the sum of their reciprocals, 3, is · Find the numbers., [CBSE 2005], 10, 14. The difference of two natural numbers is 3 and the difference of their, 3 ·, reciprocals is, Find the numbers., [CBSE 2014], 28, 15. The difference of two natural numbers is 5 and the difference of their, 5, reciprocals is · Find the numbers., [CBSE 2014], 14, 16. The sum of the squares of two consecutive multiples of 7 is 1225. Find, the multiples., 65 ·, 17. The sum of a natural number and its reciprocal is, Find the number., 8, 18. Divide 57 into two parts whose product is 680., 3 ·, 19. Divide 27 into two parts such that the sum of their reciprocals is, 20, 20. Divide 16 into two parts such that twice the square of the larger part, exceeds the square of the smaller part by 164., 21. Find two natural numbers, the sum of whose squares is 25 times their, sum and also equal to 50 times their difference., 22. The difference of the squares of two natural numbers is 45. The square, of the smaller number is four times the larger number. Find the, numbers., [CBSE 2007], 23. Three consecutive positive integers are such that the sum of the square, of the first and the product of the other two is 46. Find the integers., [CBSE 2010], , 24. A two-digit number is 4 times the sum of its digits and twice the, product of its digits. Find the number., 25. A two-digit number is such that the product of its digits is 14. If 45, is added to the number, the digits interchange their places. Find the, number., [CBSE 2012], 26. The denominator of a fraction is 3 more than its numerator. The sum of, 9, the fraction and its reciprocal is 2 · Find the fraction., 10, 27. The numerator of a fraction is 3 less than its denominator. If 1 is added, 1, to the denominator, the fraction is decreased by · Find the fraction., 15, [CBSE 2012]
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Quadratic Equations, , 28. The sum of a number and its reciprocal is 2, , 223, , 1 ·, Find the number., 30, , SOME GENERAL PROBLEMS, , 29. A teacher on attempting to arrange the students for mass drill in, the form of a solid square found that 24 students were left. When he, increased the size of the square by one student, he found that he was, short of 25 students. Find the number of students., 30. 300 apples are distributed equally among a certain number of students., Had there been 10 more students, each would have received one apple, less. Find the number of students., 31. In a class test, the sum of Kamal’s marks in mathematics and English, is 40. Had he got 3 marks more in mathematics and 4 marks less in, English, the product of the marks would have been 360. Find his marks, in two subjects separately., [CBSE 2008C], 32. Some students planned a picnic. The total budget for food was ` 2000., But, 5 students failed to attend the picnic and thus the cost for food, for each member increased by ` 20. How many students attended the, picnic and how much did each student pay for the food?, [CBSE 2010], 33. If the price of a book is reduced by ` 5, a person can buy 4 more books, for ` 600. Find the original price of the book., 34. A person on tour has ` 10800 for his expenses. If he extends his tour by, 4 days, he has to cut down his daily expenses by ` 90. Find the original, duration of the tour., 35. In a class test, the sum of the marks obtained by P in mathematics and, science is 28. Had he got 3 more marks in mathematics and 4 marks less, in science, the product of marks obtained in the two subjects would, have been 180. Find the marks obtained by him in the two subjects, separately., [CBSE 2008], 36. A man buys a number of pens for ` 180. If he had bought 3 more pens, for the same amount, each pen would have cost him ` 3 less. How, many pens did he buy?, 37. A dealer sells an article for ` 75 and gains as much per cent as the cost, price of the article. Find the cost price of the article., [CBSE 2011], PROBLEMS ON AGES, , 38. (i) One year ago, a man was 8 times as old as his son. Now, his age is, equal to the square of his son‘s age. Find their present ages.
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224, , Secondary School Mathematics for Class 10, , 1, times as old as his son. If the sum of the squares of, 2, their ages is 1325, find the ages of the father and the son. [CBSE 2017], , (ii) A man is 3, , 39. The sum of the reciprocals of Meena‘s ages (in years) 3 years ago and, 1, 5 years hence is · Find her present age., 3, 40. The sum of the ages of a boy and his brother is 25 years, and the product, of their ages in years is 126. Find their ages., 41. The product of Tanvy‘s age (in years) 5 years ago and her age 8 years, later is 30. Find her present age., 42. Two years ago, a man‘s age was three times the square of his son‘s age., In three years time, his age will be four times his son‘s age. Find their, present ages., PROBLEMS ON TIME AND DISTANCE, , 43. A truck covers a distance of 150 km at a certain average speed and, then covers another 200 km at an average speed which is 20 km per, hour more than the first speed. If the truck covers the total distance in, 5 hours, find the first speed of the truck., [CBSE 2015], 44. While boarding an aeroplane, a passenger got hurt. The pilot showing, promptness and concern, made arrangements to hospitalise the injured, and so the plane started late by 30 minutes. To reach the destination,, 1500 km away, in time, the pilot increased the speed by 100 km/hour., Find the original speed of the plane., Do you appreciate the values shown by the pilot, namely promptness, in providing help to the injured and his efforts to reach in time?, [CBSE 2013], , 45. A train covers a distance of 480 km at a uniform speed. If the speed had, been 8 km/hr less then it would have taken 3 hours more to cover the, [CBSE 2013C], same distance. Find the usual speed of the train., 46. A train travels at a certain average speed for a distance of 54 km and, then travels a distance of 63 km at an average speed of 6 km/hr more, than the first speed. If it takes 3 hours to complete the total journey,, [CBSE 2015], what is its first speed?, 47. A train travels 180 km at a uniform speed. If the speed had been, 9 km/hr more, it would have taken 1 hour less for the same journey., [CBSE 2013], Find the speed of the train.
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Quadratic Equations, , 225, , 48. A train covers a distance of 300 km at a uniform speed. If the speed of, the train is increased by 5 km/hour, it takes 2 hours less in the journey., [CBSE 2017], Find the original speed of the train., 49. A passenger train takes 2 hours less for a journey of 300 km if its speed, is increased by 5 km/hr from its usual speed. Find its usual speed., [CBSE 2007], , 50. The distance between Mumbai and Pune is 192 km. Travelling by the, Deccan Queen, it takes 48 minutes less than another train. Calculate, the speed of the Deccan Queen if the speeds of the two trains differ by, [CBSE 2003], 20 km/hr., 51. A motor boat whose speed in still water is 18 km/hr, takes 1 hour more, to go 24 km upstream than to return to the same spot. Find the speed of, [CBSE 2014], the stream., 52. The speed of a boat in still water is 15 km/hr. It goes 30 km upstream, and returns back at the same point in 4 hours 30 minutes. Find the, [CBSE 2017], speed of the stream., 53. A motorboat whose speed is 9 km/hr in still water, goes 15 km, downstream and comes back in a total time of 3 hours 45 minutes. Find, [CBSE 2007C], the speed of the stream., PROBLEMS ON TIME AND WORK AND PIPES AND CISTERN, , 54. A takes 10 days less than the time taken by B to finish a piece of work., If both A and B together can finish the work in 12 days, find the time, taken by B to finish the work., 1, 55. Two taps running together can fill a tank in 3, hours. If one tap takes, 13, 3 hours more than the other to fill the tank then how much time will, each tap take to fill the tank?, , [CBSE 2017], , 1, 56. Two pipes running together can fill a tank in 11 9 minutes. If one pipe, takes 5 minutes more than the other to fill the tank separately, find the, time in which each pipe would fill the tank separately., [CBSE 2010], 57. Two water taps together can fill a tank in 6 hours. The tap of larger, diameter takes 9 hours less than the smaller one to fill the tank, separately. Find the time in which each tap can separately fill the tank., [CBSE 2011], , PROBLEMS ON AREA AND GEOMETRY, , 58. The length of a rectangle is twice its breadth and its area is 288 cm 2 ., Find the dimensions of the rectangle.
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226, , Secondary School Mathematics for Class 10, , 59. The length of a rectangular field is three times its breadth. If the area of, the field be 147 sq metres, find the length of the field., 60. The length of a hall is 3 metres more than its breadth. If the area of the, hall is 238 sq metres, calculate its length and breadth., 61. The perimeter of a rectangular plot is 62 m and its area is 228 sq metres., Find the dimensions of the plot., 62. A rectangular field is 16 m long and 10 m wide. There is a path, of uniform width all around it, having an area of 120 m 2 . Find the, width of the path., 63. The sum of the areas of two squares is 640 m2. If the difference in their, perimeters be 64 m, find the sides of the two squares., [CBSE 2008], 64. The length of a rectangle is thrice as long as the side of a square. The, side of the square is 4 cm more than the width of the rectangle. Their, areas being equal, find their dimensions., 65. A farmer prepares a rectangular vegetable garden of area 180 sq metres., With 39 metres of barbed wire, he can fence the three sides of the, garden, leaving one of the longer sides unfenced. Find the dimensions, of the garden., 66. The area of a right triangle is 600 cm 2 . If the base of the triangle exceeds, the altitude by 10 cm, find the dimensions of the triangle., 67. The area of a right-angled triangle is 96 sq metres. If the base is three, times the altitude, find the base., 68. The area of a right-angled triangle is 165 sq metres. Determine its base, and altitude if the latter exceeds the former by 7 metres., 69. The hypotenuse of a right-angled triangle is 20 metres. If the difference, between the lengths of the other sides be 4 metres, find the other sides., 70. The length of the hypotenuse of a right-angled triangle exceeds the, length of the base by 2 cm and exceeds twice the length of the altitude, by 1 cm. Find the length of each side of the triangle., 71. The hypotenuse of a right-angled triangle is 1 metre less than twice the, shortest side. If the third side is 1 metre more than the shortest side,, find the sides of the triangle., ANSWERS (EXERCISE 4E), , 1. 12, , 2. 121, , 6. 14 and 16, , 7. 17 and 18, , 11. 16 and 18 12. 3 and 6, , 3. 16 and 12, 8. 24 and 21, 13. 10 and 5, , 4. 13 and 14, , 5. 15 and 17, , 9. 24 and 27 10. 21 and 23, 14. 7 and 4, , 15. 7 and 2
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Quadratic Equations, , 16. 21 and 28, , 17. 8, , 20. 10 and 6, , 21. 30 and 10, , 18. 17 and 40, , 2, 2, 24. 36 25. 27 26. 5 27. 5, 31. (21 and 19) or (12 and 28), , 227, , 19. 12 and 15, , 22. 9 and 6, , 23. 4, 5, 6, , 5, 6, 28. 6 or 5, 32. 20, ` 100, , 29. 600, , 30. 50, , 33. ` 30, , 34. 20 days, , 35. Either he got 12 marks in maths and 16 in science or he got 9 marks in, , maths and 19 in science, 36. 12 pens, , 37. ` 50, , 38. (i) 49 years, 7 years, , (ii) 35 years, 10 years, , 40. 18 years and 7 years, , 41. 7 years, , 39. 7 years, 42. 29 years, 5 years, , 43. 60 km/hr, , 44. 500 km/hr 45. 40 km/hr, , 46. 36 km/hr, , 47. 36 km/hr, , 48. 25 km/hr, , 49. 25 km/hr, , 50. 80 km/hr, , 51. 6 km/hr, , 52. 5 km/hr, , 53. 3 km/hr, , 54. 30 days, , 55. 5 hours, 8 hours, , 56. 20 min, 25 min, , 57. 9 hours, 18 hours, , 58. length = 24 cm, breadth = 12 cm, , 59. 21 m, , 60. breadth = 14 m, length = 17 m, , 61. length = 19 m, breadth = 12 m, , 62. 2 m, , 63. 24 m, 8 m, , 64. (l = 18 cm, b = 2 cm), side = 6 cm, , 65. Either (l = 24 m, b = 7.5 m) or (l = 15 m, b = 12 m), 66. 30 cm, 40 cm, 50 cm, , 67. 24 m, , 68. base = 15 m, altitude = 22 m, , 69. 16 m, 12 m 70. base = 15 cm, altitude = 8 cm, hypotenuse = 17 cm, 71. 3 m, 4 m, 5 m, HINTS TO SOME SELECTED QUESTIONS, 1. x x 2 156 & x 2 x 156 0 & x 2 13x 12x 156 0., 2. x x 132 & y 2 y 132, where x y, , , y 2 y 132 0 & y 2 12y 11y 132 0., , 3. Let the required numbers be x and (28 – x)., Then, x (28 x) = 192., 4. x 2 (x 1) 2 365 & 2x 2 2x 364 0 & x 2 x 182 0, , , x 2 + 14x 13x 182 = 0., , 5. x 2 (x 2) 2 514 & 2x 2 4x 510 0 & x 2 2x 255 0, & x 2 17x 15x 255 0., 6. x 2 (x 2) 2 452 & 2x 2 4x 448 0 & x 2 2x 224 0, & x 2 16x 14x 224 0.
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228, , Secondary School Mathematics for Class 10, , 7. x (x 1) 306 & x 2 x 306 0 & x 2 18x 17x 306 0., 8. x (x 3) 504 & x 2 3x 504 0 & x 2 24x 21x 504 0., 9. (3x) {3 (x 1)} 648 & x (x 1) 72 & x 2 9x 8x 72 0., 10. x (x 2) 483 & x 2 2x 483 0 & x 2 23x 21x 483 0., 11. x (x 2) 288 & x 2 2x 288 0 & x 2 18x 16x 288 0., 14. Let the required numbers be x and (x 3) . Then,, 1 1 = 3, $, (x 3) x 28, 18. x (57 x) 680 & x 2 57x 680 0 & x 2 40x 17x 680 0., , 1, 1, 3 & 27 x x 3 & x 2 27x 180 0, 19. x , (27 x) 20, x (27 x) 20, & x 2 15x 12x 180 0., 20. Let the two parts be x and (16 – x) such that x is larger. Then,, 2x 2 (16 x) 2 164 & x 2 32x 420 0 & x 2 42x 10x 420 0., 21. Let the required natural numbers be a and b. Then,, a 2 b 2 25 (a b) and (a 2 b 2) 50 (a b) ., , , 25 (a b) 50 (a b) & a b 2 (a b) & a 3b., , , , (3b) 2 b 2 25 # 4b & 10b 2 100b 0, & 10b (b 10) 0 & b 10., , 22. Let the required natural numbers be a and b. Then,, (a 2 b 2 45 and b 2 4a) & (a 2 4a 45) 0., 23. Let the required numbers be x, (x + 1) and (x + 2). Then,, x 2 (x 1) (x 2) 46 & 2x 2 3x 44 0 & 2x 2 11x 8x 44 0, & (2x 11) (x 4) ., 11, x!, , since x is an integer., 2, 24. Let the tens digit be x and units digit be y. Then,, (10x y) 4 (x y) & 6x 3y & y 2x., (10x y) 2xy & (10x 2x) 4x 2 & 4x 2 12x 0, & 4x (x 3) 0 & x 3 [a x ! 0] ., 25. Let the tens digit be x and units digit be y. Then,, xy 14 and 10x y 45 10y x, , , xy 14 and 9 (y x) 45 & xy 14 and y x 5, , , , x (x 5) 14 and y (5 x) ., , 26. Let the required fraction be, , a, $ Then,, (a + 3), , a (a 3) 29, & 10 [a 2 (a 3) 2] 29a (a 3), a, 10, (a 3), & 9a 2 27a 90 0 & a 2 3a 10 0.
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Quadratic Equations, x3·, Then,, x, 1, 1, x3 x3 1, 1, & (x 3) ;x x 1E , x, x 1 15, 15, , 27. Let the required fraction in simplest form be, , , , 15 (x 3) x (x 1) & x 2 14x 45 0 & x 5 or x 9., , So, the required fraction is, , 2·, 5, , 6, ;neglecting E, 9, , 1 61 ·, 28. Let the required number be x. Then, x x , 30, , , 30x 2 61x 30 0 & 30x 2 36x 25x 30 0., , 29. Let there be x rows and the number of students in each row be x., Then, total number of students = (x 2 + 24) ., x 2 + 24 = (x + 1) 2 25., 30. Let the total number of students be x. Then,, 300 300 , 1 & x 2 10x 3000 0, x, (x 10), & (x 50) (x 60) 0 & x 50., 31. Suppose he gets x marks in mathematics and y marks in English. Then,, x + y = 40 and (x + 3) (y 4) = 360., (x 3) (40 x 4) 360 & (x 3)(36 x) 360, , , x 2 33x 252 0 & (x 21)(x 12) 0., , 32. Let x students attended the picnic. Then, (x + 5) planned it., , , 2000 2000 , 20 1, 1, 1, 20 & x , & x 2 5x 500 0., x, x5, x 5 2000 100, , , , x 2 + 25x 20x 500 = 0., , 33. Let the original price per book be ` x. Then, reduced price = ` (x – 5)., 600 600 , 1, 1, 4 1 ·, 4& x , x, x5, x 5, 600 150, 34. Let the original duration of the tour be x days. Then,, , , 10800 10800 , 1, 1, 1, 90 & x , & x 2 4x 480 0, x, x 4 120, (x 4), , , x 2 + 24x 20x 480 = 0., , 35. Suppose P gets x marks in mathematics and y marks in science. Then,, x y 28 & y 28 x., And (x 3) (y 4) 180 & (x 3)(28 x 4) 180, , , (x 3)(24 x) 180 & x 2 21x 108 0., , 36. Suppose he bought x pens., 180 180 , 1, 1, 1, , 3& x , & x (x 3) 180 0, x, x 3 60, (x 3), , , x 2 15x 12x 180 0., , 37. Let the CP of the article be ` x. Then, gain = x%., 2, (100 x) x, 2 ` ( x 100x 2, SP = ` (, 100, 100, , 229
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230, , Secondary School Mathematics for Class 10, , x 2 100x , 75 & x 2 100x 7500 0 & x 2 150x 50x 7500 0., 100, 38. (i) Let the son‘s present age be x years., , , Son‘s age 1 year ago = (x – 1) years., Man‘s age 1 year ago = 8(x – 1) years = (8x – 8) years., Man's present age = (8x – 8 + 1) years = (8x – 7) years., , , , , (8x 7) x 2 & x 2 8x 7 0., , 39. Let Meena's present age be x years. Then,, (x 5) (x 3) 1, 1 1 1, & x 2 2x 15 3 (2x 2), &, 3, (x 3) (x 5) 3, (x 3)(x 5), , , x 2 4x 21 0., , 40. x (25 x) 126 & x 2 25x 126 0 & x 2 18x 7x 126 0., 41. Let Tanvy‘s present age be x years., Then, (x 5) (x 8) 30 & x 2 3x 70 0 & x 2 10x 7x 70 0., 42. Let son‘s age 2 years ago be x years., Then, man's age 2 years ago = 3x 2 ., , , 3x 2 5 4 (x 5) & 3x 2 4x 15 0 & 3x 2 9x 5x 15 0 & x 3., , Son‘s age = (x + 2) years = 5 years, man‘s age = (3x 2 + 2) years = 29 years., 43. Let the first speed be x km/hr. Then,, 3, 4, 150 200 , F 5, 5 & 50 <x , x, (x 20), (x 20), , , 10 [3 (x 20) 4x] x (x 20) & x 2 50x 600 0., , 44. Let the original speed of the plane be x km/hr. Then,, 1500 1500 30, 1, 1, 1 & x 2 100x 300000 0, &x, x, (x 100) 60, (x 100) 3000, , x 2 600x 500x 300000 0., His promptness is appreciable., 45. Let the usual speed of the train be x km/hr. Then,, 3 1 ·, 480 480 , 1, 1, 3& x , x, x 8, 480 160, (x 8), , , 160 [x (x 8)] x (x 8) & x 2 8x 1280 0, , , , x 2 40x + 32x 1280 = 0., , 46. Let the first speed be x km/hr. Then,, , , , 6 7, 54 63 , , x (x 6) 3 & 9 <x (x 6)F 3, 3 [6 (x 6) 7x] x (x 6) & x (x 6) 3 (13x 36), , , , x 2 33x 108 0., , 47. Let the speed be x km/hr. Then,, 180 180 , 1, x, (x 9), , , x 2 9x 1620 0 & x 2 45x 36x 1620 0.
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Quadratic Equations, 49., , 300 300 , 1, 1, 1, 2& x , & x 2 5x 750 0., x, x5, x 5 150, , , x 2 + 30x 25x 750 = 0., , 50. Let the speed of the Deccan queen be x km/hr. Then,, 192 192 48, 1 1 1 ·, & , x, x 20 x 240, 60, (x 20), , , x 2 20x 4800 0 & x 2 80x 60x 4800 0., , 51. Let the speed of the stream be x km/hr., Then, speed downstream = (18 + x) km/hr, and speed upstream = (18 – x) km/hr., 24 24 , 1, 1, , 1, , 1&, (18 x) (18 x), (18 x) (18 x) 24, , , x 2 48x 324 0 & x 2 54x 6x 324 0., , 52. Let the speed of the stream be x km/hr., 30 km, 30, Time to go 30 km upstream , hr , hr., (15 x) km, (15 x), Time to go 30 km downstream , , 30 km, 30, hr , hr., (15 x) km, (15 x), , , , 30 30 1 9, 4, 2 2, (15 x) (15 x), , , , 30 (15 x) 30 (15 x) , , , , 450 30x 450 30x , , , , 200 225 x 2 & x 2 25 & x 5., , 9, (225 x 2), 2, , 9, (225 x 2), 2, , 53. Let the speed of the stream be x km/hr., Then, speed downstream = (9 + x) km/hr, and speed upstream = (9 – x) km/hr., , , , , 15 15 45 3 15, 3, 3, 9x 9x, 60, 4, 4, 1 1 1, & 4 [(9 x) (9 x)] (9 x) (9 x), 9x 9x 4, 81 x 2 72 & x 2 9 & x 3., , 54. Suppose B takes x days to finish the work. Then, A takes (x – 10) days., 1, 1, 2, 2, 1, , , , , , x (x 10) 12 & x 34x 120 0 & x 30x 4x 120 0., 55. Let the faster pipe take x hr to fill it. Then, the other takes (x + 3) hr., (x 3) x 13, 1 1 13, , , x (x 3) 40 & x (x 3), 40, , , 13x 2 41x 120 0, , , , x, , 41 ! 1681 6240 41 ! 7921 (41 ! 89), , , & x 5., 26, 26, 26, , 231
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232, , Secondary School Mathematics for Class 10, , 56. Let the faster pipe take x min to fill it. Then, the other takes (x + 5) min., , , 1 1 9, 2, , x (x 5) 100 & 100 (2x 5) 9 (x 5x), , , , 9x 2 155x 500 0 & 9x 2 180x 25x 500 0., , HINT, , Quadratic formula may be used., , 57. Let the faster tap take x hours to fill the tank., Then, the slower tap takes (x + 9) hours to fill it., 1 1 1, , , , x (x 9) 6 & 6 (2x 9) x (x 9), , , x 2 3x 54 0 & x 2 9x 6x 54 0., , 58. 2x # x 288 & x 2 144 & x 12 cm., 59. 3x # x 147 & x 2 49 & x 7 m. So, length = 21 m., 60. x (x 3) 238 & x 2 3x 238 0 & x 2 17x 14x 238 0., breadth = 14 m and length = 17 m., 61. 2 (l b) 62 & l b 31. Let b = x m. Then, l = (31 x) m, , , (31 x) x 228 & x 2 31x 228 0 & x 2 19x 12x 228 0., , 62. Let the width of the path be x m., So, the length of the field with path (16 2x) m, and breadth of the field with path (10 2x) m., , , (16 2x) (10 2x) 16 #10 120, , 160 52x 4x 2 160 120, 4x 2 52x 120 0 & x 2 13x 30 0, x 2 15x 2x 30 0 & x (x 15) 2 (x 15) 0, (x 15) (x 2) 0 & x 15 or 2., So, width of the path = 2 m., 63. Let the sides of the squares be a metres and b metres., Then, 4a 4b 64 & 4 (a b) 64 & (a b) 16., And, a 2 b 2 640 & a 2 (a 16) 2 640, & 2a 2 32a 384 0 & a 2 16a 192 0, & a 2 24a 8a 192 0 & (a 24) (a 8) 0 & a 24., a = 24 m, b = 8 m., 64. Let each side of the square be x cm., Then, length of the rectangle = 3x cm., Width of the rectangle = (x – 4) cm., , , 3x (x 4) x 2 & 2x 2 12x 0 & 2x (x 6) 0 & x 6., , 180, 65. Let the length be x metres. Then, breadth = x m., 180 180, 360, , x x x 39 & x x 39, , , x 2 39x 360 0 & x 2 24x 15x 360 0
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Quadratic Equations, , , 233, , (x 24) (x 15) 0 & x 24 or x 15., , Either (l 24 m, b 7.5 m) or (l 15 m, b 12 m) ., 66. Let the altitude be x cm. Then, base = (x + 10) cm., 1 , , (x 10) x 600 & x 2 10x 1200 0, 2, (x 40) (x 30) 0 & x 30., , , base = 40 cm, altitude = 30 cm., , , , hypotenuse (40) 2 (30) 2 1600 900 2500 50 cm., , 67. Let the altitude be x metres. Then, base = 3x metres., 1, # 3x # x 96 & x 2 64 & x 8 m. Hence, base = 24 m., , 2, 68. Let the base be x metres. Then, altitude = (x + 7) m., 1, # x #(x 7) 165 & x 2 7x 330 0, , 2, , x 2 22x 15x 330 0 & (x 15) (x 22) 0 & x 15., 69. Let the other sides be x metres and (x – 4) metres., , , x 2 (x 4) 2 (20) 2 & 2x 2 8x 384 0, , , , x 2 4x 192 0 & (x 16) (x 12) 0 & x 16., , 70. Let the base be x cm. Then, hypotenuse = (x + 2) cm., , , , , (x 2) (2 # altitude) 1 & 2 # altitude (x 1), 1, & altitude (x 1) cm., 2, 1, 2, 2, 2, 2, (x 2) x (x 1) & 4 (x 2) 4x 2 (x 1) 2, 4, 4 (x 2 4x 4) 5x 2 2x 1 & x 2 14x 15 0, , (x 15) (x 1) 0 & x 15., base = 15 cm, hypotenuse = 17 cm, altitude = 8 cm., 71. Let the shortest side be x metres., Then, its hypotenuse (2x 1) m and third side (x 1) m., , , (2x 1) 2 x 2 (x 1) 2 & 2x 2 6x 0 & 2x (x 3) 0., , SUMMARY OF RESULTS, 1. If and are the roots of a quadratic equation ax 2 + bx + c = 0 then, b, c, (i) + = a ,, (ii) = a $, 2. The roots of the equation ax 2 + bx + c = 0 are given by, =, , b + b 2 4ac, b b 2 4ac, =, , and, 2a, 2a, , = 0 the discriminant is given by, 3. For the equation ax 2 + bx + c = 0, a Y, D = (b 2 4ac) .
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234, , Secondary School Mathematics for Class 10, , = 0., 4. Nature of Roots of quadratic equation ax 2 + bx + c = 0, a Y, Nature of Roots, , Value of D, , (i) D > 0 and D is a perfect Real,, unequal, square., rational, , and, , b ! D, 2a, , (ii) D > 0 and D is not a Real,, unequal, perfect square., irrational, , and, , b ! D, 2a, b, Each , 2a, , (iii) D = 0, , Real and equal, , (iv) D < 0, , Imaginary, , None, , TEST YOURSELF, MCQ, 1. Which of the following is a quadratic equation?, 1, (b) x + x = x 2, , (a) x 2 3 x + 2 = 0, (c) x 2 +, , 1 =, 5, x2, , (d) 2x 2 5x = (x 1) 2, , 2. Which of the following is a quadratic equation?, (a) (x 2 + 1) = (2 x) 2 + 3, (b) x 3 x 2 = (x 1) 3, (c) 2x 2 + 3 = (5 + x) (2x 3), , (d) None of these, , 3. Which of the following is not a quadratic equation?, (a) 3x x 2 = x 2 + 5, (b) (x + 2) 2 = 2 (x 2 5), (d) (x 1) 2 = 3x 2 + x 2, , (c) ( 2 x + 3) 2 = 2x 2 + 6, , 4. If x = 3 is a solution of the equation 3x 2 + (k 1) x + 9 = 0 then k = ?, (a) 11, , (b) –11, , (c) 13, , (d) –13, , 5. If one root of the equation 2x + ax + 6 = 0 is 2 then a = ?, 2, , (a) 7, , (b) –7, , 7, (c) 2, , (d), , 7, 2, , 6. The sum of the roots of the equation x 2 6x + 2 = 0 is, (a) 2, , (b) –2, , (c) 6, , (d) –6, , 7. If the product of the roots of the equation x 3x + k = 10 is –2 then the, value of k is, 2, , (a) –2, , (b) –8, , (c) 8, , (d) 12
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Quadratic Equations, , 235, , 8. The ratio of the sum and product of the roots of the equation, 7x 2 12x + 18 = 0 is, (a) 7 : 12, , (b) 7 : 18, , (c) 3 : 2, , (d) 2 : 3, , 1, 9. If one root of the equation 3x 2 10x + 3 = 0 is 3 then the other root is, 1, 1, (a) 3, (b) 3, (c) –3, (d) 3, 10. If one root of 5x 2 + 13x + k = 0 be the reciprocal of the other root then, the value of k is, (a) 0, , (b) 1, , (c) 2, , (d) 5, , 11. If the sum of the roots of the equation kx + 2x + 3k = 0 is equal to their, product then the value of k is, 2, 1, 2, 1, (b) 3, (c) 3, (d) 3, (a) 3, 2, , 12. The roots of a quadratic equation are 5 and –2. Then, the equation is, (a) x 2 3x + 10 = 0, (b) x 2 3x 10 = 0, (c) x 2 + 3x 10 = 0, , (d) x 2 + 3x + 10 = 0, , 13. If the sum of the roots of a quadratic equation is 6 and their product is, 6, the equation is, (a) x 2 6x + 6 = 0, (b) x 2 + 6x 6 = 0, (c) x 2 6x 6 = 0, , (d) x 2 + 6x + 6 = 0, , 1 1, 14. If and are the roots of the equation 3x 2 + 8x + 2 = 0 then c + m = ?, 3, 2, (a) 8, (b) 3, (c) –4, (d) 4, 15. The roots of the equation ax 2 + bx + c = 0 will be reciprocal of each, other if, (a) a = b, (b) b = c, (c) c = a, (d) none of these, 2, 16. If the roots of the equation ax + bx + c = 0 are equal then c = ?, b, (a) 2a, , b, (b) 2a, , (c), , b2, 4a, , b2, (d) 4a, , 17. If the equation 9x 2 + 6kx + 4 = 0 has equal roots then k = ?, (a) 2 or 0, , (b) –2 or 0, (c) 2 or –2, (d) 0 only, 18. If the equation x + 2 (k + 2) x + 9k = 0 has equal roots then k = ?, 2, , (a) 1 or 4, , (b) –1 or 4, (c) 1 or – 4, (d) –1 or – 4, , +, =, 19. If the equation 4x 3kx 1 0 has equal roots then k = ?, 2, , 2, (a) ! 3, , 1, (b) ! 3, , 3, (c) ! 4, , 4, (d) ! 3
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236, , Secondary School Mathematics for Class 10, , = 0 are real and unequal, if (b 2 4ac) is, 20. The roots of ax 2 + bx + c = 0, a Y, (a) > 0, , (b) = 0, (c) < 0, (d) none of these, 2, =, +, +, =, 21. In the equation ax bx c 0, it is given that D (b 4ac) > 0. Then,, the roots of the equation are, (a) real and equal, (b) real and unequal, (c) imaginary, (d) none of these, 22. The roots of the equation 2x 2 6x + 7 = 0 are, (a) real, unequal and rational, (b) real, unequal and irrational, (c) real and equal, (d) imaginary, 23. The roots of the equation 2x 2 6x + 3 = 0 are, 2, , (a) real, unequal and rational, (c) real and equal, , (b) real, unequal and irrational, (d) imaginary, , 24. If the roots of 5x 2 kx + 1 = 0 are real and distinct then, (a) 2 5 < k < 2 5, , (b) k > 2 5 only, , (c) k < 2 5 only, , (d) either k > 2 5 or k < 2 5, , 25. If the equation x 2 + 5kx + 16 = 0 has no real roots then, 8, 8, (a) k > 5, (b) k < 5, 8, 8, (c) 5 < k < 5, (d) none of these, 26. If the equation x 2 kx + 1 = 0 has no real roots then, (a) k < –2, (b) k > 2, (c) –2 < k < 2, (d) none of these, 27. For what values of k, the equation kx 2 6x 2 = 0 has real roots?, 9, (a) k # 2, (c) k # 2, , 9, (b) k $ 2, (d) None of these, , 1, 28. The sum of a number and its reciprocal is 2 20 $ The number is, 5, 4, 4, 3, 5, 6, 1, (a) 4 or 5, (b) 3 or 4, (c) 6 or 5, (d) 6 or 6, 29. The perimeter of a rectangle is 82 m and its area is 400 m 2 . The breadth, of the rectangle is, (a) 25 m, (b) 20 m, (c) 16 m, (d) 9 m, 30. The length of a rectangular field exceeds its breadth by 8 m and the area, of the field is 240 m 2 . The breadth of the field is, [CBSE 2014], (a) 20 m, (b) 30 m, (c) 12 m, (d) 16 m
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Quadratic Equations, , 31. The roots of the quadratic equation 2x 2 x 6 0 are, , 237, [CBSE 2012], , 3, 3, 3, 3, (a) 2, 2, (b) 2, 2, (c) 2, 2, (d) 2, 2, 32. The sum of two natural numbers is 8 and their product is 15. Find the, numbers., [CBSE 2012], Very-Short-Answer Questions, 33. Show that x = –3 is a solution of x 2 + 6x + 9 = 0., 34. Show that x = –2 is a solution of 3x 2 + 13x + 14 = 0., , [CBSE 2008], [CBSE 2008], , 1, 35. If x = 2 is a solution of the quadratic equation 3x 2 + 2kx 3 = 0, find, the value of k., [CBSE 2015], 2, 36. Find the roots of the quadratic equation 2x x 6 = 0., [CBSE 2012], 37. Find the solution of the quadratic equation 3 3 x 2 + 10x + 3 = 0., [CBSE 2009], , 38., 39., 40., 41., 42., 43., 44., 45., , If the roots of the quadratic equation 2x + 8x + k = 0 are equal then, find the value of k., [CBSE 2014], 2, , +, =, If the quadratic equation px 2 5 px 15 0 has two equal roots then, find the value of p., [CBSE 2015], 2, 2, +, +, =, +, +, =, If 1 is a root of the equation ay ay 3 0 and y y b 0 then find, the value of ab., [CBSE 2012], 2, , +, +, If one zero of the polynomial x 4x 1 is (2, 3 ), write the other, zero., [CBSE 2010], 2, , +, =, If one root of the quadratic equation 3x 10x k 0 is reciprocal of, the other, find the value of k., [CBSE 2014], , +, =, If the roots of the quadratic equation px (x 2) 6 0 are equal, find, the value of p., [CBSE 2013], 2, , +, Find the values of k so that the quadratic equation x 4kx k = 0 has, equal roots., [CBSE 2012], 2, , Find the values of k for which the quadratic equation 9x 3kx + k = 0, has equal roots., [CBSE 2014], 2, , Short-Answer Questions, 46. Solve: x 2 ( 3 + 1) x + 3 = 0., , [CBSE 2015], , 47. Solve: 2x + ax a = 0., , [CBSE 2014], , 48. Solve: 3x + 5 5 x 10 = 0., , [CBSE 2014], , 2, , 2, , 2, , 49. Solve:, , 3 x 2 + 10x 8 3 = 0., , [CBSE 2014]
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238, , Secondary School Mathematics for Class 10, , 50. Solve:, , 3 x 2 2 2 x 2 3 = 0., , [CBSE 2014], , 51. Solve: 4 3 x + 5x 2 3 = 0., , [CBSE 2013], , 52. Solve: 4x + 4bx (a b ) = 0., , [CBSE 2015], , 53. Solve: x + 5x (a + a 6) = 0., , [CBSE 2015], , 54. x + 6x (a + 2a 8) = 0, , [CBSE 2015], , 55. x 2 4ax + 4a 2 b 2 = 0, , [CBSE 2012], , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , ANSWERS (TEST YOURSELF), , 1. (d), , 2. (b), , 3. (c), , 4. (b), , 10. (d) 11. (d) 12. (b) 13. (a), , 5. (b), , 6. (c), , 7. (c), , 8. (d), , 9. (d), , 14. (c) 15. (c) 16. (d) 17. (c), , 18. (a), , 19. (d) 20. (a), , 21. (b) 22. (d) 23. (b) 24. (d) 25. (c) 26. (c), , 27. (b), , 28. (a) 29. (c), , 30. (c) 31. (b), , 9, , 32. 3 and 5, , 3, , 35. k = 4, , 36. x = 2 or x = 2, , 39. p = 3, , 40. ab = 3, , 1, , 44. k = 0 or k = 4, , 51. x =, , (b + a), (a b), or x = 2, 2, , 54. x = (a + 4) or x = (a 2), , 43. p = 6, , 49. x = 4 3 or x =, , 3, 2, or x = 4, 3, 53. x = (a + 3) or x = (a 2), 55. x = (2a + b) or x = (2a b), , HINTS TO SOME SELECTED QUESTIONS, 7. Given equation is x 2 3x + (k 10) = 0., Product of roots = (k 10) . So, k 10 2 & k 8., 12 18, 8. Required ratio = 7 : 7 = 2 : 3., 3, 1, 9. Let the other root be . Then, # 1 & 3., 3, 3, So, the other root is 3., 1, 10. Let the roots be and · Then,, k, 1, product of roots = b # l = 1. So, 1 & k 5., 5, , 38. k = 8, , 46. x = 3 or x = 1, , 48. x = 2 5 or x = 3, , 2, 3, , 1, 3 3, , 42. k = 3, , 5, , a, , 52. x =, , 41. (2 3 ), , 45. k = 0 or k = 1, , 47. x = a or x = 2, 50. x = 6 or x =, , 37. x 3 or x , , 2, 3
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Quadratic Equations, 2, 3k =, and product of roots =, 3., k, k, 2, , 3 & k 2·, 3, k, , 11. Sum of roots =, , , 12. Sum of the roots = 5 + (2) = 3, product of roots = 5 # (2) = 10., , , x 2 (sum of roots) x + product of roots = 0., , Hence, x 2 3x 10 = 0., 13. Required equation is x 2 6x + 6 = 0., 8, 2, and ·, 3, 3, ( ) 8 3, 1 1, , d, n, # 4., , 3, 2, , , 14. We have , , , c, 1, 15. Product of roots a · Also, a # k 1., c, , , a 1 & c a., 16. Since the roots are equal, we have D = 0., b2 ·, 4a, 17. Since the roots are equal, we have D = 0., , , b 2 4ac 0 & 4ac b 2 & c , , , , 36k 2 4 # 9 # 4 0 & 36k 2 144 & k 2 4 & k 2 or 2., , 18. Since the roots are equal, we have D = 0., , , , 4 (k 2) 2 36k 0 & (k 2) 2 9k 0, k 2 5k 4 0 & k 2 4k k 4 0 & k (k 4) (k 4) 0, & (k 4) (k 1) 0 & k 4 or k 1., , 19. Since the roots are equal, we have D = 0., 4, 16, 4, ·, , &k, 9k 2 16 0 & k 2 , or k , 9, 3, 3, = 0 are real and unequal only when (b 2 4ac) > 0., 20. The roots of ax 2 + bx + c = 0, a Y, 21. When D > 0, the roots of the given quadratic equation are real and unequal., 22. D = ( 6) 2 4 # 2 # 7 = (36 56) = 20 < 0., So, the roots of the given equation are imaginary., 23. Given equation is 2x 2 6x + 3 = 0., , , D = (6) 2 4 # 2 # 3 = (36 24) = 12,, which is greater than 0 and not a perfect square., , , , the roots are real, unequal and irrational., , 24. The roots of 5x 2 kx + 1 = 0 are real and distinct., , , (k 2 4 # 5 #1) 0 & k 2 20 & k 20 or k 20, & k 2 5 or k 2 5 ., , 25. For no real roots, we must have b 2 4ac < 0., 8, 64, 8, , &, (25k 2 4 #16) 0 & 25k 2 64 & k 2 , k ·, 25, 5, 5, , 239
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240, , Secondary School Mathematics for Class 10, , 26. For no real roots, we must have: b 2 4ac < 0., k 2 4 0 & k 2 4 & 2 k 2., 27. For real roots, we must have b 2 4ac $ 0., , , (6) 2 4 # k #( 2) $ 0 & 36 8k $ 0, & 8k $ 36 & k $, , 9, ·, 2, , 28. Let the required number be x. Then,, 1 41, x x , & 20x 2 41x 20 0, 20, & 20x 2 25x 16x 20 0, & 5x (4x 5) 4 (4x 5) 0, & (4x 5) (5x 4) 0, 5, 4, &x, or x ·, 5, 4, 29. 2 (l b) 82 & l b 41 & l (41 b) ., And, lb 400 & (41 b) b 400, & b 2 41b 400 0 & b 2 25b 16b 400 0, & b (b 25) 16 (b 25) 0, & (b 25)(b 16) 0, , , b 25 or b 16., , But b 25 & l (41 25) 16 b., , , breadth = 16 m., , 30. Let the breadth of the field be x m. Then, length = (x + 8) m., , , (x 8)# x 240 & x 2 8x 240 0 & x 2 20x 12x 240 0, & x (x 20) 12 (x 20) 0, & (x 20)(x 12) 0., , 32. + = 8 and = 15., The quadratic equation whose roots are and is given by, x 2 8x + 15 = 0, , , x 2 5x 3x + 15 = 0, , , , x (x 5) 3 (x 5) = 0, , , , (x 3) (x 5) = 0, , , , x = 3 or x = 5., , 33. Putting x = 3 in the given equation, we get, LHS = (3) 2 + 6 # (3) + 9 = (9 18 + 9) = 0 = RHS., , Hence, x = 3 is a solution of x 2 + 6x + 9 = 0., 34. Putting x = 2 in the given equation, we get, LHS = 3 # (2) 2 + 13 # (2) + 14 = (12 26 + 14) = 0 = RHS., , Hence, x = 2 is a solution of 3x 2 + 13x + 14 = 0.
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Quadratic Equations, , 241, , 1, 35. Since x = 2 is a solution of 3x 2 2kx 3 0, we have, 3 #c, , , 1 2, , m + 2k # c 1 m 3 = 0, 2, 2, , 9, 3, 3 , ·, k 3 0 & k c 3m , 4, 4, 4, , 36. 2x 2 x 6 0 & 2x 2 4x 3x 6 0 & 2x (x 2) 3 (x 2) 0, 3, ·, & (x 2) (2x 3) 0 & x 2 or x , 2, 37. 3 3 x 2 10x 3 0 & 3 3 x 2 9x x 3 0, & 3 3 x (x 3 ) (x 3 ) 0 & (x 3 ) (3 3 x 1) 0, 1, ·, & x 3 or x , 3 3, 38. D = (8) 2 4 # 2 # k = (64 8k) ., , , D 0 & 64 8k 0 & 8k 64 & k 8., , 39. The given equation is of the form ax 2 + bx + c = 0, where, a = p, b = 2 5 p and c = 15., , , D = (b 2 4ac) = (20p 2 60p) ., , Y 0], So, D 0 & 20p 2 60p 0 & 20p (p 3) 0 & p 3. [a p , 40. Since 1 is a root of each of the equations ay 2 + ay + 3 = 0 and y 2 + y + b = 0, we have, , , , , a # (1) 2 + a # 1 + 3 = 0 and 1 2 + 1 + b = 0, 3, a a 3 0 and 2 b 0 & a , and b 2., 2, 3, ab = c m # (2) = 3., 2, , 41. Let the other zero be ., Sum of the zeros of x 2 4x + 1 is 4., , , (2 3 ) 4 & 4 (2 3 ) (2 3 ) ., , 1, 42. Let the roots of 3x 2 10x + k = 0 be and ·, k, Then, product of roots = 3 ·, k, k, b # 1 l & 1 & k 3., , , 3, 3, 43. Given equation is px 2 2px + 6 = 0., This is of the form ax 2 + bx + c = 0, where a = p, b = 2p and c = 6., , , D = (b 2 4ac) = 4p 2 24p., , For equal roots, we have D = 0., , , Y 0], 4p 2 24p 0 & 4p (p 6) 0 & p 6. [a p , , 44. D = ( 4k) 2 4k = 16k 2 4k = 4k (4k 1) .
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242, , Secondary School Mathematics for Class 10, For equal roots, we must have D = 0., Now, D 0 & 4k (4k 1) 0 & k 0 or 4k 1 0 & k 0 or k , , 45. D = (3k) 2 9k = 9k 2 9k = 9k (k 1) ., For equal roots, we must have D = 0., Now, D 0 & 9k (k 1) 0 & k 0 or k 1 0 & k 0 or k 1., 46. x 2 ( 3 + 1) x + 3 = 0, x 2 3 x x 3 0 & x (x 3 ) (x 3 ) 0, , , , (x 3 ) (x 1) 0 & x 3 or x 1., , , 47. 2x, , 2, , ax a 2 0 & 2x 2 2ax ax a 2 0, & 2x (x a) a (x a) 0, & (x a)(2x a) 0 & x a or x , , 48. 3x 2 + 5 5 x 10 = 0, , 49., , , , 3x 2 + 6 5 x 5 x 10 = 0, , , , 3x (x + 2 5 ) 5 (x + 2 5 ) = 0, , , , (x 2 5 )(3x 5 ) 0 & x 2 5 or x , , 3 x 2 + 10x 8 3 = 0, , , 3 x 2 + 12x 2x 8 3 = 0, , , , 3 x (x + 4 3 ) 2 (x + 4 3 ) = 0, , , 50., , 5, ·, 3, , (x 4 3 )( 3 x 2) 0 & x 4 3 or x , , 2 ·, 3, , 3 x2 2 2 x 2 3 = 0, , , 3 x2 3 2 x 2 x 2 3 = 0, , , , 3 x (x 6 ) 2 (x 6 ) = 0, , , , (x 6 )( 3 x 2 ) 0 & x 6 or x , , 2, 3, , ·, , 51. 4 3 x 2 + 5x 2 3 = 0, , , 4 3 x 2 + 8x 3x 2 3 = 0, , , , 4x ( 3 x + 2) 3 ( 3 x 2) = 0, , , , ( 3 x 2)(4x 3 ) 0 & x , , 3, 2, ·, or x , 4, 3, , 52. 4x 2 + 4bx + (b 2 a 2) = 0, , , 4x 2 + 2 (b + a) x + 2 (b a) x + (b 2 a 2) = 0, , , , 2x {2x + (b + a)} + (b a) {2x + (b + a)} = 0, , , , {2x + (b + a)} {2x + (b a)} = 0, (b a), ab·, x, or x , 2, 0, , , , a·, 2, , 1·, 4
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Quadratic Equations, 53. x 2 + 5x (a + 3) (a 2) = 0, , , x 2 + (a + 3) x (a 2) x (a + 3) (a 2) = 0, , , , x {x + (a + 3)} (a 2) {x + (a + 3)} = 0, , , , {x + (a + 3)} {x (a 2)} = 0, , , , x = (a + 3) or x = (a 2) ., , 54. x 2 + 6x (a + 4) (a 2) = 0, , , x 2 + (a + 4) x (a 2) x (a + 4) (a 2) = 0, , , , x {x (a + 4)} (a 2) {x + (a + 4)} = 0, , , , {x + (a + 4)} {x (a 2)} = 0, , , , x = (a + 4) or x = (a 2) ., , 55. x 2 4ax + (2a + b) (2a b) = 0, , , x 2 (2a + b) x (2a b) x + (2a + b) (2a b) = 0, , , , x {x (2a + b)} (2a b) {x (2a + b)} = 0, , , , {x (2a + b)} {x (2a b)} = 0, , , , x = (2a + b) or x = (2a b) ., , , , 243
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244, , Secondary School Mathematics for Class 10, , 5, , Arithmetic Progression, , S RAMANUJAN (1887–1920), Srinivasa Aaiyangar Ramanujan was, born in Erode, a small village near, Chennai in 1887. His father was a, clerk at a cloth shop and his mother, was a housewife. On the basis of his, good school work, he was admitted to, the University of Madras. But, in first, year he passed in mathematics and, failed in other subjects. What earned, him fame, reputation and success was, his research work on numbers., HIS RESEARCH WORK AND CONTRIBUTION ON NUMBERS, , (i) In 1900, he began to work on Geometric and Arithmetic Series., He gave new methods to solve Cubic and Quadratic Equations., 1, (ii) He investigated the series / n and found Bernoulli’s Numbers., (iii) In 1908, he studied Continued Fractions and Divergent Series., (iv) In 1911, he produced a research paper on Bernoulli’s Numbers., In 1918, he was given a fellowship by Trinity College, Cambridge for, 7 years. He was a mathematical genius, despite his lack of university, education., He died at an early age of 33 years only., SEQUENCE Some numbers arranged in a definite order, according to a definite, rule, are said to form a sequence., , The number occurring at the nth place of a sequence is called its nth, term, denoted by Tn or an ., Example, , Consider the rule, Tn (2n 1) ., Putting n = 1, 2, 3, 4, 5, …, we get, T1 3, T2 5, T3 7, T4 9, T5 11, and so on., 244
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Arithmetic Progression, , 245, , Thus, the numbers 3, 5, 7, 9, 11, … form a sequence., In this sequence, the first term is 3, the second term is 5, the third term, is 7, and so on., ARITHMETIC PROGRESSION (AP), , A sequence in which each term differs from its preceding term by a constant is, called an arithmetic progression, written as AP., This constant is called the common difference of the AP., EXAMPLE 1, , Show that the progression 8, 11, 14, 17, 20, ... is an AP. Find its first, term and the common difference., , SOLUTION, , The given progression is 8, 11, 14, 17, 20, ... ., Clearly, (11 8) (14 11) (17 14) (20 17) 3 (constant)., Thus, each term differs from its preceding term by 3., So, the given progression is an AP., Its first term= 8 and common difference = 3., , EXAMPLE 2, , SOLUTION, , Show that the progression 11, 6, 1, –4, –9, ... is an AP. Find its first, term and the common difference., Clearly, (6 11) (1 6) (4 1) (9 4) 5 (constant)., Thus, each term differs from its preceding term by –5., So, the given progression is an AP., Its first term = 11 and common difference = –5., , EXAMPLE 3, , Show that each of the following progressions is an AP. Find the, common difference and the next term of each., (i) 7 , 28 , 63 , …, , SOLUTION, , [CBSE 2014], , (ii) 18 , 50 , 98 , …, , [CBSE 2012], , (i) We may write the given terms as, 7 , 4 #7 , 9#7 , …, i.e., 7 , 2 7 , 3 7 , …, Clearly, 2 7 7 3 7 2 7 7 ., So, the given progression is an AP with common, difference 7 ., Next term is 4 7 16 #7 112 .
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246, , Secondary School Mathematics for Class 10, , (ii) We may write the given terms as, 9 # 2 , 25 # 2 , 49 # 2 , …, i.e., 3 2 , 5 2 , 7 2 , … ., Clearly, (5 2 3 2 ) (7 2 5 2 ) 2 2 ., So, the given progression is an AP with common, difference 2 2 ., It is clear that the next term is 9 2 9 # 9 # 2 162 ., EXAMPLE 4, SOLUTION, , Find a and b such that the numbers a, 9, b, 25 form an AP. [CBSE 2014], Since the numbers a, 9, b, 25 form an AP, we have, 9 a b 9 25 b., Now, b 9 25 b & 2b 34 & b 17., And, 9 a b 9 & a b 18 & a 17 18 & a 1., Hence, a 1 and b 17., , ARITHMETIC SERIES, , By adding the terms of an AP, we get the corresponding arithmetic series., On adding the terms of the AP 3, 7, 11, 15, 19, ..., we get the, arithmetic series (3 + 7 + 11 + 15 + 19 + ...)., , Example, , TO FIND THE GENERAL TERM OF AN AP, THEOREM 1, , If the first term of an AP is a and its common difference is d then, show that its nth term is given by, Tn a (n 1) d., , PROOF, , In the given AP, we have, first term = a and common difference = d., So, the given AP may be written as, a, a d, a 2d, a 3d, a 4d, … ., In this AP, we have, first term, T1 a a (1 1) d;, second term, T2 a d a (2 1) d;, third term, T3 a 2d a (3 1) d;, fourth term, T4 a 3d a (4 1) d;, …, , …, , …, , …, , …, , …, , …, , …, , …, , …, , …, , …
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Arithmetic Progression, , 247, , nth term, Tn a (n 1) d., Hence, Tn a (n 1) d., , , NOTE, , The nth term of an AP is called its general term., General term of an AP, , In an AP with first term a and common difference d,, the nth term is given by, Tn a (n 1) d., , SOLVED EXAMPLES, EXAMPLE 1, , Find the (i) nth term and (ii) 16th term of the AP 3, 5, 7, 9, 11, ... ., , SOLUTION, , The given AP is 3, 5, 7, 9, 11, ... ., Its first term = 3 and common difference (5 3) 2., , , a 3 and d (5 3) 2., , (i) Its nth term is given by, Tn a (n 1) d, 3 (n 1)# 2 (2n 1), , , [a a 3 and d 2] ., , nth term (2n 1) ., , (ii) 16th term of the given AP is, T16 a (16 1) d (a 15d), (3 15 # 2) 33, , , [a a 3 and d 2] ., , 16th term = 33., , EXAMPLE 2, , Find the (i) nth term and (ii) 12th term of the AP 14, 9, 4, –1, –6, … ., , SOLUTION, , The given AP is 14, 9, 4, –1, –6, … ., Its first term = 14 and common difference (9 14) 5., , , a 14 and d 5., , (i) The nth term of the AP is given by, Tn a (n 1) d, 14 (n 1)#(5) (19 5n) [a a 14 and d 5]., , , nth term (19 5n)., , (ii) 12th term of the given AP is, T12 a (12 1) d (a 11d)
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248, , Secondary School Mathematics for Class 10, , 14 11#(5) 41, , , [a a 14 and d 5]., , 12th term 41., , EXAMPLE 3, , 1, 1, Find the 105th term of the AP 4, 4 , 5, 5 , 6, … ., 2, 2, , SOLUTION, , 9, 11, The given AP is 4, , 5, , 6, … ., 2, 2, 9, 1, Its first term = 4 and common difference a 4k ·, 2, 2, 1, a 4 and d ·, 2, The nth term of the AP is given by, Tn a (n 1) d, T105 a (105 1) d (a 104d), 1, 1, 4 a104 # k 56 :a a 4 and d D ·, 2, 2, Hence, 105th term = 56., , EXAMPLE 4, , Find the 25th term of the AP 5,, , SOLUTION, , The given AP is 5,, , 5, 5, , 0, , … ., 2, 2, , [CBSE 2015], , 5, 5, , 0, , … ., 2, 2, , 5, 5, Its first term 5 and common difference a 0k ·, 2, 2, 5, a 5 and d ·, 2, T25 a (25 1) d, a 24d (5) a24 # 5 k 5 60 55., 2, Hence, 25th term = 55., EXAMPLE 5, , SOLUTION, , If the nth term of an AP is (5n 2), find its (i) first term, (ii) common, difference and (iii) 19th term., Tn (5n 2) (given), T1 (5 #1 2) 3 and T2 (5 # 2 2) 8., Thus, we have, (i) first term = 3., (ii) common difference (T2 T1) (8 3) 5.
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Arithmetic Progression, , 249, , (iii) 19th term a (19 1) d, where a 3 and d 5, (3 18 # 5) 93., EXAMPLE 6, , If the seventh term of an AP is, 63rd term., , SOLUTION, , 1, 1, and its ninth term is , find its, 7, 9, [CBSE 2014], , Let a be the first term and d be the common difference of the, given AP. Then,, 1, & a 6d 19, 9, 1, 1, T9 & a 8d , 7, 7, T7 , , … (i), … (ii), , On subtracting (i) from (ii), we get, 1 1, 2, 2 1 ·, 2d a k , & d a12 # 63, k, 7 9, 63, 63, 1, Putting d , in (i), we get, 63, 1 1, 2 1, 2 76 1 ·, k, & a 21, & a a19 21, k, k a, 63, 9, 9, 63, 63, 1, 1, ·, Thus, a , and d , 63, 63, a a6 #, , , , T63 a (63 1) d (a 62d), a 1 62 # 1 k a 1 62 k 1., 63, 63, 63 63, , Hence, 63rd term of the given AP is 1., EXAMPLE 7, , The sum of the 4th and 8th terms of an AP is 24 and the sum of its, 6th and 10th terms is 44. Find the first three terms of the AP., [CBSE 2008, ’09, ’12], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, T4 T8 24 & (a 3d) (a 7d) 24, , & 2a 10d 24, & a 5d 12, and T6 T10 44 & (a 5d) (a 9d) 44, & 2a 14d 44, & a 7d 22., , … (i), , … (ii)
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250, , Secondary School Mathematics for Class 10, , On solving (i) and (ii), we get a 13 and d 5., , , first three terms of the given AP are –13, –8 and –3., , EXAMPLE 8, , Which term of the AP 5, 9, 13, 17, ... is 81?, , SOLUTION, , In the given AP, we have, , [CBSE 2005C], , first term = 5 and common difference (9 5) 4., a 5 and d 4., Let its nth term be 81. Then,, Tn 81 & a (n 1) d 81, , & 5 (n 1)# 4 81, & 4n 80 & n 20., , [a a 5 and d 4], , Hence, the 20th term of the given AP is 81., EXAMPLE 9, , SOLUTION, , Which term of the AP 3, 15, 27, 39, ... will be 120 more than its, 21st term?, [CBSE 2009], Here a 3 and d (15 3) 12., 21st term is given by, T21 a (21 1) d a 20d (3 20 #12) 243., Required term (243 120) 363., Let it be nth term. Then,, Tn 363 & a (n 1) d 363, , & 3 (n 1)#12 363, & 12n 372 & n 31., Hence, 31st term is the required term., EXAMPLE 10, , Is 51 a term of the AP 5, 8, 11, 14, …?, , SOLUTION, , Here a 5 and d (8 5) 3., Let the nth term of the given AP be 51. Then,, Tn 51 & a (n 1) d 51, , & 5 (n 1)# 3 51, , [a a 5 and d 3], , & 3n 49 & n 16 13 ·, But, the number of terms cannot be a fraction., , , 51 is not a term of the given AP.
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Arithmetic Progression, , 251, , EXAMPLE 11, , How many terms are there in the AP 7, 11, 15, …, 139?, , SOLUTION, , In the given AP, we have a 7 and d (11 7) 4., Suppose there are n terms in the given AP. Then,, Tn 139 & a (n 1) d 139, , & 7 (n 1)# 4 139, & 4n 136 & n 34., Hence, there are 34 terms in the given AP., EXAMPLE 12, , Find the middle term of the AP 213, 205, 197, …, 37., , SOLUTION, , Here a 213 and d (205 213) 8., , [CBSE 2015], , Let the given AP contain n terms. Then,, Tn 37 & a (n 1) d 37, , & 213 (n 1)#(8) 37, & 221 8n 37, & 8n 221 37 184 & n 23., Thus, the given AP contains 23 terms., 1, (23 1)th term = 12th term., 2, So, middle term T12 (a 11d), , , its middle term , , 213 11#(8) 213 88 125., Hence, the middle term is 125., EXAMPLE 13, , Which term of the AP 24, 21, 18, 15, ... is the first negative term?, , SOLUTION, , Here a 24 and d (21 24) 3., Let the nth term of the given AP be the first negative term., Then, Tn 0 & {a (n 1) d} 0, , & {24 (n 1)#(3)} 0, & (27 3n) 0 & 27 3n, & 3n 27 & n 9., , , n 10., , Hence, the 10th term is the first negative term of the given AP., EXAMPLE 14, , For what value of n are the nth terms of the following two APs the, same 13, 19, 25, … and 69, 68, 67, …? Also, find this term., [CBSE 2006C, ’11]
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252, SOLUTION, , Secondary School Mathematics for Class 10, , Let nth terms of the given progressions be tn and Tn respectively., The first AP is 13, 19, 25, … ., Let its first term be a and common difference be d. Then,, a 13 and d (19 13) 6., So, its nth term is given by, tn a (n 1) d, tn 13 (n 1)# 6, tn 6n 7, , … (i), , The second AP is 69, 68, 67, … ., Let its first term be A and common difference be D. Then,, A 69 and D (68 69) 1., So, it nth term is given by, Tn A (n 1)# D, Tn 69 (n 1)#(1), Tn 70 n, , … (ii), , Now, tn Tn & 6n 7 70 n, , & 7n 63 & n 9., Hence, the 9th term of each AP is the same., This term 70 9 61 [a Tn (70 n)] ., EXAMPLE 15, , If seven times the 7th term of an AP is equal to eleven times the 11th, term then what will be its 18th term?, [CBSE 2017], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, 7T7 11T11, 7 (a 6d) 11(a 10d) 7a 42d 11a 110d, , , , 4a 68d 0 & a 17d 0, a (18 1) d 0 & T18 0., , Hence, the 18th term of the given AP is zero., EXAMPLE 16, , If the nth term of a progression be a linear expression in n then, prove that this progression is an AP., , SOLUTION, , Let the nth term of a given progression be given by, Tn an b, where a and b are constants.
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Arithmetic Progression, , 253, , Then, Tn 1 a (n 1) b [(an b) a] ., , , Tn Tn 1 (an b) [(an b a] a, which is a constant., , Hence, the given progression is an AP., EXAMPLE 17, , In a given AP if pth term is q and the qth term is p then show that the, nth term is (p q n) ., [CBSE 2008], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, Tp a (p 1) d and Tq a (q 1) d., Now, Tp q and Tq p (given)., a (p 1) d q, , … (i), , and a (q 1) d p, , … (ii), , , , On subtracting (i) from (ii), we get, (q p) d (p q) & d 1., Putting d 1 in (i), we get a (p q 1) ., Thus,, , , a (p q 1) and d 1., , nth term a (n 1) d (p q 1) (n 1)#(1), p q n., , Hence, nth term (p q n) ., EXAMPLE 18, , If m times the mth term of an AP is equal to n times the nth term and, [CBSE 2004C, ’08], m ! n, show that its (m n)th term is zero., , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, Tm a (m 1) d and Tn a (n 1) d., Now, (m · Tm) (n · Tn) & m · {a (m 1) d} n · {a (n 1) d}, , & a · (m n) {(m 2 n 2) (m n)} · d 0, & (m n) · {a (m n 1)} d., & (m n) · Tm n 0, & Tm n 0 [a (m n) ! 0] ., Hence, the (m n)th term is zero., EXAMPLE 19, , If the pth, qth and rth terms of an AP be a, b, c respectively then, show that, a (q r) b (r p) c (p q) 0.
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254, SOLUTION, , Secondary School Mathematics for Class 10, , Let x be the first term and d be the common difference of the, given AP. Then,, Tp x (p 1) d, Tq x (q 1) d and Tr x (r 1) d., , , x (p 1) d a, , … (i), , x (q 1) d b, , … (ii), , x (r 1) d c., , … (iii), , On multiplying (i) by (q r), (ii) by (r p) and (iii) by (p q),, and adding, we get, a (q r) b (r p) c (p q), x · {(q r) (r p) (p q)}, d · {(p 1) (q r) (q 1) (r p) (r 1) (p q)}, (x # 0) (d # 0) 0., Hence, a (q r) b (r p) c (p q) 0., EXAMPLE 20, , SOLUTION, , 1, 1, If the mth term of an AP be n and its nth term be m then show that, its (mn)th term is 1., Let a be the first term and d be the common difference of the, given AP. Then,, Tm a (m 1) d and Tn a (n 1) d., 1, 1, Now, Tm n and Tn m (given)., 1, a (m 1) d n, 1, and a (n 1) d m, On subtracting (ii) from (i), we get, mn, 1 1, 1, (m n) d an m k ` mn j & d mn ·, 1, Putting d mn in (i), we get, (m 1) 1, 1 (m 1), 1, a mn n & a 'n mn 1 mn ·, 1, 1, Thus, a mn and d mn ·, (mn)th term a (mn 1) d, (mn 1), 1 :a a 1 D, ' 1 , mn, mn, mn, 1., Hence, the (mn)th term of the given AP is 1., , … (i), … (ii)
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Arithmetic Progression, , 255, , TO FIND THE nTH TERM FROM THE END OF AN AP, THEOREM 2, , If a be the first term, d be the common difference and l be the last, term of a given AP then show that its nth term from the end is, {l (n 1) d} ., , PROOF, , We may write the given AP as, a, (a d), (a 2d), …, (l 2d), (l d), l., Thus, we have, last term l l (1 1) d;, 2nd term from the end (l d) {l (2 1) d};, 3rd term from the end (l 2d) {l (3 1) d};, 4th term from the end (l 3d) {l (4 1) d} ., , , nth term from the end {l (n 1) d} ., , nth term from the end of an AP, , Let a be the first term, d be the common difference, and l be the last term of an AP. Then,, nth term from the end {l (n 1) d} ., EXAMPLE 21, , Find the 11th term from the end of the AP 10, 7, 4, …, 62., , SOLUTION, , We have, a 10, d (7 10) 3, l 62 and n 11., , , 11th term from the end {l (n 1)# d}, {62 (11 1)#(3)}, (62 30) 32., , Hence, the 11th term from the end of the given AP is –32., WORD PROBLEMS, EXAMPLE 22, , How many three-digit numbers are divisible by 7?, , SOLUTION, , All 3-digit numbers divisible by 7 are, 105, 112, 119, …, 994., Clearly, these numbers form an AP with, a 105, d (112 105) 7 and l 994., Let it contain n terms. Then,, Tn 994 & a (n 1)# d 994, , [CBSE 2013]
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256, , Secondary School Mathematics for Class 10, , & 105 (n 1)#7 994, & 98 7n 994 & 7n 896 & n 128., Hence, there are 128 three-digit numbers divisible by 7., EXAMPLE 23, , How many multiples of 4 lie between 10 and 250?, , SOLUTION, , All multiples of 4 lying between 10 and 250 are, , [CBSE 2011], , 12, 16, 20, 24, …, 248., Clearly, these numbers form an AP with, a 12, d (16 12) 4 and l 248., Let it contain n terms. Then,, Tn 248 & a (n 1) d 248, , & 12 (n 1)# 4 248 & 4n 240 & n 60., Hence, there are 60 multiples of 4 lying between 10 and 250., EXAMPLE 24, , SOLUTION, , A sum of ` 1000 is invested at 8% per annum simple interest., Calculate the interest at the end of 1, 2, 3, … years. Is the sequence, of interest an AP? Find the interest at the end of 30 years., Here P ` 1000, R 8% per annum. Let In be the SI at the end, of n years., Then, In , , P # R # n 1000 # 8 # n , `a, k ` (80n) ., 100, 100, , … (i), , Putting n 1, 2, 3, … in (i), we get, I1 ` (80 #1) ` 80, I2 ` (80 # 2) ` 160,, I3 ` (80 # 3) ` 240, …, Thus, I1 ` 80, I2 ` 160, I3 ` 240, I4 ` 320, … ., Thus, ` 80, ` 160, ` 240, ` 320, … form an AP with, a 80 and d (160 80) 80., , , T30 a (30 1)# d, 80 29 # 80 80 2320 2400., , Hence, the interest at the end of 30 years is ` 2400., EXAMPLE 25, , Tanvy joined her job in a company in the year 2015 on a monthly, salary of ` 40000 with an annual increment of ` 2500. In which, year will she get ` 65000 as monthly salary?
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Arithmetic Progression, , 257, , Monthly salary received by Tanvy in 2015, 2016, 2017, 2018, …, is respectively ` 40000, ` 42500, ` 45000, ` 47500, … ., This is an AP with a 40000, d 2500 and l 65000., Let the number of terms of this AP be n. Then,, , SOLUTION, , Tn 65000 & a (n 1) d 65000, & 40000 (n 1)# 2500 65000, & (n 1)# 2500 65000 40000 25000, 25000, & (n 1) 2500 10 & n 11., Thus, the 11th annual salary received by Tanvy will be ` 65000., Thus, after 10 years, i.e., in the year 2025, her annual salary, will be ` 65000., EXAMPLE 26, , In a new year, Reenu saved ` 50 in the first week and then increased, her weekly savings by ` 17.50. If in the nth week, her weekly saving, becomes ` 207.50, find the value of n., , SOLUTION, , Weekly savings by Reenu in successive weeks are ` 50, ` 67.50,, ` 85, ` 102.50, …, ` 207.50., This is clearly an AP with a 50, d 17.50 and l 207.50., Let the number of terms of this AP be n. Then,, Tn 207.50 & a (n 1) d 207.50, & 50 (n 1)#17.50 207.50, , & (n 1)#17.50 207.50 50 157.50, 157.50 15750, & (n 1) 17.50 1750 9 & n 10., Hence, Reenu’s weekly savings will be ` 207.50 in 10th week., f, , EXERCISE 5A, , 1. Show that each of the progressions given below is an AP. Find the first, term, common difference and next term of each., (i) 9, 15, 21, 27, … ., (ii) 11, 6, 1, –4, … ., 5 2 1, (iii) 1, , , , … ., 6 3 2, (iv), , 2 , 8 , 18 , 32 , … ., , (v), , 20 , 45 , 80 , 125 , … .
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258, , Secondary School Mathematics for Class 10, , 2. Find:, (i) the 20th term of the AP 9, 13, 17, 21, … ., (ii) the 35th term of the AP 20, 17, 14, 11, … ., (iii) the 18th term of the AP 2 , 18 , 50 , 98 , … ., 3 5 7 9, (iv) the 9th term of the AP , , , , … ., 4 4 4 4, (v) the 15th term of the AP –40, –15, 10, 35, … ., 3 1, 1, 3. (i) Find the 37th term of the AP 6, 7 , 9 , 11 , … ., 4 2, 4, 1, 1, (ii) Find the 25th term of the AP 5, 4 , 4, 3 , 3, … ., 2, 2, 4. Find the value of p for which the numbers 2p 1, 3p 1, 11 are in AP., Hence, find the numbers., [CBSE 2017], 5. Find the nth term of each of the following APs:, (i) 5, 11, 17, 23, … ., , (ii) 16, 9, 2, –5, … ., 6. If the nth term of a progression is (4n 10) show that it is an AP. Find its, (i) first term, (ii) common difference, and (iii) 16th term., 7. How many terms are there in the AP 6, 10, 14, 18, …, 174?, 8. How many terms are there in the AP 41, 38, 35, …, 8?, 1, 9. How many terms are there in the AP 18, 15 , 13, …, 47?, 2, 10. Which term of the AP 3, 8, 13, 18, … is 88?, 11. Which term of the AP 72, 68, 64, 60, … is 0?, 5, 1 1, 12. Which term of the AP , 1, 1 , 1 , … is 3?, 6, 6 3, 13. Which term of the AP 21, 18, 15, … is –81?, 14. Which term of the AP 8, 14, 20, 26, ... will be 72 more than its 41st term?, [CBSE 2017], , 15. Which term of the AP 5, 15, 25, … will be 130 more than its 31st term?, [CBSE 2006C], , 16. If the 10th term of an AP is 52 and 17th term is 20 more than its 13th, term, find the AP., [CBSE 2009C, 2017], 17. Find the middle term of the AP 6, 13, 20, …, 216., , [CBSE 2015], , 18. Find the middle term of the AP 10, 7, 4, …, (–62)., , [CBSE 2009C], , 2, 4, 1, 19. Find the sum of two middle most terms of the AP , 1, , …, 4 ·, 3, 3, 3, [CBSE 2013C], , 20. Find the 8th term from the end of the AP 7, 10, 13, ..., 184., , [CBSE 2005]
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Arithmetic Progression, , 21. Find the 6th term from the end of the AP 17, 14, 11, ..., (–40)., , 259, [CBSE 2005], , 22. Is 184 a term of the AP 3, 7, 11, 15, …?, 23. Is –150 a term of the AP 11, 8, 5, 2, …?, 24. Which term of the AP 121, 117, 113, … is its first negative term?, 3, 1, 1, 25. Which term of the AP 20, 19 , 18 , 17 , … is its first negative term?, 2, 4, 4, , [CBSE 2017], , 26. The 7th term of an AP is – 4 and its 13th term is –16. Find the AP., [CBSE 2014], , 27. The 4th term of an AP is zero. Prove that its 25th term is triple its 11th, term., [CBSE 2005], 28. If the sixth term of an AP is zero then show that its 33rd term is three, times its 15th term., [CBSE 2017], 29. The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP, is 34. Find its common difference., [CBSE 2015], 30. The 9th term of an AP is –32 and the sum of its 11th and 13th terms, is –94. Find the common difference of the AP., [CBSE 2015], 31. Determine the nth term of the AP whose 7th term is –1 and 16th term, is 17., [CBSE 2014], 32. If 4 times the 4th term of an AP is equal to 18 times its 18th term then, find its 22nd term., [CBSE 2012], 33. If 10 times the 10th term of an AP is equal to 15 times the 15th term,, show that its 25th term is zero., 34. Find the common difference of an AP whose first term is 5 and the sum, of its first four terms is half the sum of the next four terms. [CBSE 2012], 35. The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is, 1 less than twice its 8th term, find the AP., [CBSE 2014], 36. For what value of n, the nth terms of the arithmetic progressions, 63, 65, 67, … and 3, 10, 17, … are equal?, [CBSE 2008, ’17], 37. The 17th term of AP is 5 more than twice its 8th term. If the 11th term of, the AP is 43, find its nth term., [CBSE 2012], 38. The 24th term of an AP is twice its 10th term. Show that its 72nd term is, 4 times its 15th term., [CBSE 2013], 39. The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is, 19, find the AP., [CBSE 2013], 40. If the pth term of an AP is q and its qth term is p then show that its, (p q)th term is zero.
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260, , Secondary School Mathematics for Class 10, , 41. The first and last terms of an AP are a and l respectively. Show that the, sum of the nth term from the beginning and the nth term from the end, is (a l) ., 42. How many two-digit numbers are divisible by 6?, , [CBSE 2011], , 43. How many two-digit numbers are divisible by 3?, , [CBSE 2012], , 44. How many three-digit numbers are divisible by 9?, , [CBSE 2013], , 45. How many numbers are there between 101 and 999, which are divisible, by both 2 and 5?, [CBSE 2014], 46. In a flower bed, there are 43 rose plants in the first row, 41 in the second,, 39 in the third, and so on. There are 11 rose plants in the last row. How, many rows are there in the flower bed?, 47. A sum of ` 2800 is to be used to award four prizes. If each prize after, the first is ` 200 less than the preceding prize, find the value of each of, the prizes., 48. Find how many integers between 200 and 500 are divisible by 8., [CBSE 2017], , ANSWERS (EXERCISE 5A), , (ii) a 11, d 5, T5 9, 1, 1, (iii) a 1, d , T5 , (iv) a 2 , d 2 , T5 50, 6, 3, (v) a 20 , d 5 , T5 180, 3, 2. (i) 85, (ii) –82, (iii) 35 2, (iv) 4, (v) 310, 4, 3. (i) 69, (ii) –7, 4. p 2; 3, 7, 11, 5. (i) (6n 1), (ii) (23 7n), 1. (i) a 9, d 6, T5 33, , 6. (i) –6, 11. 19th, , (ii) 4, 12. 14th, , 17. 111, , (iii) 54, 13. 35th, , 18. –26, , 22. No 23. No 24. 32nd, 29. 3, , 30. –5, , 36. n 13, , 7. 43, , 8. 12, , 9. 27, , 14. 53rd 15. 44th 16. 7, 12, 17, 22, …, 19. 3, , 20. 163, , 25. 28th, , 26. 8, 6, 4, 2, 0, …, , 21. –25, , 31. Tn (2n 15), , 32. 0, , 37. Tn (4n 1), , 39. 3, 5, 7, 9, … 42. 15, , 43. 30 44. 100 45. 89 46. 17, , 34. 2, , 1. (iv) Given progression is 2 , 4 # 2 , 9 # 2 , 16 # 2 , …, T5 5 2 5 # 5 # 2 50 ., , 35. 1, 5, 9, 13, …, , 47. ` 1000, ` 800, ` 600, ` 400, , HINTS TO SOME SELECTED QUESTIONS, , i.e., 2 , 2 2 , 3 2 , 4 2 , …, , 10. 18th, , 48. 37
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262, , Secondary School Mathematics for Class 10, Also, a 10d 43., , … (ii), , Solve (i) and (ii), we get a 3 and d 4., 38. T24 2T10 & a 23d 2 (a 9d) & a 5d 0 & a 5d., , … (i), , T72 a 71d 5d 71d 76d., T15 a 14d 5d 14d 19d., Hence, T72 4 #T15 ., 40. Tp q & a (p 1) d q., , … (i), , Tq p & a (q 1) d p., , … (ii), , On subtracting (ii) from (i), we get (p q) d (q p) & d 1., Putting d 1 in (i), we get a p q 1., , , Tp q a (p q 1) d (p q 1) (p q 1) 0., , 41. Required sum {a (n 1) d} {l (n 1) d} (a l) ., 42. The given numbers are 12, 18, 24, 30, …, 96., Let their number be n. Then, 12 (n 1)# 6 96. Find n., 44. The given numbers are 108, 117, 126, …, 999., 45. Clearly, each given number is divisible by 10., So, these numbers are 110, 120, 130, …, 990., 46. Let there be n rows. Then, the number of plants in various rows are 43, 41, 39, …, 11, respectively., Here, a 43, d 2 and Tn 11. Find n., 47. Let the values of these prizes be ` x, ` (x 200), ` (x 400) and ` (x 600) . Their sum, is ` 2800., , ARITHMETIC MEAN, ARITHMETIC MEAN (AM) BETWEEN TWO NUMBERS, , If a, A, b are in AP we say that A is the arithmetic mean between a and b and it is, abbreviated as AM., TO FIND THE ARITHMETIC MEAN BETWEEN TWO NUMBERS, , Let the given numbers be a and b., Let A be the AM between a and b. Then,, a, A, b are in AP & (A a) (b A), , & A 12 (a b) .
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Arithmetic Progression, AM between a and b , , , , 263, , 1 , (a b) ., 2, , Arithmetic mean between a and b is, , 1 , (a b)., 2, , SOLVED EXAMPLES, EXAMPLE 1, , Find the AM between, (i) 13 and 19, , (ii) (a b) and (a b), , (i) AM between 13 and 19 , , SOLUTION, , 1, (13 19) 16., 2, , (ii) AM between (a b) and (a b) , EXAMPLE 2, , 1, [(a b) (a b)] a., 2, , If the numbers (2n 1), (3n 2) and (6n 1) are in AP, find n and, hence find these numbers., [CBSE 2013C], Since (2n 1), (3n 2) and (6n 1) are in AP, we have, , SOLUTION, , (3n 2) (2n 1) (6n 1) (3n 2), n 3 3n 3 & 2n 6 & n 3., Hence, n 3 and these numbers are 5, 11 and 17., AN IMPORTANT RESULT, , It is always convenient to make a choice of, (i) 3 numbers in AP as (a d), a, (a d);, (ii) 4 numbers in AP as (a 3d), (a d), (a d), (a 3d);, (iii) 5 numbers in AP as (a 2d), (a d), a, (a d), (a 2d) ., EXAMPLE 3, , SOLUTION, , The sum of three numbers in AP is 21 and their product is 231. Find, the numbers., Let the required numbers be (a d), a and (a d) ., Then, (a d) a (a d) 21 & 3a 21 & a 7., And, (a d) · a · (a d) 231 & a (a 2 d 2) 231, , & 7 (49 d 2) 231, [a a 7], 2, & 7d 343 231 112, & d 2 16 & d !4.
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264, , Secondary School Mathematics for Class 10, , Thus, a 7 and d !4., Hence, the required numbers are (3, 7, 11) or (11, 7, 3)., EXAMPLE 4, , SOLUTION, , Find four numbers in AP whose sum is 20 and the sum of whose, squares is 120., Let the required numbers be (a 3d), (a d), (a d) and (a 3d) ., Then, (a 3d) (a d) (a d) (a 3d) 20, , , 4a 20 & a 5., , And, (a 3d) 2 (a d) 2 (a d) 2 (a 3d) 2 120, , , 4 (a 2 5d 2) 120, , (a 2 5d 2) 30 & 25 5d 2 30, , , [a a 5], , 5d 2 5 & d 2 1 & d !1., , Thus, a 5 and d !1., Hence, the required numbers are (2, 4, 6, 8) or (8, 6, 4, 2)., , f, , EXERCISE 5B, , 1. Determine k so that (3k 2), (4k 6) and (k 2) are three consecutive, terms of an AP., [CBSE 2009C], 2. Find the value of x for which the numbers (5x 2), (4x 1) and (x 2) are, in AP., 3. If (3y 1), (3y 5) and (5y 1) are three consecutive terms of an AP then, find the value of y., [CBSE 2014], 4. Find the value of x for which (x 2), 2x, (2x 3) are three consecutive, terms of an AP., [CBSE 2009C], 5. Show that (a b) 2, (a 2 b 2) and (a b) 2 are in AP., 6. Find three numbers in AP whose sum is 15 and product is 80., HINT, , Let the numbers be (a d), a, (a d) ., , 7. The sum of three numbers in AP is 3 and their product is –35. Find the, numbers., 8. Divide 24 in three parts such that they are in AP and their product is 440., 9. The sum of three consecutive terms of an AP is 21 and the sum of the, squares of these terms is 165. Find these terms.
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Arithmetic Progression, , 265, , 10. The angles of a quadrilateral are in AP whose common difference is 10., Find the angles., Let these angles be xc, (x 10)c, (x 20)c and (x 30)c ., , HINT, , Their sum is 360., , 11. Find four numbers in AP whose sum is 28 and the sum of whose squares, is 216., 12. Divide 32 into four parts which are the four terms of an AP such that the, product of the first and the fourth terms is to the product of the second, and the third terms as 7 : 15., [CBSE 2014], Let these parts be (a 3d), (a d), (a d) and (a 3d) ., , HINT, , 13. The sum of first three terms of an AP is 48. If the product of first and, second terms exceeds 4 times the third term by 12. Find the AP., [CBSE 2013C], , Let these terms be (a d), a, (a d) ., , HINT, , ANSWERS (EXERCISE 5B), , 1. k 3, , 2. x 3, , 6. (2, 5, 8) or (8, 5, 2), 9. (4, 7, 10) or (10, 7, 4), , 3. y 5, , 4. x 5, , 7. (–5, 1, 7) or (7, 1, –5) 8. (5, 8, 11) or (11, 8, 5), 10. 75, 85, 95, 105, , 11. (4, 6, 8, 10) or (10, 8, 6, 4) 12. (2, 6, 10, 14) or (14, 10, 6, 2), , 13. 7, 16, 25, , SUM OF n TERMS OF AN AP, TO FIND THE SUM OF n TERMS OF AN AP, THEOREM 3, , PROOF, , Prove that the sum of n terms of an AP in which first term a,, common difference d and last term l, is given by, n, n, Sn (a l) and Sn · {2a (n 1) d} ., 2, 2, , Consider an AP having n terms in which, first term a, common difference d and last term l., Then, l a (n 1) d, , … (i) [a l nth term], , Also, we may write the given AP as, a, (a d), (a 2d), …, (l 2d), (l d), l., Let Sn be the sum of the first n terms of this AP. Then,, Sn a (a d) (a 2d) … (l 2d) (l d) l, , … (ii)
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266, , Secondary School Mathematics for Class 10, , Writing the above series in the reverse order, we get, Sn l (l d) (l 2d) … (a 2d) (a d) a, , … (iii), , Adding the corresponding terms of (ii) and (iii), we get, 2Sn (a l) (a l) (a l) … n times n (a l), n, , Sn (a l), 2, n , n, , , Sn, [a a (n 1) d] [2a (n 1) d] [using (i)]., 2, 2, n, n, Hence, Sn (a l) and Sn [2a (n 1) d] ., 2, 2, SUMMARY, , Sum of n terms of an AP a, a d, a 2d, …, l is given by, n, n, Sn (a l) or Sn #{2a (n 1) d} ., 2, 2, , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , Find the sum of first 24 terms of the AP 5, 8, 11, 14, … ., Here a 5, d (8 5) 3 and n 24., n, Using the formula, Sn · {2a (n 1) d}, we get, 2, 24 ·, , , S24, {2 # 5 (24 1)# 3} [a a 5, d 3 and n 24], 2, 12 #(10 69) 948., Hence, the sum of first 24 terms of the given AP is 948., , EXAMPLE 2, , Find the sum 25 + 28 + 31 + … + 100., , SOLUTION, , Here a 25, d (28 25) 3 and l 100., Let the total number of terms be n. Then,, Tn 100 & a (n 1) d 100, & 25 (n 1)# 3 100, & 3n 22 100, & 3n 78 & n 26., n, Required sum · (a l), 2, 26 · (25 100) 13 #125 1625., 2, Hence, the required sum is 1625., , [CBSE 2006C]
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Arithmetic Progression, EXAMPLE 3, , SOLUTION, , 267, , 1, 1, Find the sum 18 15 13 … a 49 k ·, 2, 2, 31 31 5 ·, We have,, 18 13, 2, 2, 2, So, the given series is an arithmetic series., 5, 99, ·, Here, a 18, d , and l , 2, 2, Let the given series contain n terms. Then,, 99, 99, Tn , & a (n 1) d 2, 2, 5, 99, & 18 (n 1)#a 2 k 2, & 36 5n 5 99, , [CBSE 2013], , & 5n 140 & n 28., , , n , (a l), 2, 28 · &18 99 0 7 (36 99) 441., 2, 2, , required sum , , Hence, the sum of the given series is –441., EXAMPLE 4, , SOLUTION, , Find the sum of first n terms of an AP whose nth term is (5n 1) ., Hence, find the sum of first 20 terms., [CBSE 2011], The nth term of the given AP is given by, Tn (5n 1) ., Let a be the first term and d be the common difference of this, AP. Then,, a T1 (5 #1 1) 4, T2 (5 # 2 1) 9., d (T2 T1) (9 4) 5., n, sum of first n terms · {2a (n 1) d}, 2, n, #{2 # 4 (n 1)# 5} 1 n (5n 3) ., 2, 2, 1, sum of first 20 terms # 20 #(5 # 20 3) 1030., 2, 1, Hence, Sn n (5n 3) and S20 1030., 2, , , EXAMPLE 5, , SOLUTION, , If the sum of first n terms of an AP is, term and hence, write its 20th term., 1, We have, Sn (3n 2 7n) ., 2, , 1, (3n 2 7n) then find its nth, 2, [CBSE 2015]
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268, , Secondary School Mathematics for Class 10, , , , 1·, 1, {3 (n 1) 2 7 (n 1)} (3n 2 n 4) ., 2, 2, 1, 1, Tn (Sn Sn 1) (3n 2 7n) (3n 2 n 4), 2, 2, 1 (6n 4) (3n 2) ., 2, , , , nth term (3n 2) ., , , , Sn 1 , , Hence, 20th term (3 # 20 2) 62., EXAMPLE 6, , SOLUTION, , How many terms of the AP 3, 5, 7, 9, … must be added to get the, sum 120?, Here, a 3 and d (5 3) 2., Let the required number of terms be n. Then,, n, Sn 120 & · {2a (n 1) d} 120, 2, n·, & 2 {2 # 3 (n 1)# 2} 120, & n2 · (2n 4) 120 & n 2 2n 120 0, & n 2 12n 10n 120 0, & n (n 12) 10 (n 12) 0 & (n 12)(n 10) 0, & n 12 0 or n 10 0 & n 12 or n 10, & n 10 [£ number of terms cannot be negative], Hence, the required number of terms is 10., , EXAMPLE 7, , SOLUTION, , How many terms of the AP 17, 15, 13, 11, ... must be added to get the, sum 72? Explain the double answer., Here, a 17 and d (15 17) 2., Let the sum of n terms be 72. Then,, n, Sn 72 & · {2a (n 1) d} 72, 2, & n · {2 #17 (n 1)#( 2)} 144, , & n (36 2n) 144 & 2n 2 36n 144 0, & n 2 18n 72 0 & n 2 12n 6n 72 0, & n (n 12) 6 (n 12) 0 & (n 12)(n 6) 0, & n 6 or n 12., , , sum of first 6 terms = sum of first 12 terms = 72., , This means that the sum of all terms from 7th to 12th is zero.
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Arithmetic Progression, , 269, , EXAMPLE 8, , The first and the last terms of an AP are 7 and 49 respectively. If sum, of all its terms is 420, find its common difference., [CBSE 2014], , SOLUTION, , Let the given AP contain n terms., Here a 7, l 49 and Sn 420., n, Sn (a l), 2, n , n, , (7 49) 420 & # 56 420, 2, 2, 420, 15., 28n 420 & n , 28, Thus, the given AP contains 15 terms and T15 49., Let d be the common difference of the given AP. Then,, T15 49 & a 14d 49, , & 7 14d 49, & 14d 42 & d 3., Hence, the common difference of the given AP is 3., EXAMPLE 9, , The sum of the first 7 terms of an AP is 63 and the sum of its next, 7 terms is 161. Find the 28th term of this AP., [CBSE 2014], , SOLUTION, , Let a be the first term and d be the common difference of the, n, given AP. Then, using Sn · [2a (n 1) d], we get, 2, 7, S7 (2a 6d) & 7 (a 3d) 63, [a S7 63], 2, … (i), & a 3d 9., Clearly, the sum of first 14 terms 63 161 224., 14, (2a 13d) 224, 2, & 7 (2a 13d) 224, … (ii), & 2a 13d 32., Multiplying (i) by 2 and subtracting the result from (ii), we get, , , , S14 224 &, , 7d 14 & d 2., Putting d 2 in (i), we get a 9 6 3., Thus, a 3 and d 2., 28th term of this AP is given by, T28 (a 27d) (3 27 # 2) 57., Hence, the 28th term of the given AP is 57.
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270, , Secondary School Mathematics for Class 10, , EXAMPLE 10, , The 14th term of an AP is twice its 8th term. If its 6th term is –8 then, find the sum of its first 20 terms., [CBSE 2015], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, T14 2 #T8 & a 13d 2 (a 7d), , & a d 0., Also, T6 8 & a 5d 8., , … (i), … (ii), , On solving (i) and (ii), we get a 2 and d 2., The sum of first 20 terms is given by, n, S20 · [2a (n 1) d], where n 20, 2, a 20 k#{(2 # 2 19 #(2)}, 2, 10 #(4 38) 10 #(34) 340., Hence, the required sum is –340., EXAMPLE 11, , The sum of 4th and 8th terms of an AP is 24 and the sum of its 6th, and 10th terms is 44. Find the sum of first ten terms of the AP., [CBSE 2012], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, T4 T8 24 & (a 3d) (a 7d) 24, , & 2a 10d 24, & a 5d 12., And, T6 T10 44 & (a 5d) (a 9d) 44, & 2a 14d 44, & a 7d 22., , … (i), , … (ii), , On solving (i) and (ii), we get a 13 and d 5., , , the sum of first 10 terms of the given AP is given by, 10, S10 a k · (2a 9d), 2, , [using Sn , , n, {(2a (n 1) d}], 2, , 5 #{2 #(13) 9 # 5} 5 (26 45) 5 #19 95., Hence, the sum of first 10 terms of the given AP is 95., EXAMPLE 12, , Sum of first 14 terms of an AP is 1505 and its first term is 10. Find, its 25th term., [CBSE 2012]
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Arithmetic Progression, SOLUTION, , 271, , Here a 10 and let d be the common difference. Then,, n, [2a (n 1) d] 1505, where n 14 and a 10, 2, · (20 13d) 1505 & (20 13d) 1505 215, & 14, 2, 7, , , & 13d 195 & d 15., , S14 1505 &, , Thus, a 10 and d 15., , , T25 (a 24d) (10 24 #15) 370., , Hence, the 25th term is 370., EXAMPLE 13, , In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of, its last 15 terms is 2565. Find the AP., [CBSE 2014], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then, the sum of first n terms is given by, n, Sn · {2a (n 1) d} ., 2, 10 ·, S10 , (2a 9d) & 5 (2a 9d) 210, 2, … (i), & 2a 9d 42., Sum of last 15 terms (S50 S35) ., (S50 S35) 2565, 50, 35, , (2a 49d) (2a 34d) 2565, 2, 2, , , 25 (2a 49d) 35 (a 17d) 2565, , , (50a 35a) (1225d 595d) 2565, 15a 630d 2565 & a 42d 171., On solving (i) and (ii), we get a 3 and d 4., , … (ii), , Hence, the required AP is 3, 7, 11, 15, 19, … ., EXAMPLE 14, , The sum of first 6 terms of an AP is 42. The ratio of its 10th term to, 30th term is 1 : 3. Find the first and the 13th term of the AP., [CBSE 2009], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, T10 a 9d and T30 a 29d, , , T10 1, , & a 9d 1, T30 3, a 29d 3
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Arithmetic Progression, , 273, , 1, 3 (S2 S1) 3 · :{2na n (2n 1) d} &na n (n 1) d0D, 2, 3, 3, n, 1, · [2a 3nd d], 3 · :na n 2 d ndD , 2, 2, 2, 3n · {2a (3n 1) d} S3 ., 2, Hence, S3 3 (S2 S1) ., , , EXAMPLE 17, , If the sum of the first p terms of an AP is the same as the sum of its, first q terms (where p ! q) then show that the sum of its first (p q), terms is zero., , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, p, q, Sp Sq & [2a (p 1) d] [2a (q 1) d], 2, 2, & (p q) (2a) (q p) (q p 1) d, , & 2a (1 p q) d, , … (i), , Sum of the first (p q) terms of the given AP, (p q), · {2a (p q 1) d}, 2, (p q), · {(1 p q) d (p q 1) d} [using (i)], , 2, 0., , , EXAMPLE 18, , If the sum of the first m terms of an AP be n and the sum of its first, n terms be m then show that the sum of its first (m n) terms is, (m n) ., , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, m, Sm n & [2a (m 1) d] n, 2, … (i), & 2am m (m 1) d 2n., n, And, Sn m & [2a (n 1) d] m, 2, … (ii), & 2an n (n 1) d 2m., On subtracting (ii) from (i), we get, 2a (m n) [(m 2 n 2) (m n)] d 2 (n m), (m n) [2a (m n 1) d] 2 (n m), , , 2a (m n 1) d 2, , … (iii)
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274, , Secondary School Mathematics for Class 10, , Sum of the first (m n) terms of the given AP, (m n), · {2a (m n 1) d}, 2, (m n), · (2) (m n) [using (iii)]., , 2, , , Hence, the sum of first (m n) terms of the given AP is (m n) ., EXAMPLE 19, , The ratio of the sums of first m and first n terms of an AP is m 2 : n 2 ., Show that the ratio of its mth and nth terms is (2m 1) : (2n 1) ., [CBSE 2017], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP. Then,, Sm sum of first m terms of the given AP;, Sn sum of first n terms of the given AP., m, [2a (m 1) d], 2, Sm m 2, m2, 2 & 2, n, Sn n, n, [2a (n 1) d], 2, 2a (m 1) d m, , &, 2a (n 1) d n, & 2an mnd nd 2am mnd md, & 2an 2am nd md, & 2a(n m) d (n m) & 2a d., , , … (i), , Tm a (m 1) d a (m 1) · 2a, , , [from (i)], Tn a (n 1) d, a (n 1) · 2a, a(2m 1) 2m 1, , , , a 2am 2a 2am a , , ·, a 2an 2a 2an a, a(2n 1) 2n 1, , EXAMPLE 20, , The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the, ratio of the 5th term to the 21st term, and also the ratio of the sum of, the first 5 terms to the sum of first 21 terms., [CBSE 2017], , SOLUTION, , Let a be the first term and d be the common difference of the, given AP., T11 2, a (11 1) d 2, &, , Then,, T18 3, a (18 1) d 3, a 10d 2, &, & 3a 30d 2a 34d, a 17d 3, … (i), & a 4d.
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276, , Secondary School Mathematics for Class 10, , WORD PROBLEMS, EXAMPLE 22, , Find the sum of all multiples of 7 lying between 500 and 900., [CBSE 2012], , SOLUTION, , All multiples of 7 lying between 500 and 900 are, 504, 511, 518, …, 896, This is an AP in which a 504, d 7 and l 896., Let the given AP contain n terms. Then,, Tn 896 & a (n 1) d 896 & 504 (n 1)#7 896, , & 497 7n 896 & 7n 399 & n 57., , , n , (a l), 2, 57 · (504 896) a 57 #1400k 39900., 2, 2, , required sum , , Hence, the required sum is 39900., EXAMPLE 23, , Find the sum of all 3-digit natural numbers which are multiples, of 11., [CBSE 2009, ’12], , SOLUTION, , All 3-digit natural numbers, which are multiples of 11 are, given as 110, 121, 132, …, 990., This is an AP in which a 110, d 11 and l 990., Let the given AP contain n terms. Then,, Tn 990 & a (n 1) d 990 & 110 (n 1)#11 990, , & 99 11n 990 & 11n 891 & n 81., , , n 81, #(110 990), (a l), 2, 2, a81 #1100k 44550., 2, , required sum , , Hence, the required sum is 44550., EXAMPLE 24, , Ramkali required ` 2500 after 12 weeks to send her daughter to, school. She saved ` 100 in the first week and increased her weekly, saving by ` 20 every week. Find whether she will be able to send her, daughter to school after 12 weeks. What value is generated in the, above situation?, [CBSE 2015], , SOLUTION, , The amounts saved by Ramkali in successive weeks are ` 100,, ` 120, ` 140, ` 160, … up to 12 terms.
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Arithmetic Progression, , 277, , These amounts form an AP in which a 100, d 20 and n 12., n, #[2a (n 1) d], we get, 2, 12, S12 #[2 # 100 11# 20] (6 # 420) 2520 2500., 2, , Using Sn , , Thus, Ramkali will be able to deposit her daughter’s fees and, so she can send her to school., The above situation shows that saving is a good habit as it, helps preserve and collect money for a good cause., EXAMPLE 25, , 200 logs are stacked in such a way that there are 20 logs in the bottom, row, 19 in the next row, 18 in the next row, and so on. In how many, rows 200 logs are placed and how many logs are there in the top row?, , SOLUTION, , Let the required number of rows be n. Then,, 20 19 18 … to n terms = 200., This is an arithmetic series in which, a 20, d (19 20) 1 and Sn 200., n, We know that Sn · {2a (n 1) d} ., 2, n·, , {2 # 20 (n 1)#( 1)} 200, 2, n (41 n) 400 & n 2 41n 400 0, n 2 25n 16n 400 0 & n (n 25) 16 (n 25) 0, (n 25) (n 16) 0 & n 25 0 or n 16 0, n 25 or n 16., Now, T25 (a 24d) 20 24 #(1) 4., This is meaningless as the number of logs cannot be negative., So, we reject the value n 25., , , n 16. Thus, there are 16 rows in the whole stack., , Now, T16 (a 15d) 20 15 #(1) 20 15 5., Hence, there are 5 logs in the top row., EXAMPLE 26, , The production of TV sets in a factory increases uniformly by a fixed, number every year. It produced 16000 sets in 6th year and 22600, in 9th year. Find the production during (i) first year (ii) 8th year, (iii) first 6 years.
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278, SOLUTION, , Secondary School Mathematics for Class 10, , Let the production during first year be a and let d be the, increase in production every year. Then,, T6 16000 & a 5d 16000, , … (i), , and T9 22600 & a 8d 22600., , … (ii), , On subtracting (i) from (ii), we get, 3d 6600 & d 2200., Putting d 2200 in (i), we get, a 5 # 2200 16000 & a 11000 16000, a 16000 11000 5000., Thus, a 5000 and d 2200., (i) Production during first year, a 5000., (ii) Production during 8th year is given by, T8 (a 7d) (5000 7 # 2200) (5000 15400) 20400., (iii) Production during first 6 years is given by, 6, {2a 5d} 3 (2 # 5000 5 # 2200), 2, 3 (10000 11000) 63000., , S6 , , EXAMPLE 27, , A spiral is made up of successive semicircles, with centres alternately, at A and B, starting with centre at A, of radii 0.5 cm, 1 cm, 1.5 cm,, 2 cm, … as shown in the given figure. What is the total length of, 22, such a spiral made up of 13 consecutive semicircles? (Take ), 7, , SOLUTION, , Let L1, L2, L3, L4, …, L13 be the lengths of semicircles of radii, 13, 0.5 cm, 1 cm, 1.5 cm, 2 cm, … and, cm respectively., 2, Then, we have, , L1 ( # 0.5) cm cm,, 2, , L2 ( #1) cm 2 ` j cm,, 2, , L3 ( #1.5) cm 3 ` j cm,, 2, , , , L4 ( # 2) cm 4 ` j cm,, 2, … … … … … …, … … … … … …, L13 a #, , 13, , k cm 13 ` j cm., 2, 2
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Arithmetic Progression, , , , 279, , total length of the spiral, L1 L2 L3 L4 … L13, $ 2 ` j 3 ` j 4 ` j … 13 ` j. cm, 2, 2, 2, 2, 2, , (1 2 3 4 … 13) cm, 2, # 13 #(1 13) cm, 2, 2, , 8Sn , , n , (a l)B, 2, , a1 # 22 # 13 #14k cm 143 cm., 2, 7, 2, Hence, the required length of the spiral is 143 cm., EXAMPLE 28, , A ladder has rungs 25 cm apart. The, rungs decrease uniformly in length from, 45 cm at the bottom to 25 cm at the top. If, the top and bottom rungs are 2.5 m apart,, what is the length of the wood required for, the rungs?, , SOLUTION, , It is given that the top and bottom, rungs are 250 cm apart and the gap, between two consecutive rungs is, 25 cm., 250 , number of rungs a, 1k 11., 25, The largest rung is 45 cm long and the smallest one is 25 cm, long., It is given that the rungs are decreasing uniformly in length, from 45 cm at the bottom to 25 cm at the top., So, the lengths of the rungs form an AP with a 45 cm and, l length of 11th rung = 25 cm., , , length of the wood required to form 11 rungs, n (a l) cm 11 (45 25) cm a11 #70k cm 385 cm., 2, 2, 2, , Hence, the required length of the wood to form these rungs, is 3.85 m.
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280, , Secondary School Mathematics for Class 10, , EXAMPLE 29, , The houses of a row in a colony are numbered consecutively from 1 to, 49. Show that there is a value of x such that the sum of the numbers, of the houses preceding the house numbered x is equal to the sum of, the numbers of the houses following it. Find the value of x., , SOLUTION, , We are given an AP, namely, 1, 2, 3, …, (x 1), x, (x 1), …, 49, such that 1 2 3 … (x 1) (x 1) (x 2) … 49., Thus, we have Sx 1 S49 Sx ., n, Using the formula, Sn (a l) in (i), we have, 2, (x 1), · {1 (x 1)} 49 · (1 49) x · (1 x), 2, 2, 2, x (x 1) x (x 1), , 1225, , 2, 2, 2x 2 2450 & x 2 1225 & x 1225 35., , … (i), , Hence, x 35., EXAMPLE 30, , Find the middle term of the sequence formed by all three-digit, numbers which leave a remainder 3 when divided by 4. Also, find the, sum of all numbers on both sides of the middle term., [CBSE 2015], , SOLUTION, , The sequence formed by the given numbers is, 103, 107, 111, 115, …, 999., This is an AP in which a 103 and d (107 103) 4., Let the total number of these terms be n. Then,, Tn 999 & a (n 1) d 999, , & 103 (n 1)# 4 999, & (n 1)# 4 896 & (n 1) 224 & n 225., , , middle term a, , 225 1, n1, kth term = 113th term., kth term a, 2, 2, , T113 (a 112d) (103 112 # 4) 551., T112 (551 4) 547., So, we have to find S112 and (S225 S113) ., Using the formula Sm , S112 , , m , (a l) for each sum, we get, 2, , 112, (103 547) (112 # 325) 36400., 2
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Arithmetic Progression, , 281, , 225, 113, (103 999) , (103 551), 2, 2, (225 # 551) (113 # 327), , (S225 S113) , , 123975 36951 87024., Sum of all numbers on LHS of the middle term is 36400., Sum of all numbers on RHS of the middle term is 87024., f, , EXERCISE 5C, , 1. Find the sum of each of the following APs:, (i) 2, 7, 12, 17, … to 19 terms., (ii) 9, 7, 5, 3, … to 14 terms., (iii) –37, –33, –29, … to 12 terms., 1 1 1, (iv), , , , … to 11 terms., 15 12 10, (v) 0.6, 1.7, 2.8, … to 100 terms., 2. Find the sum of each of the following arithmetic series:, 1, (i) 7 10 14 … 84., 2, , (ii) 34 32 30 … 10., (iii) (5) (8) (11) … (230), (iv) 5 ( 41) 9 ( 39) 13 ( 37) 17 … ( 5) 81 ( 3), 3. Find the sum of first n terms of an AP whose nth term is (5 6n) . Hence,, find the sum of its first 20 terms., 4. The sum of the first n terms of an AP is (3n 2 6n) . Find the nth term and, the 15th term of this AP., [CBSE 2014], 2, 5. The sum of the first n terms of an AP is given by Sn (3n n) . Find its, (i) nth term, (ii) first term and (iii) common difference., [CBSE 2005C], 6. (i) The sum of the first n terms of an AP is c, term and the 20th term of this AP., , 5n 2 3n ·, m Find the nth, 2, 2, [CBSE 2006C], , 3n 2 5n ·, m Find its nth term, (ii) The sum of the first n terms of an AP is c, 2, 2, and the 25th term., [CBSE 2006C], 1, 1, 7. If mth term of an AP is n and nth term is m then find the sum of its, first mn terms., [CBSE 2017], 8. How many terms of the AP 21, 18, 15, … must be added to get the, sum 0?
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282, , Secondary School Mathematics for Class 10, , 9. How many terms of the AP 9, 17, 25, … must be taken so that their sum, is 636?, [CBSE 2017], 10. How many terms of the AP 63, 60, 57, 54, … must be taken so that their, sum is 693? Explain the double answer., 1, 2, 11. How many terms of the AP 20, 19 , 18 , … must be taken so that their, 3, 3, sum is 300? Explain the double answer., 12. Find the sum of all odd numbers between 0 and 50., , [CBSE 20013C}, , 13. Find the sum of all natural numbers between 200 and 400 which are, divisible by 7., [CBSE 2012], 14. Find the sum of first forty positive integers divisible by 6., , [CBSE 2012], , 15. Find the sum of the first 15 multiples of 8., 16. Find the sum of all multiples of 9 lying between 300 and 700., 17. Find the sum of all three-digit natural numbers which are divisible, by 13., [CBSE 2011], 18. Find the sum of first 100 even natural numbers which are divisible by 5., [CBSE 2007C], , 19. Find the sum of n terms of the following series:, 1 2 a 3k , a4 n, k a4 n k 4 n … ., [CBSE 2017], 20. In an AP, it is given that S5 S7 167 and S10 235, then find the AP,, where Sn denotes the sum of its first n terms., [CBSE 2015], 21. In an AP, the first term is 2, the last term is 29 and the sum of all the, terms is 155. Find the common difference., [CBSE 2010], 22. In an AP, the first term is – 4, the last term is 29 and the sum of all its, terms is 150. Find its common difference., [CBSE 2011], 23. The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is, their sum?, 24. The first and the last terms of an AP are 5 and 45 respectively. If the, sum of all its terms is 400, find the common difference and the number, of terms., [CBSE 2012, ’14, ’17], 25. In an AP, the first term is 22, nth term is –11 and sum of first n terms is, 66. Find n and hence find the common difference., [CBSE 2008], 26. The 12th term of an AP is –13 and the sum of its first four terms is 24., Find the sum of its first 10 terms., [CBSE 2015], 27. The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are, in the ratio 1 : 5, find the AP., [CBSE 2014]
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Arithmetic Progression, , 283, , 28. The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is, 400. Find the first term and the common difference of the AP. [CBSE 2009], 29. The sum of the first 7 terms of an AP is 49 and the sum of its first 17, [CBSE 2008C], terms is 289. Find the sum of its first n terms., 30. Two APs have the same common difference. If the first terms of these, APs be 3 and 8 respectively, find the difference between the sums of, [CBSE 2011], their first 50 terms., 31. The sum of first 10 terms of an AP is –150 and the sum of its next 10, [CBSE 2010], terms is –550. Find the AP., 32. The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find, [CBSE 2015], the sum of its first 10 terms., 33. The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find, [CBSE 2015], the sum of its first 15 terms., 34. (i) An AP 5, 12, 19, … has 50 terms. Find its last term. Hence, find the, sum of its last 15 terms., [CBSE 2015], (ii) An AP 8, 10, 12, … has 60 terms. Find its last term. Hence, find the, sum of its last 10 terms., [CBSE 2015], 35. The sum of first n terms of two APs are in the ratio (3n 8) : (7n 15) ., Find the ratio of their 12th terms., 36. The sum of the 4th and the 8th terms of an AP is 24 and the sum of its, 6th and 10th terms is 44. Find the sum of its first 10 terms. [CBSE 2013C], 37. The sum of first m terms of an AP is (4m 2 m) . If its nth term is 107, find, the value of n. Also, find the 21st term of this AP., [CBSE 2013], 2, 38. The sum of first q terms of an AP is (63q 3q ) . If its pth term is –60, find, the value of p. Also, find the 11th term of its AP., [CBSE 2013], 39. Find the number of terms of the AP –12, –9, –6, …, 21. If 1 is added, to each term of this AP then find the sum of all terms of the AP thus, [CBSE 2013], obtained., 40. Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its, [CBSE 2012], 25th term., 41. Find the sum of first 51 terms of an AP whose second and third terms, are 14 and 18 respectively., 42. In a school, students decided to plant trees in and around the school to, reduce air pollution. It was decided that the number of trees that each, section of each class will plant will be double of the class in which they, are studying. If there are 1 to 12 classes in the school and each class has, two sections, find how many trees were planted by students. Which, [CBSE 2014], value is shown in the question?
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284, , Secondary School Mathematics for Class 10, , 43. In a potato race, a bucket is placed at the starting point, which is 5 m, from the first potato, and the other potatoes are placed 3 m apart in a, straight line. There are 10 potatoes in the line. A competitor starts from, the bucket, picks up the nearest potato, runs back with it, drops it in the, bucket, runs back to pick up the next potato, runs to the bucket to drop, it in, and he continues in the same way until all the potatoes are in the, bucket. What is the total distance the competitor has to run?, , 44. There are 25 trees at equal distances of 5 m in a line with a water tank,, the distance of the water tank from the nearest tree being 10 m. A, gardener waters all the trees separately, starting from the water tank, and returning back to the water tank after watering each tree to get, water for the next. Find the total distance covered by the gardener in, order to water all the trees., , 45. A sum of ` 700 is to be used to give seven cash prizes to students of a, school for their overall academic performance. If each prize is ` 20 less, than its preceding prize, find the value of each prize., 46. A man saved ` 33000 in 10 months. In each month after the first, he, saved ` 100 more than he did in the preceding month. How much did, he save in the first month?, 47. A man arranges to pay off a debt of ` 36000 by 40 monthly instalments, which form an arithmetic series. When 30 of the instalments are paid,, he dies leaving one-third of the debt unpaid. Find the value of the first, instalment., 48. A contract on construction job specifies a penalty for delay of completion, beyond a certain date as follows: ` 200 for the first day, ` 250 for the, second day, ` 300 for the third day, etc., the penalty for each succeeding, day being ` 50 more than for the preceding day. How much money the, contractor has to pay as penalty, if he has delayed the work by 30 days?, 49. A child puts one five-rupee coin of her savings in the piggy bank on the, first day. She increases her saving by one five-rupee coin daily. If the, piggy bank can hold 190 coins of five rupees in all, find the number of, days she can contribute to put the five-rupee coins into it and find the, [CBSE 2017], total money she saved.
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Arithmetic Progression, , 285, , ANSWERS (EXERCISE 5C), , 1. (i) 893, 2. (i) 1046, , 1, 2, , 33, 20, , (ii) –56, , (iii) –180, , (iv), , (ii) 286, , (iii) –8930, , (iv) 420, , 4. Tn (6n 3), T15 93, , (v) 5505, 3. Sn n (2 3n), S20 1160, , 5. (i) Tn (6n 4), , (ii) a 2 (iii) d 6, 1 mn, 6. (i) Tn (5n 1), T20 99 (ii) Tn (3n 1), T25 76, 7., 2, 8. 15, 9. 12, 10. n 21 or n 22, 22nd term is 0, 11. n 25 or n 36, sum of last 11th terms is 0 12. 625 13. 8729, 14. 4920, , 15. 960, 16. 21978, 17. 37674, 1, 19. (7n 1), 20. 1, 6, 11, 16, …, 21. d 3, 2, 8, 23. n 38 and Sn 6973, 24. d and n 16, 3, 26. S10 0, 27. 2, 10, 18, 26, …, 28. a 1, d 2, 30. 250, , 31. 3, –1, –5, –9, …, , 32. S10 175, , 18. 50500, 22. d 3, 25. n 12, d 3, 29. Sn n 2, 33. S15 495, , (ii) T60 126, sum 1170, 35. 7 : 16, 36. S10 95, 37. n 14, T21 163, 38. p 21, T11 0, 39. n 12, sum 66, 40. T25 130, 41. 5610, 42. 312, 34. (i) l 348, sum 4485, , 43. 370 m, , 44. 3500 m, , 45. ` 160, ` 140, ` 120, ` 100, ` 80, ` 60, ` 40, , 46. ` 2850, , 47. ` 510, , 48. ` 27750, , 49. 19, ` 950, , HINTS TO SOME SELECTED QUESTIONS, 2. (i) a 7, d a, , 21 7, 7k, and l 84., 2, 2, , Tn 84 & a (n 1) d 84 & 7 (n 1)#, , 7, 84 & n 23., 2, , n 23 , 1, (a l), (7 84) 1046 ·, 2, 2, 2, 3. Tn (5 6n) & T1 (5 6) 1 and T2 (5 6 # 2) 7, S23 , , a 1 and d ( 7 1) 6., n, Sn #{2 #( 1) (n 1)#( 6)} n (2 3n) ., 2, S20 20 #(2 3 # 20) 1160., , , 9. 4n 2 5n 636 0 & 4n 2 53n 48n 636 0., 10. n 2 43n 462 0 & n 2 21n 22n 462 0., 11. n 2 61n 900 0 & n 2 25n 36n 900 0., 12. Required sum (1 3 5 7 … 49) .
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286, , Secondary School Mathematics for Class 10, , Tn 49 & 1 (n 1)# 2 49 & n 25., n, Use Sn (a l) ., 2, 13. Required sum (203 210 217 … 399) ., Tn 399 & 203 (n 1)#7 399 & n 29., n, Use Sn (a l) ., 2, 14. Required sum (6 12 18 … 240) ., Here n 40., n, Use Sn (a l) ., 2, 15. Required sum (8 16 24 … 120) ., n, Here n 15. Use Sn (a l) ., 2, 16. Required sum (306 315 324 … 693) ., 17. Required sum (104 117 130 … 988) ., 18. Required sum (10 20 30 … 1000) ., n, 1 2 3, 19. Required sum (4 4 … to n terms) an n n … n k, 4n n a 1 n k :sum n (a l)D, 2 n n, 2, (1 n) 1, 4n , (7n 1) ., 2, 2, 5, 7, (2a 4d) (2a 6d) 167, 2, 2, & 12a 31d 167., 10, S10 235 &, (2a 9d) 235 & 2a 9d 47., 2, Solve (i) and (ii), we get a 1 and d 5., n, n, 21. Sn (a l) & (2 29) 155 & n 10., 2, 2, T10 29 & 2 9d 29 & d 3., 20. S5 S7 167 &, , … (i), … (ii), , 23. Let the number of terms be n. Then,, Tn 350 & a (n 1) d 350 & 17 (n 1)# 9 350 & n 38., n, Sn (a l) ., 2, n, n, 24. Sn (a l) & (5 45) 400 & n 16., 2, 2, 8, T16 45 & a 15d 45 & 5 15d 45 & d ·, 3, n, n, 25. Sn (a l) & [22 (11)] 66 & n 12., 2, 2, T12 11 & a 11d 11 & 22 11d 11 & d 3., 26. a 11d 13., , … (i), , and, , 4, (2a 3d) 24 & 2a 3d 12., 2, , … (ii)
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288, 40., , Secondary School Mathematics for Class 10, , n , 14, (a l) 1505 &, (10 l) 1505 & l 205., 2, 2, T14 205 & a 13d 205 & 10 13d 205 & d 15., T25 (a 24d) (10 24 # 5) 130., , 41. Let a be the first term of the given AP., Then, 14 a 18 14 & 14 a 4 & a 10., , , a 10, d 4 and n 51., , Now, find Sn ., 42. Required sum 2 #[2 4 6 8 … 24] 2 #, Planting trees helps in reducing air pollution., , 12 , (2 24) 312., 2, , 43. Total distance run to pick up the 1st, 2nd, 3rd, …, 10th potato, [(2 # 5) 2 #(5 3) 2 #(5 3 # 2) … 2 #(5 3 # 9)] m, 2 #[5 8 11 … 32] m 2 # 10 (5 32) m 370 m., 2, 44. Total distance covered, [(2 #10) 2 #(10 5 #1) 2 #(10 5 # 2) … 2 #(10 5 # 24)] m, 2 #[10 15 20 … 130] m 2 # 25 (10 130) m 3500 m., 2, 45. Let the values of these prizes be ` x, ` (x 20), ` (x 40), … and ` (x 20 # 6) respectively., This is an AP with a x, l (x 120) and n 7., n, 7, sum (a l) (x x 120) 7 (x 60) ., 2, 2, 7x 420 700 & 7x 1120 & x 160., 46. Suppose he saved ` x, ` (x 100), ` (x 200), …, ` (x 900) in 1st, 2nd, 3rd, … and 10th, month respectively. Then,, x (x 100) (x 200) … (x 900) 33000, , , , , 10x (100 200 300 … 900) 33000, 9, n, 10x (100 900) 33000 :a Sn (a l)D, 2, 2, 10x 33000 4500 28500 & x 2850., , 47. Let the first instalment be ` x and let it be increased by ` d each month. Then,, 2, x (x d) (x 2d) … (x 29d) # 36000, 3, 30x (1 2 3 … 29) d 24000, 29 , (1 29) d 24000 & 2x 29d 1600., 2, Also, x (x d) (x 2d) … (x 39d) 36000 gives, , … (i), , 39 , (1 39) d 36000 & 2x 39d 1800., 2, On solving (i) and (ii), we get d 20 and x 510., , … (ii), , , , 30x , , 40x , , 48. Total penalty ` (200 250 300 … up to 30 terms)., T30 (200 29 # 50) 1650.
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Arithmetic Progression, , , f, , required sum = ` &, , 289, , 30, (200 1650)0 ` 27750., 2, , EXERCISE 5D, , Very-Short and Short-Answer Questions, 1. The first three terms of an AP are respectively (3y 1), (3y 5) and, [CBSE 2014], (5y 1), find the value of y., 2. If k, (2k 1) and (2k 1) are the three successive terms of an AP, find the, value of k., [CBSE 2014], 3. If 18, a, (b 3) are in AP, then find the value of (2a b) ., [CBSE 2014], 4. If the numbers a, 9, b, 25 form an AP, find a and b., [CBSE 2014], 5. If the numbers (2n 1), (3n 2) and (6n 1) are in AP, find the value of n, and the numbers., [CBSE 2013C], 6. How many three-digit natural numbers are divisible by 7?, , [CBSE 2013], , 7. How many three-digit natural numbers are divisible by 9?, , [CBSE 2013], , 8. If the sum of first m terms of an AP is (2m 2 3m) then what is its second, term?, [CBSE 2011], 9. What is the sum of first n terms of the AP a, 3a, 5a, … ., , [CBSE 2012], , 10. What is the 5th term from the end of the AP 2, 7, 12, …, 47?, , [CBSE 2011], , 11. If an denotes the nth term of the AP 2, 7, 12, 17, …, find the value of, [CBSE 2011], (a30 a20) ., 12. The nth term of an AP is (3n 5) . Find its common difference., [CBSE 2009C], , 13. The nth term of an AP is (7 4n) . Find its common difference., , [CBSE 2008], , 14. Write the next term of the AP 8 , 18 , 32 , … ., , [CBSE 2008], , 15. Write the next term of the AP 2 , 8 , 18 , … ., , [CBSE 2008C], , 16. Which term of the AP 21, 18, 15, … is zero?, , [CBSE 2008C], , 17. Find the sum of first n natural numbers., 18. Find the sum of first n even natural numbers., 19. The first term of an AP is p and its common difference is q. Find its, 10th term., [CBSE 2008], 4, 20. If , a, 2 are in AP, find the value of a., [CBSE 2009], 5, 21. If (2p 1), 13, (5p 3) are in AP, find the value of p., [CBSE 2009]
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290, , Secondary School Mathematics for Class 10, , 22. If (2p 1), 7, 3p are in AP, find the value of p., , [CBSE 2009], , 23. If the sum of first p terms of an AP is (ap bp), find its common, difference., [CBSE 2010], 2, , 24. If the sum of first n terms is (3n 2 5n), find its common difference., 25. Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms, is 40., [CBSE 2011], 26. What is the common difference of an AP in which a27 a7 84?, [CBSE 2017], , 27. If 1 4 7 10 … x 287, find the value of x., , [CBSE 2017], , ANSWERS (EXERCISE 5D), , 1. y 5, , 2. k 3, , 3. 15, , 4. a 1, b 17, , 5. n 3, (5, 11, 17), , 6. 128, , 7. 100, , 8. 9, , 11. 50, , 12. 3, , 13. –4, , 14., , 50, , 9. n 2 a, 15., , 32, , 10. 27, 16. 8th, , 1, 17. n (n 1), 2, , 18. n (n 1) 19. (p 9q), , 7, 20. a , 5, , 22. p 3, , 23. d 2a, , 25. 3, 5, 7, 9, 11, …, , 26. d = 4.2, , 27. x 40, , 24. d 6, , HINTS TO SOME SELECTED QUESTIONS, 1. We have, (3y 5) (3y 1) (5y 1) (3y 5), , , 2y 4 6 & 2y 10 & y 5., , 2. We have, (2k 1) k (2k 1) (2k 1), , , k 1 2 & k 3., , 3. We have, a 18 (b 3 a) & 2a b (18 3) 15., 4. We have, 9 a b 9 25 b, , , 2b 34 & b 17., , Also, 9 a 17 9 8 & a 9 8 1., 5. We have, (3n 2) (2n 1) (6n 1) (3n 2), , , n 3 3n 3 & 2n 6 & n 3., , The numbers are (2 # 3 1), (3 # 3 2) and (6 # 3 1), i.e., 5, 11, 17., 6. These numbers are 105, 112, 119, 126, …, 994., Let Tn 994. Then, 105 (n 1)#7 994 & n 128., 7. These numbers are 108, 117, 126, …, 999., Let Tn 999. Then, 108 (n 1)# 9 999 & n 100., , 21. p 4
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Arithmetic Progression, 8. Sm (2m 2 3m) & S1 2 3 5 and S2 2 # 4 3 # 2 14., T2 S2 S1 14 5 9., n, 9. Required sum [2a (n 1)# 2a] n (na) n 2 a., 2, , , 10. nth term from the end l (n 1) d., , , 5th term from the end 47 4 # 5 47 20 27., , 11. (a30 a20) (2 29 # 5) (2 19 # 5) 147 97 50., 12. Tn (3n 5) & T1 (3 #1 5) 8, T2 (3 # 2 5) 11., , , d (T2 T1) 3., , 13. Tn (7 4n) & T1 (7 4 #1) 3, T2 (7 4 # 2) 1., , , d (T2 T1) (1 3) 4., , 14. Given terms are 4 # 2 , 9 # 2 , 16 # 2 , …, i.e., 2 2 , 3 2 , 4 2 , … ., So, the next term is 5 2 5 # 5 # 2 50 ., 15. Given terms are 2 , 4 # 2 , 9 # 2 , …, i.e., 2 , 2 2 , 3 2 , … ., The next term is 4 2 4 # 4 # 2 32 ., 16. Let Tn 0. Then, 21 (n 1)#(3) 0 & n 8. So, 8th term is zero., 17. Sn 1 2 3 … n , , n , (1 n) ., 2, , 18. Sn 2 4 6 … 2n , , :Sn , , n , (a l)D, 2, , n , n, (2 2n) n (n 1) . :Sn (a l)D, 2, 2, , 19. Here a p and d q., , , T10 a 9d (p 9q) ., , 20. We have, a , , 4 , 4, 14, 7, 2 a & 2a a2 k , &a ·, 5, 5, 5, 5, , 21. We have, 13 (2p 1) (5p 3) 13., , , (5p 3) (2p 1) 26 & 7p 28 & p 4., , 22. We have, 7 (2p 1) 3p 7., , , (2p 1) 3p 14 & 5p 15 & p 3., , 23. T1 S1 a b [putting p 1] ., Now, S2 (a # 4 b # 2) (4a 2b) ., , , , T2 S2 S1 (4a 2b) (a b) (3a b) ., d T2 T1 (3a b) (a b) 2a., , 24. Sn (3n 2 5n), , , T1 S1 (3 #1 5 #1) 8., , Now, S2 3 # 4 5 # 2 22., , , T2 S2 S1 22 8 14., , , , d T2 T1 14 8 6., , 291
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292, , Secondary School Mathematics for Class 10, , 25. T4 9 & a 3d 9., , … (i), , T6 T13 40 & (a 5d) (a 12d) 40 & 2a 17d 40., , … (ii), , From (i) and (ii), we get a 3 and d 2., , , AP is 3, 5, 7, 9, 11, … ., , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1 1 p 1 2p, 1. The common difference of the AP p , p , p , … is, (a) p, , (b) –p, , (c) –1, , [CBSE 2013], , (d) 1, , 1 1 3b 1 6b, 2. The common difference of the AP ,, ,, , … is, 3, 3, 3, 1, 1, (a), (b), (c) b, (d) b, 3, 3, 3. The next term of the AP 7 , 28 , 63 , … is, (a), , 70, , (b), , 84, , (c), , [CBSE 2013], , [CBSE 2014], , 98, , (d), , 112, , 4. If 4, x1, x2, x3, 28 are in AP then x3 ?, (a) 19, , (b) 23, , (c) 22, , (d) Cannot be determined, , 5. If the nth term of an AP is (2n 1) then the sum of its first three terms is, [CBSE 2012], , (a) 6n 3, , (b) 15, , (a) 6, , (b) 9, , (c) 12, (d) 21, 2, 6. The sum of first n terms of an AP is (3n 6n) . The common difference, of the AP is, [CBSE 2014], (c) 15, , (d) –3, , 7. The sum of first n terms of an AP is (5n n 2) . The nth term of the AP is, (a) (5 2n), (b) (6 2n), (c) (2n 5), (d) (2n 6), 8. The sum of first n terms of an AP is (4n 2 2n) . The nth term of this AP is, [CBSE 2014], , (a) (6n 2), (b) (7n 3), (c) (8n 2), (d) (8n 2), 9. The 7th term of an AP is –1 and its 16th term is 17. The nth term of the, AP is, [CBSE 2014], (a) (3n 8), (b) (4n 7), (c) (15 2n), (d) (2n 15)
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Arithmetic Progression, , 293, , 10. The 5th term of an AP is –3 and its common difference is –4. The sum of, its first 10 terms is, [CBSE 2011], (a) 50, , (b) –50, , (c) 30, , (d) –30, , 11. The 5th term of an AP is 20 and the sum of its 7th and 11th terms is 64., The common difference of the AP is, [CBSE 2015], (a) 4, , (b) 5, , (c) 3, , (d) 2, , 12. The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16 then, the sum of its first ten terms is, [CBSE 2015], (a) 150, , (b) 175, , (c) 160, , (d) 135, , 13. An AP 5, 12, 19, … has 50 terms. Its last term is, (a) 343, , (b) 353, , (c) 348, , [CBSE 2015], , (d) 362, , 14. The sum of first 20 odd natural numbers is, (a) 100, , (b) 210, , (c) 400, , [CBSE 2012], , (d) 420, , 15. The sum of first 40 positive integers divisible by 6 is, (a) 2460, , (b) 3640, , (c) 4920, , [CBSE 2014], , (d) 4860, , 16. How many two-digit numbers are divisible by 3?, (a) 25, , (b) 30, , (c) 32, , [CBSE 2012], , (d) 36, , 17. How many three-digit numbers are divisible by 9?, (a) 86, , (b) 90, , (c) 96, , [CBSE 2013], , (d) 100, , 18. What is the common difference of an AP in which a18 a14 32?, (a) 8, , (b) –8, , (c) 4, , (d) –4, , 19. If an denotes the nth term of the AP 3, 8, 13, 18, … then what is the value, of (a30 a20) ?, (a) 40, , (b) 36, , (c) 50, , (d) 56, , 20. Which term of the AP 72, 63, 54, … is 0?, (a) 8th, , (b) 9th, , (c) 10th, , (d) 11th, , 21. Which term of the AP 25, 20, 15, … is the first negative term?, (a) 10th, , (b) 9th, , (c) 8th, , (d) 7th, , 22. Which term of the AP 21, 42, 63, 84, … is 210?, (a) 9th, , (b) 10th, , (c) 11th, , (d) 12th, , 23. What is 20th term from the end of the AP 3, 8, 13, …, 253?, (a) 163, (b) 158, , , , , 24. (5 13 21 … 181) ?, (a) 2476, , (b) 2337, , (c) 153, , (d) 148, , (c) 2219, , (d) 2139
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294, , Secondary School Mathematics for Class 10, , 25. The sum of first 16 terms of the AP 10, 6, 2, … is, (a) 320, , (b) –320, , (c) –352, , (d) –400, , 26. How many terms of the AP 3, 7, 11, 15, … will make the sum 406?, (a) 10, , (b) 12, , (c) 14, , (d) 20, , 27. The 2nd term of an AP is 13 and its 5th term is 25. What is its 17th term?, (a) 69, , (b) 73, , (c) 77, , (d) 81, , 28. The 17th term of an AP exceeds its 10th term by 21. The common, difference of the AP is, (a) 3, , (b) 2, , (c) –3, , (d) –2, , 29. The 8th term of an AP is 17 and its 14th term is 29. The common, difference of the AP is, (a) 3, , (b) 2, , (c) 5, , (d) –2, , 30. The 7th term of an AP is 4 and its common difference is –4. What is its, first term?, (a) 16, , (b) 20, , (c) 24, , (d) 28, , ANSWERS (MCQ), , 1. (c), , 2. (d), , 10. (b) 11. (c), , 3. (d), , 4. (c), , 5. (b), , 6. (a), , 12. (b) 13. (c), , 14. (c), , 15. (c), , 7. (b), , 8. (c), , 16. (b) 17. (d) 18. (a), , 19. (c), , 20. (b) 21. (d) 22. (b) 23. (b) 24. (d) 25. (b) 26. (c), , 28. (a), , 29. (b) 30. (d), HINTS TO SOME SELECTED QUESTIONS, , p, 1p 1, 1p1, n, 1., 1. d ( p p 2 d, p, p, 1 3b 1 1 3b 1 3b , m, 1 c, 2. d ', b., 3, 3, 3, 3, 3. Given terms are 7 , 4 #7 , 9 #7 , …, i.e., 7 , 2 7 , 3 7 , … ., So, the next term is 4 7 4 # 4 #7 112 ., 4. Given AP is 4, x1, x2, x3, 28. Clearly, a 4 and T5 28., Now, T5 28 & a 4d 28 & 4 4d 28 & 4d 24 & d 6., , , 9. (d), , x3 T4 (a 3d) (4 3 # 6) 22., , 5. (T1 T2 T3) (2 #1 1) (2 # 2 1) (2 # 3 1) 3 5 7 15., , 27. (b)
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Arithmetic Progression, , 295, , 6. Given Sn (3n 2 6n) ., , , T1 S1 (3 #1 2 6 #1) 9, S2 (3 # 2 2 6 # 2) 24., , , , T2 (S2 S1) (24 9) 15., , Hence, d (T2 T1) 15 9 6., 7. Tn (Sn Sn 1) (5n n 2) {5 (n 1) (n 1) 2} (5n n 2) (7n n 2 6) (6 2n) ., 8. Tn (Sn Sn 1) (4n 2 2n) {4 (n 1) 2 2 (n 1)}, (4n 2 2n) (4n 2 6n 2) (8n 2) ., 9. T7 1 & a 6d 1., T16 17 & a 15d 17., , … (i), … (ii), , On solving (i) and (ii), we get a 13 and d 2., , , Tn a (n 1) d 13 (n 1)# 2 (2n 15) ., , 10. a 4d 3 and d 4. So, a 13., 10, S10 , [2a 9d] 5 [2 #13 9 #(4)] 50., 2, 11. T5 20 & a 4d 20., (T7 T11) 64 & (a 6d) (a 10d) 64 & a 8d 32., , … (i), … (ii), , On solving (i) and (ii), we get d 3., 12. T13 4 #T3 & a 12d 4 (a 2d) & 3a 4d 0., T5 16 & a 4d 16., On solving (i) and (ii), we get a 4 and d 3., 10, S10 , (2a 9d) 5 (2 # 4 9 # 3) 175., 2, 13. 50th term a (n 1) d (5 49 #7) 348., 14. S20 1 3 5 7 … up to 20 terms., Here a 1, d 2. So, T20 (a 19d) (1 19 # 2) 38 l., 20 , n, S20 (a l) , (1 39) 400., 2, 2, 40 , 15. Required sum 6 12 18 … 240 , (6 240) 4920., 2, 16. Two-digit numbers divisible by 3 are 12, 15, 18, …, 99., Let Tn 99. Then, 12 (n 1)# 3 99 & (n 1)# 3 87 & n 30., 17. Three-digit numbers divisible by 9 are 108, 117, 126, …, 999., Let Tn 999. Then, 108 (n 1)# 9 999., , , (n 1)# 9 891 & (n 1) 99 & n 100., , 18. a18 a14 32 & (a 17d) (a 13d) 32 & 4d 32 & d 8., 19. Here a 3 and d 5., , , (a30 a20) (3 29 # 5) (3 19 # 5) 148 98 50., , 20. Let the nth term of 72, 63, 54, … be 0. Then,, Tn 0 & 72 (n 1)#( 9) 0 & 9n 81 & n 9., , … (i), … (ii)
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296, , Secondary School Mathematics for Class 10, , 21. Let Tn 0. Then, 25 (n 1)#(5) 0., , , 30 5n & 5n 30 & n 6., , So, the required term is 7th., 22. Let Tn 210. Then, 21 (n 1)# 21 210, , , (n 1)# 21 189 & n 1 9 & n 10., , 23. 20th term from the end (253 19 # 5) (253 95) 158. [l (n 1) d], 24. Let the number of terms be n. Then, Tn 181., , , 5 (n 1)# 8 181 & (n 1)# 8 176 & (n 1) 22 & n 23., , n 23 , (a l), (5 181) 23 # 93 2139., 2, 2, 25. l T16 10 15 #(4) 50., , , sum , , n 16, (a l), (10 50) 8 #( 40) 320., 2, 2, n, 26. Let Sn 406. Then, [2 # 3 (n 1)# 4] 406, 2, n , n, , (6 4n 4) 406 & (4n 2) 406, 2, 2, , , sum , , , , n (2n 1) 406 & 2n 2 n 406 0., , , , n, , , , 1 ! 1 3248 1 3249, , 4, 4, 1 57 56, , 14., n, 4, 4, , 27. a d 13, , … (i), , [neglecting negative value], , and a 4d 25, , From (i) and (ii), we get a 9 and d 4., , , T17 (a 16d) (9 16 # 4) 73., , 28. (T17 T10) 21 & (a 16d) (a 9d) 21 & 7d 21 & d 3., 29. (T14 T8) (29 17) 12 & (a 13d) (a 7d) 12 & 6d 12 & d 2., 30. a 6d 4 & a 6 #(4) 4 & a 28., , , , … (ii).
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298, , Secondary School Mathematics for Class 10, , SUMMARY, , 1. The distance between the points A( x1 , y1 ) and B( x2 , y2 ) is given by, AB = ( x2 − x1 ) 2 + ( y2 − y1 ) 2, 2. The distance of the point P( x , y) from the origin O(0, 0) is given by, OP = x 2 + y 2 ., , SOLVED, EXAMPLE 1, SOLUTION, , EXAMPLES, , Find the distance between the points A(7, 13) and B(10, 9)., The given points are A(7 , 13) and B(10 , 9)., Then, x1 = 7 , y1 = 13 and x2 = 10 , y2 = 9., ∴, , AB = ( x2 − x1 ) 2 + ( y2 − y1 ) 2, = (10 − 7 ) 2 + ( 9 − 13) 2 =, , 3 2 + ( −4) 2, , = 9 + 16 = 25 = 5 units., EXAMPLE 2, SOLUTION, , Find the distance between the points P( −4 , 7 ) and Q( 2 , − 5)., The given points are P( −4 , 7 ) and Q( 2 , − 5)., Then, x1 = −4 , y1 = 7 and x2 = 2 , y2 = −5., ∴, , PQ = ( x2 − x1 ) 2 + ( y2 − y1 ) 2, = {2 − ( −4)} 2 + ( −5 − 7 ) 2 = 6 2 + ( −12) 2, =, , EXAMPLE 3, SOLUTION, , 36 + 144 = 180 =, , 36 × 5 = 6 5 units., , Find the distance of the point P( 6 , − 6) from the origin., Let P( 6 , − 6) be the given point and O( 0 , 0) be the origin., Then, OP = ( 6 − 0) 2 + ( −6 − 0) 2 = 6 2 + ( −6) 2, =, , 36 + 36 = 72 =, , 36 × 2 = 6 2 units., , EXAMPLE 4, , Find the values of y for which the distance between the points, [CBSE 2011], A( 3 , − 1) and B(11 , y) is 10 units., , SOLUTION, , We have, AB = 10 ⇒ AB2 = 100, ⇒ (11 − 3) 2 + ( y + 1) 2 = 100, ⇒ 8 2 + ( y + 1) 2 = 100, ⇒ ( y + 1) 2 = 100 − 64 = 36 = 6 2
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Coordinate Geometry, , 299, , ⇒ y + 1 = ±6, ⇒ y + 1 = 6 or y + 1 = −6, ⇒ y = 5 or y = −7 ., Hence, the required values of y are 5 and –7., EXAMPLE 5, , If the point P( k − 1 , 2) is equidistant from the points A( 3 , k) and, [CBSE 2014], B( k , 5), find the values of k., , SOLUTION, , It is being given that P( k − 1 , 2) is equidistant from the points, A( 3 , k) and B( k , 5). So, we have, PA = PB ⇒ PA 2 = PB2, ⇒ ( k − 1 − 3) 2 + ( 2 − k) 2 = ( k − 1 − k) 2 + ( 2 − 5) 2, ⇒ ( k − 4) 2 + ( 2 − k) 2 = ( −1) 2 + ( −3) 2, ⇒ 2 k 2 − 12 k + 20 = 1 + 9 ⇒ 2 k 2 − 12 k + 10 = 0, ⇒ k 2 − 6k + 5 = 0 ⇒ k 2 − 5k − k + 5 = 0, ⇒ k( k − 5) − ( k − 5) = 0 ⇒ ( k − 5)( k − 1) = 0, ⇒ k = 1 or k = 5., Hence, the required values of k are 1 and 5., , EXAMPLE 6, , Find the relation between x and y such that the point P( x , y) is, equidistant from the points A(1 , 4) and B( −1 , 2)., [CBSE 2012], , SOLUTION, , We have, PA = PB ⇒ PA 2 = PB2, ⇒ ( x − 1) 2 + ( y − 4) 2 = ( x + 1) 2 + ( y − 2) 2, ⇒ x 2 + y 2 − 2 x − 8 y + 17 = x 2 + y 2 + 2 x − 4 y + 5, ⇒ 4 x + 4 y − 12 = 0 ⇒ x + y = 3 ⇒ y = 3 − x., Hence, y = 3 − x is the desired relation between x and y., TWO IMPORTANT NOTES, , (i) Any point on the x-axis is of the form (x, 0)., (ii) Any point on the y-axis is of the form (0, y)., EXAMPLE 7, , Find those points on x-axis, each of which is at a distance of 5 units, from the point A(5, –3)., [CBSE 2011], , SOLUTION, , Let the required point on the x-axis be P( x , 0). Then,, PA = 5 ⇒ PA 2 = 25, ⇒ ( x − 5) 2 + ( 0 + 3) 2 = 25, ⇒ x 2 − 10 x + 25 + 9 = 25 ⇒ x 2 − 10 x + 9 = 0, ⇒ x 2 − x − 9 x + 9 = 0 ⇒ x( x − 1) − 9( x − 1) = 0
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300, , Secondary School Mathematics for Class 10, , ⇒ ( x − 1)( x − 9) = 0 ⇒ x − 1 = 0 or x − 9 = 0, ⇒ x = 1 or x = 9., Hence, the required points on the x-axis are B(1, 0) and C(9, 0)., EXAMPLE 8, SOLUTION, , Find those points on the y-axis, each of which is at a distance of, 13 units from the point A(–5, 7)., Let the required point on the y-axis be P( 0 , y). Then,, PA = 13 ⇒ PA 2 = 169, ⇒ ( 0 + 5) 2 + ( y − 7 ) 2 = 169, ⇒ y 2 − 14 y + 74 = 169 ⇒ y 2 − 14 y − 95 = 0, ⇒ y 2 − 19 y + 5 y − 95 = 0 ⇒ y( y − 19) + 5( y − 19) = 0, ⇒ ( y − 19)( y + 5) = 0 ⇒ y − 19 = 0 or y + 5 = 0, ⇒ y = 19 or y = −5., Hence, the required points on the y-axis are B( 0 , 19) and, C( 0 , − 5)., , EXAMPLE 9, , Find the point on x-axis which is equidistant from the points (5 , − 2), and ( −3 , 2)., [CBSE 2009], , SOLUTION, , Let the given points be A(5 , − 2) and B( −3 , 2) and let the, required point be P( x , 0). Then,, PA = PB ⇒ PA 2 = PB2, ⇒ ( x − 5) 2 + ( 0 + 2) 2 = ( x + 3) 2 + ( 0 − 2) 2, ⇒ ( x − 5) 2 + 4 = ( x + 3) 2 + 4, ⇒ ( x − 5) 2 − ( x + 3) 2 = 0, ⇒ ( x 2 − 10 x + 25) − ( x 2 + 6 x + 9) = 0, ⇒ −16 x + 16 = 0, ⇒ 16 x = 16 ⇒ x = 1., Hence, the required point is P(1 , 0)., , EXAMPLE 10, , Find the point on y-axis which is equidistant from the points ( −5 , 2), [CBSE 2009C], and ( 9 , − 2)., , SOLUTION, , Let the given points be A( −5 , 2) and B( 9 , − 2), and let the, required point be P( 0 , y). Then,, ⇒, ⇒, ⇒, ⇒, , PA = PB ⇒ PA 2 = PB2, ( 0 + 5) 2 + ( y − 2) 2 = ( 0 − 9) 2 + ( y + 2) 2, 5 2 + ( y − 2) 2 = ( −9) 2 + ( y + 2) 2, 25 + ( y − 2) 2 = 81 + ( y + 2) 2, ( y + 2) 2 − ( y − 2) 2 = 25 − 81
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Coordinate Geometry, , 301, , ⇒ 8 y = −56 ⇒ y = −7 ., Hence, the required point is P( 0 , − 7 )., EXAMPLE 11, , Find the coordinates of the point equidistant from three given points, [CBSE 2006], A(5 , 1), B( −3 , − 7 ) and C(7 , −1)., , SOLUTION, , Let the required point be P( x , y). Then,, PA = PB = PC ⇒ PA 2 = PB2 = PC 2, ⇒ PA 2 = PB2 and PB2 = PC 2, ⎧( x − 5) 2 + ( y − 1) 2 = ( x + 3) 2 + ( y + 7 ) 2 ,, ⇒ ⎨, 2, 2, 2, 2, ⎩⎪( x + 3) + ( y + 7 ) = ( x − 7 ) + ( y + 1), ⎧ x 2 + y 2 − 10 x − 2 y + 26 = x 2 + y 2 + 6 x + 14 y + 58 ,, ⇒ ⎨ 2, 2, 2, 2, ⎩⎪ x + y + 6 x + 14 y + 58 = x + y − 14 x + 2 y + 50., Now, x 2 + y 2 − 10 x − 2 y + 26 = x 2 + y 2 + 6 x + 14 y + 58, ⇒ 16 x + 16 y = −32 ⇒ x + y = −2., , ... (i), , And, x + y + 6 x + 14 y + 58 = x + y − 14 x + 2 y + 50, 2, , 2, , 2, , 2, , ⇒ 20 x + 12 y = −8 ⇒ 5 x + 3 y = −2 ., , ... (ii), , Multiplying (i) by 5 and subtracting (ii) from the result, we get, 2 y = −8 ⇒ y = −4., Putting y = −4 in (i), we get x = 2., ∴, , x = 2 and y = −4., , Hence, the required point is P( 2 , − 4)., EXAMPLE 12, , Points A( −1 , y) and B(5 , 7 ) lie on a circle with centre O( 2 , − 3 y)., Find the values of y. Hence, find the radius of the circle. [CBSE 2014], , SOLUTION, , Since the radii of a circle are equal, we have, OA = OB ⇒ OA 2 = OB2, ⇒ ( 2 + 1) 2 + ( −3 y − y) 2 = ( 2 − 5) 2 + ( −3 y − 7 ) 2, ⇒ 3 2 + ( −4 y) 2 = ( −3) 2 + ( 3 y + 7 ) 2, ⇒ 9 + 16 y 2 = 9 + 9 y 2 + 49 + 42 y, ⇒ 7 y 2 − 42 y − 49 = 0, ⇒ y2 − 6y − 7 = 0, ⇒ y2 − 7 y + y − 7 = 0, ⇒ y( y − 7 ) + ( y − 7 ) = 0 ⇒ ( y − 7 )( y + 1) = 0, ⇒ y − 7 = 0 or y + 1 = 0 ⇒ y = 7 or y = −1., ∴ centre is either (2, 3) or (2, –21).
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302, , Secondary School Mathematics for Class 10, CASE I, , When centre is (2, 3), In this case, radius = (5 − 2) 2 + (7 − 3) 2 =, , 32 + 42, , = 9 + 16 = 25 = 5 units., CASE II, , When centre is (2, –21), In this case, radius = (5 − 2) 2 + {7 − ( −21)} 2, =, , 3 2 + ( 28) 2 = 9 + 784, , = 793 = 28.1 units., EXAMPLE 13, SOLUTION, , Find the centre of a circle passing through the points ( 6 , − 6),, ( 3 , −7 ) and ( 3 , 3)., Let P( x , y) be the centre of a circle passing through the points, A( 6 , − 6), B( 3 , − 7 ) and C(3, 3). Then,, PA 2 = PB2 and PB2 = PC 2, , ⎧( x − 6) 2 + ( y + 6) 2 = ( x − 3) 2 + ( y + 7 ) 2 ,, ⇒ ⎨, 2, 2, 2, 2, ⎩( x − 3) + ( y + 7 ) = ( x − 3) + ( y − 3), ⎧ x 2 + y 2 − 12 x + 12 y + 72 = x 2 + y 2 − 6 x + 14 y + 58 ,, ⇒ ⎨ 2, 2, 2, 2, ⎩ x + y − 6 x + 14 y + 58 = x + y − 6 x − 6 y + 18, ⇒ 6 x + 2 y = 14 and 20 y = −40, ⇒ 3 x + y = 7 and y = −2, ⇒ x = 3 and y = −2., Hence, the centre of the circle is P(3, –2)., EXAMPLE 14, , The points A(4, 7), B( p , 3) and C(7 , 3) are the vertices of a right, triangle, right angled at B. Find the value of p., [CBSE 2015], , SOLUTION, , Since SABC is a right angled at B, we have, AB2 + BC 2 = AC 2, ⇒ {( p − 4) 2 + ( 3 − 7 ) 2} + {(7 − p) 2 + ( 3 − 3) 2} = (7 − 4) 2 + ( 3 − 7 ) 2, ⇒ ( p − 4) 2 + ( −4) 2 + (7 − p) 2 + 0 = 3 2 + ( −4) 2
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Coordinate Geometry, , 303, , ⇒ 2 p 2 − 22 p + (16 + 49 + 16) = 9 + 16, ⇒ 2 p 2 − 22 p + 56 = 0, ⇒ p 2 − 11 p + 28 = 0, ⇒ p 2 − 7 p − 4 p + 28 = 0, ⇒ p( p − 7 ) − 4( p − 7 ) = 0, ⇒ ( p − 7 )( p − 4) = 0, ⇒ p − 7 = 0 or p − 4 = 0, ⇒ p = 7 or p = 4., When p = 7 then the points B and C coincide and so no triangle, is formed in this case., So, p ≠ 7. Hence, p = 4., EXAMPLE 15, , SOLUTION, , Show that the points ( a , a), ( − a, − a) and ( − 3 a,, vertices of an equilateral triangle. Find its area., , 3 a) are the, [CBSE 2015], , Let the given points be A( a, a), B( − a, − a) and C ( − 3 a,, Then, AB2 = ( − a − a) 2 + ( − a − a) 2 = ( −2 a) 2 + ( −2 a) 2 = 8 a2 ,, , 3 a)., , BC 2 = ( − 3 a + a) 2 + ( 3 a + a) 2, = ( a − 3 a) 2 + ( a + 3 a) 2 = 2( a2 + 3 a2 ) = 8 a2, and AC 2 = ( − 3 a − a) 2 + ( 3 a − a) 2 = ( 3 a + a) 2 + ( 3 a − a) 2, = 2( 3 a2 + a2 ) = 8 a2 ., ∴, , AB = BC = AC = 8 a2 = 2 2 a., , Hence, SABC is equilateral and each of its sides is 2 2a., Area of equilateral SABC, ⎧ 3, ⎫, ⎧ 3, ⎫, =⎨, × ( side) 2 ⎬ sq units = ⎨, × ( 2 2 a) 2 ⎬ sq units, ⎩ 4, ⎭, ⎩ 4, ⎭, ⎧ 3, ⎫, =⎨, × 8 a2 ⎬ sq units = 2 3 a2 sq units., 4, ⎩, ⎭, EXAMPLE 16, , Prove that the points A( −3 , 0), B(1 , − 3) and C( 4 , 1) are the vertices, of an isosceles right-angled triangle. Find the area of this triangle., , SOLUTION, , Let A( −3 , 0), B(1 , − 3) and C( 4 , 1) be the given points. Then,, AB = [1 − ( −3)]2 + ( −3 − 0) 2 = 4 2 + ( −3) 2 = 25 = 5 units,, BC = ( 4 − 1) 2 + [1 − ( −3)]2 =, , 3 2 + 4 2 = 25 = 5 units, , and AC = [4 − ( −3)]2 + (1 − 0) 2 = 7 2 + 1 2 = 50 = 5 2 units.
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304, , Secondary School Mathematics for Class 10, , Thus, AB = BC = 5 units., \, , CABC is isosceles., , Also, ( AB2 + BC 2 ) = (5 2 + 5 2 ) = 50 and, AC 2 = (5 2 ) 2 = 50., Thus, AB2 + BC 2 = AC 2 ., This shows that CABC is right-angled at B., In CABC, we have, \, , base = BC = 5 units and height = AB = 5 units., 1, 1, area of CABC =æç ´ base ´ height ö÷ =æç ´ 5 ´ 5 ö÷ sq units, è2, ø è2, ø, = 12.5 sq units., , PROPERTIES OF VARIOUS TYPES OF QUADRILATERALS, , A quadrilateral is a, (i), (ii), (iii), (iv), , rectangle if its opposite sides are equal and the diagonals are equal., square if all its sides are equal and the diagonals are equal., parallelogram if its opposite sides are equal., parallelogram but not a rectangle if its opposite sides are equal and the, diagonals are not equal., (v) rhombus but not a square if all its sides are equal and the diagonals are, not equal., , EXAMPLE 17, , Show that the points (1 , 1), ( -1 , 5), (7 , 9) and ( 9 , 5) taken in that, order are the vertices of a rectangle. Also, find the area of the, rectangle., [CBSE 2009C], , SOLUTION, , Let A(1 , 1), B( -1 , 5), C (7 , 9) and D( 9 , 5) be the vertices of, quad. ABCD. Then,, AB2 = ( -1 - 1) 2 + (5 - 1) 2, \, , = ( -2) 2 + 4 2 = 4 + 16 = 20., AB = 20 units = 4 ´ 5 units, , \, , BC = (7 + 1) 2 + ( 9 - 5) 2, = 8 2 + 4 2 = 64 + 16 = 80., BC = 80 units = 16 ´ 5 units = 4 5 units., , = 2 5 units., 2, , CD 2 = 9 - 7 2 + 5 - 9 2, = 2 2 + ( -4) 2 = 4 + 16 = 20., \, , CD = 20 units = 4 ´ 5 units = 2 5 units.
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Coordinate Geometry, , 305, , AD 2 = ( 9 − 1) 2 + (5 − 1) 2, = 8 2 + 4 2 = 64 + 16 = 80., ∴, , AD = 80 units = 16 × 5 units = = 4 5 units., , Thus, AB = CD = 2 5 units and BC = AD = 4 5 units., Also, AC 2 = (7 − 1) 2 + ( 9 − 1) 2, = 6 2 + 8 2 = 36 + 64 = 100., ∴, , AC = 100 units = 10 units., , And, BD 2 = ( 9 + 1) 2 + (5 − 5) 2 = 10 2 + 0 2 = 100., ∴, , BD = 100 units = 10 units., , ∴, , diagonal AC = diagonal BD., , Thus, ABCD is a quadrilateral whose opposite sides are, equal and the diagonals are equal., Hence, quad. ABCD is a rectangle., Area of rectangle ABCD = length × breadth, = 4 5 × 2 5 sq units, = 40 sq units., EXAMPLE 18, , Show that the points A( 3 , 5), B( 6 , 0), C (1 , − 3) and D( −2 , 2) are the, vertices of a square ABCD., [CBSE 2012], , SOLUTION, , Let A( 3 , 5), B( 6 , 0), C (1 , − 3) and D( −2 , 2) be the angular points, of a quadrilateral ABCD. Join AC and BD., Now, AB = ( 6 − 3) 2 + ( 0 − 5) 2, =, , 3 2 + ( −5) 2, , = 9 + 25 =, , 34 units,, , BC = (1 − 6) + ( −3 − 0) 2 = ( −5) 2 + ( −3) 2, 2, , = 25 + 9 =, , 34 units,, , CD = ( −2 − 1) 2 + ( 2 + 3) 2 = ( −3) 2 + 5 2, = 9 + 25 =, , 34 units,, , and DA = ( 3 + 2) + (5 − 2) 2 = 5 2 + 3 2, 2, , = 25 + 9 =, , 34 units., , Thus, AB = BC = CD = DA., Diag. AC = (1 − 3) 2 + ( −3 − 5) 2 = ( −2) 2 + ( −8) 2, = 4 + 64 = 68 = 2 17 units.
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306, , Secondary School Mathematics for Class 10, , Diag. BD = ( −2 − 6) 2 + ( 2 − 0) 2, = ( −8) 2 + 2 2 = 64 + 4, ∴, , = 68 = 2 17 units., diag. AC = diag. BD., , Thus, ABCD is a quadrilateral in which all sides are equal and, the diagonals are equal., Hence, quad. ABCD is a square., EXAMPLE 19, , If A( 2 , − 1), B( 3 , 4), C ( −2 , 3) and D( −3 , − 2) be four points in a, plane, show that ABCD is a rhombus but not a square. Find the area, of the rhombus., [CBSE 2013], , SOLUTION, , Let A( 2 , −1), B( 3 , 4), C( −2 , 3) and, D( −3 , − 2) be the angular points of a, quad. ABCD. Join AC and BD. Then,, AB = ( 3 − 2) 2 + ( 4 + 1) 2, = 1 2 + 5 2 = 26 units,, BC = ( −2 − 3) 2 + ( 3 − 4) 2 = ( −5) 2 + ( −1) 2 = 26 units,, CD = ( −3 + 2) 2 + ( −2 − 3) 2 = ( −1) 2 + ( −5) 2 = 26 units,, DA = ( −3 − 2) 2 + ( −2 + 1) 2 = ( −5) 2 + ( −1) 2 = 26 units., ∴, , AB = BC = CD = DA = 26 units., , Now, AC = ( −2 − 2) 2 + ( 3 + 1) 2 = ( − 4) 2 + 4 2, =, , 32 = 4 2 units, , and BD = ( −3 − 3) 2 + ( −2 − 4) 2 = ( −6) 2 + ( − 6) 2, ∴, , = 72 = 6 2 units., diagonal AC ≠ diagonal BD., , Thus, ABCD is a quadrilateral in which all sides are equal but, diagonals are not equal., ∴ ABCD is a rhombus but not a square., 1, ∴ area of the rhombus ABCD = × (product of diagonals), 2, 1, ⎛, = ⎜ × AC × BD ⎞⎟, ⎝2, ⎠, 1, = ⎛⎜ × 4 2 × 6 2 ⎞⎟ sq units, ⎝2, ⎠, = 24 sq units., Hence, the area of rhombus ABCD is 24 sq units.
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Coordinate Geometry, COLLINEAR POINTS, , 307, , Three points A, B, C are said to be collinear if they lie on the, , same straight line., TEST FOR COLLINEARITY OF THREE POINTS, , In order to show that three given points A, B, C are collinear, we find the, distances AB, BC and AC. If the sum of any two of these distances is equal to, the third distance then the given points are collinear., EXAMPLE 20, SOLUTION, , Prove that the points A(1 , 1), B( −2 , 7 ) and C( 3 , − 3) are collinear., Let A(1 , 1), B( −2 , 7 ) and C( 3 , − 3) be the given points. Then,, AB = ( −2 − 1) 2 + (7 − 1) 2 = ( −3) 2 + 6 2 = 45 = 3 5 units,, BC = ( 3 + 2) 2 + ( −3 − 7 ) 2 = 5 2 + ( −10) 2 = 125 = 5 5 units,, AC = ( 3 − 1) 2 + ( −3 − 1) 2 = 2 2 + ( −4) 2 = 20 = 2 5 units., ∴ AB + AC = ( 3 5 + 2 5 ) units = 5 5 units = BC., Thus, AB + AC = BC ., Hence, the given points A, B, C are collinear., , EXERCISE 6A, 1. Find the distance between the points:, (i) A(9, 3) and B(15, 11), (iii) A( −6 , − 4) and B( 9 , −12), , (ii) A(7 , − 4) and B( −5 , 1), (iv) A(1 , − 3) and B( 4 , − 6), , (v) P( a + b , a − b) and Q( a − b , a + b), (vi) P( a sin α , a cos α) and Q( a cos α , − a sin α), 2. Find the distance of each of the following points from the origin:, (i) A(5 , −12), (ii) B( −5 , 5), (iii) C( −4 , − 6) ., 3. Find all possible values of x for which the distance between the points, A( x , − 1) and B(5, 3) is 5 units., 4. Find all possible values of y for which the distance between the points, A(2, –3) and B(10, y) is 10 units., 5. Find the values of x for which the distance between the points P( x , 4), [CBSE 2011], and Q(9, 10) is 10 units., 6. If the point A( x , 2) is equidistant from the points B( 8 , − 2) and C( 2 , − 2),, find the value of x. Also, find the length of AB., 7. If the point A( 0 , 2) is equidistant from the points B( 3 , p) and C ( p , 5),, find the value of p. Also, find the length of AB., [CBSE 2014]
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308, , Secondary School Mathematics for Class 10, , 8. Find the coordinates of the point on x-axis which is equidistant from the, [CBSE 2017], points (–2, 5) and (2, –3)., 9. Find points on the x-axis, each of which is at a distance of 10 units from, the point A(11 , − 8) ., 10. Find the point on the y-axis which is equidistant from the points A(6, 5), and B(–4, 3)., 11. If the point P( x , y) is equidistant from the points A(5, 1) and B(–1, 5),, [CBSE 2005], prove that 3 x = 2 y., 12. If P( x , y) is a point equidistant from the points A( 6 , −1) and B(2, 3),, [CBSE 2008C], show that x − y = 3., 13. Find the coordinates of the point equidistant from three given points, [CBSE 2006], A(5 , 3), B(5 , − 5) and C(1 , − 5)., 14. If the points A( 4 , 3) and B( x , 5) lie on a circle with the centre O( 2 , 3),, find the value of x., [CBSE 2009], HINT, , OA 2 = OB 2 ., , 15. If the point C(–2, 3) is equidistant from the points A(3, –1) and B(x, 8),, [CBSE 2013C], find the values of x. Also, find the distance BC., 16. If the point P(2, 2) is equidistant from the points A( −2 , k) and, [CBSE 2014], B( −2 k , − 3), find k. Also, find the length of AP., 17., , (i) If the point ( x , y) is equidistant from the points ( a + b , b − a) and, [CBSE 2011], ( a − b , a + b), prove that bx = ay., , (ii) If the distances of P( x , y) from A(5, 1) and B(–1, 5) are equal then, [CBSE 2017], prove that 3 x = 2 y., 18. Using the distance formula, show that the given points are collinear:, (i) (1, –1), (5, 2) and (9, 5), , (ii) (6, 9), (0, 1) and (–6, –7), , (iii) (–1, –1), (2, 3) and (8, 11), , (iv) (–2, 5), (0, 1) and (2, –3)., , 19. Show that the points A(7, 10), B(–2, 5) and C(3, –4) are the vertices of an, isosceles right triangle., [CBSE 2007, ’13], 20. Show that the points A(3, 0), B(6, 4) and C(–1, 3) are the vertices of an, isosceles right triangle., 21. If A(5, 2), B(2, –2) and C(–2, t) are the vertices of a right triangle with, [CBSE 2015], ∠B = 90 ° then find the value of t., 22. Prove that the points A(2, 4), B(2, 6) and C( 2 + 3 , 5) are the vertices of, an equilateral triangle., [CBSE 2013C], 23. Show that the points ( −3 , − 3), (3, 3) and ( −3 3 , 3 3 ) are the vertices of, [CBSE 2012], an equilateral triangle., 24. Show that the points A(−5, 6), B(3, 0) and C(9, 8) are the vertices of an, isosceles right-angled triangle. Calculate its area.
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Coordinate Geometry, , 309, , 25. Show that the points O(0, 0), A(3, 3) and B(3, − 3) are the vertices of, an equilateral triangle. Find the area of this triangle., 26. Show that the following points are the vertices of a square:, (i) A( 3 , 2), B( 0 , 5), C( −3 , 2) and D( 0 , − 1), , [CBSE 2008C], , (ii) A( 6 , 2), B( 2 , 1), C (1 , 5) and D(5 , 6), , [CBSE 2006], , (iii) A( 0 , − 2), B( 3 , 1), C ( 0 , 4) and D( −3 , 1), 27. Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the, vertices of a rhombus. Find the area of this rhombus., HINT, , Area of a rhombus =, , 1, × (product of its diagonals)., 2, , 28. Show that the points A(3, 0), B(4, 5), C(–1, 4) and D(–2, –1) are the, vertices of a rhombus. Find its area., 29. Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices, of a rhombus. Find its area., 30. Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular, points of a parallelogram. Is this figure a rectangle?, 31. Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a, parallelogram. Show that ABCD is not a rectangle., 32. Show that the following points are the vertices of a rectangle:, (i) A( −4 , − 1), B( −2 , − 4), C ( 4 , 0) and D(2, 3), (ii) A( 2 , − 2), B(14 , 10), C (11 , 13) and D( −1 , 1), (iii) A( 0 , − 4), B( 6 , 2), C ( 3 , 5) and D( −3 , − 1), 33. Show that a ABC with vertices A( −2 , 0), B( 0 , 2) and C(2, 0) is similar to, [CBSE 2017], aDEF with vertices D( − 4 , 0), F( 4 , 0) and E(0, 4)., 34. Show that a ABC, where A(22, 0), B(2, 0), C(0, 2) and a PQR where, [CBSE 2017], P( − 4 , 0), Q( 4 , 0), R ( 0 , 4) are similar triangles., ANSWERS (EXERCISE 6A), , 1. (i) 10 units, (iv) 3 2 units, 2. (i) 13 units, , (ii) 13 units, , (iii) 17 units, , (v) 2 2 b units, , (vi) 2 a units, , (ii) 5 2 units, , (iii) 5 2 units, , 3. x = 8 or x = 2, , 4. y = 3 or y = −9, , 5. x = 1 or x = 17, , 6. x = 5 , AB = 5 units, , 7. p = 1 , AB = 10 units, , 8. (–2, 0), , 9. (5 , 0) and (17 , 0), , 10. (0, 9), , 13. P( 3 , − 1), , 14. x = 2, 15. x = 2 or x = −6 , BC = 41 units, 16. ( k = −3 ⇒ AP = 41 ), ( k = −1 ⇒ AP = 5), 21. t = 1
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310, , Secondary School Mathematics for Class 10, , 25. 3 3 sq units, 29. 3 sq units, , 24. 50 sq units, 28. 24 sq units, , 27. 45 sq units, 30. No, , HINTS TO SOME SELECTED QUESTIONS, 15. AC = BC ⇒ AC 2 = BC 2, ⇒ ( 3 + 2 ) 2 + ( −4 ) 2 = ( x + 2 ) 2 + 5 2, ⇒ ( x + 2 ) 2 = 16 ⇒ x + 2 = 4 or x + 2 = −4, ⇒ x = 2 or x = −6., Distance between B(2, 8) and C(–2, 3) = 4 2 + 5 2 = 41 units., Distance between B(–6, 8) and C(–2, 3) = 4 2 + ( −5 ) 2 = 41 units., 16. PA = PB ⇒ PA 2 = PB 2, ⇒ ( 2 + 2)2 + ( 2 − k )2 = ( 2 + 2k )2 + ( 2 + 3)2, ⇒ ( 2 − k ) 2 − ( 2 + 2 k ) 2 = 25 − 16 = 9, ⇒ ( 4 + k 2 − 4k ) − ( 4 + 4k 2 + 8k ) = 9, ⇒ 3 k 2 + 12 k + 9 = 0, ⇒ k 2 + 4 k + 3 = 0 ⇒ ( k + 3 )( k + 1) = 0, ⇒ k = −3 or k = −1., Case 1. When k = −3, we have, , AP = distance between A( −2 , − 3 ) and P( 2 , 2 ) = 41 units., Case 2. When k = −1, we have, , AP = distance between A( −2 , − 1) and P( 2 , 2 ) = 25 = 5 units., 17. (i) It is given that the point P( x , y ) is equidistant from the points A( a + b , b − a) and, B( a − b , a + b ). Then,, PA = PB ⇒ PA 2 = PB 2, ⇒ {x − ( a + b )} 2 + {y − ( b − a)} 2 = {x − ( a − b )} 2 + {y − ( a + b )} 2, ⇒ {( x − a) + b} 2 − {( x − a) − b} 2 = {( y − b ) + a} 2 − {( y − b ) − a} 2, ⇒ 4( x − a)b = 4( y − b )a ⇒ ( x − a)b = ( y − b )a ⇒ bx = ay., 21. AB 2 + BC 2 = AC 2 ⇒ 25 + {16 + (t + 2 ) 2 } = 49 + (t − 2 ) 2, ⇒ (t + 2 ) 2 − (t − 2 ) 2 = 8 ⇒ 4 × t × 2 = 8 ⇒ t = 1., 23. Let the given points be A( −3 , − 3 ), B( 3 , 3 ) and C( −3 3 , 3 3 ). Then,, AB 2 = ( 6 2 + 6 2 ) = 72,, BC 2 = ( −3 3 − 3 ) 2 + ( 3 3 − 3 ) 2 = ( −3 ) 2 ( 3 + 1) 2 + 3 2 ( 3 − 1) 2, = 9 × [( 3 + 1) 2 + ( 3 − 1) 2 ] = 9 × 2 × [( 3 ) 2 + 12 ] = 72 ,, AC 2 = ( −3 3 + 3 ) 2 + ( 3 3 + 3 ) 2 = ( −3 ) 2 × ( 3 − 1) 2 + 3 2 × ( 3 + 1) 2, = 9 × [( 3 − 1) 2 + ( 3 + 1) 2 ] = 9 × 2 × [( 3 ) 2 + 12 ] = 72., ∴, , AB = BC = AC = 72 = 6 2 units., , ⎧ 3, ⎫ ⎛ 3, ⎞, 25. Here, a = 12. So, area = ⎨, × a2 ⎬ = ⎜, × 12 ⎟ = 3 3 sq units., ⎠, ⎩ 4, ⎭ ⎝ 4
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Coordinate Geometry, , 311, , 27. Diagonal AC = ( −5 ) 2 + 5 2 = 50 = 5 2 , and, diagonal BD = 9 2 + 9 2 = 162 = 9 2 ., ∴, , 1, 1, area = ⎛⎜ × AC × BD ⎞⎟ = ⎛⎜ × 5 2 × 9 2 ⎞⎟ sq units = 45 sq units., ⎝2, ⎠ ⎝2, ⎠, , 33. Find the length of sides AB, BC and CA of a ABC and of sides DE, EF and FD of a DEF., AB BC CA, Then, show that, =, =, ⋅, DE EF FD, , ................................................................, , SECTION FORMULA, The coordinates of the point P( x , y) which, divides the line segment joining A( x1 , y1 ) and B( x2 , y2 ) internally in, the ratio m : n are given by, mx2 + nx1, my2 + ny1, x=, ,y=, ⋅, m+n, m+n, Let X′ OX and YOY′ be the coordinate axes., Let A( x1 , y1 ) and B( x2 , y2 ) be the end points of the given line, segment AB., Let P( x , y) be the point which divides AB in the ratio m : n., , THEOREM 2, , PROOF, , (Section formula), , Then,, , AP m, = ⋅, PB n, , Draw AL ⊥ OX; BM ⊥ OX; PN ⊥ OX; AR ⊥ PN; and PS ⊥ BM., Now, AR = LN = ON − OL = ( x − x1 );, PS = NM = OM − ON = ( x2 − x);, PR = PN − RN = PN − AL = ( y − y1 );, BS = BM − SM = BM − PN = ( y2 − y)., Clearly, CARP and CPSB are similar and, therefore, their sides are, proportional.
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Coordinate Geometry, SOLUTION, , 313, , The end points of AB are A( 4 , − 3) and B( 9 , 7 )., ∴, , ( x1 = 4 , y1 = −3) and ( x2 = 9 , y2 = 7 )., , Also, m = 3 and n = 2., Let the required point be P( x , y)., By the section formula, we have, (mx2 + nx1 ), (my2 + ny1 ), ,, x=, y=, (m + n), (m + n), ⇒ x=, , ( 3 × 9 + 2 × 4), [ 3 × 7 + 2 × ( −3)], , y=, ( 3 + 2), ( 3 + 2), , ⇒ x = 7 , y = 3., Hence, the required point is P(7 , 3)., EXAMPLE 2, SOLUTION, , Find the coordinates of the midpoint of the line segment joining the, points A( −5 , 4) and B(7 , − 8)., Let M( x , y) be the midpoint of AB. Then,, [( −5) + 7 ], [4 + ( −8)], x=, = 1 and y =, = −2., 2, 2, Hence, the required point is M(1 , − 2)., , EXAMPLE 3, , Find the coordinates of the points of trisection of the line segment, joining the points A( −5 , 6) and B( 4 , − 3)., [CBSE 2009C], , SOLUTION, , Let P and Q be the points of trisection of AB., , Then, P divides AB in the ratio 1 : 2., So, the coordinates of P are, ⎛ 1 × 4 + 2 × ( −5) 1 × ( −3) + 2 × 6 ⎞, P⎜, ,, ⎟ , i.e., P( −2 , 3)., ⎝, ⎠, 1+2, 1+2, Also, Q divides AB in the ratio 2 : 1., So, the coordinates of Q are, ⎛ 2 × 4 + 1 × ( −5) 2 × ( −3) + 1 × 6 ⎞, Q⎜, ,, ⎟ , i.e., Q(1 , 0)., ⎝, ⎠, 2 +1, 2 +1, Hence, the points of trisection of AB are P( −2 , 3) and Q(1 , 0)., EXAMPLE 4, , SOLUTION, , Find the coordinates of a point P on the line segment joining A(1 , 2), 2, and B( 6 , 7 ) such that AP = AB., [CBSE 2015], 5, 2, AP 2, AP = AB ⇒, =, 5, AB 5
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314, , Secondary School Mathematics for Class 10, , AP, 2, = ⇒ 5 AP = 2 AP + 2 PB, AP + PB 5, AP 2, ⇒ 3 AP = 2 PB ⇒, = ⇒ AP : PB = 2 : 3., PB 3, ⇒, , So, P divides AB in the ratio 2 : 3., ∴ coordinates of P are, ⎛⎜ 2 × 6 + 3 × 1 , 2 × 7 + 3 × 2 ⎞⎟ = ⎛⎜ 15 , 20 ⎞⎟ = ( 3 , 4)., ⎝, ⎠ ⎝ 5 5 ⎠, 2+ 3, 2+ 3, EXAMPLE 5, , SOLUTION, , Point P divides the line segment joining the points A( 2 , 1) and, AP 1, = ⋅ If P lies on the line 2 x − y + k = 0 , find, B(5 , − 8) such that, AB 3, [CBSE 2010], the value of k., AP 1, AP, 1, = ⇒, =, AB 3, AP + PB 3, ⇒ 3 AP = AP + PB ⇒ 2 AP = PB, AP 1, ⇒, = ⇒ AP : PB = 1 : 2., PB 2, So, P divides AB in the ratio 1 : 2., ∴, , coordinates of P are, ⎛ 1 × 5 + 2 × 2 1 × ( −8) + 2 × 1 ⎞ ⎛⎜ 9 −6 ⎞⎟, ,, ,, = ( 3 , − 2)., ⎜, ⎟ =, ⎝, ⎠ ⎝3 3⎠, 1+2, 1+2, , Since P( 3 , − 2) lies on 2 x − y + k = 0 , we have, 2 × 3 − ( −2) + k = 0 ⇒ 6 + 2 + k = 0 ⇒ k = −8., Hence, the required value of k is –8., EXAMPLE 6, SOLUTION, , In what ratio does the point P( 2 , −5) divide the line segment joining, A( −3 , 5) and B( 4 , − 9)?, Let the required ratio be k : 1., Then, by the section formula, the coordinates of P are, ⎛ 4 k − 3 −9 k + 5 ⎞, P⎜, ,, ⎟⋅, ⎝ k +1, k +1 ⎠, ∴, , 4k − 3, −9 k + 5, = 2 and, = −5, k +1, k +1, , ⇒ 4 k − 3 = 2 k + 2 and −9 k + 5 = −5 k − 5, ⇒ 2 k = 5 and 4 k = 10, 5, ⇒ k = in each case., 2, , [Q P( 2 , − 5) is given]
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Coordinate Geometry, , 315, , 5, : 1 , which is 5 : 2., 2, Hence, P divides AB in the ratio 5 : 2., , So, the required ratio is, , EXAMPLE 7, , Find the ratio in which the point P( x , 2) divides the line segment, joining the points A(12 , 5) and B( 4 , − 3). Also, find the value of x., [CBSE 2014], , SOLUTION, , Let the required ratio be k : 1., Then, by section formula, the coordinates of P are, ⎛ 4 k + 12 −3 k + 5 ⎞, P⎜, ,, ⎟⋅, ⎝ k +1, k +1 ⎠, But, this point is given as P( x , 2)., −3 k + 5, 3, = 2 ⇒ −3 k + 5 = 2 k + 2 ⇒ 5 k = 3 ⇒ k = ⋅, ∴, k +1, 5, So, the required ratio is 3 : 5., 3, Putting k = in P, we get, 5, ⎧ 4 × 3 + 12 ⎫, ⎨, ⎬ 72, 5, ⎭=, x= ⎩, = 9., 8, ⎛⎜ 3 + 1 ⎞⎟, ⎝5, ⎠, Hence, x = 9., , EXAMPLE 8, , Find the ratio in which the point P(11 , y) divides the line segment, joining the points A(15 , 5) and B( 9 , 20). Also, find the value of y., , SOLUTION, , Let the required ratio be k : 1., Then, by the section formula, the coordinates of P are, , [CBSE 2013C], , ⎛ 9 k + 15 20 k + 5 ⎞, P⎜, ,, ⎟⋅, ⎝ k +1, k +1 ⎠, But, this point is given as P(11 , y)., 9 k + 15, = 11 ⇒ 9 k + 15 = 11 k + 11 ⇒ 2 k = 4 ⇒ k = 2., ∴, k +1, So, the required ratio is 2 : 1., Putting k = 2 in P, we get, 20 × 2 + 5 45, y=, =, = 15., 3, (2 + 1), Hence, y = 15.
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316, , Secondary School Mathematics for Class 10, , EXAMPLE 9, , If the point P( −1 , 2) divides the line segment joining the points, A( 2 , 5) and B in the ratio 3 : 4, find the coordinates of B., , SOLUTION, , Let the coordinates of B be ( x1 , y1 )., Then, P divides the join of A( 2 , 5), and B( x1 , y1 ) in the ratio 3 : 4., So, by the section formula, the coordinates of P are, 3 x + 8 3 y1 + 20 ⎞, 3 x + 8 3 y1 + 20 ⎞, ,, ,, P ⎛⎜ 1, ⎟ , i.e., P ⎛⎜ 1, ⎟⋅, ⎝ 3+4, ⎝ 7, ⎠, 3+4 ⎠, 7, But, the coordinates of P are (–1, 2)., 3 x1 + 8, 3 y + 20, = −1 and 1, =2, ∴, 7, 7, ⇒ 3 x1 + 8 = −7 and 3 y1 + 20 = 14, ⇒ 3 x1 = −15 and 3 y1 = −6, ⇒ x1 = −5 and y1 = −2., Hence, the coordinates of B are ( −5 , − 2)., , EXAMPLE 10, , Find the lengths of the medians AD and BE of SABC whose vertices, are A(7 , − 3), B(5 , 3) and C( 3 , − 1)., [CBSE 2014], , SOLUTION, , Since D and E are the midpoints of BC and AC respectively,, so their coordinates are, ⎛ 5 + 3 3 + ( −1) ⎞, D⎜, ,, ⎟, ⎝ 2, ⎠, 2, and E ⎛⎜, ⎝, , 7 + 3 −3 − 1 ⎞, ⎟,, ,, 2, 2 ⎠, , i.e., D(4, 1) and E(5, –2)., ∴, , AD = (7 − 4) 2 + ( −3 − 1) 2 =, , 3 2 + ( −4) 2, , = 9 + 16 = 25 = 5 units., ∴, , BE = (5 − 5) 2 + ( −2 − 3) 2 = 0 + ( −5) 2 = 25 = 5 units., , EXAMPLE 11, , The three vertices of a parallelogram ABCD taken in order are, A( 3 , − 4), B( −1 , − 3) and C( −6 , 2). Find the coordinates of the, fourth vertex D., [CBSE 2013], , SOLUTION, , Let A( 3 , − 4), B( −1 , − 3) and C( −6 , 2) be the three given vertices, of a||gm ABCD and let its fourth vertex be D( a, b)., Joint AC and BD.
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Coordinate Geometry, , 317, , Let AC and BD intersect at the, point O., We know that the diagonals of, a||gm bisect each other., So, O is the midpoint of each, one of AC and BD., −3, ⎛ 3 + ( −6) −4 + 2 ⎞, Midpoint of AC is ⎜, , − 1 ⎞⎟, ,, ⎟ , i.e., O ⎛⎜, ⎝ 2, ⎠, ⎝, 2, 2 ⎠, ⎛ −1 + a −3 + b ⎞, and midpoint of BD is O ⎜, ,, ⎟., ⎝ 2, 2 ⎠, −1 + a −3, =, ⇒ −1 + a = −3 ⇒ a = −2, 2, 2, −3 + b, and, = −1 ⇒ −3 + b = −2 ⇒ b = 1., 2, , ∴, , Hence, the fourth vertex of||gm ABCD is D(–2, 1)., EXAMPLE 12, , If ( 3 , 3), ( 6 , y), ( x , 7 ) and (5, 6) are the vertices of a parallelogram, taken in order, find the values of x and y., [CBSE 2011], , SOLUTION, , Let A( 3 , 3), B( 6 , y), C ( x , 7 ) and, D(5 , 6) be the vertices of a ||gm, ABCD. Join AC and BD,, intersecting each other at the, point O., We know that the diagonals of, ||gm bisect each other., Therefore, O is the midpoint of AC as well as that of BD., x + 3 7 + 3⎞, x+ 3 ⎞, ⎟ , i.e. , ⎛⎜, ,, , 5⎟, ∴ midpoint of AC is ⎛⎜, ⎝ 2, ⎝ 2, ⎠, 2 ⎠, 5 + 6 6 + 4⎞, 11 6 + y ⎞, ⎟ , i.e. , ⎛⎜ ,, ⎟., and midpoint of BD is ⎛⎜, ,, ⎝ 2, ⎝ 2, 2 ⎠, 2 ⎠, But, these points coincide at the point O., x + 3 11, 6+y, and, =, =5, ∴, 2, 2, 2, ⇒ x + 3 = 11 and 6 + y = 10, ⇒ x = 8 and y = 4., Hence, x = 8 and y = 4.
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318, , Secondary School Mathematics for Class 10, , EXAMPLE 13, , Let D( 3 , − 2), E( −3 , 1) and F( 4 , − 3) be the midpoints of the sides, BC, CA and AB respectively of CABC. Then, find the coordinates of, the vertices A, B and C., , SOLUTION, , Let A( x1 , y1 ), B( x2 , y2 ) and C ( x3 , y3 ), be the vertices of the given CABC,, and let D( 3 , − 2), E( −3 , 1) and, F( 4 , − 3) be the midpoints of BC, CA, and AB respectively., D is the midpoint of BC, x + x3, y + y3, = 3, 2, = −2, ⇒ 2, 2, 2, ⎧ x + x3 = 6, ⇒ ⎨ 2, ⎩ y2 + y3 = − 4., , ... (i), ... (ii), , E is the midpoint of CA, x + x3, y + y3, ⇒ 1, = −3 , 1, =1, 2, 2, ⎧ x + x3 = −6, ⇒ ⎨ 1, ⎩ y1 + y3 = 2., , ... (iii), ... (iv), , F is the midpoint of AB, x + x2, y + y2, ⇒ 1, = 4, 1, = −3, 2, 2, ⎧ x + x2 = −8, ⇒ ⎨ 1, ⎩ y1 + y2 = −6., , ... (v), ... (vi), , Adding (i), (iii) and (v), we get, 2( x1 + x2 + x3 ) = 8 ⇒ ( x1 + x2 + x3 ) = 4., , ... (vii), , Using (i), (iii) and (v) with (vii), we get, x1 = −2 , x2 = 10 and x3 = −4., Adding (ii), (iv) and (vi), we get, 2( y1 + y2 + y3 ) = −8 ⇒ y1 + y2 + y3 = − 4., , ... (viii), , Using (ii), (iv) and (vi) with (viii), we get, y1 = 0 , y2 = −6 and y3 = 2., Hence, the vertices of CABC are A( −2 , 0), B(10 , − 6) and, C( − 4 , 2)., EXAMPLE 14, , The coordinates of one end point of a diameter AB of a circle are, A( 4 , − 1) and the coordinates of the centre of the circle are C(1 , − 3)., Find the coordinates of B.
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Coordinate Geometry, SOLUTION, , 319, , Let C(1 , − 3) be the centre of the given circle and let A( 4 , − 1), and B( a, b) be the two end points of the given diameter AB., Then, clearly C is the midpoint of AB., ⎛ 4 + a −1 + b ⎞, ∴ the coordinates of C are C ⎜, ,, ⎟⋅, ⎝ 2, 2 ⎠, But, the coordinates of C are (1, –3)., 4+a, −1 + b, ∴, = 1 and, = −3, 2, 2, ⇒ 4 + a = 2 and −1 + b = −6, ⇒ a = −2 and b = −5., Hence, the coordinates of B are B( −2 , − 5)., , CENTROID OF A TRIANGLE, , The point of intersection of the medians of a triangle is, , called its centroid., TO FIND THE COORDINATES OF THE CENTROID OF A TRIANGLE, , Let A( x1 , y1 ), B( x2 , y2 ) and C ( x3 , y3 ) be the vertices of a CABC., Let D be the midpoint of BC., x + x3 y2 + y3 ⎞, Then, the coordinates of D are D ⎛⎜ 2, ,, ⎟⋅, ⎝ 2, 2 ⎠, Let G( x , y) be the centroid of CABC., , Then, G divides AD in the ratio 2 : 1., ⎧ 2( x2 + x3 ), ⎫, + x1 ⎬, ⎨, 2, ⎩, ⎭ = ( x1 + x2 + x3 ) ,, ∴ x=, 3, ( 2 + 1), ⎧ 2( y2 + y3 ), ⎫, + y1 ⎬, ⎨, 2, ⎩, ⎭ = ( y1 + y2 + y3 ) ⋅, y=, 3, ( 2 + 1)
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320, , Secondary School Mathematics for Class 10, , x + x2 + x3 y1 + y2 + y3 ⎞, ∴ the coordinates of G are G ⎛⎜ 1, ,, ⎟⋅, ⎝, ⎠, 3, 3, EXAMPLE 15, SOLUTION, , Find the centroid of CABC whose vertices are A( −3 , 0), B(5 , − 2), and C( −8 , 5)., Here, ( x1 = −3 , y1 = 0), ( x2 = 5 , y2 = −2) and ( x3 = −8 , y3 = 5)., Let G( x , y) be the centroid of CABC. Then,, 1, 1, x = ( x1 + x2 + x3 ) = ( −3 + 5 − 8) = −2,, 3, 3, 1, 1, y = ( y1 + y2 + y3 ) = ( 0 − 2 + 5) = 1., 3, 3, Hence, the centroid of CABC is G( −2 , 1)., , EXAMPLE 16, , Two vertices of a CABC are given by A( 6 , 4) and B( −2 , 2), and its, centroid is G( 3 , 4). Find the coordinates of the third vertex C of, [CBSE 2005C], CABC., , SOLUTION, , It is being given that the two vertices of CABC are A(6, 4) and, B(–2, 2). Let its third vertex be C ( x , y)., Then, the coordinates of the centroid of CABC are, 4 + x 6 + y⎞, ⎛ 6 + ( −2) + x 4 + 2 + y ⎞, ⎟⋅, G⎜, ,, ,, ⎟ , i.e., G ⎛⎜, ⎝ 3, ⎝, ⎠, 3, 3, 3 ⎠, But, it is given that the centroid of CABC is G( 3 , 4)., 4+x, 6+y, ∴, = 3 and, =4, 3, 3, ⇒ 4 + x = 9 and 6 + y = 12., ⇒ x = 5 and y = 6., Hence, the coordinates of the third vertex is C(5 , 6)., , EXERCISE 6B, 1., , (i) Find the coordinates of the point which divides the join of A( −1 , 7 ), and B( 4 , − 3) in the ratio 2 : 3., (ii) Find the coordinates of the point which divides the join of, A( −5 , 11) and B( 4 , − 7 ) in the ratio 7 : 2., , 2. Find the coordinates of the points of trisection of the line segment, joining the points A(7 , − 2) and B(1, –5)., [CBSE 2017], 3. If the coordinates of points A and B are (–2, –2) and (2, –4) respectively,, 3, find the coordinates of the point P such that AP = AB, where P lies on, 7, [CBSE 2015], the line segment AB.
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Coordinate Geometry, , 321, , 4. Point A lies on the line segment PQ joining P(6, –6) and Q( −4 , − 1), PA 2, in such a way that, = ⋅ If the point A also lies on the line, PQ 5, [CBSE 2015], 3 x + k( y + 1) = 0 , find the value of k., 5. Points P, Q, R and S divide the line segment joining the points A(1, 2), and B(6, 7) in five equal parts. Find the coordinates of the points P, Q, [CBSE 2014], and R., 6. Points P, Q and R in that order are dividing a line segment joining, A(1 , 6) and B(5 , − 2) in four equal parts. Find the coordinates of P, Q, [CBSE 2009C], and R., 7. The line segment joining the points A( 3 , − 4) and B(1 , 2) is trisected at, 5, the points P( p , − 2) and Q ⎛⎜ , q⎞⎟ ⋅ Find the values of p and q. [CBSE 2005], ⎝3 ⎠, 8. Find the coordinates of the midpoint of the line segment joining, (i) A( 3 , 0) and B( −5 , 4) (ii) P( −11 , − 8) and Q( 8 , − 2)., 9. If ( 2 , p) is the midpoint of the line segment joining the points A( 6 , − 5), and B( −2 , 11), find the value of p., [CBSE 2010], 10. The midpoint of the line segment joining A( 2 a, 4) and B( −2 , 3b) is, C (1 , 2 a + 1). Find the values of a and b., 11. The line segment joining A( −2 , 9) and B( 6 , 3) is a diameter of a circle, with centre C. Find the coordinates of C., 12. Find the coordinates of a point A, where AB is a diameter of a circle with, centre C( 2 , − 3) and the other end of the diameter is B(1 , 4)., 13. In what ratio does the point P(2, 5) divide the join of A(8, 2) and, B( −6 , 9)?, 3 5, 14. Find the ratio in which the point P ⎛⎜ , ⎞⎟ divides the line segment, ⎝ 4 12 ⎠, 1 3, [CBSE 2015], joining the points A ⎛⎜ , ⎞⎟ and B(2, –5)., ⎝ 2 2⎠, 15. Find the ratio in which the point P(m, 6) divides the join of A( −4 , 3) and, [CBSE 2004], B( 2 , 8). Also, find the value of m., 16. Find the ratio in which the point (–3, k) divides the join of A(–5, –4) and, [CBSE 2007], B(–2, 3). Also, find the value of k., 17. In what ratio is the line segment joining A( 2 , − 3) and B(5 , 6) divided by, the x-axis? Also, find the coordinates of the point of division.
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322, , Secondary School Mathematics for Class 10, , 18. In what ratio is the line segment joining the points A( −2 , − 3) and, B( 3 , 7 ) divided by the y-axis? Also, find the coordinates of the point of, division., [CBSE 2006C], 19. In what ratio does the line x − y − 2 = 0 divide the line segment joining, the points A(3, –1) and B(8, 9)?, [CBSE 2007], 20. Find the lengths of the medians of a CABC whose vertices are, A( 0 , − 1), B( 2 , 1) and C ( 0 , 3)., 21. Find the centroid of CABC whose vertices are A( −1 , 0), B(5 , − 2) and, [CBSE 2003], C( 8 , 2)., 22. If G( −2 , 1) is the centroid of a CABC and two of its vertices are A(1 , − 6), and B( −5 , 2), find the third vertex of the triangle., 23. Find the third vertex of a CABC if two of its vertices are B( −3 , 1) and, [CBSE 2004C], C( 0 , − 2) , and its centroid is at the origin., 24. Show that the points A( 3 , 1), B( 0 , − 2), C (1 , 1) and D( 4 , 4) are the, vertices of a parallelogram ABCD., 25. If the points P( a, −11), Q(5 , b), R ( 2 , 15) and S(1 , 1) are the vertices of a, parallelogram PQRS, find the values of a and b., 26. If three consecutive vertices of a parallelogram ABCD are A(1 , − 2) ,, B( 3 , 6) and C(5 , 10), find its fourth vertex D., 27. In what ratio does y-axis divide the line segment joining the points, (–4, 7) and (3, –7)?, [CBSE 2012], 1, 28. If the point P ⎛⎜ , y ⎞⎟ lies on the line segment joining the points A( 3 , − 5), ⎝2 ⎠, and B( −7 , 9) then find the ratio in which P divides AB. Also, find the, [CBSE 2010], value of y., 29. Find the ratio in which the line segment joining the points A( 3 , − 3) and, B( −2 , 7 ) is divided by x-axis. Also, find the point of division. [CBSE 2014], 30. The base QR of an equilateral triangle PQR lies on x-axis. The, coordinates of the point Q are (–4, 0) and origin is the midpoint of the, base. Find the coordinates of the points P and R., [CBSE 2015], 31. The base BC of an equilateral triangle ABC lies on y-axis. The, coordinates of point C are (0, –3). The origin is the midpoint of the base., Find the coordinates of the points A and B. Also, find the coordinates of, another point D such that ABCD is a rhombus., [CBSE 2015], 32. Find the ratio in which the point P(–1, y) lying on the line segment, joining points A( −3 , 10) and B( 6 , − 8) divides it. Also, find the value of y., [CBSE 2013]
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Coordinate Geometry, , 323, , 33. ABCD is a rectangle formed by the points A( −1 , − 1), B( −1 , 4), C(5 , 4), and D(5 , − 1). If P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively, show that PQRS is a rhombus., 34. The midpoint P of the line segment joining the points A( −10 , 4) and, B( −2 , 0) lies on the line segment joining the points C( −9 , − 4) and, D( −4 , y). Find the ratio in which P divides CD. Also find the value of y., , [CBSE 2014], , 35. A line intersects the y-axis and x-axis at the points P and Q respectively., If (2, –5) is the midpoint of PQ then find the coordinates of P and Q., [CBSE 2017], , 24, 36. In what ratio does the point ⎛⎜ , y ⎞⎟ divide the line segment joining the, ⎝ 11 ⎠, points P(2, –2) and Q(3, 7)? Also, find the value of y., , [CBSE 2017], , 37. The midpoints of the sides BC, CA and AB of a a ABC are D(3, 4), E(8, 9), and F(6, 7) respectively. Find the coordinates of the vertices of the, [CBSE 2017], triangle., 38. If two adjacent vertices of a parallelogram are (3, 2) and (–1, 0) and the, diagonals intersect at (2, –5) then find the coordinates of the other two, [CBSE 2017], vertices., ANSWERS (EXERCISE 6B), , −2 −20 ⎞, ⎟, 4. k = 2, 3. P ⎛⎜, ,, ⎝ 7, 7 ⎠, 5. P(2, 3), Q(3, 4), R(4, 5), 6. P(2, 4), Q(3, 2), R(4, 0), −3, 7, ⎛, 7. p = , q = 0, 9. p = 3 10. a = 2 , b = 2, 8. (i) (–1, 2) (ii) ⎜, , − 5 ⎞⎟, ⎝ 2, ⎠, 3, −2, 11. C( 2 , 6), 12. A( 3 , − 10), 13. 3 : 4, 14. 1 : 5, 15. 3 : 2 , m =, 5, 2, 17. (1 : 2), (3, 0) 18. (2 : 3), (0, 1), 19. 2 : 3, 16. 2 : 1 , k =, 3, 21. G( 4 , 0), 20. AD = 10 units, BE = 2 units, CF = 10 units, 22. C( −2 , 7 ), 23. A( 3 , 1), 25. a = 4 , b = 3, 26. D( 3 , 2), 3, −3, 29. (3 : 7), ⎛⎜ , 0 ⎞⎟, 27. 4 : 3, 28. 1 : 3 , y =, ⎝2 ⎠, 2, 30. P( 0 , 4 3 ) or P( 0 , − 4 3 ) and R( 4 , 0), 31. A( 3 3 , 0) or A( −3 3 , 0) and B( 0 , 3);, [a( 3 3 , 0), b( 0 , 3), d( 3 3 , − 6)] or [a( −3 3 , 0), b( 0 , 3), d( −3 3 , − 6)], 4, 32. 2 : 7 , y = 6 34. 3 : 2 , y = 6 35. P(0, –10), Q(4, 0), 36. 2 : 9; y = −, 11, 37. A(11, 12), B(1, 2), C(5, 8), 38. (1, –12); (5, –10), 1. (i) (1 , 3), , (ii) ( 2 , − 3), , 2. (5, –3), (3, –4)
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324, , Secondary School Mathematics for Class 10, HINTS TO SOME SELECTED QUESTIONS, , 3., , AP 3, AB 7, AP + PB, = ⇒, = ⇒, AB 7, AP 3, AP, AP, PB 7, ⇒, +, = ⇒ 1+, AP AP 3, PB ⎛ 7, 4, ⇒, = ⎜ − 1⎞⎟ = ⇒, ⎠ 3, AP ⎝ 3, , 7, 3, PB 7, =, AP 3, AP 3, = ⇒ AP : PB = 3 : 4., PB 4, , =, , mx + nx1 my2 + ny1 ⎞, ,, ∴ coordinates of P are ⎛⎜ 2, ⎟ , where m = 3 , n = 4,, ⎝ m+n, m+n ⎠, 3 × 2 + 4 × ( −2 ) 3 × ( −4 ) + 4 × ( −2 ) ⎞, ⎛ −2 , −20 ⎞ ⋅, i.e., P ⎛⎜, ,, ⎟, ⎟ =P⎜, ⎝ 7, ⎝, ⎠, 3+ 4, 3+ 4, 7 ⎠, PA 2, PQ 5, PA + AQ 5, 4., = ⇒, = ⇒, =, PQ 5, PA 2, PA, 2, ⇒ 1+, ⇒, , AQ 5, AQ ⎛ 5, 3, = ⇒, = ⎜ − 1⎞⎟ =, ⎠ 2, PA 2, PA ⎝ 2, , PA 2, = ⇒ PA : AQ = 2 : 3., AQ 3, , 2 × ( −4 ) + 3 × 6 2 × ( −1) + 3 × ( −6 ) ⎞, ∴ coordinates of A are ⎛⎜, ,, ⎟ = A( 2 , − 4 )., ⎝, ⎠, 2+ 3, 2+ 3, Since the point A( 2 , − 4 ) lies on the line 3 x + k( y + 1) = 0 , we have, ( 3 × 2 ) + k( −4 + 1) = 0 ⇒ 3 k = 6 ⇒ k = 2., 5. P divides AB in the ratio 1 : 4., Q divides AB in the ratio 2 : 3., R divides AB in the ratio 3 : 2., 6. P divides AB in the ratio 1 : 3., Q divides AB in the ratio 2 : 2 = 1 : 1., R divides AB in the ratio 3 : 1., 7. P divides AB in the ratio 1 : 2., 1 × 1 + 2 × 3 1 × 2 + 2 × ( −4 ) ⎞, This point is P ⎛⎜, ,, ⎟., ⎝, ⎠, 1+ 2, 1+ 2, 7, 7, ⎛, ⎞, ∴ ⎜ , − 2⎟ ≡ ( p, − 2) ⇒ p = ⋅, ⎝3, ⎠, 3, Q divides AB in the ratio 2 : 1. So, we get, 2 × 1 + 1 × 3 2 × 2 + 1 × ( −4 ) ⎞, ⎛5 ⎞, Q ⎛⎜, ,, ⎟ , i.e., ⎜ , 0 ⎟, ⎝3 ⎠, ⎝, ⎠, 2+ 1, 2+ 1, ∴, , ⎛5 ,, ⎜, ⎝3, , 5, 0 ⎞⎟ ≡ ⎛⎜ ,, ⎠ ⎝3, , q⎞⎟ ⇒ q = 0., ⎠, , 14. Let the required ratio be k : 1., ⎛ 2 k + 1 −5 k + 3 ⎞, ⎜, 2,, 2 ⎟., Then, the coordinates of P are ⎜, ⎟, k, +, 1, k, +, 1, ⎜, ⎟, ⎝, ⎠
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Coordinate Geometry, , ∴, , 1, 2 = 3 ⇒ 4 k + 1 = 3 ⇒ 16 k + 4 = 6 k + 6, k+1, 4, 2( k + 1) 4, , 2k +, , ⇒ 10 k = 2 ⇒ k =, ∴, , 325, , 1, ⋅, 5, , 1, required ratio is : 1, i.e., 1 : 5., 5, , 15. Let the required ratio be k : 1., 2k − 4 8k + 3 ⎞, Then, the coordinates of P are ⎛⎜, ,, ⎟., ⎝ k+1 k+1 ⎠, 8k + 3, 3, = 6 ⇒ 8k + 3 = 6k + 6 ⇒ 2k = 3 ⇒ k = ⋅, k+1, 2, 3, So, the required ratio is : 1, i.e., 3 : 2., 2, ⎛ 2 × 3 − 4⎞, ⎜, ⎟, ⎠ −2, 2k − 4 ⎝, 2, =, ⋅, =, ∴ m=, 5, k+1, ⎛ 3 + 1⎞, ⎜, ⎟, ⎝2, ⎠, ∴, , 17. Let the x-axis be cut by the join of A( 2 , − 3 ) and B(5 , 6 ) in the ratio k : 1 at the point, P( x , 0 )., Then,, ∴, , 6k − 3, 3 1, = 0 ⇒ 6k − 3 = 0 ⇒ 6k = 3 ⇒ k = = ⋅, k+1, 6 2, , required ratio is, , 1, : 1, i.e., 1 : 2., 2, , 1× 5 + 2 × 2 ⎞, Point of division is P ⎛⎜, , 0 ⎟ , i.e., (3, 0)., ⎝, ⎠, 1+ 2, 19. Let the required ratio be k : 1., 8 k + 3 9 k − 1⎞, Then, the point P ⎛⎜, ,, ⎟ lies on the line x − y − 2 = 0., ⎝ k+ 1 k+ 1⎠, ∴, , 8k + 3 9k − 1, −, − 2 = 0 ⇒ ( 8 k + 3 ) − ( 9 k − 1) − 2( k + 1) = 0, k+1, k+1, ⇒ ( 8k − 9k − 2k ) + ( 3 + 1 − 2) = 0, ⇒ 3k = 2 ⇒ k =, , 2, ⋅, 3, , 2, Hence, the required ratio is ⎛⎜ : 1⎞⎟ , i.e., 2 : 3., ⎝3 ⎠, 27. Let the required ratio be k : 1., Then, x-coordinate of the point of division is 0., ∴, , 3k − 4, 4, = 0 ⇒ 3k − 4 = 0 ⇒ k = ⋅, k+1, 3
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326, , Secondary School Mathematics for Class 10, 4, Hence, the required ratio is ⎛⎜ : 1⎞⎟ , i.e., 4 : 3., ⎝3 ⎠, , 28. Let the required ratio be k : 1., −7 k + 3 9 k − 5 ⎞, Then, we get P ⎛⎜, ,, ⎟., ⎝ k+1, k+ 1⎠, ∴, , −7 k + 3 1, = ⇒ −14 k + 6 = k + 1, k+1, 2, ⇒ 15 k = 5 ⇒ k =, , 1, ⋅, 3, , 1, So, the required ratio is ⎛⎜ : 1⎞⎟ , i.e., 1 : 3., ⎝3 ⎠, Putting k =, , 1, , we get, 3, , ⎛ 9× 1⎞ − 5, ⎜, ⎟, 9k − 5 ⎝, −6 −3, 3⎠, =, =, =, ⋅, y=, 4, 2, k+1, ⎛ 1 + 1⎞, ⎜, ⎟, ⎝3, ⎠, 29. Let the required ratio be k : 1., −2 k + 3 7 k − 3 ⎞, Then, the point of division is P ⎛⎜, ,, ⎟⋅, ⎝ k+1, k+1⎠, Since this point lies on the x-axis, its y-coordinates must be 0., 7k − 3, 3, = 0 ⇒ 7k − 3 = 0 ⇒ k = ⋅, ∴, k+1, 7, 3, So, the required ratio is ⎛⎜ : 1⎞⎟ , i.e., 3 : 7., ⎝7 ⎠, 3, , to get, 7, −2 k + 3 7 k − 3 ⎞, ⎛3 ⎞, P ⎛⎜, ,, ⎟ as P ⎜ , 0 ⎟ ., ⎝ k+1, ⎝2 ⎠, k+1⎠, , Now, put k =, , 30. Since R lies on the x-axis, its coordinates, are of the form R( a, 0 ) and clearly O , the, origin is the midpoint of QR. So,, −4 + a, = 0 ⇒ − 4 + a = 0 ⇒ a = 4., 2, So, the coordinates of R are R(4, 0)., Clearly, P lies on the, , y-axis. Let the, , coordinates of P be P( 0 , c )., Since SPQR is equilateral, we have, PQ 2 = PR 2 = QR 2 ., ∴, , 16 + c 2 = 64 ⇒ c 2 = 48 ⇒ x = ± 48 = ±4 3.
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Coordinate Geometry, , 327, , 31. Since B lies on the y-axis, its coordinates, are of the form B( 0 , b ). Since O( 0 , 0 ) is, the midpoint of BC , we have, −3 + b, = 0 ⇒ −3 + b = 0 ⇒ b = 3., 2, So, the coordinates of B are B( 0 , 3 )., Clearly, the vertex A lies on the x-axis., Let this point be A( a, 0 )., Since SABC is equilateral, we have, AC 2 = AB 2 = BC 2 ., ∴, , a2 + 9 = 36 ⇒ a2 = 27, ⇒ a = ±3 3., , So, the coordinates of A are ( 3 3 , 0 ) or ( −3 3 , 0 )., 32. Let the required ratio be k : 1., 6 k − 3 −8 k + 10 ⎞, Then, we get P ⎛⎜, ,, ⎟., ⎝ k+1, k+1 ⎠, But,, , 6k − 3, 2, = −1 ⇒ 6 k − 3 = − k − 1 ⇒ 7 k = 2 ⇒ k = ⋅, k+1, 7, , 2, So, the required ratio is ⎛⎜ : 1⎞⎟ = ( 2 : 7 )., ⎝7 ⎠, 34. The point P is P ⎛⎜, ⎝, , −10 − 2 4 +, ,, 2, 2, , 0⎞, ⎟ , i.e., (–6, 2)., ⎠, , Let P( −6 , 2 ) divide CD in the ratio k : 1., ky − 4, −4 k − 9, Then,, =2, = −6 and, k+1, k+1, ⇒ − 4 k − 9 = −6 k − 6 and ky − 4 = 2 k + 2, ⇒ 2 k = 3 and k( y − 2 ) = 6, 3, 6, 2, ⇒ k = and ( y − 2 ) = = ⎛⎜ 6 × ⎞⎟ = 4, 2, k ⎝, 3⎠, ⇒ k=, , 3, and y = 4 + 2 = 6., 2, , Hence, the ratio is 3 : 2 and y = 6., , ................................................................, , AREA OF A TRIANGLE, THEOREM 1, , The area of a CABC with vertices A( x1 , y1 ), B( x2 , y2 ) and C ( x3 , y3 ), is given by, 1, area(C ABC ) =⏐ {x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )}⏐⋅, ⏐2, ⏐
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330, , Secondary School Mathematics for Class 10, , But, it is given that ar( SABC ) = 24 sq units., 1, 3, 21, 2 k 2 + 3 k + 21 = 24 ⇒ k 2 + k +, = 24 ., ∴, 2, 2, 2, , ... (i), , 3, 21 ⎫ ⎛ 2 3, 9 ⎞ ⎛ 21 9 ⎞, But ⎧⎨ k 2 + k +, − ⎟, ⎬ = ⎜ k + k + ⎟⎠ + ⎜⎝, 2, 2 ⎭ ⎝, 2, 16, 2 16 ⎠, ⎩, 2, ⎧, 3, 159 ⎫, = ⎨ ⎛⎜ k + ⎞⎟ +, ⎬ > 0., ⎝, 4⎠, 16 ⎭, ⎩, So, we may write (i) as, 3, 21, = 24 ⇒ 2 k 2 + 3 k + 21 = 48, k2 + k +, 2, 2, ⇒ 2 k 2 + 3 k − 27 = 0 ⇒ 2 k 2 + 9 k − 6 k − 27 = 0, ⇒ k( 2 k + 9) − 3( 2 k + 9) = 0 ⇒ ( 2 k + 9)( k − 3) = 0, 9, ⇒ k − 3 = 0 or 2 k + 9 = 9 ⇒ k = 3 or k = − ⋅, 2, 9, Hence, k = 3 or k = − ⋅, 2, EXAMPLE 3, , If A( 4 , − 6), B( 3 , − 2) and C(5 , 2) are the vertices of a SABC and, AD is its median. Prove that the median AD divides SABC into two, triangles of equal areas., [CBSE 2014], , SOLUTION, , It is being given that A(4, –6),, B(3, –2) and C(5, 2) are the vertices, of a SABC and AD is its median., Clearly, D is the midpoint of BC., ∴, , the coordinates of D are, ⎛⎜ 3 + 5 , −2 + 2 ⎞⎟ = ( 4 , 0)., ⎝ 2, 2 ⎠, , For SABD, we have, A( x1 = 4 , y1 = −6), B( x2 = 3 , y2 = −2) andD( x3 = 4 , y3 = 0)., 1, ar( SABD) = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), 2, 1, =, 4 ⋅ ( −2 − 0) + 3 ⋅ ( 0 + 6) + 4 ⋅ ( −6 + 2), 2, 1, =, 4 × ( −2) + 3 × 6 + 4 × ( −4), 2, 1, 1, = − 8 + 18 − 16 =, −6, 2, 2, 1, = ⎛⎜ × 6 ⎞⎟ = 3 sq units., ⎝2, ⎠
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Coordinate Geometry, , 331, , For SADC , we have, ( x1 = 4 , y1 = −6), ( x2 = 4 , y2 = 0) and ( x3 = 5 , y3 = 2)., 1, ∴ ar (SADC ) = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), 2, 1, = 4 ⋅ ( 0 − 2) + 4 ⋅ ( 2 + 6) + 5 ⋅ ( −6 − 0), 2, 1, = 4 × ( −2) + 4 × 8 + 5 × ( −6), 2, 1, 1, = − 8 + 32 − 30 =, − 6|, 2, 2, 1, = ⎛⎜ × 6 ⎞⎟ = 3 sq units., ⎝2, ⎠, ∴ ar( SABD) = ar( SADC )., Hence, the median AD divides SABC into two triangles of, equal areas., EXAMPLE 4, , Find the area of the triangle formed by joining the midpoints of the, sides of the triangle whose vertices are A( 2 , 2), B( 4 , 4) and C( 2 , 6)., [CBSE 2009C], , SOLUTION, , Let A( 2 , 2), B( 4 , 4) and C( 2 , 6), be the vertices of the given, SABC. Let D, E and F be the, midpoints of AB, BC and CA, respectively., Then, the coordinates of D, E, and F are, 2 + 4 2 + 4⎞, 4 + 2 4 + 6⎞, 2 + 2 2 + 6⎞, ⎟ , E ⎛⎜, ⎟ and F ⎛⎜, ⎟,, D ⎛⎜, ,, ,, ,, ⎝ 2, ⎝ 2, ⎝ 2, 2 ⎠, 2 ⎠, 2 ⎠, i.e., D(3, 3), E(3, 5) and F(2, 4)., For SDEF, we have, ( x1 = 3 , y1 = 3), ( x2 = 3 , y2 = 5) and ( x3 = 2 , y3 = 4)., 1, ∴ ar( SDEF) = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), 2, 1, =, 3 ⋅ (5 − 4) + 3 ⋅ ( 4 − 3) + 2 ⋅ ( 3 − 5), 2, 1, = ( 3 × 1) + ( 3 × 1) + 2 × ( −2), 2, 1, 1, = 3 + 3 − 4 = ⎛⎜ × 2 ⎞⎟ = 1 sq unit., ⎝2, ⎠, 2, Hence, the area of SDEF is 1 sq unit.
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334, , Secondary School Mathematics for Class 10, , =, ∴, , 1, 1 + 8 + 15 = 12 sq units., 2, , ar(||gm ABCD) = 2 × ar( SABC ) = 2 × 12 = 24 sq units., , Base AB = ( 2 − 1) 2 + ( 3 + 2) 2 = 1 2 + 5 2 = 26 units., Let h be the height of the||gm ABCD. Then,, ar(||gm ABCD) = (base × height) = ( 26 × h) sq units., ∴, , ⎧ 24, 26 ⎫, 26 × h = 24 ⇒ h = ⎨, ×, ⎬=, 26 ⎭, ⎩ 26, , ⎛⎜ 24 × 26 ⎞⎟ units, ⎝ 26, ⎠, , 12, 61.08, ⇒ h = ⎛⎜ × 5.09 ⎞⎟ =, = 4.69 units., ⎝ 13, ⎠, 13, Hence, the height of the parallelogram is 4.69 units., EXAMPLE 8, , Show that the points A( −1 , 1), B(5 , 7 ) and C ( 8 , 10) are collinear., , SOLUTION, , Let A( −1 , 1), B(5 , 7 ) and C ( 8 , 10) be the given points. Then,, ( x1 = −1 , y1 = 1), ( x2 = 5 , y2 = 7 ) and ( x3 = 8 , y3 = 10)., ∴, , x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), = ( −1)(7 − 10) + 5(10 − 1) + 8(1 − 7 ), , = ( 3 + 45 − 48) = 0., Hence, the given points are collinear., EXAMPLE 9, , Show that the points A( a, b + c), B(b , c + a) and C ( c , a + b) are, collinear., [CBSE 2010], , SOLUTION, , Let A( a, b + c), B(b , c + a) and C ( c , a + b) be the given points., Then, ( x1 = a, y1 = b + c), ( x2 = b , y2 = c + a) and, ( x3 = c , y3 = a + b)., ∴, , x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), = a( c + a − a − b) + b( a + b − b − c) + c(b + c − c − a), = a( c − b) + b( a − c) + c(b − a) = 0., , Hence, the given points are collinear., EXAMPLE 10, , If the area of SABC with vertices A( x , y), B(1 , 2) and C( 2 , 1) is 6 sq, units then prove that x + y = 15 or x + y + 9 = 0., [CBSE 2013], , SOLUTION, , The vertices of SABC are A( x , y), B(1 , 2) and C( 2 , 1)., Here, ( x1 = x , y1 = y), ( x2 = 1 , y2 = 2) and ( x3 = 2 , y3 = 1)., 1, ∴ ar( SABC ) = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), 2
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Coordinate Geometry, , 335, , 1, x ⋅ ( 2 − 1) + 1 ⋅ (1 − y) + 2 ⋅ ( y − 2), 2, 1, 1, = x + 1 − y + 2y − 4 =, x+y− 3 ., 2, 2, =, , ∴, , 1, x + y − 3 = 6 ⇒ x + y − 3 = 12, 2, ⇒ x + y − 3 = 12 or x + y − 3 = −12, ⇒ x + y = 15 or x + y + 9 = 0., , Hence, x + y = 15 or x + y + 9 = 0., EXAMPLE 11, , Find the value of k for which the points A( k + 1 , 2 k), B( 3 k , 2 k + 3), [CBSE 2015, ’17], and C (5 k − 1 , 5 k) are collinear., , SOLUTION, , The given points, C (5 k − 1 , 5 k)., , are, , A( k + 1 , 2 k),, , B( 3 k , 2 k + 3), , and, , Here, ( x1 = k + 1 , y1 = 2 k), ( x2 = 3 k , y2 = 2 k + 3) and, ( x3 = 5 k − 1 , y3 = 5 k)., Let the given points be collinear. Then,, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒ ( k + 1)( 2 k + 3 − 5 k) + 3 k(5 k − 2 k) + (5 k − 1)[2 k − ( 2 k + 3)] = 0, ⇒ ( k + 1)( 3 − 3 k) + 3 k × 3 k + (5 k − 1) × ( −3) = 0, ⇒ 3 − 3 k 2 + 9 k 2 − 15 k + 3 = 0 ⇒ 6 k 2 − 15 k + 6 = 0, ⇒ 2k 2 − 5k + 2 = 0 ⇒ 2k 2 − 4k − k + 2 = 0, ⇒ 2 k( k − 2) − ( k − 2) = 0 ⇒ ( k − 2)( 2 k − 1) = 0, 1, ⇒ k − 2 = 0 or 2 k − 1 = 0 ⇒ k = 2 or k = ⋅, 2, 1, Hence, k = 2 or k = ⋅, 2, EXAMPLE 12, , If the points A( −1 , − 4), B(b , c) and C(5 , − 1) are collinear and, [CBSE 2014], 2b + c = 4 , find the values of b and c., , SOLUTION, , Let A( x1 = −1 , y1 = −4), B( x2 = b , y2 = c) and C ( x3 = 5 , y3 = −1), be the given points., Since these points A, B and C are collinear, we have, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒ ( −1) ⋅ ( c + 1) + b ⋅ ( −1 + 4) + 5 ⋅ ( −4 − c) = 0, ⇒ − c − 1 + 3b − 20 − 5 c = 0, ⇒ 3b − 6 c = 21 ⇒ b − 2 c = 7 ., , ... (i), , Also, it is given that 2b + c = 4., , ... (ii)
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336, , Secondary School Mathematics for Class 10, , On solving (i) and (ii), we get b = 3 and c = −2., Hence, b = 3 and c = −2., EXAMPLE 13, , If R ( x , y) is a point on the line segment joining the points P( a, b) and, [CBSE 2010], Q(b , a) then prove that x + y = a + b., , SOLUTION, , It is being given that the points P( a, b), R ( x , y) and Q(b , a) lie, on the same line segment and, therefore, these points are, collinear., Let P( x1 = a, y1 = b), R ( x2 = x , y2 = y) and Q( x3 = b , y3 = a) be, the given collinear points. Then,, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒ a( y − a) + x( a − b) + b(b − y) = 0, ⇒ ay − a2 + ax − bx + b 2 − by = 0, ⇒ ( a − b) y − ( a 2 − b 2 ) + ( a − b) x = 0, ⇒ y − ( a + b) + x = 0 ⇒ x + y = a + b., Hence, x + y = a + b., , EXERCISE 6C, 1. Find the area of SABC whose vertices are:, (i) A(1, 2), B(−2, 3) and C(−3, −4), (ii) A(–5, 7), B(−4, –5) and C(4, 5), (iii) A(3, 8), B(–4, 2) and C(5, –1), , [CBSE 2008], [CBSE 2008C], , (iv) A(10, −6), B(2, 5) and C(−1, 3), 2. Find the area of quadrilateral ABCD whose vertices are A( 3 , − 1),, [CBSE 2012], B( 9 , − 5), C(14 , 0) and D( 9 , 19)., 3. Find the area of quadrilateral PQRS whose vertices are P( −5 , − 3),, [CBSE 2015], Q( −4 , − 6), R( 2 , − 3) and S(1 , 2)., 4. Find the area of quadrilateral ABCD whose vertices are A( −3 , −1),, [CBSE 2013C], B( −2 , − 4), C( 4 , − 1) and D( 3 , 4)., 5. If A(–7, 5), B(– 6, –7), C(–3, –8) and D(2, 3) are the vertices of a, quadrilateral ABCD then find the area of the quadrilateral. [CBSE 2017], 6. Find the area of the triangle formed by joining the midpoints of the, sides of the triangle whose vertices are A( 2 , 1), B( 4 , 3) and C( 2 , 5)., [CBSE 2011]
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Coordinate Geometry, , 337, , 7. A(7 , − 3), B(5 , 3) and C( 3 , −1) are the vertices of a SABC and AD is its, median. Prove that the median AD divides SABC into two triangles of, equal areas., 8. Find the area of SABC with A(1 , − 4) and midpoints of sides through A, [CBSE 2015], being (2, –1) and (0, –1)., 9. A( 6 , 1), B( 8 , 2) and C( 9 , 4) are the vertices of a parallelogram ABCD. If, [CBSE 2013C], E is the midpoint of DC, find the area of SADE., 10., , (i) If the vertices of SABC be A(1, –3), B(4, p) and C(–9, 7) and its area, [CBSE 2012], is 15 square units, find the values of p., (ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and, 7, (3, –2). If the third vertex is ⎛⎜ , y ⎞⎟ , find the value of y. [CBSE 2017], ⎝2 ⎠, , 11. Find the value of k so that the area of the triangle with vertices, [CBSE 2015], A( k + 1 , 1), B( 4 , − 3) and C (7 , − k) is 6 square units., 12. For what value of k( k > 0) is the area of the triangle with vertices ( −2 , 5),, [CBSE 2012], ( k , − 4) and ( 2 k + 1 , 10) equal to 53 square units?, 13. Show that the following points are collinear:, (i) A( 2 , − 2), B( −3 , 8) and C ( −1 , 4), (ii) A( −5 , 1), B(5 , 5) and C (10 , 7 ), (iii) A(5 , 1), B(1 , −1) and C (11 , 4), (iv) A(8, 1), B(3, –4) and C(2, –5), 14. Find the value of x for which the points A( x , 2), B( −3 , − 4) and C(7 , −5), [CBSE 2015], are collinear., 15. For what value of x are the points A( −3 , 12), B(7 , 6) and C ( x , 9), collinear?, 16. For what value of y are the points P(1 , 4), Q( 3 , y) and R( −3 , 16) are, collinear?, 17. Find the value of y for which the points A( −3 , 9), B( 2 , y) and C( 4 , −5) are, collinear., 18. For what values of k are the points A( 8 , 1), B( 3 , − 2 k) and C ( k , −5), collinear., [CBSE 2015], 19. Find a relation between x and y, if the points A(2, 1), B( x , y) and C(7 , 5), [CBSE 2009C], are collinear., 20. Find a relation between x and y, if the points A( x , y), B( −5 , 7 ) and, [CBSE 2015], C( −4 , 5) are collinear., 1 1, 21. Prove that the points A( a, 0), B( 0 , b) and C(1 , 1) are collinear, if + = 1., a b
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338, , Secondary School Mathematics for Class 10, , 22. If the points P( −3 , 9), Q( a, b) and R( 4 , −5) are collinear and a + b = 1,, [CBSE 2014], find the values of a and b., 23. Find the area of SABC with vertices A( 0 , − 1), B( 2 , 1) and C( 0 , 3). Also,, find the area of the triangle formed by joining the midpoints of its sides., [CBSE 2014], Show that the ratio of the areas of two triangles is 4 : 1., 24. If a ≠ b ≠ c , prove that ( a, a2 ), (b , b 2 ), ( 0 , 0) will not be collinear., , [CBSE 2017], , ANSWERS (EXERCISE 6C), , 1. (i) 11 sq units, 2. 132 sq units, 8. 12 sq units, , (ii) 53 sq units, , (iii) 37.5 sq units, , (iv) 24.5 sq units, , 3. 28 sq units 4. 28 sq units 5. 77 sq units 6. 1 sq unit, 3, 13, 9. sq unit 10. (i) p = −3 or p = −9 (ii) y =, 4, 2, , 15. x = 2, 11, 19. 4 x − 5 y − 3 = 0, 16. y = −2, 17. y = −1, 18. k = 2 or k =, 2, 20. 2 x + y + 3 = 0 22. a = 2 , b = −1 23. 4 sq units, 1 sq unit, 11. k = 3, , 12. k = 3, , 14. x = −63, , HINTS TO SOME SELECTED QUESTIONS, 8. Let the vertices of SABC be A( 1, − 4 ), B( x2 , y2 ) and C ( x3 , y3 )., Let D( 2 , − 1) and E( 0 , − 1) be the midpoints of AB and AC respectively. Then,, −4 + y 2, 1 + x2, = 2,, = −1, 2, 2, ⇒ 1 + x 2 = 4 , − 4 + y 2 = −2, x2 = 3 , y2 = 2., −4 + y 3, 1 + x3, And,, = 0,, = −1, 2, 2, ⇒ 1 + x 3 = 0 , − 4 + y 3 = −2, ⇒, , ⇒, , x3 = −1, y3 = 2., , In SABC, we have, A( x1 = 1, y1 = −4 ), B( x2 = 3 , y2 = 2 ) and C ( x3 = −1, y3 = 2 )., Now, find ar( SABC )., 9. Let the fourth vertex be D( x , y )., Midpoint of AC is ⎛⎜, ⎝, , 6+ 9 4+, ,, 2, 2, , 1⎞, ⎛ 15 5 ⎞, ⎟ , i.e., ⎜ , ⎟ ⋅, ⎠, ⎝ 2 2⎠, , 8 + x 2 + y⎞, Midpoint of BD is ⎛⎜, ,, ⎟⋅, ⎝ 2, 2 ⎠
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Coordinate Geometry, ∴, , 339, , 2+ y 5, 8 + x 15, =, and, = ⇒ x = 7 and y = 3., 2, 2, 2, 2, , So, we get the point D(7, 3)., 7 + 9 3 + 4⎞, ⎛ 7⎞, Midpoint of DC is E ⎛⎜, ,, ⎟ , i.e., E ⎜ 8 , ⎟ ⋅, ⎝ 2, ⎝ 2⎠, 2 ⎠, 7, Now, A(6, 1), D(7, 3) and E ⎛⎜ 8 , ⎞⎟ are vertices of SADE. Now find its area., ⎝ 2⎠, 10. (i) Taking A( x1 = 1, y1 = −3 ), B( x2 = 4 , y2 = p ) and C ( x3 = −9 , y3 = 7 ), we get, ar( SABC ) = 15 sq units, ⇒, , 1, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 15, 2, , ⇒, , 1 ⋅ ( p − 7 ) + 4 ⋅ (7 + 3 ) − 9 ⋅ ( −3 − p ) = 30, , ⇒, , ( p − 7 ) + 40 + 27 + 9 p = 30 ⇒ 10 p + 60 = 30, , ⇒, , 10 p + 60 = 30 or 10 p + 60 = −30 ⇒ p = −3 or p = −9., , 11. Let A( x1 = k + 1, y1 = 1), B( x2 = 4 , y2 = −3 ) and C ( x3 = 7 , y3 = − k ) be the vertices of, SABC. Then,, 1, ar( SABC ) = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), 2, 1, = ( k + 1)( −3 + k ) + 4( − k − 1) + 7 ( 1 + 3 ), 2, 1, 1, = ( k 2 − 6 k + 21) = ( k 2 − 6 k + 9 ) + 12, 2, 2, 1, 1, = ( k − 3 ) 2 + 12 = [( k − 3 ) 2 + 12]., 2, 2, 1, [( k − 3 ) 2 + 12] = 6 ⇒ ( k − 3 ) 2 + 12 = 12 ⇒ ( k − 3 ) 2 = 0., ∴, 2, Hence, k = 3., 12. Let A( x1 = −2 , y1 = 5 ), B( x2 = k , y2 = −4 ) and C ( x3 = 2 k + 1, y3 = 10 ). Then,, ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , ar( SABC ) = 53 sq units, 1, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 53, 2, 1, ( −2 )( −4 − 10 ) + k( 10 − 5 ) + ( 2 k + 1)(5 + 4 ) = 53, 2, 28 + 5 k + 18 k + 9 = 106, 23 k + 37 = 106, 23 k + 37 = 106 or 23 k + 37 = −106, 143, ⋅, k = 3 or k = −, 23, , 14. Here, A( x1 = x , y1 = 2 ), B( x2 = −3 , y2 = −4 ) and C ( x3 = 7 , y3 = −5 )., ∴, , S = 0 ⇒ x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0
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340, , Secondary School Mathematics for Class 10, Þ x( -4 + 5 ) - 3( -5 - 2 ) + 7 ( 2 + 4 ) = 0, Þ x + 21 + 42 = 0 Þ x = -63., , 20. Here, A( x1 = x , y1 = y ), B( x2 = -5 , y2 = 7 ) and C ( x3 = -4 , y3 = 5 )., \, , S = 0 Þ x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) = 0, Þ x(7 - 5 ) - 5(5 - y ) - 4( y - 7 ) = 0, Þ 2 x - 25 + 5 y - 4 y + 28 = 0, Þ 2 x + y + 3 = 0., , 21. Here, A( x1 = a, y1 = 0 ), B( x2 = 0 , y2 = b ) and C ( x3 = 1, y3 = 1)., \, , S = 0 Þ x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) = 0, Þ a( b - 1) + 0 × ( 1 - 0 ) + 1 × ( 0 - b ) = 0, Þ ab - a - b = 0 Þ a + b = ab, 1 1, Þ + = 1., a b, , ................................................................, , EXERCISE 6D, Very-Short-Answer Questions, 1. Points A( -1 , y) and B(5 , 7 ) lie on a circle with centre O( 2 , - 3 y). Find the, values of y., [CBSE 2014], 2. If the point A(0, 2) is equidistant from the points B( 3 , p) and C ( p , 5),, [CBSE 2014], find p., 3. ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3)., [CBSE 2014], Find the length of one of its diagonal., 4. If the point P( k -1 , 2) is equidistant from the points A( 3 , k) and B( k , 5),, [CBSE 2014], find the values of k., 5. Find the ratio in which the point P( x , 2) divides the join of A(12 , 5) and, [CBSE 2014], B( 4 , - 3)., 6. Prove that the diagonals of a rectangle ABCD with vertices A(2, –1),, B(5, –1), C(5, 6) and D(2, 6) are equal and bisect each other. [CBSE 2014], 7. Find the lengths of the medians AD and BE of SABC whose vertices are, [CBSE 2014], A(7, –3), B(5, 3) and C(3, –1)., 8. If the point C ( k , 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3, [CBSE 2013C], then find the value of k., 9. Find the point on x-axis which is equidistant from points A(–1, 0) and, [CBSE 2013C], B(5, 0)., æ2 ö, æ -8 ö, 10. Find the distance between the points ç, , 2 ÷ and ç , 2 ÷ ×, è5 ø, è 5, ø, , [CBSE 2009C]
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Coordinate Geometry, , 341, , 11. Find the value of a, so that the point ( 3 , a) lies on the line represented by, [CBSE 2009], 2 x − 3 y = 5., 12. If the points A( 4 , 3) and B( x , 5) lie on the circle with centre O( 2 , 3), find, [CBSE 2009], the value of x., 13. If P( x , y) is equidistant from the points A(7 , 1) and B( 3 , 5), find the, relation between x and y., 14. If the centroid of SABC having vertices A( a, b), B(b , c) and C ( c , a) is the, origin, then find the value of ( a + b + c)., 15. Find the centroid of SABC whose vertices are A( 2 , 2), B( − 4 , − 4) and, C(5 , − 8)., 16. In what ratio does the point C( 4 , 5) divide the join of A( 2 , 3) and B(7 , 8)?, 17. If the points A( 2 , 3), B( 4 , k) and C( 6 , − 3) are collinear, find the value, of k., ANSWERS (EXERCISE 6D), , 1. y = 7 or y = −1, , 2. p = 1, , 7. AD = 5 units, BE = 5 units, 11. a =, , 1, 3, , 16. 2 : 3, , 12. x = 2, , 3. 5 units, 16, 8. k =, 5, , 13. x − y = 2, , 4. k = 1 or k = 5, 9. P(2, 0), , 14. a + b + c = 0, , 17. k = 0, HINTS TO SOME SELECTED QUESTIONS, , 1. We know that the radii of a circle are equal., ∴, , OA = OB ⇒ OA 2 = OB 2, ⇒ ( 2 + 1) 2 + ( −3 y − y ) 2 = ( 2 − 5 ) 2 + ( −3 y − 7 ) 2, ⇒ 3 2 + 16 y 2 = ( −3 ) 2 + ( 9 y 2 + 49 + 42 y ), ⇒ 7 y 2 − 42 y − 49 = 0 ⇒ y 2 − 6 y − 7 = 0, ⇒ ( y − 7 )( y + 1) = 0 ⇒ y = 7 or y = −1., , 3. Diagonal AC = diagonal BD, = ( 4 − 0 ) 2 + ( 0 − 3 ) 2 = 25 = 5 units., , 5. 3 : 5, , 10. 2 units, −10 ⎞, ⎟, 15. ⎛⎜ 1 ,, ⎝, 3 ⎠
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342, , Secondary School Mathematics for Class 10, , 4. PA = PB ⇒ PA 2 = PB 2, ⇒ ( k − 1 − 3)2 + ( 2 − k )2 = ( k − 1 − k )2 + ( 2 − 5)2, ⇒ ( k − 4 ) 2 + ( 2 − k ) 2 = 1 + 9 = 10, ⇒ 2 k 2 − 12 k + 10 = 0 ⇒ k 2 − 6 k + 5 = 0, ⇒ ( k − 1)( k − 5 ) = 0, ⇒ k = 1 or k = 5., 5. Let the required ratio be k : 1. Then,, −3 k + 5, 3, = 2 ⇒ −3 k + 5 = 2 k + 2 ⇒ 5 k = 3 ⇒ k = ⋅, k+1, 5, 3, Required ratio = ⎛⎜ : 1⎞⎟ = 3 : 5., ⎝5 ⎠, 6. Diagonal AC = (5 − 2 ) 2 + ( 6 + 1) 2, =, , 3 2 + 7 2 = 9 + 49 = 58, , Diagonal BD = (5 − 2 ) 2 + ( −1 − 6 ) 2, =, , 3 2 + ( −7 ) 2 = 9 + 49 = 58 ., , ∴, , Diag. AC = Diag. BD., 5 + 2 6 − 1⎞ ⎛ 7 5 ⎞, Midpoint of AC = ⎛⎜, ,, ⎟ =⎜ , ⎟⋅, ⎝ 2, 2 ⎠ ⎝ 2 2⎠, , Midpoint of BD = ⎛⎜, ⎝, , 5 + 2 −1 + 6 ⎞ ⎛ 7 5 ⎞, ,, ⎟ =⎜ , ⎟⋅, 2, 2 ⎠ ⎝ 2 2⎠, , 5 + 3 3 − 1⎞, 7. Midpoint of BC is D ⎛⎜, ,, ⎟ , i.e., D(4, 1)., ⎝ 2, 2 ⎠, 7 + 3 −3 − 1 ⎞, Midpoint of AC is E⎛⎜, ,, ⎟ , i.e., E(5, –2)., ⎝ 2, 2 ⎠, AD = (7 − 4 ) 2 + ( −3 − 1) 2 =, , 3 2 + ( −4 ) 2 = 25 = 5 units., , BE = (5 − 5 ) 2 + ( 3 + 2 ) 2 = 0 + 25 = 25 = 5 units., 8. k =, , ( 2 × 5 + 3 × 2 ) 16, =, ⋅, ( 2 + 3), 5, , 9. Let the required point be P( x , 0 ). Then,, PA 2 = PB 2 ⇒ ( x + 1) 2 + ( 0 − 0 ) 2 = ( x − 5 ) 2 + ( 0 − 0 ) 2, ⇒ x 2 + 1 + 2 x = x 2 + 25 − 10 x, ⇒ 12 x = 24 ⇒ x = 2., ∴ the required point is P(2, 0)., 2, , −8 2 ⎞, 10. Given distance = ⎛⎜, − ⎟ − ( 2 − 2 ) 2 = ( −2 ) 2 − 0 2 = 4 = 2 units., ⎝ 5, 5⎠, 11. Since ( 3 , a) lies on the line 2 x − 3 y = 5 , we have, 1, 2 × 3 − 3a = 5 ⇒ 3a = 1 ⇒ a = ⋅, 3
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Coordinate Geometry, , 343, , 12. ( x − 2 ) 2 + (5 − 3 ) 2 = ( 4 − 2 ) 2 + ( 3 − 3 ) 2, ⇒, , ( x − 2)2 + 2 2 = 2 2 + 0 2, , ⇒, , ( x − 2 ) 2 = 0 ⇒ x − 2 = 0 ⇒ x = 2., , 13. (7 − x ) 2 + ( 1 − y ) 2 = ( 3 − x ) 2 + (5 − y ) 2, ⇒, , ( 49 + x 2 − 14 x ) + ( 1 + y 2 − 2 y ) = ( 9 + x 2 − 6 x ) + ( 25 + y 2 − 10 y ), , ⇒, , x 2 + y 2 − 14 x − 2 y + 50 = x 2 + y 2 − 6 x − 10 y + 34, , ⇒, , 8 x − 8 y = 16 ⇒ x − y = 2., , a + b + c a + b + c⎞, 14. Centroid is ⎛⎜, ,, ⎟⋅, ⎝, ⎠, 3, 3, a+ b + c, So,, = 0 ⇒ a + b + c = 0., 3, 7 k + 2 8k + 3 ⎞, 16. Let the required ratio be k : 1. Then, C is ⎛⎜, ,, ⎟⋅, ⎝ k+1 k+1 ⎠, 7k + 2, 2, = 4 ⇒ 7 k + 2 = 4k + 4 ⇒ 3k = 2 ⇒ k = ⋅, ∴, k+1, 3, 2, Required ratio is : 1, i.e., 2 : 3., 3, 17. Let ( x1 = 2 , y1 = 3 ), ( x2 = 4 , y2 = k ) and ( x3 = 6 , y3 = −3 )., ∴, , S = 0 ⇒ x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒ 2( k + 3 ) + 4( −3 − 3 ) + 6( 3 − k ) = 0, ⇒ − 4 k = 0 ⇒ k = 0., , ................................................................, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. The distance of the point P(–6, 8) from the origin is, (a) 8, , (b) 2 7, , (c) 6, , [CBSE 2013C], , (d) 10, , 2. The distance of the point (–3, 4) from x-axis is, (a) 3, , (b) –3, , (c) 4, , [CBSE 2012], , (d) 5, , 3. The point on x-axis which is equidistant from points A( −1 , 0) and, [CBSE 2013], B(5 , 0) is, (a) (0, 2), , (b) (2, 0), , (c) (3, 0), , (d) (0, 3), , 4. If R(5 , 6) is the midpoint of the line segment AB joining the points, [CASE 2014], A( 6 , 5) and B(4, 4) then y equals, (a) 5, , (b) 7, , (c) 12, , (d) 6, , 5. If the point C ( k , 4) divides the join of the points A( 2 , 6) and B(5 , 1) in the, ratio 2 : 3 then the value of k is, [CBSE 2013C], 28, 16, 8, (c), (d), (a) 16, (b), 5, 5, 5
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344, , Secondary School Mathematics for Class 10, , 6. The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is, [CBSE 2014], , (a) (7 + 5 ), , (b) 5, , (c) 10, , (d) 12, , 7. If A(1 , 3), B(–1, 2), C(2, 5) and D(x, 4) are the vertices of a||gm ABCD, [CBSE 2012], then the value of x is, 3, (a) 3, (b) 4, (c) 0, (d), 2, 8. If the points A( x , 2), B( −3 , − 4) and C(7 , −5) are collinear then the value, [CBSE 2014], of x is, (a) –63, , (b) 63, , (c) 60, , (d) –60, , 9. The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in, square units is, [CBSE 2012], (a) 20, , (b) 12, , (c) 6, , (d) 16, , 10. The area of SABC with vertices A( a, 0), O( 0 , 0) and B( 0 , b) in square, units is, [CBSE 2011], 1, 1, 1, (c) a2b 2, (d) b 2, (a) ab, (b) ab, 2, 2, 2, a ⎞, ⎛, 11. If P ⎜ , 4 ⎟ is the midpoint of the line segment joining the points, ⎝2 ⎠, A( −6 , 5) and B( −2 , 3) then the value of a is, (a) –8, , (b) 3, , (c) –4, , [CBSE 2011], , (d) 4, , 12. ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3)., The length of one of its diagonals is, [CBSE 2014], (a) 5, , (b) 4, , (c) 3, , (d) 25, , 13. The coordinates of the point P dividing the line segment joining the, points A(1 , 3) and B( 4 , 6) in the ratio 2 : 1 is, [CBSE 2012], (a) (2, 4), , (b) (3, 5), , (c) (4, 2), , (d) (5, 3), , 14. If the coordinates of one end of a diameter of a circle are (2, 3) and the, coordinates of its centre are (–2, 5), then the coordinates of the other end, [CBSE 2012], of the diameter are, (a) (–6, 7), , (b) (6, –7), , (c) (4, 2), , (d) (5, 3), , 15. In the given figure P(5 , − 3) and Q( 3 , y) are the points of trisection of the, line segment joining A(7 , −2) and B(1, –5). Then, y equals, [CBSE 2012], (a) 2, (c) –4, , (b) 4, −5, (d), 2
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Coordinate Geometry, , 345, , 16. The midpoint of segment AB is P(0, 4). If the coordinates of B are (–2, 3),, [CBSE 2011], then the coordinates of A are, (a) (2, 5), , (b) (–2, –5), , (c) (2, 9), , (d) (–2, 11), , 17. The point P which divides the line segment joining the points A(2, –5), and B(5, 2) in the ratio 2 : 3 lies in the quadrant, [CBSE 2011], (a) I, , (b) II, , (c) III, , (d) IV, , 18. If A(–6, 7) and B(–1, –5) are two given points then the distance 2AB is, [CBSE 2011], , (a) 13, , (b) 26, , (c) 169, , (d) 238, , 19. Which point on x-axis is equidistant from the points A(7, 6) and, B( −3 , 4)?, (a) (0, 4), , (b) (–4, 0), , (c) (3, 0), , (d) (0, 3), , 20. The distance of P(3, 4) from the x-axis is, (a) 3 units, , (b) 4 units, , (c) 5 units, , (d) 1 unit, , 21. In what ratio does the x-axis divide the join of A(2, –3) and B(5, 6)?, (a) 2 : 3, , (b) 3 : 5, , (c) 1 : 2, , (d) 2 : 1, , 22. In what ratio does the y-axis divide the join of P(–4, 2) and Q(8, 3)?, (a) 3 : 1, , (b) 1 : 3, , (c) 2 : 1, , (d) 1 : 2, , 23. If P(–1, 1) is the midpoint of the line segment joining A(–3, b) and, B(1 , b + 4) then b = ?, (a) 1, (b) –1, (c) 2, (d) 0, 24. The line 2 x + y − 4 = 0 divides the line segment joining A(2, –2) and, B( 3 , 7 ) in the ratio, (a) 2 : 5, (b) 2 : 9, (c) 2 : 7, (d) 2 : 3, 25. If A( 4 , 2), B( 6 , 5) and C(1 , 4) be the vertices of SABC and AD is a, median, then the coordinates of D are, 5, 7 9, 7, (a) ⎛⎜ , 3 ⎞⎟, (b) ⎛⎜ 5 , ⎞⎟, (d) none of these, (c) ⎛⎜ , ⎞⎟, ⎝2 ⎠, ⎝ 2 2⎠, ⎝ 2⎠, 26. If A( −1 , 0), B(5 , − 2) and C( 8 , 2) are the vertices of a SABC then its, centroid is, (a) (12, 0), (b) (6, 0), (c) (0, 6), (d) (4, 0), 27. Two vertices of SABC are A( −1 , 4) and B(5 , 2) and its centroid is, G( 0 , − 3). Then, the coordinates of C are, (a) (4, 3), , (b) (4, 15), , (c) (–4, –15), , (d) (–15, –4), , 28. The points A( − 4 , 0), B( 4 , 0) and C( 0 , 3) are the vertices of a triangle,, which is, (a) isosceles, (b) equilateral (c) scalene, (d) right angled
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346, , Secondary School Mathematics for Class 10, , 29. The points P( 0 , 6), Q( −5 , 3) and R( 3 , 1) are the vertices of a triangle,, which is, (a) equilateral (b) isosceles, (c) scalene, (d) right angled, 30. If the points A( 2 , 3), B(5 , k) and C( 6 , 7 ) are collinear then, −3, 11, (a) k = 4, (b) k = 6, (c) k =, (d) k =, 2, 4, 31. If the points A(1 , 2), O( 0 , 0) and C ( a, b) are collinear then, (a) a = b, , (b) a = 2b, , (c) 2a = b, , (d) a + b = 0, , 32. The area of SABC with vertices A(3, 0), B(7, 0) and C(8, 4) is, (a) 14 sq units, , (b) 28 sq units, , (c) 8 sq units, , (d) 6 sq units, , 33. AOBC is a rectangle whose three vertices are A( 0 , 3), O( 0 , 0) and B(5 , 0)., The length of each of its diagonals is, (a) 5 units, , (b) 3 units, , (c) 4 units, , (d), , 34 units, , 34. If the distance between the points A( 4 , p) and B(1 , 0) is 5 then, (a) p = 4 only, , (b) p = −4 only, , (c) p = ±4, , (d) p = 0, , ANSWERS (MCQ), , 1. (d), 9. (c), 17. (d), 25. (c), 33. (d), , 2. (c), 10. (b), 18. (b), 26. (d), 34. (c), , 3. (b), 11. (a), 19. (c), 27. (c), , 4. (b), 12. (a), 20. (b), 28. (a), , 5. (c), 13. (b), 21. (c), 29. (d), , 6. (d), 14. (a), 22. (d), 30. (b), , 7. (b), 15. (c), 23. (b), 31. (c), , HINTS TO SOME SELECTED QUESTIONS, 1. OP 2 = ( −6 ) 2 + 8 2 = 36 + 64 = 100 ⇒ OP = 100 = 10 units., 2. Clearly, the distance of the point P( −3 , 4 ) from x-axis is 4 units., , 8. (a), 16. (a), 24. (b), 32. (c)
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Coordinate Geometry, 3. Let the required point be P( x , 0 ). Then,, PA 2 = PB 2 ⇒ ( x + 1) 2 = ( x − 5 ) 2, ⇒ x 2 + 2 x + 1 = x 2 − 10 x + 25, ⇒ 12 x = 24 ⇒ x = 2., So, the required point is P(2, 0)., 4., , 5+ y, = 6 ⇒ 5 + y = 12 ⇒ y = 7 ., 2, , 5. k =, , ( 2 × 5 ) + ( 3 × 2 ) 16, =, ⋅, ( 2 + 3), 5, , 6. AB = 4 units, BC = 3 units., AC 2 = ( 3 − 0 ) 2 + ( 0 − 4 ) 2, = ( 9 + 16 ) = 25., ⇒, , AC = 5 units., , ∴, , perimeter = (4 + 3 + 5) units, = 12 units., , 7. Midpoint of BD = midpoint of AC, −1 + x 1 + 2, =, ⇒ − 1 + x = 3 ⇒ x = 4., ⇒, 2, 2, , 8. Here, ( x1 = x , y1 = 2 ), ( x2 = −3 , y2 = −4 ) and ( x3 = 7 , y3 = −5 )., ∴, , x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, , ⇒, , x( −4 + 5 ) − 3( −5 − 2 ) + 7 ( 2 + 4 ) = 0, , ⇒, , x + 21 + 42 = 0 ⇒ x = −63., , 9. Here, ( x1 = 5 , y1 = 0 ), ( x2 = 8 , y2 = 0 ) and ( x3 = 8 , y3 = 4 )., 1, ∴ S = x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ), 2, 1, = 5( 0 − 4 ) + 8( 4 − 0 ) + 8( 0 − 0 ), 2, 1, 1, =, − 20 + 32 = ⎛⎜ × 12 ⎞⎟ = 6 sq units., ⎝2, ⎠, 2, 10. Here, base = a units and height = b units., 1, ∴ area = ⎛⎜ × base × height ⎞⎟, ⎝2, ⎠, =, , 1, ab sq units., 2, , 347
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348, , 11., , Secondary School Mathematics for Class 10, a ( −6 − 2 ), =, = −4 ⇒ a = −8., 2, 2, , 12. Diagonal BD = ( 4 − 0 ) 2 + ( 0 − 3 ) 2 = 16 + 9, = 25 = 5 units., , ( 2 × 4 + 1 × 1) ( 2 × 6 + 1 × 3 ) ⎞, 13. Coordinates of P are ⎛⎜, ,, ⎟ = ( 3 , 5 )., ⎝, ⎠, 2+ 1, 2+ 1, 14. Let C( −2 , 5 ) be the centre of the given circle and A(2, 3) and B( x , y ) be the end points of, a diameter ACB., Then, C is the midpoint of AB., 3+ y, 2+ x, = −2 and, =5, ∴, 2, 2, ⇒, , 2 + x = −4 and 3 + y = 10, , ⇒, , x = −6 and y = 7 ., , So, the coordinates of B are (–6, 7)., 15. Q ( 3 , y ) divides AB in the ratio 2 : 1., 2 × 1 + 1 × 7 2 × ( −5 ) + 1 × ( −2 ) ⎞, So, Q is ⎛⎜, ,, ⎟ , i.e., (3, –4)., ⎝, ⎠, 2+ 1, 2+ 1, Hence, y = −4., 16. Let the point A be ( a, b ). Then,, a + ( −2 ), b+ 3, = 0 and, =4, 2, 2, ⇒, , a − 2 = 0 and b = 8 − 3 ⇒ a = 2 , b = 5., , ∴, , the point A is (2, 5)., , 2× 5 + 3× 2 2× 2 − 3× 5⎞, ⎛ −11 ⎞ ., 17. The point P is given by P ⎛⎜, ,, ⎟ = P ⎜ 3,, ⎟, ⎝, ⎠, ⎝, 5 ⎠, 2+ 3, 2+ 3, So, P lies in IV quadrant.
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Coordinate Geometry, 18. 2 AB = 2 × ( −1 + 6 ) 2 + ( −5 − 7 ) 2 = 2 × (5 ) 2 + ( −12 ) 2, = 2 × 169 = 2 × 13 = 26., 19. Let the required point be P( x , 0 ). Then,, AP 2 = BP 2 ⇒ ( x − 7 ) 2 + ( 0 − 6 ) 2 = ( x + 3 ) 2 + ( 0 − 4 ) 2, ⇒ x 2 − 14 x + 85 = x 2 + 6 x + 25, ⇒ 20 x = 60 ⇒ x = 3., ∴, , the required point is P(3, 0)., , 21. Let the x-axis cut AB at P( x , 0 ) in the ratio k : 1., Then,, ∴, , 6k − 3, 1, = 0 ⇒ 6k − 3 − 0 ⇒ 6k = 3 ⇒ k = ⋅, k+1, 2, , 1, required ratio = ⎛⎜ : 1⎞⎟ = 1 : 2., ⎝2 ⎠, , 22. Let the y-axis cut AB at P( 0 , y ) in the ratio k : 1. Then,, 8k − 4 3k + 2 ⎞, 8k − 4, P ⎛⎜, ,, =0, ⎟ = P( 0 , y ) ⇒, ⎝ k+1 k+1 ⎠, k+1, ⇒, ∴, , 1, ⋅, 2, 1, required ratio = ⎛⎜ : 1⎞⎟ = 1 : 2., ⎝2 ⎠, 8k − 4 = 0 ⇒ k =, , 23. We have, , b + ( b + 4), = 1 ⇒ 2 b + 4 = 2 ⇒ 2 b = −2 ⇒ b = −1., 2, , 24. Let the required ratio be k : 1., 3k + 2 7 k − 2 ⎞, Then, the point of division is P ⎛⎜, ,, ⎟., ⎝ k+1 k+1⎠, This point lies on the line 2 x + y − 4 = 0., 2( 3 k + 2 ) (7 k − 2 ), − 4 = 0 ⇒ 6k + 4 + 7 k − 2 − 4k − 4 = 0, +, k+1, k+1, 2, ⇒ 9k = 2 ⇒ k = ⋅, 9, 2 ⎞, ⎛, So, the required ratio is ⎜ : 1⎟ , i. e. , ( 2 : 9 )., ⎝9 ⎠, , ∴, , 6 + 1 5 + 4⎞, ⎛7 9⎞, 25. Midpoint of BC is D ⎛⎜, ,, ⎟ = D⎜ , ⎟., ⎝ 2, ⎝ 2 2⎠, 2 ⎠, x + x2 + x3 y1 + y2 + y3 ⎞, ⎛ −1 + 5 + 8 , 0 − 2 + 2 ⎞ = ( 4 , 0 )., 26. Centroid is G ⎛⎜ 1, ,, ⎟, ⎟ =G⎜, ⎝, ⎠, ⎝, ⎠, 3, 3, 3, 3, 27. Let the vertex C be C ( x , y ). Then,, , ∴, , 4+ 2+ y, −1 + 5 + x, = 0 and, = −3 ⇒ x + 4 = 0 and 6 + y = −9, 3, 3, x = −4 and y = −15., , So, the coordinates of C are (–4, –15)., , 349
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350, , Secondary School Mathematics for Class 10, , 28. AB 2 = ( 4 + 4 ) 2 + ( 0 − 0 ) 2 = 8 2 + 0 2 = 64 + 0 = 64 ⇒ AB = 64 = 8 units., BC 2 = ( 0 − 4 ) 2 + ( 3 − 0 ) 2 = ( −4 ) 2 + 3 2 = 16 + 9 = 25 ⇒ BC = 25 = 5 units., AC 2 = ( 0 + 4 ) 2 + ( 3 − 0 ) 2 = 4 2 + 3 2 = 16 + 9 = 25 ⇒ AC = 25 = 5 units., ∴, , SABC is isosceles., , 29. PQ 2 = ( −5 − 0 ) 2 + ( 3 − 6 ) 2 = ( −5 ) 2 + ( −3 ) 2 = 25 + 9 = 34 ,, QR 2 = ( 3 + 5 ) 2 + ( 1 − 3 ) 2 = 8 2 + ( −2 ) 2 = 64 + 4 = 68 ,, PR 2 = ( 3 − 0 ) 2 + ( 1 − 6 ) 2 = 3 2 + ( −5 ) 2 = ( 9 + 25 ) = 34., ∴, , PQ 2 + PR 2 = QR 2 ., , Hence, SPQR is right-angled., 30. Here, ( x1 = 2 , y1 = 3 ), ( x2 = 5 , y2 = k ) and ( x3 = 6 , y3 = 7 )., Since the given points are collinear, we must have:, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒, , 2( k − 7 ) + 5(7 − 3 ) + 6( 3 − k ) = 0 ⇒ 2 k − 14 + 20 + 18 − 6 k = 0, , ⇒, , 4 k = 24 ⇒ k = 6., , 31. Here, ( x1 = 1, y1 = 2 ), ( x2 = 0 , y2 = 0 ) and ( x3 = a, y3 = b )., Since the given points are collinear, we have, x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0, ⇒, , 1 ⋅ ( 0 − b ) + 0 ⋅ ( b − 2 ) + a ⋅ ( 2 − 0 ) = 0 ⇒ − b + 0 + 2 a = 0 ⇒ 2a = b., , 32. Here, ( x1 = 3 , y1 = 0 ), ( x2 = 7 , y2 = 0 ) and ( x3 = 8 , y3 = 4 )., 1, ∴ area of S ABC = {x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )}, 2, 1, 1, = { 3( 0 − 4 ) + 7 ( 4 − 0 ) + 8( 0 − 0 )} = {−12 + 28 + 0}, 2, 2, 1, = ⎛⎜ × 16 ⎞⎟ = 8 sq units., ⎝2, ⎠, 33. Diagonal OC, = diagonal AB, = (5 − 0 ) 2 + ( 0 − 3 ) 2, = 25 + 9 =, , 34 units., , 34. ( 4 − 1) 2 + ( p − 0 ) 2 = 5 2 ⇒ p 2 = ( 25 − 9 ) = 16 ⇒ p = ±4., , _
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Triangles, , 351, , Triangles, , 7, , Two geometric figures which have the same shape and size, are known as congruent figures., CONGRUENT FIGURES, , Congruent figures are alike in every respect., SIMILAR FIGURES Geometric figures which have the same shape but different sizes, are known as similar figures., , Two congruent figures are always similar but two similar figures need, not be congruent., Examples, (i) Any two line segments are similar., (ii) Any two equivalent triangles are similar., (iii) Any two squares are similar., (iv) Any two circles are similar., , SIMILAR POLYGONS, , Two polygons having the same number of sides are said to be similar, if, (i) their corresponding angles are equal, and, (ii) the lengths of their corresponding sides are proportional., If two polygons ABCDE and PQRST are similar, we write,, ABCDE ~ PQRST, where the symbol ‘~’ stands for ‘is similar to’., The constant ratio between the corresponding sides of two similar, figures is known as the scale factor, or the representative fraction. Since, triangles are also polygons, so the same set of conditions apply for the, similarity of triangles., EQUIANGULAR TRIANGLES, , Two triangles are said to be equiangular if their corresponding angles are equal., 351
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352, , Secondary School Mathematics for Class 10, , SIMILAR TRIANGLES, , Two triangles are said to be similar to each other if, (i) their corresponding angles are equal, and, (ii) their corresponding sides are proportional., , RESULTS ON SIMILAR TRIANGLES, (BASIC-PROPORTIONALITY THEOREM) OR (THALES’ THEOREM), THEOREM 1, , GIVEN, , If a line is drawn parallel to one side of a triangle to intersect the, other two sides in distinct points then the other two sides are divided, in the same ratio., [CBSE 2002C, ’04C, ’05, ’06C, ’07, ’09, ’10], , A 3ABC in which DE BC and DE intersects AB and AC at D and, E respectively., , TO PROVE, , AD AE ·, DB EC, , CONSTRUCTION, , Join BE and CD., Draw EL = AB and DM = AC., , PROOF, , We have, 1, 1, ar(3ADE) 2 # AD # EL, [a 3 2 # base # height], 1, and ar(3DBE) 2 # DB# EL., 1, ar(3ADE) 2 # AD # EL AD, ·, , , , … (i), DB, 1, ar(3DBE), #, #, DB, EL, 2, 1, Again, ar(3ADE) ar(3AED) 2 # AE # DM, 1, and ar(3ECD) 2 # EC # DM., 1, ar(3ADE) 2 # AE # DM AE, ·, , , , ar(3ECD) 1 # EC # DM EC, 2, , … (ii), , Now, 3DBE and 3ECD being on the same base DE and between, the same parallels DE and BC, we have, … (iii), ar(3DBE) ar(3ECD), From (i), (ii) and (iii), we have, AD AE ·, DB EC
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Triangles, COROLLARY, , PROOF, , 353, , In a 3ABC, a line DE BC intersects AB in D and AC in E, then, prove that, AB AC, AD AE, (i) DB EC, (ii) AB AC ·, , (i) From Basic-Proportionality theorem, we have, AD AE, AD AE , DB EC & DB 1 EC 1, , , AB AC ·, & ADDBDB AEECEC & DB, EC, (ii) From Basic-Proportionality theorem, we have, AD AE, DB EC, DB EC & AD AE, DB EC, k a1 AE k, & a1 AD, (AD DB) (AE EC), , &, AD, AE, AB AC, AD AE ·, & AD AE & AB AC, , SUMMARY, , In 3ABC, let DE BC. Then,, AD AE, (i) DB EC (B.P.T.), AB AC, (ii) DB EC, AD AE, (iii) AB AC, THEOREM 2, , (Converse of Thales’ theorem) If a line divides any two sides of, , a triangle in the same ratio then the line must be parallel to the, third side., GIVEN, , A 3ABC and a line l intersecting AB at D and, AD AE, AC at E, such that DB EC ·, , TO PROVE, PROOF, , DE BC., , If possible, let DE not be parallel to BC. Then, there must be another, line through D, which is parallel to BC. Let DF BC., Then, by Thales’ theorem, we have, AD AF ·, DB FC, AD AE, But, DB EC (given)., , … (i), … (ii)
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354, , Secondary School Mathematics for Class 10, , From (i) and (ii), we get, AF AE, AF AE , AF FC AE EC, FC EC & FC 1 EC 1 & FC, EC, 1 1, AC AC, & FC EC & FC EC & FC EC., This is possible only when E and F coincide., Hence, DE < BC., , SOLVED EXAMPLES, EXAMPLE 1, , In the given figure, MN AB,, BC 7.5 cm, AM 4 cm and, MC 2 cm. Find the length, of BN., [CBSE 2010], , SOLUTION, , In 3ABC, MN AB., MC NC, AC BC, [by Thales’ theorem], MC, NC, , AM MC BC, 2 x, , , where NC x cm, 4 2 7.5, 2 #7.5 15 , x, 6, 6 2.5, NC 2.5 cm., Hence, BN BC NC (7.5 2.5) cm 5 cm., , EXAMPLE 2, , AD 3, In the given figure, DE BC and DB 5 ·, If AC 4.8 cm, find the length of AE., [CBSE 2008C], , SOLUTION, , Let AE x cm., Then, EC (AC AE) (4.8 x) cm., Now, in 3ABC, DE BC., , , AD AE, 3, x, DB EC & 5 (4.8 x), , , , 3(4.8 x) 5x 8x 14.4, , , , x 1.8., , Hence, AE 1.8 cm.
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Triangles, EXAMPLE 3, , In the given figure, in 3ABC, DE BC, so that AD (4x 3) cm, AE (8x 7) cm,, BD (3x 1) cm and CE (5x 3) cm. Find, the value of x., [CBSE 2002C], , SOLUTION, , In 3ABC, DE BC., AD AE, BD CE [by Thales’ theorem], 4x 3 8x 7, (4x 3)(5x 3) (3x 1)(8x 7), , 3x 1 5x 3, 20x 2 27x 9 24x 2 29x 7, 4x 2 2x 2 0 2x 2 x 1 0, 2x 2 2x x 1 0 2x(x 1) (x 1) 0, (x 1)(2x 1) 0, (x 1) 0, , or (2x 1) 0, 1, x 1 or x 2 ·, 1, 1, But, x 2 & AD :4 #a 2 k 3D 5., , 1, And, distance can never be negative. So, x ! 2 ·, Hence, x 1., EXAMPLE 4, , SOLUTION, , If D and E are points on the sides AB, and AC respectively of 3ABC such that, AB 5.6 cm, AD 1.4 cm, AC 7.2 cm, and AE 1.8 cm, show that DE BC., Given, AB 5.6 cm, AD 1.4 cm,, AC 7.2 cm and AE 1.8 cm., AD 1.4 1, AE 1.8 1, AB 5.6 4 and AC 7.2 4, AD AE, AB AC ·, Hence, by the converse of Thales’ theorem, DE BC., , EXAMPLE 5, , In the adjoining figure, MN QR., Find (i) PN and (ii) PR., , 355
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356, SOLUTION, , Secondary School Mathematics for Class 10, , In 3PQR, MN QR., PM PN, MQ NR [by Thales’ theorem], 1.9 x, , 5.7 6.9 , where PN x cm, 1.9 # 6.9 , x 5.7, 2.3., Hence, (i) PN x cm 2.3 cm, and (ii) PR PN NR (2.3 6.9) cm 9.2 cm., , , EXAMPLE 6, , AD AE, In the given figure, DB EC and, +ADE +ACB. Prove that 3ABC is an, isosceles triangle., , SOLUTION, , We have, AD AE, DB EC & DE BC [by the converse of Thales’ theorem], +ADE +ABC (corresponding O)., But, +ADE +ACB (given)., +ABC +ACB., So, AB AC [sides opposite to equal angles]., Hence, 3ABC is an isosceles triangle., , EXAMPLE 7, , M and N are points on the sides AC and BC respectively of a 3ABC., In each of the following cases, state whether MN AB., (i) CM 4.2 cm, MA 2.8 cm, NB 3.6 cm, CN 5.7 cm, (ii) CB 6.92 cm, CN 1.04 cm, CA 1.73 cm, CM 0.26 cm, (iii) CM 5.1 cm, CA 6.8 cm, CB 5.6 cm, NB 1.4 cm, , SOLUTION, , (i) We have, CM 4.2 3, CN 5.7 19 ·, MA 2.8 2 and NB 3.6 12, CM CN, Since MA ! NB ·, So, MN is not parallel to AB., (ii) We have, MA CA CM (1.73 0.26) cm 1.47 cm, and NB CB CN (6.92 1.04) cm 5.88 cm.
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Triangles, , 357, , CM 0.26 26, CN 1.04 26 ·, MA 1.47 147 and NB 5.88 147, CM CN, Clearly, MA NB and so MN AB, [by the converse of Thales’ theorem]., , , , (iii) We have, MA CA CM (6.8 5.1) cm 1.7 cm, and CN CB NB (5.6 1.4) cm 4.2 cm., , , CM 5.1 3, CN 4.2 3 ·, MA 1.7 1 and NB 1.4 1, , CM CN, Clearly, MA NB and so MN AB, [by the converse of Thales’ theorem]., EXAMPLE 8, , In the given figure, DE AC and DF AE., BF BE, Prove that FE EC ·, , SOLUTION, , [CBSE 2005, ’07], , In 3BAE, DF AE., , , BD BF ·, DA FE, , … (i) [by Thales’ theorem], , In 3BAC, DE AC., , , BD BE ·, DA EC, , … (ii) [by Thales’ theorem], , From (i) and (ii), we get, BF BE, FE EC, , :each equal to BD D ·, DA, , EXAMPLE 9, , In the figure given along side, DE OQ, and DF OR. Show that EF QR., , SOLUTION, , In 3POQ, DE OQ., PD PE, DO EQ ·, In 3POR, DF OR., , … (i) [by Thales’ theorem]
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358, , Secondary School Mathematics for Class 10, , PD PF ·, DO FR, From (i) and (ii), we get, PD D ·, PE PF :, EQ FR each equal to DO, , , , … (ii), , Thus, in 3PQR, E and F are points on PQ and PR respectively, PE PF, such that EQ FR ·, Hence, EF QR [by the converse of Thales’ theorem]., EXAMPLE 10, , In the given figure, LM CB and LN CD., AM AN, Prove that AB AD ·, , SOLUTION, , In 3ALM, LM CB., AB AC, AM AL, AM AL & AB AC ·, In 3ALN, LN CD., AC AD, AL AN, AL AN & AC AD ·, From (i) and (ii), we get, AM AN ·, AB AD, , EXAMPLE 11, , … (i), , [by Thales’ theorem], , … (ii), , [by Thales’ theorem], , In the given figure, AB DE and BD EF., Prove that DC 2 CF # AC., [CBSE 2004C, ’10], , SOLUTION, , In 3ABC, AB DE., CD CE, DA EB ·, In 3CDB, BD EF., CF CE, FD EB ·, From (i) and (ii), we get, CD CF, DA FD, , … (i) [by Thales’ theorem], , … (ii) [by Thales’ theorem]
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Triangles, , , , , , , EXAMPLE 12, , DA FD, [taking reciprocals], DC CF, DA FD , DC 1 CF 1, DA DC FD CF, DC, CF, AC DC, DC CF, DC 2 CF # AC., , In the given figure, PQ AB and PR AC., Prove that QR BC., , SOLUTION, , 359, , In 3OAB, PQ AB., OP OQ, PA QB ·, In 3AOC, PR AC., OP OR, PA RC ·, From (i) and (ii), we get, , [CBSE 2002, ’05C], , … (i) [by Thales’ theorem], , … (ii) [by Thales’ theorem], , OQ OR, , QB RC in 3OBC., Thus, in 3OBC, Q and R are points on OB and OC respectively, OQ OR, such that QB RC ·, Hence, by the converse of Thales’ theorem, QR BC., EXAMPLE 13, , SOLUTION, , In the given figure, in 3ABC, +B +C, and BD CE. Prove that DE BC., , A 3ABC in which +B +C and BD CE., TO PROVE DE BC., GIVEN, , PROOF, , In 3ABC, +B +C & AB AC, [sides opposite equal O are equal]., , Now, AB AC & (AD BD) (AE CE), & AD AE, [a BD CE (given)], AE, & AD, [a BD CE]., BD CE
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360, , Secondary School Mathematics for Class 10, , AD AE, Thus, BD CE ·, Hence, DE BC [by the converse of Thales’ theorem]., EXAMPLE 14, , In 3ABC, D and E are two points on, AB such that AD BE. If DP BC and, EQ AC, prove that PQ AB., , SOLUTION, , GIVEN A 3ABC and D, E are two, points on AB such that AD BE., Also, DP BC, and EQ AC., TO PROVE, , PQ AB., , In 3ABC, DP BC., AD AP ·, DB PC, In 3CBA, EQ AC., PROOF, , , , , … (i), , [by Thales’ theorem], , BE BQ, [by Thales’ theorem], EA QC, AD BQ ·, … (ii), DB QC, [a BE AD, EA (ED DA) (DE EB) DB], , From (i) and (ii), we get, AP BQ ·, PC QC, , , PQ AB, , [by the converse of Thales’ theorem]., , EXAMPLE 15, , If three or more parallel lines are intersected by two transversals,, prove that the intercepts made by them on the transversals are, proportional., , SOLUTION, , GIVEN, , Three lines l, m, n such that l m n., , These lines are cut by the, transversals AB and CD in P, Q, R, and E, F, G respectively., TO PROVE, , PQ EF, ·, , QR FG, , Draw PM CD,, meeting the lines m and n at L and, M respectively., , CONSTRUCTION
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Triangles, PROOF, , 361, , PL EF and PE LF & PLFE is a gm, , & PL EF., , … (i), (opp. sides of a gm), , Also, LM FG and LF MG & LMGF is a gm, , & LM FG., , … (ii), (opp. sides of a gm), , In 3PRM, QL RM and therefore, by Thales’ theorem, we have, PQ PL EF, , , QR LM FG [using (i) and (ii)]., , , PQ EF, ·, , QR FG, , EXAMPLE 16, , ABCD is a quadrilateral and P, Q, R, S are the points of trisection, of the sides AB, BC, CD and DA respectively and are adjacent to A, and C. Prove that PQRS is a parallelogram., , SOLUTION, , GIVEN A quadrilateral ABCD in which, P, Q, R, S are the points of trisection of, AB, BC, CD and DA respectively (as, shown)., TO PROVE, , PQRS is a parallelogram., , CONSTRUCTION, PROOF, , , , Join AC., , BP BQ 2, In 3BAC, we have PA QC 1 ·, , PQ AC., , … (i) [by the converse of Thales’ theorem], , Also, in 3DAC, we have, , , , DS DR 2 ·, SA RC 1, … (ii) [by the converse of Thales’ theorem], SR AC., , Thus, PQ SR [from (i) and (ii)]., Similarly, by joining BD we can prove that SP RQ., Hence, PQRS is a parallelogram., EXAMPLE 17, , ABCD is a trapezium with AB DC. E and F are points on nonparallel sides AD and BC respectively such that EF AB. Show that, AE BF ·, ED FC
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362, SOLUTION, , Secondary School Mathematics for Class 10, GIVEN A trap. ABCD in which AB DC. E and F are points on, AD and BC respectively such that EF AB., TO PROVE, , AE BF ·, ED FC, , CONSTRUCTION, PROOF, , Join AC, intersecting EF at G., , EF AB and AB DC & EF DC., , Now, in 3ADC, EG DC., AE AG ·, ED GC, Similarly, in 3CAB, GF AB., , , CG CF, GA FB, AG BF, GC FC ·, From (i) and (ii), we get, , … (i) [by Thales’ theorem], , , , [by Thales’ theorem], … (ii), , AE BF ·, ED FC, EXAMPLE 18, , ABCD is a trapezium in which AB DC and its diagonals intersect, each other at the point O., AO BO, Prove that OC OD ·, [CBSE 2004], , SOLUTION, , A trapezium ABCD in which AB DC and its diagonals, AC and BD intersect at O., AO BO ·, TO PROVE, OC OD, CONSTRUCTION Through O, draw, EO AB, meeting AD at E., GIVEN, , In 3ADC, EO DC, AE AO, ED OC, In 3DAB, EO AB., DE DO, EA OB, AE BO, ED OD ·, From (i) and (ii), we get, AO BO ·, OC OD, PROOF, , [a EO AB DC] ., … (i) [by Thales’ theorem], , [by Thales’ theorem], … (ii)
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364, , Secondary School Mathematics for Class 10, , And, the distance cannot be negative., 1, x! 2 ·, Hence, x 2., MIDPOINT THEOREM, EXAMPLE 21, , Prove that the line segment joining the midpoints of any two sides of, a triangle is parallel to the third side., , SOLUTION, , A 3ABC in which D and E are the, midpoints of AB and AC respectively., GIVEN, , DE BC., Since D and E are the midpoints, of AB and AC respectively, we have, AD DB and AE EC., AD AE, DB EC [each equal to 1]., Hence, by the converse of Thales’ theorem, DE BC., TO PROVE, PROOF, , EXAMPLE 22, , Prove that a line drawn through the midpoint, of one side of a triangle parallel to another, side bisects the third side., , SOLUTION, , A 3ABC in which D is the, midpoint of AB and DE BC, meeting, AC at E., TO PROVE AE EC., PROOF Since DE BC, by Thales’ theorem, we have, AE AD , [a AD DB (given)], EC DB 1, AE, EC 1 & AE EC., , EXAMPLE 23, , In 3ABC, AD is a median and E is the midpoint of AD. If BE is, 1, produced, it meets AC in F. Show that AF 3 AC., [CBSE 2006C], , SOLUTION, , A 3ABC in which AD is a, median and E is the midpoint of AD., Also, BE is produced to meet AC at F., 1, TO PROVE AF AC., 3, , GIVEN, , GIVEN
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Triangles, CONSTRUCTION, PROOF, , 365, , From D, draw DG EF, meeting AC at G., , In 3BCF, D is the midpoint of BC and DG BF., , G is the midpoint of CF., So, FG GC., In 3ADG, E is the midpoint of AD and EF DG., F is the midpoint of AG., So, AF FG., Thus, AF FG GC., AC (AF FG GC) 3AF., 1, Hence, AF 3 AC., EXAMPLE 24, , Prove that the line segments joining the midpoints of the adjacent, sides of a quadrilateral form a parallelogram., , SOLUTION, , A quadrilateral ABCD in which P, Q, R, S are the, midpoints of AB, BC, CD and DA respectively., GIVEN, , TO PROVE, , PQRS is a parallelogram., , CONSTRUCTION, , Join AC., , In 3ABC, P and Q are, the midpoints of AB and BC, respectively., , PROOF, , , , PQ AC., , … (i), , [by midpoint theorem], , In3DAC, S and R are the midpoints of AD and CD respectively., , , SR AC., , … (ii), , [by midpoint theorem], , From (i) and (ii), we get PQ SR., Similarly, by joining BD, we can prove that PS QR., Hence, PQRS is a parallelogram., ANGLE-BISECTOR THEOREM, EXAMPLE 25, , SOLUTION, , Prove that the internal bisector of an angle of a triangle divides the, opposite side internally in the ratio of the sides containing the angle., GIVEN, , A 3ABC in which AD, the bisector of +A, meets BC, , in D., BD AB ·, DC AC, CONSTRUCTION Draw CE DA, meeting BA produced at E., TO PROVE
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366, , Secondary School Mathematics for Class 10, PROOF, , Since DA CE, we have, , +2 +3, , [alternate int. O], , and +1 +4, , [corresponding O]., , But, +1 +2, , [a AD is the bisector of +A], , +3 +4., So, AE AC., Now, in 3BCE, DA CE., BD AB, DC AE, BD AB, DC AC, [a AE AC] ., BD AB, Hence, DC AC ·, EXAMPLE 26, , SOLUTION, , BD AB, In a 3ABC, let D be a point on BC such that DC AC · Prove that, AD is the bisector of +A., A 3ABC in which D is a point on BC such that, BD AB ·, DC AC, TO PROVE AD is the bisector of +A., GIVEN, , Produce BA to E such that, , AE AC. Join EC., , CONSTRUCTION, , BD AB, DC AC (given), BD AB, DC AE [a AC AE], DA < CE, [by the converse of Thales’ theorem]., PROOF, , +2 +3, , … (i), , and +1 +4, , … (ii) [corresponding O], , Also, AE AC & +3 +4., , … (iii), , [alternate int. O], , +1 +2 [from (i), (ii) and (iii)]., Hence, AD is the bisector of +A., EXAMPLE 27, , SOLUTION, , In the given figure, AD is the bisector of +BAC. If AB 10 cm,, [CBSE 2001], AC 6 cm and BC 12 cm, find BD and DC., Let BD x cm. Then,, DC (BC BD) (12 x) cm.
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Triangles, , 367, , In 3ABC, AD is the bisector of +BAC., So, by the angle-bisector theorem, we have, BD AB, x 10, DC AC & 12 x 6, & 6x 10(12 x), , & 16x 120 & x 7.5., Hence, BD 7.5 cm, and DC (12 7.5) cm 4.5 cm., EXAMPLE 28, , If the bisector of an angle of a triangle bisects the opposite side, prove, that the triangle is isosceles., [CBSE 2000, ’01, ’02], , SOLUTION, , GIVEN A 3ABC in which AD is the bisector of +BAC such that, BD DC., TO PROVE, , AB AC., , PROOF Since AD is the bisector of +A,, by the angle-bisector theorem, we have, AB BD , [a BD DC], AC DC 1, AB, AC 1 & AB AC., Hence, 3ABC is an isosceles triangle., , EXAMPLE 29, , SOLUTION, , If the diagonal BD of a quadrilateral ABCD bisects both +B and +D,, AB AD, prove that BC CD ·, GIVEN A quad. ABCD in which diagonal BD bisects both +B, and +D., AB AD ·, TO PROVE, BC CD, CONSTRUCTION Join AC, intersecting BD, at E., PROOF In 3CBA, BE is the bisector of, +ABC., AE AB, EC BC ·, … (i), [by the angle-bisector theorem], In 3ADC, DE is the bisector of +ADC., AE AD, EC CD ·, … (ii) [by the angle-bisector theorem], AB AD, From (i) and (ii), we get BC CD ·
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368, , Secondary School Mathematics for Class 10, , f, , EXERCISE 7A, , 1. D and E are points on the sides AB and AC respectively of a 3ABC such, that DE BC., (i) If AD 3.6 cm, AB 10 cm and AE 4.5 cm,, find EC and AC., (ii) If AB 13.3 cm, AC 11.9 cm and EC 5.1 cm,, find AD., AD 4, (iii) If DB 7 and AC 6.6 cm, find AE., AD 8, (iv) If AB 15 and EC 3.5 cm, find AE., 2. D and E are points on the sides AB and AC respectively of a 3ABC such, that DE BC., Find the value of x, when, (i) AD x cm, DB (x 2) cm,, AE (x 2) cm and EC (x 1) cm., (ii) AD 4 cm, DB (x 4) cm, AE 8 cm, and EC (3x 19) cm., (iii) AD (7x 4) cm, AE (5x 2) cm, DB (3x 4) cm, and EC 3x cm., 3. D and E are points on the sides AB and AC respectively of a 3ABC. In, each of the following cases, determine whether DE BC or not., (i) AD 5.7 cm, DB 9.5 cm, AE 4.8 cm and, EC 8 cm., (ii) AB 11.7 cm, AC 11.2 cm, BD 6.5 cm and, AE 4.2 cm., (iii) AB 10.8 cm, AD 6.3 cm, AC 9.6 cm and, EC 4 cm., (iv) AD 7.2 cm, AE 6.4 cm, AB 12 cm and AC 10 cm., 4. In a 3ABC, AD is the bisector of +A., (i) If AB 6.4 cm, AC 8 cm and BD 5.6 cm,, find DC., (ii) If AB 10 cm, AC 14 cm and BC 6 cm,, find BD and DC., (iii) If AB 5.6 cm, BD 3.2 cm and BC 6 cm, find AC., (iv) If AB 5.6 cm, AC 4 cm and DC 3 cm, find BC., , [CBSE 2001C], [CBSE 1999C]
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Triangles, , 369, , 5. M is a point on the side BC of a parallelogram, ABCD. DM when produced meets AB, produced at N. Prove that, DM DC, DN AN, (i) MN BN ,, (ii) DM DC ·, 6. Show that the line segment which joins the midpoints of the oblique, sides of a trapezium is parallel to the parallel sides., 7. In the adjoining figure, ABCD is a trapezium, in which CD AB and its diagonals intersect, at O. If AO (2x 1) cm, OC (5x 7) cm,, DO (7x 5) cm and OB (7x 1) cm, find, the value of x., 8. In a 3ABC, M and N are points on the sides AB and AC respectively, such that BM CN. If +B +C then show that MN BC., 9. 3ABC and 3DBC lie on the same side of BC,, as shown in the figure. From a point P on, BC, PQ AB and PR BD are drawn, meeting, AC at Q and CD at R respectively. Prove that, QR AD., 10. In the given figure, side BC of 3ABC is, bisected at D and O is any point on AD. BO, and CO produced meet AC and AB at E and, F respectively, and AD is produced to X so, that D is the midpoint of OX. Prove that, AO : AX AF : AB and show that EF BC., 11. ABCD is a parallelogram in which P is the, midpoint of DC and Q is a point on AC such, 1, that CQ 4 AC. If PQ produced meets BC at R,, prove that R is the midpoint of BC., 12. In the adjoining figure, ABC is a triangle in, which AB AC. If D and E are points on AB, and AC respectively such that AD AE, show, that the points B, C, E and D are concyclic.
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370, , Secondary School Mathematics for Class 10, , 13. In 3ABC, the bisector of +B meets AC at D., A line PQ AC meets AB, BC and BD at P, Q, and R respectively., Show that PR # BQ QR # BP., , ANSWERS (EXERCISE 7A), , 1. (i) EC 8 cm, AC 12.5 cm, , (ii) AD 7.6 cm (iii) AE 2.4 cm, , (iv) AE 4 cm, 2. (i) x 4, 3. (i) Yes, , (ii) x 11 (iii) x 4, (ii) No, , (iii) Yes (iv) No, 4. (i) DC 7 cm (ii) BD 2.5 cm, DC 3.5 cm (iii) AC 4.9 cm, (iv) BC 7.2 cm, 7. x 2, HINTS TO SOME SELECTED QUESTIONS, 5. (i) In 3NDA, MB DA, NM NB, DM AB, DM DC, MD BA & MN BN & MN BN [a AB DC]., NM NB, NM, NB, NM MD NB BA, (ii) MD BA & MD 1 BA 1 &, MD, BA, DN AN, DN AN, & DM AB & DM DC [a AB DC] ., 6. Let E and F be the midpoints of the sides AD and BC of a, trapezium ABCD having AB CD. Produce AD and BC to, meet at P., In 3PAB, DC AB., , , PD PC, PD PC, PD PC, DA CB & 2DE 2CF & DE CF, & DC EF., , 8. +B +C & AB AC & AM BM AN CN & AM AN [a BM CN] ., AM AN, Now, AM AN, BM CN & BM CN, AM AN, & MB NC & MN BC., , CP CQ, 9. In 3CAB, QP AB & PB QA ·
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Triangles, , 371, , CP CR, In 3CDB, RP DB & PB RD ·, CQ CR, QA RD & QR AD (in 3CDA)., 10. Join BX and CX., Clearly, BD DC and OD DX (given)., Thus, the diagonals of quad. OBXC bisect each other., , , OBXC is a parallelogram., , , , BX CF and so, OF BX., , Similarly, OE XC., In 3ABX, OF BX., AO AF ·, , AX AB, In 3ACX, OE XC., AO AE ·, , AX AC, , … (i), , … (ii), , AF AE, From (i) and (ii), we get AB AC ·, Hence, FE BC., 11. Join BD. Suppose it meets AC at S., 1, Since the diagonals of gm bisect each other, CS 2 AC., 1, 1, 1, Now, CS 2 AC and CQ 4 AC & CQ 2 CS., Q is the midpoint of CS., So, PQ DS and therefore, QR SB., In 3CSB, Q is the midpoint of CS and QR SB, so R is the midpoint of BC., 12. AB AC and AD AE, , , , AB AD AC AE DB EC, AD DB, [each equal to 1], AE EC, [by the converse of Thales‘ theorem], DE BC, , , , , , +DEC +ECB 180c, , , , +DEC +CBD 180c [a AB AC & +C +B], , , , quad. BCED is cyclic., , Hence, the points B, C, E, D are concyclic., 13. In 3BQP, BR is the bisector of +B., , , QR BQ, , PR BP, , [by angle-bisector theorem]., , CRITERIA FOR SIMILARITY OF TWO TRIANGLES, Two triangles are similar, if (i) their corresponding angles are equal and (ii) their, corresponding sides are proportional.
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372, , Secondary School Mathematics for Class 10, , Thus, 3ABC +3DEF, if, (i) +A +D, +B +E, +C +F, AB BC CA, and (ii) DE EF FD ·, THEOREM 1, , (AAA-similarity) If in two triangles, the corresponding angles are, , equal, then their corresponding sides are proportional and hence the, triangles are similar., GIVEN, , 3ABC and 3DEF such that +A +D, +B +E and +C +F., , 3ABC +3DEF., CONSTRUCTION Cut DP AB and DQ AC. Join PQ., TO PROVE, , PROOF, , In 3ABC and 3DPQ, we have, [by construction], AB DP, [by construction], AC DQ, [given], +A +D, , [by, SAS-congruence], 3ABC ,3DPQ, +B +P, +E +P, [a +B +E (given)], , [a corresponding O are equal], PQ EF, DP DQ, DE DF, AB CA, , [a DP AB and DQ AC], DE FD, AB BC, Similarly, DE EF ·, AB BC CA ·, , DE EF FD, AB BC CA, Thus, +A +D, +B +E, +C +F and DE EF FD ·, Hence, 3ABC +3DEF., , , IMPORTANT REMARK, COROLLARY, , Two 3s are similar they are equiangular., , (AA-similarity) If two angles of one triangle are respectively equal, to two angles of another triangle then the two triangles are similar.
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Triangles, , 373, , In 3ABC and 3DEF, let +A +D and +B +E., Then, 3rd +C 3rd +F., , PROOF, , Thus, the two triangles are equiangular and hence similar., REMARK, , AA-similarity is the same as AAA-similarity., (SSS-similarity) If the corresponding sides of two triangles are, , THEOREM 2, , proportional then their corresponding angles are equal, and hence, the two triangles are similar., GIVEN, , AB BC AC, 3ABC and 3DEF in which DE EF DF ·, , TO PROVE, , 3ABC +3DEF., , Let us take 3ABC and 3DEF such that, AB BC AC, DE EF DF ( 1)., Cut DP AB and DQ AC. Join PQ., , CONSTRUCTION, , PROOF, , AB AC, DP DQ, DE DF & DE DF, , [a AB DP and AC DQ]., , So, by the converse of Thales’ theorem, PQ EF., +P +E, [corresponding O], +Q +F, , , [corresponding O], , 3DPQ +3DEF [by AA-similarity], , DP PQ, DE EF, AB PQ ·, , … (i) [a DP AB], DE EF, AB BC, But, DE EF ·, … (ii) [given], PQ BC, , [from (i) and (ii)], , EF EF, , BC PQ., Thus, AB DP, AC DQ and BC PQ.,
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374, , Secondary School Mathematics for Class 10, , , , 3ABC ,3DPQ [by SSS-congruence], , , , +A +D, +B +P +E and +C +Q +F, +A +D, +B +E and +C +F., , , , Thus, the given triangles are equiangular and hence similar., (SAS-similarity) If one angle of a triangle is equal to one angle of the, , THEOREM 3, , other triangle and the sides including these angles are proportional, then the two triangles are similar., GIVEN, , AB AC, 3ABC and 3DEF in which +A +D and DE DF ·, , TO PROVE, , 3ABC +3DEF., , CONSTRUCTION, , PROOF, , Let us take 3ABC and 3DEF such that, AB AC, , DE DF (< 1) and +A +D., Cut DP AB and DQ AC. Join PQ., , In 3ABC and 3DPQ, we have, [by construction], AB DP, +A +D, AC DQ, , , , (given), [by construction], , 3ABC ,3DPQ [by SAS-congruence], +A +D, +B +P and +C +Q., , AB AC, Now, DE DF, (given), DP DQ, , [a AB DP and AC DQ], DE DF, , [by the converse of Thales’ theorem], PQ EF, +P +E and +Q +F, [corresponding O], +A +D, +B +P +E and +C +Q +F., Thus, +A +D, +B +E and +C +F., , , , So, the given triangles are equiangular and hence similar.
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376, , Secondary School Mathematics for Class 10, , SOME MORE RESULTS, THEOREM 1, , GIVEN, , 3ABC and 3DEF in which +A +D, +B +E and +C +F and, AL = BC and DM = EF., , TO PROVE, PROOF, , If two triangles are equiangular, prove that the ratio of their, corresponding sides is the same as the ratio of the corresponding, altitudes., , BC AL ·, EF DM, , Since 3ABC and 3DEF are equiangular, 3ABC +3DEF., AB BC ·, , DE EF, In 3ALB and 3DME, we have, +ALB +DME 90c and +B +E (given)., , … (i), , , , 3ALB +3DME [by AA-similarity], AB AL ·, , DE DM, BC, AL, From (i) and (ii), we get EF DM ·, , THEOREM 2, , GIVEN, , If two triangles are equiangular, prove that the ratio of their, corresponding sides is the same as the ratio of the corresponding, medians., , 3ABC and 3DEF in which +A +D, +B +E and +C +F and, AL and DM are the medians., , TO PROVE, PROOF, , … (ii), , BC AL ·, EF DM, , Since 3ABC and 3DEF are equiangular, we have 3ABC +3DEF.
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Triangles, , AB BC ·, DE EF, AB BC 2BL, BL, But, DE EF 2EM EM ·, , , , 377, , … (i), , Now, in 3ABL and 3DEM, we have, AB BL, , DE EM and +B +E (given)., , , 3ABL +3DEM [by SAS-similarity], , , , AB AL ·, DE DM, , … (ii), , BC, AL, From (i) and (ii), we get EF DM ·, THEOREM 3, , GIVEN, , 3ABC and 3DEF in which +A +D, +B +E and +C +F, and, AX and DY are the bisectors of +A and +D respectively., , TO PROVE, PROOF, , If two triangles are equiangular, show that the ratio of the, corresponding sides is the same as the ratio of the corresponding, angle-bisector segments., , BC AX ·, EF DY, , Since 3ABC and 3DEF are equiangular, we have 3ABC +3DEF., , , AB BC ·, DE EF, , … (i), , 1, 1, Now, +A +D & 2+A 2+D & +BAX +EDY., Thus, in 3ABX and 3DEY, we have, +BAX +EDY, , (proved), , +B +E, , (given), , , , 3ABX +3DEY [by AA-similarity], , , , AB AX ·, DE DY, , BC AX, From (i) and (ii), we get EF DY ·, , … (ii)
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378, , Secondary School Mathematics for Class 10, , SOLVED EXAMPLES, EXAMPLE 1, , In the adjoining figure, 3AHK is similar, to 3ABC. If AK 10 cm, BC 3.5 cm, and HK 7 cm, find AC., [CBSE 2010], , SOLUTION, , 3AHK +3ABC, AK HK, 10, 7, AC BC & x 3.5 , where AC x cm, 10 # 3.5 , x, 5., 7, AC 5 cm., , EXAMPLE 2, , SOLUTION, , In the given figure, DE BC, AD 2 cm,, BD 2.5 cm, AE 3.2 cm and DE 4 cm., Find AC and BC., [CBSE 2001C], Since DE BC, we have, +ADE +ABC (corresponding O), and +AED +ACB (corresponding O)., 3ADE +3ABC [by AA-similarity]., So, the corresponding sides of 3ADE and 3ABC are, proportional., AD DE AE, AB BC AC ·, … (i), AD DE, 2, 4, Now, AB BC & 4.5 BC, [a AB AD BD 4.5 cm], & BC a 4 #24.5 k cm 9 cm., DE AE, Again, BC AC [from (i)], 4 3.2, 9 AC [a BC 9 cm], 9 # 3.2, AC a 4 k cm 7.2 cm., Hence, AC 7.2 cm and BC 9 cm., , EXAMPLE 3, , In the given figure, AB CD., Prove that 3AOB +3DOC., , SOLUTION, , AB CD (given).
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380, , Secondary School Mathematics for Class 10, , +OQP +BOQ +OPQ 132c 56c 76c., Now, 3OQP +3OAB, +OQP +OAB [corresponding angles of similar triangles]., +OAB +OQP 76c., EXAMPLE 7, , SOLUTION, , In the given figure, AP · AR AS · AQ., Prove that +P +S and +Q +R., , We have, +PAQ +SAR., , … (i), , [vertically opposite angles], , Also, AP · AR AS · AQ (given), AP AQ ·, … (ii), AS AR, From (i) and (ii), we have, , , 3PAQ +3SAR [by SAS-similarity], +P +S and +Q +R, [corresponding angles of similar triangles]., EXAMPLE 8, , A vertical stick which is 15 cm long casts a 12-cm-long shadow on, the ground. At the same time, a vertical tower casts a 50-m-long, shadow on the ground. Find the height of the tower., , SOLUTION, , Let AB be the vertical stick and let AC be its shadow., Then, AB 0.15 m and AC 0.12 m., Let DE be the vertical tower and let DF be its shadow., Then, DF 50 m. Let DE x m., Now, in 3BAC and 3EDF,, we have, +BAC +EDF 90c, +ACB +DFE, [angular elevation of the sun at the same time], 3BAC +3EDF, AB AC, 0.15 0.12, DE DF & x 50, (0.15 # 50), 62.5., x, 0.12, Hence, the height of the tower is 62.5 m.
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Triangles, , 381, , EXAMPLE 9, , Prove that the ratio of the perimeters of two similar triangles is the, same as the ratio of their corresponding sides., , SOLUTION, , GIVEN, , 3ABC and 3PQR in, , which BC a, CA b, AB c, and QR p, RP q, PQ r., Also, 3ABC +3PQR., TO PROVE, , a b c abc, p q r pqr$, , Since 3ABC and 3PQR are similar, therefore their, corresponding sides are proportional., , PROOF, , , , a bc, p q r k (say), , , , a kp, b kq and c kr., , , , perimeter of 3ABC a b c kp kq kr, , , perimeter of 3PQR p q r, pqr, , … (i), , , , k (p q r), k., (p q r), , … (ii), , From (i) and (ii), we get, a b c a b c perimeter of 3ABC, p q r p q r perimeter of 3PQR, , [each equal to k]., , EXAMPLE 10, , The perimeters of two similar triangles are 25 cm and 15 cm, respectively. If one side of the first triangle is 9 cm, find the, corresponding side of the second triangle., [CBSE 2002C], , SOLUTION, , Let 3ABC +3DEF given in such a way that perimeter of, 3ABC 25 cm, perimeter of 3DEF 15 cm and AB 9 cm., Then, we have to find DE. Let DE x cm., We know that the ratio of the perimeters of two similar, triangles is the same as the ratio of their corresponding sides., , , perimeter of 3ABC AB, , perimeter of 3DEF DE, , , , 25 9, 9 #15 , 15 x & x a 25 k 5.4., , , , DE 5.4 cm., , Hence, the corresponding side of the second triangle is 5.4 cm.
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384, , Secondary School Mathematics for Class 10, , and +OBA +ODC [alternate angles, since AB DC]., 3OAB +3OCD [by AA-similarity]., OA OB, And so, OC OD ·, EXAMPLE 16, , In 3ABC, AD = BC, and AD 2 BD · CD., Prove that +BAC 90c., , SOLUTION, , A 3ABC in which, AD = BC and AD 2 BD · CD., , GIVEN, , TO PROVE, , +BAC 90c., , BD AD, AD 2 BD · CD & AD CD ·, Now, in 3DBA and 3DAC, we have, BD AD, +BDA +ADC 90c and AD CD ·, 3DBA +3DAC [by SAS-similarity], PROOF, , +B +2 and +1 +C, +1 +2 +B +C & +A +B +C, , & 2+A +A +B +C 180c, & +A +BAC 90c., EXAMPLE 17, , In the given figure PA, QB and RC, each is perpendicular to AC such that, PA x, RC y, QB z, AB a and, BC b., 1 1 1, Prove that x y z ·, [CBSE 2000C], , SOLUTION, , PA = AC and QB = AC & QB PA., Thus, in 3PAC, QB PA. So, 3QBC +3PAC., QB BC, z, b, PA AC & x · … (i) [by the property of similar A], a b, In 3RAC, QB RC. So, 3QBA +3RCA., QB AB, z, a, RC AC & y · … (ii) [by the property of similar A], a b, From (i) and (ii), we get, zz b a , x y aa b a b k 1
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386, EXAMPLE 20, , Secondary School Mathematics for Class 10, , In the given figure, 3ABC and 3AMP, are right-angled at B and M respectively., Prove that: (i) 3ABC +3AMP, CA BC, (ii) PA MP, , SOLUTION, , 3ABC and 3AMP such that, +B 90c and +M 90c., CA BC ·, TO PROVE (i) 3ABC +3AMP (ii), PA MP, PROOF (i) In 3ABC and 3AMP, we have, +ABC +AMP 90c, +A +A (common), 3ABC +3AMP [by AA-similarity]., GIVEN, , (ii) Since 3ABC +3AMP, their corresponding sides are, proportional., CA BC, PA MP ·, EXAMPLE 21, , SOLUTION, , EXAMPLE 22, , In a 3ABC, AB AC and D is a point on AC such that, BC 2 AC # DC. Prove that BD BC., GIVEN A 3ABC in which AB AC and D, is a point on AC such that BC 2 AC # DC., TO PROVE BD BC., 2, PROOF BC AC # DC (given), BC AC, DC BC ·, Thus, in 3ABC and 3BDC, we have, BC AC, , DC BC and +C +C (common)., 3ABC +3BDC [by SAS-similarity], AC AB, BC BD, AC AC, BC BD [a AB AC (given)], BD BC., Hence, BD BC., , In the given figure, CD and GH are respectively the bisectors, of +ACB and +FGE of 3ABC and 3FEG respectively. If, 3ABC +3FEG, prove that:, CD AC, (a) 3ADC +3FHG, (b) 3BCD +3EGH, (c) GH FG
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388, SOLUTION, , Secondary School Mathematics for Class 10, , 3ABC +3DEF, , (given), , +A +D, +B +E and +C +F, and, , AB BC CA ·, DE EF FD, , … (i), … (ii), , Since BM and EN are medians, we have, CA 2AM 2CM and FD 2DN 2FN., , , , , from (ii), we have, AB BC CA 2AM 2CM, DE EF FD 2DN 2FN, AB BC CA AM CM ·, DE EF FD DN FN, , … (iii), , (a) In 3AMB and 3DNE, we have, +BAM +EDN [a +A +D from (i)], AB AM, and, [from (iii)]., DE DN, 3AMB +3DNE [by SAS-similarity]., (b) In 3CMB and 3FNE, we have, +BCM +EFN [a +C +F from (i)], BC CM, and, [from (iii)]., EF FN, 3CMB +3FNE [by SAS-similarity]., (c) As proved above, 3AMB +3DNE and so, AB BM ·, DE EN, From (ii) and (iv), we get, , … (iv), , BM AC ·, EN FD, EXAMPLE 24, , In a 3ABC, P and Q are points on AB and AC respectively such, that PQ BC. Prove that the median AD, drawn from A to BC,, bisects PQ., , SOLUTION, , GIVEN A 3ABC in which P and Q are, points on AB and AC respectively such, that PQ BC and AD is the median,, cutting PQ at E., TO PROVE, , PE EQ.
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Triangles, PROOF, , 389, , In 3APE and 3ABD, we have, , +PAE +BAD, , [common], , +APE +ABD [corresponding O], 3APE +3ABD [by AA-similarity]., But, in similar triangles, the corresponding sides are, proportional., , , AE PE ·, AD BD, , … (i), , In 3AEQ and 3ADC, we have, +QAE +CAD [common], +AQE +ACD [corresponding angles], 3AEQ +3ADC [by AA-similarity]., But, in similar triangles, the corresponding sides are, proportional., , , AE EQ ·, AD DC, , … (ii), AE ·, PE EQ, :each equal to, D, BD DC, AD, [a AD is the median], , From (i) and (ii), we get, But, BD DC, PE EQ., EXAMPLE 25, , SOLUTION, , In the given figure, E is a point on side CB produced of an isosceles, 3ABC with AB AC. If AD = BC and EF = AC, prove that, 3ABD +3ECF., A 3ABC in which AB AC, and AD = BC. Side CB is produced, to E and EF = AC., GIVEN, , TO PROVE, , 3ABD +3ECF., , We know that the angles, opposite to equal sides of a triangle, are equal., PROOF, , +B +C [a AB AC] ., Now, in 3ABD and 3ECF, we have, +B +C [proved above], +ADB +EFC 90c., 3ABD +3ECF [by AA-similarity].
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390, , Secondary School Mathematics for Class 10, , EXAMPLE 26, , Prove that the line segments joining the midpoints of the sides of a, triangle form four triangles, each of which is similar to the original, triangle., , SOLUTION, , GIVEN A 3ABC in which D, E, F, are the midpoints of BC, CA and AB, respectively., , 3AFE +3ABC,, 3FBD +3ABC,, 3EDC +3ABC., and 3DEF +3ABC., , TO PROVE, , We shall first show that 3AFE +3ABC., Since F and E are the midpoints of AB and AC respectively, so, by the midpoint theorem, we have FE BC., +AFE +B, [corresponding O], Now, in 3AFE and 3ABC, we have, [corresponding O], +AFE +B, and +A +A, (common)., 3AFE +3ABC [by AA-similarity]., Similarly, 3FBD +3ABC and 3EDC +3ABC., Now, we shall show that 3DEF +3ABC., In the same manner as above, we can prove that, ED AF and DF EA., AFDE is a gm., [opposite angles of a gm]., +EDF +A, Similarly, BDEF is a gm., [opposite angles of a gm]., +DEF +B, Thus, in 3DEF and 3ABC, we have, +EDF +A and +DEF +B., 3DEF +3ABC [by AA-similarity]., Hence, the result follows., PROOF, , EXAMPLE 27, , In the given figure, DEFG is, a square and +BAC 90c., Prove that, (i) 3AGF +3DBG, (ii) 3AGF +3EFC, (iii) 3DBG +3EFC, (iv) DE 2 BD # EC, , [CBSE 2009]
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392, , Secondary School Mathematics for Class 10, , AP #CP BP # DP AP # PC BP # PD., Hence, AP # PC BP # PD., EXAMPLE 29, , Through the midpoint M of the side CD of a parallelogram ABCD,, the line BM is drawn, intersecting AC in L and AD produced in E., Prove that EL 2BL., [CBSE 2006C, ’08, ’09], , SOLUTION, , A gm ABCD and M is the, midpoint of CD. Line BM is, drawn, intersecting AC in L and, AD produced in E., TO PROVE, , EL 2BL., , In 3BMC and 3EMD, we have, (vert. opp. O), +1 +2, [M is the midpoint of CD], MC MD, +BCM +EDM [alternate interior O], 3BMC ,3EMD, BC DE., But, BC AD [opposite sides of a gm], PROOF, , , , BC AD DE & AE (AD DE) 2BC., , … (i), , Now, in 3AEL and 3CBL, we have, (vert. opp. O), +6 +5, , [alternate interior O], +3 +4, 3AEL +3CBL [AA-similarity], EL AE 2BC , , 2 [using (i)], BL BC, BC, EL 2BL., Hence, EL 2BL., EXAMPLE 30, , A lamp is 3.3 m above the, ground. A boy 110 cm tall, walks away from the base of, this lamp post at a speed of, 0.8 m/s. Find the length of, the shadow of the boy after, 4 seconds., , SOLUTION, , Let AB be the lamp post and PQ be the boy, where P is the, position of the boy after 4 seconds.
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Triangles, , 393, , AP = distance moved in 4 s at 0.8 m/s (4 # 0.8) m 3.2 m., PM is the length of shadow of the boy., Let PM x m., In 3AMB and 3PMQ, we have, +MAB +MPQ 90c, [a both the lamp post and the boy stand vertically erect], +AMB +PMQ (common), 3AMB +3PMQ [by AA-similarity]., AM AB, PM PQ, [corres. sides of similar triangles are proportional], 3.2 x 3.3, AP PM AB, & x 1.1, PM, PQ, , And so,, , , , [a AB 3.3 m, PQ 110 cm 1.1 m], , , , 3.2 x 3x & 2x 3.2 & x 1.6., the length of the shadow of the boy after 4 seconds is 1.6 m., , EXAMPLE 31, , Sides AB and BC and median AD of a triangle ABC are respectively, proportional to sides PQ and QR and median PM of another triangle, PQR, as shown in the figure. Prove that 3ABC +3PQR., , SOLUTION, , We have, AB BC AD, PQ QR PM, 1, BC, AB AD BC 2, BD ·, , PQ PM QR 1, QM, QR, 2, In 3ABD and 3PQM, we have, AB AD BD, PQ PM QM, 3ABD +3PQM, , … (i), , [from (i)], [by SSS-similarity]., , And so,+B +Q [corres. angles of similar triangles are equal].
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394, , Secondary School Mathematics for Class 10, , Now, in 3ABC and 3PQR, we have, +B +Q, AB BD, and, PQ QM, , [proved above], [from (i)]., , 3ABC +3PQR [by SAS-similarity]., EXAMPLE 32, , Sides AB and AC and median AD of a triangle ABC are respectively, proportional to sides PQ and PR and median PM of another triangle, PQR. Prove that 3ABC +3PQR., , SOLUTION, , GIVEN AD and PM are medians of 3ABC and 3PQR, respectively such that, , AB AC AD ·, PQ PR PM, 3ABC +3PQR., CONSTRUCTION Produce AD to E such that AD DE and, produce PM to N such that PM MN., Join EC and NR., TO PROVE, , In 3ABD and 3ECD, we have, [a D is the midpoint of BC], BD CD, [by construction], AD ED, +BDA +CDE [vertically opposite angles], 3ABD ,3ECD [by SAS-congruency]., , PROOF
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Triangles, , 397, , 4. In the given figure, if +ADE +B, show that, 3ADE +3ABC. If AD 3.8 cm, AE 3.6 cm,, BE 2.1 cm and BC 4.2 cm, find DE., , 5. The perimeters of two similar triangles ABC and PQR are 32 cm and, [CBSE 2001], 24 cm respectively. If PQ 12 cm, find AB., 6. The corresponding sides of two similar triangles ABC and DEF are, BC 9.1 cm and EF 6.5 cm. If the perimeter of 3DEF is 25 cm, find the, perimeter of 3ABC., 7. In the given figure, +CAB 90c and, AD = BC. Show that 3BDA +3BAC. If, AC 75 cm, AB 1 m and BC 1.25 m,, find AD., , 8. In the given figure, +ABC 90c and, BD = AC. If AB 5.7 cm, BD 3.8 cm, and CD 5.4 cm, find BC., , 9. In the given figure, +ABC 90c and, BD = AC. If BD 8 cm, AD 4 cm,, find CD., , 10. P and Q are points on the sides AB and AC respectively of a 3ABC., If AP 2 cm, PB 4 cm, AQ 3 cm and QC 6 cm, show that, BC 3PQ.
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398, , Secondary School Mathematics for Class 10, , 11. ABCD is a parallelogram and E is a point, on BC. If the diagonal BD intersects AE at F,, prove that AF # FB EF # FD., , 12. In the given figure, DB = BC, DE = AB and, AC = BC., Prove that, , BE AC ·, DE BC, , [CBSE 2008], , 13. A vertical pole of length 7.5 m casts a shadow 5 m long on the ground, and at the same time a tower casts a shadow 24 m long. Find the height, of the tower., 14. In an isosceles 3ABC, the base AB is, produced both ways in P and Q such that, AP # BQ AC 2 ., Prove that 3ACP +3BCQ., , 15. In the given figure, +1 +2 and, , AC CB ·, BD CE, , Prove that 3ACB +3DCE., , 16. ABCD is a quadrilateral in which AD BC., If P, Q, R, S be the midpoints of AB, AC, CD, and BD respectively, show that PQRS is a, rhombus., , 17. In a circle, two chords AB and CD intersect at, a point P inside the circle. Prove that, (a) 3PAC +3PDB, (b) PA · PB PC · PD.
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406, , Secondary School Mathematics for Class 10, , SOLVED EXAMPLES, EXAMPLE 1, , If the areas of two similar triangles are in the ratio 25 : 64, find the, ratio of their corresponding sides., [CBSE 2009], , SOLUTION, , Let 3ABC and 3DEF be similar. Then,, ar (3ABC) AB 2 BC 2 AC 2, , , , ar (3DEF) DE 2 EF 2 DF 2, AB 2 BC 2 AC 2 25 5 2, a k, k a k a k, DE, EF, DF, 8, 64, AB BC AC 5 ·, , DE EF DF 8, Hence, the ratio of their corresponding sides is 5 : 8., a, , EXAMPLE 2, , In the adjoining figure, S and T are points, on the sides PQ and PR respectively of, 3PQR such that PT 2 cm, TR 4 cm, and ST is parallel to QR. Find the ratio, of the areas of 3PST and 3PQR., [CBSE 2010], , SOLUTION, , ST < QR (given), +PST +PQR, and +PTS +PRQ, , [corresponding angles], [corresponding angles]., And so, 3PST +3PQR [by AA-similarity]., , , , ar (3PST) PT 2, PT 2, , , ar (3PQR) PR 2 (PT TR) 2, , , EXAMPLE 3, , SOLUTION, , 22 4 1 ·, , (2 4) 2 36 9, , The areas of two similar triangles 3ABC and 3PQR are, 25 cm 2 and 49 cm 2 respectively. If QR 9.8 cm, find BC. [CBSE 2006], We know that the ratio of the areas of two similar triangles is, equal to the ratio of the squares of their corresponding sides., , , ar (3ABC) BC 2, , ar (3PQR) QR 2, , c, , , BC 2 ar (3ABC) 25 5 2, m, a k, 7, QR, ar (3PQR) 49, , BC 5, BC 5, & 9.8 7, QR 7
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Triangles, , 407, , 5 # 9.8, cm 5 #1.4 cm 7 cm., 7, Hence, BC 7 cm., BC , , EXAMPLE 4, , The areas of two similar triangles are 81 cm 2 and 49 cm 2 respectively., If the altitude of the bigger triangle is 4.5 cm, find the corresponding, altitude of the smaller triangle., [CBSE 2002], , SOLUTION, , Let the given triangles be 3ABC and 3DEF such that, ar (3ABC) 81 cm 2 and ar (3DEF) 49 cm 2 . Let AL and DM be, the corresponding altitudes of 3ABC and 3DEF respectively., Then, AL 4.5 cm., We know that the ratio of the areas of two similar triangles is, equal to the ratio of the squares of their corresponding altitudes., 2, ar (3ABC), AL 2, , ar (3DEF) DM, AL 2 ar (3ABC) 81 9 2, a, a k, k, 7, DM, ar (3DEF) 49, 4.5 9, AL 9, , & DM 7, DM 7, 4.5 #7, 7, DM , cm cm 3.5 cm., 9, 2, Hence, the altitude of the smaller triangle is 3.5 cm., , EXAMPLE 5, , The areas of two similar triangles are 121 cm 2 and 64 cm 2, respectively. If the median of first triangle is 12.1 cm, find the, corresponding median of the other., [CBSE 2001], , SOLUTION, , Let 3ABC +3DEF such that ar (3ABC) 121 cm 2 and, ar (3DEF) 64 cm 2 ., Let AP and DQ be the corresponding medians of 3ABC and, 3DEF respectively such that AP 12.1 cm., We know that the ratio of the areas of two similar triangles, is equal to the ratio of the squares of their corresponding, medians., ar (3ABC) AP 2, , , ar (3DEF) DQ 2, AP 2 ar (3ABC) 121 11 2, m, c, a k, 8, 64, DQ, ar (3DEF), AP 11, .1 11, , & 12, [a AP 12.1 cm], 8, DQ 8, DQ
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408, , Secondary School Mathematics for Class 10, , (12.1# 8), cm (1.1# 8) cm 8.8 cm., 11, Hence, the corresponding median is 8.8 cm., DQ , , EXAMPLE 6, , 3ABC +3DEF in which AX and DY are the bisectors of +A and, +D respectively. If AX 6.5 cm and DY 5.2 cm, find the ratio of, the areas of 3ABC and 3DEF., , SOLUTION, , We know that the ratio of the areas of two similar triangles is, equal to the ratio of the squares of the corresponding anglebisector segments., , , , 2, 2, ar (3ABC) AX 2 (6.5) 2, , , a6.5 k a 5 k 25 ·, 5.2, 16, 4, ar (3DEF) DY 2 (5.2) 2, ar (3ABC) : ar (3DEF) 25 : 16., , EXAMPLE 7, , In the given figure, the line segment, XY is parallel to side AC of 3ABC, and it divides the triangle into, two parts of equal area. Prove that, AX : AB (2 2 ) : 2., , SOLUTION, , Since XY AC, we have, +A +BXY and +C +BYX [corres. O]., 3ABC +3XBY, ar (3ABC) AB 2, ·, , , ar (3XBY) XB 2, But, ar (3ABC) 2 # ar (3XBY) [given], , , ar (3ABC), 2., ar (3XBY), , From (i) and (ii), we get, , , , , , , AB 2 , AB 2, 2&a k 2, 2, XB, XB, AB , 2 & AB 2 (XB), XB, AB 2 (AB AX), 2 AX ( 2 1) AB, 2 (2 2 ), AX ( 2 1), ·, , #, 2, AB, 2, 2, , Hence, AX : AB (2 2 ) : 2., , … (i), , … (ii)
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414, , Secondary School Mathematics for Class 10, , 7. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively., If the altitude of the first triangle is 6.3 cm, find the corresponding, altitude of the other., [CBSE 2001], 8. The areas of two similar triangles are 100 cm2 and 64 cm2 respectively., If a median of the smaller triangle is 5.6 cm, find the corresponding, median of the other., 9. In the given figure, ABC is a triangle and PQ, is a straight line meeting AB in P and AC, in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm,, 1, QC = 4.5 cm, prove that area of 3APQ is, 16, of the area of 3ABC., [CBSE 2005], 10. In the given figure, DE BC. If DE 3 cm,, BC 6 cm and ar (3ADE) 15 cm 2, find the, area of 3ABC., , 11. 3ABC is right-angled at A and AD = BC. If, BC 13 cm and AC 5 cm, find the ratio of, the areas of 3ABC and 3ADC. [CBSE 2000C], , 12. In the given figure, DE BC and DE : BC 3 : 5., Calculate the ratio of the areas of 3ADE and, the trapezium BCED., , 13. In 3ABC, D and E are the midpoints of AB, and AC respectively. Find the ratio of the, areas of 3ADE and 3ABC.
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416, , Secondary School Mathematics for Class 10, , PYTHAGORAS’ THEOREM, We have proved earlier in this chapter that:, ”If a perpendicular is drawn from the vertex of the right angle, of a right triangle to the hypotenuse then the triangles on both sides, of the perpendicular are similar to the whole triangle and also to, each other.”, Now, we shall use this theorem to prove the Pythagoras’ theorem., THEOREM 1, , (Pythagoras’ theorem) In a right triangle, the square of the, , hypotenuse is equal to the sum of the squares of the other two sides., [CBSE 2001, ’02, ’03, ’04, ’04C, ’05, ’06, ’06C, ’07, ’07C, ’09], GIVEN, , A 3ABC in which +ABC 90c., , TO PROVE, , AC 2 AB 2 BC 2 ., , CONSTRUCTION, PROOF, , Draw BD = AC., , In 3ADB and 3ABC, we have, (common), +A +A, , +ADB +ABC [each equal to 90], , 3ADB +3ABC [by AA-similarity], AD AB, , AB AC, , AD # AC AB 2., , … (i), , In 3BDC and 3ABC, we have, (common), +C +C, , +BDC +ABC [each equal to 90], , 3BDC +3ABC [by AA-similarity], DC BC, , BC AC, , DC # AC BC 2., , … (ii), , From (i) and (ii), we get, AD # AC DC # AC AB 2 BC 2, (AD DC)# AC AB 2 BC 2, , THEOREM 2, , AC # AC AB 2 BC 2 AC 2 AB 2 BC 2 ., (Converse of Pythagoras’ theorem) In a triangle, if the square of, one side is equal to the sum of the squares of the other two sides then, the angle opposite to the first side is a right angle., [CBSE 2001, ’03, ’05C, ’06, ’06C, ’07, ’09, ’09C]
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Triangles, , 417, , A 3ABC in which AC 2 AB 2 BC 2 ., TO PROVE +B 90c., GIVEN, , CONSTRUCTION, , PROOF, , Draw a 3DEF such that DE AB, EF BC and +E 90c., , In 3DEF, we have +E 90c., So, by Pythagoras’ theorem, we have, DF 2 DE 2 EF 2, DF 2 AB 2 BC 2., But, AC 2 AB 2 BC 2., , … (i) [a DE AB and EF BC], , , , … (ii) [given], , From (i) and (ii), we get AC DF 2 & AC DF., 2, , Now, in 3ABC and 3DEF, we have, AB DE, BC EF and AC DF., , , 3ABC ,3DEF., Hence, +B +E 90c., , SOME IMPORTANT RESULTS BASED UPON, PYTHAGORAS’ THEOREM, THEOREM 1, , In a 3ABC, AD is perpendicular to BC., Prove that (AB 2 CD 2) (AC 2 BD 2) ., [CBSE 2003, ’05C, ’09], , GIVEN, , A 3ABC in which AD = BC., (AB 2 CD 2) (AC 2 BD 2)., , TO PROVE, PROOF, , From right 3ADB, we have, , , AB 2 AD 2 BD 2 [by Pythagoras’ theorem], (AB 2 BD 2) AD 2., , … (i), , From right 3ADC, we have, , , AC 2 AD 2 CD 2, (AC 2 CD 2) AD 2., , … (ii), , From (i) and (ii), we get (AB BD ) (AC CD ) ., Hence, (AB 2 CD 2) (AC 2 BD 2) ., 2, , 2, , 2, , 2
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418, , Secondary School Mathematics for Class 10, , THEOREM 2, , Given a 3ABC in which +A 90c and AD = BC. Prove that, AD 2 BD · CD., , [CBSE 2004C, ’06, ’09], , A 3ABC in which +A 90c and AD = BC., 2, TO PROVE AD BD · CD., GIVEN, , PROOF, , In 3BAC, +A 90c., , BC 2 AB 2 AC 2, , … (i), , [by Pythagoras’ theorem]., In 3ADB, +ADB 90c., , , AB 2 AD 2 BD 2., , … (ii) [by Pythagoras’ theorem], , In 3ADC, +ADC 90c., , , AC 2 AD 2 CD 2., , … (iii) [by Pythagoras’ theorem], , From (ii) and (iii), we get, (AB 2 AC 2) 2AD 2 BD 2 CD 2, , , , , , [using (i)], BC 2 2AD 2 BD 2 CD 2, (BD CD) 2 2AD 2 BD 2 CD 2, BD 2 CD 2 2BD · CD 2AD 2 BD 2 CD 2, 2AD 2 2BD · CD AD 2 BD · CD., , Hence, AD 2 BD · CD., THEOREM 3, , In 3ABC, AD = BC such that AD 2 BD · CD. Prove that 3ABC is, right-angled at A., [CBSE 2006], , A 3ABC in which AD = BC and AD 2 BD · CD., TO PROVE +A 90c., GIVEN, , PROOF, , In right 3ADB, +ADB 90c., , AB 2 AD 2 BD 2., , … (i), , [by Pythagoras’ theorem], In right 3ADC, +ADC 90c., , … (ii) [by Pythagoras’ theorem], AC 2 AD 2 CD 2., Adding (i) and (ii), we get, (AB 2 AC 2) BD 2 CD 2 2AD 2, BD 2 CD 2 2BD · CD [a AD 2 BD · CD], (BD CD) 2 BC 2 ., Thus, (AB 2 AC 2) BC 2 ., Hence, 3ABC is right-angled at A.
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Triangles, THEOREM 4, , 419, , In a 3ABC, +ABC 90c (i.e., +B is acute) and AD = BC. Prove that, AC 2 AB 2 BC 2 2BC · BD., , A 3ABC in which +ABC 90c, and AD = BC., 2, 2, 2, TO PROVE AC AB BC 2BC · BD., GIVEN, , PROOF, , In 3ADB, +ADB 90c., , AB 2 AD 2 BD 2., , … (i), , In 3ADC, +ADC 90c., , AC 2 AD 2 CD 2, AD (BC BD), 2, , [by Pythagoras’ theorem], [by Pythagoras’ theorem], , 2, , AD 2 BC 2 BD 2 2BC · BD, (AD 2 BD 2) BC 2 2BC · BD, AB 2 BC 2 2BC · BD [using (i)]., Hence, AC 2 AB 2 BC 2 2BC · BD., Note, , BD is known as the projection of AB on BC. So, this theorem is stated as:, , ”In an acute-angled triangle, the square of the side opposite to an acute angle is, equal to the sum of the squares of the other two sides minus twice the product of, one side and the projection of the other on the first.”, THEOREM 5, , GIVEN, , In a 3ABC, +ABC 90c (i.e., +B is obtuse) and AD = (CB, produced). Prove that AC 2 AB 2 BC 2 2BC · BD., , A 3ABC in which +ABC 90c and, , AD = (CB produced)., TO PROVE, PROOF, , AC 2 AB 2 BC 2 2BC · BD., , In 3ADB, +ADB 90c., , … (i) [by Pythagoras’ theorem], AB 2 AD 2 BD 2., In 3ADC, +ADC 90c., , [by Pythagoras’ theorem], AC 2 AD 2 CD 2, AD 2 (BC BD) 2, [a CD BC BD], AD 2 BC 2 BD 2 2BC · BD, (AD 2 BD 2) BC 2 2BC · BD, AB 2 BC 2 2BC · BD [using (i)]., , Note, , BD is the projection of AB on BC. So, this theorem is stated as:, , ”In an obtuse-angled triangle, the square of the side opposite to the obtuse angle, is equal to the sum of the squares of the other two sides plus twice the product of, one side and the projection of the other on the first.”
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420, , Secondary School Mathematics for Class 10, , THEOREM 6, , In 3ABC, if AD is the median, then prove that, (AB 2 AC 2) 2(AD 2 BD 2) ., , GIVEN, , A 3ABC in which AD is the median., (AB 2 AC 2) 2(AD 2 BD 2) ., , TO PROVE, , Draw AL = BC., In 3ALD, +ALD 90c., , CONSTRUCTION, PROOF, , , , +ADL 90c and therefore,, +ADB 90c., , Thus, in 3ADB, +ADB 90c and AL = (BD produced)., , AB 2 AD 2 BD 2 2BD · DL., , … (i), , In 3ADC, +ADC 90c and AL = DC., , AC 2 AD 2 CD 2 2CD · DL, … (ii) [a CD BD], AC 2 AD 2 BD 2 2BD · DL., 2, 2, Adding (i) and (ii), we get (AB AC ) 2(AD 2 BD 2) ., , Note, , This theorem can be stated as:, ”In any triangle, the sum of the squares of any two sides is equal to twice the, square of half of the third side together with twice the square of the median which, bisects the third side.”, , SOLVED EXAMPLES, EXAMPLE 1, , Sides of some triangles are given below. Determine which of them are, right triangles., (i) 8 cm, 15 cm, 17 cm, (ii) 9 cm, 11 cm, 6 cm, (iii) (2a 1) cm, 2 2a cm and (2a 1) cm, , SOLUTION, , For the given triangle to be right-angled, the sum of the, squares of the two smaller sides must be equal to the square of, the largest side., (i) Let a = 8 cm, b = 15 cm and c = 17 cm. Then,, (a 2 b 2) {(8) 2 (15) 2} cm 2 (64 225) cm 2 289 cm 2, and c 2 (17) 2 cm 2 289 cm 2 ., (a 2 b 2) c 2 ., Hence, the given triangle is right-angled., (ii) Let a = 9 cm, b = 6 cm and c = 11 cm. Then,, (a 2 b 2) {(9) 2 (6) 2} cm 2 (81 36) cm 2 117 cm 2
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Triangles, , 421, , and c 2 (11) 2 cm 2 121 cm 2 ., (a 2 b 2) ! c 2 ., Hence, the given triangle is not right-angled., (iii) Let p (2a 1) cm, q 2 2a cm and r (2a 1) cm. Then,, (p 2 q 2) (2a 1) 2 cm 2 (2 2a ) 2 cm 2, {(4a 2 1 4a) 8a} cm 2 (4a 2 4a 1) cm 2, (2a 1) 2 cm 2 r 2 ., (p q ) r 2 ., 2, , 2, , Hence, the given triangle is right-angled., EXAMPLE 2, , A man goes 15 m due west and then 8 m due north. How far is he, from the starting point?, , SOLUTION, , Starting from A, let the man go from A to B and then from B to, C, as shown in the figure. Then,, AB 15 m, BC 8 m and +ABC 90c., From right 3ABC, we have, AC 2 AB 2 BC 2, {(15) 2 (8) 2} m 2, (225 64) m 2, 289 m 2, , , AC 289 m = 17 m., , Hence, the man is 17 m away from the starting position., EXAMPLE 3, , A ladder 25 m long just reaches the top of a building 24 m high, from the ground. Find the distance of the foot of the ladder from the, building., , SOLUTION, , Let AB be the building and CB be the ladder. Then,, AB 24 m, CB 25 m and +CAB 90c., By Pythagoras’ theorem, we have, CB 2 AB 2 AC 2, AC 2 CB 2 AB 2 [(25) 2 (24) 2] m 2, (625 576) m 2 49 m 2, , , AC 49 m 7 m., , Hence, the distance of the foot of the ladder from the building, is 7 m.
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422, , Secondary School Mathematics for Class 10, , EXAMPLE 4, , A ladder 15 m long reaches a window which is 9 m above the ground, on one side of a street. Keeping its foot at the same point, the ladder, is turned to the other side of the street to reach a window 12 m high., Find the width of the street., , SOLUTION, , Let AB be the street and let C be the foot of the ladder. Let D and, E be the given windows such that AD 9 m and BE 12 m., Then, CD and CE are the two positions of the ladder., Clearly, +CAD 90c, +CBE 90c, and CD CE 15 m., From right 3CAD, we have, CD 2 AC 2 AD 2, [by Pythagoras’ theorem], , , AC 2 CD 2 AD 2, [(15) 2 (9) 2] m 2 (225 81) m 2 144 m 2 ., , , , AC 144 m 12 m., , From right 3CBE, we have, CE 2 CB 2 BE 2 [by Pythagoras’ theorem], CB 2 CE 2 BE 2, [(15) 2 12) 2] m 2 (225 144) m 2 81 m 2, CB 81 m 9 m., Width of the street AC CB 12 m 9 m 21 m., EXAMPLE 5, , Two poles of heights 6 metres and 11 metres stand vertically on a, plane ground. If the distance between their feet is 12 metres, find the, distance between their tops., [CBSE 2002], , SOLUTION, , Let AB and CD be the given vertical poles. Then,, AB = 6 m, CD = 11 m and AC = 12 m., Draw BE AC. Then,, CE = AB = 6 m, BE = AC = 12 m., , , DE CD CE 11 m 6 m 5 m., , In right 3BED, we have:, BD 2 BE 2 DE 2, {(12) 2 (5) 2} m 2, (144 25) m 2 169 m 2
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Triangles, , 423, , BD 169 m 13 m., Hence, the distance between the tops of the poles = 13 m., EXAMPLE 6, , In a rhombus of side 10 cm, one of the diagonals is 12 cm long. Find, the length of the second diagonal., [CBSE 2001], , SOLUTION, , Let ABCD be the given rhombus, whose diagonals intersect at O., Then, AB 10 cm., Let AC = 12 cm and BD = 2x cm., We know that the diagonals of, a rhombus bisect each other at, right angles., , , OA , , 1, 1, AC 6 cm, OB BD x cm, and +AOB 90c., 2, 2, , From right 3AOB, we have, AB 2 OA 2 OB 2, OB 2 AB 2 OA 2, {(10) 2 (6) 2} cm 2 (100 36) cm 2 64 cm 2, , , x 2 64 & x 64 8., , OB 8 cm., BD 2 #OB 2 # 8 cm 16 cm., Hence, the length of the second diagonal is 16 cm., EXAMPLE 7, , 3ABD is a right triangle in which +A 90c and AC = BD., Prove that:, (i) AB 2 BC · BD, , SOLUTION, , (ii) AC 2 BC · DC, , We know that:, If a perpendicular is drawn from the vertex, of the right angle of a right triangle to the, hypotenuse then the triangles on both sides, of the perpendicular are similar to the, whole triangle and also to each other., 3ABC +3DBA, 3ABC +3DAC, 3DBA +3DAC., , (iii) AD 2 BD · CD
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424, , Secondary School Mathematics for Class 10, , (i) 3ABC +3DBA, AB BC, AB BC, , & BD, & AB 2 BC · BD., DB BA, AB, (ii) 3ABC +3DAC, AC BC, , & AC 2 BC · DC., DC AC, (iii) 3DBA +3DAC, AD BD, , & AD 2 BD · CD., CD AD, EXAMPLE 8, , BL and CM are medians of a 3ABC, right-angled at A. Prove that, 4(BL2 CM 2) 5BC 2 ., , SOLUTION, , [CBSE 2006C, ’10], , A 3ABC in which BL and CM are medians and, +A 90c., GIVEN, , TO PROVE, , In 3BAC, +A 90c., , PROOF, , , , 4(BL2 CM 2) 5BC 2 ., , BC AB 2 AC 2., 2, , … (i), [by Pythagoras’ theorem], , In 3BAL, +A 90c., , , BL2 AL2 AB 2, , [by Pythagoras’ theorem], , 1, 1, BL2 a ACk AB 2 BL2 AC 2 AB 2, 2, 4, 2, , , , 4BL2 AC 2 4AB 2., , … (ii), , In 3CAM, +A 90c., , , CM 2 AM 2 AC 2, , 2, 1, 1, CM 2 a ABk AC 2 CM 2 AB 2 AC 2, 2, 4, , , , 4CM 2 AB 2 4AC 2., , … (iii), , On adding (ii) and (iii), we get 4(BL CM ) 5(AB AC 2) ., 2, , Hence, 4(BL2 CM 2) 5BC 2 [using (i)]., EXAMPLE 9, , In the given figure, the, perpendicular from A on side BC, of a 3ABC, intersects BC at D, such that DB 3CD. Prove that, 2AB 2 2AC 2 BC 2 ., [CBSE 2003, ’05, ’09], , 2, , 2
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Triangles, SOLUTION, , GIVEN, , A 3ABC in which AD = BC and BD 3CD., , TO PROVE, PROOF, , , , 425, , 2AB 2 2AC 2 BC 2 ., , We have BD 3CD., , BC BD CD 3CD CD 4CD, , CD , , 1, BC., 4, , … (i), , In 3ADB, +ADB 90c., AB 2 AD 2 BD 2., In 3ADC, +ADC 90c., , … (ii) [by Pythagoras‘ theorem], , AC 2 AD 2 CD 2 ., , … (iii) [by Pythagoras’ theorem], , , , , On subtracting (iii) from (ii), we get, AB 2 AC 2 BD 2 CD 2, [(3CD) 2 (CD 2)] 8CD 2 [a BD 3CD], 8 # 1 BC 2 1 BC 2, 2, 16, , , [using (i)]., , 2AB 2 2AC 2 BC 2 ., , Hence, 2AB 2 2AC 2 BC 2 ., EXAMPLE 10, , In 3ABC, +B 90c and D is the midpoint of BC. Prove that, AC 2 AD 2 3CD 2 ., , SOLUTION, , GIVEN, , A 3ABC in which +B 90c, , and D is the midpoint of BC., TO PROVE, , AC 2 AD 2 3CD 2 ., Join AD., , CONSTRUCTION, PROOF, , , , In 3ABC, +B 90c., , AC 2 AB 2 BC 2., , … (i) [by Pythagoras’ theorem], , In 3ABD, +B 90c., , , AD 2 AB 2 BD 2, , , , AB AD 2 BD 2 ., , , , AC 2 (AD 2 BD 2) BC 2 [using (i)], , , , AC 2 AD 2 CD 2 (2CD) 2, , , , AC AD 3CD ., , … (ii) [by Pythagoras’ theorem], , 2, , 2, , 2, , 2, , Hence, AC 2 AD 2 3CD 2 ., , [a BD CD and BC 2CD]
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426, , Secondary School Mathematics for Class 10, , EXAMPLE 11, , 3ABC is right-angled at B and D is the midpoint of BC. Prove that, [CBSE 2008C, ’10], AC 2 (4AD 2 3AB 2) ., , SOLUTION, , GIVEN, , A 3ABC in which+B 90c and D is the midpoint of BC., , TO PROVE, PROOF, , , , AC 2 (4AD 2 3AB 2) ., , In 3ABC, +B 90c., , AC AB 2 BC 2, [by Pythagoras’ theorem], 2, , AB 2 (2BD) 2 [a BC 2BD], AB 2 4BD 2, AB 2 4 (AD 2 AB 2) [a AB 2 BD 2 AD 2], (4AD 2 3AB 2) ., Hence, AC 2 (4AD 2 3AB 2) ., EXAMPLE 12, , In an isosceles 3ABC, AB AC and BD = AC. Prove that, (BD 2 CD 2) 2CD · AD., , SOLUTION, , A 3ABC in which, AB AC and BD = AC., , GIVEN, , TO PROVE, PROOF, , (BD 2 CD 2) 2CD · AD., , From right 3ADB, we have, , AB AD 2 BD 2, [by Pythagoras’ theorem], 2, , AC 2 AD 2 BD 2, [a AB AC], (CD AD) 2 AD 2 BD 2 [a AC CD AD], CD 2 AD 2 2CD · AD AD 2 BD 2, (BD 2 CD 2) 2CD · AD., Hence, (BD 2 CD 2) 2CD · AD., EXAMPLE 13, , 3ABC is an isosceles triangle, right-angled at C. Prove that, AB 2 2AC 2 ., , SOLUTION, , 3ABC is an isosceles triangle, rightangled at C, … (i), BC AC, Now, by Pythagoras’ theorem,, AB 2 BC 2 AC 2
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Triangles, , EXAMPLE 14, , SOLUTION, , 427, , , , AB 2 AC 2 AC 2 [using (i)], , , , AB 2 2AC 2 ., , 3ABC is an isosceles triangle with AC BC. If AB 2 2AC 2, prove, that 3ABC is a right triangle., [CBSE 2000C], In 3ABC, we have, AC BC., , … (i), , Now, AB 2 2AC 2, , (given), , , , AB AC AC, , , , AB 2 AC 2 BC 2, , 2, , 2, , (given), , 2, , [using (i)], , +C 90c, [by converse of Pythagoras’ theorem], Hence, 3ABC is a right triangle., EXAMPLE 15, , 3ABC is a right triangle in which +C 90c and CD = AB. If, BC a, CA b, AB c and CD p then prove that, (i) cp ab, , SOLUTION, , (ii), , 1 1 1·, p2 a2 b2, , [CBSE 1997C, ’98, ’99, ’02], , (i) We have, ar (3ABC) , , 1, 1, # AB#CD cp, 2, 2, [taking AB as base], , and ar (3ABC) , , 1, 1, # BC # AC ab, 2, 2, [taking BC as base]., , 1 1, cp, ab & cp ab., 2, 2, Hence, cp ab., , , c, 1, (ii) cp ab & p ·, ab, , , 1 c2 b2 a2, p2 a2 b2, a2 b2, , 2, 2, d b2 2 a2 2 n c 12 12 m ·, a b, a b, a b, 1 1 1·, Hence, 2, p, a2 b2, , [a AB 2 AC 2 BC 2]
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428, , Secondary School Mathematics for Class 10, , EXAMPLE 16, , In an equilateral triangle, prove that three times the square of one, side is equal to four times the square of one of its altitudes., , SOLUTION, , GIVEN, , [CBSE 2002, ’07C], , A 3ABC in which AB BC CA and AD = BC., , 3AB 2 4AD 2 ., PROOF In 3ADB and 3 ADC, we have, AB AC (given), +B +C 60c, and +ADB +ADC 90c., 3ADB ,3ADC [AAS-congruence], 1, BD DC BC., 2, From right 3ADB, we have, TO PROVE, , AB 2 AD 2 BD 2, , [by Pythagoras’ theorem], , , , AD 2 a1 BCk AD 2 1 BC 2, 2, 4, 4AB 2 4AD 2 BC 2, , , , 3AB 2 4AD 2 [a BC AB]., , 2, , Hence, 3AB 2 4AD 2 ., EXAMPLE 17, , In an equilateral triangle with side a, prove that, (i) altitude , , SOLUTION, , 3, 3 2, a (ii) area , a ., 2, 4, , [CBSE 1997, ’99, ’01C, ’02C], , Let 3ABC be an equilateral triangle with side a., Then, AB AC BC a., Draw AD = BC., In 3ADB and 3ADC, we have, AB AC (given), +B +C 60c, and +ADB +ADC 90c., 3ADB ,3ADC., a, BD DC ·, 2, (i) From right 3ADB, we have, AB 2 AD 2 BD 2 [by Pythagoras’ theorem], AD AB 2 BD 2, , , a 2, a2 ` j , 2, , Hence, altitude , , a2 , 3, a., 2, , a2 , 4, , 3a 2 3, a., 2, 4
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Triangles, , 429, , 1, 1, # base # altitude a # BC # ADk, 2, 2, 3, d1 # a#, an [using (i)], 2, 2, , (ii) Area of 3ABC , , d, , 3 2, a n sq units., 4, , Hence, area(3ABC) d, EXAMPLE 18, , 3 2, a n sq units., 4, , O is a point in the interior of 3ABC, OD = BC, OE = AC and, OF = AB, as shown in the figure., , Prove that:, (i) OA 2 OB 2 OC 2 OD 2 OE 2 OF 2 AF 2 BD 2 CE 2, (ii) AF 2 BD 2 CE 2 AE 2 BF 2 CD 2, SOLUTION, , (i) Using Pythagoras’ theorem for each of the right triangles, namely 3OFA, 3ODB and 3OEC, we get, … (i), OA 2 OF 2 AF 2, … (ii), OB 2 OD 2 BD 2, … (iii), OC 2 OE 2 CE 2, Adding (i), (ii) and (iii), we get, OA 2 OB 2 OC 2, OD 2 OE 2 OF 2 AF 2 BD 2 CE 2 ., 2, Hence, OA OB 2 OC 2 OD 2 OE 2 OF 2, AF 2 BD 2 CE 2 ., (ii) Using Pythagoras’ theorem for each of the right triangles,, namely 3ODB and 3ODC, we get, OB 2 OD 2 BD 2 and OC 2 OD 2 CD 2 ., OB 2 OC 2 BD 2 CD 2, … (iv), Similarly, we have, OC 2 OA 2 CE 2 AE 2, and OA 2 OB 2 AF 2 BF 2., , … (v), … (vi)
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430, , Secondary School Mathematics for Class 10, , Adding the corresponding sides of (iv), (v) and (vi),, we get, AF 2 BD 2 CE 2 AE 2 BF 2 CD 2 0., Hence, AF 2 BD 2 CE 2 AE 2 BF 2 CD 2 ., EXAMPLE 19, , O is any point inside a rectangle ABCD., Prove that OB 2 OD 2 OA 2 OC 2 ., DEDUCTION In the given figure, O is a, point inside a rectangle ABCD such that, OB 6 cm, OD 8 cm and OA 5 cm,, find the length of OC., [CBSE 2009C], , SOLUTION, , GIVEN, , [CBSE 2006C], , O is a point inside a rectangle, , ABCD., TO PROVE, , OB 2 OD 2 OA 2 OC 2 ., , Through O, draw, POQ BC so that P lies on AB and Q, lies on DC., , CONSTRUCTION, , PROOF, , we have, , POQ BC & PQ = AB and QP = DC, +BPQ 90c and +CQP 90c., , , BPQC and APQD are both rectangles., , , , BP CQ, , [opposite sides of a rectangle], , DQ AP, , [opposite sides of a rectangle], , From right 3OPB, we have OB 2 OP 2 BP 2 ., From right 3OQD, we have OD 2 OQ 2 DQ 2 ., From right 3OPA, we have OA 2 OP 2 AP 2 ., From right 3OQC, we have OC 2 OQ 2 CQ 2 ., , , OB 2 OD 2 OP 2 OQ 2 BP 2 DQ 2, OP 2 OQ 2 CQ 2 AP 2, [a BP CQ and DQ AP], (OP AP ) (OQ 2 CQ 2), 2, , 2, , OA 2 OC 2 ., Hence, OB 2 OD 2 OA 2 OC 2 .
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Triangles, DEDUCTION, , 431, , Let OC x cm. Then,, , OB 2 OD 2 OA 2 OC 2, 62 82 52 x2, , , x 2 36 64 25 75, , , , x 75 5 3 (5 #1.732) 8.66 & OC 8.66 cm., , EXAMPLE 20, , Prove that the sum of the squares on the sides of a rhombus is equal, to the sum of the squares on its diagonals., [CBSE 2005, ’06, ’08C], , SOLUTION, , GIVEN A rhombus ABCD whose, diagonals AC and BD intersect at O., , (AB 2 BC 2 CD 2 DA 2), , TO PROVE, , (AC 2 BD 2) ., We know that the diagonals, of a rhombus bisect each other at, right angles., PROOF, , +AOB +BOC +COD +DOA 90c,, OA , , 1, 1, AC and OB BD., 2, 2, , From right 3AOB, we have, AB 2 OA 2 OB 2 [by Pythagoras’ theorem], 2, 2, a1 ACk a1 BDk 1 (AC 2 BD 2), 2, 2, 4, , , , 4AB 2 (AC 2 BD 2) ., , … (i), , Similarly, we have:, 4BC 2 (AC 2 BD 2), , … (ii), , 4CD (AC 2 BD 2), , … (iii), , 4DA (AC 2 BD 2), , … (iv), , 2, , 2, , On adding (i), (ii), (iii) and (iv), we get, (AB 2 BC 2 CD 2 DA 2) (AC 2 BD 2) ., REMARK, , In a rhombus ABCD, we have AB BC CD DA, so the above, result may be given as 4AB 2 (AC 2 BD 2) ., , EXAMPLE 21, , P and Q are points on the sides CA and CB of a 3ABC, right-angled, at C. Prove that (AQ 2 BP 2) (AB 2 PQ 2) ., [CBSE 2007, ’08]
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Triangles, , 433, , Thus, 3PR 2 5PS 2 8PT 2 0., Hence, 8PT 2 3PR 2 5PS 2 ., EXAMPLE 23, , In an isosceles 3ABC, AB AC and D is a point on BC. Prove that, AB 2 AD 2 BD · CD., , SOLUTION, , GIVEN, , A 3ABC in which AB AC and D is a point on BC., , TO PROVE, , (AB 2 AD 2) BD $ CD., Draw AL = BC., , CONSTRUCTION, PROOF, , In right A ALB and ALC, we have:, , hyp. AB = hyp. AC (given), AL AL, , (common), , 3ALB ,3ALC, , , [by RHS-congruence], , BL CL., , From right A ALB and ALD, by Pythagoras’ theorem, we have:, AB 2 AL2 BL2, , … (i), , AD AL DL, 2, , , , 2, , … (ii), , 2, , AB AD BL DL, 2, , 2, , 2, , 2, , (BL DL)(BL DL), BD · (CL DL), , [a BL CL], , BD · (CL DL), [a BL DL BD and BL CL], BD · CD., Hence, AB AD 2 BD · CD., 2, , EXAMPLE 24, , In an equilateral3ABC, D is a point on side BC such that BD , Prove that 9AD 2 7AB 2 ., , SOLUTION, , GIVEN, , A 3ABC in which AB BC CA, , and D is a point on BC such that, 1, BD BC., 3, TO PROVE, , 9AD 2 7AB 2 ., , CONSTRUCTION, , Draw AL = BC., , 1, BC., 3
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Triangles, , 435, , Thus, in 3ACD, we have AD 2 AC 2 CD 2 ., Hence, +ACD 90c [by converse of Pythagoras’ theorem]., EXAMPLE 26, , SOLUTION, , Equilateral triangles are drawn on the sides of a right triangle. Prove, that the area of the triangle on the hypotenuse is equal to the sum of, the areas of the triangles on the other two sides., [CBSE 2002], GIVEN A 3ABC in which +B 90c. Equilateral triangles, 3BCD, 3CAE and 3ABF are drawn on the sides BC, CA and, AB respectively., TO PROVE, , ar (3CAE) ar (3BCD) ar (3ABF)., , PROOF 3BCD, 3CAE and 3ABF are, equiangular and hence similar., , We have, , , , ar (3BCD) ar (3ABF) BC 2 AB 2, , , , ar (3CAE) ar (3CAE) CA 2 CA 2, ar (3BCD) ar (3ABF) BC 2 AB 2 CA 2, , , 1, ar (3CAE), CA 2, CA 2, [by Pythagoras’ theorem], , ar (3BCD) ar (3ABF) ar (3CAE)., EXAMPLE 27, , Given a right-angled 3ABC. The lengths of the sides containing the, right angle are 6 cm and 8 cm. A circle is inscribed in 3ABC. Find, the radius of the circle., , SOLUTION, , In 3ABC, we have, +B 90c, AB 6 cm, and BC 8 cm., A circle is inscribed in 3ABC., Let O be its centre and M, N, and P be the points where it, touches the sides AB, BC and, CA respectively. Then,, OM = AB, ON = BC, OP = CA., Let r cm be the radius of the circle., Then, OM ON OP r cm., Now, AB 2 BC 2 CA 2, , [by Pythagoras’ theorem], , 6 cm 8 cm CA 2 & CA 10 cm., 2, , 2, , 2, , 2
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436, , Secondary School Mathematics for Class 10, , Now, ar (3ABC) ar (3AOB) ar (3BOC) ar (3COA), 1, 1, 1, # AB# BC a # AB #OMk a # BC #ONk, , 2, 2, 2, a1 #CA#OPk, 2, 1, 1, 1, 1, # 6 # 8 a # 6 # rk a # 8 # rk a #10 # rk, , 2, 2, 2, 2, , r 2 & radius 2 cm., EXAMPLE 28, , Prove that the sum of the squares of the diagonals of a parallelogram, is equal to the sum of the squares of its sides., , SOLUTION, , GIVEN, , A parallelogram PQRS., , TO PROVE, , PR 2 QS 2, PQ 2 QR 2 RS 2 SP 2 ., , We have proved earlier that if, AD is a median of a 3ABC, then, PROOF, , AB 2 AC 2 2(AD 2 BD 2), 1, i.e., AB 2 AC 2 2AD 2 BC 2., 2, Now, let O be the point of intersection, of the diagonal PR and QS., , … (i), , The diagonals of a parallelogram bisect, each other., O is the midpoint of PR as well, as QS., Applying result (i) to 3PQR and 3RSP, we get, 1, PQ 2 QR 2 2OQ 2 PR 2, 2, 1, and RS 2 SP 2 2OS 2 PR 2., 2, Adding (ii) and (iii), we get, 2, 2, 1, 1, PQ 2 QR 2 RS 2 SP 2 2 a QSk 2 a QSk PR 2, 2, 2, PQ 2 QR 2 RS 2 SP 2 PR 2 QS 2 ., EXAMPLE 29, , SOLUTION, , … (ii), … (iii), , Given a 3ABC in which +B 90c and AB 3 BC. Prove that, +C 60c., Let D be the midpoint of the hypotenuse AC., Join BD.
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Triangles, , 437, , We have, AC 2 AB 2 BC 2 [by Pythagoras’ theorem], , , AC 2 ( 3 BC) 2 BC 2, , , , AC 4BC & AC 2BC, , , , 2CD 2BC [a D is the midpoint of AC], , [a AB 3 BC (given)], 2, , 2, , CD BC., , … (i), , Also, we know that the midpoint of the hypotenuse of a right, triangle is equidistant from the vertices., , , BD CD, , … (ii), , From (i) and (ii), we get, BC BD CD., 3BCD is equilateral and hence +C 60c., f, , EXERCISE 7D, , 1. The sides of certain triangles are given below. Determine which of, them are right triangles., (i) 9 cm, 16 cm, 18 cm, (iii) 1.4 cm, 4.8 cm, 5 cm, (v) (a 1) cm, 2 a cm, (a 1) cm, , (ii) 7 cm, 24 cm, 25 cm, (iv) 1.6 cm, 3.8 cm, 4 cm, , 2. A man goes 80 m due east and then 150 m due north. How far is he, from the starting point?, 3. A man goes 10 m due south and then 24 m due west. How far is he from, the starting point?, 4. A 13-m-long ladder reaches a window of a building 12 m above the, ground. Determine the distance of the foot of the ladder from the, building., 5. A ladder is placed in such a way that its foot is at a distance of 15 m, from a wall and its top reaches a window 20 m above the ground. Find, the length of the ladder., 6. Two vertical poles of height 9 m and 14 m stand on a plane ground., If the distance between their feet is 12 m, find the distance between, their tops.
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438, , Secondary School Mathematics for Class 10, , 7. A guy wire attached to a vertical pole of height 18 m is 24 m long and, has a stake attached to the other end. How far from the base of the pole, should the stake be driven so that the wire will be taut?, 8. In the given figure, O is a point inside a, 3PQR such that +POR 90c, OP 6 cm and, OR 8 cm. If PQ 24 cm and QR 26 cm,, prove that 3PQR is right-angled., [CBSE 2006, ’09C], , 9. 3ABC is an isosceles triangle with AB AC 13 cm. The length of, altitude from A on BC is 5 cm. Find BC., [CBSE 2000C], 10. Find the length of altitude AD of an isosceles 3ABC in which, AB AC 2a units and BC a units., 11. 3ABC is an equilateral triangle of side 2a units. Find each of its, altitudes., 12. Find the height of an equilateral triangle of side 12 cm., 13. Find the length of a diagonal of a rectangle whose adjacent sides are, 30 cm and 16 cm., 14. Find the length of each side of a rhombus whose diagonals are 24 cm, and 10 cm long., 15. In 3ABC, D is the midpoint of BC and AE = BC. If AC AB, show that, AB 2 AD 2 BC · DE , , 1, BC 2 ., 4, , [CBSE 2006C], , 16. In the given figure, +ACB 90c and CD = AB., Prove that, , BC 2 BD ·, AC 2 AD, , 17. In the given figure, D is the midpoint of side BC and AE = BC. If, BC a, AC b, AB c, ED x, AD p and AE h, prove that, (i) b 2 p 2 ax , , a2, 4, , (ii) c 2 p 2 ax , , a2, 4, , 1, (iii) (b 2 c 2) 2p 2 a 2, 2, 2, 2, (iv) (b c ) 2ax
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Triangles, , 439, , 18. In 3ABC, AB AC. Side BC is produced, to D. Prove that, (AD 2 AC 2) BD · CD., , 19. ABC is an isosceles triangle, rightangled at B. Similar triangles ACD and, ABE are constructed on sides AC and, AB. Find the ratio between the areas of, [CBSE 2002], 3ABE and 3ACD., , 20. An aeroplane leaves an airport and flies due north at a speed of 1000 km, per hour. At the same time, another aeroplane leaves the same airport, and flies due west at a speed of 1200 km per hour. How far apart will be, 1, the two planes after 1 hours?, 2, 21. In a 3ABC, AD is a median and AL = BC., Prove that:, (a) AC 2 AD 2 BC · DL a, , BC 2, k, 2, BC 2, (b) AB 2 AD 2 BC · DL a k, 2, 1, 2, 2, 2, (c) AC AB 2AD BC 2, 2, 22. Naman is doing fly-fishing in a, stream. The tip of his fishing rod is, 1.8 m above the surface of the water, and the fly at the end of the string rests, on the water 3.6 m away from him and, 2.4 m from the point directly under, the tip of the rod. Assuming that the, string (from the tip of his rod to the fly) is taut, how much string does, he have out (see the adjoining figure)? If he pulls in the string at the rate, of 5 cm per second, what will be the horizontal distance of the fly from, him after 12 seconds?
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440, , Secondary School Mathematics for Class 10, ANSWERS (EXERCISE 7D), , 1. (ii), (iii), (v), , 2. 170 m, , 3. 26 m, , 6. 13 m, , 7. 6 7 m, , 9. 24 m, , 12. 6 3 cm, , 13. 34 m, , 11. a 3 units, , 4. 5 m, 10., , a 15, units, 2, , 14. 13 cm, , 20. 300 61 km 22. 2.8 m (approx.), HINTS TO SOME SELECTED QUESTIONS, 7. Let AB be the vertical pole and C be the position of the, stake so that BC is a taut wire., In rt. 3ABC, +A 90c, AB 18 m, BC 24 m., Now, AC 2 AB 2 BC 2 & AC 2 (24 2 18 2) m 2 252 m 2, & AC 252 m 6 7 m., 8. PR OP 2 OR 2 6 2 8 2 10., Now, 24 2 10 2 26 2, i.e., PQ 2 PR 2 QR 2 & +QPR 90c., 10. In 3ABC, AB AC 2a and BC a., Draw AD = BC. Then, D is the midpoint of BC, a, and so BD $, 2, a 15, a 2 15a 2, AD 2 AB 2 BD 2 4a 2 , & AD , $, 2, 4, 4, , 15. In 3AEB, +AEB 90c., , , AB 2 AE 2 BE 2, , … (i), , In 3AED, +AED 90c., , , AD 2 (AE 2 DE 2), , , , AE 2 (AD 2 DE 2)., , , , AB 2 (AD 2 DE 2) BE 2, (AD, , 2, , [using (i)], , DE ) (BD DE) 2, 2, , 2, , (AD 2 DE 2) a1 BC DEk, 2, AD 2 1 BC 2 BC $ DE., 4, 16. 3ABC +3CBD, BC AB, , & BC 2 AB $ BD., BD CB, , … (i), , 5. 25 m, , 19. 1 : 2
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Triangles, , 441, , 3ABC +3ACD, AC AB, , & AC 2 AB · AD. … (ii), AD AC, Dividing (i) by (ii), we get, , BC 2 BD ·, AC 2 AD, , 17. (i) AC 2 AE 2 EC 2, , , , , a2, a2, b 2 h 2 ax k (h 2 x 2) ax , 2, 4, 2, , p 2 ax a ·, 4, (ii) AB 2 AE 2 BE 2, 2, a, a2, , c 2 h 2 a xk (h 2 x 2) ax , 2, 4, 2, a, 2, p ax ·, 4, (iii) Add (i) and (ii)., (iv) Subtract (ii) from (i)., 18. Draw AE = BC. Then, BE CE., AD 2 AE 2 DE 2 and AC 2 AE 2 CE 2, , , (AD 2 AC 2) DE 2 CE 2, (DE CE)(DE CE), (DE BE)(DE CE) [a CE BE], BD · CD., , 2, ar (3ABC) AB 2, , AB 1, 19., ar (3ACD) AC 2 2AB 2 2, , [a AC 2 AB 2 BC 2 2AB 2]., , 20. Distance covered by first plane in 1, OP 1500 km (north)., , 1, hours, 2, , Distance covered by second plane in 1, OQ 1800 km (west)., Distance between the two planes after 1, , 1, hours, 2, , 1, hours, 2, , PQ OP 2 OQ 2 300 61 km., 21. (a) In right 3ACL, AC 2 AL2 LC 2., , … (i), , In right 3 ALD, AL AD 2 DL2., 2, , , , … (ii), , AC 2 (AD 2 DL2) LC 2 [from (i) and (ii)], (AD 2 DL2) (DL DC) 2, 2, , (AD 2 DL2) aDL 1 BCk, 2, , 2, , AD 2 BC · DL 1 BC 2 AD 2 BC · DL aBC k ·, 2, 4
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442, , Secondary School Mathematics for Class 10, (b) In right 3ABL, AB 2 AL2 LB 2., , … (iii), , In right 3ALD, AL AD 2 DL2 .., 2, , , , AB 2 (AD 2 DL2) LB 2, , … (iv), , [from (iii) and (iv)], , (AD 2 DL2) (BD DL) 2, 2, , 2, , (AD 2 DL2) a1 BC DLk AD 2 BC · DL aBC k ·, 2, 2, (c) Adding the results of (a) and (b), we get, 1, AC 2 AB 2 2AD 2 BC 2 ., 2, 22. Let of the string that was out of the rod, BC BM 2 CM 2 (1.8) 2 (2.4) 2 m, 3.24 5.76 m 9 m 3 m., He pulls the string at a rate of 5 cm per second., , , length of string pulled in 12 s, (5 #12) cm 60 cm 0.6 m., , So, after 12 s, we have BCl (3 0.6) m 2.4 m and BM 1.8 m., , , ClM (BCl) 2 BM 2 (2.4) 2 (1.8) 2 m 2.52 m . 1.6 m., , Horizontal distance of the fly from him after 12 s ClA, ClM MA (1.6 1.2) m, 2.8 m., , f, , EXERCISE 7E, , Very-Short-Answer and Short-Answer Questions:, 1. State the two properties which are necessary for given two triangles to, be similar., 2. State the basic proportionality theorem., 3. State the converse of Thales’ theorem., 4. State the midpoint theorem., 5. State the AAA-similarity criterion., 6. State the AA-similarity criterion., 7. State the SSS-criterion for similarity of triangles., 8. State the SAS-similarity criterion., 9. State Pythagoras’ theorem., 10. State the converse of Pythagoras’ theorem., 11. If D, E and F are respectively the midpoints of sides AB, BC and CA of, 3ABC then what is the ratio of the areas of 3DEF and 3ABC?
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Triangles, , 443, , 12. Two triangles ABC and PQR are such that AB 3 cm, AC 6 cm,, +A 70c, PR 9 cm, +P 70c and PQ 4.5 cm. Show that, 3ABC +3PQR and state the similarity criterion., 13. If 3ABC +3DEF such that 2AB DE and BC 6 cm, find EF., 14. In the given figure, DE BC such that, AD x cm, DB (3x 4) cm, AE (x 3) cm, and EC (3x 19) cm. Find the value of x., 15. A ladder 10 m long reaches the window of a house 8 m above the, ground. Find the distance of the foot of the ladder from the base of, the wall., 16. Find the length of the altitude of an equilateral triangle of side 2a cm., 17. 3ABC +3DEF such that ar (3ABC) 64 cm 2 and ar (3DEF) 169 cm 2 ., If BC 4 cm, find EF., 18. In a trapezium ABCD, it is given that AB CD and AB 2CD. Its diagonals, AC and BD intersect at the point O such that ar (3AOB) 84 cm 2 . Find, ar (3COD) ., 19. The corresponding sides of two similar triangles are in the ratio 2 : 3., If the area of the smaller triangle is 48 cm2, find the area of the larger, triangle., 20. In an equilateral triangle with side a, prove that area , , 3 2, a ., 4, , 21. Find the length of each side of a rhombus whose diagonals are 24 cm, and 10 cm long., 22. Two triangles DEF and GHK are such that +D 48c and +H 57c. If, 3DEF +3GHK then find the measure of +F., 23. In the given figure, MN BC and AM : MB 1 : 2., Find, , area (3AMN), ·, area (3ABC), , 24. In triangles BMP and CNR it is given that PB 5 cm, MP 6 cm,, BM 9 cm and NR 9 cm. If 3BMP +3CNR then find the perimeter, of 3CNR., 25. Each of the equal sides of an isosceles triangle is 25 cm. Find the length, of its altitude if the base is 14 cm.
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444, , Secondary School Mathematics for Class 10, , 26. A man goes 12 m due south and then 35 m due west. How far is he from, the starting point?, 27. If the lengths of the sides BC, CA and AB of a 3ABC are a, b and c, respectively and AD is the bisector of +A then find the lengths of BD, and DC., 28. In the given figure, +AMN +MBC 76c., If p, q and r are the lengths of AM, MB and, BC respectively then express the length of, MN in terms of p, q and r., , 29. The lengths of the diagonals of a rhombus are 40 cm and 42 cm. Find, the length of each side of the rhombus., For each of the following statements state whether true (T) or false (F):, 30., , (i) Two circles with different radii are similar., (ii) Any two rectangles are similar., (iii) If two triangles are similar then their corresponding angles are, equal and their corresponding sides are equal., (iv) The length of the line segment joining the midpoints of any two, sides of a triangle is equal to half the length of the third side., (v) In a 3ABC, AB 6 cm, +A 45c and AC 8 cm and in a 3DEF,, DF 9 cm, +D 45c and DE 12 cm, then 3ABC +3DEF., (vi) The polygon formed by joining the midpoints of the sides of a, quadrilateral is a rhombus., (vii) The ratio of the areas of two similar triangles is equal to the ratio, of their corresponding angle-bisector segments., , (viii) The ratio of the perimeters of two similar triangles is the same as, the ratio of their corresponding medians., (ix) If O is any point, OA 2 OC 2 OB 2 OD 2 ., , inside, , a, , rectangle, , ABCD, , then, , (x) The sum of the squares on the sides of a rhombus is equal to the, sum of the squares on its diagonals.
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Triangles, , 445, , ANSWERS (EXERCISE 7E), , 11. 1 : 4, , 12. SAS-similarity, , 13. 12 cm, , 14. x 2, , 16., , 17. 6.5 cm, , 18. 21 cm, , 19. 108 cm, , 1, 23., 9, , 24. 30 cm, , 3 a cm, , 22. 75, 27. BD , , , , ab, ac, ; DC , bc, bc, , 28. MN , , 2, , 25. 24 cm, , ac, ab, , 15. 6 m, 2, , 21. 13 cm, 26. 37 cm, , 29. 29 cm, , 30. (i) T (ii) F (iii) F (iv) T (v) F (vi) F (vii) T (viii) T (ix) T (x) T, , HINTS TO SOME SELECTED QUESTIONS, 14., , AD AE, [by Thales’ theorem], DB EC, , 17., , 2, ar (3AOB) AB 2, , a AB k 4, 1, CD, ar (3COD) CD 2, , , , ar (3COD) , , [a 3AOB +3COD], , 1, # ar (3AOB)., 4, , 22. +D 48c, +E +H 57c [a 3DEF +3GHK], , , +F 180c (+D +E) 180c (48c 57c) 75c., , 23. MN BC & 3AMN +3ABC., , , 24., , 2, 2, area (3AMN) AM 2, , c AM m a x k 1, x 2x, 9, AM MB, area (3ABC), AB 2, , [AM : MB 1 : 2, so, let AM x, then MB 2x]., , Perimeter of 3BMP MP, Perimeter of 3CNR NR, , , perimeter of 3CNR , , [a 3BMP +3CNR], NR, NR, # perimeter of 3BMP , #(PB MP BM), MP, MP, 9 #(5 6 9) cm 30 cm., 6, , 25. Let the given triangle be 3ABC having AB AC 25 cm,, base BC 14 cm., Let AD = BC. Then, D is the midpoint of BC., Altitude AD , , AC 2 DC 2 25 2 7 2 cm 24 cm.
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446, , Secondary School Mathematics for Class 10, , 27. Let BD x. Then, DC BC BD a x., BD AB, [by angle-bisector theorem], DC AC, ab ·, x c, ac, and so, a x , &x, ax b, bc, bc, , Now,, , , 28. +AMN +MBC & MN < BC & 3AMN +3ABC., , , AM MN, a MN, ac ·, &, c & MN a b, AB, BC, ab, , 29. The diagonals of a rhombus intersect at right, angles and bisect each other., In the figure,, 40, 42, OA , cm 20 cm, OB , cm 21 cm., 2, 2, AB OA 2 OB 2 20 2 21 2 cm 841 cm 29 cm., 30. (ii) Two rectangles are similar only if their corresponding sides are proportional., (iii) If two triangles are similar, then, (a) their corresponding angles are equal, (b) their corresponding sides are proportional (but not necessarily equal), (iv) Let D and E be the midpoints of sides AB and AC, respectively of a 3ABC., Then, by the midpoint theorem,, DE < BC & 3ADE +3ABC., , , AD DE, 1 DE, 1, & , & DE BC., 2 BC, 2, AB BC, , (v) We have, Clearly,, Note, , AB 6 1 AC 8 ·, ,, DE 12 2 DF 9, , AB AC, and so 3ABC is not similar to 3DEF., !, DE DF, , Here 3ABC +3DFE., , (vi) The polygon formed by joining the midpoints of the sides of a quadrilateral is a, parallelogram (not necessarily a rhombus)., (vii) The ratio of the perimeters of two similar triangles is the same as the ratio of, their corresponding sides which is the same as the ratio of the corresponding, medians.
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Triangles, , 447, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. A man goes 24 m due west and then 10 m due north. How far is he from, the starting point?, (a) 34 m, , (b) 17 m, , (c) 26 m, , (d) 28 m, , 2. Two poles of height 13 m and 7 m respectively stand vertically on a, plane ground at a distance of 8 m from each other. The distance between, their tops is, (a) 9 m, , (b) 10 m, , (c) 11 m, , (d) 12 m, , 3. A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At, the same time, what is the length of the shadow of a pole 6 m high?, (a) 2.4 m, , (b) 1.35 m, , (c) 1.5 m, , (d) 13.5 m, , 4. A vertical pole 6 m long casts a shadow of length 3.6 m on the ground., What is the height of a tower which casts a shadow of length 18 m at, the same time?, (a) 10.8 m, , (b) 28.8 m, , (c) 32.4 m, , (d) 30 m, , 5. The shadow of a 5-m-long stick is 2 m long. At the same time the length, of the shadow of a 12.5-m-high tree (in m) is, [CBSE 2011], (a) 3.0, , (b) 3.5, , (c) 4.5, , (d) 5.0, , 6. A ladder 25 m long just reaches the top of a building 24 m high from the, ground. What is the distance of the foot of the ladder from the building?, (a) 7 m, , (b) 14 m, , (c) 21 m, , (d) 24.5 m, , 7. In the given figure, O is a point inside, a 3MNP such that +MOP 90c,, and, If, OM 16 cm, OP 12 cm., MN 21 cm and +NMP 90c then, NP ?, (a) 25 cm, , (b) 29 cm, , (c) 33 cm, , (d) 35 cm, , 8. The hypotenuse of a right triangle is 25 cm. The other two sides are, such that one is 5 cm longer than the other. The lengths of these, sides are, (a) 10 cm, 15 cm, , (b) 15 cm, 20 cm, , (c) 12 cm, 17 cm, , (d) 13 cm, 18 cm, , 9. The height of an equilateral triangle having each side 12 cm, is, (a) 6 2 cm, , (b) 6 3 cm, , (c) 3 6 cm, , (d) 6 6 cm
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448, , Secondary School Mathematics for Class 10, , 10. 3ABC is an isosceles triangle with AB AC 13 cm and the length of, altitude from A on BC is 5 cm. Then, BC ?, (a) 12 cm, (b) 16 cm, (c) 18 cm, (d) 24 cm, 11. In a 3ABC it is given that AB 6 cm, AC 8 cm, and AD is the bisector of +A. Then, BD : DC ?, (a) 3 : 4, (b) 9 : 16, (c) 4 : 3, , (d), , 3 :2, , 12. In a 3ABC it is given that AD is the internal, bisector of +A. If BD 4 cm, DC 5 cm and, AB 6 cm, then AC ?, (a) 4.5 cm, (b) 8 cm, (c) 9 cm, (d) 7.5 cm, 13. In a 3ABC, it is given that AD, is the internal bisector of +A., If AB 10 cm, AC 14 cm and, BC 6 cm, then CD ?, (a) 4.8 cm, (b) 3.5 cm, (c) 7 cm, (d) 10.5 cm, 14. In a triangle, the perpendicular from the vertex to the base bisects the, base. The triangle is, (a) right-angled, (b) isosceles, (c) scalene, (d) obtuse-angled, 15. In an equilateral triangle ABC, if AD = BC then which, of the following is true?, (a) 2AB 2 3AD 2, (b) 4AB 2 3AD 2, (c) 3AB 2 4AD 2, , (d) 3AB 2 2AD 2, , 16. In a rhombus of side 10 cm, one of the diagonals is 12 cm long. The, length of the second diagonal is, (a) 20 cm, (b) 18 cm, (c) 16 cm, (d) 22 cm, 17. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. The, length of each side of the rhombus is, (a) 12 cm, (b) 13 cm, (c) 14 cm, (d) 17 cm
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Triangles, , 449, , 18. If the diagonals of a quadrilateral divide each other proportionally then, it is a, (a) parallelogram, , (b) trapezium, , (c) rectangle, , (d) square, , 19. In the given figure, ABCD is a trapezium, whose diagonals AC and BD intersect at O, such that OA (3x 1) cm, OB (2x 1) cm,, OC (5x 3) cm and OD (6x 5) cm. Then,, x?, (a) 2, , (b) 3, , (c) 2.5, , (d) 4, , 20. The line segments joining the midpoints of the adjacent sides of a, quadrilateral form, (a) a parallelogram, , (b) a rectangle, , (c) a square, , (d) a rhombus, , 21. If the bisector of an angle of a triangle bisects the opposite side then the, triangle is, (a) scalene, , (b) equilateral, , (c) isosceles, , (d) right-angled, , 22. In 3ABC it is given that, +C 50c then +BAD ?, , AB BD ·, If +B 70c and, AC DC, , (a) 30, , (b) 40, , (c) 45, , (d) 50, , 23. In 3ABC, DE BC so that AD 2.4 cm, AE 3.2 cm, and EC 4.8 cm. Then, AB ?, (a) 3.6 cm, (c) 6.4 cm, , (b) 6 cm, (d) 7.2 cm, , 24. In a 3ABC, if DE is drawn parallel to BC, cutting, AB and AC at D and E respectively such that, AB 7.2 cm, AC 6.4 cm and AD 4.5 cm. Then,, AE ?, (a) 5.4 cm, (c) 3.6 cm, , (b) 4 cm, (d) 3.2 cm
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450, , Secondary School Mathematics for Class 10, , 25. In 3ABC, DE BC so that AD (7x 4) cm,, AE (5x 2) cm, DB (3x 4) cm and EC 3x cm., Then, we have, (a) x 3, (b) x 5, (c) x 4, (d) x 2.5, 26. In 3ABC, DE < BC such that, , AD 3 ·, If AC 5.6 cm, DB 5, , then AE ?, (a) 4.2 cm, , (b) 3.1 cm, , (c) 2.8 cm, , (d) 2.1 cm, , 27. 3ABC +3DEF and the perimeters of 3ABC and 3DEF are 30 cm and, 18 cm respectively. If BC 9 cm then EF ?, (a) 6.3 cm, , (b) 5.4 cm, , (c) 7.2 cm, , (d) 4.5 cm, , 28. 3ABC +3DEF such that AB 9.1 cm and DE 6.5 cm. If the perimeter, of 3DEF is 25 cm, what is the perimeter of 3ABC?, (a) 35 cm, , (b) 28 cm, , (c) 42 cm, , (d) 40 cm, , 29. In 3ABC, it is given that AB 9 cm, BC 6 cm and CA 7.5 cm. Also,, 3DEF is given such that EF 8 cm and 3DEF +3ABC. Then, perimeter, of 3DEF is, (a) 22.5 cm, , (b) 25 cm, , (c) 27 cm, , (d) 30 cm, , 30. ABC and BDE are two equilateral triangles such that D is the midpoint, of BC. Ratio of the areas of triangles ABC and BDE is, (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 4, , (d) 4 : 1, , 31. It is given that 3ABC +3DFE. If +A 30c, +C 50c, AB 5 cm,, AC 8 cm and DF 7.5 cm then which of the following is true?, (a) DE 12 cm, +F 50c, (b) DE 12 cm, +F 100c, (c) EF 12 cm, +D 100c, , (d) EF 12 cm, +D 30c, , 32. In the given figure, +BAC 90c and AD = BC. Then,, (a) BC $ CD BC 2, (b) AB $ AC BC 2, (c) BD $ CD AD 2, (d) AB $ AC AD 2, 33. In 3ABC, AB 6 3 cm, AC 12 cm and BC 6 cm. Then, +B is, (a) 45c, , (b) 60c, , (c) 90c, , (d) 120c
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Triangles, , 34. In 3ABC and 3DEF, it is given that, (a) +B +E, , 451, , AB BC, then, DE FD, , (b) +A +D, , (c) +B +D, (d) +A +F, 35. In 3DEF and 3PQR, it is given that +D +Q and +R +E, then which, of the following is not true?, EF DF, EF DE, DE EF, DE DF, (a), (b), (c), (d), PR PQ, RP QR, PQ RP, QR PQ, 36. If 3ABC +3EDF and 3ABC is not similar to 3DEF then which of the, following is not true?, (a) BC · EF AC · FD, (b) AB · EF AC · DE, (c) BC · DE AB · EF, (d) BC · DE AB · FD, 37. In 3ABC and 3DEF, it is given that +B +E, +F +C and AB 3DE,, then the two triangles are, (a) congruent but not similar, (c) neither congruent nor similar, 38. If in 3ABC and 3PQR, we have, (a) 3PQR +3CAB, (c) 3CBA +3PQR, , (b) similar but not congruent, (d) similar as well as congruent, , AB BC CA, then, QR PR PQ, (b) 3PQR +3ABC, (d) 3BCA +3PQR, , 39. In the given figure, two line segments, AC and BD intersect each other at the, point P such that PA 6 cm, PB 3 cm,, PC 2.5 cm, PD 5 cm, +APB 50c and, +CDP 30c then +PBA ?, (a) 50, , (b) 30, , (c) 60, , (d) 100, , 40. Corresponding sides of two similar triangles are in the ratio 4 : 9. Areas, of these triangles are in the ratio, (a) 2 : 3, , (d) 16 : 81, ar (3PQR), BC 2, ?, 41. It is given that 3ABC +3PQR and, then, QR 3, ar (3ABC), (a), , 2, 3, , (b) 4 : 9, , (b), , 3, 2, , (c) 9 : 4, , (c), , 4, 9, , 42. In an equilateral 3ABC, D is the midpoint, of AB and E is the midpoint of AC. Then,, ar (3ABC) : ar (3ADE) ?, (a) 2 : 1, , (b) 4 : 1, , (c) 1 : 2, , (d) 1 : 4, , (d), , 9, 4
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452, , Secondary School Mathematics for Class 10, , 43. In 3ABC and 3DEF, we have, ar (3ABC) : ar (3DEF) ?, (a) 5 : 7, , (b) 25 : 49, , (a) 36 : 49, , (b) 6 : 7, , AB BC AC 5, , then, DE EF DF 7, , (c) 49 : 25, (d) 125 : 343, 2, 44. 3ABC +3DEF such that ar(3ABC) 36 cm and ar (3DEF) 49 cm 2 ., Then, the ratio of their corresponding sides is, (c) 7 : 6, , (d), , 6: 7, , 45. Two isosceles triangles have their corresponding angles equal and their, areas are in the ratio 25 : 36. The ratio of their corresponding heights is, (a) 25 : 36, , (b) 36 : 25, , (c) 5 : 6, , (d) 6 : 5, , 46. The line segments joining the midpoints of the sides of a triangle form, four triangles, each of which is, (a) congruent to the original triangle, (b) similar to the original triangle, (c) an isosceles triangle, (d) an equilateral triangle, 47. If 3ABC +3QRP,, PR ?, (a) 8 cm, , ar (3ABC) 9, , AB 18 cm and BC 15 cm then, ar (3PQR) 4, (b) 10 cm, , (c) 12 cm, , (d), , 20, cm, 3, , 48. In the given figure, O is the point of intersection, of two chords AB and CD such that OB OD, and +AOC 45c. Then, 3OAC and 3ODB are, (a), (b), (c), (d), , equilateral and similar, equilateral but not similar, isosceles and similar, isosceles but not similar, 49. In an isosceles 3ABC, if AC BC and AB 2 2AC 2 then +C ?, (a) 30, , (c) 60, (d) 90, 50. In 3ABC, if AB 16 cm, BC 12 cm and AC 20 cm, then 3ABC is, (a) acute-angled, (c) obtuse-angled, , (b) 45, , (b) right-angled, (d) not possible, , True/False Type, , 51. Which of the following is a true statement?, (a) Two similar triangles are always congruent., (b) Two figures are similar if they have the same shape and size.
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Triangles, , 453, , (c) Two triangles are similar if their corresponding sides are, proportional., (d) Two polygons are similar if their corresponding sides are, proportional., 52. Which of the following is a false statement?, (a) If the areas of two similar triangles are equal then the triangles are, congruent., (b) The ratio of the areas of two similar triangles is equal to the ratio, of their corresponding sides., (c) The ratio of the areas of two similar triangles is equal to the ratio, of squares of their corresponding medians., (d) The ratio of the areas of two similar triangles is equal to the ratio, of squares of their corresponding altitudes., Matching of columns, 53. Match the following columns:, Column I, , Column II, , (a) In a given 3ABC, DE BC and (p) 6, AD 3 ·, If AC 5.6 cm then, DB 5, AE …… cm., (b) If 3ABC +3DEF such that (q) 4, 2AB 3DE and BC 6 cm then, EF …… cm., (c) If 3ABC +3PQR such that, ar (3ABC) : ar (3PQR) 9 : 16 and, BC 4.5 cm then QR …… cm., , (r) 3, , (d) In the given figure, AB CD and (s) 2.1, OA (2x 4) cm, OB (9x 21) cm,, OC (2x 1) cm and OD 3 cm., Then x ?, , The correct answer is, (a) —……,, (b) —……,, , (c) —……,, , (d) —……
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454, , Secondary School Mathematics for Class 10, , 54. Match the following columns:, Column I, , Column II, , (a) A man goes 10 m due east and then (p) 25 3, 20 m due north. His distance from, the starting point is …… m., (b) In an equilateral triangle with each (q) 5 3, side 10 cm, the altitude is …… cm., (c) The area of an equilateral triangle, having each side 10 cm is …… cm2., , (r) 10 5, , (d) The length of diagonal of a, rectangle having length 8 m and, breadth 6 m is …… m., , (s) 10, , The correct answer is, (a) —……,, , (b) —……,, , (c) —……,, , (d) —……, ANSWERS (MCQ), , 1. (c), , 2. (b), , 6. (a), , 7. (b), , 10. (d) 11. (a), , 12. (d) 13. (b) 14. (b) 15. (c), , 16. (c), , 17. (b) 18. (b), , 19. (a), , 20. (a), , 21. (c), , 23. (b) 24. (b) 25. (c), , 26. (d) 27. (b), , 28. (a), , 29. (d) 30. (d) 31. (b) 32. (c), , 37. (b) 38. (a), , 3. (c), , 4. (d), 22. (a), , 5. (d), , 33. (c), , 34. (c), , 9. (b), , 35. (b) 36. (c), , 39. (d) 40. (d) 41. (d) 42. (b) 43. (b) 44. (b) 45. (c), , 46. (b) 47. (b) 48. (c), , 49. (d) 50. (b) 51. (c), , 53. (a)–(s), (b)–(q), (c)–(p), (d)–(r), , 52. (b), , 54. (a)–(r), (b)–(q), (c)–(p), (d)–(s), , HINTS TO SOME SELECTED QUESTIONS, 5. Let AB be the stick and AC be its shadow., Let DE be the tree and DF be its shadow., 3ABC +3DEF., , 11., , 8. (b), , , , 5 2, AB AC, &, DE DF, 12.5 x, , , , 12.5 # 2 , xa, k 5., 5, , BD AB 6 3, DC AC 8 4, , [by angle-bisector theorem]
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456, , Secondary School Mathematics for Class 10, , 32. In 3DBA and 3DAC, we have, +ADB +CDA 90c, +ABD +CAD 90c +C, and +BAD +ACD 90c +B., BD AD, 3DBA +3DAC &, AD CD, & BD $ CD AD 2 ., 33. In 3ABC, AC is the longest side., AB 2 BC 2 {(6 3 ) 2 6 2} cm 2 (108 36) cm 2 144 cm 2 (12 cm) 2 AC 2 ., , , by the converse of Pythagoras’ theorem, we have +B 90c., , 34. Clearly, B ) D, A ) E and C ) F, , , +B +D., , 35. +D +Q, +E +R and +F +P., DE DF EF ·, QR PQ RP, , , , 3DEF +3QRP &, , , , DE EF, is not true., PQ RP, , 36. Since 3ABC +3EDF and 3ABC is not similar to 3DEF, so, AB BC, !, & BC $ DE AB $ EF is not true., DE EF, 37. 3ABC +3DEF, , [by AA-similarity]., , But 3ABC and 3DEF are not congruent since their corresponding sides AB and DE, are not equal., 38., , AB BC CA, & B ) R, A ) Q and C ) P., QR PR PQ, , , 3PQR +3CAB., , ar (3PQR) QR 2, QR 2 3 2 9, m a k ·, , c, 2, 2, 4, BC, ar (3ABC) BC, AD AE 1, 42. Clearly,, and +A +A (common)., AB AC 2, 41., , , , 3ABC +3ADE [by SAS-similarity]., , , , ar (3ABC) : ar (3ADE) a, , 44. a, , AB 2 2 2 4 , 4 : 1., k a k, 1, 1, AD, , AB 2 ar (3ABC) 36 6 2, AB 6 ·, a k &, k, 7, DE, DE 7, ar (3DEF) 49, , , , the ratio of the corresponding sides is 6 : 7., , 45. The two triangles have corresponding angles equal and so they are similar., , , the ratio of their areas is equal to the ratio of the squares of their corresponding, sides but the ratio of their corresponding sides is equal to the ratio of their, corresponding altitudes (or heights).
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Triangles, , 457, , So, the ratio of their areas is equal to the ratio of the squares of their heights., 25 5 2, 36 6 2, 5, ratio of their heights ·, 6, 46. ar (3QRP) ar (3PQR) ., Now, ratio of their areas , , ar (3ABC) 9 3 2, BC 3, , &, [a 3ABC +3QRP], RP 2, ar (3QRP) 4 2 2, 2, 2, & PR # BC #15 cm 10 cm., 3, 3, 48. In 3OAC and 3ODB, we have, , , +AOC +DOB (ver. opp. O) and +OAC +ODB (O in the same segment)., OC OA AC, 3OAC +3ODB &, OB OD BD, OC OB , , 1 & OC OA [a OB OD (given)]., OA OD, Clearly, OA ! OD., OA, AC, !1 &, ! 1 & AC ! BD., BD, OD, , , , , 3OAC and 3ODB are isosceles and similar., , 49. AB 2AC 2 AC 2 AC 2 BC 2 AC 2 [a AC BC] ., 2, , , , by converse of Pythagoras’ theorem, we have +C 90c., , 50. AC is the longest side of 3ABC., AB 2 BC 2 16 2 12 2 256 144 400 20 2 AC 2., , , +B 90c [by the converse of Pythagoras’ theorem]., , 51. (a) Two similar triangles need not be congruent., (b) Similar figures need not be of the same size., (c) It is clearly true., (d) Two polygons are similar only when their corresponding angles are equal and, their corresponding sides are proportional., 53. (a) Let AE x cm. Then, EC (5.6 x) cm., , , , , 3, x, AD AE, & , DB EC, 5 5.6 x, 3(5.6 x) 5x & 8x 3 # 5.6, 3 # 5.6 16.8 , x, 2.1., 8, 8, , 3 6, AB BC, & x & 3x 12 & x 4., 2, DE EF, ar (3ABC) BC 2, 9 BC 2, 32, BC 2, , m, (c), &, &a k c, 16 QR 2, 4, QR, ar (3PQR) QR 2, , (b), , , , BC 3, 4, 4, & QR # BC a # 4.5k cm 6 cm., 3, 3, QR 4
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458, , Secondary School Mathematics for Class 10, (d) 3OAB +3OCD &, , OA OB, 2x 4 9x 21, &, 3, 2x 1, OC OD, , , , 6x 12 18x 2 51x 21 & 18x 2 57x 9 0, , , , 6x 2 19x 3 0 & (x 3)(6x 1) 0 & x 3 or x , , 1·, 6, , 1, makes (2x 1) 0. So, we reject it., 6, x 3., , But, x , , , The correct answer is: (a)–(s), (b)–(q), (c)–(p), (d)–(r)., 54. (a) Let OA 10 m and AB 20 m. Then,, OB 2 OA 2 AB 2 {(10) 2 (20) 2} m 2 500 m 2, , , OB 500 m 5 #100 m 10 5 m., , (b) Altitude , (c) Area , , 3, 3, a d #10n cm 5 3 cm., 2, 2, , 3 2, 3, a d #10 #10n cm 2 25 3 cm 2 ., 4, 4, , (d) d 2 (8 2 6 2) m 2 (64 36) m 2 100 m 2 & d 100 m 10 m., Then, the correct answer is: (a)–(r), (b)–(q), (c)–(p), (d)–(s)., , TEST YOURSELF, MCQ, 1. 3ABC +3DEF and their perimeters are 32 cm and 24 cm respectively., If AB 10 cm then DE ?, (a) 8 cm, , (b) 7.5 cm, , (c) 15 cm, , (d) 5 3 cm, , 2. In the given figure, DE BC. If DE 5 cm,, BC 8 cm and AD 3.5 cm then AB ?, (a) 5.6 cm, , (b) 4.8 cm, , (c) 5.2 cm, , (d) 6.4 cm, , 3. Two poles of height 6 m and 11 m stand vertically upright on a plane, ground. If the distance between their feet is 12 m then the distance, between their tops is, (a) 12 m, , (b) 13 m, , (c) 14 m, , (d) 15 m, , cm2, , 4. The areas of two similar triangles are 25, and 36 cm2 respectively. If, the altitude of the first triangle is 3.5 cm then the corresponding altitude, of the other triangle is, (a) 5.6 cm, , (b) 6.3 cm, , (c) 4.2 cm, , (d) 7 cm
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Triangles, , 459, , Short-Answer Questions, 5. If 3ABC +3DEF such that 2AB DE and BC 6 cm, find EF., 6. In the given figure, DE BC such that, AD x cm, DB (3x 4) cm, AE (x 3) cm, and EC (3x 19) cm. Find the value of x., , 7. A ladder 10 m long reaches the window of a house 8 m above the, ground. Find the distance of the foot of the ladder from the base of, the wall., 8. Find the length of the altitude of an equilateral triangle of side 2a cm., 9. 3ABC +3DEF such that ar (3ABC) 64 cm 2 and ar (3DEF) 169 cm 2 ., If BC 4 cm, find EF., 10. In a trapezium ABCD, it is given that AB CD and AB 2CD. Its, diagonals AC and BD intersect at the point O such that, ar (3AOB) 84 cm 2 . Find ar (3COD) ., 11. The corresponding sides of two similar triangles are in the ratio 2 : 3., If the area of the smaller triangle is 48 cm2, find the area of the larger, triangle., 12. In the given figure, LM CB and LN CD., Prove that, , AM AN ·, AB AD, , 13. Prove that the internal bisector of an angle of a triangle divides the, opposite side internally in the ratio of the sides containing the angle., 14. In an equilateral triangle with side a, prove that area , , 3 2, a ., 4, , 15. Find the length of each side of a rhombus whose diagonals are 24 cm, and 10 cm long., 16. Prove that the ratio of the perimeters of two similar triangles is the, same as the ratio of their corresponding sides.
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460, , Secondary School Mathematics for Class 10, , Long-Answer Questions, 17. In the given figure, 3ABC and 3DBC have, the same base BC. If AD and BC intersect at, ar (3ABC) AO, ·, , O, prove that, ar (3DBC) DO, , 18. In the given figure, XY AC and XY divides, 3ABC into two regions, equal in area. Show, that, , AX (2 2 ) ·, 2, AB, , 19. In the given figure, 3ABC is an obtuse triangle,, obtuse-angled at B. If AD = CB (produced), prove that AC 2 AB 2 BC 2 2BC · BD., , 20. In the given figure, each one of PA, QB and, RC is perpendicular to AC. If AP x, QB z,, RC y, AB a, 1 1 1·, x y z, , and, , BC b,, , show, , that, , ANSWERS (TEST YOURSELF), , 1. (b), 8., , 2. (a), , 3 a cm, , 3. (b), , 4. (c), , 9. 6.5 cm, , 5. 12 cm, , 6. x 2, , 10. 21 cm2 11. 108 cm2, , , , 7. 6 m, 15. 13 cm
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Circles, , 461, , Circles, , 8, , A circle is a collection of all points in a plane which are at a constant, distance from a fixed point. The constant distance is called the radius and the fixed, point is called the centre of the circle., , CIRCLE, , SECANT A line which intersects a circle in two distinct, points is called a secant to the circle., , A line meeting a circle only in one point is, called a tangent to the circle at that point., , TANGENT, , The point at which the tangent line intersects the circle is called the point, of contact., NOTE, , The tangent to a circle is a special case of the secant, when the two end, points of its corresponding chord coincide., , Consider a line l which is a secant to a circle, with centre O. Let PQ be the corresponding chord, of this secant. If we rotate the secant anticlockwise, about the point P then point Q comes closer to the, point P. Let Q1, Q2, Q3 etc., be the positions of Q, as the secant goes on rotating then we find these, positions getting closer to P sequentially. Finally,, as the position of Q coincides with P, the secant is, reduced to a tangent at point P., NUMBER OF TANGENTS TO A CIRCLE, , (i) There is no tangent passing through a, point lying inside the circle., (ii) There is one and only one tangent passing, through a point lying on a circle., (iii) There are exactly two tangents through a, point lying outside a circle. In the figure,, PT and PTl are two tangents from a point, P lying outside the circle., 461
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462, , Secondary School Mathematics for Class 10, , LENGTH OF A TANGENT The length of the segment of the tangent from the external, point P to the point of contact with the circle is called the length of the tangent from, the point P to the circle., , RESULTS ON TANGENTS, , The tangent at any point of a circle is perpendicular to the radius, through the point of contact., [CBSE 2007, ’09, ’11, ’12, ’13, ’14, ’15], , THEOREM 1, , GIVEN, , A circle with centre O and a tangent AB, at a point P of the circle., OP = AB., , TO PROVE, , CONSTRUCTION, , Take a point Q, other than P, on, AB. Join OQ., , Q is a point on the tangent AB, other than the point of contact P., , PROOF, , , , Q lies outside the circle., , Let OQ intersect the circle at R., Then, OR OQ, But, OP OR, , [a part is less than the whole]., , … (i), , [radii of the same circle]., , … (ii), , , , [from (i) and (ii)]., , OP OQ, , Thus, OP is shorter than any other line segment joining O to any, point of AB, other than P., In other words, OP is the shortest distance between the point O, and the line AB., But, the shortest distance between a point and a line is the, perpendicular distance., , , OP = AB., , (i) From this theorem we also conclude that at any point on a, circle, one and only one tangent can be drawn to the circle., , REMARKS, , (ii) The line containing the radius through the point of contact, is called the ’normal‘ to the circle at the point of contact., THEOREM 2, , GIVEN, , (Converse of Theorem 1) A line drawn through the end of a radius, and perpendicular to it is a tangent to the circle., , A circle with centre O in which OP is, a radius and AB is a line through P, such that OP = AB., , TO PROVE, , AB is a tangent to the circle at the, point P.
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Circles, CONSTRUCTION, PROOF, , 463, , Take a point Q, different from P, on AB. Join OQ., , We know that the perpendicular distance from a point to a line is, the shortest distance between them., , , OP = AB, , , , OP OQ., , , , Q lies outside the circle [a OP is the radius and OP OQ]., , & OP is the shortest distance from O to AB., , Thus, every point on AB, other than P, lies outside the circle., , , AB meets the circle at the point P only., , Hence, AB is the tangent to the circle at the point P., REMARK, , This theorem gives us a method of constructing a tangent at a, point P on the circle. We first draw the radius OP and then draw, PT = OP. Then, PT is the tangent at P., , THEOREM 3, , The lengths of tangents drawn from an external point to a circle, are equal. [CBSE 2007, ’08, ’08C, ’09, ’09C, ’10, ’11, ’12, ’13, ’13C, ’14, ’15, ’17], , Two tangents AP and AQ are drawn, from a point A to a circle with centre O., TO PROVE AP AQ., , GIVEN, , CONSTRUCTION, PROOF, , Join OP, OQ and OA., , AP is a tangent at P and OP is the radius through P., , , OP = AP., , Similarly, OQ = AQ., In the right AOPA and OQA, we have:, [radii of the same circle], OP OQ, OA OA, , [common], , , , 3OPA ,3OQA [by RHS-congruence]., Hence, AP AQ, [cpct]., REMARK, , The above theorem can also be proved by using the Pythagoras’, theorem:, AP 2 OA 2 OP 2 OA 2 OQ 2 AQ 2 [a OP OQ radius], , , THEOREM 4, , AP AQ., If two tangents are drawn from an external point then, (i) they subtend equal angles at the centre, and, (ii) they are equally inclined to the line segment joining the centre, to that point.
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464, , Secondary School Mathematics for Class 10, , A circle with centre O and a point A, outside it. Also, AP and AQ are the two, tangents to the circle., TO PROVE +AOP +AOQ and +OAP +OAQ., GIVEN, , PROOF, , In 3AOP and 3AOQ, we have, [tangents from an external point are equal], AP AQ, OP OQ, OA OA, , , REMARK, , [radii of the same circle], [common], , 3AOP ,3AOQ [by SSS-congruence]., , Hence, +AOP +AOQ and +OAP +OAQ., Since +OAP +OAQ, i.e., AO is the bisector of +PAQ, so the centre, lies on the bisector of the angle between the two tangents., , THEOREM 5, , Prove that in two concentric circles, the chord of the larger circle,, which touches the smaller circle, is bisected at the point of contact., [CBSE 2009, ’12], , Two circles with the same centre O and AB is a, chord of the larger circle touching the smaller, circle at C., TO PROVE AC BC., GIVEN, , CONSTRUCTION, PROOF, , Join OC., , AB is a tangent to the smaller circle at the point C and OC is the, radius through C., , , OC = AB., , But, the perpendicular drawn from the centre of a circle to a chord, bisects the chord., , , OC bisects AB, Hence, AC BC., , THEOREM 6, , GIVEN, , Prove that the tangents drawn at the ends of a diameter of a circle are, parallel., [CBSE 2011, ’12, ’14, ’17], , CD and EF are the tangents at the end points, A and B of the diameter AB of a circle with, centre O., , TO PROVE, PROOF, , [a AB is a chord of larger circle]., , CD EF., , CD is the tangent to the circle at the point A.
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Circles, , 465, , , CD = OA & +OAD 90c & +BAD 90c., EF is the tangent to the circle at the point B., , EF = OB & +OBE 90c & +ABE 90c., Thus, +BAD +ABE (each equal to 90)., But these are alternate interior angles., , THEOREM 7, , GIVEN, , CD EF., Prove that the line segment joining the points of contact of two, parallel tangents to a circle is a diameter of the circle., , CD and EF are two parallel tangents at the points A and B of a, circle with centre O., , TO PROVE, , AOB is a diameter of the circle., , CONSTRUCTION, PROOF, , Join OA and OB. Draw OG CD., , OG CD and AO cuts them., +CAO +GOA 180c, +90c +GOA 180c [a OA = CD], +GOA 90c., Similarly, +GOB 90c., , , +GOA +GOB 90c 90c 180c, , , , AOB is a straight line., , Hence, AOB is a diameter of the circle with centre O., THEOREM 8, , Prove that the angle between the two tangents drawn from an, external point to a circle is supplementary to the angle subtended by, the line segments joining the points of contact to the centre., [CBSE 2008C, ’14], , PA and PB are the tangents drawn from a point P to a circle with, centre O. Also, the line segments OA and OB are drawn., TO PROVE +APB +AOB 180c., GIVEN, , PROOF, , We know that the tangent to a circle is, perpendicular to the radius through, the point of contact., , PA = OA & +OAP 90c, and, , , PB = OB & +OBP 90c., +OAP +OBP 90c 90c 180c., , … (i), , But, we know that the sum of all the angles of a quadrilateral, is 360.
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466, , Secondary School Mathematics for Class 10, , , , +OAP +OBP +APB +AOB 360c., , … (ii), , From (i) and (ii), we get, +APB +AOB 180c., THEOREM 9, , PROOF, , Prove that there is one and only one tangent at any point on the, circumference of a circle., , Let P be a point on the circumference of a circle with centre O., If possible, let PT and PTl be two tangents at a point P of the circle., Now, the tangent at any point of a circle is perpendicular to the, radius through the point of contact., , , , OP = PT and similarly, OP = PTl, +OPT 90c and +OPTl 90c., , , , +OPT +OPTl, , This is possible only when PT and PTl coincide., Hence, there is one and only one tangent at any point on the, circumference of a circle., THEOREM 10, , PROOF, , Prove that the perpendicular at the point of contact of the tangent to, a circle passes through the centre., , Let PT be a tangent to a circle with centre O,, where P is the point of contact., Let PQ = PT, where Q lies on the circle, i.e.,, +QPT 90c., If possible, let PQ not pass through the, centre O., Join PO and produce it to meet the circle at R., Then PO being the radius through the point of contact, we have, , , , PO = PT [a the tangent is perpendicular to the radius, through the point of contact], +OPT 90c & +RPT 90c., , Thus, we have +QPT +RPT 90c., This is possible only if P, Q and R are collinear., But a straight line cuts a circle in at the most two points., So, the points Q and R coincide., Hence, PQ passes through the centre O, i.e., the perpendicular at, the point of contact to the tangent passes through the centre.
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Circles, , 467, , SOLVED EXAMPLES, EXAMPLE 1, , From a point P, 10 cm away from the centre of a circle, a tangent PT, of length 8 cm is drawn. Find the radius of the circle., [CBSE 2002], , SOLUTION, , Let O be the centre of the given, circle., Then, OP 10 cm. Also, PT 8 cm., Join OT., Now, PT is a tangent at T and OT is the radius through the, point of contact T., , , OT = PT., , In the right 3OTP, we have, OP 2 OT 2 PT 2, , [by Pythagoras’ theorem], , OT OP PT (10) 2 (8) 2 cm 36 cm 6 cm., 2, , 2, , Hence, the radius of the circle is 6 cm., EXAMPLE 2, , SOLUTION, , A tangent PQ at a point P of a circle of radius 5 cm meets a line, through the centre O at a point Q so that OQ 13 cm. Find the, length of PQ., [CBSE 2010], We have OP radius = 5 cm, OQ 13 cm., PQ is a tangent at P and OP is the, radius through the point of contact P., OP = PQ., In right 3 OPQ, we have, OQ 2 OP 2 PQ 2, , , EXAMPLE 3, , SOLUTION, , [by Pythagoras’ theorem], , PQ OQ OP 13 2 5 2 cm 144 cm 12 cm., 2, , 2, , In the given figure, AB is the, diameter of a circle with centre O and, AT is a tangent. If +AOQ 58c,, find +ATQ., [CBSE 2015], +AOQ 58c, +ABQ , , 1, +AOQ 29c, 2, [a the angle subtended by an arc at the centre is, double the angle subtended by it at any point, on the remaining part of the circle]
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468, , Secondary School Mathematics for Class 10, , +ABT 29c., Now, AT is a tangent at A and OA is the radius through the, point of contact A., , , OA = AT, i.e., +OAT 90c & +BAT 90c., , In 3BAT, we have, +BAT +ABT +ATB 180c, , , 90c 29c +ATB 180c & +ATB 61c., , +ATQ +ATB 61c., EXAMPLE 4, , Tangents PA and PB are drawn from an, external point P to two concentric circles, with centre O and radii 8 cm and 5 cm, respectively, as shown in the figure. If, AP 15 cm then find the length of BP., , SOLUTION, , We have, OA = AP and OB = BP, [a the tangent at any point of a circle is, perpendicular to the radius through the, point of contact]., Join OP., , [CBSE 2012], , In right 3OAP, we have, OA 8 cm, AP 15 cm., , , OP 2 OA 2 AP 2, [by Pythagoras’ theorem], , OP OA 2 AP 2 8 2 15 2 cm 289 cm 17 cm., In right 3OBP, we have, OB 5 cm, OP 17 cm., OP 2 OB 2 BP 2 [by Pythagoras’ theorem], BP OP 2 OB 2 17 2 5 2 cm 264 cm., Thus, the length of BP 264 cm 16.25 cm (approx)., , , , EXAMPLE 5, , In the given figure, two circles touch, each other at the point C. Prove that, the common tangent to the circles at C,, bisects the common tangent at P and Q., [CBSE 2013]
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Circles, SOLUTION, , 469, , In the given figure, PR and CR are both tangents drawn to the, same circle from an external point R., , , PR CR., , … (i), , Also, QR and CR are both tangents drawn to the same circle, (second circle) from an external point R., , , QR CR., , … (ii), , From (i) and (ii), we get, PR QR [each equal to CR]., , , R is the midpoint of PQ,, , i.e., the common tangent to the circles at C, bisects the common, tangent at P and Q., EXAMPLE 6, , Two concentric circles of radii a and b (a b) are given. Find the, length of the chord of the larger circle which touches the smaller, circle., [CBSE 2015], , SOLUTION, , Let O be the common centre of the two, circles and AB be the chord of the larger, circle which touches the smaller circle at C., Join OA and OC. Then, OA a and OC b., Now, OC = AB and OC bisects AB, [a the chord of the larger circle touching the smaller, circle, is bisected at the point of contact]., In right 3ACO, we have, OA 2 OC 2 AC 2, , [by Pythagoras’ theorem], , , , AC OA OC a 2 b 2 ., , , , AB 2AC 2 a 2 b 2, , 2, , 2, , [a C is the midpoint of AB], , i.e., required length of the chord AB 2 a 2 b 2 ., EXAMPLE 7, , Two concentric circles are of radii 7 cm and r cm respectively,, where r 7. A chord of the larger circle of length 46 cm, touches, the smaller circle. Find the value of r., [CBSE 2011], , SOLUTION, , Let O be the common centre of the two, circles and AB be the chord of the larger, circle which touches the smaller circle at C., Then, AB 46 cm.
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470, , Secondary School Mathematics for Class 10, , Join OA and OC. Then, OA r cm and OC 7 cm., Now, OC = AB and OC bisects AB., , [See Theorem 5.], , In right 3ACO, we have, OA 2 OC 2 AC 2, , [by Pythagoras’ theorem], , OA OC 2 AC 2 , , 2, 1, OC 2 a ABk, 2, [a C is the midpoint of AB], , r cm 7 2 23 2 cm 578 cm 17 2 & r 17 2 cm., EXAMPLE 8, , SOLUTION, , In the given figure, the radii of two concentric, circles are 13 cm and 8 cm. AB is a diameter, of the bigger circle and BD is a tangent to the, smaller circle touching it at D. Find the length, of AD., [HOTS] [CBSE 2014], We have +AEB 90c [angle in a semicircle]., Also, OD = BE and OD bisects BE., , [See Theorem 5.], , In right 3OBD, we have, OB 2 OD 2 BD 2, , [by Pythagoras’ theorem], , BD OB 2 OD 2 13 2 8 2 cm, 105 cm, [a OB 13 cm, OD 8 cm]., BE 2BD 2 105 cm, In right 3AEB, we have, AB 2 AE 2 BE 2, , , [a D is the midpoint of BE]., , [by Pythagoras’ theorem], , AE AB BE 26 2 (2 105 ) 2 cm 256 cm 16 cm, 2, , 2, , [a AB diameter 2 #OB 2 #13 cm 26 cm] ., In right 3AED, we have, AD 2 AE 2 DE 2, , , [by Pythagoras’ theorem], , AD AE DE 16 2 ( 105 ) 2 cm, 2, , 2, , 19 cm [a DE BD 105 cm] ., [Note We can also find AE by using midpoint theorem, since, in 3ABE, O is the midpoint of AB and D is the midpoint of BE, and so OD AE and AE 2 #OD 16 cm.], EXAMPLE 9, , From a point P outside a circle with centre O, tangents PA and PB, are drawn to the circle. Prove that OP is the right bisector of the line, segment AB., [CBSE 2015]
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Circles, SOLUTION, , 471, , GIVEN PA and PB are tangents to a circle with centre O from, an external point P., TO PROVE, , OP is the right bisector, , of AB., Join AB. Let AB, intersect OP at M., CONSTRUCTION, , In 3MAP and 3MBP, we have, PA PB [a tangents to a circle from an external point, are equal], , MP MP [common], , PROOF, , +MPA +MPB [a tangents from an external point are, equally inclined to the line segment, joining the centre to that point, i.e.,, +OPA +OPB], 3MAP ,3MBP, [by SAS-congruence]., And so, MA MB, [cpct], and +AMP +BMP, But, +AMP +BMP 180c, , [cpct]., [linear pair], , +AMP +BMP 90c., Hence, OP is the right bisector of AB., EXAMPLE 10, , Prove that the tangents at the extremities of any chord of a circle,, make equal angles with the chord., [CBSE 2014, ’17], , SOLUTION, , AB is any chord of a circle with, centre O. Tangents at the extremities A, and B of this chord meet at an external, point P. Chord AB intersects the line, segment OP at M., GIVEN, , TO PROVE, PROOF, , +MAP +MBP., , In 3MAP and 3MBP, we have, , PA PB [a tangents from an external point are equal], MP MP [common], +MPA +MPB [a +OPA +OPB since tangents from, an external point are equally, inclined to the line segment joining, the point to the centre]
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472, , Secondary School Mathematics for Class 10, , 3MAP , 3MBP, [by SAS-congruence]., And so, +MAP +MBP [cpct]., EXAMPLE 11, , SOLUTION, , Prove that the tangent drawn at the midpoint of an arc of a circle is, parallel to the chord joining the end points of the arc., [CBSE 2015], %, GIVEN Point P is the midpoint of arc QR of a circle with centre, O. ST is the tangent to the circle at point P., TO PROVE, , Chord QR ST., , %, P is the midpoint of QR, % %, QP PR, chord QP chord PR, [a in a circle, if two arcs are equal, then their, corresponding chords are equal], PROOF, , +PQR +PRQ, +TPR +PRQ, [Note +PQR +TPR, angles in alternate segments], QR ST. [a +TPR and +PRQ are alternate int. O], EXAMPLE 12, , In the given figure, AB is a chord, of length 9.6 cm of a circle with, centre O and radius 6 cm. The, tangents at A and B intersect at, P. Find the length of PA., [CBSE 2009C, ’13C, ’15], , SOLUTION, , A circle with centre O and radius 6 cm. AB is a chord of, length 9.6 cm. The tangents at A and B intersect at P., GIVEN, , TO FIND, , The length PA., , CONSTRUCTION, , Join OP and OA. Let OP and AB intersect at M., , Let PA x cm and PM y cm., Now, PA PB [a tangents from an external point are equal], and OP is the bisector of +APB, [a two tangents to a circle from an external point, are equally inclined to the line segment, joining the centre to that point]., Also, OP = AB and OP bisects AB at M, [a OP is the right bisector of AB].
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Circles, , , , AM MB , , 473, , 9.6, cm 4.8 cm., 2, , In right 3AMO, we have OA 6 cm and AM 4.8 cm., , , OM OA 2 AM 2 6 2 4.8 2 12.96 3.6 cm., , In right 3AMP, we have, AP 2 PM 2 AM 2 & x 2 y 2 (4.8) 2, , , x 2 y 2 23.04., , … (i), , In right 3PAO, we have, OP 2 PA 2 OA 2, , [Note +PAO 90c, since AO is the, radius at the point of contact], , (y 3.6) 2 x 2 6 2, , [a OP PM MO (y 3.6) cm], , , , y 7.2y 12.96 x 36 & 7.2y 46.08, , , , y 6.4 cm., , 2, , 2, , [using (i)], , Putting this value of y in (i), we get, x 2 (6.4) 2 23.04 40.96 23.04 64 x 64 8., , , PA 8 cm., , EXAMPLE 13, , Two tangents PA and PB are, drawn to a circle with centre O, from an external point P. Prove that, +APB 2+OAB. [CBSE 2009, ’14], , SOLUTION, , GIVEN, , A circle with centre O, and PA, PB are the tangents on, it from a point P outside it., , TO PROVE, PROOF, , +APB 2+OAB., , Let +APB xc., , We know that the tangents to a circle from an external point, are equal. So, PA PB., Since the angles opposite to the equal sides of a triangle are, equal, so, PA PB & +PBA +PAB., Also, the sum of the angles of a triangle is 180., +APB +PAB +PBA 180c, , , xc 2+PAB 180c, , [a +PBA +PAB]
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Circles, SOLUTION, , 475, , We know that the lengths of the tangents drawn from an, external point to a circle are equal., Let PL PN x; QL QM y; RM RN z., Now, PL QL PQ & x y 10,, , … (i), , QM RM QR & y z 8,, , … (ii), , RN PN PR & z x 12., , … (iii), , Subtracting (ii) from (iii), we get x y 4., , … (iv), , Solving (i) and (iv), we get x 7, y 3., Substituting y 3 in (ii), we get z 5., , , QM y 3 cm, RN z 5 cm, PL x 7 cm., , EXAMPLE 16, , A circle is inscribed in a 3ABC, touching, BC, CA and AB at P, Q and R respectively,, as shown in the given figure. If AB 10 cm,, AQ 7 cm and CQ 5 cm then find the, length of BC., [CBSE 2009C], , SOLUTION, , We know that the lengths of tangents drawn from an external, point to a circle are equal., , , AR AQ 7 cm., BR (AB AR) (10 7) cm 3 cm., , , , BP BR 3 cm,, CP CQ 5 cm., , , EXAMPLE 17, , BC (BP CP) (3 5) cm 8 cm., , A circle is touching the side BC of 3ABC, at P and touching AB and AC produced, at Q and R respectively. Prove that, 1, AQ (perimeter of 3ABC)., 2, [CBSE 2001, ’02, ’06, ’14, ’17], , SOLUTION, , We know that the lengths of tangents drawn from an external, point to a circle are equal., , , AQ AR,, , … (i), , [tangents from A]
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476, , Secondary School Mathematics for Class 10, , BP BQ,, , … (ii) [tangents from B], , CP CR., , … (iii) [tangents from C], , Perimeter of 3ABC, AB BC AC, AB BP CP AC, AB BQ CR AC [using (ii) and (iii)], AQ AR, 2AQ [using (i)]., , , AQ , , 1, (perimeter of 3ABC)., 2, , EXAMPLE 18, , PA and PB are tangents to the circle with centre O from an external, point P, touching the circle at A and B respectively. Show that the, quadrilateral AOBP is cyclic., [CBSE 2014], , SOLUTION, , GIVEN PA and PB are tangents to, the circle with centre O from an, external point P., TO PROVE, , Quadrilateral AOBP is, , cyclic., We know that the tangent at any point of a circle is, perpendicular to radius through the point of contact., PROOF, , , , PA = OA, i.e., +OAP 90c, , … (i), , and PB = OB, i.e., +OBP 90c, , … (ii), , Now, the sum of all the angles of a quadrilateral is 360., +AOB +OAP +APB +OBP 360c, +AOB +APB 180c [using (i) and (ii)], , , EXAMPLE 19, , quadrilateral OAPB is cyclic, , In the given figure, tangents, PQ and PR are drawn from, an external point P to a circle, with centre O, such that, +RPQ 30c. A chord RS is, drawn parallel to the tangent, PQ. Find +RQS., , [since both pairs of opposite, angles have the sum 180.].
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478, , Secondary School Mathematics for Class 10, , (b) Join OA, OB and OC., Area (3ABC) area (3OAB), area (3OBC), area (3OCA), a1 # AB#OPk, 2, a1 # BC #OQk, 2, 1, a #CA#OR)k, 2, 1, a # AB# rk a1 # BC # rk a1 #CA# rk, 2, 2, 2, 1, (AB BC CA)# r, 2, 1, (perimeter of 3ABC)# r., 2, EXAMPLE 21, , In the given figure, ABC is a right-angled, triangle with AB 6 cm and AC 8 cm., A circle with centre O has been inscribed, inside the triangle. Calculate the value of, r, the radius of the inscribed circle., [CBSE 2006C, ’13], , SOLUTION, , Join OA, OB and OC., Draw OD = AB, OE = BC and OF = CA., Then, OD OE OF r cm., , , ar (3ABC) , , 1, 1, # AB# AC a # 6 # 8k cm 2 24 cm 2 ., 2, 2, , Now, ar (3ABC) , , 1, #(perimeter of 3ABC)# r, 2, , , , 24 , , 1, #(AB BC CA)# r, 2, , , , 24 , , 1, #(6 10 8)# r & r 2, 2, [a BC 2 AB 2 AC 2 & BC 6 2 8 2 10] ., , Hence, the radius of the inscribed circle is 2 cm.
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Circles, , 479, , EXAMPLE 22, , A triangle ABC is drawn to, circumscribe a circle of radius, 4 cm such that the segments BD, and DC into which BC is divided, by the point of contact D are of, lengths 6 cm and 8 cm respectively., Find the lengths of the sides AB, and AC., [HOTS] [CBSE 2014], , SOLUTION, , We know that the lengths of tangents drawn from an exterior, point to a circle are equal., AE AF x cm (say), BD BF 6 cm, CD CE 8 cm., And so, AB AF BF (x 6) cm, BC BD CD 14 cm,, AC CE AE (x 8) cm., Perimeter, 2s AB BC AC [(x 6) 14 (x 8)] cm, (2x 28) cm, s (x 14) cm., , , ar(3ABC) s(s AB)(s BC)(s AC), , (x 14){(x 14) (x 6)}{(x 14) 14}{(x 14) (x 8)} cm 2, 48x (x 14) cm 2., Join OE and OF and also OA, OB and OC., ar (3ABC) ar (3OAB) ar (3OBC), ar (3OCA), , … (i), , a1 # AB#OFk, 2, a1 # BC #ODk, 2, 1, a # AC #OEk, 2, 1, : #(x 6)# 4D :1 #14 # 4D :1 #(x 8)# 4D, 2, 2, 2, 2 [(x 6) 14 (x 8)] 4 (x 14) cm 2., … (ii), From (i) and (ii), we get, 48x (x 14) 4(x 14), 48x (x 14) 16 (x 14) 2 [on squaring both sides], 16 #14 , 48x 16 (x 14) & x , 7., 32, AB (x 6) cm (7 6) cm 13 cm, and AC (x 8) cm (7 8) cm 15 cm.,
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480, EXAMPLE 23, , Secondary School Mathematics for Class 10, , In the given figure, a triangle, ABC is drawn to circumscribe a, circle of radius 3 cm, such that, the segments BD and DC into, which BC is divided by the point of, contact D are of lengths 6 cm and, 8 cm respectively. Find the side, AB, if the area of 3ABC is 63 cm 2 ., [CBSE 2010, ’11, ’15], , SOLUTION, , We know that the lengths of, tangents drawn from an exterior, point to a circle are equal., AE AF x cm (say);, BD BF 6 cm;, CD CE 8 cm., And so, AB AF BF (x 6) cm; BC BD CD 14 cm;, CA CE AE (x 8) cm., , , , Join OE and OF and also OA, OB and OC., ar (3ABC) ar (3OAB) ar (3OBC) ar (3OCA), 1, 1, 1, 63 a # AB#OFk a # BC #ODk a #CA#OEk, 2, 2, 2, 1, 1, 1, 63 & #(x 6)# 30 a #14 # 3k & #(x 8)# 30, 2, 2, 2, 3, 63 #(2x 28) & x 7., 2, AB (x 6) cm (7 6) cm 13 cm., , , , EXAMPLE 24, , In the given figure, XP and XQ, are two tangents to the circle with, centre O, drawn from an external, point X. ARB is another tangent,, touching the circle at R. Prove that, [CBSE 2014], XA AR XB BR., , SOLUTION, , We know that the lengths of tangents drawn from an exterior, point to a circle are equal., , , XP XQ,, AP AR,, BR BQ ., , … (i), … (ii), … (iii), , [tangents from X], [tangents from A], [tangents from B]
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Circles, , 481, , Now, XP XQ & XA AP XB BQ, , EXAMPLE 25, , XA AR XB BR [using (ii) and (iii)]., , If from an external point P of a circle with centre O, two tangents PQ, and PR are drawn such that +QPR 120c, prove that 2PQ PO., [HOTS] [CBSE 2014], , SOLUTION, , In 3OPQ, we have, +PQO 90c [a the tangent at any point, is perpendicular to the, radius through the point, of contact], and +QPO , , 1, 1, #+QPR #120c 60c., 2, 2, [a the two tangents drawn from an external, point are equally inclined to the line segment, joining the centre to that point and so, +QPO +RPO], , In right 3OPQ, we have, PQ, cos (+QPO) , PO, PQ, PQ, cos 60c , & 12 PO & 2PQ PO., PO, EXAMPLE 26, , A quadrilateral ABCD is drawn to, circumscribe a circle, as shown in the, figure. Prove that AB CD AD BC., [CBSE 2008, ’08C, ’09, ’12, ’13, ’14, ’17], , SOLUTION, , We know that the lengths of tangents, drawn from an exterior point to a, circle are equal., … (i) [tangents from A], AP AS,, … (ii) [tangents from B], BP BQ,, … (iii) [tangents from C], CR CQ,, … (iv) [tangents from D], DR DS., AB CD (AP BP) (CR DR), (AS BQ) (CQ DS) [using (i), (ii), (iii), (iv)], (AS DS) (BQ CQ), AD BC., Hence, AB CD AD BC., ,
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482, EXAMPLE 27, , SOLUTION, , Secondary School Mathematics for Class 10, , In the given figure, ABCD is a quadrilateral, such that +D 90c. A circle with centre O, and radius r, touches the sides AB, BC, CD, and DA at P, Q, R and S respectively. If, BC 40 cm, CD 25 cm and BP 28 cm,, find r., It is given that +D 90c., Also,+ORD +OSD 90c. [a tangent at a point is perpendicular to the radius through, the point of contact], +ROS 180c +D 90c. [a angle between the tangents, from an external point is, supplementary to the angle, subtended by the line, segments joining the points, of contact to the centre], And, OR OS r., , , ROSD is a square and so OR DR, i.e., r DR., , … (i), , We know that the lengths of tangents drawn from an exterior, point to a circle are equal., , , BP BQ and CQ CR., , Now, CQ BC BQ BC BP (40 28) cm 12 cm., And so, DR CD CR CD CQ (25 12) cm 13 cm., , EXAMPLE 28, , r DR 13 cm, , [using (i)]., , Prove that the parallelogram circumscribing a circle is a rhombus., [CBSE 2008, ’09, ’10, ’12, ’13, ’14], , SOLUTION, , A parallelogram ABCD circumscribes a circle with centre O., TO PROVE AB BC CD AD., PROOF We know that the lengths of, tangents drawn from an exterior point, to a circle are equal., AP AS,, … (i) [tangents from A], … (ii) [tangents from B], BP BQ,, … (iii) [tangents from C], CR CQ,, , … (iv) [tangents from D], DR DS., GIVEN
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Circles, , , , 483, , AB CD AP BP CR DR, AS BQ CQ DS [from (i), (ii), (iii) and (iv)], (AS DS) (BQ CQ), AD BC., , Thus, AB CD AD BC, , , 2AB 2AD [a opposite sides of a gm are equal], , , , AB AD., , , , CD AB AD BC., , Hence, ABCD is a rhombus., EXAMPLE 29, , Prove that the opposite sides of a quadrilateral circumscribing a, circle subtend supplementary angles at the centre of the circle., [CBSE 2012, ’13C, ’14, ’17], , SOLUTION, , GIVEN A quad. ABCD circumscribes a, circle with centre O., TO PROVE, , +AOB +COD 180c,, , and +AOD +BOC 180c., CONSTRUCTION, , Join OP, OQ, OR and OS., , We know that the tangents drawn from an external, point of a circle subtend equal angles at the centre., PROOF, , +1 +2, +3 +4, +5 +6 and +7 +8., And, +1 +2 +3 +4 +5 +6 +7 +8 360c, [O at a point], 2(+2 +3) 2(+6 +7) 360c, and, 2(+1 +8) 2(+4 +5) 360c, +2 +3 +6 +7 180c and +1 +8 +4 +5 180c, +AOB +COD 180c and +AOD +BOC 180c., EXAMPLE 30, , In the given figure, PA is a tangent, from an external point P to a circle, with centre O. If +POB 115c, find, [CBSE 2009C], +APO., , SOLUTION, , We know that the tangent at a, point to a circle is perpendicular, to the radius passing through the, point of contact.
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Circles, SOLUTION, , 485, , In quad. APQB, we have, +APO 90c and +BQO 90c, [a tangent at any point is perpendicular to the, radius through the point of contact]., Now, +APO +BQO +QBC +PAC 360c, +PAC +QBC 360c (+APO +BQO) 180c., , … (i), , We have, +CAO , , 1, 1, +PAC and +CBO +QBC, 2, 2, [a tangents from an external point are equally, inclined to the line segment joining the centre, to that point]., , +CAO +CBO , , 1, 1, (+PAC +QBC) #180c 90c. … (ii), 2, 2, , In 3AOB, we have, +CAO +AOB +CBO 180c, +AOB 180c (+CAO +CBO) 90c, EXAMPLE 33, , SOLUTION, , f, , [using (ii)]., , The incircle of an isosceles triangle ABC, with AB AC, touches, the sides AB, BC, CA at D, E and F respectively. Prove that E, bisects BC., [CBSE 2008, ’12, ’13C, ’14], We know that the tangents drawn from, an external point to a circle are equal., AD AF, … (i) [tangents from A], BD BE, … (ii) [tangents from B], CE CF. … (iii) [tangents from C], [given], Now, AB AC, AD BD AF CF, BD CF, BE CE, [using (ii) and (iii)], E bisects BC., , EXERCISE 8A, , 1. A point P is at a distance of 29 cm from the centre of a circle of radius, 20 cm. Find the length of the tangent drawn from P to the circle., [CBSE 2017]
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486, , Secondary School Mathematics for Class 10, , 2. A point P is 25 cm away from the centre of a circle and the length, of tangent drawn from P to the circle is 24 cm. Find the radius of, the circle., 3. Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length, of the chord of the larger circle which touches the smaller circle., [CBSE 2011], , 4. In the given figure, a circle inscribed, in a triangle ABC, touches the sides, AB, BC and AC at points D, E and F, respectively. If AB 12 cm, BC 8 cm, and AC 10 cm, find the lengths of AD,, BE and CF., [CBSE 2013], 5. In the given figure, PA and PB are the, tangent segments to a circle with centre, O. Show that the points A, O, B and P, are concyclic., , 6. In the given figure, the chord AB of, the larger of the two concentric circles,, with centre O, touches the smaller, circle at C. Prove that AC CB., , 7. From an external point P, tangents, PA and PB are drawn to a circle with, centre O. If CD is the tangent to the, circle at a point E and PA 14 cm, find, the perimeter of 3PCD., [CBSE 2002], 8. A circle is inscribed in a 3ABC, touching AB, BC and AC at P, Q and R, respectively. If AB 10 cm, AR 7 cm, and CR 5 cm, find the length of BC., [CBSE 2009C]
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Circles, , 487, , 9. In the given figure, a circle touches all the, four sides of a quadrilateral ABCD whose three, sides are AB 6 cm, BC 7 cm and CD 4 cm., Find AD., , 10. In the given figure, an isosceles triangle ABC,, with AB AC, circumscribes a circle. Prove that, the point of contact P bisects the base BC., [CBSE 2012, ’17], , 11. In the given figure, O is the centre of, two concentric circles of radii 4 cm, and 6 cm respectively. PA and PB are, tangents to the outer and inner circle, respectively. If PA 10 cm, find the, length of PB up to one place of decimal., 12. In the given figure, a triangle ABC, is drawn to circumscribe a circle of, radius 3 cm such that the segments, BD and DC into which BC is divided, by the point of contact D, are of, lengths 6 cm and 9 cm respectively. If, the area of 3ABC 54 cm 2 then find, the lengths of sides AB and AC., [CBSE 2011, ’15], , 13. PQ is a chord of length 4.8 cm of a, circle of radius 3 cm. The tangents at P, and Q intersect at a point T as shown, in the figure. Find the length of TP., [CBSE 2013C], , 14. Prove that the line segment joining the points of contact of two parallel, tangents of a circle, passes through its centre., [CBSE 2014]
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488, , Secondary School Mathematics for Class 10, , 15. In the given figure, a circle with centre O,, is inscribed in a quadrilateral ABCD such, that it touches the side BC, AB, AD and, CD at points P, Q, R and S respectively., If AB 29 cm, AD 23 cm, +B 90c and, DS 5 cm then find the radius of the, circle., [CBSE 2008, ’13], 16. In the given figure, O is the centre of the, circle and TP is the tangent to the circle from, an external point T. If +PBT 30c, prove that, [HOTS] [CBSE 2015], BA : AT 2 : 1., , ANSWERS (EXERCISE 8A), , 1. 21 cm, , 2. 7 cm, , 3. 12 cm, , 4. AD 7 cm, BE 5 cm, CF 3 cm, 9. AD 3 cm 11. 10.9 cm, 13. TP 4 cm, , 7. 28 cm, , 12. AB 10 cm, AC 12 cm, , 15. 11 cm, HINTS TO SOME SELECTED QUESTIONS, , 4. x y 12, y z 8, z x 10., Solving, we get x 7, y 5, z 3., , , 8. BC 8 cm, , AD x 7 cm, BE y 5 cm, CF z 3 cm., , 5. OA = AP and OB = BP, , , +OAP 90c and +OBP 90c, , , , +OAP +OBP 180c, , , , quad. AOBP is cyclic, , [a a quad. is cyclic if the, sum of a pair of opposite, angles is 180].
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Circles, , 489, , 7. PA PB, CA CE, DB DE, [£ tangents from an external point are equal]., Perimeter of 3PCD PC CD PD, PC CE DE PD, PC CA DB PD, PA PB 2PA [a PB PA], (2 #14) cm 28 cm., 8. We know that the tangents to a circle from an exterior, point are equal., , , AP AR 7 cm;, BQ BP AB AP (10 7) cm 3 cm;, CQ CR 5 cm., , And so, BC BQ CQ (3 5) cm 8 cm., 9. When a qudrilateral ABCD is drawn to circumscribe a circle then AB CD AD BC., , , AD AB CD BC (6 4 7) cm 3 cm., , 10. We know that the tangents to a circle from an external point, are equal., , , AQ AR, BP BR, CP CQ., , Now, AB AC & AR BR AQ CQ, , , AR BP AR CP & BP CP, , , , P bisects the base BC., , 1, 13. R is the midpoint of PQ, i.e., PR QR a # 4.8k cm 2.4 cm., 2, Also, TO = PQ., In right 3PRO, we have, PO 2 PR 2 RO 2 & RO PO 2 PR 2, Let TR x and TP y., , 3 2 2.4 2 3.24 1.8., , In right 3PTR, we have TP 2 TR 2 PR 2 & y 2 x 2 (2.4) 2., , … (i), , In right 3OTP, we have, TO 2 TP 2 PO 2 & (TR RO) 2 TP 2 PO 2, , , (x 1.8) 2 y 2 3 2 & x 2 3.6x 3.24 y 2 9., , Solving (i) and (ii), we get x 3.2, y 4., , , TP 4 cm., , 14. Let APB and CQD be two parallel tangents to a circle, with centre O., Join OP and OQ. Draw RO AB., Now, +APO +ROP 180c [co-interior angles], , , +ROP 90c, , [a +APO 90c]., , … (ii)
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Circles, , 491, , 3. In the given figure, O is the centre of a circle., PT and PQ are tangents to the circle from an, external point P. If +TPQ 70c, find +TRQ., [CBSE 2015], , 4. In the given figure, common, tangents AB and CD to the two, circles with centres O1 and O2, intersect at E. Prove that, [CBSE 2014], AB CD., 5. If PT is a tangent to a circle with centre O, and PQ is a chord of the circle such that, +QPT 70c, then find the measure of +POQ., , Short-Answer Questions, 6. In the given figure, a triangle ABC is drawn, to circumscribe a circle of radius 2 cm such, that the segments BD and DC into which, BC is divided by the point of contact D, are, of lengths 4 cm and 3 cm respectively. If the, area of 3ABC 21 cm 2 then find the lengths of, sides AB and AC., [CBSE 2011], 7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the, chord of the larger circle (in cm) which touches the smaller circle., [CBSE 2012, ’14], , 8. Prove that the perpendicular at the point of contact of the tangent to a, circle passes through the centre., 9. In the given figure, two tangents RQ and RP, are drawn from an external point R to the circle, with centre O. If +PRQ 120c, then prove that, [CBSE 2015], OR PR RQ.
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492, , Secondary School Mathematics for Class 10, , 10. In the given figure, a circle inscribed in a triangle, ABC touches the sides AB, BC and CA at points D,, E and F respectively. If AB 14 cm, BC 8 cm and, CA 12 cm. Find the lengths AD, BE and CF., [CBSE 2013], , 11. In the given figure, O is the centre of the, circle. PA and PB are tangents. Show that, AOBP is a cyclic quadrilateral. [CBSE 2014], , 12. In two concentric circles, a chord of length 8 cm of the larger circle, touches the smaller circle. If the radius of the larger circle is 5 cm then, [CBSE 2013C], find the radius of the smaller circle., 13. In the given figure, PQ is a chord of a, circle with centre O and PT is a tangent. If, +QPT 60c, find +PRQ. [HOTS] [CBSE 2015], , 14. In the given figure, PA and PB are two, tangents to the circle with centre O. If, +APB 60c then find the measure of +OAB., , 15. If the angle between two tangents drawn from an external point P to a, circle of radius a and centre O, is 60 then find the length of OP., [CBSE 2017], , ANSWERS (EXERCISE 8B), , 1. AD 5 cm, , 2. 25, , 6. AB 7.5 cm, AC 6.5 cm, 10. AD 9 cm, BE 5 cm, CF 3 cm, 14. 30, , 3. 55, , 5. 140c, , 7. 8 cm, 12. 3 cm, , 13. 120, , 15. a 3, HINTS TO SOME SELECTED QUESTIONS, , 1. When a quadrilateral ABCD circumscribes a circle then AB CD AD BC., , , AD AB CD BC (6 8 9) cm 5 cm.
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Circles, 2. +APB 2+OAB & +OAB , , 50c , 1, #+APB , 25c., 2, 2, , 493, [See Solved Example 13.], , 3. Join OT and OQ., Then, +TOQ +TPQ 180c & +TOQ 180c 70c 110c., [See Theorem 8.], Now, +TRQ , , 1, 1, #+TOQ #110c 55c, 2, 2, [a the angle subtended by an arc of a circle at the centre is double, the angle subtended by it at any point on the remaining part of, the circle]., , 4. We know that the tangents to a circle from an external point are equal., , , AE CE, BE DE, , , , AE BE CE DE & AB CD., , 5. Mark a point R in the alternate segment., Join RP and RQ., Then, +PRQ +QPT 70c, [angles in the alternate segments]., , , +POQ 2+PRQ 2 #70c 140c., [a angle subtended by an arc on the centre is double the angle, subtended at any point on the remaining part of the circle.], , 6. Mark the points of contact E and F of tangents AC and, AB respectively., We know that the tangents to a circle from an external, point are equal., , , AE AF x cm (say);, BD BF 4 cm; CD CE 3 cm., , Now, ar (3ABC) , , , , 1, #(perimeter of 3ABC)# r, 2, , 1, #{(x 4) (4 3) (3 x)}# 2 & x 3.5., 2, , AB (x 4) cm 7.5 cm; AC (x 3) cm 6.5 cm., , 21 , , 7. Let AB be the chord, P be the point where it touches the, smaller circle and O be the common centre., AB is tangent to smaller circle and so, OP = AB., , , P is the midpoint of AB, i.e., AP PB, [a a perpendicular from the centre on any chord, bisects the chord]., , Now, OP 3 cm, OA 5 cm & AP OA 2 OP 2 5 2 3 2 4., , , AB 2AP 8 cm.
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494, , Secondary School Mathematics for Class 10, , 9. Join OP and OQ., 1, #+PRQ 60c., 2, [a two tangents from an external point are, , +PRQ 120c & +PRO +QRO , , equally inclined to the line segment joining, the centre to that point., And so, +PRO +QRO , , 1, +PRQ.], 2, , In right 3OPR, we have, cos +PRO , Similarly, RQ , , , PR, PR, 1 PR, 1, & cos 60c , & , & PR #OR., 2 OR, 2, OR, OR, , 1, #OR., 2, , PR RQ OR., , 10. x y 14, y z 8, z x 12., Solving, x 9, y 5, z 3., , , AD x 9 cm, BE y 5 cm, CF z 3 cm., , 12. Let AB be the chord, P be the point where it touches the, smaller circle and O be the common centre., Then, AB is the tangent to smaller circle at P., , , OP = AB., , And so, P is the midpoint of AB,, i.e., AP PB , , 1, 1, AB a # 8k cm 4 cm, 2, 2, [a perpendicular from the centre on any chord bisects the chord]., , Now, OA 5 cm, AP 4 cm & OP OA 2 AP 2 5 2 4 2 3., , , radius of smaller circle OP 3 cm., , 13. Mark a point M in the alternate segment., Join MP and MQ., Then, +PMQ +QPT 60c, [angles in the alternate segments]., Now, +PMQ +PRQ 180c, [a PMQR is a cyclic quadrilateral], , , +PRQ 180c +PMQ 180c 60c 120c.
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Circles, , 495, , 14. PA PB [a tangents from an external point are equal], 1, 1, +PAB +PBA #(180c +APB) #(180c 60c) 60c., 2, 2, Now, +PAO 90c, [a radius through the point of contact is perpendicular, to the tangent], , , +OAB +PAB 90c & +OAB 90c +PAB 90c 60c 30c., , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. The number of tangents that can be drawn from an external point to a, circle is, [CBSE 2011, ’12], (a) 1, , (b) 2, , (c) 3, , (d) 4, , 2. In the given figure, RQ is a tangent to the circle with, centre O. If SQ 6 cm and QR 4 cm, then OR is, equal to, [CBSE 2014], (a) 2.5 cm, , (b) 3 cm, , (c) 5 cm, , (d) 8 cm, , 3. In a circle of radius 7 cm, tangent PT is, drawn from a point P such that PT 24 cm., If O is the centre of the circle, then length, OP ?, (a) 30 cm, , (b) 28 cm, , (c) 25 cm, , (d) 18 cm, , 4. Which of the following pairs of lines in a circle cannot be parallel?, [CBSE 2011], , (a) Two chords, , (b) A chord and a tangent, , (c) Two tangents, , (d) Two diameters, , 5. The chord of a circle of radius 10 cm subtends a right angle at its centre., The length of the chord (in cm) is, [CBSE 2014], (a), , 5, 2, , (b) 5 2, , (c) 10 2, , 6. In the given figure, PT is a tangent to the circle, with centre O. If OT 6 cm and OP 10 cm, then, the length of tangent PT is, (a) 8 cm, , (b) 10 cm, , (c) 12 cm, , (d) 16 cm, , (d) 10 3
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496, , Secondary School Mathematics for Class 10, , 7. In the given figure, point P is 26 cm away from, the centre O of a circle and the length PT of the, tangent drawn from P to the circle is 24 cm., Then, the radius of the circle is, [CBSE 2011, ’12], (a) 10 cm, , (b) 12 cm, , (c) 13 cm, , (d) 15 cm, , 8. PQ is a tangent to a circle with centre O at the point P. If 3OPQ is an, isosceles triangle, then +OQP is equal to, [CBSE 2014], (a) 30, , (b) 45, , (c) 60, , (d) 90, , 9. In the given figure, AB and AC are tangents to, the circle with centre O such that +BAC 40c., Then, +BOC is equal to, [CBSE 2011, ’14], (a) 80, , (b) 100, , (c) 120, , (d) 140, , 10. If a chord AB subtends an angle of 60 at the centre of a circle, then the, angle between the tangents to the circle drawn from A and B is, [CBSE 2013C], , (a) 30, , (b) 60, , (c) 90, , 11. In the given figure, O is the centre of two, concentric circles of radii 6 cm and 10 cm. AB is a, chord of outer circle which touches the inner circle., The length of chord AB is, (a) 8 cm, , (b) 14 cm, , (c) 16 cm, , (d), , 136 cm, , 12. In the given figure, AB and AC are tangents, to a circle with centre O and radius 8 cm. If, OA 17 cm, then the length of AC (in cm) is, [CBSE 2012], , (a) 9, (c), , (b) 15, 353, , (d) 25, , 13. In the given figure, O is the centre of a circle,, AOC is its diameter such that +ACB 50c. If AT, is the tangent to the circle at the point A then, +BAT ?, (a) 40, , (b) 50, , (c) 60, , (d) 65, , (d) 120
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Circles, , 497, , 14. In the given figure, O is the centre of a circle, PQ is, a chord and PT is the tangent at P. If +POQ 70c,, then +TPQ is equal to, [CBSE 2011], (a) 35, , (b) 45, , (c) 55, , (d) 70, , 15. In the given figure, AT is a tangent to the, circle with centre O such that OT 4 cm and, +OTA 30c. Then, AT ?, (a) 4 cm, , (b) 2 cm, , (c) 2 3 cm, , (d) 4 3 cm, , 16. If PA and PB are two tangents to a circle with, centre O such that +AOB 110c then +APB is, equal to, [CBSE 2011, ’14], (a) 55, , (b) 60, , (c) 70, , (d) 90, , 17. In the given figure, the length of BC is, [CBSE 2012, ’14], , (a) 7 cm, (b) 10 cm, (c) 14 cm, (d) 15 cm, 18. In the given figure, if +AOD 135c then, [CBSE 2013C], +BOC is equal to, (a) 25, (c) 52.5, , , , (b) 45, (d) 62.5, , 19. In the given figure, O is the centre of a circle and, PT is the tangent to the circle. If PQ is a chord, such that +QPT 50c then +POQ ?, (a) 100, , (b) 90, , (c) 80, , (d) 75, , 20. In the given figure, PA and PB are two tangents, to the circle with centre O. If +APB 60c then, [CBSE 2011], +OAB is, (a) 15, , (b) 30, , (c) 60, , (d) 90
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498, , Secondary School Mathematics for Class 10, , 21. If two tangents inclined at an angle of 60 are drawn to a circle of radius, 3 cm then the length of each tangent is, (a) 3 cm, , (b), , 3 3, cm, 2, , (c) 3 3 cm, , (d) 6 cm, , 22. In the given figure, PQ and PR are tangents, to a circle with centre A. If +QPA 27c then, [CBSE 2012], +QAR equals, (a) 63, , (b) 117, , (c) 126, , (d) 153, , 23. In the given figure, PA and PB are two tangents, drawn from an external point P to a circle with, centre C and radius 4 cm. If PA = PB, then the, length of each tangent is, [CBSE 2013], (a) 3 cm, , (b) 4 cm, , (c) 5 cm, , (d) 6 cm, , 24. If PA and PB are two tangents to a circle, with centre O such that +APB 80c. Then,, +AOP ?, (a) 40, , (b) 50, , (c) 60, , (d) 70, , 25. In the given figure, O is the centre of, the circle. AB is the tangent to the circle, at the point P. If +APQ 58c then the, measure of +PQB is, [CBSE 2014], (a) 32, (b) 58, (c) 122, , (d) 132, , 26. In the given figure, O is the centre of the circle., AB is the tangent to the circle at the point P. If, +PAO 30c then +CPB +ACP is equal to, (a) 60, , (b) 90, , (c) 120, , (d) 150, , 27. In the given figure, PQ is a tangent to a circle, with centre O. A is the point of contact. If, +PAB 67c, then the measure of +AQB is, (a) 73, , (b) 64, , (c) 53, , (d) 44, ,
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Circles, , 499, , 28. In the given figure, two circles touch, each other at C and AB is a tangent, to both the circles. The measure of, [HOTS] [CBSE 2013C], +ACB is, (a) 45, , (b) 60, , (c) 90, , (d) 120, , 29. O is the centre of a circle of radius 5 cm. At a, distance of 13 cm from O, a point P is taken. From, this point, two tangents PQ and PR are drawn to, the circle. Then, the area of quad. PQOR is, (a) 60 cm2, , (b) 32.5 cm2, , (c) 65 cm2, , (d) 30 cm2, , 30. In the given figure, PQR is a tangent to the circle, at Q, whose centre is O and AB is a chord parallel, to PR such that +BQR 70c. Then, +AQB ?, (a) 20, , (b) 35, , (c) 40, , (d) 45, , 31. The length of the tangent from an external point P to a circle of radius, 5 cm is 10 cm. The distance of the point from the centre of the circle is, (a) 8 cm, (b) 104 cm, (c) 12 cm, (d) 125 cm, [CBSE 2013C], , 32. In the given figure, O is the centre of a circle,, BOA is its diameter and the tangent at the, point P meets BA extended at T. If +PBO 30c, then +PTA ?, (a) 60, , (b) 30, , (c) 15, , (d) 45, , 33. In the given figure, a circle touches the side DF, of 3EDF at H and touches ED and EF produced, at K and M respectively. If EK 9 cm then the, perimeter of 3EDF is, [CBSE 2012], (a) 9 cm, , (b) 12 cm, , (c) 13.5 cm, , (d) 18 cm, , 34. To draw a pair of tangents to a circle, which are inclined to each other, at an angle of 45, we have to draw tangents at the end points of those
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500, , Secondary School Mathematics for Class 10, , two radii, the angle between which is, (a) 105, , (b) 135, , [CBSE 2011], , (c) 140, , (d) 145, , 35. In the given figure, O is the centre of a, circle; PQL and PRM are the tangents at, the points Q and R respectively and S is, a point on the circle such that +SQL 50c, and +SRM 60c. Then, +QSR ?, (a) 40, , (b) 50, , (c) 60, , (d) 70, , 36. In the given figure, a triangle PQR is, drawn to circumscribe a circle of radius, 6 cm such that the segments QT and, TR into which QR is divided by the, point of contact T, are of lengths 12 cm, and 9 cm respectively. If the area of, 3PQR 189 cm 2 then the length of side, PQ is, [CBSE 2011], (a) 17.5 cm, , (b) 20 cm, , (c) 22.5 cm, , 37. In the given figure, QR is a common, tangent to the given circles, touching, externally at the point T. The tangent at, T meets QR at P. If PT 3.8 cm then the, length of QR is, [CBSE 2014], (a) 1.9 cm, , (b) 3.8 cm, , (c) 5.7 cm, , (d) 7.6 cm, , 38. In the given figure, quad. ABCD is circumscribed,, touching the circle at P, Q, R and S. If AP 5 cm,, BC 7 cm and CS 3 cm. Then, the length AB ?, (a) 9 cm, , (b) 10 cm, , (c) 12 cm, , (d) 8 cm, , 39. In the given figure, quad. ABCD is, circumscribed, touching the circle at P, Q, R, and S. If AP 6 cm, BP 5 cm, CQ 3 cm and, DR 4 cm then perimeter of quad. ABCD is, (a) 18 cm, , (b) 27 cm, , (c) 36 cm, , (d) 32 cm, , (d) 25 cm
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Circles, , 501, , 40. In the given figure, O is the centre of a circle,, AB is a chord and AT is the tangent at A. If, +AOB 100c then +BAT is equal to [CBSE 2011], (a) 40, , (b) 50, , (c) 90, , (d) 100, , 41. In a right triangle ABC, right-angled at B, BC 12 cm and AB 5 cm., The radius of the circle inscribed in the triangle is, [CBSE 2014], (a) 1 cm, , (b) 2 cm, , (c) 3 cm, , (d) 4 cm, , 42. In the given figure, a circle is inscribed in a, quadrilateral ABCD touching its sides AB,, BC, CD and AD at P, Q, R and S respectively., If the radius of the circle is 10 cm, BC 38 cm,, PB 27 cm and AD = CD then the length of, CD is, [CBSE 2013], (a) 11 cm, , (b) 15 cm, , (c) 20 cm, , (d) 21 cm, , 43. In the given figure, 3ABC is right-angled at B, such that BC 6 cm and AB 8 cm. A circle, with centre O has been inscribed inside the, triangle. OP = AB, OQ = BC and OR = AC. If, OP OQ OR x cm then x ?, (a) 2 cm, , (b) 2.5 cm, , (c) 3 cm, , (d) 3.5 cm, , 44. Quadrilateral ABCD is circumscribed to a circle. If AB 6 cm, BC 7 cm, and CD 4 cm then the length of AD is, [CBSE 2012], (a) 3 cm, , (b) 4 cm, , (c) 6 cm, , (d) 7 cm, , 45. In the given figure, PA and PB are tangents, to the given circle such that PA 5 cm and, +APB 60c. The length of chord AB is, (a) 5 2 cm, , (b) 5 cm, , (c) 5 3 cm, , (d) 7.5 cm
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502, , Secondary School Mathematics for Class 10, , 46. In the given figure, DE and DF are tangents from, an external point D to a circle with centre A. If, DE 5 cm and DE = DF then the radius of the, circle is, [CBSE 2013], (a) 3 cm, , (b) 4 cm, , (c) 5 cm, , (d) 6 cm, , 47. In the given figure, three circles with centres A, B,, C respectively touch each other externally. If AB 5, cm, BC 7 cm and CA 6 cm then the radius of the, circle with centre A is, (a) 1.5 cm, , (b) 2 cm, , (c) 2.5 cm, , (d) 3 cm, , 48. In the given figure, AP, AQ and BC are tangents to, the circle. If AB 5 cm, AC 6 cm and BC 4 cm, then the length of AP is, [CBSE 2012], (a) 15 cm, (b) 10 cm, (c) 9 cm, , (d) 7.5 cm, , 49. In the given figure, O is the centre of two, concentric circles of radii 5 cm and 3 cm. From, an external point P tangents PA and PB are, drawn to these circles. If PA 12 cm then PB, is equal to, (a) 5 2 cm, , (b) 3 5 cm, , (c) 4 10 cm, , (d) 5 10 cm, , True/False Type, , 50. Which of the following statements is not true?, (a) If a point P lies inside a circle, no tangent can be drawn to the, circle, passing through P., (b) If a point P lies on the circle, then one and only one tangent can be, drawn to the circle at P., (c) If a point P lies outside the circle, then only two tangents can be, drawn to the circle from P., (d) A circle can have more than two parallel tangents, parallel to a, given line.
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Circles, , 503, , 51. Which of the following statements is not true?, (a) A tangent to a circle intersects the circle exactly at one point., (b) The point common to the circle and its tangent is called the point, of contact., (c) The tangent at any point of a circle is perpendicular to the radius, of the circle through the point of contact., (d) A straight line can meet a circle at one point only., 52. Which of the following statements is not true?, (a) A line which intersects a circle in two points, is called a secant of, the circle., (b) A line intersecting a circle at one point only, is called a tangent to, the circle., (c) The point at which a line touches the circle, is called the point of, contact., (d) A tangent to the circle can be drawn from a point inside the circle., Assertion-and-Reason Type, , Each question consists of two statements, namely, Assertion (A) and, Reason (R). For selecting the correct answer, use the following code:, (a) Both Assertion (A) and Reason (R) are true and Reason (R) is a, correct explanation of Assertion (A)., (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a, correct explanation of Assertion (A)., (c) Assertion (A) is true and Reason (R) is false., (d) Assertion (A) is false and Reason (R) is true., 53., , Assertion (A), , Reason (R), , At a point P of a circle with centre The tangent at any point of a circle is, O and radius 12 cm, a tangent PQ perpendicular to the radius through, of length 16 cm is drawn. Then, the point of contact., , OQ 20 cm., , The correct answer is (a)/(b)/(c)/(d)., 54., , Assertion (A), , Reason (R), , If two tangents are drawn to a circle A parallelogram circumscribing a, from an external point then they circle is a rhombus., subtend equal angles at the centre., , The correct answer is (a)/(b)/(c)/(d).
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504, , Secondary School Mathematics for Class 10, , Assertion (A), , 55., , Reason (R), , In the given figure, a quad. ABCD is In two concentric circles, the chord, drawn to circumscribe a given circle, of the larger circle, which touches, the smaller circle, is bisected at the, as shown., point of contact., Then, AB BC AD DC., , The correct answer is (a)/(b)/(c)/(d)., ANSWERS (MCQ), , 1., 10., 19., 28., 37., 46., 55., , (b), (d), (a), (c), (d), (c), (d), , 2., 11., 20., 29., 38., 47., , (c), (c), (b), (a), (a), (b), , 3., 12., 21., 30., 39., 48., , (c), (b), (c), (c), (c), (d), , 4., 13., 22., 31., 40., 49., , (d), (b), (c), (d), (b), (c), , 5., 14., 23., 32., 41., 50., , (c), (a), (b), (b), (b), (d), , 6., 15., 24., 33., 42., 51., , (a), (c), (b), (d), (d), (d), , 7., 16., 25., 34., 43., 52., , (a), (c), (a), (b), (a), (d), , 8., 17., 26., 35., 44., 53., , (b), (b), (b), (d), (a), (a), , 9., 18., 27., 36., 45., 54., , (d), (b), (d), (c), (b), (b), , HINTS TO SOME SELECTED QUESTIONS, 4. Every diameter passes through the centre and so no two diameters of a circle can be, parallel., 5. AB OA 2 OB 2 10 2 10 2 10 2 cm., , 8. We have OP = PQ, i.e., +OPQ 90c., 3OPQ is isosceles & OP PQ & +OQP +POQ 45c, [a in a triangle, angles opposite equal sides are equal]., 9. ABOC is a cyclic quadrilateral, , , [see Solved Example 18]., , +BAC +BOC 180c & +BOC 180c +BAC 180c 40c 140c.
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508, , Secondary School Mathematics for Class 10, , 41. In right 3ABC, we have, AC , ar (3ABC) , , , AB 2 BC 2 5 2 12 2 cm 13 cm., 1, #(perimeter of 3ABC)# r, 2, , 1, 1, # 5 #12 #(5 12 13)# r & r 2 cm., 2, 2, 1, :a ar (3ABC) # BC # ABD, 2, , 42. BQ PB 27 cm; CQ BC BQ (38 27) cm 11 cm; CR CQ 11 cm., Join OR. Then, SDRO is a square., , , DR SO radius 10 cm., , And so, CD DR CR (10 11) cm 21 cm., 43. AC 2 AB 2 BC 2 8 2 6 2 64 36 100, , , AC 100 cm 10 cm., , CR CQ (BC BQ) (6 x) cm., AR AP (AB BP) (8 x) cm., AC (AR CR) [(8 x) (6 x)] cm (14 2x) cm, &, , (14 2x) 10 & 2x 4 & x 2 cm., , 44. AB CD AD BC [see Solved Example 26], &, , AD AB CD BC (6 4 7) cm 3 cm., , 45. PA PB & +PBA +PAB xc (say), Then, xc xc 60c 180 & x 60c., , , 3APB is equilateral and so, AB PA 5 cm., , 46. Join AE and AF. Then, AFDE is a square., , , radius of circle AE DE 5 cm., , 47. Let the radii of the three circles be x, y, z respectively. Then,, x y 5, y z 7 and z x 6 & 2 (x y z) 18 & x y z 9., , , x (x y z) (y z) (9 7) 2 cm., , 49. Join OB. Then, OA 5 cm, OB 3 cm., OP OA 2 PA 2 5 2 12 2 13 cm., PB OP 2 OB 2 13 2 3 2 160 cm 4 10 cm., 53. OQ OP 2 PQ 2 12 2 16 2 20 cm., , , A is true. Also, R is true and is a correct explanation of A., , Hence, the correct answer is (a)., 54. A and R are both true (see Theorem 4(i) and Solved Example 28) but R is not a correct, explanation of A., Hence, the correct answer is (b).
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Circles, , 509, , 55. A is false. Correct relation is AB CD AD BC., R is true (see Theorem 5)., Hence, the correct answer is (d)., , TEST YOURSELF, MCQ, 1. In the given figure, O is the centre of a circle,, PQ is a chord and the tangent PT at P makes, an angle of 50 with PQ. Then, +POQ ?, (a) 130, , (b) 100, , (c) 90, , (d) 75, , 2. If the angle between two radii of a circle is 130 then the angle between, the tangents at the ends of the radii is, (a) 65, , (b) 40, , (c) 50, , 3. If tangents PA and PB from a point P to a circle, with centre O are drawn so that +APB 80c, then +POA ?, (a) 40, , (b) 50, , (c) 80, , (d) 60, , 4. In the given figure, AD and AE are the tangents, to a circle with centre O and BC touches the, circle at F. If AE 5 cm then perimeter of, 3ABC is, (a) 15 cm, , (b) 10 cm, , (c) 22.5 cm, , (d) 20 cm, , Short-Answer Questions, 5. In the given figure, a quadrilateral ABCD, is drawn to circumscribe a circle such, that its sides AB, BC, CD and AD touch, the circle at P, Q, R and S respectively., If AB x cm, BC 7 cm, CR 3 cm and, AS 5 cm, find x., , (d) 90
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510, , Secondary School Mathematics for Class 10, , 6. In the given figure, PA and PB are the tangents, to a circle with centre O. Show that the points, A, O, B, P are concyclic., , 7. In the given figure, PA and PB are two tangents, from an external point P to a circle with centre, O. If +PBA 65c, find +OAB and +APB., , 8. Two tangent segments BC and BD are drawn, to a circle with centre O such that+CBD 120c., Prove that OB 2BC., , 9. Fill in the blanks., (i) A line intersecting a circle in two distinct points is called a ...... ., (ii) A circle can have ...... parallel tangents at the most., (iii) The common point of a tangent to a circle and the circle is called the, …… ., (iv) A circle can have …… tangents., 10. Prove that the lengths of two tangents drawn from an external point to, a circle are equal., 11. Prove that the tangents drawn at the ends of the diameter of a circle are, parallel., 12. In the given figure, if AB AC, prove that BE CE., , 13. If two tangents are drawn to a circle from an external point, show that, they subtend equal angles at the centre., 14. Prove that the tangents drawn at the ends of a chord of a circle make, equal angles with the chord.
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Circles, , 511, , 15. Prove that the parallelogram circumscribing a circle, is a rhombus., 16. Two concentric circles are of radii 5 cm and 3 cm respectively. Find the, length of the chord of the larger circle which touches the smaller circle., Long-Answer Questions, 17. A quadrilateral is drawn to circumscribe a circle. Prove that the sums of, opposite sides are equal., 18. Prove that the opposite sides of a quadrilateral circumscribing a circle, subtend supplementary angles at the centre of the circle., 19. Prove that the angle between the two tangents drawn from an external, point to a circle is supplementary to the angle subtended by the line, segments joining the points of contact at the centre., 20. PQ is a chord of length 16 cm of a circle, of radius 10 cm. The tangents at P and Q, intersect at a point T as shown in the figure., Find the length of TP., , ANSWERS (TEST YOURSELF), , 1. (b), , 2. (c), , 3. (b), , 4. (b), , 5. 9 cm, , 7. +OAB 25c, +APB 50c, 9. (i) secant, 16. 8 cm, , (ii) two, , (iii) point of contact, , 20. 10.7 cm (approx.), , , , (iv) infinitely many
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514, , Secondary School Mathematics for Class 10, , STEPS OF CONSTRUCTION, , Step 1. Draw a line segment BC 5.5 cm., Step 2. With B as centre and radius equal, , to 5 cm, draw an arc., Step 3. With C as centre and radius equal, , to 6.5 cm, draw another arc cutting, the previous arc at A., Step 4. Join AB and AC., , Thus, 3ABC with given lengths of, sides is obtained., Step 5. Below BC, make an acute angle +CBX., Step 6. Along BX, mark off five points B1, B2, B3, B4 and B5 such that, , BB1 B1 B2 B2 B3 B3 B4 B4 B5 ., Step 7. Join B5 C., Step 8. From B3, draw B3 D B5 C (by making an angle+BB3 D +BB5 C),, , meeting BC at D., Step 9. From D, draw DE CA (by making an angle +BDE +BCA),, , meeting AB at E., Then, 3EBD is the required triangle, each of whose sides is, of the corresponding side of 3ABC., PROOF, , 3, 5, , Since DE CA, we have 3ABC +3EBD., , , EXAMPLE 3, , EB DE BD 3 ·, AB CA BC 5, Draw a triangle ABC in which AB 5 cm, BC 6 cm and, 5, +ABC 60c. Then, construct a triangle whose sides are times the, 7, corresponding sides of 3ABC., [CBSE 2011, ’15], , STEPS OF CONSTRUCTION, , Step 1. Draw a line segment BC 6 cm., Step 2. At B, construct +CBX 60c., Step 3. With B as centre and radius 5 cm, draw an arc cutting ray BX, , at A., Step 4. Join AC., , Thus, 3ABC is obtained.
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516, , Secondary School Mathematics for Class 10, , Step 3. Below BC, make an acute angle +CBZ., Step 4. Along BZ, mark off four points B1, B2, B3 and B4 such that, , BB1 B1 B2 B2 B3 B3 B4 ., Step 5. Join B3 C., Step 6. From B4, draw B4 D B3 C, meeting BC (produced) at D., Step 7. From D, draw DE CA, meeting ray BX at E., , Then, 3EBD is the required triangle, each of whose sides is, times the corresponding side of 3ABC., PROOF, , 4, 3, , Since CA DE, we have 3ABC +3EBD., , , EXAMPLE 5, , EB DE BD 4 ·, AB CA BC 3, Construct an isosceles triangle whose base is 6 cm and altitude, 3, 4 cm. Then, construct another triangle whose sides are times the, 4, corresponding sides of the first triangle., [CBSE 2010, ’15], , STEPS OF CONSTRUCTION, , Step 1. Draw a line segment BC 6 cm., Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D., Step 3. With D as centre and radius 4 cm, draw an arc cutting XY, , at A., Step 4. Join AB and AC., , Thus, isosceles 3ABC having base 6 cm and altitude 4 cm is, obtained.
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Constructions, , 517, , Step 5. Below BC, make an acute angle +CBZ., Step 6. Along BZ, mark off four points B1, B2, B3 and B4 such that, , BB1 B1 B2 B2 B3 B3 B4 ., Step 7. Join B4 C., Step 8. From B3, draw B3 E B4 C, meeting BC at E., Step 9. From E, draw EF CA, meeting BA at F., , Then, 3FBE is the required triangle, each of whose sides is, times the corresponding side of 3ABC., PROOF, , 3, 4, , Since EF CA, we have 3FBE +3ABC., , , EXAMPLE 6, , FB EF BE 3 ·, AB CA BC 4, Construct a 3ABC in which AB 4 cm, +B 60c and altitude CL, 3 cm. Construct a 3ADE similar to 3ABC such that each side of, 3, 3ADE is times that of the corresponding side of 3ABC., 2, , STEPS OF CONSTRUCTION, , Step 1. Draw a line segment AB 4 cm., Step 2. Construct +ABP 60c., Step 3. Draw a line GH AB at a distance of 3 cm from AB, intersecting, , BP at C.
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518, , Secondary School Mathematics for Class 10, Step 4. Join CA., , Thus, 3ABC is obtained., Step 5. Extend AB to D such that AD , , 3, 3, AB a # 4k cm 6 cm., 2, 2, , Step 6. Draw DE BC, cutting AC produced at E., , Then, 3ADE is the required triangle similar to 3ABC such, 3, that each side of 3ADE is times the corresponding side of, 2, 3ABC., PROOF, , Since DE BC, we have 3ADE +3ABC., , , f, , AD DE AE 3 ·, AB BC AC 2, , EXERCISE 9A, , 1. Draw a line segment AB of length 7 cm. Using ruler and compasses,, AP 3 ·, find a point P on AB such that, [CBSE 2011], AB 5, 2. (i) Draw a line segment of length 8 cm and divide it internally in the, ratio 4 : 5., [CBSE 2017], (ii) Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8., Measure the two parts., 3. Construct a3PQR, in which PQ 6 cm, QR 7 cm and PR 8 cm. Then,, 4, construct another triangle whose sides are times the corresponding, 5, [CBSE 2013, ’14], sides of 3PQR., 4. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another, 7, triangle whose sides are, of the corresponding sides of the first, 5, triangle., [CBSE 2008C]
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Constructions, , 519, , 5. Construct a 3ABC with BC 7 cm, +B 60c and AB 6 cm. Construct, 3, another triangle whose sides are times the corresponding sides of, 4, [CBSE 2008, ’09, ’15], 3ABC., 6. Construct a3ABC in which AB 6 cm, +A 30c and+B 60c. Construct, another 3ABlCl similar to 3ABC with base ABl 8 cm., [CBSE 2015], 7. Construct a 3ABC in which BC 8 cm, +B 45c and+C 60c. Construct, 3, another triangle similar to 3ABC such that its sides are, of the, 5, corresponding sides of 3ABC., [CBSE 2010, ’12, ’14], 8. To construct a triangle similar to 3ABC in which BC 4.5 cm, +B 45c, 3, and +C 60c, using a scale factor of , BC will be divided in the ratio, 7, [CBSE 2012], , (a) 3 : 4, , (b) 4 : 7, , (c) 3 : 10, , (d) 3 : 7, , 9. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm, 1, and then another triangle whose sides are 1 times the corresponding, 2, sides of the isosceles triangle., 10. Draw a right triangle in which the sides (other than hypotenuse) are of, lengths 4 cm and 3 cm. Then, construct another triangle whose sides are, 5, times the corresponding sides of the given triangle., [CBSE 2011], 3, HINTS TO SOME SELECTED QUESTIONS, 1. Divide the line segment AB in the ratio 3 : 2., 6. Construct 3ABC. Produce AB to Bl such that ABl 8 cm., Draw BlCl BC such that BlCl meets AC produced at Cl., , CONSTRUCTION OF TANGENTS TO A CIRCLE, CONSTRUCTION OF A TANGENT TO A CIRCLE AT A GIVEN POINT ON THE CIRCLE, EXAMPLE 1, , Draw a circle of radius 2.5 cm. Take a point P on it. Construct a, tangent at the point P., , STEPS OF CONSTRUCTION, , Step 1. Draw a circle of radius 2.5 cm taking a point O as its centre., Step 2. Mark a point P on this circle.
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520, , Secondary School Mathematics for Class 10, Step 3. Join OP., Step 4. Construct +OPT 90c., Step 5. Produce TP to Tl., , Then TlPT is the required, tangent., CONSTRUCTION OF A TANGENT TO A CIRCLE AT A POINT ON IT WITHOUT, USING THE CENTRE, EXAMPLE 2, , Draw a circle of radius 5 cm. Take a point P on it. Without using the, [CBSE 2002], centre of the circle construct a tangent at the point P., , STEPS OF CONSTRUCTION, , Step 1. Draw a circle of radius 5 cm., Step 2. Mark a point P on it., Step 3. Draw any chord PQ., Step 4. Take a point R on the major arc QP., Step 5. Join PR and RQ., Step 6. Make +QPT +PRQ., Step 7. Produce TP to Tl, as shown in the figure., , Then, TlPT is the required tangent at P., CONSTRUCTION OF TANGENTS TO A CIRCLE FROM A POINT OUTSIDE IT WHEN, THE CENTRE OF THE CIRCLE IS KNOWN, EXAMPLE 3, , Draw a circle of radius 2 cm with centre O and take a point P, outside the circle such that OP 6.5 cm. From P, draw two tangents, to the circle., , STEPS OF CONSTRUCTION, , Step 1. Draw a circle with O as centre, , and radius 2 cm., Step 2. Mark a point P outside the circle, , such that OP 6.5 cm., Step 3. Join OP and bisect it at M., Step 4. Draw a circle with M as centre, , and radius equal to MP, to, intersect the given circle at the, points T and Tl.
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Constructions, , 521, , Step 5. Join PT and PTl., , Then, PT and PTl are the required tangents., EXAMPLE 4, , Write the steps of construction for drawing two tangents to a circle, of radius 6 cm from a point 10 cm away from its centre. [CBSE 2013C], , STEPS OF CONSTRUCTION, , Step 1. Draw a circle with O as centre, , and radius 6 cm., Step 2. Mark a point P outside the circle, , such that OP 10 cm., Step 3. Join OP and bisect it at a point M., Step 4. Draw a circle with M as the, , centre and radius equal to MP,, to intersect the given circle at, the points T and Tl., Step 5. Join PT and PTl., , Then, PT and PTl are the required tangents., EXAMPLE 5, , Draw a line segment AB of length 7 cm. Taking A as centre, draw, a circle of radius 3 cm and taking B as centre, draw another circle of, radius 2 cm. Construct tangents to each circle from the centre of the, other circle., [CBSE 2011, ’15], , STEPS OF CONSTRUCTION, , Step 1. Draw, , a line, AB 7 cm., , segment, , Step 2. With A as centre and radius, , 3 cm, draw a circle., Step 3. With B as centre and radius, 2 cm, draw another circle., Step 4. Bisect the line segment AB, , at M., Step 5. With M as centre and radius equal to AM, draw a circle, , intersecting the first circle at Q and Ql and the second circle at, P and Pl., Step 6. Join AP and APl. Also, join BQ and BQl.
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522, , Secondary School Mathematics for Class 10, , Then, BQ and BQl are the required tangents to the first circle, and AP and APl are the required tangents to the second circle., CONSTRUCTION OF TANGENTS TO A CIRCLE FROM A POINT OUTSIDE IT, WITHOUT USING THE CENTRE, EXAMPLE 6, , Draw a circle of radius 2.5 cm, and take a point P outside it., Without using the centre of, the circle, draw two tangents, to the circle from the point P., , STEPS OF CONSTRUCTION, , Step 1. Draw a circle of radius, , 2.5 cm and take a point p, outside it., Step 2. Through P draw a secant, , PAB to intersect the circle at, A and B., Step 3. Produce AP to a point C such that PA PC., Step 4. Draw a semicircle with CB as diameter., Step 5. Draw PD = CB, intersecting the semicircle at D., Step 6. With P as centre and PD as radius, draw arcs to intersect the, , circle at T and Tl., Step 7. Join PT and PTl., , Then, PT and PTl are the required tangents., CONSTRUCTION OF TANGENTS TO A CIRCLE INCLINED TO EACH OTHER AT A, GIVEN ANGLE, EXAMPLE 7, , Draw a circle of radius 3 cm. Draw a pair of tangents to this circle,, which are inclined to each other at an angle of 60c., , STEPS OF CONSTRUCTION, , Step 1. Draw a circle with O as centre and radius = 3 cm., Step 2. Draw any diameter AOB of this circle., Step 3. Construct +BOC 60c such that radius OC meets the circle, , at C.
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Constructions, , 523, , Step 4. Draw AM = AB and CN = OC., , Let AM and CN intersect each, other at P., Then, PA and PC are the, desired tangents to the given, circle, inclined at an angle, of 60., PROOF, , +AOC (180c 60c) 120c., In quad. OAPC, we have, +OAP 90c, +AOC 120c, +OCP 90c., , , EXAMPLE 8, , +APC [360c (90c 120c 90c)] 60c., Draw a circle of radius 3.5 cm. Draw two tangents to the circle, which are perpendicular to each other., [CBSE 2015], , STEPS OF CONSTRUCTION, , Step 1. Draw a circle with O as centre, , and radius = 3.5 cm., Step 2. Draw any diameter AOB of, , this circle., Step 3. Construct +BOX 90c such, , that ray OX meets the circle, at C., Step 4. Draw AM = AB and CN =OC., , Let AM and CN intersect each, other at a point P., Then, PA and PC are the required tangents (which are, perpendicular to each other)., PROOF, , +AOC (180c +BOC) 180c 90c 90c., In quad. OAPC, we have, +OAP 90c, +AOC 90c, +OCP 90c., , , +APC [360c (90c 90c 90c)] 90c., , CONSTRUCTION OF TANGENTS TO A CIRCLE FROM A POINT ON A LARGER, CIRCLE CONCENTRIC WITH THE FIRST ONE, EXAMPLE 9, , Draw two concentric circles of radii 2 cm and 5 cm. Taking a point, on the outer circle, construct a pair of tangents to the other. Measure
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524, , Secondary School Mathematics for Class 10, , the lengths of the two tangents. Also, verify the measurement by, actual calculation., [CBSE 2013C], STEPS OF CONSTRUCTION, , Step 1. Mark a point O on the paper., Step 2. With O as centre and radii, , 2 cm and 5 cm, draw two, concentric circles., Step 3. Mark a point P on the outer, , circle., Step 4. Join OP and bisect it at a, , point M., Step 5. Draw a circle with M as the centre and radius equal to MP, to, , intersect the inner circle in points T and Tl., Step 6. Join PT and PTl., , Then, PT and PTl are the required tangents., Measurement Upon measuring the two tangents, we find PT PTl 4.6 cm., Calculation of the length of tangent, , Join OT to form a right 3OTP [a PT is a tangent at point T]., Then, PT OP 2 OT 2 5 2 2 2 25 4 21 . 4.58 cm., [a OP radius of larger circle = 5 cm,, OT radius of smaller circle = 2 cm.], EXAMPLE 10, , Construct a right triangle ABC with AB 6 cm, BC 8 cm and, +B 90c. Draw BD, the perpendicular from B on AC. Draw the, circle through B, C and D and construct the tangents from A to this, circle., [CBSE 2014, ’15], , STEPS OF CONSTRUCTION, , Step 1. Draw a line segment AB 6 cm., Step 2. At B, construct +ABX 90c., Step 3. With B as centre and radius 8 cm, draw an arc cutting ray BX, , at C., Step 4. Join AC., , Thus, 3ABC is obtained.
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Constructions, , 525, , Step 5. From B, draw BD = AC., Step 6. Bisect BC at point O., Step 7. With O as centre and radius OB, draw a circle. This circle, , passes through B, C and D., Thus, the required circle is obtained., Step 8. Join AO and bisect it at M., Step 9. With M as the centre and radius equal to AM, draw a circle, , cutting the previous circle at the points T and B., Step 10. Join AT., , Then, AT and AB are the required tangents., f, , EXERCISE 9B, , 1. Draw a circle of radius 3 cm. From a point P, 7 cm away from the, centre of the circle, draw two tangents to the circle. Also, measure, the lengths of the tangents., [CBSE 2010], 2. Draw two tangents to a circle of radius 3.5 cm from a point P at a, distance of 6.2 cm from its centre., [CBSE 2013], 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its, diameters extended on both sides, each at a distance of 7 cm on opposite, sides of its centre. Draw tangents to the circle from these two points P, and Q., [CBSE 2017], 4. Draw a circle with centre O and radius 4 cm. Draw any diameter AB of, this circle. Construct tangents to the circle at each of the two end points, of the diameter AB.
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526, , Secondary School Mathematics for Class 10, , 5. Draw a circle with the help of a bangle. Take any point P outside the, circle. Construct the pair of tangents from the point P to the circle., HINT, , See Example 6., , 6. Draw a line segment AB of length 8 cm. Taking A as centre, draw a, circle of radius 4 cm and taking B as centre, draw another circle of, radius 3 cm. Construct tangents to each circle from the centre of the, other circle., [CBSE 2014], 7. Draw a circle of radius 4.2 cm. Draw a pair of tangents to this circle, inclined to each other at an angle of 45., 8. Write the steps of construction for drawing a pair of tangents to a circle, of radius 3 cm, which are inclined to each other at an angle of 60., [CBSE 2011, ’12, ’14], , 9. Draw a circle of radius 3 cm. Draw a tangent to the circle making an, angle of 30 with a line passing through the centre., HINT, , Draw a circle with centre O and, radius 3 cm. Draw a radius OA and, produce it to B. Make +AOP 60c., Draw PQ = OP, meeting OB at Q., Then, PQ is the desired tangent such, that +OQP 30c., , 10. Construct a tangent to a circle of radius 4 cm from a point on the, concentric circle of radius 6 cm and measure its length. Also, verify the, measurement by actual calculation., 11. Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on, the outer circle, construct the pair of tangents to the inner circle., [CBSE 2017], , 12. Write the steps of construction to construct the tangents to a circle from, an external point., [CBSE 2017], , TEST YOURSELF, Long-Answer Questions, 1. Draw a line segment AB of length 5.4 cm. Divide it into six equal parts., Write the steps of construction., 2. Draw a line segment AB of length 6.5 cm and divide it in the ratio 4 : 7., Measure each of the two parts.
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Constructions, , 527, , 3. Construct a 3ABC in which BC 6.5 cm, AB 4.5 cm and +ABC 60c., 3, Construct a triangle similar to this triangle whose sides are of the, 4, corresponding sides of 3ABC., [CBSE 2009], 4. Construct a 3ABC in which BC 5 cm, +C 60c and altitude from A, equal to 3 cm. Construct a 3ADE similar to 3ABC such that each side, 3, of 3ADE is times the corresponding side of 3ABC. Write the steps of, 2, construction., 5. Construct an isosceles triangle whose base is 9 cm and altitude 5 cm., 3, Construct another triangle whose sides are, of the corresponding, 4, sides of the first isosceles triangle., [CBSE 2009C], 6. Draw a 3ABC, right-angled at B such that AB 3 cm and BC 4 cm., 7, Now, construct a triangle similar to 3ABC, each of whose sides is, 5, times the corresponding side of 3ABC., 7. Draw a circle of radius 4.8 cm. Take a point P on it. Without using the, centre of the circle, construct a tangent at the point P. Write the steps of, construction., 8. Draw a circle of radius 3.5 cm. Draw a pair of tangents to this circle, which are inclined to each other at an angle of 60. Write the steps of, construction., 9. Draw a circle of radius 4 cm. Draw tangent to the circle making an, angle of 60 with a line passing through the centre., 10. Draw two concentric circles of radii 4 cm and 6 cm. Construct a tangent, to the smaller circle from a point on the larger circle. Measure the length, of this tangent.,
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528, , Secondary School Mathematics for Class 10, , Trigonometric Ratios, , 10, , It is that branch of mathematics which deals with the, measurement of angles and the problems allied with angles., , TRIGONOMETRY, , TRIGONOMETRIC RATIOS (T-RATIOS) OF AN ACUTE ANGLE OF A RIGHT, TRIANGLE, , Let +BAC be an acute angle of a right-angled 3ABC., In right-angled 3ABC, let, base AB x units,, perpendicular BC y units, and hypotenuse AC r units., We define the following ratios, known as Trigonometric Ratios for ., perpendicular y, (i) sine hypotenuse r , and is written as sin ., base, x, (ii) cosine hypotenuse r , and is written as cos ., (iii) tangent , , perpendicular y, , and is written as tan ., x, base, , hypotenuse, r, (iv) cosecant perpendicular y , and is written as cosec ., (v) secant , , hypotenuse r, , and is written as sec ., x, base, , base, x, (vi) cotangent perpendicular y , and is written as cot ., RECIPROCAL RELATION, , Clearly, we have, 1, (i) cosec sin , , 1, (ii) sec cos , 528, , 1, (iii) cot tan
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Trigonometric Ratios, , 529, , SUMMARY, , Consider a 3 ABC in which +B 90c and +A ., Let AB x, BC y and AC r. Then,, y, y, x, (i) sin r, (ii) cos r, (iii) tan x, 1, r, 1, r, (iv) cosec sin y, (v) sec cos x, 1, x, (vi) cot tan y, Also, we have, 1, (i) cosec sin , , 1, (ii) sec cos , , 1, (iii) cot tan , , T-RATIOS OF AN ANGLE ARE WELL-DEFINED, THEOREM, , PROOF, , Show that the value of each of the trigonometric ratios of an angle, does not depend on the size of the triangle. It only depends on the, angle., Consider a 3ABC in which +B 90c and +A c., Take a point P on AC and draw PQ = AB., Then, 3AQP is similar to 3ABC., AQ AP PQ, , , , AB AC CB, PQ CB, , , , AP AC sin ., AQ AB, Similarly, AP AC cos , PQ CB, and AQ AB tan ., Similarly, if we produce AC to R and draw RS = AB produced, then 3ASR is similar to 3ABC., AS AR RS, , AB AC CB, RS CB , , AR AC sin ., AS AB, Similarly, AR AC cos , RS CB , and, AS AB tan ., Hence, the trigonometric ratios of an angle do not depend on the, size of the triangle. They only depend on the angle.
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530, REMARK, , Secondary School Mathematics for Class 10, , Consider a 3ABC in which +B 90c and +A c., We know that in a right triangle, the, hypotenuse is the longest side., , , BC, AB, sin AC 1 and cos AC 1., , Thus, sin 1 and cos 1., POWER OF T-RATIOS, , We write, , (sin )2 sin 2 ;, , (sin )3 sin 3 ;, , (cos )3 cos 3 ; etc., , QUOTIENT RELATION OF T-RATIOS, THEOREM 1, , For any acute angle , prove that, sin , (i) tan cos ;, , PROOF, , cos , (ii) cot sin ;, , (iii) tan · cot 1., , Consider a right-angled 3ABC in which, +B 90c and +A c. Let AB x units,, BC y units and AC r units. Then,, y (y/r), (i) tan x , (x/r), [dividing num. and denom. by r], sin ·, cos , , , sin , tan cos ·, , x (x/r), (ii) cot y , (y/r), , [dividing num. and denom. by r], , cos ·, sin , , , cos , cot sin ·, , sin cos , (iii) tan · cot cos · sin 1., SUMMARY, , sin , (i) tan cos , , cos , (ii) cot sin , , (iii) tan · cot 1
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Trigonometric Ratios, , 531, , SQUARE RELATION, THEOREM 2, , For any acute angle , prove that, (i) sin 2 cos 2 1;, (ii) 1 tan 2 sec 2 ;, (iii) 1 cot 2 cosec 2 ., , PROOF, , Consider a right-angled 3ABC in which, +B 90c and +A c. Let AB x units,, BC y units and AC r units., Then, by Pythagoras‘ theorem, we have, x2 y2 r2 ., y2 x2, y2 x2, (i) sin 2 cos 2 a r k ` r j d 2 2 n, r, r, (x 2 y 2) r 2, 2, , r2, r, 1., sin 2 cos 2 1., y2, y2 y2 x2 r2, 2, (ii) 1 tan 2 1 a x k 1 2 , x, x2, x, 2, ` r j sec 2 ., x, 1 tan 2 sec 2 ., 2, 2, x2, x2 x y r2, (iii) 1 cot 2 1 a y k 1 2 , 2, y, y, y2, , [a x 2 y 2 r 2], , [a x 2 y 2 r 2], , [a x 2 y 2 r 2], , a r k cosec 2 ., y, 2, , 1 cot 2 cosec 2 ., SUMMARY, , (i) sin 2 cos 2 1, , (ii) 1 tan 2 sec 2 , , (iii) 1 cot 2 cosec 2 , , SOLVED EXAMPLES, EXAMPLE 1, SOLUTION, , 8, If sin A 17 , find other trigonometric ratios of +A., Let us draw a 3ABC in which +B 90c., BC, 8, Then, sin A AC 17 ·, Let BC 8k and AC 17k, where k is positive.
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532, , Secondary School Mathematics for Class 10, , By Pythagoras‘ theorem, we have, , , AC 2 AB 2 BC 2, AB 2 AC 2 BC 2, , , , AB 2 (17k) 2 (8k) 2 289k 2 64k 2, 225k 2, , AB 225k 2 15k., BC, 8k 8, AB 15k 15, sin A AC , ; cos A AC , ;, 17k 17, 17k 17, sin A, 8, 17, 8, tan A cos A a17 # 15 k 15 ;, 1, 17, 1, 17, cosec A sin A 8 ; sec A cos A 15, 1, 15, and cot A tan A 8 ·, , , EXAMPLE 2, SOLUTION, , 9, If cos A 41 , find other trigonometric ratios of +A., Let us draw a 3ABC in which +B 90c., AB, 9, Then, cos A AC 41 ·, Let AB 9k and AC 41k, where k is positive., By Pythagoras‘ theorem, we have, AC 2 AB 2 BC 2, BC 2 AC 2 AB 2, BC 2 (41k) 2 (9k) 2 1681k 2 81k 2 1600k 2, BC 1600k 2 40k., BC 40k 40, 9, sin A AC , ; cos A 41 (given);, 41k 41, sin A, 40, 41, 40, tan A cos A a 41 # 9 k 9 ;, 1, 41, 1, 41, cosec A sin A 40 ; sec A cos A 9, 1, 9, and cot A tan A 40 ·, , EXAMPLE 3, , If tan A 3 , find other trigonometric ratios of +A., , SOLUTION, , Let us draw a 3ABC in which +B 90c., 3, BC, Then, tan A AB 1 ·, Let BC 3 k and AB k, where k is positive.
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Trigonometric Ratios, , By Pythagoras‘ theorem, we have, AC 2 AB 2 BC 2 k 2 ( 3 k) 2, k 2 3k 2 4k 2 ., , , AC 4k 2 2k., , , , BC, sin A AC , AB, cos A AC , , 3k, 3, , 2 ;, 2k, k 1, ;, 2k 2, , 1, 2, tan A 3 (given); cosec A sin A , ;, 3, 1, 1, 1 ·, sec A cos A 2 and cot A tan A , 3, EXAMPLE 4, SOLUTION, , 25, If sec 7 , find all trigonometric ratios of ., Let us draw a 3ABC in which +B 90c. Let +A c., AC 25, Then, sec AB 7 ·, Let AC 25k and AB 7k, where k is positive., By Pythagoras‘ theorem, we have, AC 2 AB 2 BC 2, BC 2 AC 2 AB 2 (25k)2 (7k)2, 625k 2 49k 2 576k 2, BC 576k 2 24k., , , BC 24k 24, 1, 7, sin AC , ; cos sec 25 ;, 25k 25, sin , 24, 25, 24, 1, 25, tan cos a 25 # 7 k 7 ; cosec sin 24 ;, 25, 1, 7, sec 7 (given) and cot tan 24 ·, , EXAMPLE 5, , 5 cosec 4 tan k ·, 3, If cos 5 , find the value of a, sec cot , , SOLUTION, , Let us draw a 3ABC in which +B 90c., Let +A c., AB 3, Then, cos AC 5 ·, Let AB 3k and AC 5k, where k is, positive., , 533
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534, , Secondary School Mathematics for Class 10, , By Pythagoras‘ theorem, we have, AC 2 AB 2 BC 2, BC 2 AC 2 AB 2 (5k) 2 (3k) 2 25k 2 9k 2 16k 2, BC 16k 2 4k., 1, 5, BC 4k 4, sec cos 3 ; tan AB , ;, 3k 3, 1, 3, AC 5k 5 ·, cot tan 4 ; and cosec BC , 4k 4, 5, 4, 16 k, a5 # 4 4 # 3 k a 25, 4 tan , 3, 4, 5, cosec, , , , k, a, 53, 53, sec cot , a3 4 k, a3 4 k, 75 64, 12 a11 # 12 k 11 ·, 29, 29, 12, 20 9, 12, EXAMPLE 6, , (2cos sin ) 12, 5, ·, , If sec 4 , show that, 7, (cot tan ), , SOLUTION, , Consider a 3ABC in which +A and +B 90c., hypotenuse AC 5 5x, , , AB 4 4x (say)., base, AC 5x and AB 4x, where x is positive., , sec , , , By Pythagoras’ theorem we have, AC 2 AB 2 BC 2, BC 2 AC 2 AB 2 (5x)2 (4x)2 9x 2, BC 3x., AB 4x 4, BC 3x 3, cos AC 5x 5 ; sin AC 5x 5 ;, AB 4x 4, BC 3x 3, cot BC 3x 3 ; and tan AB 4x 4 ·, 4 3, 8 3, 5, (2cos sin ) a2 # 5 5 k a 5 5 k a 5 k, , , , 1 12 ·, , 7, 7, 43, 43, (cot tan ), k 7, a3 4 k, a 3 4 k a12 k a12, EXAMPLE 7, , SOLUTION, , In a 3ABC it is given that +B 90c and AB : AC 1 : 2 . Find the, 2tan A ·, value of a, k, 1 tan 2 A, Consider a 3ABC in which +B 90c and AB : AC 1 : 2 .
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Trigonometric Ratios, , 535, , Let AB x. Then, AC 2 x., By Pythagoras‘ theorem, we have, AC 2 AB 2 BC 2 & BC 2 AC 2 AB 2, BC 2 ( 2 x)2 (x) 2 2x 2 x 2 x 2, BC x., , , BC x, tan A AB x 1., , So, the given expression c, , EXAMPLE 8, , SOLUTION, , EXAMPLE 9, , SOLUTION, , If 3tan 4, evaluate, , 2 tan A 2 #1 2 , m a k 2 1., 1 1, 1 tan 2 A, , 3 sin 2 cos ·, 3 sin 2 cos , , 4, 3 tan 4 & tan 3 ·, Given expression, , 3 sin 2 cos , 3 sin 2 cos , , 3 tan 2, [dividing num. and denom. by cos ], 3 tan 2, a3 # 34 2k 6, 3., , :a tan 4D, 3, 2, 4, a3 # 3 2k, If 5 cot 3, find the value of a, , 5 sin 3 cos k ·, 4 sin 3 cos , , 3, 5 cot 3 & cot 5 ·, 5 sin 3 cos k (5 3 cot ), Given expression a, 4 sin 3 cos , (4 3 cot ), [dividing num. and denom. by sin ], , , a5 3 # 35 k, , a4 3 # 35 k, , , , a5 95 k, , a4 95 k, , EXAMPLE 10, , If 7 sin 2 3 cos 2 4, show that tan , , SOLUTION, , 7 sin 2 3cos 2 4, , , 4 sin 2 3 sin 2 3 cos 2 4, , , , 4 sin 2 3(sin 2 cos 2 ) 4, , a16 # 5 k 16 ·, 5, 29, 29, , 1 ·, 3, , [CBSE 2008]
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536, , Secondary School Mathematics for Class 10, , , , 4 sin 2 3 # 1 4, , , , 1, 4 sin 2 1 & sin 2 4 ·, , [a sin 2 cos 2 1], , 1, 3, cos 2 (1 sin 2 ) a1 4 k 4 ·, sin 2 1, a 4 # 34 k 13 ·, tan 2 , cos 2 , 1 ·, Hence, tan , 3, , , EXAMPLE 11, , (2 2 sin )(1 sin ), 15, ·, If cot 8 then evaluate , (1 cos )(2 2 cos ), , SOLUTION, , Given expression , , (2 2 sin )(1 sin ), (1 cos )(2 2 cos ), , , , 2(1 sin )(1 sin ), 2(1 cos )(1 cos ), , , , (1 sin 2 ) cos 2 , , cot 2 , (1 cos 2 ) sin 2 , , [CBSE 2009], , 2, (cot ) 2 a15 k 225 ·, 8, 64, , 225, Hence, the value of the given expression is 64 ·, EXAMPLE 12, , SOLUTION, , In 3ABC, right-angled at B, AB 5 cm and BC 12 cm. Find the, values of sin A, sec A, sin C and sec C., In 3ABC, +B 90c, AB 5 cm and BC 12 cm., By Pythagoras‘ theorem, we have, AC 2 (AB 2 BC 2) {(5)2 (12)2} cm 2, (25 144) cm 2 169 cm 2 ., , , AC 169 cm 2 13 cm., , For T-ratios of +A, we have, base = AB = 5 cm,, perpendicular = BC = 12 cm, and hypotenuse = AC = 13 cm., BC 12, AC 13, sin A AC 13 and sec A AB 5 ·, For T-ratios of +C, we have, , , , base = BC = 12 cm,
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Trigonometric Ratios, , 537, , perpendicular = AB = 5 cm, and hypotenuse = AC = 13 cm., , EXAMPLE 13, , SOLUTION, , AB, 5, AC 13, sin C AC 13 and sec C BC 12 ·, , In a 3ABC, +B 90c, AB 5 cm and (BC AC) 25 cm. Find, the values of sin A, cos A, cosec C and sec C., Let BC x cm. Then, AC (25 x) cm., By Pythagoras‘ theorem, we have, , , , , AB 2 BC 2 AC 2, 5 2 x 2 (25 x) 2, 25 x 2 625 50x x 2, 50x 600, , x 12., BC 12 cm, AC 13 cm and AB 5 cm., For T-ratios of +A, we have, AB, 5, BC 12, sin A AC 13 and cos A AC 13 ·, For T-ratios of +C, we have, base, BC = 12 cm,, perpendicular, AB = 5 cm, and hypotenuse, AC = 13 cm., , EXAMPLE 14, , SOLUTION, , AC 13, AC 13, cosec C AB 5 and sec C BC 12 ·, , In a 3ABC, +B 90c, AB 7 cm and (AC BC) 1 cm. Find the, values of sin A, cos A, sin C and cos C., Let BC x cm. Then, AC (x 1) cm., By Pythagoras‘ theorem, we have, AB 2 BC 2 AC 2, 7 2 x 2 (x 1) 2, , , 49 x 2 x 2 2x 1, , , , 2x 48, , , , x 24., , , , BC 24 cm, AC 25 cm and AB 7 cm.
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538, , Secondary School Mathematics for Class 10, , For T-ratios of +A, we have, BC 24, AB, 7, sin A AC 25 and cos A AC 25 ·, For T-ratios of +C, we have, AB, 7, BC 24, sin C AC 25 and cos C AC 25 ·, , EXAMPLE 15, , In a 3ABC, +C 90c and tan A , , 1 ·, Find the values of:, 3, , (i) (sin A $ cos B cos A $ sin B) (ii) (cos A · cos B sin A · sin B), [CBSE 2008], SOLUTION, , Consider a 3ABC in which +C 90c and tan A , Then, tan A , , 1, BC 1 ·, & AC, 3, 3, , 1 ·, 3, , Let BC x. Then, AC 3 x., By Pythagoras‘ theorem, we have, AB 2 AC 2 BC 2, ( 3 x)2 x 2 (3x 2 x 2) 4x 2, , , AB 4x 2 2x., , For T-ratios of +A, we have, base AC 3 x, perpendicular BC x and, hypotenuse AB 2x., 3x, 3, BC, x 1, AC, sin A AB 2x 2 and cos A AB 2x 2 ·, For T-ratios of +B, we have, , , base BC x, perpendicular AC 3 x and, hypotenuse AB 2x., , , 3x, 3, AC, BC, x 1, sin B AB 2x 2 and cos B AB 2x 2 ·, , 3, 3, 1 1, (i) (sin A · cos B cos A · sin B) c 2 # 2 2 # 2 m, a1 3 k 1., 4 4, (sin A · cos B cos A · sin B) 1.
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Trigonometric Ratios, , 539, , 3, 3, 1 1, (ii) (cos A cos B sin A sin B) c 2 # 2 2 # 2 m 0., (cos A cos B sin A sin B) 0., EXAMPLE 16, , SOLUTION, , If +A and +B are acute angles such that cos A cos B then prove, that +A +B., Let 3ACD and 3BEF be two right triangles given in such a, way that cos A cos B. Then,, cos A cos B, AC BE, , AD BF, AC AD, BE BF k (say), AC k (BE), and AD k (BF) ., , , CD , EF, , , AD 2 AC 2, BF 2 BE 2, k BF 2 BE 2, k, BF 2 BE 2, , … (i), [using Pythagoras’ theorem], [using (i)]., , AC AD CD, Thus, we have: BE BF EF ·, 3ACD + 3BEF and hence, +A +B., f, , EXERCISE 10, , 3, 1. If sin 2 , find the value of all T-ratios of ., 7, 2. If cos 25 , find the values of all T-ratios of ., 15, 3. If tan 8 , find the values of all T-ratios of ., 4. If cot 2, find the values of all T-ratios of ., 5. If cosec 10 , find the values of all T-ratios of ., a2 b2, , find the values of all T-ratios of ., a2 b2, 7. If 15 cot A 8, find the values of sin A and sec A., 6. If sin , , 9, 8. If sin A 41 , find the values of cos A and tan A.
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540, , Secondary School Mathematics for Class 10, , 9. If cos 0.6, show that (5sin 3tan ) 0., sin , 10. If cosec 2, show that acot , 2., 1 cos k, (cosec 2 sec 2 ) 3, 1, ·, 11. If tan , , show that, (cosec 2 sec 2 ) 4, 7, (1 sin cos ) 3, 20, ·, 12. If tan 21 , show that , (1 sin cos ) 7, (sin 2 cos ) 12, 5, ·, , 13. If sec 4 , show that, 7, (tan cot ), sec cosec 1 ·, 3, 14. If cot 4 , show that, sec cosec , 7, cosec 2 cot 2 7 ·, 3, sec 2 1, a·, b, a, If sin , show that (sec tan ) , b, ba, (sin cot ), 3, 3, ·, , If cos 5 , show that, 160, 2 tan , 7, 4, If tan 3 , show that (sin cos ) 5 ·, (a sin b cos ) (a 2 b 2), a, ·, , If tan , show that, b, (a sin b cos ) (a 2 b 2), (4 cos sin ) 4, ·, If 3 tan 4, show that, (2 cos sin ) 5, (4 sin 3 cos ) 1, ·, If 3 cot 2, show that, (2 sin 6 cos ) 3, (1 tan 2 ), (cos 2 sin 2 ) ., If 3 cot 4, show that, (1 tan 2 ), (3 4 sin 2 ), (3 tan 2 ), 17, ·, , If sec 8 , verify that, 2, , (4 cos 3) (1 3 tan 2 ), , 3, 15. If sin 4 , show that, 16., 17., 18., 19., 20., 21., 22., 23., , 24. In the adjoining figure, +B 90c, +BAC c, BC CD 4 cm and, AD 10 cm. Find (i) sin and (ii) cos ., HINT, , , , AB 2 (AD 2 BD 2) 36 cm 2, AB 6 cm., , AC 2 (AB 2 BC 2) 52 cm 2, , , AC 2 13 cm., , Thus, AB 6 cm and AC 2 13 cm., , 25. In a 3ABC, +B 90c, AB 24 cm and BC 7 cm., Find (i) sin A, , (ii) cos A, , (iii) sin C, , (iv) cos C.
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Trigonometric Ratios, , 541, , 26. In a 3ABC, +C 90c, +ABC c, BC 21 units and AB 29 units., 41, Show that (cos 2 sin 2 ) 841 ·, 27. In a 3ABC, +B 90c, AB 12 cm and BC 5 cm., Find (i) cos A, , (ii) cosec A, , (iii) cos C, , (iv) cosec C., , 1, 28. If sin 2 , prove that (3 cos 4 cos 3 ) 0., 1 ·, 29. In a 3ABC, +B 90c and tan A , Prove that, 3, (i) sin A · cos C cos A · sin C 1, , (ii) cos A · cos C sin A · sin C 0, , 30. If +A and +B are acute angles such that sin A sin B then prove that, +A +B., 31. If +A and +B are acute angles such that tan A tan B then prove that, +A +B., 32. In a right 3ABC, right-angled at B, if tan A 1 then verify that, 2 sin A · cos A 1., 33. In the figure of 3PQR, +P c and +R c., Find (i) ( x 1 ) cot , (ii) ( x 3 x 2 ) tan , (iii) cos , 34. If x cosec A cos A and y cosec A cos A then prove that, 2, 2 2 x ym , c m c, 1 0., x y, 2, 35. If x cot A cos A and y cot A cos A, prove that, xy 2 xy 2, c, m c, m 1., xy, 2, ANSWERS (EXERCISE 10), , 3, , 1, , 1, 2, , cosec , , sec 2, 3, 3, 24, 7, 24, 7, 25, 25, 2. sin 25 , cos 25 , tan 7 , cot 24 , cosec 24 , sec 7, 15, 8, 15, 8, 17, 17, 3. sin 17 , cos 17 , tan 8 , cot 15 , cosec 15 , sec 8, 1. sin 2 , cos 2 , tan 3 , cot , , 4. sin , , 5, 1, 2, 1, , cos , , tan 2 , cot 2, cosec 5 , sec 2, 5, 5
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542, , Secondary School Mathematics for Class 10, , 5. sin , , 10, 1, 3, 1, , cos , , tan 3 , cot 3, cosec 10 , sec 3, 10, 10, , a2 b2, 2ab, a2 b2, 2, 2 , cos 2, 2 , tan 2ab ,, a b, a b, 2, b2, a, a2 b2, 2ab, cosec 2 2 , sec , , cot 2 2, 2ab, a b, a b, 40, 9, 15, 17, 7. sin A 17 , sec A 8, 8. cos A 41 , tan A 40, 6. sin , , 24. (i) sin , , 7, , 25. (i) 25, , 12, , 27. (i) 13, 33. (i), , x, 2, , 2 13, 13, 24, (ii) 25, 13, (ii) 5, (ii), , x2, 2, , 3 13, (ii) cos 13, 24, 7, (iii) 25, (iv) 25, 5, 13, (iii) 13, (iv) 12, 2 x1, (iii), (x 2), , HINTS TO SOME SELECTED QUESTIONS, 6. In right 3ABC, +B 90c and +BAC ., a 2 b 2 BC ·, a 2 b 2 AC, 2, Let BC (a b 2) and AC (a 2 b 2) . Then,, Given: sin , , AB 2 (AC 2 BC 2) (a 2 b 2) 2 (a 2 b 2) 2 4a 2 b 2, , , AB 2ab., , Now, we can find all the T-ratios of ., 8, 15 BC, 7. cot A 15 & tan A 8 AB ·, AC 2 AB 2 BC 2 (8) 2 (15) 2 64 225 289, , , , AC 289 17., BC 15, AC 17 ·, sin A , and sec A , 8, AC 17, AB, , 9, BC, 8. sin A 41 AC ·, AB 2 AC 2 BC 2 (41) 2 (9) 2 1681 81 1600, , , , AB 1600 40., AB 40, BC, 9, cos A AC 41 and tan A AB 40 ·
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Trigonometric Ratios, 6, 3 AB, 9. cos 10 5 AC ·, BC 2 AC 2 AB 2 (5) 2 (3) 2 25 9 16 & BC 4., , , 4, 4, sin 5 and tan 3 ·, , 1 BC, 10. cosec 2 & sin 2 AC ·, AB 2 AC 2 BC 2 (2) 2 (1) 2 4 1 3 & AB 3 ., 3, 1, sin 2 , cos 2 and cot 3 ., (2 3 ), 1, 3 2., #, Given expression ) 3 , (2 3 ), (2 3 ), , , 11. tan , , 1 BC ·, 7 AB, , , , AC 2 AB 2 BC 2 ( 7 ) 2 (1) 2 (7 1) 8 & AC 8 2 2 ., , , , AC 2 2, cosec BC 1 2 2 & cosec 2 8, , AC 2 2, and sec AB , & sec 2 78 ·, 7, 20 BC, 12. tan 21 AB ·, , , AC 2 AB 2 BC 2 (21) 2 (20) 2 441 400 841, , , , AC 841 29., , 5, 4 AB, 13. sec 4 & cos 5 AC ·, , , BC 2 AC 2 AB 2 (5) 2 (4) 2 25 16 9 & BC 3., , , , 3, 4, 3, 4, sin 5 , cos 5 , tan 4 , cot 3 ·, , 16. sin , , a BC ·, b AC, , , , AB 2 AC 2 BC 2 b 2 a 2, , , , AB b 2 a 2 ., b, a, sec , and tan , b2 a2, b2 a2, (b a), (b a), , (sec tan ) , (b a) · (b a), b2 a2, , , , , , , b a· b a, (b a) · (b a), , , , ba·, ba, , 543
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544, , Secondary School Mathematics for Class 10, , 4 BC, 18. tan 3 AB ·, AC 2 AB 2 BC 2 9 16 25 & AC 5., 4 3, 7, (sin cos ) a 5 5 k 5 ·, , 19., , (a sin b cos ) (a tan b), , [dividing num. and denom. by cos ], (a sin b cos ) (a tan b), a, `a # b bj (a 2 b 2), , 2 2 ·, a, `a # b bj (a b ), , (4 cos sin ) (4 tan ), , [dividing num. and denom. by cos ], (2 cos sin ) (2 tan ), a4 34 k 8 4, , , ·, a2 34 k 10 5, (4 sin 3 cos ) (4 3 cot ), , 21., [dividing num. and denom. by sin ], (2 sin 6 cos ) (2 6 cot ), a4 3 # 23 k 2 1, , ·, a2 6 # 23 k 6 3, 20., , 4 AB, 22. cot 3 BC ·, AC 2 AB 2 BC 2 (4) 2 (3) 2 16 9 25 & AC 5., , , 3, 4, 3, sin 5 , cos 5 and tan 4 ·, , 17, 8, AB, 23. sec 8 & cos 17 AC ·, , , BC 2 AC 2 AB 2 (17) 2 (8) 2 (289 64) 225, , BC 225 15., 15, 8, 15, sin 17 , cos 17 and tan 8 ·, 33, Then, LHS RHS 611 ·, , , 25. Given, 3ABC in which +B 90c, BC 7 cm and AB 24 cm., , , AC 2 (24) 2 (7) 2 576 49 625 & AC 25 cm., , BC, 7, AB 24, AB 24, (i) sin A AC 25 · (ii) cos A AC 25 · (iii) sin C AC 25 ·, BC, 7, (iv) cos C AC 25 ·, 3, 1, 3, 28. cos 2 (1 sin 2 ) a1 4 k 4 & cos 2 ·, , , 3, 3 3, 3 3 3 3, (3 cos 4 cos 3 ) c3 # 2 4 # 8 m c 2 2 m 0.
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Trigonometric Ratios, , 545, , 30. Let 3ACD and 3BEF be two right triangles, such that sin A sin B., Then, sin A sin B, CD EF, AD BF, CD AD, EF BF k (say), CD k (EF) and AD k (BF) ., , , AC , BE, , 2, , 2, , AD CD, BF 2 EF 2, , … (i), , [using Pythagoras’ theorem], , k BF 2 EF 2, k [using (i)]., BF 2 EF 2, CD AD AC ·, EF, BF, BE, 3ACD + 3BEF and hence +A +B., , , , , , 31. Let 3ACD and 3BEF be two right, triangles such that tan A tan B., Then, tan A tan B, CD EF, AC BE, CD AC, EF BE and +C +E 90c., 3ACD + 3BEF and hence, +A +B., BC, 32. tan A 1 & AB 1 & BC AB k (say)., AC 2 (AB 2 BC 2) (k 2 k 2) 2k 2 & AC 2 k., k 1, 2k, 2, AB, k 1 ·, and cos A AC , 2k, 2, 1, 1 , , 2 sin A cos A c2 #, #, m 1., 2, 2, , , BC, sin A AC , , 33. PQ 2 (PR 2 QR 2) (x 2) 2 x 2 4 (x 1), PQ 2 x 1 ., QR, , (i) cot , PQ 2, QR, , (ii) tan , PQ 2, , , , x, x, & ( x 1) cot ·, 2, x1, x, x2, x2, , & ( x 3 x 2 ) tan ·, 3, 2, 2, x1 2 x x, PQ 2 x 1, ·, , (iii) cos , PR, (x 2), (x y), 2 ·, 34. Adding, we get, cosec A , & sin A , 2, (x y), xy, ·, Subtracting, we get, cos A , 2, 2, 2 2 x ym , sin 2 A cos 2 A 1 & c m c, 1 0., x y, 2, ,
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546, , Secondary School Mathematics for Class 10, , T-Ratios of Some, Particular Angles, , 11, , TRIGONOMETRIC RATIOS OF 45, 60 AND 30, (GEOMETRICALLY), TRIGONOMETRIC RATIOS OF 45, , Let 3ABC be a right-angled triangle in which, +B 90c and +A 45c. Then, clearly, +C 45c., +A +C & AB BC., Let AB BC a units. Then,, AC AB 2 BC 2 a 2 a 2 2a 2 2 a units., Base AB a; perpendicular BC a and hypotenuse AC 2 a., , , BC, sin 45c AC , , a 1, ;, 2a, 2, , AB, cos 45c AC , , BC a, tan 45c AB a 1;, 1, 2;, sec 45c , cos 45c, , a 1, ;, 2a, 2, , 1, 2;, sin 45c, 1, 1., cot 45c , tan 45c, , cosec 45c , , TRIGONOMETRIC RATIOS OF 60 AND 30, , Consider an equilateral 3ABC with each side equal to 2a., Then, each angle of 3ABC is 60., From A, draw AD = BC., Then, clearly, BD DC a., Also, +ADB 90c., +BAD 30c., From right-angled 3ADB, we have, AD AB 2 BD 2 (2a) 2 a 2 4a 2 a 2 3a 2 3 a., T-RATIOS OF 60, , In right-angled 3ADB, we have, base BD a, perpendicular AD 3 a and hypotenuse AB 2a., 546
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548, , Secondary School Mathematics for Class 10, , SOLVED EXAMPLES, EXAMPLE 1, , Evaluate:, (i) sin 60c · cos 30c cos 60c · sin 30c, (ii) tan 30c · cosec 60c tan 60c · sec 30c, , SOLUTION, , On substituting the values of various T-ratios, we get, (i) sin 60c · cos 30c cos 60c · sin 30c, c, , 3, 3 1, # 1 m a3 1 k 2 1 ·, 2 # 2, 2, 2, 4 4, 4 2, (ii) tan 30c · cosec 60c tan 60c · sec 30c, c 1 # 2 3 # 2 m a 2 2k 2 2 ·, 3, 3, 3, 3, 3, EXAMPLE 2, , Evaluate:, 1, 1, (i) sin 2 30c · cos 2 45c 4 tan 2 30c 2 sin 2 90c 8 cot 2 60c, (ii), , SOLUTION, , tan 2 60c 4 sin 2 45c 3 sec 2 30c 5 cos 2 90c, cosec 30c sec 60c cot 2 30c, , On substituting the values of various T-ratios, we get, 1, 1, (i) sin 2 30c · cos 2 45c 4 tan 2 30c 2 sin 2 90c 8 cot 2 60c, , EXAMPLE 3, , 2, 2, 2, 2, a1 k · c 1 m 4 # c 1 m 1 # (1) 2 1 # c 1 m, 2, 2, 8, 2, 3, 3, 1, 1, 1, 1, 1, 1, a # 4# #1 # k, 2, 3 2, 8, 3, 4, 1, 4, 1, 1, 48, a k, , 8 3 2 24, 24 2., tan 2 60c 4 sin 2 45c 3 sec 2 30c 5 cos 2 90c, (ii), cosec 30c sec 60c cot 2 30c, 1 2, 2 2, ( 3 ) 2 4 # c m 3 # c m 5 # 02, 2, 3, , 2 2 ( 3) 2, JK, N, 1, 4, KK3 4 # 2 3 # 3 5 # 0OOO 3 2 4 0, K, k 9., Oa, 1, 43, L, P, Show that, , (i) cos 60c · cos 30c sin 60c · sin 30c cos 90c, (ii) cos 60c 1 2 sin 2 30c 2 cos 2 30c 1, tan 60c tan 30c , (iii) , tan 30c, 1 tan 60c · tan 30c
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T-Ratios of Some Particular Angles, SOLUTION, , 549, , We have, (i) cos 60c cos 30c sin 60c sin 30c, 3, 3, 1m c 3 3 m , , c1 #, 0., 2, 2, 2 #2, 4, 4, Also, cos 90c 0., cos 60c cos 30c sin 60c sin 30c cos 90c., 1, (ii) cos 60c 2 ;, 1 2, 1, 1, 1, 1 2 sin 2 30c :1 2 # a 2 k D a1 2 # 4 k a1 2 k 2 ;, 2, , 3, 3, 3, 1, 2 cos 2 30c 1 <2 # c 2 m 1F a2 # 4 1k a 2 1k 2 ·, , , cos 60c 1 2 sin 2 30c 2 cos 2 30c 1., , 1, c 3, m, (3 1), 3, , tan 60c tan 30c , , 1 ·, (iii) , 1 tan 60c tan 30c 1 3 # 1, 3#2, 3, c, m, 3, 1 ·, Also, tan 30c , 3, , tan 60c tan 30c , , tan 30c., 1 tan 60c tan 30c, EXAMPLE 4, , Taking 30c, verify each of the following:, (i) sin 2 2 sin cos , (ii) cos 2 2 cos 2 1 1 2 sin 2 , 2 tan , (iii) tan 2 , 1 tan 2 , , SOLUTION, , When 30c, we have 2 60c., 3, (i) sin 2 sin 60c 2 ·, 3, 3, 1, 2 sin cos 2 sin 30c cos 30c c2 # 2 # 2 m 2 ·, sin 2 2 sin cos ., 1, (ii) cos 2 cos 60c 2 ·, 2 cos 2 1 2 # cos 2 30c 1, <2 # c, , 2, , 3m, 1F a2 # 3 1k a 3 1k 1 ·, 2, 2, 2, 4
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550, , Secondary School Mathematics for Class 10, , 1 2 sin 2 1 2 sin 2 30c, 2, :1 2 # a1 k D a1 2 # 1 k a1 1 k 1 ·, 2, 2, 2, 4, 2, 2, , , , , cos 2 2 cos 1 1 2 sin ., (iii) tan 2 tan 60c 3 ., 1, 2, 2#, c m, 3, 3, 2 tan 2 tan 30c , , 1 tan 2 1 tan 2 30c 1 1 2 a1 1 k, c m, 3, 3, c 2 # 3m 3 ., 2, 3, 2 tan ·, tan 2 , 1 tan 2 , EXAMPLE 5, , Taking 30c, verify that:, (i) sin 3 3 sin 4 sin 3 , , SOLUTION, , (ii) cos 3 4 cos 3 3 cos , , When 30c, we have 3 90c., (i) sin 3 sin 90c 1., And, 3 sin 4 sin 3 3 sin 30c 4 sin 3 30c, , , , 3, :3 # 1 4 # a1 k D a 3 4 # 1 k, 2, 2, 2, 8, 3, 1, a k 1., 2 2, sin 3 3 sin 4 sin 3 ., , (ii) cos 3 cos 90c 0., And, 4 cos 3 3 cos 4 cos 3 30c 3 cos 30c, <4 # c, c4 #, c, , EXAMPLE 6, , 3, , 3m, 3, 3# F, 2, 2, , 3 3 3 3m, , 8, 2, , 3 3 3 3m, , 0., 2, 2, , cos 3 4 cos 3 3 cos ., , Taking A 60c and B 30c, verify that:, sin (A B) sin A cos B cos A sin B., , SOLUTION, , A 60c and B 30c & A B (60c 30c) 30c., 1, sin (A B) sin 30c 2 ·
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T-Ratios of Some Particular Angles, , 551, , And, sin A · cos B cos A · sin B, sin 60c · cos 30c cos 60c · sin 30c, c, , , , 3, 3 1, # 1m, 2 # 2, 2, 2, a3 1 k 2 1 ·, 4 4, 4 2, sin (A B) sin A cos B cos A sin B., , EXAMPLE 7, , 3, If sin (A B) 1 and cos (A B) 2 , find A and B., , SOLUTION, , sin (A B) 1 & sin (A B) sin 90c, , [a sin 90c 1], , & A B 90c., , … (i), , 3, cos (A B) 2 & cos (A B) cos 30c, & A B 30c., Solving (i) and (ii), we get A 60c and B 30c., , … (ii), , Hence, A 60c and B 30c., EXAMPLE 8, , If tan (A B) 3 and tan (A B) , A B, find the values of A and B., , SOLUTION, , 1, , 0c A B 90c and, 3, [CBSE 2008C], , tan (A B) 3 & tan (A B) tan 60c, [a tan 60c 3 ], , , … (i), & A B 60c., 1, 1, & tan (A B) tan 30c, tan (A B) , ;a tan 30c , E, 3, 3, … (ii), & A B 30c., , , Solving (i) and (ii), we get A 45c and B 15c., Hence, A 45c and B 15c., , EXAMPLE 9, , SOLUTION, , If tan (A B) 3 and tan (A B) 1, 0c (A B) 90c and, [CBSE 2009C], A B then find A and B., tan (A B) 3 & tan (A B) tan 60c, , & A B 60c., tan (A B) 1 & tan (A B) tan 45c, & A B 45c., , [a, , 3 tan 60c], … (i), , [a 1 tan 45c], , Solving (i) and (ii), we get A (52.5)c and B (7.5)c ., Hence, A (52.5)c and B (7.5)c ., , … (ii)
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552, , Secondary School Mathematics for Class 10, , EXAMPLE 10, , In the adjoining figure, 3ABC is right-angled at, B. If +A 30c and AB 9 cm, find (i) BC and, (ii) AC., , SOLUTION, , From right-angled 3ABC, we have, BC, BC, 1, (i) AB tan 30c & 9 cm , 3, 3, 9 3, 9, #, o cm 3 cm 3 3 cm., 3, 3, AC, AC , (ii) BC cosec 30c 2 &, 2 & AC 6 3 cm., 3 3 cm, Hence, BC 3 3 cm and AC 6 3 cm., , , EXAMPLE 11, , SOLUTION, , EXAMPLE 12, SOLUTION, , BC e, , In right 3ABC, +B 90c, AB 3 cm and, AC 6 cm. Find +C and +A., AB , 31, AC sin C & sin C 6 2, sin C sin 30c & +C 30c, +A 180c (90c 30c) (180c 120c) 60c., 1, 6, If 6x sec and x tan , find the value of 9 ax 2 2 k ·, x, We have, 6x sec , , … (i), , 6, and x tan , , [CBSE 2010], , … (ii), , Adding (i) and (ii), we get, 1, … (iii), 6 ax x k (sec tan )., Subtracting (ii) from (i), we get, 1, … (iv), 6 ax x k (sec tan )., Multiplying the corresponding sides of (iii) and (iv), we get, 1, 1, 1, 36 ax 2 2 k (sec 2 tan 2 ) 1 & 9 ax 2 2 k 4 ·, x, x, 1, 1, Hence, 9 ax 2 2 k 4 ·, x, f, , EXERCISE 11, , Evaluate each of the following:, , 1. sin 60c cos 30c cos 60c sin 30c 2. cos 60c cos 30c sin 60c sin 30c, sin 30c cot 45c sin 60c cos 30c, 3. cos 45c cos 30c sin 45c sin 30c 4., cos 45c sec 60c tan 45c sin 90c
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T-Ratios of Some Particular Angles, , 553, , 5 cos 2 60c 4 sec 2 30c tan 2 45c, sin 2 30c cos 2 30c, 2, 6. 2 cos 60c 3 sin 2 45c 3 sin 2 30c 2 cos 2 90c, 3, 1, 7. cot 2 30c 2 cos 2 30c 4 sec 2 45c 4 cosec 2 30c, 8. (sin 2 30c 4 cot 2 45c sec 2 60c)(cosec 2 45c sec 2 30c), 5., , 9., , 4, 1, , 2 cos 2 45c sin 2 0c, cot 2 30c sin 2 30c, , 10. Show that:, 1 sin 60c tan 60c 1, (i), cos 60c, tan 60c 1, , (ii), , cos 30c sin 60c , cos 30c, 1 sin 30c cos 60c, , 11. Verify each of the following:, (i) sin 60c cos 30c cos 60c sin 30c sin 30c, (ii) cos 60c cos 30c sin 60c sin 30c cos 30c, (iii) 2 sin 30c cos 30c sin 60c, (iv) 2 sin 45c cos 45c sin 90c, 12. If A 45c, verify that:, (i) sin 2A 2 sin A cos A, (ii) cos 2A 2 cos 2 A 1 1 2 sin 2 A, 13. If A 30c, verify that:, 2 tan A, 1 tan 2 A, 2 tan A, (ii) cos 2A , (iii) tan 2A , 2, 1 tan A, 1 tan 2 A, 1 tan 2 A, 14. If A 60c and B 30c, verify that:, (i) sin (A B) sin A cos B cos A sin B, (ii) cos (A B) cos A cos B sin A sin B, 15. If A 60c and B 30c, verify that:, (i) sin (A B) sin A cos B cos A sin B, (ii) cos (A B) cos A cos B sin A sin B, tan A tan B, (iii) tan (A B) , 1 tan A tan B, 1, 1, 16. If A and B are acute angles such that tan A 3 , tan B 2 and, tan A tan B, tan (A B) , , show that A B 45c., 1 tan A tan B, 2 tan A, 17. Using the formula, tan 2A , , find the value of tan 60, it, 1 tan 2 A, 1, ·, being given that tan 30c , 3, (i) sin 2A , , 18. Using the formula, cos A , 1, being given that cos 60c 2 ·, , 1 cos 2A, , find the value of cos 30c, it, 2
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554, , Secondary School Mathematics for Class 10, , 19. Using the formula, sin A , , 1 cos 2A, , find the value of sin 30c, it, 2, , 1, being given that cos 60c 2 ·, 20. In the adjoining figure, 3ABC is a right-angled, triangle in which +B 90c, +A 30c and AC 20 cm., Find (i) BC, (ii) AB., , 21. In the adjoining figure, 3ABC is right-angled at B, and +A 30c. If BC 6 cm, find (i) AB, (ii) AC., , 22. In the adjoining figure, 3ABC is right-angled at B, and +A 45c. If AC 3 2 cm, find (i) BC, (ii) AB., , 23. If sin (A B) 1 and cos (A B) 1, 0c# (A B) # 90c and A B then, find A and B., 1, 1, 24. If sin (A B) 2 and cos (A B) 2 , 0c (A B) 90c and A B then, find A and B., 1, and tan (A B) 3 , 0c (A B) 90c and A B, 3, then find A and B., , 25. If tan (A B) , , 1, 3, 26. If 3x cosec and x cot , find the value of 3 ax 2 2 k ·, x, 27. If sin (A B) sin A cos B cos A sin B, and cos (A B) cos A cos B sin A sin B,, find the values of (i) sin 75 and (ii) cos 15., , [CBSE 2010], , ANSWERS (EXERCISE 11), , 1. 1, , 2. 0, , 3., , ( 3 1), 2 2, , 4., , ( 2 1), 2, , 67, , 5. 12, , 5, , 6. 4
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T-Ratios of Some Particular Angles, , 7. 1, , 2, , 13, , 8. 3, , 9. 3, , 17., , 3, , 21. (i) AB 6 3 cm, , (ii) AB 10 3 cm, (ii) AC 12 cm, , 22. (i) BC 3 cm, , (ii) AB 3 cm, , 20. (i) BC = 10 cm, , ( 3 1), , (ii), , 2 2, , 3, , 18. 2, , 1, , 19. 2, , 23. +A +B 45c, , 24. +A 45c, +B 15c 25. +A 45c, +B 15c, 27. (i), , 555, , ( 3 1), , 1, , 26. 3, , 2 2, , HINTS TO SOME SELECTED QUESTIONS, x, 1, 20. Let BC x cm. Then, 20 sin 30c 2 & x 10., AB 2 (AC 2 BC 2) {(20) 2 (10) 2} cm 2 300 cm 2 & AB 300 cm 10 3 cm., , , BC 10 cm and AB 10 3 cm., , x, 21. Let AB x cm. Then, 6 cot 30c 3 & x 6 3 ., AC 2 (AB 2 BC 2) {(6 3 ) 2 (6) 2} cm 2 (108 36) cm 2 144 cm 2 & AC 12 cm., 22. Let BC x cm. Then,, , x , 1, sin 45c , & x 3., 3 2, 2, , AB , AB , , , BC tan 45c 1 & 3 cm 1 & AB 3 cm., 23. sin (A B) 1 & sin (A B) sin 90c., cos (A B) 1 & cos (A B) cos 0c., Solve A B 90c, A B 0c for A and B., 1, 24. sin (A B) 2 & sin (A B) sin 30c., 1, cos (A B) 2 & cos (A B) cos 60c., Solve A B 30c and A B 60c for A and B., 25. tan (A B) , , 1, & tan (A B) tan 30c., 3, , tan (A B) 3 & tan (A B) tan 60c., Solve A B 30c and A B 60c for A and B., 1, 26. Adding, we get 3 ax x k (cosec cot ) ., 1, Subtracting, we get 3 ax x k (cosec cot ) ., , , 9 cx 2 , , 1 , 1, 1, 2, 2, 2, m (cosec cot ) 1 & 3 cx 2 m 3 ·, x2, x, , 27. Take A 45c and B 30c., ,
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556, , Secondary School Mathematics for Class 10, , Trigonometric Ratios of, Complementary Angles, , 12, , TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES, COMPLEMENTARY ANGLES, , Two angles are said to be complementary if their sum, , is 90c., Thus, and (90c ) are complementary angles., TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES, , Consider a 3OMP, right-angled at M in which, +MOP and therefore, +OPM (90c ) ., Let OM x, MP y and OP r. Then,, y, y, x, sin r , cos r , tan x ,, r, r, x, cosec y , sec x , cot y ·, For considering the trigonometric ratios of (90c ), we have, base = MP, perpendicular = OM and hypotenuse = OP., , , OM x, sin (90c ) OP r cos , PM y, cos (90c ) OP r sin , OM x, tan (90c ) PM y cot , , , , cosec (90c ) , , 1, 1 sec , sin (90c ) cos , , sec (90c ) , , 1, 1 cosec , cos (90c ) sin , , cot (90c ) , , 1, 1 tan , tan (90c ) cot , , SUMMARY, , (i) sin (90c ) cos , , (ii) cos (90c ) sin , , (iii) tan (90c ) cot , (v) sec (90c ) cosec , , (iv) cot (90c ) tan , (vi) cosec (90c ) sec , 556
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Trigonometric Identities, , 567, , Trigonometric Identities, , 13, TRIGONOMETRIC IDENTITY, , An equation involving trigonometric ratios of an angle is called a trigonometric, identity if it is true for all values of the angle., SOME TRIGONOMETRIC IDENTITIES, , In Chapter 10, we proved some identities,, , summarised below., SUMMARY, , (i) sin 2 cos 2 1, , (ii) 1 tan 2 sec 2 , , (iii) 1 cot 2 cosec 2 , , (iv) tan , , sin , cos , and cot , cos , sin , , (v) tan # cot 1, , SOLVED EXAMPLES, EXAMPLE 1, , Prove that, (i) (1 sin 2 ) sec 2 1, , (ii) (1 tan 2 ) cos 2 1, , (iii) (1 tan 2 ) (1 sin )(1 sin ) 1, (iv) 2cos 2 , SOLUTION, , 2, 2, (1 cot 2 ), , [CBSE 2009C], , We have, (i) LHS (1 sin 2 ) sec 2 , [a 1 sin 2 cos 2 ], , cos 2 · sec 2 , cos 2 ·, , , 1 , 1 RHS., cos 2 , , LHS = RHS., , (ii) LHS (1 tan 2 ) cos 2 , [a 1 tan 2 sec 2 ], , sec 2 · cos 2 , , , , 1 ·, cos 2 1 RHS., cos 2 , , LHS = RHS., 567
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568, , Secondary School Mathematics for Class 10, , (iii) LHS (1 tan 2 ) (1 sin ) (1 sin ), (1 tan 2 )(1 sin 2 ) sec 2 · cos 2 , [a (1 tan 2 ) sec 2 , (1 sin 2 ) cos 2 ], , , 1 ·, cos 2 , cos 2 , , 2, ;a sec , , 1, E, cos 2 , , 1 RHS., , , LHS = RHS., , (iv) LHS 2 cos 2 , 2 cos 2 , , 2, (1 cot 2 ), 2, cosec 2 , , [a 1 cot 2 cosec 2 ], ;a, , 2 cos 2 2 sin 2 , 2(cos 2 sin 2 ) 2 #1, , 1, sin 2 E, cosec 2 , , [a cos 2 sin 2 1], , 2 RHS., , EXAMPLE 2, , LHS = RHS., , Prove that, (i) (cosec 2 1) tan 2 1, , SOLUTION, , (ii) (sec 2 1)(1 cosec 2 ) 1, , We have, (i) LHS (cosec 2 1) tan 2 , (1 cot 2 1) tan 2 , , [a cosec 2 1 cot 2 ], , cot 2 · tan 2 , 1 ·, tan 2 , tan 2 , 1 RHS., , , :a cot , , 1, D, tan , , (ii) LHS (sec 2 1)(1 cosec 2 ), (1 tan 2 1)[1 (1 cot 2 )], [a sec 2 1 tan 2 and cosec 2 1 cot 2 ], tan 2 · (cot 2 ), tan 2 · c, , 1, m, tan 2 , , 1 RHS., , :a cot , , 1, D, tan
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Trigonometric Identities, EXAMPLE 3, , Prove that, , SOLUTION, , sin 3 cos 3 , sin cos 1., sin cos , We have, LHS , , , , 569, , [CBSE 2009], , sin 3 cos 3 , sin cos , sin cos , (sin cos )(sin 2 cos 2 sin cos ), sin cos , (sin cos ), [a (a 3 b 3) (a b) (a 2 b 2 ab)], , (1 sin cos ) sin cos 1 [a sin 2 cos 2 1], RHS., , EXAMPLE 4, , LHS = RHS., , Prove that, (sin cosec ) (cos sec ) , , SOLUTION, , We have, , 1, · [CBSE 2009C, ’10], tan cot , , LHS (sin cosec )(cos sec ), , asin , , , , (sin 2 1) (cos 2 1), 1, 1, ·, kacos , k, sin , cos , sin , cos , , {(1 sin 2 )} · [(1 cos 2 )}, sin cos , (cos 2 )(sin 2 ), sin cos , [a 1 sin 2 cos 2 and 1 cos 2 sin 2 ], , (cos 2 sin 2 ), cos sin ., sin cos , 1, RHS , tan cot , 1, 1, , cos sin , , sin cos sin 2 cos 2 (sin 2 cos 2 ), cos sin , cos sin , cos sin , [a sin 2 cos 2 1]., , , , EXAMPLE 5, , LHS = RHS., , Prove that, (1 cot cosec )(1 tan sec ) 2., , [CBSE 2005, ’07, ’08]
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570, SOLUTION, , Secondary School Mathematics for Class 10, , We have, LHS (1 cot cosec )(1 tan sec ), , , EXAMPLE 6, , SOLUTION, , a1 cos 1 ka1 sin 1 k, sin sin , cos cos , (sin cos 1) (cos sin 1), ·, , sin , cos , [(sin cos ) 1] · [(sin cos ) 1], , sin cos , (sin cos ) 2 1, , sin cos , 2, 2, , , , sin cos 2 sin cos 1, sin cos , , , 1 2 sin cos 1, [a sin 2 cos 2 1], sin cos , 2 sin cos 2 RHS., sin cos , LHS = RHS., , Prove that, cos sin , (cos sin ) ., (1 tan ) (1 cot ), , [CBSE 2007], , We have, cos sin , (1 tan ) (1 cot ), , cos sin , sin , cos , a1 , k a1 , k, cos , sin , cos 2 , sin 2 , , , , (cos sin ) (sin cos ), , LHS , , cos 2 , sin 2 , , cos sin cos sin , 2, 2, (cos sin )(cos sin ), , cos sin , (cos sin ), (cos sin ), , , (cos sin ) RHS., , EXAMPLE 7, , SOLUTION, , LHS = RHS., , Prove that, sin 2 sin 3 , tan ., 2 cos 3 cos , We have, LHS , , 2, sin 2 sin 3 sin (1 2 sin ), 2 cos 3 cos cos (2 cos 2 1), , [CBSE 2010]
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Trigonometric Identities, , , , , EXAMPLE 8, , SOLUTION, , 571, , sin (1 2 sin 2 ), cos (2 cos 2 1), , tan ·, , [1 2(1 cos 2 )], (2 cos 2 1), , tan ·, , (2 cos 2 1), tan RHS., (2 cos 2 1), , [a sin 2 1 cos 2 ], , LHS = RHS., , Prove that, tan A cot A , (1 tan A cot A) ., (1 cot A) (1 tan A), , [CBSE 2010], , We have, tan A cot A , LHS , (1 cot A) (1 tan A), , sin A, cos A, cos A sin A, cos A, sin A, c1 , m c1 , m, sin A, cos A, , , , sin 2 A, cos 2 A, , , cos A (sin A cos A) sin A (cos A sin A), , , , sin 2 A, cos 2 A, , cos A (sin A cos A) sin A (sin A cos A), , , , sin 3 A cos 3 A, sin A cos A (sin A cos A), , , , (sin A cos A)(sin 2 A cos 2 A sin A cos A), sin A cos A (sin A cos A), , , 1 sin A cos A ·, sin A cos A, RHS (1 tan A cot A), , [a (a 3 b 3) (a b)(a 2 b 2 ab)], , 2, 2, , , c1 sin A cos A m sin A cos A sin A cos A, cos A sin A, sin A cos A, , , , EXAMPLE 9, , (1 sin A cos A), ·, sin A cos A, , LHS = RHS., , Prove that, (cosec A sin A) (sec A cos A) , , 1, ·, (tan A cot A), [CBSE 2010]
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572, SOLUTION, , Secondary School Mathematics for Class 10, , We have, LHS (cosec A sin A)(sec A cos A), 2, 2, , , a 1 sin Aka 1 cos Ak c1 sin A m · c1 cos A m, sin A, cos A, sin A, cos A, 2, 2, cos A sin A cos A sin A., cos A sin A, 1, 1, , RHS , sin A cos A, (tan A cot A), c, m, cos A sin A, , , , cos A sin A , cos A sin A [a sin 2 A cos 2 A 1] ., (sin 2 A cos 2 A), , Hence, LHS = RHS., EXAMPLE 10, , Prove that, d, , SOLUTION, , 1 tan 2 A 1 tan A 2 , k tan 2 A., n a , 1 cot A, 1 cot 2 A, , [CBSE 2008C], , We have, d, , 1 tan 2 A sec 2 A, n, 1 cot 2 A, cosec 2 A, [a 1 tan 2 A sec 2 A and 1 cot 2 A cosec 2 A], , , , sin 2 A , 1 ·, sin 2 A , tan 2 A., 2, cos A, cos 2 A, , sin A 2, m, cos A, 1 tan A , k, And, a , 1 cot A, cos A 2, c1 , m, sin A, c1 , , 2, , (cos A sin A) 2, , , 2, cos 2 A, sin 2 A tan 2 A., 2, (sin A cos A), cos A, , sin 2 A, , EXAMPLE 11, , d, , 1 tan 2 A 1 tan A 2 , k tan 2 A., n a , 1 cot A, 1 cot 2 A, , Prove that, tan A sin A sec A 1, (i), tan A sin A sec A 1, cot A cos A cosec A 1, (ii), cot A cos A cosec A 1, , [CBSE 2008]
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Trigonometric Identities, SOLUTION, , 573, , We have, tan A sin A, tan A sin A, sin A , 1 , 1k, sin A sin A a, cos A, , cos A, 1 , sin A , 1k, sin A sin A a, cos A, cos A, 1 , 1k, a, , A, cos, , sec A 1 RHS., 1 , sec A 1, 1k, a, cos A, , (i) LHS , , , , LHS = RHS., , cot A cos A, cot A cos A, cos A , 1 , 1k, cos A cos A a, sin, A, A, sin, , , 1 , cos A , 1k, cos A cos A a, sin A, sin A, 1 , 1k (cosec A 1), a, A, sin, , , RHS., 1 , (cosec A 1), 1k, a, sin A, , (ii) LHS , , , EXAMPLE 12, , LHS = RHS., , Prove that, (i) sec 2 cosec 2 sec 2 cosec 2 , (ii) tan 2 sin 2 tan 2 sin 2 , (iii) tan 2 cot 2 2 sec 2 cosec 2 , , SOLUTION, , We have, (i) LHS sec 2 cosec 2 , 1 1 sin 2 cos 2 , cos 2 sin 2 , cos 2 sin 2 , 1, , [a sin 2 cos 2 1], cos 2 sin 2 , sec 2 cosec 2 RHS., LHS RHS., , , , , (ii) LHS tan 2 sin 2 , 2, 2, 2, 2, , sin 2 sin 2 sin sin2 cos , cos , cos
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574, , Secondary School Mathematics for Class 10, , sin 2 (1 cos 2 ), , 2, sin 2 · sin 2 , cos , cos , tan 2 sin 2 RHS., LHS = RHS., , , , , , 2, , (iii) LHS tan 2 cot 2 2, (1 tan 2 ) (1 cot 2 ) sec 2 cosec 2 , 1 1 sin 2 cos 2 , cos 2 sin 2 , cos 2 sin 2 , 1, , sec 2 cosec 2 RHS., 2, cos sin 2 , LHS = RHS., , , , EXAMPLE 13, , Prove that, (cosec cot ) 2 , , SOLUTION, , 1 cos ·, 1 cos , , [CBSE 2000C], , LHS (cosec cot ) 2, 2, 2, , a 1 cos k a1 cos k, sin sin , sin , , (1 cos ) 2, , (1 cos )(1 cos ), (1 cos 2 ), , [a sin 2 1 cos 2 ], sin , (1 cos )(1 cos ) (1 cos ), , , RHS., (1 cos )(1 cos ) (1 cos ), , , , EXAMPLE 14, , 2, , , , LHS = RHS., , Prove that, (sec 4 sec 2 ) (tan 2 tan 4 ) ., , SOLUTION, , We have, LHS (sec 4 sec 2 ) sec 2 (sec 2 1), , (1 tan 2 )(1 tan 2 1) (1 tan 2 ) tan 2 , (tan 2 tan 4 ) RHS., , EXAMPLE 15, , LHS = RHS., , Prove that, c1 , , 1, 1, 1, ·, mc1 , m, tan 2 A, cot 2 A, (sin 2 A sin 4 A), , [CBSE 2006C]
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Trigonometric Identities, SOLUTION, , 575, , We have, 1, 1, mc1 , m, tan 2 A, cot 2 A, (1 cot 2 A)(1 tan 2 A) cosec 2 A · sec 2 A, 1, 12 · 12 , sin A cos A sin 2 A cos 2 A, 1, 1, , , RHS., sin 2 A (1 sin 2 A) (sin 2 A sin 4 A), , LHS c1 , , , EXAMPLE 16, , LHS = RHS., , Prove that, (sin cosec ) 2 (cos sec ) 2 (7 tan 2 cot 2 ) ., [CBSE 2008, ’09C], , SOLUTION, , We have, LHS (sin cosec ) 2 (cos sec ) 2, , (sin 2 cosec 2 2 sin cosec ), (cos 2 sec 2 2 cos sec ), (sin 2 cosec 2 2) (cos 2 sec 2 2), [a sin cosec 1 and cos sec 1], (sin 2 cos 2 ) 4 (cosec 2 sec 2 ), 1 4 (1 cot 2 ) (1 tan 2 ), [a sin 2 cos 2 1, cosec 2 1 cot 2 , and sec 2 1 tan 2 ], 2, 2, (7 tan cot ) RHS., , EXAMPLE 17, , SOLUTION, , LHS = RHS., , Prove that, 1 sin , (i) , (sec tan ) 2, 1 sin , (1 cos ), (cosec cot ) 2, (ii), (1 cos ), , [CBSE 2008C], , [CBSE 2007C], , We have, (i) RHS (sec tan ) 2, 2, 2, (1 sin ) 2, , a 1 sin k a1 sin k , cos cos , cos , cos 2 , (1 sin ) 2 (1 sin )(1 sin ), , , (1 sin 2 ) (1 sin )(1 sin )
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576, , Secondary School Mathematics for Class 10, , , , , (1 sin ), LHS., (1 sin ), , RHS = LHS., , 1 cos 2, k, sin sin , 2, (1 cos ) 2 (1 cos )(1 cos ), , a1 cos k , , sin , sin 2 , 1 cos 2 , (1 cos )(1 cos ) (1 cos ), , , LHS., (1 cos )(1 cos ) (1 cos ), , (ii) RHS (cosec cot ) 2 a, , , EXAMPLE 18, , SOLUTION, , LHS = RHS., , Prove that, (i), , 1, (cosec cot ), (cosec cot ), , (ii), , (sec tan ), (1 2 tan 2 2 sec tan ), (sec tan ), , We have, (i) LHS , , , (cosec cot ), 1, 1, , #, (cosec cot ) (cosec cot ) (cosec cot ), (cosec cot ), (cosec 2 cot 2 ), , cosec cot RHS, [a cosec 2 cot 2 1] ., , , , LHS = RHS., , (ii) LHS , , (sec tan ) (sec tan ) (sec tan ), , #, (sec tan ) (sec tan ) (sec tan ), , (sec tan ) 2, (sec tan ) 2, (sec 2 tan 2 ), [a sec 2 tan 2 1], 2, 2, sec tan 2 sec tan , (1 tan 2 ) tan 2 2 sec tan , , , 1 2 tan 2 2 sec tan RHS., , EXAMPLE 19, , LHS = RHS., , Prove that, 1, 1, ·, 1 1 , (cosec cot ) sin sin (cosec cot ), [CBSE 2002C, ’06]
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Trigonometric Identities, SOLUTION, , EXAMPLE 20, , SOLUTION, , 577, , 1, 1, (cosec cot ) sin , (cosec cot ), 1, 1, , #, (cosec cot ) (cosec cot ) sin , (cosec cot ), 1 , cosec , , :a, cosec D, sin , (cosec 2 cot 2 ), (cosec cot ) cosec , [a cosec 2 cot 2 1], cot ., 1 , 1, RHS , sin (cosec cot ), (cosec cot ), 1, cosec , #, , (cosec cot ) (cosec cot ), (cosec cot ), cosec , (cosec 2 cot 2 ), cosec (cosec cot ), [a cosec 2 cot 2 1], cot ., LHS = RHS., LHS , , Prove that, 1, 1, ·, 1 1 , (sec tan ) cos cos (sec tan ), , [CBSE 2005, ’08], , We have, LHS , , 1, 1, (sec tan ) cos , , (sec tan ), 1, sec , #, (sec tan ) (sec tan ), (sec tan ), sec , , (sec 2 tan 2 ), , , (sec tan ) sec , , [a sec 2 tan 2 1], , tan ., 1 , 1, RHS , cos (sec tan ), sec , sec , , , , (sec tan ), 1, #, (sec tan ) (sec tan ), (sec tan ), , (sec 2 tan 2 ), sec (sec tan ), tan ., LHS = RHS., , [a sec 2 tan 2 1]
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578, EXAMPLE 21, , Secondary School Mathematics for Class 10, , Prove that, sin , (i), (cosec cot ), (1 cos ), (ii), , SOLUTION, , 1, (sec tan ), (sec tan ), , We have, (i) LHS , , , , , , , sin , (1 cos ), (1 cos ), sin , #, (1 cos ) (1 cos ), [multiplying num. and denom. by (1 cos )], sin (1 cos ) sin (1 cos ) (1 cos ), , , sin , (1 cos 2 ), sin 2 , , a 1 cos k (cosec cot ) RHS., sin sin , LHS = RHS., , (ii) LHS , , , 1, (sec tan ), (sec tan ), 1, #, , (sec tan ) (sec tan ), [multiplying num. and denom. by (sec tan )], , (sec tan ), (sec 2 tan 2 ), (sec tan ), , , [a sec 2 tan 2 1], , RHS., , EXAMPLE 22, , SOLUTION, , LHS RHS., , Prove that, sec tan 1 cos ·, [CBSE 2002, ’03, ’05C], tan sec 1 (1 sin ), sec tan 1, LHS , tan sec 1, (sec tan ) (sec 2 tan 2 ), , [a sec 2 tan 2 1], (tan sec 1), (sec tan )[1 (sec tan )], , (tan sec 1), (sec tan )(tan sec 1), , (sec tan ), (tan sec 1)
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Trigonometric Identities, , 579, , (1 sin ) (1 sin ) (1 sin ), a 1 sin k , , #, cos cos , cos , cos , (1 sin ), , , (1 sin 2 ), cos 2 , , cos RHS., , cos (1 sin ) cos (1 sin ) (1 sin ), , Hence,, EXAMPLE 23, , SOLUTION, , sec tan 1 cos ·, tan sec 1 (1 sin ), , Prove that, sin cos 1 , 1, ·, sin cos 1 (sec tan ), sin cos 1, LHS , sin cos 1, sin 1, 1, , cos, cos , , sin 1, 1, cos , cos , [on dividing num. and denom. by cos ], (sec tan 1), , , tan 1 sec , tan 1 sec (tan sec 1), (sec tan ) (sec 2 tan 2 ), [a 1 sec 2 tan 2 ], (tan sec 1), (sec tan )[1 (sec tan )], , (tan sec 1), (sec tan )(tan sec 1), , (sec tan ) ., (tan sec 1), , , RHS , , , , 1, (sec tan ), (sec tan ) (sec tan ), 1, , #, (sec tan ) (sec tan ) (sec 2 tan 2 ), , (sec tan ), Hence, LHS = RHS., EXAMPLE 24, , SOLUTION, , Prove that, cot cosec 1 1 cos ·, cot cosec 1, sin , We have, (cot cosec ) 1, LHS , (cot cosec 1), , [a sec 2 tan 2 1] .
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580, , Secondary School Mathematics for Class 10, , (cosec cot ) (cosec 2 cot 2 ), (cot cosec 1), [a 1 cosec 2 cot 2 ], (cosec cot )[1 (cosec cot )], , (cot cosec 1), (cosec cot )(cot cosec 1), , (cot cosec 1), , (cosec cot ) a 1 cos k 1 cos RHS., sin sin , sin , cot cosec 1 1 cos ·, Hence,, cot cosec 1, sin , , , EXAMPLE 25, , Prove that, sec 2 cosec 2 tan cot ., , SOLUTION, , We have, sec 2 cosec 2 , , LHS , , (1 tan 2 ) (1 cot 2 ) tan 2 cot 2 2, tan 2 cot 2 2 tan · cot , , [a tan $ cot 1], , (tan cot ) tan cot RHS., 2, , , EXAMPLE 26, , SOLUTION, , LHS = RHS., , Prove that, 1 sin , (i), sec tan , 1 sin , 1 cos , (ii), cosec cot , 1 cos , We have, 1 sin , 1 sin 1 sin , #, (i) LHS , , 1 sin , 1 sin , 1 sin , sin , sin , 1, 1, , , cos , 1 sin 2 , 1 sin , sec tan RHS., cos cos , LHS = RHS., , , , , 1 cos , 1 cos 1 cos , #, 1 cos , 1 cos , 1 cos , cos , cos , 1, 1, , , sin , 1 cos 2 , , (ii) LHS
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Trigonometric Identities, , 581, , 1 cos , cosec cot RHS., sin sin , LHS = RHS., , , , EXAMPLE 27, , Prove that, 1 sin , 1 sin , , SOLUTION, , 1 sin , 2 sec ., 1 sin , , [CBSE 2001C], , We have, LHS , , 1 sin , 1 sin , , 1 sin , 1 sin , , 1 sin , 1 sin , , , , 1 sin 1 sin , 1 sin , 1 sin , (1 sin )(1 sin ), 2, 2, , 2 2 sec RHS., , 2, 2, (1 sin ), cos cos , LHS = RHS., , , , EXAMPLE 28, , SOLUTION, , Prove that, sec 1 , sec 1, , sec 1 , 2 cosec ., sec 1, , [CBSE 2006C], , We have, LHS , , sec 1 , sec 1, , sec 1 , sec 1, , sec 1, sec 1, , sec 1, sec 1, , , , sec 1 sec 1 , (sec 1) (sec 1), , , EXAMPLE 29, , SOLUTION, , 2 sec 2 sec , tan , sec 2 1, [a sec 2 1 tan 2 ], 2 sec cot 2 # cos 2 2 cosec RHS., cos sin sin , LHS = RHS., , Prove that, cosec cosec , 2 sec 2 ., (cosec 1) (cosec 1), , [CBSE 2004C, ’09], , We have, cosec cosec , (cosec 1) (cosec 1), cosec (cosec 1) cosec (cosec 1), , (cosec 2 1), , LHS , , , , 2 cosec 2 , (1 cot 2 1), , [a cosec 2 1 cot 2 ]
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582, , Secondary School Mathematics for Class 10, 2, , 2 cosec, 2 cosec 2 tan 2 , 2, cot , , 2#, , sin 2 2 , 1, 2 sec 2 RHS., 2 #, sin cos 2 cos 2 , , AN IMPORTANT REMARK, , In order to show that a given trigonometric equation is not an identity, it is, sufficient to find by hit and trial, a single value of which does not satisfy it., EXAMPLE 30, , Show that (cos 2 sin 2 ) , , SOLUTION, , Putting 30c, we find:, , 2 tan , is not an identity., (1 tan 2 ), , LHS (cos 2 30c sin 2 30c), 2, , )d, , 2, 3, n a1 k 3 a 3 1 k 2 1 , and, 2, 2, 4 4, 4 2, 2, 1, 2#, 3, 3, 2 tan 30c , , d 2 # 3n 3 ., RHS , 2, 2, 1, 1, (1 tan 30c), 3 2, , <1 d n F a1 3 k, 3, LHS ! RHS., , Hence, the given equation is not an identity., EXAMPLE 31, , Prove that, (i) (sin 2 A cos 2 B cos 2 A sin 2 B) (sin 2 A sin 2 B), (ii) (tan 2 A sec 2 B sec 2 A tan 2 B) (tan 2 A tan 2 B), , SOLUTION, , We have, (i) LHS (sin 2 A cos 2 B cos 2 A sin 2 B), sin 2 A (1 sin 2 B) (1 sin 2 A) sin 2 B, sin 2 A sin 2 B RHS., (ii) LHS (tan 2 A sec 2 B sec 2 A tan 2 B), tan 2 A (1 tan 2 B) (1 tan 2 A) tan 2 B, (tan 2 A tan 2 B) RHS., , EXAMPLE 32, , Prove that, (tan 2 A tan 2 B) , , (sin 2 A sin 2 B), 2, , 2, , cos A cos B, , , , (cos 2 B cos 2 A), cos 2 B cos 2 A, , ·, , [CBSE 2005]
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Trigonometric Identities, SOLUTION, , 583, , We have, sin 2 A sin 2 B, cos 2 A cos 2 B, , (tan 2 A tan 2 B) , , 2, 2, 2, 2, , sin A cos 2B cos2 A sin B, cos A cos B, , sin 2 A (1 sin 2 B) (1 sin 2 A) sin 2 B, , , , cos 2 A cos 2 B, (sin 2 A sin 2 B), , , , cos 2 A cos 2 B, , Also, (tan 2 A tan 2 B) , , ·, , sin 2 A sin 2 B, cos 2 A cos 2 B, , 2, 2, 2, 2, , sin A cos 2B cos2 A sin B, cos A cos B, (1 cos 2 A) cos 2 B cos 2 A (1 cos 2 B), , cos 2 B cos 2 A, (cos 2 B cos 2 A), ·, , cos 2 B cos 2 A, , Hence, (tan 2 A tan 2 B) , f, , (sin 2 A sin 2 B), 2, , 2, , cos A cos B, , , , (cos 2 B cos 2 A), cos 2 B cos 2 A, , EXERCISE 13A, , Prove each of the following identities:, 1. (i) (1 cos 2 ) cosec 2 1, , (ii) (1 cot 2 ) sin 2 1, , 2. (i) (sec 2 1) cot 2 1, , (ii) (sec 2 1)(cosec 2 1) 1, , (iii) (1 cos 2 ) sec 2 tan 2 , 3. (i) sin 2 , , 1, 1, (1 tan 2 ), , (ii), , 1, 1, , 1, (1 tan 2 ) (1 cot 2 ), , 4. (i) (1 cos )(1 cos )(1 cot 2 ) 1, (ii) cosec (1 cos )(cosec cot ) 1, 1 , 1, sin 2 , 1, 1, (iii) cos 2 , (1 cot 2 ), , 5. (i) cot 2 , , 6., , (ii) tan 2 , , 1, 1, , 2 sec 2 , (1 sin ) (1 sin ), , 1 , 1, cos 2 , , ·
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584, , Secondary School Mathematics for Class 10, , 7. (i) sec (1 sin )(sec tan ) 1, (ii) sin (1 tan ) cos (1 cot ) (sec cosec ), 8. (i) 1 , (ii) 1 , 9., , [CBSE 2008], , cot , cosec , (1 cosec ), 2, , tan 2 , sec , (1 sec ), , (1 tan 2 ) cot , cosec 2 , , tan , , tan 2 cot 2 , 1, (1 tan 2 ) (1 cot 2 ), sin (1 cos ) , 11., 2 cosec , sin , (1 cos ), tan cot , 12., (1 sec cosec ), (1 cot ) (1 tan ), 10., , 13., , cos 2 , sin 3 , (1 sin cos ), (1 tan ) (sin cos ), , 14., , cos , sin 2 , (cos sin ), (1 tan ) (cos sin ), , 15. (1 tan 2 ) (1 cot 2 ) , 16., , [CBSE 2008C], , 1, (sin 2 sin 4 ), , cot , tan , , sin cos , (1 tan 2 ) 2 (1 cot 2 ) 2, , 17. (i) sin6 cos6 1 3 sin 2 cos 2 , (ii) sin 2 cos 4 cos 2 sin 4 , (iii) cosec 4 cosec 2 cot 4 cot 2 , 18. (i), (ii), 19. (i), (ii), 20. (i), (ii), , 1 tan 2 , (cos 2 sin 2 ), 1 tan 2 , 1 tan 2 , tan 2 , cot 2 1, tan tan , 2 cosec , (sec 1) (sec 1), (cosec 1), cot , , 2 sec , cot , (cosec 1), sec 1 , sin 2 , sec 1 (1 cos ) 2, sec tan cos 2 , sec tan (1 sin ) 2, , [CBSE 2007]
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Trigonometric Identities, , 21. (i), (ii), (iii), , 585, , 1 sin , (sec tan ), 1 sin , 1 cos , (cosec cot ), 1 cos , 1 cos 1 cos , 2 cosec , 1 cos , 1 cos , , cos 3 sin 3 cos 3 sin 3 , 2, cos sin , cos sin , sin , sin , , 2, 23., (cot cosec ) (cot cosec ), 22., , [CBSE 2009C], , sin cos sin cos , 2, sin cos sin cos (2 sin 2 1), sin cos sin cos , 2, (ii), sin cos sin cos (1 2 cos 2 ), , 24. (i), , 25., , 1 cos sin 2 , cot , sin (1 cos ), , 26. (i), , cosec cot , (cosec cot ) 2 1 2 cot 2 2 cosec cot , cosec cot , , sec tan , (ii), (sec tan ) 2 1 2 tan 2 2 sec tan , sec tan , 1 cos sin 1 sin , 27. (i) , 1 cos sin , cos , sin 1 cos 1 sin , (ii), cos 1 sin , cos , sin , cos , , 1, 28., (sec tan 1) (cosec cot 1), sin cos sin cos , 2, 2, , sin cos sin cos (sin 2 cos 2 ) (2 sin 2 1), cos cosec sin sec cosec sec , 30., cos sin , sec cosec , 31. (1 tan cot )(sin cos ) c, m, cosec 2 sec 2 , 29., , 32., , [CBSE 2003], , [CBSE 2003], , [CBSE 2001C], , [CBSE 2007], , cot 2 (sec 1) sec 2 (sin 1), , 0, (1 sin ), (1 sec ), , 1 sin 2 cos 2 , 1, 1, 2, 2, , 33. (, 2 (sin cos ) , 2, 2, 2, 2, (sec cos ) (cosec sin ), 2 sin 2 cos 2 , (sin A sin B) (cos A cos B), , 0, 34., (cos A cos B) (sin A sin B)
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586, , Secondary School Mathematics for Class 10, , tan A tan B , tan A tan B, cot A cot B, 36. Show that none of the following is an identity:, (i) cos 2 cos 1 (ii) sin 2 sin 2 (iii) tan 2 sin cos 2 , 37. Prove that: (sin 2 sin 3 ) (2 cos 3 cos ) tan ., 35., , 38. If 1 sin 2 3 sin cos then prove that tan 1 or, , 1·, 2, , HINTS TO SOME SELECTED QUESTIONS, 8. (i) Write cot 2 (cosec 2 1) ., (ii) Write tan 2 (sec 2 1) ., cos 2 , tan 2 cot 2 sin 2 , # cos 2 n d 2 # sin 2 n (sin 2 cos 2 ) 1., d, sec 2 cosec 2 , cos 2 , sin , sin, cos, , , ·, Write tan , and cot , cos , sin , sin , Write tan , and simplify., cos , sin , Write tan , and simplify., cos , 1, LHS sec 2 · cosec 2 , cos 2 sin 2 , 1, 1, ·, , , sin 2 (1 sin 2 ) (sin 2 sin 4 ), , 10. LHS , 12., 13., 14., 15., , 16. LHS , , cos , tan cot sin , # cos 4 k a, # sin 4 k, a, cos , sin , sec 4 cosec 4 , , (sin cos 3 sin 3 cos ) sin cos (cos 2 sin 2 ) sin cos ., 17. (i) sin6 cos6 (sin 2 ) 3 (cos 2 ) 3, (sin 2 cos 2 )(sin 4 cos 4 sin 2 cos 2 ), (sin 2 cos 2 ) 2 3 sin 2 cos 2 (1 3 sin 2 cos 2 ) ., (ii) (cos sin ) (cos 2 sin 2 )(cos 2 sin 2 ), 4, , 4, , , , (cos 4 sin 4 ) (cos 2 sin 2 ), , , , (sin 2 cos 4 ) (cos 2 sin 4 ) ., , (iii) (cosec 4 cot 4 ) (cosec 2 cot 2 )(cosec 2 cot 2 ), , , (cosec 4 cot 4 ) (cosec 2 cot 2 ), , , , (cosec 4 cosec 2 ) (cot 4 cot 2 ) ., , [a cosec 2 cot 2 1], , 1 , 1k (1 cos ) (1 cos ) (1 cos 2 ), cos , sin 2 ·, , , , #, 1 , (1 cos ) (1 cos ) (1 cos ) 2 (1 cos ) 2, 1k, a, cos , 1 sin , a, k, cos cos (1 sin ) (1 sin ) 1 sin 2 cos 2 ·, , (ii) LHS, #, 1 sin , (1 sin ) (1 sin ) (1 sin ) 2 (1 sin ) 2, a, k, cos cos , , 20. (i) LHS , , a
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Trigonometric Identities, 21. (i) LHS , , 1 sin , 1 sin (1 sin ) (1 sin ), , , a 1 sin k, #, cos , cos cos , 1 sin , 1 sin , 1 sin 2 , (sec tan ) ., , 22. (cos 3 sin 3 ) (cos sin )(cos 2 sin 2 cos sin ) ., And, (cos 3 sin 3 ) (cos sin ) (cos 2 sin 2 cos sin ) ., cos , 1 ·, 23. Write cot , and cosec , sin , sin , 25. Write sin 2 (1 cos 2 ) ., 26. (i) LHS , , (cosec cot ) (cosec cot ), (cosec cot ) 2 ., #, (cosec cot ) (cosec cot ), , And, (cosec cot ) 2 cosec 2 cot 2 2 cosec cot , (1 cot 2 ) cot 2 2 cosec cot ., (ii) LHS , , (sec tan ) (sec tan ), ·, #, (sec tan ) (sec tan ), , 27. (i) On dividing num. and denom. by cos , we get, (sec tan ) 1 (sec tan ) (sec 2 tan 2 ), , (sec 1 tan ), (sec 1 tan ), (sec tan )(1 sec tan ), , (sec tan ) a 1 sin k ·, cos cos , (1 sec tan ), (ii) On dividing num. and denom. by cos , we get, LHS , , LHS , , 2, 2, tan sec 1 (sec tan ) (sec tan ), 1 sec tan , (tan sec 1), , , , (sec tan ) [1 (sec tan )], (sec tan ) ., (tan sec 1), , cos , sin , , 1 sin , 1 cos , 1k a, 1k, a, sin sin , cos cos , & sin cos sin cos 0, 1 sin cos 1 cos sin , 1, 1, (sin cos ) · ', , 1, 1 (sin cos ) 1 (sin cos ), (1 sin cos ) (1 sin cos ), (sin cos ) · ), 3, 1 (sin cos ) 2, , 28. LHS , , (sin cos )#, 31. LHS a1 , , 2, 1., 2 sin cos , , (sin cos sin 2 cos 2 ) (sin cos ), sin cos , k (sin cos ) , cos sin , sin cos , , 3, 3, 2, 2, (sin 3 cos 3 ), c sin cos m c sin cos m, sin cos , sin cos sin cos , cos sin , , c sec 2 cosec, m RHS., cosec sec 2 , , , , 32. LHS , , cot 2 (sec 2 1) sec 2 (sin 2 1) (cot 2 tan 2 ) (sec 2 cos 2 ), , 0., (1 sin )(1 sec ), (1 sin )(1 sec ), , 587
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588, , Secondary School Mathematics for Class 10, , _b, Z], bb, ]], bb, ]], 1, 1, , `b (sin 2 cos 2 ), 33. LHS [], 1, b, ]] 1 , cos 2 m c 2 sin 2 mbb, ]c, 2, b, ] cos , sin , a, \, 2, 2, ) cos 4 sin 4 3 sin 2 cos 2 , (1 cos ) (1 sin ), ), , , , 34. LHS , , cos 2 , sin 2 , , 3 sin 2 cos 2 , 2, 2, , sin (1 cos ) cos (1 sin 2 ), 2, , cos 4 (1 sin 2 ) sin 4 (1 cos 2 ), (1 cos 2 )(1 sin 2 ), cos 4 sin 4 cos 2 sin 2 (cos 2 sin 2 ), 1 (sin 2 cos 2 ) sin 2 cos 2 , (cos sin 2 ) 2 sin 2 cos 2 , 2, , 2 sin 2 cos 2 , , 2, 2, , 1 sin 2 cos 2 ·, , 2 sin cos , , (sin A sin B)(sin A sin B) (cos A cos B)(cos A cos B), (cos A cos B)(sin A sin B), , , , (sin 2 A sin 2 B) (cos 2 A cos 2 B) (sin 2 A cos 2 A) (sin 2 B cos 2 B), , (cos A cos B)(sin A sin B), (cos A cos B)(sin A sin B), , , , (1 1), 0., (cos A cos B)(sin A sin B), , 35. LHS , , tan A tan B tan A tan B (tan A tan B) , tan A tan B., 1 1, (tan A tan B), tan A tan B, , 36. (i) cos 2 30c cos 30c d, , 2, 3 32 3, 3, 3 3, n , , , ! 1., 2, 2, 2, 4, 4, , 12 1 1 1 3, (ii) sin 2 30c sin 30c a k ! 2., 2, 2 4 2 4, (iii) LHS tan 2 45c sin 45c (1) 2 , RHS cos 2 45c d, , , , 1 1 , 1, 2, 2, , 21, 2, , ·, , 1 2 1·, n 2, 2, , LHS ! RHS., , 38. 1 sin 3 sin cos & sec 2 tan 2 3 tan [dividing throughout by cos 2 ], 2, , & (1 tan 2 ) tan 2 3 tan , & 2 tan 2 3 tan 1 0, & 2 tan 2 2 tan tan 1 0, & 2 tan (tan 1) (tan 1) 0, & (tan 1) (2 tan 1) 0, 1, & tan 1 or ·, 2
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Trigonometric Identities, , 589, , ELIMINATION OF TRIGONOMETRIC RATIOS, In order to eliminate the T-ratios from given relations, we make use of, the fundamental trigonometrical identities, as shown in the examples, given below., , SOLVED EXAMPLES, EXAMPLE 1, , SOLUTION, , If x a sin b cos , x2 y2 a2 b2 ., , and, , y a cos b sin ,, , prove, , that, , We have, x 2 y 2 (a sin b cos ) 2 (a cos b sin ) 2, a 2 (sin 2 cos 2 ) b 2 (cos 2 sin 2 ), a2 b2, , [a sin 2 cos 2 1]., , Hence, x 2 y 2 a 2 b 2 ., EXAMPLE 2, , If x a sin and y b tan , prove that e, , SOLUTION, , We have, , a2 b2 , o 1., x2 y2, , a, 1, x a sin & x , & xa cosec , sin , b, 1, and y b tan & y , & by cot ., tan , , … (i), … (ii), , Squaring (i) and (ii) and subtracting, we get, , e, , a2 b2 , o (cosec 2 cot 2 ) 1., x2 y2, , Hence, e, EXAMPLE 3, , a2 b2 , o 1., x2 y2, , If tan sin m and tan sin n, prove that, (m 2 n 2) 4 mn ., , SOLUTION, , [CBSE 2010], , We have, LHS (m 2 n 2), , (tan sin ) 2 (tan sin ) 2 4 tan sin , [a (a b) 2 (a b) 2 4ab] ., RHS 4 mn 4 (tan sin )(tan sin ), , 4 (tan 2 sin 2 ) 4 ·, , d, , sin 2 , sin 2 n, cos 2
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590, , Secondary School Mathematics for Class 10, , sin 2 sin 2 cos 2 , 4 · sin · 1 cos 2 , cos , cos , 2, 4 tan · sin 4 tan sin ., 4·, , Thus, LHS = RHS., Hence, (m 2 n 2) 4 mn ., EXAMPLE 4, , If sec tan m, show that, , SOLUTION, , We have, , (m 2 1), sin ., (m 2 1), , [CBSE 2004], , (m 2 1) (sec tan ) 2 1, sec 2 tan 2 2 sec tan 1, (sec 2 1) tan 2 2 sec tan , 2 tan 2 2 sec tan , 2 tan (tan sec ) ., , [a sec 2 1 tan 2 ], … (i), , (m 2 1) (sec tan ) 2 1, sec 2 tan 2 2 sec tan 1, (1 tan 2 ) sec 2 2 sec tan , 2 sec 2 2 sec tan , 2 sec (sec tan ) ., , [a 1 tan 2 sec 2 ], … (ii), , From (i) and (ii), we get, (m 2 1) tan , , a sin # cos k sin ., cos , (m 2 1) sec , Hence,, EXAMPLE 5, , (m 2 1), sin ., (m 2 1), , If sin cos m and sec cosec n, prove that, n(m 2 1) 2m., , SOLUTION, , We have, n(m 2 1) (sec cosec )[(sin cos ) 2 1], a 1 1 k [(sin 2 cos 2 ) 2 sin cos 1], cos sin , (sin cos ), · 2 sin cos , , sin cos , 2 (sin cos ) 2m., Hence, n(m 2 1) 2m., , EXAMPLE 6, , If cos sin 2 cos , show that (cos sin ) 2 sin ., [CBSE 2008C]
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Trigonometric Identities, SOLUTION, , 591, , We have, cos sin 2 cos , (cos sin ) 2 2 cos 2 , cos 2 sin 2 2 cos sin 2 cos 2 , cos 2 2 cos sin sin 2 , cos 2 2 cos sin sin 2 sin 2 sin 2 , [adding sin 2 on both sides], 2, 2, (cos sin ) 2 sin , , , , , , , (cos sin ) 2 sin ., Hence, (cos sin ) 2 sin ., EXAMPLE 7, , If sin sin 2 1, prove that cos 2 cos 4 1., , SOLUTION, , sin sin 2 1, sin 1 sin 2 , [a 1 sin 2 cos 2 ], , sin cos 2 , sin 2 cos 4 , 1 cos 2 cos 4 , , [a sin 2 1 cos 2 ], , cos 2 cos 4 1., Hence, cos 2 cos 4 1., EXAMPLE 8, , If cos sin 1, prove that cos sin !1., , SOLUTION, , We have, (cos sin ) 2 (cos sin ) 2 2 (cos 2 sin 2 ), (cos sin ) 2 (cos sin ) 2 2 [a (cos 2 sin 2 ) 1], 1 (cos sin ) 2 2, , [a cos sin 1 (given)], , (cos sin ) 1, 2, , (cos sin ) !1, , [taking square root on both sides]., , Hence, (cos sin ) !1., EXAMPLE 9, , SOLUTION, , If x sin 3 y cos 3 sin cos and x sin y cos , prove that, x 2 y 2 1., x sin 3 y cos 3 sin cos , (x sin ) sin 2 (y cos ) cos 2 sin cos , (x sin ) sin 2 (x sin ) cos 2 sin cos , [a y cos x sin ]
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592, , Secondary School Mathematics for Class 10, , (x sin ) (sin 2 cos 2 ) sin cos , x sin sin cos , x cos ., Now, x sin y cos , , [a sin 2 cos 2 1], … (i), , cos sin y cos , , [a x cos ], , , , y sin ., , … (ii), , On squaring (i) and (ii) and adding, we get x y 1., 2, , 2, , Hence, x 2 y 2 1., EXAMPLE 10, , SOLUTION, , If x a sec cos , y b sec sin and z c tan then prove that, 2, 2, y2, e x2 2 o d1 z2 n ·, a, b, c, We have, x, a sec cos , , y, sec sin , b, Squaring (i) and (ii) and adding, we get, and, , … (ii)., , 2, y2, e x2 2 o sec 2 (cos 2 sin 2 ), a, b, sec 2 , , [a cos 2 sin 2 1], , 2, (1 tan 2 ) d1 z2 n, c, , 8a tan zcB, , Hence, e, EXAMPLE 11, , … (i), , 2, x2 y o z2 ·, d1, n, a2 b2, c2, , If x r sin cos , y r sin sin and z r cos , prove that, x2 y2 z2 r2 ., , SOLUTION, , We have, x 2 y 2 z 2 r 2 sin 2 cos 2 r 2 sin 2 sin 2 r 2 cos 2 , r 2 sin 2 (cos 2 sin 2 ) r 2 cos 2 , r 2 sin 2 r 2 cos 2 , , [a cos 2 sin 2 1], , r 2 (sin 2 cos 2 ) r 2 [a sin 2 cos 2 1] ., Hence, (x 2 y 2 z 2) r 2 ., EXAMPLE 12, , If cosec sin m and sec cos n, prove that, (m 2 n), , 2, , 3, , (mn 2), , 2, , 3, , 1.
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Trigonometric Identities, SOLUTION, , 593, , We have, m 2 n (cosec sin ) 2 · (sec cos ), 2, a 1 sin k · a 1 cos k, sin , cos , , , , , (m 2 n), , 1, , (1 sin 2 ) 2 (1 cos 2 ) cos 4 sin 2 , ·, , cos 3 , #, cos , sin 2 , sin 2 cos , 3, , cos ., , … (i), , Again, mn 2 (cosec sin ) · (sec cos ) 2, 2, a 1 sin k · a 1 cos k, sin , cos , , , , (1 sin 2 ) (1 cos 2 ) 2, ·, sin , cos 2 , , 2, 4, d cos # sin 2 n sin 3 , sin cos , , , , (mn 2), , 1, , 3, , sin ., , … (ii), , On squaring (i) and (ii) and adding the results, we get, (m 2 n), , 2, , 3, , (mn 2), , Hence, (m 2 n), EXAMPLE 13, , 2, , 3, , 2, , 3, , 1, , (mn 2), , 2, , 3, , [a cos 2 sin 2 1] ., , 1., , If a cos b sin c, prove that, (a sin b cos ) ! a 2 b 2 c 2 ., , SOLUTION, , [CBSE 2006C], , Given, a cos b sin c., Now, (a cos b sin ) 2 (a sin b cos ) 2, a 2 (cos 2 sin 2 ) b 2 (sin 2 cos 2 ) (a 2 b 2) ., Thus, (a cos b sin ) 2 (a sin b cos ) 2 (a 2 b 2), , , c 2 (a sin b cos ) 2 (a 2 b 2), , , , (a sin b cos ) 2 (a 2 b 2 c 2), , , , (a sin b cos ) ! a 2 b 2 c 2 ., , Hence, (a sin b cos ) ! a 2 b 2 c 2 ., EXAMPLE 14, , If (3 sin 5 cos ) 5, prove that (5 sin 3 cos ) !3., , … (i)
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594, , Secondary School Mathematics for Class 10, , SOLUTION, , We have, (3 sin 5 cos ) 2 (5 sin 3 cos ) 2, 9 (sin 2 cos 2 ) 25 (sin 2 cos 2 ), (9 25) 34., (3 sin 5 cos ) 2 (5 sin 3 cos ) 2 34, 5 2 (5 sin 3 cos ) 2 34, [a 3 sin 5 cos 5], (5 sin 3 cos ) !3 [taking square root on each side], Hence, (5 sin 3 cos ) !3., , , f, , EXERCISE 13B, , 1. If a cos b sin m and a sin b cos n, prove that, (m 2 n 2) (a 2 b 2) ., 2. If x a sec b tan and y a tan b sec , prove that, (x 2 y 2) (a 2 b 2) ., y, y, x, x, 3. If c a sin cos m 1 and c a cos sin m 1, prove that, b, b, 2, 2, y, e x2 2 o 2., a, b, 4. If (sec tan ) m and (sec tan ) n, show that mn 1., 5. If (cosec cot ) m and (cosec cot ) n, show that mn 1., 2, , x 3 y 3, 6. If x a cos 3 and y b sin 3 , prove that ` a j c m 1., b, 7. If (tan sin ) m and (tan sin ) n, prove that, (m 2 n 2) 2 16mn., 2, , 8. If (cot tan ) m and (sec cos ) n, prove that, (m 2 n), , 2, , 3, , (mn 2), , 2, , 3, , 1., , 9. If (cosec sin ) a 3 and (sec cos ) b 3, prove that, a 2 b 2 (a 2 b 2) 1., 10. If (2 sin 3 cos ) 2, prove that (3 sin 2 cos ) !3., 11. If (sin cos ) 2 cos , show that cot ( 2 1) ., 12. If (cos sin ) 2 sin , prove that (sin cos ) 2 cos ., 13. If sec tan p, prove that, 1, 1, (i) sec cp p m, 2, , 1, 1, (ii) tan cp p m, 2, , (iii) sin , , p2 1, ·, p2 1
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Trigonometric Identities, , 14. If tan A n tan B and sin A m sin B, prove that cos 2 A , , 595, , (m 2 1), ·, (n 2 1), , 15. If m (cos sin ) and n (cos sin ) then show that, m n , 2, ·, n, m, 1 tan 2 , HINTS TO SOME SELECTED QUESTIONS, 7. See Example 3., 8. m a, , cos sin c cos 2 sin 2 m , 1, ·, k, sin cos , sin cos , sin cos , , na, , 1 cos 2 m sin 2 ·, 1 , cos k c, cos , cos , cos , , , , m2 n d, , and mn 2 e, , , (m 2 n), , 2, , sin 2 1 , 1, #, sec 3 , n, 2, sin cos cos , cos 3 , 2, , sin 4 sin 3 , 1, #, tan 3 ., o, sin cos cos 2 , cos 3 , , 3, , (mn 2), , 2, , 3, , (sec 3 ), , 2, , 3, , (tan 3 ), , 2, , 3, , (sec 2 tan 2 ) 1., 2, , 9. a 3 a, , 1 sin 2 m cos 2 , cos 3 ·, 1 , sin k c, &a, 1, sin , sin , sin , sin 3 , , b3 a, , 1 cos 2 m sin 2 , sin 3 ·, 1 , cos k c, &b, 1, cos , cos , cos , cos 3 , , , , 2, , a 2 b 2 (a 2 b 2) a 4 b 2 a 2 b 4 a 3 (ab 2) (a 2 b) b 3, 4, 2, SRS 2 3, WV SR 4 3, WV, 2, 2, 3 W S, 3 W, cos · SScos 1 · sin 2 WW SScos 2 · sin 1 WW · sin , sin SS, cos, , W S, W, sin 3 cos 3 W Ssin 3 cos 3 W, T, X T, X, 2, 2, cos · sin cos · sin (cos 2 sin 2 ) 1., sin , cos , , 10. (2 sin 3 cos ) 2 (3 sin 2 cos ) 2 13 (sin 2 cos 2 ) 13, , , 2 2 (3 sin 2 cos ) 2 13 & (3 sin 2 cos ) 2 9, , , , (3 sin 2 cos ) !3., , 11. sin cos 2 cos , , , 1 cot 2 cot [dividing both side by sin ], , , , ( 2 1) cot 1 & cot , , ( 2 1), 1, 1, , ( 2 1) ., #, ( 2 1) ( 2 1 ) ( 2 1 ), , 12. (cos sin ) 2 2 sin 2 cos 2 sin 2 2 cos sin 2 sin 2 , , , sin 2 2 cos sin cos 2 sin 2 2 cos sin cos 2 2 cos 2 , , , , (sin cos ) 2 2 cos 2 & (sin cos ) 2 cos .
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596, , Secondary School Mathematics for Class 10, , 13. sec tan p., , … (i), , sec tan 1., , … (ii), , On dividing (ii) by (i), we get, 1, sec tan p ·, , … (iii), , 2, , , , 2, , 1, 1, 1, 1, from (i) and (iii), we get sec cp p m, tan cp p m ·, 2, 2, , 1 c 1m, p p, (p 2 1), 2, , tan, ·, , 2, Also, sin , sec 1, 1, cp m (p 1), p, 2, 14. We have to eliminate B. So, we find cosec B and cot B from the given relations and use, the identity cosec 2 B cot 2 B 1., m ·, sin A, n ·, tan A n tan B & cot B , tan A, Squaring (i) and (ii), and subtracting the results, we get, sin A m sin B & cosec B , , m2 n2 , 1 & m 2 n 2 cos 2 A sin 2 A & m 2 n 2 cos 2 A 1 cos 2 A, sin 2 A tan 2 A, (m 2 1), ·, & (n 2 1) cos 2 A (m 2 1) & cos 2 A 2, (n 1), (m n), ·, 15. LHS , mn, Now, (m n) 2 cos and mn (cos 2 sin 2 ) cos 2 (1 tan 2 ) ., , , mn cos 1 tan 2 ., , Hence, LHS , , f, , (m n), mn, , , , 2 cos , cos 1 tan 2 , , , , 2, 1 tan 2 , , RHS., , EXERCISE 13C, , Very-Short and Short-Answer Questions, 1. Write the value of (1 sin 2 ) sec 2 ., 2. Write the value of (1 cos 2 ) cosec 2 ., 3. Write the value of (1 tan 2 ) cos 2 ., 4. Write the value of (1 cot 2 ) sin 2 ., 1, 5. Write the value of csin 2 , m·, 1 tan 2 , 1 ·, 6. Write the value of ccot 2 , m, sin 2 , 7. Write the value of sin cos (90c ) cos sin (90c ) ., 8. Write the value of cosec 2 (90c ) tan 2 ., , … (i), … (ii)
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Trigonometric Identities, , 9. Write the value of sec 2 (1 sin )(1 sin ) ., 10. Write the value of cosec 2 (1 cos )(1 sin ) ., 11. Write the value of sin 2 cos 2 (1 tan 2 )(1 cot 2 ) ., 12. Write the value of (1 tan 2 )(1 sin )(1 sin ) ., 13. Write the value of 3 cot 3 cosec ., 4 ·, 14. Write the value of 4 tan 2 , cos 2 , tan 2 sec 2 ·, 15. Write the value of, cot 2 cosec 2 , 1, 16. If sin , write the value of (3 cot 2 3) ., 2, 2, 17. If cos , write the value of (4 4 tan 2 ) ., 3, 7, 18. If cos , write the value of (tan cot ) ., 25, (sec 1), 2, ·, 19. If cos , write the value of, 3, (sec 1), (cos sin ), ·, 20. If 5 tan 4, write the value of, (cos sin ), (2 cos sin ), ·, 21. If 3 cot 4, write the value of, (4 cos sin ), (1 cos 2 ), 1, ·, 22. If cot , , write the value of, (2 sin 2 ), 3, 2, , 597, [CBSE 2009], , [CBSE 2008], , 2, , [CBSE 2009], , [CBSE 2008], , (cosec 2 sec 2 ), 1, ·, , write the value of, (cosec 2 sec 2 ), 5, 4, 24. If cot A and (A B) 90c, what is the value of tan B?, 3, 3, 25. If cos B and (A B) 90c, find the value of sin A., 5, 26. If 3 sin cos and is an acute angle, find the value of ., 23. If tan , , 27. Write the value of tan 10c tan 20c tan 70c tan 80c., 28. Write the value of tan 1c tan 2c … tan 89c., 29. Write the value of cos 1c cos 2c … cos 180c., 5, 30. If tan A , find the value of (sin A cos A) sec A., 12, 31. If sin cos ( 45c), where is acute, find the value of ., sin 50c cosec 40c , 4 cos 50c cosec 40c., cos 40c sec 50c, 33. Find the value of sin 48c sec 42c cos 48c cosec 42c., 32. Find the value of, , [CBSE 2008]
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598, , Secondary School Mathematics for Class 10, , 34. If x a sin and y b cos , write the value of (b 2 x 2 a 2 y 2) ., 5, 1, 35. If 5x sec and x tan , find the value of 5 cx 2 2 m ·, x, , [CBSE 2010], , 2, 1, 36. If cosec 2x and cot x , find the value of 2 cx 2 2 m ·, x, , [CBSE 2010], , 37. If sec tan x, find the value of sec ., cos 38c cosec 52c, ·, tan 18c tan 35c tan 60c tan 72c tan 55c, 39. If sin x, write the value of cot ., 40. If sec x, write the value of tan ., 38. Find the value of, , ANSWERS (EXERCISE 13C), , 1. 1, , 2. 1, , 3. 1, , 4. 1, , 5. 1, , 10. 1, , 11. 1, , 12. 1, , 13. –3, , 14. –4, , 19., , 1, 5, , 28. 1, 36., , 1, 2, , 20., , 1, 9, , 29. 0, 37., , 6. –1, , 3, 11, 2, 22., 23., 5, 3, 13, 17, 30., 31. (67.5), 12, 21., , x2 1, 2x, , 38., , 1, 39., 3, , 15. 1, 24., , 4, 3, , 32. –2, , 1 x2, x, , 7. 1, , 8. 1, , 16. 12, , 17. 9, , 25., , 3, 5, , 18., , 625, 168, , 26. 30 27. 1, , 33. 2, 40., , 9. 1, , 34. a 2 b 2 35., , 1, 5, , x2 1, , HINTS TO SOME SELECTED QUESTIONS, 2. (1 cos 2 ) cosec 2 sin 2 · cosec 2 1., 4. (1 cot 2 ) sin 2 cosec 2 sin 2 1., 1, 1, 2, 2, 2, m csin , m (sin cos ) 1., 1 tan 2 , sec 2 , 1, 2, 2, 2, 2, 6. ccot 2 , m (cot cosec ) cot (1 cot ) 1., sin 2 , 7. sin cos (90c ) cos sin (90c ) sin · sin cos · cos sin 2 cos 2 1., 5. csin 2 , , 8. cosec 2 (90c ) tan 2 sec 2 tan 2 1., 9. Given expression sec 2 (1 sin 2 ) sec 2 cos 2 1., 10. Given expression cosec 2 (1 cos 2 ) cosec 2 sin 2 1., 11. Given expression sin 2 (1 cot 2 ) cos 2 (1 tan 2 ), (sin 2 cosec 2 )(cos 2 sec 2 ) (1#1) 1., 12. Given expression (1 tan 2 )(1 sin 2 ) sec 2 cos 2 1.
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Trigonometric Identities, , 599, , 13. Given expression 3 cot 2 3 (1 cot 2 ) 3., 14. Given expression 4 tan 2 4 sec 2 4 tan 2 4 (1 tan 2 ) 4., 15. Given expression , , (sec 2 tan 2 ), (cosec 2 cot 2 ), , 1 1., 1, , 16. 3 cot 2 3 3 (cot 2 1) 3 cosec 2 (3 # 4) 12 [a cosec 2] ., 9, 3, 17. (4 4 tan 2 ) 4 (1 tan 2 ) 4 sec 2 a4 # k 9 :a sec D ·, 2, 4, , , , 18. Consider a right 3ABC in which +B 90c, AB 7 and AC 25., BC 2 AC 2 AB 2 (25) 2 (7) 2 (625 49) 576., , , BC 576 24., , , , (tan cot ) a, , 24 7 576 49 625 ·, k, 7 24, 168, 168, , 19., , 3, (sec 1) a 2 1k 1 2, , a # k 1·, 3, 2 5, 5, sec 1), 1k, a, 2, , 20., , (cos sin ) (1 tan ), , (cos sin ) (1 tan ), , [dividing num. and denom. by cos ], , a1 k, , 4, 5 1, 4, :a tan D ·, 5, 9, 4, a1 k, 5, (2 cos sin ) (2 cot 1), , 21., [dividing num. and denom. by sin ], (4 cos sin ) (4 cot 1), , , JK, N, 4, KK2 # 1OOO, 3, O 11 ·, KK, KK, 4 OOO 13, K4 # 3 1 O, L, P, 3, 2, 2, (1 cos ), sin, , , 4, 22., (2 sin 2 ) (2 sin 2 ) a2 3 k, 4, 3, 3, 4, a # k ·, 5, 4 5, , 3, 1, 4, :cosec 2 (1 cot 2 ) a1 k & sin 2 D, 3, 3, 4, , 6, 1, 23. cosec 2 (1 cot 2 ) (1 5) 6, sec 2 (1 tan 2 ) a1 k ·, 5, 5, 6, (cosec 2 sec 2 ) a6 5 k 24 2, , , ·, , (cosec 2 sec 2 ) a6 6 k 36 3, 5, 24. A 90c B & cot A cot (90c B) tan B., 4·, 3, 25. B 90c A & cos B cos (90c A) sin A, , , tan B cot A , , , , sin A cos B , , 3·, 5
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602, , Secondary School Mathematics for Class 10, , 8. sec 70c sin 20c cos 20c cosec 70c ?, (a) 0, , (b) 1, , (c) –1, , (d) 2, , 9. If sin 3A cos (A 10c) and 3A is acute then +A ?, (a) 35, , (b) 25, (c) 20, (d) 45, , , , 10. If sec 4A cosec (A 10c) and 4A is acute then +A ?, (a) 20, , (b) 30, , (c) 40, (d) 50, 11. If A and B are acute angles such that sin A cos B then (A B) ?, (a) 45, , (b) 60, , , 12. If cos ( ) 0 then sin ( ) ?, , (c) 90, , (d) 180, , (a) sin , (b) cos , 13. sin (45c ) cos (45c ) ?, , (c) sin 2, , (d) cos 2, , (a) 2 sin , (b) 2 cos , 2, 2, , , 14. sec 10c cot 80c ?, , (c) 0, , (d) 1, , (a) 1, , (b) 0, , (c), , 3, 2, , (d), , 1, 2, , 15. cosec 2 57c tan 2 33c ?, (a) 0, , (b) 1, , 2, , 2, , (c) –1, , 2 tan 30c sec 52c sin 38c , ?, cosec 2 70c tan 2 20c, 1, 2, (a) 2, (b), (c), 2, 3, 2, 2, , (sin 22c sin 68c), sin 2 63c cos 63c sin 27c3 ?, 17. ), (cos 2 22c cos 2 68c), 16., , (a) 0, 18., , (b) 1, , (c) 2, , (d), , 3, 2, , (d) 3, , cot (90c ) · sin (90c ) cot 40c, , (cos 2 20c cos 2 70c) ?, sin , tan 50c, (a) 0, , 19., , (d) 2, , 2, , (b) 1, , (c) –1, , (d) none of these, , cos 38c cosec 52c, ?, tan 18c tan 35c tan 60c tan 72c tan 55c, (a), , 3, , (b), , 1, 3, , (c), , 1, 3, , (d), , 2, 3, , 20. If 2 sin 2 3 then ?, (a) 30, , (b) 45, 21. If 2 cos 3 1 then ?, (a) 10, , (b) 15, , (c) 60, , (d) 90, , (c) 20, , (d) 30
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Trigonometric Identities, , 22. If 3 tan 2 3 0 then ?, (a) 15, (b) 30, 23. If tan x 3 cot x then x ?, , 603, , (c) 45, , (d) 60, , (b) 60, (c) 30, , 24. If x tan 45c cos 60c sin 60c cot 60c then x ?, 1, 1, (a) 1, (b), (c), 2, 2, , (d) 15, , (a) 45, , (d), , 3, , 25. If tan 2 45c cos 2 30c x sin 45c cos 45c then x ?, 1, 1, (a) 2, (b) –2, (c), (d), 2, 2, 2, , , 26. sec 60c 1 ?, (a) 2, (b) 3, (c) 4, (d) 0, , , , , , 27. (cos 0c sin 30c sin 45c) (sin 90c cos 60c cos 45c) ?, (a), , 5, 6, , (b), , 5, 8, , (c), , 28. sin 2 30c 4 cot 2 45c sec 2 60c ?, 1, (a) 0, (b), 4, 29. 3 cos 2 60c 2 cot 2 30c 5 sin 2 45c ?, 13, 17, (a), (b), 6, 4, 30. cos 2 30c cos 2 45c 4 sec 2 60c , (a), , 73, 8, , (b), , 75, 8, , 3, 5, , (d), , 7, 4, , (c) 4, , (d) 1, , (c) 1, , (d) 4, , 1, cos 2 90c 2 tan 2 60c ?, 2, 81, 83, (c), (d), 8, 8, , 31. If cosec 10 then sec ?, (a), , 3, 10, , (b), , 10, 3, , 8, then cosec ?, 15, 8, 17, (a), (b), 8, 17, a, 33. If sin then cos ?, b, b2 a2, b, (a), (b), 2, 2, b, b a, , 1, 10, , (c), , 2, 10, , (d), , 32. If tan , , 34. If tan 3 then sec ?, 3, 2, (a), (b), 2, 3, , (c), , 17, 15, , a, b2 a2, , (c), , (c), , 1, 2, , (d), , 15, 17, , b, (d) a, , (d) 2
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604, , Secondary School Mathematics for Class 10, , 25, then sin ?, 7, 24, 7, (a), (b), 7, 24, , 35. If sec , , 36. If sin , (a), , 1, 3, , 37. If cos , (a), , 3, 4, , (c), , 24, 25, , (c), , 3, 2, , (d) none of these, , 1, then cot ?, 2, (b), , 3, , 4, then tan ?, 5, 4, (b), 3, , (c), , 3, 5, , (d) 1, , (d), , 5, 3, , (d), , 1, 9, , 3, 1, 38. If 3x cosec and x cot then 3 cx 2 2 m ?, x, (a), , 1, 27, , (b), , 1, 81, , (c), , 1, 3, , 2, 1, 39. If 2x sec A and x tan A then 2 cx 2 2 m ?, x, (a), , 1, 2, , 40. If tan , (a), , 7, 3, , (b), , 1, 4, , 4, then (sin cos ) ?, 3, 7, (b), 4, , (c), , 1, 8, , (d), , 1, 16, , (c), , 7, 5, , (c), , 5, 7, , 41. If (tan cot ) 5 then (tan 2 cot 2 ) ?, (a) 27, , (b) 25, , (c) 24, , (d) 23, , 5, 42. If (cos sec ) then (cos 2 sec 2 ) ?, 2, (a), , 21, 4, , (b), , 17, 4, , (c), , 29, 4, , (cosec 2 sec 2 ), 1, ?, then, (cosec 2 sec 2 ), 7, 3, 2, 2, (a), (b), (c), 3, 3, 4, , (7 sin 3 cos ), ?, 44. If 7 tan 4 then, (7 sin 3 cos ), , (d), , 33, 4, , (d), , 3, 4, , (d), , 5, 14, , 43. If tan , , (a), , 1, 7, , (b), , 5, 7, , (c), , 3, 7
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Trigonometric Identities, , 45. If 3 cot 4 then, (a), , (a), , (5 sin 3 cos ), ?, (5 sin 3 cos ), , 1, 3, , 46. If tan , , 605, , (b) 3, , (c), , 1, 9, , (c), , a2, (a b 2), , (d) 9, , (a sin b cos ), a, ?, then, b, (a sin b cos ), , (a 2 b 2), (a 2 b 2), , (b), , (a 2 b 2), (a 2 b 2), , 2, , (d), , b2, (a b 2), 2, , 47. If sin A sin 2 A 1 then cos 2 A cos 4 A ?, (a), , 1, 2, , (b) 1, , (c) 2, , (d) 3, , 48. If cos A cos 2 A 1 then sin 2 A sin 4 A ?, (a) 1, 49., , 50., , (b) 2, , (c) 4, , (d) 3, , 1 sin A , ?, 1 sin A, (a) sec A tan A, , (b) sec A tan A, , (c) sec A tan A, , (d) none of these, , 1 cos A , ?, 1 cos A, (a) cosec A cot A, , (b) cosec A cot A, , (c) cosec A cot A, , (d) none of these, , 51. If tan , (a), , (cos sin ), a, ?, then, (cos sin ), b, , ab, ab, , (b), , ab, ab, , (c), , ba, ba, , (d), , ba, ba, , 1 cos , 1 cos , , (c), , 1 sin , 1 sin , , (d) none of these, , 52. (cosec cot ) 2 ?, (a), , 1 cos , 1 cos , , (b), , 53. (sec A tan A)(1 sin A) ?, (a) sin A, , (b) cos A, , (c) sec A, , (d) cosec A, , ANSWERS (MCQ), , 1. (d) 2. (c), 10. (a) 11. (c), 19. (c) 20. (a), , 3. (d) 4. (b) 5. (c), 6. (d) 7. (c), 12. (d) 13. (c) 14. (a) 15. (b) 16. (c), 21. (c) 22. (b) 23. (b) 24. (a) 25. (c), , 8. (d) 9. (b), 17. (c) 18. (b), 26. (b) 27. (d)
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608, , Secondary School Mathematics for Class 10, , 31. cosec , , AC 10 10 x, 10x and BC x., x & AC, 1, BC, , , , AB 2 AC 2 BC 2 ( 10 x) 2 (x 2) 10x 2 x 2 9x 2, , , , AB 9x 2 3x., , , , sec , , 32. tan , , AC 10 x 10 ·, 3x, 3, AB, , BC 8 8x, & BC 8x and AB 15x., AB 15 15x, , , , AC 2 AB 2 BC 2 (225x 2 64x 2) 289x 2, , , , AC 289x 2 17x., , , , cosec , , AC 17x 17 ·, 8x, 8, BC, , 33. cos 2 (1 sin 2 ) d1 , 34. tan , , BC 3 , 1, AB, , a2 b2 a2, & cos , n, b2, b2, , b2 a2, ·, b, , 3x, 3 x and AB x., x & BC, , , , AC 2 AB 2 BC 2 x 2 3x 2 4x 2 & AC 2x., , , , sec , , 35. sec , , AC 2x , x 2., AB, , AC 25 25x, & AC 25x and AB 7x., 7, 7x, AB, , , , BC 2 AC 2 AB 2 625x 2 49x 2 576x 2 & BC 24x., , , , sin , , 36. sin , , BC 24x 24 ·, AC 25x 25, , 1 BC ·, 2 AC, , , , AB 2 AC 2 BC 2 2 2 1 2 3, , , , AB 3 ., , cot , 37. cos , , AB 3 , 3., 1, BC, 4 AB ·, 5 AC, , , , BC 2 AC 2 AB 2 25 16 9, , , , BC 3., , , , tan , , BC 3 ·, AB 4
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Trigonometric Identities, 38. We know that cosec 2 cot 2 1., , , , 32, 9, 1, (3x) 2 a x k 1 & 9x 2 2 1 & 9 cx 2 2 m 1, x, x, 1, 1, 1, 1 1, 2, 2, cx 2 m 9 & 3 cx 2 m 3 # 9 3 ·, x, x, , 39. We know that sec 2 A tan 2 A 1., 4, 1, 22, (2x) 2 a x k 1 & 4x 2 2 1 & 4 cx 2 2 m 1, x, x, 1, 1, 1, 1 1, & cx 2 2 m & 2 cx 2 2 m 2 # ·, 4, 4 2, x, x, BC, 4, ·, 40. tan , 3 AB, , , , , AC 2 AB 2 BC 2 (3) 2 (4) 2 25, , , , AC 25 5., , , , 4 3, 7, (sin cos ) a k ·, 5 5, 5, , 41. (tan cot ) 2 5 2 & tan 2 cot 2 2 25, , , (tan 2 cot 2 ) 23., , 42. (cos sec ) 2 , , 25, 25, 25 17 ·, & cos 2 sec 2 2 , & cos 2 sec 2 , 2, 4, 4, 4, 4, , 1, & cot 7 ., 7, 8, 1, sec 2 (1 tan 2 ) a1 k and cosec 2 (1 cot 2 ) (1 7) 8., 7, 7, 8, a8 k, cosec 2 sec 2 , 7 48 3 ·, , cosec 2 sec 2 a8 8 k 64 4, 7, 4, 44. Given, tan ·, 7, (7 sin 3 cos ) (7 tan 3), , , [dividing num. and denom. by cos ], (7 sin 3 cos ) (7 tan 3), 4, a7 # 3k (4 3), 7, , , 1·, 4, (4 3) 7, a7 # 3k, 7, 4, 45. Given, cot ·, 3, 5 sin 3 cos 5 3 cot , k, a, [dividing num. and denom. by sin ], 5 sin 3 cos , 5 3 cot , 4, a5 3 # k (5 4), 3 , , 9 9., 4, (5 4) 1, a5 3 # k, 3, 46. Dividing num. and denom. by cos , we get, a, (a sin b cos ) (a tan b) aa # b bk (a 2 b 2), ·, , , , (a sin b cos ) (a tan b) aa # a bk (a 2 b 2), b, 43. tan , , 609
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610, , Secondary School Mathematics for Class 10, , 47. sin A sin 2 A 1 & sin A (1 sin 2 A) & sin A cos 2 A., (cos 2 A cos 4 A) (sin A sin 2 A) 1., , , , 48. cos A cos 2 A 1 & cos A (1 cos 2 A) & cos A sin 2 A., (sin 2 A sin 4 A) (cos A cos 2 A) 1., , , , (1 sin A) (1 sin A) (1 sin A) (1 sin A), , , #, (1 sin A) (1 sin A), 1 sin 2 A, cos 2 A, (1 sin A), , c 1 sin A m (sec A tan A) ., cos A, cos A cos A, , 49., , 1 sin A , 1 sin A, , 50., , 1 cos A , 1 cos A, , 51., , (1 cos A) (1 cos A) (1 cos A) (1 cos A), , , #, (1 cos A) (1 cos A), 1 cos 2 A, sin 2 A, , (1 cos A), , c 1 cos A m (cosec A cot A) ., sin A, sin A sin A, , (cos sin ) (1 tan ), , (cos sin ) (1 tan ), , , [dividing num. and denom. by cos ], , a1 k, , a, b (b a) ·, a, , a1 k (b a), b, , 52. (cosec cot ) 2 a, , 2, 2, 1 cos 2 (1 cos ) (1 cos ) (1 cos ) ·, k, sin sin , (1 cos 2 ) (1 cos ), sin 2 , , 53. (sec A tan A) (1 sin A) c, , , 1 sin A, m (1 sin A), cos A cos A, , (1 sin A) (1 sin A) 1 sin 2 A cos 2 A, , , cos A., cos A, cos A, cos A, , TEST YOURSELF, MCQ, cos 2 56c cos 2 34c , 3 tan 2 56c tan 2 34c ?, sin 2 56c sin 2 34c, 1, (a) 3, (b) 4, (c) 6, (d) 5, 2, 1, 1, 2, 2, 2, 2, 2. The value of asin 30c cos 45c 4 tan 30c sin 90c cot 2 60ck ?, 2, 8, 3, 5, (a), (b), (c) 6, (d) 2, 8, 8, 1., , 3. If cos A cos 2 A 1 then (sin 2 A sin 4 A) ?, 1, (a), (b) 2, (c) 1, 2, , (d) 4
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Trigonometric Identities, , 4. If sin , , 611, , 3, then (cosec cot ) ?, 2, , (a) (2 3 ), , (b) 2 3, , (c), , (d), , 2, , 3, , Short-Answer Questions, , (sin A cos A), 4, 9., 5. If cot A , prove that, 5, (sin A cos A), 2, 1, 1, 6. If 2x sec A and x tan A, prove that cx 2 2 m ·, 4, x, 7. If 3 tan 3 sin , prove that (sin 2 cos 2 ) , 8. Prove that, , 1·, 3, , (sin 2 73c sin 2 17c), 1., (cos 2 28c cos 2 62c), , 9. If 2 sin 2 3 , prove that 30c., 1 cos A , (cosec A cot A) ., 1 cos A, (p 2 1), ·, 11. If cosec cot p, prove that cos 2, (p 1), 10. Prove that, , 12. Prove that (cosec A cot A) 2 , , (1 cos A), ·, (1 cos A), , 13. If 5 cot 3, show that the value of a, , 16, 5 sin 3 cos , k is ·, 29, 4 sin 3 cos , , 14. Prove that (sin 32c cos 58c cos 32c sin 58c) 1., 15. If x a sin b cos and y a cos b sin , prove that x 2 y 2 a 2 b 2 ., 16. Prove that, , (1 sin ), (sec tan ) 2 ., (1 sin ), , Long-Answer Questions, 17. Prove that, , 1, 1, ·, 1 1 , (sec tan ) cos cos (sec tan ), , 18. Prove that, , (sin A 2 sin 3 A), tan A., (2 cos 3 A cos A), , 19. Prove that, , tan A cot A , (1 tan A cot A) ., (1 cot A) (1 tan A), , 20. If sec 5A cosec (A 36c) and 5A is an acute angle, show that A 21c., ANSWERS (TEST YOURSELF), , 1. (b), , 2. (d), , 3. (c), , , , 4. (d)
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LINE OF SIGHT When an observer looks from a point O at an object P then the line, OP is called the line of sight., , ANGLE OF ELEVATION, , Suppose that from a point O, you look up at an, object P, placed above the level of your eye. Then,, the angle which the line of sight makes with, the horizontal line through O is called the angle, of elevation of P, as seen from O., Example, , Let OX be a horizontal line on the level ground and let a, person at O be looking up towards an object P, say an, aeroplane or the top of a tree or the top of a tower or a flag at, the top of a house., Then, ∠XOP is the angle of elevation of P from O., , ANGLE OF DEPRESSION, , Now, suppose that from a point O, you, look down at an object P, placed below, the level of your eye., Then, the angle which the line of, sight makes with the horizontal line, through O is called the angle of, depression of P, as seen from O., , SOLVED, , EXAMPLES, , EXAMPLE 1, , A vertical pole stands on the level ground. From a point on, the ground, 25 m away from the foot of the pole, the angle of, elevation of its top is found to be 60°. Find the height of the pole., [Take 3 = 1.732.], , SOLUTION, , Let AB be the pole standing on a level ground and let O be the, position of the observer. Then, OA = 25 m, ∠OAB = 90 ° and, ∠AOB = 60 °., 612
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Heights and Distances, , 613, , Let AB = h metres., From the right COAB, we have, AB, = tan 60 ° = 3, OA, h, ⇒, = 3, 25, ⇒ h = 25 × 3 = 25 × 1.732 = 43. 3., Hence, the height of the pole is 43.3 m., EXAMPLE 2, , SOLUTION, , A kite is flying, attached to a thread which is 165 m long. The thread, makes an angle of 30° with the ground. Find the height of the kite, from the ground, assuming that there is no slack in the thread., Let OX be the horizontal ground and let A be the position of, the kite. Let O be the position of the observer and OA be the, thread. Draw AB ⊥ OX., Then, ∠BOA = 30 °, OA = 165 m and ∠OBA = 90 °., Height of the kite from the ground = AB., Let AB = h metres., From right COBA, we have, AB, 1, = sin 30 ° =, OA, 2, h, 1, 165, ⇒, =, ⇒ h=, = 82.5., 165 2, 2, Hence, the height of the kite from the ground = 82.5 m., , EXAMPLE 3, , SOLUTION., , The length of a string between a kite and a point on the ground is, 85 m. If the string makes an angle θ with the ground level such that, tan θ = 15 8 then find the height of the kite from the ground. Assume, that there is no slack in the string., [CBSE 2014], Let OX be the horizontal, ground and let A be the, position of the kite. Let O be, the position of the observer, and OA be the string. Draw, AB ⊥ OX., Then, ∠BOA = θ such that, 15, tan θ =, , OA = 85 m and, 8, ∠OBA = 90 ° ., Let AB = h metres.
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614, , Secondary School Mathematics for Class 10, , From right SOBA, we have, AB, 15, = sin θ =, OA, 17, ⇒, , ⎡Q tan θ = 15 ⇒ sin θ = 15 ⎤, ⎢⎣, 8, 17 ⎥⎦, , 15, 15, h, =, ⇒h=, × 85 = 75., 85 17, 17, , Hence, the height of the kite from the ground is 75 m., EXAMPLE 4, , A ladder 15 m long just reaches the top of a vertical wall. If the ladder, makes an angle of 60° with the wall, find the height of the wall., [CBSE 2013], , SOLUTION, , Let OX be the horizontal ground and let OA be the ladder, leaning against the wall AB. Then,, OA = 15 m, ∠OAB = 60 ° and, ∠OBA = 90 ° ., Let AB = h metres., Now, ∠AOB = ( 90 ° − 60 ° ) = 30 °., From right SOBA, we have, AB, 1, h 1, = sin 30 ° = ⇒, =, OA, 2, 15 2, ⇒ h = 15 ×, , 1, = 7.5., 2, , Hence, the height of the wall is 7.5 metres., EXAMPLE 5, , If a tower 30 m high, casts a shadow 10 3 m long on the ground,, then what is the angle of elevation of the sun?, [CBSE 2017], , SOLUTION, , Let AB be the pole and let AC be its shadow., Let the angle of elevation of the sun be θ°., Then, ∠ACB = θ, ∠CAB = 90 °,, AB = 30 m and AC = 10 3 m., From right CCAB, we have, AB, 30, tan θ =, =, = 3, AC 10 3, ⇒ θ = 60 ° ., Hence, the angular elevation of the sun is 60°., , EXAMPLE 6, , If a 1.5-m-tall girl stands at a distance of 3 m from a lamp-post and, casts a shadow of length 4.5 m on the ground then find the height of, the lamp-post.
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Heights and Distances, SOLUTION, , 615, , Let AB be the lamp-post and CD be the girl., Let CE be the shadow of CD. Then,, CD = 1 .5 m , CE = 4 .5 m and AC = 3 m., Let AB = h metres., Now, SAEB and SCED are similar., 1 .5 1, AB CD, h, =, ⇒, =, =, ∴, AE CE, ( 3 + 4 .5) 4 .5 3, ⇒h=, , 1, × 7.5 = 2.5., 3, , Hence, the height of the lamp-post is 2.5 metres., EXAMPLE 7, , The shadow of a tower, when the angle of elevation of the sun is 45°,, is found to be 10 metres longer than when the angle of elevation is, 60°. Find the height of the tower. [Given 3 = 1.732.], , SOLUTION, , Let AB be the tower and let AC and AD, be its shadows when the angles of, elevation of the sun are 60° and 45°, respectively., ∴ ∠ACB = 60 °, ∠ADB = 45 °,, ∠DAB = 90 ° and CD = 10 m., Let AB = h metres and AC = x metres., From right CCAB, we get, AC, 1, = cot 60 ° =, AB, 3, ⇒, , x, 1, h, =, ⇒x=, ⋅, h, 3, 3, , K (i), , From right CDAB, we get, AD, x + 10, = cot 45 ° = 1 ⇒, =1, AB, h, ⇒ x + 10 = h ⇒ x = h − 10., Equating the values of x from (i) and (ii), we get, h, = h − 10 ⇒ h = 3 h − 10 3 ⇒ ( 3 − 1) h = 10 3, 3, ⇒ h=, , ⎧ 10 3, ( 3 + 1) ⎫, 10 3, =⎨, ×, ⎬, ( 3 − 1) ⎩ ( 3 − 1) ( 3 + 1) ⎭, = 5 3 ( 3 + 1) = 15 + 5 3, , K (ii)
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616, , Secondary School Mathematics for Class 10, , ⇒ h = (5 + 5 × 1 .732) = (15 + 8 . 66) = 23 . 66., Hence, the height of the tower is 23.66 m., The angle of elevation of the sun from the earth is called the, sun’s altitude., SUN’S ALTITUDE, , The sun’s altitudes are different at different times of the day., EXAMPLE 8, , SOLUTION, , A tower is 50 m high. Its shadow is x metres shorter when, the sun’s altitude is 45° than when it is 30°. Find the value of x., [Given 3 = 1.732.], Let PQ be the tower and let PA and PB be its shadows when, the altitudes of the sun are 45° and 30° respectively. Then,, ∠PAQ = 45 °, ∠PBQ = 30 °, ∠BPQ = 90 °, PQ = 50 m., Let AB = x metres., From right CAPQ, we have, AP, = cot 45 ° = 1, PQ, ⇒, , AP, = 1 ⇒ AP = 50 m., 50 m, , From right CBPQ, we have, BP, x + 50, = cot 30 ° = 3 ⇒, =, PQ, 50, , 3 ⇒ x = 50( 3 − 1)., , ⇒ x = 50(1 .732 − 1) = (50 × 0 .732) = 36.6, Hence, x = 36.6., EXAMPLE 9, , SOLUTION, , The shadow of a tower standing on a level ground is found to be 30 m, longer when the sun’s altitude is 30°, than when it was 60°. Find the, height of the tower. [Take 3 = 1 .732.], [CBSE 2012], Let AB be the tower and let AC, and AD be the lengths of its, shadows when ∠ACB = 60 ° and, ∠ADB = 30 ° ., Let AB = h metres, and, AC = x metres., From right SCAB, we have, AC, 1, x, 1, = cot 60 ° =, ⇒ =, AB, 3, h, 3, ⇒, , h, 1 ⎞, x = ⎛⎜ h ×, ⋅, ⎟ =, ⎝, 3⎠, 3, , ... (i)
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Heights and Distances, , From right SDAB, we have, AD, x + 30, = cot 30 ° = 3 ⇒, =, AB, h, , 617, , 3, , ⇒ x = ( 3 h − 30)., , ... (ii), , Equating the values of x from (i) and (ii), we get, h, = ( 3 h − 30) ⇒ h = 3 h − 30 3, 3, ⇒ 2 h = 30 3 ⇒ h = 15 3 = (15 × 1 .732) = 25 . 98., Hence, the height of the tower is 25.98 m., EXAMPLE 10, , SOLUTION, , From a point O on the ground, the angle of elevation of the top of a, tower is 30° and that of the top of the flagstaff on the top of the tower, is 60°. If the length of the flagstaff is 5 metres, find the height of the, tower., [ CBSE 2015], Let AB be the tower and BC be the, flagstaff., Let O be a point on the ground such, that ∠AOB = 30 ° and ∠AOC = 60 °, BC = 5 m (given)., Let AB = h metres and OA = x metres., From right SOAB, we have, OA, x, = cot 30 ° = 3 ⇒ =, AB, h, , 3, , ⇒ x = h 3., , ... (i), , From right SOAC , we have, OA, 1, = cot 60 ° =, AC, 3, x, 1, h +5, =, ⇒ x=, ⋅, h +5, 3, 3, Equating the values of x from (i) and (ii), we get, h +5, h 3=, ⇒ 3 h = h + 5 ⇒ 2 h = 5 ⇒ h = 2 .5 ., 3, ⇒, , ... (ii), , Hence, the height of the tower is 2.5 metres., EXAMPLE 11, , Two pillars of equal heights stand on either side of a road which is, 100 m wide. At a point on the road between the pillars, the angles of, elevation of the tops of the pillars are 60° and 30°. Find the height of, each pillar and position of the point on the road. [Take 3 = 1 .732.], [CBSE 2011, ’13]
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618, SOLUTION, , Secondary School Mathematics for Class 10, , Let AB and CD be two pillars, each of height h metres and let, AC be the road such that AC = 100 m., Let O be the point of observation on AC., Let OA = x metres and OC = (100 − x) m., Also, ∠AOB = 60 ° and ∠COD = 30 ° ., AB ⊥ AC and CD ⊥ AC ., From right SOAB, we have, AB, = tan 60 ° = 3, OA, h, ⇒, = 3 ⇒ h = 3 x., x, From right SOCD, we have, CD, 1, = tan 30 ° =, OC, 3, (100 − x), h, 1, ⇒, =, ⇒ h=, ⋅, (100 − x), 3, 3, , ... (i), , ... (ii), , Equating the values of h from (i) and (ii), we get, (100 − x), 3x =, ⇒ 3 x = (100 − x) ⇒ 4 x = 100 ⇒ x = 25., 3, Putting x = 25 m in (i), we get, h = ( 25 × 3 ) = ( 25 × 1 .732) = 43. 3., Hence, the height of each pillar is 43.3 m and the point of, observation is 25 m away from the first pillar., EXAMPLE 12, , SOLUTION, , The angle of elevation of an aeroplane from a point on the ground is, 60°. After a flight of 15 seconds, the angle of elevation changes to 30°., If the aeroplane is flying at a constant height of 1500 3 m, find the, speed of the plane in km/hr., [CBSE 2015], Let OX be the horizontal ground and let O be the point of, observation. Let A and B be the two positions of the aeroplane., Let AC ⊥ OX and BD ⊥ OX. Then,, ∠COA = 60 ° , ∠DOB = 30 °, and AC = BD = 1500 3 m., From right SOCA, we have, OC, 1, = cot 60 ° =, AC, 3
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Heights and Distances, , ⇒, , 619, , OC, 1, =, ⇒ OC = 1500 m., 1500 3, 3, , From right SODB, we have, OD, OD, = cot 30 ° = 3 ⇒, =, BD, 1500 3 m, , 3, , ⇒ OD = (1500 × 3) m = 4500 m., ∴, , CD = (OD − OC ) = ( 4500 − 1500) m = 3000 m., , Thus, the aeroplane covers 300 m in 15 seconds., 3000 60 × 60 ⎞, ∴ speed of the aeroplane = ⎛⎜, ×, ⎟ km/hr, ⎝ 15, 1000 ⎠, = 720 km/hr., Hence, the speed of the aeroplane is 720 km/hr., EXAMPLE 13, , SOLUTION, , The angles of elevation and depression of the top and bottom of, a tower from the top of a building 60 m high are 30° and 60°, respectively. Find the difference between the heights of the building, and the tower and also the distance between them. [CBSE 2013C, ’14], Let AB be the tower and CD be the building. Then, CD = 60 m., Let CAX be the horizontal ground., Draw DE ⊥ AB. Now, ∠EDB = 30 °, and ∠CAD = ∠ADE = 60 ° ., Also, AE = CD = 60 m., Let BE = h metres., From right SACD, we have, CA, 1, = cot 60 ° =, CD, 3, CA, 1, 60, ⇒, =, ⇒ CA =, ⋅ ... (i), 60 m, 3, 3, From right SEDB, we have, DE, DE, = cot 30 ° = 3 ⇒, =, BE, hm, , 3, , ⇒ DE = h 3 m., But, CA = DE., ∴ from (i) and (ii), we get, 60, = h 3 ⇒ 3 h = 60 ⇒ h = 20., 3, , ... (ii)
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620, , Secondary School Mathematics for Class 10, , Hence, the difference between the heights of the tower and the, building is 20 metres., ⎛ 60, From (i), we get CA = ⎜, ×, ⎝ 3, , 3⎞, ⎟ m = 20 3 metres., 3⎠, , Hence, the distance between the tower and the building is, 20 3 m., EXAMPLE 14, , SOLUTION, , A man on a cliff observes a boat at an angle of depression of 30°, which is approaching the shore to the point immediately beneath the, observer with a uniform speed. Six minutes later, the angle of, depression of the boat is found to be 60°. Find the total time taken by, the boat to reach the shore., [CBSE 2014], Let AB be the cliff with the man at B., Let C and D be the two positions of, the boat approaching the shore at A., Then,, ∠ACB = 30 ° and ∠ADB = 60 ° ., Let AB = h metres, CD = x metres, and DA = y metres., From right SDAB, we have, AD, 1, y, 1, h, = cot 60 ° =, ⇒ =, ⇒y=, ⋅, AB, 3, h, 3, 3, From right SCAB, we have, AC, x+y, = cot 30 ° = 3 ⇒, = 3 ⇒ x + y = h 3., AB, h, On subtracting (i) from (ii), we get, h ⎞, 2h, ⎛, x = ⎜h 3 −, ⋅, ⎟ ⇒x=, ⎝, 3⎠, 3, , ... (i), , ... (ii), , ... (iii), , From (iii) and (i), we get, 2h h, x:y =, :, = 2 : 1., 3, 3, Let the time taken to cover y units be t minutes. Then,, ratio of distances = ratio of times taken to cover them., 2 6, So, 2 : 1 = 6 : t ⇒ = ⇒ 2t = 6 ⇒ t = 3., 1 t, Thus, the boat takes 3 minutes to cover the distance DA., Hence, the total time taken by the boat to reach the shore is, ( 6 + 3) minutes = 9 minutes.
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Heights and Distances, EXAMPLE 15, , SOLUTION, , 621, , Observed from the top of a 75-m-high lighthouse (from sea level), the, angles of depression of two ships are 30° and 45°. If one ship is, exactly behind the other on the same side of the lighthouse, find the, distance between the ships. (Use 3 = 1 .73.), [CBSE 2014], Let AB be the lighthouse and C and D be the positions of two, ships. Then,, AB = 75 m, ∠ACB = 30 °, and ∠ADB = 45 ° ., Let CD = x metres., From right SDAB, we have, AD, = cot 45 ° = 1, AB, AD, ⇒, = 1 ⇒ AD = 75 m., 75 m, From right SCAB, we have, CA, x + 75, = cot 30 ° = 3 ⇒, =, AB, 75, , 3, , ⇒ x + 75 = 75 3 ⇒ x = 75( 3 − 1) = 75(1 .73 − 1), ⇒ x = (75 × 0 .73) = 54 .75., Hence, the distance between the ships is 54.75 metres., EXAMPLE 16, , SOLUTION, , Two ships are approaching a lighthouse from opposite directions. The, angles of depression of the two ships from the top of a lighthouse are, 30° and 45°. If the distance between the two ships is 100 metres, find, the height of the lighthouse. (Use 3 = 1 .732.), [CBSE 2014], Let AB be the lighthouse and C and D be the positions of the, two ships., Then, ∠ACB = 30 ° and ∠ADB = 45 ° ., Let AB = h metres, CA = x metres., Then, AD = (100 − x) m., , From right SBAD, we have, AD, = cot 45 ° = 1, AB
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622, , Secondary School Mathematics for Class 10, , ⇒, , (100 − x), = 1 ⇒ x = (100 − h)., h, , From right SBAC , we have, AC, x, = cot 30 ° = 3 ⇒ =, AB, h, , ... (i), , 3 ⇒ x = h 3., , ... (ii), , Equating the values of x from (i) and (ii), we get, 100 − h = h 3 ⇒ h( 3 + 1) = 100, ⎧ 100, ( 3 − 1) ⎫, ×, ⇒ h= ⎨, ⎬ = 50( 3 − 1), (, 3, 1, ), ( 3 − 1) ⎭, +, ⎩, ⇒ h = 50(1.732 − 1) = (50 × 0.732) = 36.6., Hence, the height of the lighthouse is 36.6 m., EXAMPLE 17, , From the top of a lighthouse, the angles of depression of two ships on, the opposite sides of it are observed to be α and β. If the height of the, lighthouse be h metres and the line joining the ships passes through, the foot of the lighthouse, show that the distance between the ships is, h(tan α + tan β), metres., [HOTS], tan α tan β, , SOLUTION, , Let AB be the lighthouse and C and D be the positions of the, two ships. Then, AB = h metres., Clearly, ∠ACB = α and ∠ADB = β., Let AC = x metres and, AD = y metres., From right CCAB, we have, AC, x, = cot α ⇒ = cot α ⇒ x = h cot α., AB, h, From right CDAB, we have, AD, y, = cot β ⇒, = cot β ⇒ y = h cot β., AB, h, , K (i), , K (ii), , Adding the corresponding sides of (i) and (ii), we get, ⎛ 1, 1 ⎞, x + y = h(cot α + cot β) = h ⎜, +, ⎟, ⎝ tan α tan β ⎠, ⇒ x+y =, , h(tan α + tan β), ⋅, tan α tan β, , Hence, the distance between the ships is, , h(tan α + tan β), m., tan α tan β
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624, , SOLUTION, , Secondary School Mathematics for Class 10, , in water of lake is 60°. Find the height of the cloud from the surface, of water., [CBSE 2010, ’15, ’17], Let AB be the surface of the lake and let P be a point vertically, above A such that AP = 60 m., Let C be the position of the cloud and let, D be its reflection in the lake., Draw PQ ⊥ CD. Then,, ∠QPC = 30 ° , ∠QPD = 60 °,, BQ = AP = 60 m., Let CQ = x metres. Then,, BD = BC = ( x + 60) m., From right CPQC , we have, PQ, = cot 30 ° = 3, CQ, ⇒, , PQ, =, xm, , 3 ⇒ PQ = x 3 m., , K (i), , From right CPQD, we have, PQ, 1, = cot 60 ° =, QD, 3, ⇒, , ( x + 120), PQ, 1, =, ⇒ PQ =, m., ( x + 60 + 60) m, 3, 3, , K (ii), , Equating the values of PQ from (i) and (ii), we get, ( x + 120), x 3=, 3, ⇒ 3 x = x + 120 ⇒ 2 x = 120 ⇒ x = 60., ∴ height of the cloud from the surface of the lake, = BC = ( 60 + x) m = ( 60 + 60) m = 120 m., Hence, the height of the cloud from the surface of the lake is, 120 metres., EXAMPLE 20, , A round balloon of radius r subtends an angle α at the eye of the, observer while the angle of elevation of its centre is β. Prove that the, height of the centre of the balloon is, ⎛⎜ r sin β cosec α ⎞⎟ ⋅, [HOTS], ⎝, 2⎠, , SOLUTION, , Let us represent the balloon by a circle with centre C and, radius r. Let OX be the horizontal ground and let O be the
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Heights and Distances, , 625, , point of observation. From O, draw tangents OA and OB to the, circle. Join CA, CB and CO. Draw CD ⊥ OX., ∴, , ∠AOB = α , ∠DOC = β and, α, ∠AOC = ∠BOC = ⋅, 2, , From right SOAC , we have, OC, α, = cosec, AC, 2, OC, α, ⇒, = cosec, r, 2, α, ⇒ OC = rcosec ⋅, 2, , ... (i), , From right SODC , we have, CD, = sin β ⇒ CD = (OC ) × sin β, OC, α, [using (i)]., ⇒ CD = r sin β cosec, 2, Hence, the height of the centre of the balloon from the ground, α, is r sin β cosec ⋅, 2, EXAMPLE 21, , A boy whose eye level is 1.3 m from the ground, spots a balloon, moving with the wind in a horizontal line at some height from the, ground. The angle of elevation of the balloon from the eyes of the boy, at an instant is 60°. After 2 seconds, the angle of elevation reduces to, 30°. If the speed of the wind is 29 3 m/s then find the height of the, [HOTS] [CBSE 2009C], balloon from the ground., , SOLUTION, , Let AB be the position of the boy, and AX be the horizontal, ground. Let C and D be the, two positions of the balloon., Draw CL ⊥ AX, DM ⊥ AX and, BN ⊥ DM, intersecting CL at P., Then, ∠CBN = 60 ° and ∠DBN = 30 °., Distance covered by the balloon in 2 seconds, = ( 29 3 × 2) m = 58 3 m., ∴, , CD = 58 3 m.
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626, , Secondary School Mathematics for Class 10, , Let BP = x metres. Then,, BN = ( BP + PN) = ( BP + CD) = ( x + 58 3 ) m and DN = CP., From right SBND, we have, DN, 1, = tan 30 ° =, BN, 3, ⇒, , ( x + 58 3 ), DN, 1, m., =, ⇒ DN =, ( x + 58 3 ) m, 3, 3, , From right SBPC, we have, CP, = tan 60 ° = 3, BP, CP, ⇒, = 3 ⇒ CP = ( x 3 ) m., xm, Now, DN = CP ⇒, , … (i), , … (ii), , ( x + 58 3 ), = (x 3), 3, , ⇒ ( 3 x − x) = 58 3 ⇒ 2 x = 58 3 ⇒ x = 29 3 ., From (ii), we get, CP = ( 29 3 × 3 ) m = ( 29 × 3) m = 87 m., ∴, , height of the balloon from the ground, = CL = CP + PL = CP + AB = 87 m + 1.3 m = 88.3 m., , Hence, the height of the balloon from the ground is, 88.3 metres., EXAMPLE 22, , An aeroplane when flying at a height of 3000 metres from the ground, passes vertically above another aeroplane at an instant when the, angles of elevation of the two planes from the same point on the ground, are 60° and 45° respectively. Find the vertical distance between the, aeroplanes at that instant. [Take 3 = 1 .73.], [CBSE 2008], , SOLUTION, , Let O be the point of observation. Let A, and B be the positions of the two planes at, the given instant when A is vertically, above B., Let AB when produced meet the ground, at C., Then,, ∠COA = 60 °,, ∠COB = 45 °,, ∠OCB = ∠OCA = 90 ° and AC = 3000 m., Let AB = x metres. Then, BC = ( 3000 − x) m.
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Heights and Distances, , 627, , From right SOCB, we have, OC, OC, = cot 45 ° = 1 ⇒, =1, BC, ( 3000 − x) m, ⇒ OC = ( 3000 − x) m., , ... (i), , From right SOCA, we have, OC, 1, OC, 1, = cot 60 ° =, ⇒, =, AC, 3, 3000 m, 3, ⎛ 3000, ⇒ OC = ⎜, ×, ⎝ 3, , 3⎞, ⎟ m = 1000 3 m., 3⎠, , ... (ii), , From (i) and (ii), we get, ( 3000 − x) = 1000 3, ⇒ x = ( 3000 − 1000 3) = ( 3000 − 1000 × 1 .73), = ( 3000 − 1730) = 1270., Hence, the required distance between the two aeroplanes is, 1270 metres., EXAMPLE 23, , A man standing on the deck of a ship, which is 10 m above the water, level, observes the angle of elevation of the top of a hill as 60°, and the, angle of depression of the base of the hill as 30 . Find the distance of, the hill from the ship and the height of the hill., [CBSE 2006, ’10], , SOLUTION, , Let AB be the deck and CD be the hill., Let the man be at B., Then, AB = 10 m., Let BE ⊥ CD and AC ⊥ CD., Then, ∠EBD = 60 ° and ∠EBC = 30 °., ∴, , ∠ACB = ∠EBC = 30 ° ., , Let CD = h metres., Then, CE = AB = 10 m and, ED = ( h − 10) m., From right CCAB, we have, AC, = cot 30 ° =, AB, , 3⇒, , ⇒ AC = 10 3 m., ∴, , BE = AC = 10 3 m., , AC, =, 10 m, , 3, ... (i)
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628, , Secondary School Mathematics for Class 10, , From right CBED, we have, DE, h − 10, = tan 60 ° = 3 ⇒, = 3 [using (i)], BE, 10 3, ⇒ h − 10 = 30 ⇒ h = 40., Hence, the distance of the ship from the hill is10 3 metres and, the height of the hill is 40 metres., EXAMPLE 24, , From a window (h metres high above the ground) of a house in a, street, the angles of elevation and depression of the top and the foot of, another house on the opposite side of the street are θ and φ, respectively. Show that the height of the opposite house is, [CBSE 2006], h(1 + tan θ cot φ ) metres., , SOLUTION, , Let AB be the house with window at B and let CD be the, another house. Then, AB = h metres., Draw BE|| AC , meeting CD at E. Then,, ∠EBD = θ and ∠ACB = ∠EBC = φ ., Let CD = H metres. Then,, CE = AB = h metres and, ED = ( H − h) m., From right CACB, we have, AC, AC, = cot φ ⇒, = cot φ ⇒ AC = h cot φ metres., AB, h, From right CBED, we have, ( H − h), DE, = tan θ ⇒, = tan θ, BE, h cot φ, , [Q BE = AC = h cot φ m], , ⇒ ( H − h) = h tan θ cot φ, ⇒ H = h(1 + tan θ cot φ)., Hence, the height of the opposite house is h(1 + tan θ cot φ ), metres., EXAMPLE 25, , From the top of a building 60 m high, the angles of depression of the, top and bottom of a tower are observed to be 30° and 60°. Find the, height of the tower., [CBSE 2005], , SOLUTION, , Let AB be the building and CD, be the tower such that, ∠BDE = 30 °, ∠BCA = 60 ° and, AB = 60 m., Let CA = DE = x metres.
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Heights and Distances, , 629, , From right CCAB, we have, CA, 1, = cot 60 ° =, AB, 3, x, 1, 1, ⇒, =, ⇒ x = 60 ×, 3, 60, 3, ⇒ x = 60 ×, , 1, ×, 3, , 3, = 20 3, 3, , ⇒ CA = DE = 20 3 m., From right CBED, we have, BE, 1, BE, 1, = tan 30 ° =, ⇒, =, DE, 3m, 20 3 m, 3, ⇒ BE = 20 3 ×, ∴, , K (i), , [using (i)], , 1, m = 20 m., 3, , CD = AE = AB − BE = 60 m − 20 m − 40 m., , Hence, the height of the tower is 40 m., EXAMPLE 26, , The angle of elevation of a jet plane from a point A on the ground is, 60°. After a flight of 30 seconds, the angle of elevation changes to, 30°. If the jet plane is flying at a constant height of 3600 3 metres,, find the speed of the jet plane., [CBSE 2008], , SOLUTION, , Let A be the point of observation and let AX be a horizontal, line through A and QC ⊥ AX. Let P and Q be the two positions, of the plane. Let PB ⊥ AX., , Then, PB = QC = 3600 3 m, ∠BAP = 60 ° and ∠BAQ = 30 ° ., From right SABP, we have, AB, 1, AB, 1, = cot 60 ° =, ⇒, =, BP, 3, 3600 3 m, 3, 1, ⇒ AB = 3600 3 ×, m = 3600 m., 3, Let BC = PQ = x metres., Then, AC = AB + BC = ( x + 3600) m [using (i)]., , ... (i)
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630, , Secondary School Mathematics for Class 10, , From right SACQ, we have, AC, x + 3600, = cot 30 ° = 3 ⇒, =, CQ, 3600 3, , 3, , ⇒ x + 3600 = ( 3600 × 3) = 10800, ⇒ x = 10800 − 3600 = 7200., Thus, PQ = 7200 metres., Now, 7200 m is covered in 30 seconds., 7200 60 × 60 ⎞, ∴ speed of the jet plane = ⎛⎜, ×, ⎟ km/hr, ⎝ 30, 1000 ⎠, = 864 km/hr., Hence, the speed of the jet plane is 864 km/hr., EXAMPLE 27, , SOLUTION, , The angle of elevation of a jet fighter from a point A on the ground is, 60°. After a flight of 15 seconds, the angle of elevation changes to 30°., If the jet is flying at a speed of 720 km/hour, find the constant, height at which the jet is flying. [Use 3 = 1.732.], [CBSE 2008], Let O be the point of observation on the ground OX., Let A and B be the two positions of the jet., Then, ∠XOA = 60 ° and ∠XOB = 30 ° ., Draw AL ⊥ OX and BM ⊥ OX., Let AL = BM = h metres., Speed of the jet, 5, = ⎛⎜ 720 × ⎞⎟ m/s, ⎝, 18 ⎠, = 200 m/s., Time taken to cover the distance AB = 15 s., Distance covered = speed × time, = 200 m/s × 15 s, , = 200 × 15 m = 3000 m., ∴, , LM = AB = 3000 m., , Let OL = x metres., From right SOLA, we have, OL, 1, = cot 60 ° =, AL, 3, ⇒, , x, 1, h, =, ⇒ x=, ⋅, h, 3, 3, , ... (i)
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Heights and Distances, , 631, , From right SOMB, we have, OM, = cot 30 ° = 3, BM, ( x + 3000), ⇒, = 3 ⇒ x = ( h 3 − 3000)., h, Equating the values of x from (i) and (ii), we get, h, = h 3 − 3000 ⇒ h = 3 h − 3000 3, 3, , ... (ii), , ⇒ 2 h = 3000 3 ⇒ h = 1500 3 = 1500 × 1 .732 = 2598., Hence, the required height is 2598 m., EXAMPLE 28, , SOLUTION, , A 1.2-m-tall girl spots a balloon moving with the wind in a, horizontal line at a height of 88.2 metres from the ground. The angle, of elevation of the balloon from the eyes of the girl at any instant is, 60°. After some time, the angle of elevation reduces to 30°. Find the, distance travelled by the balloon during the interval., Let AB be the position of the girl and AX be the horizontal, ground. Let C and D be the two positions of the balloon., Draw CL ⊥ AX , DM ⊥ AX and BN ⊥ DM, intersecting CL at P., , Then, ∠CBP = 60 ° , ∠DBN = 30 °, AB = PL = NM = 1.2 m and, CL = DM = 88.2 m., ∴, , CP = 88.2 m − 1.2 m = 87 m., , From right SBPC, we have, 87 m, BP, 1, BP, 1, = cot 60 ° =, ⇒, =, ⇒ BP =, CP, 3, 87 m, 3, 3, ⇒ BP =, , 87 m, ×, 3, , 3, = 29 3 m., 3, , From right SBND, we have, BN, BP + PN, = cot 30 ° = 3 ⇒, =, DN, CP, ⇒, , 29 3 m + CD, =, 87 m, , 3, , 3, , [Q PN = CD and DN = CP]
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632, , Secondary School Mathematics for Class 10, , ⇒ 29 3 m + CD = 87 3 m, ⇒ CD = 87 3 m − 29 3 m = 58 3 m., Hence, the required distance travelled by the balloon is, 58 3 m., EXAMPLE 29, , A tree breaks due to storm and the broken part bends so that the top of, the tree touches the ground making an angle 30° with it. The distance, between the foot of the tree to the point where the top touches the, ground is 9 m. Find the height of the tree., , SOLUTION, , Let AB be the original height of the tree., Suppose it got bent at a point C and let the, part CB take the position CD, meeting the, ground at D. Then,, AD = 9 m, ∠ADC = 30 ° and CD = CB., Let AC = x metres and CD = CB = y metres., From right ADAC , we have, AC, 1, x, 1, = tan 30 ° =, ⇒, =, AD, 3, 9, 3, ⇒ x=, , 9, 9, ⇒ x=, ×, 3, 3, , 3, = 3 3., 3, , Also, from right SDAC , we have, CD, 2, y, 2, = sec 30 ° =, ⇒, =, AD, 3, 9, 3, ⇒ y=, ∴, , 18, 18, ⇒ y=, ×, 3, 3, , 3, = 6 3., 3, , AC = 3 3 m and CB = 6 3 m., , Total height of the tree = 3 3 m + 6 3 m = 9 3 metres., EXAMPLE 30, , A person standing on the bank of a river observes that the angle of, elevation of the top of a tree standing on the opposite bank is 60°., When he moves 30 metres away from the bank, he finds the angle of, elevation to be 30°. Find the height of the tree and the width of, the river. [Take 3 = 1.732.], [CBSE 2008C], , SOLUTION, , Let AB be the tree and AC be the river., Let C and D be the two positions of the person., Then, ∠ACB = 60 °, ∠ADB = 30 °, ∠DAB = 90 ° and CD = 30 m., Let AB = h metres and AC = x metres.
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Heights and Distances, , 633, , From right CCAB, we have, AC, 1, = cot 60 ° =, AB, 3, ⇒, , x, 1, h, =, ⇒x=, ⋅, h, 3, 3, , K (i), , From right CDAB, we have, AD, = cot 30 ° = 3, AB, x + 30, ⇒, = 3 ⇒ x = 3 h − 30., h, , K (ii), , Equating the values of x from (i) and (ii), we get, h, = 3 h − 30 ⇒ h = 3 h − 30 3, 3, ⇒ 2 h = 30 3 ⇒ h = 15 3 = 15 × 1 .732 = 25 . 98., Putting h = 15 3 in (i), we get x =, , 15 3, = 15., 3, , Hence, the height of the tree is 25.98 m and the width of the, river is 15 metres., EXAMPLE 31, , The angles of elevation of the top of a tower from two points on the, ground at distances a metres and b metres from the base of the tower, and in the same straight line are complementary. Prove that the, [HOTS] [CBSE 2000C, ’05C], height of the tower is ab metres., , SOLUTION, , Let AB be the tower and let C and D be the two positions of the, observer. Then,, AC = a metres and AD = b metres., Let ∠ACB = θ. Then, ∠ADB = ( 90 ° − θ)., Let AB = h metres., From right CDAB, we have, AB, h, = tan ( 90 °−θ) ⇒ = cot θ, AD, b, ⇒ h = b tan θ., From right CCAB, we have, AB, h, = tan θ ⇒ = tan θ ⇒ h = a tan θ., AC, a, From (i) and (ii), we get h 2 = ab ⇒ h = ab ., Hence, the height of the tower is ab metres., , L (i), L (ii)
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634, EXAMPLE 32, , Secondary School Mathematics for Class 10, , A boy standing on a horizontal plane finds a bird flying at a distance, of 100 m from him at an elevation of 30°. A girl standing on the roof, of a 20-m-high building, finds the angle of elevation of the same bird, to be 45°. The boy and the girl are on the opposite sides of the bird., Find the distance of the bird from the girl.[Given 2 = 1.41.] [HOTS], [CBSE 2007], , SOLUTION, , Let O be the position of the bird, B be the position of the boy, and FG be the building at which G is the position of the girl., Let OL, , BF and GM, , OL. Then,, , BO = 100 m, ∠OBL = 30 ° ,, FG = 20 m and ∠OGM = 45 ° ., From right COLB, we have, OL, OL, 1, = sin 30 ° ⇒, =, BO, 100 m 2, ⇒ OL = 100 m ×, , 1, = 50 m., 2, , OM = OL − ML = OL − FG = 50 m − 20 m = 30 m., From right COMG, we have, OM, 1, = sin 45 ° =, ⇒ OG = 2 × OM = 2 × 30 m, OG, 2, ⇒ OG = 30 × 1.41 m = 42.3 m., Distance of the bird from the girl = 42.3 m., EXAMPLE 33, , A 1.5-m-tall boy is standing at some distance from a 30-m-tall, building. The angle of elevation from his eyes to the top of the, building increases from 30° to 60° as he walks towards the building., Find the distance he walked towards the building., [HOTS], , SOLUTION, , Let AB be the building and let CD and EF be the two positions, of the boy. Draw DFG||CEA. Then,, CD = EF = 1.5 m, ∠GDB = 30 ° and ∠GFB = 60 °,, AB = 30 m, GB = 30 m − 1.5 m = 28.5 m., From right SDGB, we have, DG, DG, = cot 30 ° ⇒, =, GB, 28.5 m, , 3 ⇒ DG =, , 57 3, m., 2, , From right SFGB, we have, FG, FG, 1, 57, = cot 60 ° ⇒, =, ⇒ FG =, m., GB, 28.5 m, 3, 2 3
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Heights and Distances, , ∴, , 635, , ⎛ 57 3, 57 ⎞, ⎛ 171 − 57 ⎞ m, DF = DG − FG = ⎜, −, ⎟, ⎟ m= ⎜, ⎝ 2 3 ⎠, 2 3⎠, ⎝ 2, =, , 114, 57, 57, m=, m=, ×, 2 3, 3, 3, , 3, m = 19 3 m., 3, , Hence, the required distance is 19 3 m., , EXERCISE 14, 1. A tower stands vertically on the ground. From a point on the ground, which is 20 m away from the foot of the tower, the angle of elevation of, its top is found to be 60°. Find the height of the tower. [Take 3 = 1.732.], 2. A kite is flying at a height of 75 m from the level ground, attached to a, string inclined at 60° to the horizontal. Find the length of the string,, assuming that there is no slack in it. [Take 3 = 1.732.], 3. An observer 1.5 m tall is 30 m away from a chimney. The angle of, elevation of the top of the chimney from his eye is 60°. Find the height, of the chimney., [CBSE 2013C], 4. The angles of elevation of the top of a tower from two points at, distances of 5 metres and 20 metres from the base of the tower and in, the same straight line with it, are complementary. Find the height of, [CBSE 2014], the tower., 5. The angle of elevation of the top of a tower at a distance of 120 m from a, point A on the ground is 45°. If the angle of elevation of the top of a, flagstaff fixed at the top of the tower, at A is 60°, then find the height of, [CBSE 2014], the flagstaff. [Use 3 = 1 .732.], 6. From a point on the ground 40 m away from the foot of a tower, the, angle of elevation of the top of the tower is 30°. The angle of elevation of
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636, , Secondary School Mathematics for Class 10, , the top of a water tank (on the top of the tower) is 45°. Find (i) the height, of the tower, (ii) the depth of the tank., 7. A vertical tower stands on a horizontal plane and is surmounted by a, vertical flagstaff of height 6 m. At a point on the plane, the angle of, elevation of the bottom of the flagstaff is 30° and that of the top of the, flagstaff is 60°. Find the height of the tower. [Use 3 = 1 .732.] [CBSE 2011], 8. A statue 1.46 m tall, stands on the top of a pedestal. From a point on the, ground, the angle of elevation of the top of the statue is 60° and from the, same point, the angle of elevation of the top of the pedestal is 45°. Find, [CBSE 2008], the height of the pedestal. [Use 3 = 1.73.], 9. The angle of elevation of the top of an unfinished tower at a distance of, 75 m from its base is 30°. How much higher must the tower be raised, so that the angle of elevation of its top at the same point may be 60°?, [Take 3 = 1.732.], 10. On a horizontal plane there is a vertical tower with a flagpole on the top, of the tower. At a point, 9 metres away from the foot of the tower, the, angle of elevation of the top and bottom of the flagpole are 60° and 30°, respectively. Find the height of the tower and the flagpole mounted, [CBSE 2005], on it. [Take 3 = 1.73.], 11. Two poles of equal heights are standing opposite to each other on either, side of the road which is 80 m wide. From a point P between them on, the road, the angle of elevation of the top of one pole is 60° and the, angle of depression from the top of another pole at P is 30°. Find the, height of each pole and distances of the point P from the poles., [CBSE 2015], , 12. Two men are on opposite sides of a tower. They measure the angles of, elevation of the top of the tower as 30° and 45° respectively. If the, height of the tower is 50 metres, find the distance between the two men., [Take 3 = 1.732.], 13. From the top of a tower 100 m high, a man observes two cars on the, opposite sides of the tower and in same straight line with its base,, with angles of depression 30° and 45° respectively. Find the distance, [CBSE 2011, ’17], between the cars. [Take 3 = 1.732.], 14. A straight highway leads to the foot of a tower. A man standing on the, top of the tower observes a car at an angle of depression of 30°, which is, approaching the foot of the tower with a uniform speed. Six seconds, later, the angle of depression of the car is found to be 60°. Find the time, taken by the car to reach the foot of the tower form this point., [HINT See Solved Example 14.]
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Heights and Distances, , 637, , 15. A TV tower stands vertically on a bank of canal. From a point on the, other bank directly opposite the tower, the angle of elevation of the top, of the tower is 60°. From another point 20 m away from this point on the, line joining this point to the foot of the tower, the angle of elevation of, the top of the tower is 30°. Find the height of the tower and the width of, the canal., 16. The angle of elevation of the top of a building from the foot of a tower is, 30°. The angle of elevation of the top of the tower from the foot of the, building is 60°. If the tower is 60 m high, find the height of the building., [CBSE 2013], , 17. The horizontal distance between two towers is 60 metres. The angle of, depression of the top of the first tower when seen from the top of the, second tower is 30°. If the height of the second tower is 90 metres, find, the height of the first tower. [Use 3 = 1.732.], 18. The angle of elevation of the top of a chimney from the foot of a tower is, 60° and the angle of depression of the foot of the chimney from the top, of the tower is 30°. If the height of the tower is 40 metres, find the height, of the chimney., According to pollution control norms, the minimum height of a smokeemitting chimney should be 100 metres. State if the height of the abovementioned chimney meets the pollution norms. What value is, [CBSE 2014], discussed in this question?, 19. From the top of a 7-metre-high building, the angle of elevation of the, top of a cable tower is 60° and the angle of depression of its foot is 45°., [CBSE 2017], Determine the height of the tower. [Use 3 = 1.732.], 20. The angle of depression from the top of a tower of a point A on the, ground is 30°. On moving a distance of 20 metres from the point, A towards the foot of the tower to a point B, the angle of elevation of the, top of the tower from the point B is 60°. Find the height of the tower and, [CBSE 2012], its distance from the point A., 21. The angle of elevation of the top of a vertical tower from a point on the, ground is 60°. From another point 10 m vertically above the first, its, [CBSE 2011], angle of elevation is 30°. Find the height of the tower., 22. The angles of depression of the top and bottom of a tower as seen from, the top of a 60 3-m-high cliff are 45° and 60° respectively. Find the, [CBSE 2012], height of the tower., 23. A man on the deck of a ship, 16 m above water level, observes that the, angles of elevation and depression respectively of the top and bottom
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638, , Secondary School Mathematics for Class 10, , of a cliff are 60° and 30°. Calculate the distance of the cliff from the ship, [CBSE 2007C], and height of the cliff. [Take 3 = 1.732.], 24. The angle of elevation of the top Q of a vertical tower PQ from a point X, on the ground is 60°. At a point Y, 40 m vertically above X, the angle of, elevation is 45°. Find the height of tower PQ. [Take 3 = 1.73.], [CBSE 2003C], , 25. The angle of elevation of an aeroplane from a point on the ground is 45°., After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is, flying at a height of 2500 metres, find the speed of the aeroplane., 26. The angle of elevation of the top of a tower from a point on the same, level as the foot of the tower is 30°. On advancing 150 m towards the, foot of the tower, the angle of elevation becomes 60°. Show that the, [CBSE 2006C], height of the tower is 129.9 metres. [Given 3 = 1.732.], 27. As observed from the top of a lighthouse, 100 m above sea level, the, angle of depression of a ship, sailing directly towards it, changes from, 30° to 60°. Determine the distance travelled by the ship during the, [CBSE 2004, ’08C], period of observation. [Use 3 = 1.732.], 28. From a point on a bridge across a river, the angles of depression of the, banks on opposite sides of the river are 30° and 45° respectively. If the, bridge is at a height of 2.5 m from the banks, find the width of the river., [Take 3 = 1.732.], 29. The angles of elevation of the top of a tower from two points at, distances of 4 m and 9 m from the base of the tower and in the same, straight line with it are complementary. Show that the height of the, tower is 6 metres., 30. A ladder of length 6 metres makes an angle of 45° with the floor while, leaning against one wall of a room. If the foot of the ladder is kept fixed, on the floor and it is made to lean against the opposite wall of the room,, it makes an angle of 60° with the floor. Find the distance between two, [CBSE 2011], walls of the room., 31. From the top of a vertical tower, the angles of depression of two cars in, the same straight line with the base of the tower, at an instant are found, to be 45° and 60°. If the cars are 100 m apart and are on the same side of, [CBSE 2011], the tower, find the height of the tower., 32. An electrician has to repair an electric fault on a pole of height 4 metres., He needs to reach a point 1 metre below the top of the pole to undertake, the repair work. What should be the length of the ladder that he should, use, which when inclined at an angle of 60° to the horizontal would, enable him to reach the required position? [Use 3 = 1.73.]
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Heights and Distances, , 639, , 33. From the top of a building AB, 60 m high, the angles of depression of the, top and bottom of a vertical lamp-post CD are observed to be 30° and, 60° respectively. Find, (i) the horizontal distance between AB and CD,, (ii) the height of the lamp-post,, (iii) the difference between the heights of the building and the, lamp-post., [CBSE 2009], 34. A man observes a car from the top of a tower, which is moving towards, the tower with a uniform speed. If the angle of depression of the car, changes from 30° to 45° in 12 minutes, find the time taken by the car, now to reach the tower., [CBSE 2017], 35. An aeroplane is flying at a height of 300 m above the ground. Flying, at this height the angles of depression from the aeroplane of two points, on both banks of a river in opposite directions are 45° and 60°, respectively. Find the width of the river. [Use 3 = 1.732.] [CBSE 2017], 36. From a point on the ground the angles of elevation of the bottom, and top of a communication tower fixed on the top of a 20-m-high, building are 45° and 60° respectively. Find the height of the tower., [Take 3 = 1.732.], [CBSE 2017], 37. From the top of a hill, the angles of depression of two consecutive, kilometre stones due east are found to be 45° and 30° respectively., Find the height of the hill., [CBSE 2017], 38. If at some time of the day the ratio of the height of a vertically standing, pole to the length of its shadow on the ground is 3 : 1 then find the, [CBSE 2017], angle of elevation of the sun at that time., ANSWERS (EXERCISE 14), , 1. 34.64 m, , 2. 86.6 m, , 6. (i) 23.1 m (ii) 16.9 m, , 3. 53.46 m, , 4. 10 m, , 5. 87.84 m, , 7. 3 m, , 8. 2 m, , 9. 86.6 m, , 10. 5.19 m, 10.38 m, 11. 20 3 m; 20 m from left pole and 60 m from right pole, 12. 136.6 m, , 13. 273.2 m, , 14. 3 seconds 15. 10 3 m, 10 m, , 16. 20 m, , 17. 55.36 m, , 18. 120 m, , 19. 19.12 m, , 21. 15 m, , 22. 43.92 m, , 23. 27.71 m, 64 m, , 24. 94.6 m, , 25. 439.2 km/hr, , 27. 115.46 m, , 28. 6.83 m, , 31. 136.6 m, , 32. 3.46 m, , 30. 7.24 m, , 20. 17.32 m, 30 m
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640, , Secondary School Mathematics for Class 10, , 34. 6( 3 + 1) minutes, , 33. (i) 34.64 m (ii) 40 m (iii) 20 m, 35. 473.2 m, , 36. 14.64 m, , 37., , 3 +1, km 38. 60°, 2, , HINTS TO SOME SELECTED QUESTIONS, 2. Let the length of the string be x metres., x, 2, Then,, = cosec 60 ° =, 75, 3, 2, ⇒ x = 75 ×, 3, 2, 3, ×, ∴ x = 75 ×, 3, 3, = 50 3., 3. Let AB be the chimney and CD be the observer., Draw DE ⊥ AB., Then, AE = CD = 1. 5 m., Let AB = x metres. Then, BE = ( x − 1. 5 ) m., ∠EDB = 60 ° ., Also, DE = CA = 30 m., From right SBED, we have, BE, = tan 60 °, DE, x − 1. 5, ⇒, = 3., 30, 4., , AB, h, = tan θ ⇒ = tan θ, AC, 5, ⇒ h = 5 tanθ., AB, = tan( 90 ° − θ) = cot θ., AD, h, = cot θ ⇒ h = 20 cot θ, ⇒, 20, Multiplying (i) and (ii), we get, ∴, , ... (i), , ... (ii), , h 2 = 100 tan θ × cot θ = 100 ⇒ h = 10., , 5. Let BC be the tower and CD be the flagstaff and let AB be, the horizontal ground such that AB = 120 m, ∠BAC = 45 °, and ∠BAD = 60 ° . Then,, BC, BC, = tan 45 ° = 1 ⇒, = 1 ⇒ BC = 120 m., AB, 120 m, Let CD = x metres. Then,, BD, 120 + x, = tan 60 ° = 3 ⇒, =, AB, 120, , 3 . Find x.
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Heights and Distances, , 641, , 6. Let BC be the tower and CD be the water tank., Let A be the point of observation. Then, ∠BAC = 30 ° , ∠BAD = 45 ° and AB = 40 m., From right SABD, we have, BD, BD, = tan 45 ° = 1 ⇒, = 1 ⇒ BD = 40 m., AB, 40 m, From right SABC , we have, BC, 1, BC, 1, = tan 30 ° =, ⇒, =, AB, 40 m, 3, 3, ⇒, , BC =, , 40 m, 3, , ⇒ BC =, , 40 m, 3, , (i) Height of the tower = BC =, , ×, , 3 40 3, =, m⋅, 3, 3, , 40 3, m = 23.1 m., 3, , (ii) Depth of the tank = CD = ( BD − BC ) = ( 40 − 23.1) m = 16.9 m., OA, OA, 7., ... (i), = cot 30 ° ⇒, = 3 ⇒ OA = h 3 m., AB, hm, ( h + 6), OA, OA, 1, = cot 60 ° ⇒, =, ⇒ OA =, m. ... (ii), AC, ( h + 6) m, 3, 3, ( h + 6), , ∴, , h 3=, , ⇒, , 2 h = 6 ⇒ h = 3., , 3, , ⇒ 3 h = h + 6., , 9. Let AB be the unfinished tower and let AC be the, complete tower. Let O be the point of observation., Then, OA = 75 m, ∠AOB = 30 ° and ∠AOC = 60 ° ., Let AB = h metres and AC = H metres., AB, h, 1, = tan 30 ° ⇒, =, ⇒, OA, 75 m, 3, And,, , AC, = tan 60 ° ⇒, OA, , H, =, 75 m, , h = 25 3 m., , 3 ⇒ H = 75 3 m., , Hence, the required height is ( H − h ) m = (50 3 ) m., 11. From right SPAB, we have, AB, = tan 60 ° = 3, AP, ⇒, , h = x 3., , From right SPCD, we have, CD, 1, = tan 30 ° =, PC, 3, ( 80 − x ), ⋅, ⇒ h=, 3, , ... (i), , .... (ii)
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642, , Secondary School Mathematics for Class 10, , \, , x 3=, , ( 80 - x ), 3, , Þ 3 x + x = 80, , Þ, , 4 x = 80 Þ x = 20, , \, , height = 20 3 m., , P is 20 m from left pole and 60 m from right pole., AC, 12., = cot 30 °= 3 Þ AC = 50 3 m. ... (i), AB, AD, ... (ii), = cot 45 °= 1 Þ AD = 50 m., AB, Required distance, = AC + AD = 50( 3 + 1) m, = (50 ´ 2.732 ) m = 136.6 m., 13. From right SBAD, we have, AD, AD, = cot 45 °= 1 Þ, Þ AD = 100 m., AB, 100 m, From right SBAC , we have, AC, AC, = cot 30 °= 3 Þ, =, AB, 100 m, Þ, , AC = ( 100 ´ 1. 732 ) m = 173.2 m., , \, , distance between the cars, , 3, , = ( 173.2 + 100 ) m = 273.2 m., 15. Let AB be the tower and AC be the canal., x, 1, h, ... (i), = cot 60 °=, Þ x=, ×, h, 3, 3, ( x + 20 ), = cot 30 °= 3 Þ x = ( h 3 - 20 ). ... (ii), h, h, From (i) and (ii), we get, = ( h 3 - 20 )., 3, \ h = ( 3 h - 20 3 ) Þ 2 h = 20 3 Þ h = 10 3 ., Putting h = 10 3 in (i), we get x m = 10 m (width of the canal)., 16. Let AB be the building and CD be the tower., From right SACD, we have, AC, 1, x, 1, 60, = cot 60 °=, Þ, =, Þ x=, CD, 60, 3, 3, 3, 60, 3, Þ x=, ´, = 20 3., 3, 3, From right SBAC , we have, , Þ, , AB, 1, h, 1, x, = tan 30 °=, Þ =, Þ h=, AC, x, 3, 3, 3, 20 3, h=, = 20., 3
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Heights and Distances, , 643, , 17. From right SBED, we have, 90 − h, 1, 60, = tan 30 ° =, ⇒ 90 − h =, 60, 3, 3, 60, ×, 3, , 3, = 20 3, 3, , ⇒, , 90 − h =, , ⇒, , h = 90 − 20 3 = 90 − 20 × 1.732, , ⇒, , h = 90 − 34.64 = 55. 36., , 18. Let AB be the tower and CD be the chimney., From right SACD, we have, AC, 1, AC, 1, h, = cot 60 ° =, ⇒, =, ⇒ AC =, ⋅ ... (i), CD, h, 3, 3, 3, From right SCAB, we have, AC, ... (ii), = cot 30 ° = 3 ⇒ AC = 40 3., AB, h, From (i) and (ii), we get, = 40 3 ⇒ h = 120., 3, Clearly, the height of the given chimney meets the, pollution norms., We should comply with the prescribed rules and contribute to the cleanliness of the, environment., 19. Let AB be the building and CD be the cable tower., Draw BE ⊥ CD. Let CD = h metres., Then, CE = AB = 7 m and DE = ( h − 7 ) m., From right SCAB, we have, AC, AC, = cot 45 ° = 1 ⇒, =1, AB, 7m, ⇒, , AC = 7., , ∴, , BE = AC = 7 m., , From right SBED, we have, DE, h−7, = tan 60 ° = 3 ⇒, = 3., BE, 7, ∴ h = 7 3 + 7 = 7 ( 3 + 1) = 7 ( 1. 732 + 1), ⇒, , h = (7 × 2 . 732 ) = 19.12., , 20. From right SAPQ , we have, PQ, h, 1, = tan 30 ° ⇒, =, PA, x + 20, 3, x + 20, ⇒ h=, ⋅, 3, From right SBPQ , we have, PQ, h, = tan 60 ° ⇒ = 3 ⇒ h = x 3., PB, x, From (i) and (ii), we get, ( x + 20 ), = x 3 ⇒ x = 10., 3, , ... (i), , ... (ii)
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644, , Secondary School Mathematics for Class 10, , ∴, , h = 10 3 = 10 × 1.732 = 17 . 32., , Distance AP = ( 20 + x ) m = ( 20 + 10 ) m = 30 m., 21. Let the height of the tower be h metres., From right SCAB, we have, CA, CA, 1, = cot 60 ° ⇒, =, AB, h, 3, h, ⇒ CA =, ⋅, 3, From right SBED, we have, DE, DE, = cot 30 ° ⇒, =, BE, ( h − 10 ), ⇒ DE =, , ... (i), , 3, , 3 ( h − 10 )., , ... (ii), , But, CA = DE., h, = 3 ( h − 10 ) ⇒ 3 h − 30 = h, ∴, 3, ⇒ 2 h = 30 ⇒ h = 15., 22. Let AB be the cliff and CD be the tower., Let AE = CD = h metres. Then, BE = ( 60 3 − h ) m., From right SBED, we have, DE, DE, = cot 45 ° ⇒, =1, BE, 60 3 − h, ⇒ DE = 60 3 − h., , ... (i), , From right SCAB, we have, CA, CA, 1, = cot 60 ° ⇒, =, AB, 60 3, 3, ⇒ CA =, , 60 3, = 60., 3, , But, CA = DE., ∴, , 60 3 − h = 60 ⇒ h = 60( 3 − 1) = 60 × ( 1. 732 − 1), = ( 60 × 0 . 732 ) = 43.92., , 23. Let the height of the cliff be h metres and the distance of, the cliff from the ship be x metres., From right SCAB, we have, AC, x, = cot 30 ° ⇒, = 3 ⇒ x = 16 3., AB, 16, From right SBLD, we have, DL, h − 16, = tan 60 ° = 3 ⇒, = 3, BL, 16 3, [Q BL = x = 16 3], ⇒ h − 16 = 48 ⇒ h = 64., , ... (ii)
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Heights and Distances, , 645, , 24. Let the height of tower PQ be h metres., From right SXPQ , we have, XP, XP, 1, h, = cot 60 ° ⇒, =, ⇒ XP =, ⋅, PQ, h, 3, 3, From right SYRQ , we have, YR, YR, = cot 45 ° ⇒, = 1⇒ YR = ( h − 40 )., QR, h − 40, , ... (i), , ... (ii), , But, YR = XP., ∴, , h − 40 =, , h, ⇒ h( 3 − 1) = 40 3, 3, ⎧ 40 3, ( 3 + 1) ⎫, ⇒ h=⎨, ×, ⎬ = 20( 3 +, −, (, 3, 1, ), ( 3 + 1) ⎭, ⎩, , 3), , ⇒ h = 20( 3 + 1. 73 ) = ( 20 × 4 . 73 ) = 94.6., 25. Let A and B be the two positions of the aeroplane, and let O be the point of observation., From right SOCA , we have, OC, OC, = cot 45 ° ⇒, = 1 ⇒ OC = 2500 m., AC, 2500 m, From right SODB, we have, OD, OD, = cot 30 ° ⇒, =, BD, 2500 m, ∴, , 3 ⇒ OD = 2500 3 m., , AB = CD = OD − OC = 2500( 3 − 1) m = ( 25000 × 0.732 ) m = 1830 m., , Thus, the aeroplane covers 1830 m in 15 seconds., 1830 60 × 60 ⎞, ×, ∴ its speed = ⎛⎜, ⎟ km/hr = 439.2 km/hr., ⎝ 15, 1000 ⎠, 27. Let AB be the lighthouse and C and D be the two positions of the ship., x, 100, Then,, = cot 60 ° ⇒ x =, m., 100 m, 3, x+ y, = cot 30 ° = 3 ⇒ x + y = ( 100 3 ) m., 100 m, 100, 200, m=, m, ∴ y = x + y − x = 100 3 m −, 3, 3, 200 m, 3 200 m, =, ×, =, × 3, 3, 3, 3, 200 m, 346.4 m, ⇒ y=, × 1 . 732 =, = 115.46 m., 3, 3, 28. Let A and B be two points on the banks on opposite sides of the river., Let P be a point on the bridge at a height of 2.5 m., Then, DP = 2 . 5 m., From right SPDB, we have, DB, DB, = cot 45 ° ⇒, = 1 ⇒ DB = 2.50 m., PD, 2 .5 m
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646, , Secondary School Mathematics for Class 10, From right SPDA , we have, AD, AD, = cot 30 ° ⇒, =, PD, 2.5 m, , ∴, , 3 ⇒ AD = 2.5 3 m., , Width of the river = AB = AD + DB = 2 .5 3 m + 2 . 5 m, = 2 . 5( 3 + 1) m =, , 5, × ( 1. 732 + 1) m = 6.83 m., 2, , 30. Let AB and CD be the two opposite walls and let the foot of the ladder be fixed at the, point O on the ground. Let OB and OD be the two positions of the ladder., From right SOAB, we have, OA, OA, 1, = cos 45 ° ⇒, =, OB, 6m, 2, 6m, 6m, 2, ⇒ OA =, ×, = 3 2 m., ⇒ OA =, 2, 2, 2, From right SOCD, we have, OC, OC 1, = cos 60 ° ⇒, = ⇒ OC = 3 m., OD, 6m 2, Distance between two walls = AC = OA + OC = 3( 2 + 1) m = 3(1. 414 + 1) m, = ( 3 × 2.414 ) m = 7 .242 m ≈ 7 .24 m., 32. Let AB be the electric pole such that AB = 4 m. Let C be a point, 1 m below B., Then, AC = 4 m − 1 m = 3 m., Let OC be the ladder. Then, ∠AOC = 60 °., Let OC = x metres., From right SOAC , we have, OC, x, 2, 6, 6, = cosec 60 ° ⇒ =, ⇒ x=, ⇒ x=, ×, AC, 3, 3, 3, 3, ⇒ x = ( 2 × 1 . 73 ) = 3.46., Hence, the length of the ladder is 3.46 m., , 3, = 2 3., 3, , ................................................................, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. If the height of a vertical pole is equal to the length of its shadow on the, ground, the angle of elevation of the sun is, [CBSE 2014], (a) 0°, (b) 30°, (c) 45°, (d) 60°, 2. If the height of a vertical pole is 3 times the length of its shadow on the, ground then the angle of elevation of the sun at that time is, [CBSE 2012, ’14], , (a) 30°, , (b) 45°, , (c) 60°, , (d) 75°
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Heights and Distances, , 647, , 3. If the length of the shadow of a tower is 3 times its height then the, [CBSE 2012], angle of elevation of the sun is, (a) 45°, (b) 30°, (c) 60°, (d) 90°, 4. If a pole 12 m high casts a shadow 4 3 m long on the ground then the, [CBSE 2013C], sun’s elevation is, (a) 60°, (b) 45°, (c) 30°, (d) 90°, 5. The shadow of a 5-m-long stick is 2 m long. At the same time, the length, [CBSE 2011], of the shadow of a 12.5-m-high tree is, (a) 3 m, , (b) 3.5 m, , (c) 4.5 m, , (d) 5 m, , 6. A ladder makes an angle of 60° with the ground when placed against a, wall. If the foot of the ladder is 2 m away from the wall, the length of the, [CBSE 2014], ladder is, 4, (a), (c) 2 2 m, (d) 4 m, m, (b) 4 3 m, 3, 7. A ladder 15 m long makes an angle of 60° with the wall. Find the height, [CBSE 2017], of the point, where the ladder touches the wall., (a) 15 3 m, , (b), , 15 3, m, 2, , (c), , 15, m, 2, , (d) 15 m, , 8. From a point on the ground, 30 m away from the foot of a tower,, the angle of elevation of the top of the tower is 30°. The height of the, [CBSE 2014], tower is, (a) 30 m, , (b) 10 3 m, , (c) 10 m, , (d) 30 3 m, , 9. The angle of depression of a car parked on the road from the top of a, 150-m-high tower is 30°. The distance of the car from the tower is, [CBSE 2014], , (a) 50 3 m, , (b) 150 3 m, , (c) 150 2 m, , (d) 75 m, , 10. A kite is flying at a height of 30 m from the ground. The length of string, from the kite to the ground is 60 m. Assuming that there is no slack in the, [CBSE 2012], string, the angle of elevation of the kite at the ground is, (a) 45°, , (b) 30°, , (c) 60°, , (d) 90°, , 11. From the top of a cliff 20 m high, the angle of elevation of the top of a, tower is found to be equal to the angle of depression of the foot of the, [CBSE 2013C], tower. The height of the tower is, (a) 20 m, , (b) 40 m, , (c) 60 m, , (d) 80 m, , 12. If a 1.5-m-tall girl stands at a distance of 3 m from a lamp-post and casts a, shadow of length 4.5 m on the ground, then the height of the lamp-post is, (a) 1.5 m, , (b) 2 m, , (c) 2.5 m, , (d) 2.8 m
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648, , Secondary School Mathematics for Class 10, , 13. The length of the shadow of a tower standing on level ground is found, to be 2x metres longer when the sun’s elevation is 30° than when it was, 45°. The height of the tower is, (a) ( 2 3x) m, , (b) ( 3 2x) m, , (c) ( 3 − 1) x m (d) ( 3 + 1) x m, , 14. The lengths of a vertical rod and its shadow are in the ratio 1 : 3 . The, angle of elevation of the sun is, (a) 30°, , (b) 45°, , (c) 60°, , (d) 90°, , 15. A pole casts a shadow of length 2 3 m on the ground when the sun’s, [CBSE 2015], elevation is 60°. The height of the pole is, (a) 4 3 m, , (b) 6 m, , (c) 12 m, , (d) 3 m, , 16. In the given figure, a tower AB is 20 m high and BC, its, shadow on the ground is 20 3 m long. The sun’s, [CBSE 2015], altitude is, (a) 30°, , (b) 45°, , (c) 60°, , (d) none of these, , 17. The tops of two towers of heights x and y, standing on a level ground, subtend angles of 30° and 60° respectively at the centre of the line, joining their feet. Then, x : y is, [CBSE 2015], (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 3, , (d) 3 : 1, , 18. The angle of elevation of the top of a tower from a point on the ground, 30 m away from the foot of the tower is 30°. The height of the tower is, (a) 30 m, , (b) 10 3 m, , (c) 20 m, , (d) 10 2 m, , 19. The string of a kite is 100 m long and it makes an angle of 60° with the, horizontal. If there is no slack in the string, the height of the kite from, the ground is, (a) 50 3 m, , (b) 100 3 m, , (c) 50 2 m, , (d) 100 m, , 20. If the angles of elevation of the top of a tower from two points at, distances a and b from the base and in the same straight line with it are, complementary then the height of the tower is, a, (b) ab, (c) a + b, (d) a − b, b, 21. On the level ground, the angle of elevation of a tower is 30°. On moving, 20 m nearer, the angle of elevation is 60°. The height of the tower is, (a), , (a) 10 m, , (b) 10 3 m, , (c) 15 m, , (d) 20 m, , 22. In a rectangle, the angle between a diagonal and a side is 30° and the, length of this diagonal is 8 cm. The area of the rectangle is, 16, (a) 16 cm 2, (b), (c) 16 3 cm 2 (d) 8 3 cm 2, cm 2, 3
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Heights and Distances, , 649, , 23. From the top of a hill, the angles of depression of two consecutive km, stones due east are found to be 30° and 45°. The height of the hill is, 1, 1, (b) ( 3 + 1) km, (a) ( 3 − 1) km, 2, 2, (c) ( 3 − 1) km, , (d) ( 3 + 1) km, , 24. If the elevation of the sun changes from 30° to 60° then the difference, between the lengths of shadows of a pole 15 m high, is, (a) 7.5 m, (b) 15 m, (c) 10 3 m, (d) 5 3 m, 25. An observer 1.5 m tall is 28.5 m away from a tower and the angle of, elevation of the top of the tower from the eye of the observer is 45°. The, height of the tower is, (a) 27 m, , (b) 30 m, , (c) 28.5 m, , (d) none of these, , ANSWERS (MCQ), , 1. (c), , 2. (c), , 3. (b), , 4. (a), , 5. (d), , 6. (d), , 7. (c), , 8. (b), , 9. (b) 10. (b), , 11. (a) 12. (c) 13. (d) 14. (a) 15. (b) 16. (a) 17. (c) 18. (b) 19. (a) 20. (b), 21. (b) 22. (c) 23. (b) 24. (c) 25. (b), HINTS TO SOME SELECTED QUESTIONS, 1. Let AB be the pole and AC be its shadow, such that AB = AC ., Let ∠ACB = θ. Then,, tan θ =, ⇒, , AB, = 1 ⇒ tan θ = tan 45 °, AC, , θ = 45 ° ., , 2. Let AB be the pole and AC be its shadow., Let AC = x m. Then, AB =, , 3 x m., , Let ∠ACB = θ. Then,, tan θ =, , AB, =, AC, , 3x, =, x, , 3 ⇒ θ = 60 ° ., , 4. Let AB be the pole and AC be its shadow., AB = 12 m and AC = 4 3 m., Let ∠ACB = θ. Then, tanθ =, ⇒, ⇒, , 12, ×, 4 3, θ = 60 ° ., tan θ =, , 3, =, 3, , AB, 12, =, AC 4 3, , 3 = tan 60 °
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650, , Secondary School Mathematics for Class 10, , 5. Ratio of lengths of objects = ratio of lengths of their shadows., Let the length of shadow of the tree be x m. Then,, 5, 2, = ⇒ 5 x = 2 × 12 . 5 = 25 ⇒ x = 5, 12 . 5 x, 6. Let AB be the ladder and BC be the wall., Then, ∠CAB = 60 ° and CA = 2 m., Let AB = x metres. Then,, AC, 1, 2 1, = cos 60 ° = ⇒ =, AB, 2, x 2, ⇒, , x = 4., , 7. Let AB be the ladder and BC be the wall., Then, ∠ABC = 60 ° ⇒ ∠CAB = ( 90 ° − 60 ° ) = 30 ° ., Let BC = h m. Then,, BC, h, 1, = sin 30 ° ⇒, =, AB, 15 2, 15, ., ⇒ h=, 2, 8. Let AB be the tower and BC = 30 m be the ground such, that ∠BCA = 30 ° ., Let AB = h metres. Then,, AB, h, 1, 30, = tan 30 ° ⇒, =, ⇒ h=, BC, 30, 3, 3, ⇒, , ⎛ 30, h=⎜, ×, ⎝ 3, , 3⎞, ⎟ = 10 3., 3⎠, , 9. Let AB be the tower and C be the position of the car, on the ground such that ∠ACB = 30 ° ., AC, x, Let AC = x metres. Then,, = cot 30 ° ⇒, = 3, AB, 150, ⇒ x = 150 3., 10. Let AB be the string and B be the kite., Let AC be the horizontal and let BC ⊥ AC ., Let ∠CAB = θ., BC = 30 m and AB = 60 m. Then,, BC, 30 1, = sin θ ⇒ sin θ =, = ⇒ sin θ = sin 30 ° ⇒ θ = 30 ° ., AB, 60 2, 11. Let AB be the cliff and CD be the tower. Draw BE ⊥ CD., Let ∠ACB = ∠EBD = α and let DE = h metres., Also, AB = 20 m. Let AC = BE = x m. Then, x, x, = cot α and, = cot α, h, 20, x x, Thus, =, ⇒ h = 20 m., h 20
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Heights and Distances, , 651, , 12. Let AB be the position of the girl and let CD be the lamp-post. Let AE be the shadow, of AB. Then, AB = 1. 5 m, AC = 3 m and AE = 4.5 m., Let CD = x metres. Now, SAEB and SCED are similar., 1. 5 1, 1, CD AB, x, =, ⇒, =, = ⇒ x = ⎛⎜ 7 . 5 × ⎞⎟ = 2 . 5., ∴, ⎝, 7 .5 4 .5 3, 3⎠, EC AE, Hence, the height of the lamp-post is 2.5 m., 13. Let h m be the height of the tower and let a m and ( a + 2 x ) m be the lengths of the, shadows of the tower when sun’s elevation is 45° and 30° respectively. Then,, h, = tan 45 ° = 1 ⇒ a = h., a, h, 1, h, 1, = tan 30 ° =, ⇒, =, ⋅, a + 2x, h + 2x, 3, 3, 2x, 3 h = h + 2 x ⇒ 2 x = ( 3 − 1)h ⇒ h =, ∴, ( 3 − 1), ∴, , ⎧ 2x, ( 3 + 1) ⎫, h=⎨, ×, ⎬ = x ( 3 + 1)., ⎩ ( 3 − 1) ( 3 + 1) ⎭, , 14. Let AB be the rod of length x metres and let AC be its, shadow of length ( 3x ) m. Let ∠ACB = θ., Then,, , AB, x, 1, = tan θ ⇒ tan θ =, =, = tan 30 °, AC, 3x, 3, ⇒ θ = 30 ° ., , 15. Let the height of the pole be h metres., h, Then,, = tan 60 ° = 3, 2 3, ⇒ h = (2 3 ×, , 3 ) = 6., , 16. Let AB = 20 m be the tower and BC = 20 3 m be the, length of its shadow. Let ∠ACB = θ., AB, 20, 1, Then, tan θ =, =, =, = tan 30 °⋅, AC 20 3, 3, ∴, , θ = 30 ° ., , 17. Let AB and CD be the given pillars and O be the midpoint of AC., Then, AB = x , CD = y , ∠AOB = 30 ° and ∠COD = 60 ° ., From right SOAB, we have, OA, OA, = cot 30 ° ⇒, = 3, AB, x, ⇒ OA = x 3., From right SOCD, we have, y, 1, OC, OC, = cot 60 ° ⇒, =, ⇒ OC =, ⋅, OD, y, 3, 3, , ... (i), , ... (ii)
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652, , Secondary School Mathematics for Class 10, , But, OA = OC ., y, x 1, ⇒ 3 x = y ⇒ = ⇒ x : y = 1 : 3., ∴ x 3=, y 3, 3, 18., , 30 m, h, 1, = tan 30 ° =, ⇒ h=, 30, 3, 3, ⇒, , 20., , h=, , 30 m, 3, , ×, , 3, = 10 3 m., 3, , h, = tan θ ⇒ h = a tan θ., a, h, = tan( 90 ° − θ) = cot θ, b, ⇒ h = b cot θ, , ... (i), , ... (ii), , From (i) and (ii), we get, h 2 = ab and hence h = ab ., x, 1, h, ... (i), 21. = cot 60 ° =, ⇒ x=, h, 3, 3, x + 20, = cot 30 ° = 3 ⇒ x = ( h 3 − 20 ) ... (ii), h, h, ∴, = ( h 3 − 20 ) ⇒ h = 3 h − 20 3, 3, ⇒ 2 h = 20 3 ⇒ h = 10 3 ., 22. Let ABCD be the rectangle., BC, 1, x, 1, = sin 30 ° = ⇒, = ⇒ x = 4 cm., AC, 2, 8 cm 2, ∴, , AB 2 = ( 8 ) 2 cm 2 − ( 4 ) 2 = ( 64 − 16 ) cm 2 = 48 cm 2, , ⇒, , AB = 48 cm 2 = 4 3 cm., , ∴, , ar( ABCD ) = ( 4 3 × 4 ) cm 2 = 16 3 cm 2 ., , 23. Let AB be the hill., From right SBAD, we have, AD, x, = cot 45 ° ⇒ = 1 ⇒ x = h., AB, h, From right SBAC , we have, AC, x+ 1, = cot 30 ° ⇒, =, AD, h, From (i) and (ii),we have, , 3 ⇒ x = ( h 3 − 1), , h = ( h 3 − 1) ⇒ h( 3 − 1) = 1, 1, ( 3 − 1), , ⇒, , h=, , ⇒, , ⎧, ( 3 + 1) ⎫ 1, 1, h=⎨, ×, ⎬ = ( 3 + 1)., ⎩ ( 3 − 1) ( 3 + 1) ⎭ 2, , ... (i), , ... (ii)
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Heights and Distances, , 653, , 24. Let AB be the pole and AC and AD be its shadows when ∠ACB = 30 ° and ∠ADB = 60 ° ., From right SCAB, we have, AC, AC, = cot 30 ° ⇒, =, AB, 15 m, ⇒, , 3, , AC = 15 3 m., , ... (i), , From right SDAB, we have, 15 m, AD, AD, 1, = cot 60 °⇒, =, ⇒ AD =, AB, 15 m, 3, 3, ⇒, , AD =, , 15 m, 3, , 3, =5 3 m, 3, , ×, , ... (ii), , Required difference = ( 15 3 − 5 3 ) m = 10 3 m., 25. Let AB be the observer and CD = h metres be the tower., BE = AC = 28 . 5 m., From right SBED, we have, DE, DE, = tan 45 ° ⇒, =1, BE, 28 . 5 m, ⇒ DE = 28 . 5 m, ∴, , h − 1. 5 = 28 . 5 ⇒ h = 30., , _
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PERIMETER, , The perimeter of a plane figure is the length of its boundary., , In case of a triangle or a polygon, the perimeter is the sum of the lengths, of its sides., The unit of perimeter is same as the unit of length., The area of a plane figure is the measure of the surface enclosed by its, boundary., , AREA, , The area of a triangle or a polygon is the measure of the surface, enclosed by its sides., Area is measured in square units such as square centimetres and, square metres, written as cm 2 and m 2 respectively., PERIMETER AND AREA OF TRIANGLES [FORMULAE], , 1., , 1, × base × corresponding height., 2, ⎛1, ⎞, ar(SABC) = ⎜ × BC × AD⎟ sq units., ⎝2, ⎠, , (i) Area of a triangle =, , (ii) Heron’s formula:, Let a, b, c be the sides of a SABC. Then,, 1, s = ( a + b + c) is called its semiperimeter., 2, ar(SABC) = s(s − a)(s − b)(s − c)., 2. In a right SABC, let ∠B = 90°. Then,, ⎛1, ⎞, ar(SABC) = ⎜ × BC × AB⎟ sq units., ⎝2, ⎠, 3. In an equilateral triangle of side a, we have, ⎛ 3 ⎞, (i) Height = ⎜, a⎟ units., ⎝ 2 ⎠, ⎛ 3 2⎞, (ii) Area = ⎜, a ⎟ sq units., ⎝ 4, ⎠, (iii) Perimeter = 3a units., 654
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Perimeter and Area of Plane Figures, , 655, , 4. For an isosceles C ABC in which AB = AC = a and, BC = b , we have, (i) Height =, , 4a 2 − b 2, units., 2, , ⎛1, ⎞, (ii) Area = ⎜ b 4a 2 − b 2 ⎟ sq units., ⎝4, ⎠, (iii) Perimeter = ( 2a + b) units., , SOLVED EXAMPLES, EXAMPLE 1, , Find the area of a triangle having base 25 cm and height 10.8 cm., , SOLUTION, , 1, Area of the given triangle = ⎛⎜ × base × height ⎞⎟, ⎝2, ⎠, 1, = ⎛⎜ × 25 × 10.8 ⎞⎟ cm 2 = 135 cm 2 ., ⎝2, ⎠, , EXAMPLE 2, , The base of a triangular field is three times its altitude. If the cost of, sowing the field at ` 960 per hectare is ` 12960, find its base and, height., , SOLUTION, , Total cost of sowing the field = ` 12960., Rate per hectare = ` 960., total cost ⎛ 12960 ⎞, Area =, =⎜, ⎟ hectares = 13.5 hectares, ⎝ 960 ⎠, rate, = (13 .5 × 10000) m 2 = 135000 m 2 ., Let the altitude of the field be x metres., Then, its base = 3x metres., ⎛ 3x2 ⎞ 2, 1, 1, Area = ⎛⎜ × base × height ⎞⎟ = ⎛⎜ × 3 x × x ⎞⎟ m 2 = ⎜, ⎟ m ., ⎝2, ⎠ ⎝2, ⎠, ⎝ 2 ⎠, ∴, , 3x2, 2, = 135000 ⇒ x 2 = ⎛⎜ 135000 × ⎞⎟ = 90000, ⎝, 2, 3⎠, ⇒ x = 90000 = 300., , ∴, EXAMPLE 3, , base = ( 3 × 300) m = 900 m and altitude = 300 m., , The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 , and its, perimeter is 144 cm. Find (i) the area of the triangle, and (ii) the, height corresponding to the longest side.
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656, SOLUTION, , Secondary School Mathematics for Class 10, , On dividing 144 cm in the ratio 3 : 4 : 5, we get, 3, 4, a = ⎛⎜ 144 × ⎞⎟ cm = 36 cm, b = ⎛⎜ 144 × ⎞⎟ cm = 48 cm, ⎝, ⎝, 12 ⎠, 12 ⎠, 5, and c = ⎛⎜ 144 × ⎞⎟ cm = 60 cm., ⎝, 12 ⎠, 1, ∴ s = ( 36 + 48 + 60) cm = 72 cm., 2, ( s − a) = (72 − 36) cm = 36 cm,, ( s − b) = (72 − 48) cm = 24 cm, and ( s − c) = (72 − 60) cm = 12 cm., (i) Area of the triangle = s( s − a)( s − b)( s − c), = 72 × 36 × 24 × 12 cm 2, = 72 × 12 cm 2 = 864 cm 2 ., (ii) Let base = 60 cm and the corresponding height = h cm., 1, Then, area of the triangle = ⎛⎜ × 60 × h ⎞⎟ cm 2 = ( 30 h) cm 2 ., ⎝2, ⎠, 864, ∴, 30 h = 864 ⇒ h =, = 28.8., 30, Longest side = 60 cm, corresponding height = 28.8 cm., , EXAMPLE 4, , Each side of an equilateral triangle measures 10 cm. Calculate (i) the, area of the triangle, and (ii) the height of the triangle. [Given,, 3 = 1.732.], , SOLUTION, , Here, a = 10 cm., ⎛ 3, ⎞, (i) Area of the triangle = ⎜, × a2 ⎟ sq units, ⎝ 4, ⎠, ⎛ 3, ⎞, =⎜, × 10 × 10 ⎟ cm 2 = ( 25 × 3 ) cm 2, ⎝ 4, ⎠, = ( 25 × 1.732) cm 2 = 43.3 cm 2 ., ⎛ 3, ⎞, (ii) Height of the triangle = ⎜, × a⎟ units, ⎝ 2, ⎠, ⎛ 3, ⎞, =⎜, × 10 ⎟ cm = (5 × 3 ) cm, ⎝ 2, ⎠, = (5 × 1.732) cm = 8.66 cm., , EXAMPLE 5, , The height of an equilateral triangle is 15 cm. Find its area. [Given,, 3 = 1 .73.], , SOLUTION, , Let each side of the triangle be a cm. Then,, ⎛ 3, ⎞, its height = ⎜, × a⎟ cm., ⎝ 2, ⎠
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Perimeter and Area of Plane Figures, , ∴, , ⎛ 30, 3, × a = 15 ⇒ a = ⎜, ×, 2, ⎝ 3, , 657, , 3⎞, ⎟ = 10 3., 3⎠, , Thus, each side of the triangle = a cm = 10 3 cm., ⎛ 3, ⎞ ⎛ 3, ⎞, Area of the triangle = ⎜, × a2 ⎟ = ⎜, × 10 3 × 10 3 ⎟ cm 2, ⎝ 4, ⎠ ⎝ 4, ⎠, 75, ×, 173, ⎞⎟ cm 2, = (75 3 ) cm 2 = ⎛⎜, ⎝ 100 ⎠, 519, =, cm 2 = 129 .75 cm 2 ., 4, EXAMPLE 6, , Find the area of an isosceles triangle, each of whose equal sides is, 13 cm and whose base is 24 cm., , SOLUTION, , Here, each equal side, a = 13 cm and base, b = 24 cm., 1, ∴ area of the triangle = ⎛⎜ b ⋅ 4 a2 − b 2 ⎞⎟ cm 2, ⎝4, ⎠, 1, = ⎡ × 24 × 4 × 169 − 24 × 24 ⎤ cm 2, ⎢⎣ 4, ⎥⎦, 2, = 60 cm ., , EXAMPLE 7, , The base of an isosceles triangle measures 24 cm and its area is, 192 cm 2 . Find its perimeter., , SOLUTION, , Here, base, b = 24 cm, and let each equal side be a cm. Then,, 1, 1, area = b 4 a2 − b 2 sq units = × 24 × 4 a2 − 576 cm 2, 4, 4, = 12 × a2 − 144 cm 2 ., But, area = 192 cm 2 [given]., ∴, , 12 × a2 − 144 = 192 ⇒, , a2 − 144 = 16, , ⇒ a2 − 144 = 256 ⇒ a2 = 400 ⇒ a = 20., ∴ perimeter of the triangle = ( 2a + b) cm, = ( 2 × 20 + 24) cm = 64 cm., EXAMPLE 8, , The difference between the sides at right angles in a right-angled, triangle is 14 cm. The area of the triangle is 120 cm 2 . Calculate the, perimeter of the triangle., , SOLUTION, , Let the sides containing the right angle be x cm and, ( x − 14) cm., 1, Then, its area = ⎡ × x × ( x − 14) ⎤ cm 2 ., ⎢⎣ 2, ⎥⎦, But, area = 120 cm 2 [given]., 1, x( x − 14) = 120 ⇒ x 2 − 14 x − 240 = 0, ∴, 2
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658, , Secondary School Mathematics for Class 10, , ⇒ x 2 − 24 x + 10 x − 240 = 0 ⇒ x( x − 24) + 10( x − 24) = 0, ⇒ ( x − 24)( x + 10) = 0 ⇒ x = 24 [neglecting x = − 10]., ∴, , one side = 24 cm, and other side = (24 – 14) cm = 10 cm., , Hypotenuse = ( 24) 2 + (10) 2 cm = 576 + 100 cm, = 676 cm = 26 cm., ∴, , perimeter of the triangle = ( 24 + 10 + 26) cm = 60 cm., , EXERCISE 15A, 1. Find the area of the triangle whose base measures 24 cm and the, corresponding height measures 14.5 cm., 2. Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm., Also, find the height corresponding to the longest side., 3. Find the area of the triangle whose sides are 18 cm, 24 cm and 30 cm., Also, find the height corresponding to the smallest side., 4. The sides of a triangle are in the ratio 5 : 12 : 13, and its perimeter is, 150 m. Find the area of the triangle., 5. The perimeter of a triangular field is 540 m, and its sides are in the ratio, 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the, field at ` 40 per 100 m 2 ., 6. The perimeter of a right triangle is 40 cm and its hypotenuse measures, 17 cm. Find the area of the triangle., 7. The difference between the sides at right angles in a right-angled, triangle is 7 cm. The area of the triangle is 60 cm 2 . Find its perimeter., 8. The lengths of the two sides of a right triangle containing the right, angle differ by 2 cm. If the area of the triangle is 24 cm 2 , find the, perimeter of the triangle., 9. Each side of an equilateral triangle is 10 cm. Find (i) the area of the, triangle and (ii) the height of the triangle., 10. The height of an equilateral triangle is 6 cm. Find its area. [Take, 3 = 1 .73.], 11. If the area of an equilateral triangle is 36 3 cm 2 , find its perimeter., 12. If the area of an equilateral triangle is 81 3 cm 2 , find its height., 13. The base of a right-angled triangle measures 48 cm and its hypotenuse, measures 50 cm. Find the area of the triangle., 14. The hypotenuse of a right-angled triangle is 65 cm and its base is 60 cm., Find the length of perpendicular and the area of the triangle.
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Perimeter and Area of Plane Figures, , 659, , 15. Find the area of a right-angled triangle, the radius of whose, circumcircle measures 8 cm and the altitude drawn to the hypotenuse, measures 6 cm., 16. Find the length of the hypotenuse of an isosceles right-angled triangle, whose area is 200 cm 2 . Also, find its perimeter. [Given, 2 = 1.41.], 17. The base of an isosceles triangle measures 80 cm and its area is, 360 cm 2 . Find the perimeter of the triangle., 18. Each of the equal sides of an isosceles triangle measures 2 cm more than, its height, and the base of the triangle measures 12 cm. Find the area of, the triangle., 19. Find the area and perimeter of an isosceles right triangle, each of whose, equal sides measures 10 cm. [Take 2 = 1.41.], 20. In the given figure, CABC is an equilateral triangle, the length of whose side is equal to 10 cm, and, CDBC is right-angled at D and BD = 8 cm. Find the, area of the shaded region. [Take 3 = 1.732.], , ANSWERS (EXERCISE 15A), 2, , 3. 216 cm 2 , 24 cm, 4. 750 m 2, 1. 174 cm, 2. 336 cm 2 , 16 cm, 5. 9000 m 2 , ` 3600, 7. 40 cm, 8. 24 cm, 6. 60 cm 2, 2, 12. 9 3 cm, 9. (i) 43.3 cm (ii) 8.66 cm 10. 20.76 cm 2 11. 36 cm, 2, 2, 2, 14. 25 cm, 750 cm, 15. 48 cm, 16. 28.2 cm, 68.2 cm, 13. 336 cm, 19. 50 cm 2 , 34.1 cm 20. 19.3 cm 2, 17. 162 cm, 18. 48 cm 2, HINTS TO SOME SELECTED QUESTIONS, 6. Sum of two sides = ( 40 − 17 ) cm = 23 cm., Let these sides be x cm and ( 23 − x ) cm., ∴, , x 2 + ( 23 − x ) 2 = ( 17 ) 2 ⇒ x 2 + x 2 − 46 x + 529 = 289, ⇒ 2 x 2 − 46 x + 240 = 0 ⇒ x 2 − 23 x + 120 = 0, ⇒ ( x − 15 )( x − 8 ) = 0 ⇒ x = 15 or x = 8., , ∴, , base = 15 cm and height = 8 cm. Find the area of the triangle., , 7. Let these sides be x cm and ( x − 7 ) cm., 1, × x × ( x − 7 ) = 60 ⇒ x 2 − 7 x − 120 = 0 ⇒ ( x − 15 )( x + 8 ) = 0, ∴, 2, ⇒ x = 15 [Q x ≠ −8]., ∴, , these sides are 15 cm and 8 cm., , Hypotenuse = ( 15 ) 2 + 8 2 cm = 289 cm = 17 cm.
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660, , 9., , Secondary School Mathematics for Class 10, ⎛ 3, ⎞ ⎛ 1. 732, ⎞, (i) Area of the triangle = ⎜, × a2 ⎟ = ⎜, × 10 × 10 ⎟ cm 2 = 43. 3 cm 2 ., ⎠, ⎝ 4, ⎠ ⎝ 4, (ii), , 43. 3 cm 2, 1, 1, × a × height = 43. 3 cm 2 ⇒ × 10 cm × h = 43. 3 cm 2 ⇒ h =, = 8.66 cm., 5 cm, 2, 2, , 10. Height =, ∴, , area =, , 3a, ⇒, 2, , 3a, 12, 12, = 6 cm ⇒ a =, ×, cm =, 2, 3, 3, , 3, cm = 4 3 cm., 3, , ⎛ 3, ⎞, 3, 3, × a2 =, × ( 4 3 ) 2 cm 2 = ⎜, × 48 ⎟ cm 2 = ( 12 × 1. 73 ) cm 2 ., 4, 4, ⎝ 4, ⎠, , 11., , 3 2, a = 36 3 cm 2 ⇒ a2 = 144 cm 2 ⇒ a = 144 cm = 12 cm., 4, , 12., , 3 2, a = 81 3 cm 2 ⇒ a2 = ( 81 × 4 ) cm 2 ⇒ a = 196 cm = 18 cm., 4, Height =, , 3a ⎛, =⎜, 2, ⎝, , 3 × 18 ⎞, ⎟ cm = 9 3 cm., 2, ⎠, , 13. Height = (50 ) 2 − ( 48 ) 2 cm = 98 × 2 cm = (7 × 2 ) cm = 14 cm., 15. The circumcentre of a right triangle is the midpoint of its, hypotenuse., ∴ hypotenuse = 2 × (radius of the circumcircle), = ( 2 × 8 ) cm = 16 cm., ∴ base = 16 cm, height = 6 cm., 16. Let base = height = a cm. Then,, 1, × a × a = 200 cm 2 ⇒ a2 = 400 cm 2 ⇒ a = 400 cm 2 = 20 cm., 2, ∴, , hypotenuse = a2 + a2 = 2 a2 = 2 × 400 cm, = 20 2 cm = ( 20 × 1. 41) cm., , 1, 17. × 80 × 4 a2 − 6400 = 360 ⇒ 80 × a2 − 1600 = 360 ⇒ a2 − 1600 = 9, 2, ⇒ a2 − 1600 = 81 ⇒ a2 = 1681 ⇒ a = 1681 = 41 cm., Perimeter = ( 41 + 41 + 80 ) cm = 162 cm., 18. Let the height be h cm. Then, a = ( h + 2 ) cm and b = 12 cm., ∴, , 1, 1, × 12 × h = × 12 × 4( h + 2 ) 2 − 144 ., 2, 4, , 20. CD = ( 10 ) 2 − 8 2 cm =, , 36 cm = 6 cm., , Required area = ar( SABC ) − ar( SDBC ), ⎧ 3, ⎫, 1, =⎨, × ( 10 ) 2 − × 8 × 6 ⎬ cm 2 ., 2, 4, ⎩, ⎭, , ................................................................
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Perimeter and Area of Plane Figures, , 661, , PERIMETER AND AREA OF QUADRILATERALS [FORMULAE], , 1. Area and perimeter of a rectangle:, (i) Area = (length × breadth) sq units., area, area, (ii) Length =, and Breadth =, ⋅, breadth, length, (iii) Perimeter = 2( l + b) units., (iv) Diagonal = l 2 + b 2 units., 2. Area of 4 walls of a room = [2( l + b) × h)] sq units., 3. Area and perimeter of a square:, (i) Area = a 2 sq units., ⎧1, ⎫, (ii) Area = ⎨ × (diagonal) 2 ⎬ sq units., 2, ⎭, ⎩, (iii) Perimeter = 4a units., (iv) Diagonal = 2 a units., 4. (i) Area of a parallelogram = {(base) × (height)} sq units., ⎛1, ⎞, (ii) Area of a rhombus = ⎜ × product of diagonals⎟ sq units., ⎝2, ⎠, 5. Area of a trapezium, 1, = × (sum of parallel sides), 2, × (distance between them)., 6. Area of a quadrilateral:, (i) Let ABCD be a quadrilateral with diagonal AC., Let BL ⊥ AC and DM ⊥ AC. Let BL = h 1 and DM = h2 ., ⎧1, ⎫, ar (quad. ABCD) = ⎨ ( h1 + h 2 ) × AC ⎬ sq units., ⎭, ⎩2, (ii) When diagonals of a quadrilateral are perpendicular, to each other:, ⎛1, ⎞, ar(quad. ABCD) = ⎜ × product of diagonals⎟ sq units., ⎝2, ⎠, , SOLVED, EXAMPLE 1, , EXAMPLES, , The length of a rectangular field exceeds its breadth by 8 m and the, area of the field is 240 m 2 . Find the dimensions of the field., [CBSE 2014]
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662, , Secondary School Mathematics for Class 10, , SOLUTION, , Let the breadth of the field be x metres., Then, its length = ( x + 8) m., ∴ area = ( x + 8) ⋅ x m 2 ., ∴ ( x + 8) x = 240 ⇒ x 2 + 8 x − 240 = 0, ⇒ x 2 + 20 x − 12 x − 240 = 0 ⇒ x( x + 20) − 12( x + 20) = 0, [Q x ≠ −20]., ⇒ ( x + 20)( x − 12) = 0 ⇒ x = 12, ∴ breadth = 12 m and length = 20 m., , EXAMPLE 2, , Sum of the areas of two squares is 400 cm 2 . If the difference of their, perimeters is 16 cm, find the sides of the two squares. [CBSE 2013], , SOLUTION, , Let the sides of the given squares be a cm and b cm, respectively. Then,, a2 + b 2 = 400, and 4 a − 4b = 16 ⇒ a − b = 4., Squaring (ii) on both sides, we get, ( a − b) 2 = 16 ⇒ ( a2 + b 2 ) − 2 ab = 16, ⇒ 400 − 2 ab = 16 [using (i)], ⇒ 2 ab = 384 ⇒ ab = 192., , ... (i), ... (ii), , ... (iii), , Now, ( a + b) − ( a − b) = 4 ab, 2, , 2, , ⇒ ( a + b) 2 − 16 = 4 × 192., ⇒ ( a + b) 2 = 768 + 16 = 784 ⇒ a + b = 784 = 28., On solving a + b = 28 and a − b = 4, we get a = 16 and b = 12., Hence, the sides of the given squares are 16 cm and 12 cm., EXAMPLE 3, , SOLUTION, , The length of the diagonal of a square is 24 cm. Find (i) the area of the, square and (ii) its perimeter. [Given, 2 = 1.41.], 1, (i) Area of the square = ⎧⎨ × ( diagonal) 2 ⎫⎬ sq units., ⎭, ⎩2, 1, ⎛, ⎞, = ⎜ × 24 × 24 ⎟ cm 2 = 288 cm 2 ., ⎝2, ⎠, (ii) Side of the square = 288 cm = 144 × 2 cm, = 12 2 cm = (12 × 1.41) cm = 16.92 cm., ∴ perimeter of the square = ( 4 × 16.92) cm = 67 .68 cm., , EXAMPLE 4, , The longer side of a rectangular hall is 24 m, and the length of its, diagonal is 26 m. Find the area of the hall., , SOLUTION, , Let ABCD be the hall in which AB = 24 m, and AC = 26 m., By Pythagoras’ theorem, we have, BC = AC 2 − AB2 units = ( 26) 2 − ( 24) 2 m
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Perimeter and Area of Plane Figures, , 663, , = ( 26 − 24)( 26 + 24) m = 50 × 2 m, = 100 m = 10 m., Thus, length = 24 m, breadth = 10 m., ∴, , area of the hall = ( 24 × 10) m 2 = 240 m 2 ., , EXAMPLE 5, , The length and the breadth of a rectangular park are in the ratio 8 : 5., A path, 1.5 m wide, running all around the outside of the park has an, area of 594 m 2 . Find the dimensions of the park., , SOLUTION, , Let the length and the breadth of the park, be 8x metres and 5x metres respectively., Then, area of the park = ( 8 x × 5 x) m 2, = ( 40 x 2 ) m 2 ., Length of the park including the path = ( 8 x + 3) m., Breadth of the park including the path = (5 x + 3) m., Area of the park including the path = ( 8 x + 3)(5 x + 3) m 2 ., Area of the path = [( 8 x + 3)(5 x + 3) − 40 x 2 ] m 2, = ( 39 x + 9) m 2 ., ∴, , 39 x + 9 = 594 ⇒ 39 x = 585, 585, ⇒ x=, = 15., 39, , ∴, , length = ( 8 × 15) m = 120 m ,, breadth = (5 × 15) m = 75 m., , EXAMPLE 6, , A rectangular lawn, 75 m by 60 m, has two, roads, each 4 m wide, running through the, middle of the lawn, one parallel to length and, the other parallel to breadth, as shown in the, figure. Find the cost of gravelling the roads, at ` 50 per m 2 ., , SOLUTION, , Area of the road ABCD = (75 × 4) m 2 = 300 m 2 ., Area of the road EFGH = ( 60 × 4) m 2 = 240 m 2 ., Clearly, area PQRS is common to both the roads., Area PQRS = ( 4 × 4) m 2 = 16 m 2 ., Area of the roads to be gravelled, = ( 300 + 240 − 16) m 2 = 524 m 2 ., Cost of gravelling the roads, = ` (524 × 50) = ` 26200.
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664, , Secondary School Mathematics for Class 10, , EXAMPLE 7, , Find the area of the quadrilateral whose sides measure 9 cm, 40 cm,, 28 cm and 15 cm, and in which the angle between the first two sides, is a right angle., , SOLUTION, , Let ABCD be the given quadrilateral in which AB = 9 cm,, BC = 40 cm, CD = 28 cm, DA = 15 cm and ∠ABC = 90 ° ., By Pythagoras’ theorem,, AC = AB2 + BC 2 units = ( 9) 2 + ( 40) 2 cm, = 1681 cm = 41 cm., 1, Area of CABC = ⎛⎜ × AB × BC ⎞⎟, ⎝2, ⎠, 1, = ⎛⎜ × 9 × 40 ⎞⎟ cm 2 = 180 cm 2 ., ⎝2, ⎠, In CACD,, ∴, ∴, , let a = AC = 41 cm, b = CD = 28 cm and c = DA = 15 cm., 1, s = ( 41 + 28 + 15) cm = 42 cm., 2, ( s − a) = 1 cm, ( s − b) = 14 cm and ( s − c) = 27 cm., , Area of CACD = s( s − a)( s − b)( s − c) sq units, = 42 × 1 × 14 × 27 cm 2, = (14 × 3 × 3) cm 2 = 126 cm 2 ., Area of the quadrilateral ABCD, = area of C ABC + area of C ACD, = (180 + 126) cm 2 = 306 cm 2 ., EXAMPLE 8, , In a four-sided field, the length of the longer diagonal is 120 m. The, lengths of perpendiculars from the opposite vertices on this diagonal, are 20 m and 15 m. Find the area of the field., , SOLUTION, , Let ABCD be the field and AC be its longer, diagonal., Let BL ⊥ AC and DM ⊥ AC . Then,, AC = 120 m , BL = 20 m, and DM = 15 m., 1, Area of the field = ⎧⎨ × ( BL + DM) × AC ⎫⎬ sq units, ⎭, ⎩2, 1, = ⎧⎨ × ( 20 + 15) × 120 ⎫⎬ m 2 = 2100 m 2 ., ⎭, ⎩2
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Perimeter and Area of Plane Figures, , 665, , EXAMPLE 9, , Find the area of a parallelogram one of whose sides measures 40 cm, and the corresponding height measures 19.5 cm., , SOLUTION, , Area of the given parallelogram, = (base × height) sq units, 195 ⎞, = ( 40 × 19 .5) cm 2 = ⎛⎜ 40 ×, ⎟ cm 2, ⎝, 10 ⎠, = 780 cm 2 ., , EXAMPLE 10, , The adjacent sides of a parallelogram are 36 cm and 27 cm in, length. If the distance between the shorter sides is 12 cm, find the, distance between the longer sides., , SOLUTION, , Longer side = 36 cm, shorter side = 27 cm., Distance between the shorter sides = 12 cm., Let the distance between the longer sides, be x cm., Area of the||gm, = (longer side × distance between the longer sides), = (shorter side × distance between the shorter sides), 27 × 12, = 9., ∴ 36 × x = 27 × 12 ⇒ x =, 36, ∴ distance between the longer sides = 9 cm., , EXAMPLE 11, , SOLUTION, , The diagonals of a rhombus are 48 cm and 20 cm long. Find (i) the, area of the rhombus, and (ii) the perimeter of the rhombus., (i) Area of the rhombus, 1, = × (product of diagonals) sq units, 2, 1, ⎛, = ⎜ × 48 × 20 ⎞⎟ cm 2, ⎝2, ⎠, 2, = 480 cm ., (ii) We know that the diagonals of a rhombus bisect each, other at right angles., ∴ OA = OC = 24 cm,OB = OD = 10 cm and ∠AOB = 90 ° ., ∴ by Pythagoras’ theorem, we have, AB = OA 2 + OB2 units = ( 24) 2 + (10) 2 cm, = 676 cm = 26 cm., ∴ perimeter of the rhombus = ( 4 × 26) cm = 104 cm., , EXAMPLE 12, , Find the area of a trapezium whose parallel sides are 35 cm and, 23 cm long and the distance between them is 15 cm.
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666, , Secondary School Mathematics for Class 10, , SOLUTION, , Area of the trapezium, 1, = × (sum of parallel sides) × (distance between them), 2, 1, = ⎧⎨ × ( 35 + 23) × 15 ⎫⎬ cm 2 = 435 cm 2 ., ⎭, ⎩2, , EXAMPLE 13, , Find the area of a trapezium ABCD in which, AB || DC , AB = 77 cm, BC = 25 cm,, [HOTS], CD = 60 cm and DA = 26 cm., , SOLUTION, , Draw DE || BC and DL ⊥ AB., Then, DE = BC = 25 cm., AE = ( AB − EB) = ( AB − DC ) = (77 − 60) cm = 17 cm., In CDAE, we have, a = AE = 17 cm, b = DE = 25 cm and c = DA = 26 cm., 1, ∴ s = (17 + 25 + 26) cm = 34 cm ,, 2, ( s − a) = 17 cm , ( s − b) = 9 cm and ( s − c) = 8 cm., ∴, , area of CDAE = s( s − a)( s − b)( s − c), =, , 34 × 17 × 9 × 8 cm 2, , = (17 × 3 × 4) cm 2 = 204 cm 2 ., 1, Also, area of CDAE = × AE × DL sq units, 2, 1, = × 17 cm × DL., 2, 1, 204 × 2 ⎞, ∴, × 17 cm × DL = 204 cm 2 ⇒ DL = ⎛⎜, ⎟ cm = 24 cm., ⎝ 17 ⎠, 2, 1, ar(trap. ABCD) = ⎧⎨ ( AB + DC ) × DL ⎫⎬ sq units, ⎭, ⎩2, 1, = ⎧⎨ (77 + 60) × 24 ⎫⎬ cm 2 = 1644 cm 2 ., ⎭, ⎩2, , EXERCISE 15B, 1. The perimeter of a rectangular plot of land is 80 m and its breadth is, 16 m. Find the length and area of the plot., 2. The length of a rectangular park is twice its breadth and its perimeter is, 840 m. Find the area of the park.
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Perimeter and Area of Plane Figures, , 667, , 3. One side of a rectangle is 12 cm long and its diagonal measures 37 cm., Find the other side and the area of the rectangle., 4. The area of a rectangular plot is 462 m 2 and its length is 28 m. Find its, perimeter., 5. A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The, area of the lawn is 3375 m 2 . Find the cost of fencing the lawn at ` 65 per, metre., 6. A room is 16 m long and 13.5 m broad. Find the cost of covering its floor, with 75-m-wide carpet at ` 60 per metre., 7. The floor of a rectangular hall is 24 m long and 18 m wide. How many, carpets, each of length 2.5 m and breadth 80 cm, will be required to, cover the floor of the hall?, 8. A 36-m-long, 15-m-broad verandah is to be paved with stones, each, measuring 6 dm by 5 dm. How many stones will be required?, 9. The area of a rectangle is 192 cm 2 and its perimeter is 56 cm. Find the, dimensions of the rectangle., 10. A rectangular park 35 m long and 18 m wide is to be covered with grass,, leaving 2.5 m uncovered all around it. Find the area to be laid with grass., 11. A rectangular plot measures 125 m by 78 m. It has a gravel path 3 m, wide all around on the outside. Find the area of the path and the cost of, gravelling it at ` 75 per m 2 ., 12., , (i) A footpath of uniform width runs all around the inside of a, rectangular field 54 m long and 35 m wide. If the area of the path is, 420 m 2 , find the width of the path., (ii) A carpet is laid on the floor of a room 8 m by 5 m. There is a border, of constant width all around the carpet. If the area of the border is, 12 m 2 , find its width., , 13. The length and the breadth of a rectangular garden are in the ratio 9 : 5., A path 3.5 m wide, running all around inside it has an area of 1911 m 2 ., Find the dimensions of the garden., 14. A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an, uncovered margin of 25 cm all around the room. If the breadth of the, carpet is 80 cm, find its cost at ` 80 per metre., 15. In a rectangular park of dimensions 50 m × 40 m, a rectangular pond, is constructed so that the area of grass strip of uniform width, surrounding the pond would be 1184 m 2 . Find the length and breadth, [CBSE 2017], of the pond., 16. A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running, through its middle, one parallel to its length and the other parallel to its, breadth. Find the cost of gravelling the roads at ` 40 per m 2 .
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668, , Secondary School Mathematics for Class 10, , 17. The dimensions of a room are 14 m × 10 m × 6.5 m.There are two doors, and 4 windows in the room. Each door measures 2.5 m × 1.2 m and, each window measures 1.5 m × 1 m. Find the cost of painting the four, walls of the room at ` 35 per m 2 ., 18. The cost of painting the four walls of a room 12 m long at ` 30 per m 2 is, ` 7560 and the cost of covering the floor with mat at ` 25 per m 2 is, ` 2700. Find the dimensions of the room., 19. Find the area and perimeter of a square plot of land whose diagonal is, 24 m long. [Take 2 = 1.41.], 20. Find the length of the diagonal of a square whose area is 128 cm 2 . Also,, find its perimeter., 21. The area of a square field is 8 hectares. How long would a man take to, cross it diagonally by walking at the rate of 4 km per hour?, 22. The cost of harvesting a square field at ` 900 per hectare is ` 8100. Find, the cost of putting a fence around it at ` 18 per metre., 23. The cost of fencing a square lawn at ` 14 per metre is ` 28000. Find the, cost of mowing the lawn at ` 54 per 100 m 2 ., 24. In the given figure, ABCD is a quadrilateral, in which diagonal BD = 24 cm, AL ⊥ BD and, CM ⊥ BD such that AL = 9 cm and, CM = 12 cm. Calculate the area of the, quadrilateral., 25. Find the area of the quadrilateral ABCD in, which AD = 24 cm, ∠BAD = 90 ° and SBCD, is an equilateral triangle having each side, equal to 26 cm. Also, find the perimeter of the, quadrilateral. [Given, 3 = 1 .73.], 26. Find the perimeter and area of the, quadrilateral ABCD in which AB = 17 cm,, AD = 9 cm, CD = 12 cm, ∠ACB = 90 ° and, AC = 15 cm., 27. Find the area of the quadrilateral ABCD in which AB = 42 cm,, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.
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Perimeter and Area of Plane Figures, , 669, , 28. Find the area of a parallelogram with base equal to 25 cm and the, corresponding height measuring 16.8 cm., 29. The adjacent sides of a parallelogram are 32 cm and 24 cm. If the, distance between the longer sides is 17.4 cm, find the distance between, the shorter sides., 30. The area of a parallelogram is 392 m 2 . If its altitude is twice the, corresponding base, determine the base and the altitude., 31. The adjacent sides of a parallelogram, ABCD measure 34 cm and 20 cm, and, the diagonal AC measures 42 cm. Find, the area of the parallelogram., 32. Find the area of the rhombus, the lengths of whose diagonals are 30 cm, and 16 cm. Also, find the perimeter of the rhombus., 33. The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long,, find (i) the length of the other diagonal, and (ii) the area of the rhombus., 34. The area of a rhombus is 480 cm 2 , and one of its diagonals measures, 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of, its sides, and (iii) its perimeter., 35. The parallel sides of a trapezium are 12 cm and 9 cm and the distance, between them is 8 cm. Find the area of the trapezium., 36. The shape of the cross section of a canal is a trapezium. If the canal is, 10 m wide at the top, 6 m wide at the bottom and the area of its cross, section is 640 m 2 , find the depth of the canal., 37. Find the area of a trapezium whose parallel sides are 11 m and 25 m, long, and the nonparallel sides are 15 m and 13 m long., [HOTS], ANSWERS (EXERCISE 15B), , 1. Length = 24 m, area = 384 m 2, 2. 39200 m 2, 3. 35 cm, 420 cm 2, 4. 89 m, 5. ` 15600, 6. ` 17280 7. 216, 8. 1800, 2, 11. 1254 m 2 , ` 94050, 9. Length = 16 cm, breadth = 12 cm, 10. 240 m, 12. (i) 2.5 m (ii) 50 cm, 13. Length = 180 m, breadth = 100 m, 14. ` 1320 15. length = 34 m, breadth = 24 m, 16. ` 27800 17. ` 10500, 18. Length = 12 m, breadth = 9 m and height = 6 m, 19. 288 m 2 , 67 .68 m, 20. 16 cm, 45.12 cm 21. 6 minutes 22. ` 21600 23. ` 135000 24. 252 cm 2, 25. 412.37 cm 2 , 86 cm, 26. 46 cm, 114 cm 2, 27. 546 cm 2, 2, 28. 420 cm 29. 23.2 cm 30. Base = 14 m, altitude = 28 m, 31. 672 cm 2 32. 240 cm 2 , 68 cm, 33. (i) 24 cm (ii) 216 cm 2, 34. (i) 20 cm (ii) 26 cm (iii) 104 cm 35. 84 cm 2 36. 80 m, 37. 216 m 2
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670, , Secondary School Mathematics for Class 10, HINTS TO SOME SELECTED QUESTIONS, , 3. Other side = ( 37 ) 2 − ( 12 ) 2 cm, = ( 37 + 12 )( 37 − 12 ) cm, = 49 × 25 cm = (7 × 5 ) cm, = 35 cm., 6. Length of the carpet =, , ⎛ 16 × 13 . 5 ⎞, area of the floor, =⎜, ⎟ m = 288 m., width of the carpet ⎝ 0 . 75 ⎠, , 7. Area of the floor = ( 24 × 18 ) m 2 ., 25 80 ⎞ 2, 2, Area of each carpet = ⎛⎜ ×, ⎟ m = 2m ., ⎝ 10 100 ⎠, Number of carpets =, 8. Number of stones =, , area of the floor, ⋅, area of each carpet, , area of the verandah ⎛ 36 × 15 ⎞, =⎜, ⎟⋅, area of each stone, ⎝ 0.6 × 0 .5 ⎠, , 9. l + b = 28 and lb = 192., ( l − b ) 2 = ( l + b ) 2 − 4 lb = ( 28 ) 2 − 4 × 192 = 16 ⇒ l − b = 4., On solving l + b = 28 and l − b = 4, we get l = 16 , b = 12., 10. Outer area = ( 35 × 18 ) m 2 = 630 m 2 ., Inner area = ( 30 × 13 ) m 2 = 390 m 2 ., Grass area = ( 630 − 390 ) m 2 = 240 m 2 ., 11. Area of the path = [( 125 + 6 ) × (78 + 6 ) − ( 125 × 78 )] m 2 ., 12. (i) Let the width of the path be x metres. Then,, (54 × 35 ) − {(54 − 2 x ) × ( 35 − 2 x )} = 420, ⇒ 2 x 2 − 89 x + 210 = 0 ⇒ 2 x 2 − 84 x − 5 x + 210 = 0, [Q width of the path ≠ 42 m]., ⇒ ( x − 42 )( 2 x − 5 ) = 0 ⇒ x = 42 or x = 2.5, (ii) Let the width of the border be x metres. Then,, ∴, , ( 8 × 5 ) − ( 8 − 2 x )(5 − 2 x ) = 12 ⇒ 2 x 2 − 13 x + 6 = 0 ⇒ ( x − 6 )( 2 x − 1) = 0., 1, x = m = 50 cm., 2, , 13. Let the length and breadth be ( 9x ) m and (5x ) m respectively., Then, ( 9 x × 5 x ) − ( 9 x − 7 )(5 x − 7 ) = 1911., 14. Area to be carpeted = ( 4.9 − 0.5 )( 3.5 − 0.5 ) m 2 = 13.2 m 2 ., 13.2 ⎞, Length of the carpet = ⎛⎜, ⎟ m = 16.5 m., ⎝ 0.80 ⎠, 16. Area of the roads = {( 80 × 5 ) + ( 64 × 5 ) − (5 × 5 )} m 2 = 695 m 2 ., 13, 17. Area of 4 walls = [2( l + b ) × h] = ⎛⎜ 2 × 24 × ⎞⎟ m 2 = 312 m 2 ., ⎝, 2⎠, 3, 5 12 ⎞ ⎛, ⎫, ⎧⎛, Area not to be painted = ⎨ ⎜ 2 × × ⎟ + ⎜ 4 × × 1⎞⎟ ⎬ m 2 = 12 m 2 ., ⎠⎭, 2, 2 10 ⎠ ⎝, ⎩⎝
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Perimeter and Area of Plane Figures, 18. Area of 4 walls =, , 671, , total cost of painting ⎛ 7560 ⎞ 2, 2, =⎜, ⎟ m = 252 m ., ⎝ 30 ⎠, rate per m 2, , Area of the floor =, , total cost of matting ⎛ 2700 ⎞ 2, 2, =⎜, ⎟ m = 108 m ., ⎝ 25 ⎠, rate per m 2, , Let breadth = b metres and height = h metres. Then,, 108, 12 × b = 108 ⇒ breadth =, m = 9 m., 12, Also, 2( l + b ) × h = 252 ⇒ 2( 12 + 9 ) × h = 252 ⇒ h = 6 m., 19. Area =, , 1 2 ⎛1, d = ⎜ × 24 × 24 ⎞⎟ m 2 = 288 m 2 ., ⎝2, ⎠, 2, , Now, a2 = 288 ⇒ a = 288 = 12 2 ⇒ 4 a = 48 2 m., 20., , 1 2, d = 128 ⇒ d 2 = 256 = ( 16 ) 2 ⇒ d = 16., 2, a2 = 128 ⇒ a = 128 = 8 2 ⇒ 4 a = 32 2 ., , 21. Area = ( 8 × 10000 ) m 2 ⇒, , 1, × ( diagonal ) 2 = 8 × 10000, 2, , ⇒ ( diagonal ) 2 = 160000 ⇒ diagonal = 400 m., 1, 1, 24. ar (quad. ABCD) = ar(SABD ) + ar( SCBD ) = ⎛⎜ × BD × AL ⎞⎟ + ⎛⎜ × BD × CM ⎞⎟, ⎠, ⎝2, ⎠ ⎝2, =, , 1, × BD × ( AL + CM ) cm 2 ., 2, , 25. AB = BD 2 − AD 2 = ( 26 ) 2 − ( 24 ) 2 cm = ( 26 + 24 )( 26 − 24 ) cm, = 50 × 2 cm = 100 cm = 10 cm., ar(quad. ABCD) = ar(right SBAD) + ar(equilateral SBCD)., 29. Area of the parallelogram = ( 32 × 17 .4 ) cm 2 ., 24 × h = 32 × 17 .4. Find h., 31. ar(||gm ABCD) = 2 × ar( SABC )., 32. We know that the diagonals of a rhombus bisect each other at right angles., So, OA = 15 cm,OB = 8 cm and ∠AOB = 90 °., ∴, , AB 2 = (OA 2 + OB 2 ) = ( 15 ) 2 + ( 8 ) 2 = 225 + 64 = 289, , ⇒, , AB = 289 = 17 cm., , So, perimeter = ( 4 × 17 ) cm = 68 cm., 33. Each side = 15 cm., ( 15 ) 2 = 9 2 + x 2 ⇒ x 2 = ( 225 − 81) = 144 ⇒ x = 144 = 12 cm., 1, ∴ other diagonal = 2 x = 24 cm. Area = ⎛⎜ × d1 × d2 ⎞⎟ cm 2 ., ⎝2, ⎠, 1, 36. ( 10 + 6 ) × d = 640 ⇒ d = 80 cm., 2, , ................................................................
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672, , Secondary School Mathematics for Class 10, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. The length of a rectangular hall is 5 m more than its breadth. If the area, of the hall is 750 m 2 then its length is, (a) 15 m, , (b) 20 m, , (c) 25 m, , (d) 30 m, , 2. The length of a rectangular field is 23 m more than its breadth. If the, perimeter of the field is 206 m, then its area is, (a) 2420 m 2, , (b) 2520 m 2, , (c) 2480 m 2, , (d) 2620 m 2, , 3. The length of a rectangular field is 12 m and the length of its diagonal is, 15 m. The area of the field is, (a) 108 m 2, , (b) 180 m 2, , (c) 30 3 m 2, , (d) 12 15 m 2, , 4. The cost of carpeting a room 15 m long with a carpet 75 cm wide, at ` 70, per metre, is ` 8400. The width of the room is, (a) 9 m, , (b) 8 m, , (c) 6 m, , (d) 12 m, , 5. The length of a rectangle is thrice its breadth and the length of its, diagonal is 8 10 cm. The perimeter of the rectangle is, (a) 15 10 cm, , (b) 16 10 cm, , (c) 24 10 cm, , (d) 64 cm, , 6. On increasing the length of a rectangle by 20% and decreasing its, breadth by 20%, what is the change in its area?, (a) 20% increase, , (b) 20% decrease, , (c) No change, , (d) 4% decrease, , 7. A rectangular ground 80 m × 50 m has a path 1 m wide outside around, it. The area of the path is, (a) 264 m 2, , (b) 284 m 2, , (c) 400 m 2, , (d) 464 m 2, , 8. The length of the diagonal of a square is 10 2 cm. Its area is, (a) 200 cm 2, , (b) 100 cm 2, , (c) 150 cm 2, , (d) 100 2 cm 2, , 9. The area of a square field is 6050 m 2 . The length of its diagonal is, (a) 135 m, , (b) 120 m, , (c) 112 m, , (d) 110 m, , 10. The area of a square field is 0.5 hectare. The length of its diagonal is, (a) 150 m, , (b) 100 2 m, , (c) 100 m, , (d) 50 2 m, , 2, , 11. The area of an equilateral triangle is 4 3 cm . Its perimeter is, (a) 9 cm, , (b) 12 cm, , (c) 12 3 cm, , (d) 6 3 cm, , 12. Each side of an equilateral triangle is 8 cm. Its area is, (a) 24 cm 2, , (b) 24 3 cm 2, , (c) 16 3 cm 2, , (d) 8 3 cm 2
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Perimeter and Area of Plane Figures, , 673, , 13. Each side of an equilateral triangle is 6 3 cm. The altitude of the, triangle is, (a) 8 cm, , (b) 9 cm, , (c) 3 3 cm, , (d) 6 cm, , 14. The height of an equilateral triangle is 3 3 cm. Its area is, (a) 6 3 cm 2, , (b) 27 cm 2, , (c) 9 3 cm 2, , (d) 27 3 cm 2, , 15. The base and height of a triangle are in the ratio 3 : 4 and its area is, 216 cm 2 . The height of the triangle is, (a) 18 cm, , (b) 24 cm, , (c) 21 cm, , (d) 28 cm, , 16. The lengths of the sides of a triangular field are 20 m, 21 m and 29 m., The cost of cultivating the field at ` 9 per m 2 is, (a) ` 2610, , (b) ` 3780, , (c) ` 1890, , (d) ` 1800, , 17. The side of a square is equal to the side of an equilateral triangle. The, ratio of their areas is, (a) 4 : 3, , (b) 2 : 3, , (c) 4 : 3, , (d) none of these, , 18. The side of an equilateral triangle is equal to the radius of a circle whose, area is 154 cm 2 . The area of the triangle is, (a) 49 cm 2, , (b), , 49 3, cm 2, 4, , (c), , 7 3, cm 2, 4, , (d) 77 cm 2, , 19. The area of a rhombus is 480 cm 2 and the length of one of its diagonals, is 20 cm. The length of each side of the rhombus is, (a) 24 cm, , (b) 30 cm, , (c) 26 cm, , (d) 28 cm, , 20. One side of a rhombus is 20 cm long and one of its diagonals measures, 24 cm. The area of the rhombus is, (a) 192 cm 2, , (b) 480 cm 2, , (c) 240 cm 2, , (d) 384 cm 2, , ANSWERS (MCQ), , 1. (d) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (a) 8. (b) 9. (d) 10. (c), 11. (b) 12. (c) 13. (b) 14. (c) 15. (b) 16. (c) 17. (c) 18. (b) 19. (c) 20. (d), HINTS TO SOME SELECTED QUESTIONS, 1. Let the length be x metres. Then, breadth = ( x − 5 ) m., ∴ x( x − 5 ) = 750 ⇒ x 2 − 5 x − 750 = 0 ⇒ ( x − 30 )( x + 25 ) = 0 ⇒ x = 30 [Q x ≠ −25 ]., Hence, length = 30 m., 2. Let the breadth be x metres. Then, length = ( x + 23 ) m., ∴ 2 [( x + 23 ) + x] = 206 ⇒ 4 x = 160 ⇒ x = 40., ∴ b = 40 m and l = 63 m., Hence, area = ( 63 × 40 ) m 2 = 2520 m 2 .
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674, , Secondary School Mathematics for Class 10, , 3. l = 12 m and diagonal ( d ) = 15 m., b 2 = ( d 2 − l 2 ) = ( 15 ) 2 − ( 12 ) 2 = ( 15 + 12 )( 15 − 12 ) = 27 × 3 = 81, ⇒ b = 81 m = 9 m., ∴ area = ( 12 × 9 ) m 2 = 108 m 2 ., total cost, 8400 ⎞, 4. Length of the carpet =, = ⎛⎜, ⎟ m = 120 m., rate per m ⎝ 70 ⎠, 3, Area of the room = area of the carpet = ⎛⎜ 120 × ⎞⎟ m 2 = 90 m 2 ., ⎝, 4⎠, area, 90, Width of the room =, = ⎛⎜ ⎞⎟ m = 6 m., length ⎝ 15 ⎠, 5. Let the breadth be x cm. Then, length = 3x cm., ∴, , diagonal = ( 3 x ) 2 + x 2 cm = x 10 cm., , Thus, x 10 = 8 10 ⇒ x = 8., Perimeter = 2( 3 x + x ) cm = 8 x cm = ( 8 × 8 ) cm = 64 cm., 6. Let the length be l m and breadth be b m., Then, original area = ( l × b ) m 2 = lb m 2 ., 120 ⎞, 6l, 80 ⎞, 4b, New length = ⎛⎜ l ×, m, new breadth = ⎛⎜ b ×, m., ⎟ m=, ⎟ m=, ⎝, ⎝, 100 ⎠, 5, 100 ⎠, 5, 24 lb ⎞ 2, 24 lb ⎞ 2 lb 2, ⎛, 2, New area = ⎛⎜, m ., ⎟ m = 96 m . Change in area = ⎜ ab −, ⎟m =, ⎝ 25 ⎠, ⎝, 25 ⎠, 25, ∴, , % change =, , lb × 100, % = 4%., 25 × lb, , 7. Area of the ground without path = ( 80 × 50 ) m 2 = 4000 m 2 ., Area of the ground with path = (82 × 52) m 2 = 4264 m 2 ., Area of the path = ( 4264 − 4000 ) m 2 = 264 m 2 ., 1, 1, 8. Area = × ( diagonal ) 2 = × ( 10 2 ) 2 cm 2 = 100 cm 2 ., 2, 2, 9., , 1, × ( diagonal ) 2 = 6050 ⇒ ( diagonal ) 2 = 12100 = ( 110 ) 2 ⇒ diagonal = 110 m., 2, , 10. Area = ( 0 . 5 × 10000 ) m 2 = 5000 m 2 ., 1, × d 2 = 5000 ⇒ d 2 = 10000 = ( 100 ) 2 ⇒ d = 100 m., 2, 11., , 3, × a2 = 4 3 ⇒ a2 = 16 ⇒ a = 4., 4, ∴, , perimeter = 3 a = ( 3 × 4 ) cm = 12 cm., , ⎛ 3, ⎞ ⎧ 3, ⎫, 13. Area of the triangle = ⎜, × a2 ⎟ = ⎨, × ( 6 3 ) 2 ⎬ cm 2, ⎝ 4, ⎠ ⎩ 4, ⎭, ⎛ 3, ⎞, 2, =⎜, × 108 ⎟ cm = ( 27 3 ) cm 2 ., ⎝ 4, ⎠, 1, 27 3, × 6 3 × h = 27 3 ⇒ h =, = 9 cm., 2, 3 3
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Perimeter and Area of Plane Figures, , 14., , 675, , 1, 3 2, 2h ⎛ 2, ⎞, × a× h =, a ⇒ a=, =⎜, × 3 3 ⎟ cm = 6 cm., ⎠, 2, 4, 3 ⎝ 3, ∴, , ⎛ 3, ⎞, area = ⎜, × 6 × 6 ⎟ cm 2 = 9 3 cm 2 ., 4, ⎝, ⎠, , 15. Let the base be 3x cm and height be 4x cm. Then,, 1, × 3 x × 4 x = 216 ⇒ x 2 = 36 ⇒ x = 6., 2, ∴, , height = ( 4 × 6 ) cm = 24 cm., 1, 16. s = ( 20 + 21 + 29 ) = 35 , (s − a) = 15 , (s − b ) = 14 and (s − c ) = 6., 2, ∴, , area =, , 35 × 15 × 14 × 6 m 2 = (7 × 5 × 3 × 2 ) m 2 = 210 m 2 ., , Required cost = ` ( 210 × 9 ) = ` 1890., 17. Let the side of the square = side of the triangle = a. Then,, area of the square, a2, 4, =, =, = 4 : 3., area of the triangle, 3 2, 3, a, 4, 18. Area of the circle = 154 cm 2 ., 22, × R 2 = 154, ∴ πR 2 = 154 ⇒, 7, 7, ⇒ R 2 = ⎛⎜ 154 × ⎞⎟ = 7 2 ⇒ R = 7 cm., ⎝, 22 ⎠, ∴, , side of the triangle = radius of the circle = 7 cm., , ⎛ 3, ⎞, 49 3, Area of the triangle = ⎜, × 7 × 7 ⎟ cm 2 =, cm 2 ., 4, ⎝ 4, ⎠, 19., , 1, 1, × d1 × d2 = 480 cm 2 ⇒ × 20 × d2 = 480 cm 2 ⇒ d2 = 48 cm., 2, 2, ∴, , OB = 10 cm, OA = 24 cm and ∠AOB = 90 ° ., , ∴, , AB 2 = OA 2 + OB 2 = ( 24 ) 2 + ( 10 ) 2 = 576 + 100 = 676, , ⇒, , AB = 676 cm = 26 cm., , Hence, each side of the rhombus is 26 cm., 20. Let ABCD be the rhombus in which AB = 20 cm and, 1, OA = ⎛⎜ × 24 ⎞⎟ cm = 12 cm., ⎝2, ⎠, ∴, , OB 2 = AB 2 − OA 2 = ( 20 ) 2 − ( 12 ) 2 = 256 = ( 16 ) 2, , ⇒, , OB = 16 cm., , ∴, , BD = ( 2 × 16 ) cm = 32 cm., 1, area = ⎛⎜ × 24 × 32 ⎞⎟ cm 2 = 384 cm 2 ., ⎝2, ⎠, , ∴, , ................................................................
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676, , Secondary School Mathematics for Class 10, , TEST YOURSELF, MCQ, 1. In the given figure ABDC is a quadrilateral in, which ∠ABC = 90 ° , ∠BDC = 90 ° , AC = 17 cm,, BC = 15 cm, BD = 12 cm and CD = 9 cm. The, area of quad. ABDC is, (a) 102 cm 2, , (b) 114 cm 2, , (c) 95 cm 2, , (d) 57 cm 2, , 2. In the given figure ABCD is a trapezium, in which AB = 40 m, BC = 15 m, CD = 28 m,, AD = 9 m and CE ⊥ AB. Area of trap. ABCD is, (a) 306 m 2, (b) 316 m 2, (c) 296 m 2, , (d) 284 m 2, , 3. The sides of a triangle are in the ratio 12 : 14 : 25 and its perimeter is, 25.5 cm. The largest side of the triangle is, (a) 7 cm, , (b) 14 cm, , (c) 12.5 cm, , (d) 18 cm, , 4. The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance, between them is 6.5 cm. The area of the trapezium is, (a) 104 cm 2, , (b) 78 cm 2, , (c) 52 cm 2, , (d) 65 cm 2, , Short-Answer Questions, 5. Find the area of an equilateral triangle having each side of length 10 cm., [Take 3 = 1.732.], 6. Find the area of an isosceles triangle each of whose equal sides is 13 cm, and whose base is 24 cm., 7. The longer side of a rectangular hall is 24 m and the length of its, diagonal is 26 m. Find the area of the hall., 8. The length of the diagonal of a square is 24 cm. Find its area., 9. Find the area of a rhombus whose diagonals are 48 cm and 20 cm long., 10. Find the area of a triangle whose sides are 42 cm, 34 cm and 20 cm., 11. A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3 and, its area is 3375 m 2 . Find the cost of fencing the lawn at ` 20 per metre., 12. Find the area of a rhombus each side of which measures 20 cm and one, of whose diagonals is 24 cm.
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Perimeter and Area of Plane Figures, , 677, , 13. Find the area of a trapezium whose parallel sides are 11 cm and 25 cm, long and nonparallel sides are 15 cm and 13 cm., 14. The adjacent sides of a ||gm ABCD measure, 34 cm and 20 cm and the diagonal AC is 42 cm, long. Find the area of the||gm., , 15. The cost of fencing a square lawn at ` 14 per metre is ` 2800. Find the, cost of mowing the lawn at ` 54 per 100 m 2 ., 16. Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm,, CD = 29 cm, DA = 34 cm and diag. BD = 20 cm., , Long-Answer Questions, 17. A parallelogram and a rhombus are equal in area. The diagonals of the, rhombus measure 120 m and 44 m. If one of the sides of the||gm is, 66 m long, find its corresponding altitude., 18. The diagonals of a rhombus are 48 cm and 20 cm long. Find the, perimeter of the rhombus., 19. The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If, the distance between the shorter sides is 12 cm, find the distance, between the longer sides., 20. In a four-sided field, the length of the longer diagonal is 128 m. The, lengths of perpendiculars from the opposite vertices upon this, diagonal are 22.7 m and 17.3 m. Find the area of the field., ANSWERS (TEST YOURSELF), , 1. (b), 6. 60 cm, , 2. (a), 2, , 7. 240 m, , 11. ` 4800, 16. 546 cm, , 3. (c), 2, , 12. 384 cm, 2, , 17. 40 m, , 2, , 5. 43.3 cm 2, , 4. (c), , 8. 288 cm, , 2, , 13. 216 cm, , 2, , 18. 104 cm, _, , 9. 480 cm, , 2, , 10. 336 cm 2, , 14. 672 cm, , 2, , 15. ` 1350, , 19. 9 cm, , 20. 2560 m 2
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Area of Circle, Sector and Segment, , The set of points which are at a constant distance of, r units from a fixed point O is called a circle with centre O, and radius = r units. The circle is denoted by C(O , r)., In other words, a circle is the locus of a point which, moves in such a way that its distance from a fixed point O remains constant, at r units., The fixed point O is called the centre and the constant distance r units, is called its radius., CIRCUMFERENCE The perimeter (or length of boundary) of a circle is called, its circumference., RADIUS A line segment joining the centre of a circle and a, point on the circle is called a radius of the circle., Plural of radius is radii. In the given figure,, OA , OB, OC are three radii of the circle., CIRCLE, , A line segment joining any two points on a circle, is called a chord of the circle., In the given figure, PQ , RS and AOB are three chords, of a circle with centre O., DIAMETER A chord of a circle passing through its centre is, called a diameter of the circle., Diameter is the longest chord of a circle. In the above figure, AOB is a, diameter., CHORD, , Diameter = 2 × radius., A line which intersects a circle at two points, is called a secant of the circle., In the given figure, line l is a secant of the circle, with centre O., ARC A continuous piece of a circle is called an arc of, the circle., In the given figure, AB is an arc of a circle, with centre, SECANT, , O, denoted by ABT. The remaining part of the circle,, shown by the dotted lines, represents BAT., 678
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Area of Circle, Sector and Segment, , 679, , CENTRAL ANGLE An angle subtended by an arc at the centre of a circle is, called its central angle., In the given figure of a circle with centre O,, , central angle of ABT = ∠AOB = θ ° ., If θ° < 180 ° then ABT is called the minor arc and, BAT is called the major arc., SEMICIRCLE A diameter divides a circle into two, equal arcs. Each of these two arcs is called a, semicircle., In the given figure of a circle with centre O, ACBQ, , and BDAQ are semicircles., An arc whose length is less than the arc of a semicircle is called a minor arc., An arc whose length is more than the arc of a semicircle is called a major arc., A segment of a circle is the region bounded, by an arc and a chord, including the arc and the chord., The segment containing the minor arc is called a, minor segment, while the segment containing the, major arc is the major segment., The centre of the circle lies in the major segment., SECTOR OF A CIRCLE The region enclosed by an arc of, a circle and its two bounding radii is called a sector, of the circle., In the given figure, OACBO is a sector of the, circle with centre O., If arc AB is a minor arc then OACBO is called the, minor sector of the circle., The remaining part of the circle is called the major sector of the circle., SEGMENT, , QUADRANT One-fourth of a circular disc is called a quadrant. The central, angle of a quadrant is 90°., , FORMULAE ON AREA OF CIRCLE, SECTOR AND SEGMENT, , 1. For a circle of radius r, we have, (i) Circumference = 2πr, (ii) Area = πr 2 ., 2. For a semicircle of radius r, we have, 1, (i) Perimeter = ( πr + 2r), (ii) Area = πr 2 ., 2, 3. For a ring having outer radius = R and inner radius = r,, we have, Area of the ring = π( R 2 − r 2 ).
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680, , Secondary School Mathematics for Class 10, , 4. Let an arc ACB make an angle θ° at the centre of a circle of radius r. Then,, we have, 2 πrθ, (i) Length of minor arc ACB =, ⋅, 360, 2 πrθ ⎞, ⎟⋅, Length of major arc BDA = ⎛⎜ 2 πr −, ⎝, 360 ⎠, (ii) Area of minor sectorOACBO =, , πr 2θ ⎛ 1, = ⎜ × radius × arc length ⎞⎟ ⋅, ⎠, 360 ⎝ 2, , ⎛, πr 2θ ⎞, Area of major sector OADBO = ⎜ πr 2 −, ⎟⋅, 360 ⎠, ⎝, ⎛ πr 2θ 1 2, ⎞, (iii) Area of minor segment ACBA = ⎜, − r sin θ ⎟ ⋅, ⎝ 360 2, ⎠, Area of major segment BDAB = [πr 2 − ( area of minor segment)]., 2πrθ ⎞, ⎟⋅, (iv) Perimeter of sector OACBO = ⎛⎜ 2r +, ⎝, 360 ⎠, 5. For Rotation of the Hands of a Clock:, (i) Angle described by minute hand in 60 minutes = 360°., (ii) Angle described by hour hand in 12 hours = 360°., 6. For Rotating Wheels:, (i) Distance moved by a wheel in 1 rotation = its cicumference., (ii) Number of rotations in 1 minute =, , distance moved in 1 minute, ⋅, circumference, , 7. Touching Circles:, (i) When two circles touch internally [see fig. (i)], then, distance between their centres = difference of their radii., (ii) When two circles touch externally [see fig. (ii)], then, distance between their centres = sum of their radii.
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Area of Circle, Sector and Segment, , SOLVED, , EXAMPLES, , EXAMPLE 1, , Find the circumference and area of a circle of diameter 28 cm., , SOLUTION, , Diameter = 28 cm ⇒ radius r = 14 cm., ∴, , 681, , circumference of the circle = 2πr, 22, = ⎛⎜ 2 ×, × 14 ⎞⎟ cm = 88 cm., ⎝, ⎠, 7, , 22, Area of the circle = πr 2 = ⎛⎜, × 14 × 14 ⎞⎟ cm 2, ⎝ 7, ⎠, = 616 cm 2 ., EXAMPLE 2, , Find the area of a circle whose circumference is 66 cm., , SOLUTION, , Let the radius of the circle be r cm., Then, its circumference = ( 2πr) cm., 22, ∴ 2 πr = 66 ⇒ 2 ×, × r = 66, 7, 7 ⎞ 21, ⎟ =, ⋅, ⇒ r = ⎛⎜ 66 ×, ⎝, 44 ⎠, 2, ∴, , 22 21 21 ⎞, area of the circle = πr 2 cm 2 = ⎛⎜, ×, × ⎟ cm 2, ⎝ 7, 2, 2⎠, = 346.5 cm 2 ., , EXAMPLE 3, , A steel wire, when bent in the form of a square, encloses an area of, 121 cm 2 .The same wire is bent in the form of a circle. Find the area of, the circle., , SOLUTION, , Area of the square = 121 cm 2 ., Side of the square = 121 cm = 11 cm., Perimeter of the square = 4 × 11 cm = 44 cm., ∴, , length of the wire = 44 cm., , ∴, , circumference of the circle = length of the wire = 44 cm., , Let the radius of the circle be r cm., 22, Then, 2 πr = 44 ⇒ 2 ×, × r = 44 ⇒ r = 7 ., 7, ∴, , area of the circle = πr 2, 22, = ⎛⎜, × 7 × 7 ⎞⎟ cm 2 = 154 cm 2 ., ⎝ 7, ⎠
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682, , Secondary School Mathematics for Class 10, , EXAMPLE 4, , A wire is looped in the form of a circle of radius 28 cm. It is rebent, into a square form. Determine the length of the side of the square., , SOLUTION, , Length of the wire = circumference of the circle, 22, = ⎛⎜ 2 ×, × 28 ⎞⎟ cm = 176 cm., ⎝, ⎠, 7, ∴, , perimeter of the square = length of the wire = 176 cm., , 176 ⎞, ⎟ cm = 44 cm., Hence, the side of the square = ⎛⎜, ⎝ 4 ⎠, EXAMPLE 5, , A circular park, 42 m in diameter, has a path 3.5 m wide running, round it on the outside. Find the cost of gravelling the path at, ` 20 per m 2 ., , SOLUTION, , Radius of the circular park = 21 m., Radius of the outer circle = 21 m + 3.5 m = 24.5 m., Area of the path = π × [( 24.5) 2 − ( 21) 2 ] m 2, 22, =, × ( 24.5 + 21)( 24.5 − 21) m 2, 7, 22, = ⎛⎜, × 45.5 × 3.5 ⎞⎟ m 2 = 500.5 m 2 ., ⎝ 7, ⎠, ∴, , cost of gravelling the path = ` (500.5 × 20) = ` 10010., , EXAMPLE 6, , A road which is 7 m wide surrounds a circular park whose, circumference is 352 m. Find the area of the road., , SOLUTION, , Let the radius of the park be r m., Then, its circumference = 2πr m., 22, × r = 352, ∴ 2 πr = 352 ⇒ 2 ×, 7, 7, ⇒ r = 352 ×, = 56., 44, Thus, inner radius = 56 m,, outer radius = (56 + 7 ) m = 63 m., Area of the road = π[( 63) 2 − (56) 2 ] m 2, 22, =, × ( 63 + 56)( 63 − 56) m 2, 7, 22, = ⎛⎜, × 119 × 7 ⎞⎟ m 2 = 2618 m 2 ., ⎝ 7, ⎠
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Area of Circle, Sector and Segment, , 683, , EXAMPLE 7, , A racetrack is in the form of a ring whose inner and outer, circumferences are 437 m and 503 m respectively. Find the width, of the track and also its area., , SOLUTION, , Let r m and R m be the radii of inner and outer boundaries., Then, 2 πr = 437 and 2 πR = 503, 437, 503, and R =, ⋅, ⇒ r=, 2π, 2π, Width of the track = ( R − r) m, 503 437 ⎞, ⎟ m, = ⎛⎜, −, ⎝ 2π, 2π ⎠, 1, × (503 − 437 ) m, 2π, 1, 7, = ⎛⎜ ×, × 66 ⎞⎟ m = 10.5 m., ⎝ 2 22, ⎠, =, , Area of the track = π( R 2 − r 2 ) m 2 = π( R + r)( R − r) m 2, ⎡ 503 437 ⎞, ⎤, ⎟ × 10.5 m 2, = ⎢ π ⎛⎜, +, ⎥⎦, 2π ⎠, ⎣ ⎝ 2π, [B ( R − r) = 10.5], 21 ⎞ 2, 940 ⎞, ⎡, ⎤, ⎟ × 10.5 m 2 = ⎛⎜ 470 ×, ⎟ m, = ⎢ ⎛⎜ π ×, ⎥, ⎝, ⎠, ⎝, 2⎠, 2, π, ⎣, ⎦, = ( 235 × 21) m 2 = 4935 m 2 ., EXAMPLE 8, , If the perimeter of a semicircular protractor is 36 cm, find its, diameter., , SOLUTION, , Let the radius of the protractor be r cm., Then, perimeter = ( πr + 2r) cm = ( π + 2)r cm, 22, 36, = ⎛⎜, + 2 ⎞⎟ r cm =, r cm., ⎝ 7, ⎠, 7, ∴, , 36, 7 ⎞, ⎟ = 7., r = 36 ⇒ r = ⎛⎜ 36 ×, ⎝, 7, 36 ⎠, , Hence, diameter = 2r cm = 14 cm., EXAMPLE 9, , A bicycle wheel makes 5000 revolutions in moving 11 km. Find the, diameter of the wheel., , SOLUTION, , Distance covered by the wheel in 1 revolution, ⎛ 11 × 1000 × 100 ⎞, =⎜, ⎟ cm = 220 cm., ⎝, ⎠, 5000
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684, , Secondary School Mathematics for Class 10, , ∴, , the circumference of the wheel = 220 cm., , Let the diameter of the wheel be d cm., 22, 7, Then, πd = 220 ⇒, × d = 220 ⇒ d = 220 ×, = 70., 7, 22, Hence, the diameter of the wheel is 70 cm., EXAMPLE 10, , The diameter of each wheel of a bus is 140 cm. How many revolutions, per minute must a wheel make in order to move at a speed of 66 km, per hour?, , SOLUTION, , Distance covered by a wheel in 1 minute, 66 × 1000 × 100 ⎞, ⎟ cm = 110000 cm., = ⎛⎜, ⎝, ⎠, 60, 22, Circumference of a wheel = ⎛⎜ 2 ×, × 70 ⎞⎟ cm = 440 cm., ⎝, ⎠, 7, 110000 ⎞, ⎟ = 250., Number of revolutions in 1 min = ⎛⎜, ⎝ 440 ⎠, , EXAMPLE 11, , Two circles touch externally. The sum of their areas is 130π sq cm, and the distance between their centres is 14 cm. Find the radii of, the circles., , SOLUTION, , Since the given circles touch externally, we have, sum of their radii = distance between their centres = 14 cm., Let the radii of the given circles be x cm and (14 − x) cm., Sum of their areas = [πx 2 + π(14 − x) 2 ] cm 2 ., ∴, , πx 2 + π (14 − x) 2 = 130 π, , ⇒ x 2 + (14 − x) 2 = 130, ⇒ 2 x 2 − 28 x + 66 = 0, ⇒ x 2 − 14 x + 33 = 0, ⇒ ( x − 11)( x − 3) = 0, ⇒ x − 11 = 0 or x − 3 = 0, ⇒ x = 11 or x = 3., Now, x = 11 ⇒ (14 − x) = (14 − 11) = 3., And, x = 3 ⇒ (14 − x) = (14 − 3) = 11., Hence, the radii of the given circles are 11 cm and 3 cm., EXAMPLE 12, , Two circles touch internally. The sum of their areas is (116 π) cm 2 and, the distance between their centres is 6 cm. Find the radii of the circles., [CBSE 2017]
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Area of Circle, Sector and Segment, SOLUTION, , 685, , The circles touch internally., ∴, , difference of their radii, = distance between their centres = 6 cm., , Let the radii of given circles be r cm and (r + 6) cm., Sum of their areas = [πr 2 + π (r + 6) 2 ] cm 2 = π [r 2 + (r + 6) 2 ] cm 2 ., ∴, , π [r 2 + (r + 6) 2 ] = 116 π, , ⇒ r 2 + (r + 6) 2 = 116, ⇒ 2r 2 + 12r − 80 = 0, ⇒ r 2 + 6r − 40 = 0, ⇒ (r + 10)(r − 4) = 0, ⇒ r + 10 = 0 or r − 4 = 0, ⇒ r = 4 [neglecting r = − 10, as radius cannot be negative], ∴, , the radii of the given circles are 4 cm and 10 cm., , EXAMPLE 13, , Find the area of a right-angled triangle, if the radius of its circumcircle is 7.5 cm and the altitude drawn to the hypotenuse is 6 cm long., , SOLUTION, , Let SABC be right-angled at B., Hypotenuse AC, = diameter of its circumcircle, = ( 2 × 7 .5) cm = 15 cm., Let BD⊥ AC . Then, BD = 6 cm., 1, 1, ∴ ar( SABC ) = ⎛⎜ AC × BD ⎞⎟ = ⎛⎜ × 15 × 6 ⎞⎟ cm 2 = 45 cm 2 ., ⎝2, ⎠ ⎝2, ⎠, , EXAMPLE 14, , SOLUTION, , In the given figure, OABC is a rhombus whose, three vertices A, B, C lie on a circle of radius, 10 cm and centre O. Find the area of the, rhombus. [Take 3 = 1.732.], Clearly, OA = OB = OC = 10 cm. Let OB and AC intersect at P., Since the diagonals of a rhombus bisect each other at right, angles, we have OP = 5 cm and ∠OPC = 90 ° ., Now, AC = 2CP and, CP = OC 2 − OP 2 cm = (10) 2 − 5 2 cm, = 75 cm = 5 3 cm., ∴, , AC = ( 2 × 5 3 ) cm = 10 3 cm, = (10 × 1 .732) cm = 17 . 32 cm.
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Area of Circle, Sector and Segment, , 687, , EXAMPLE 16, , A chord of a circle of radius 14 cm makes a right angle at the centre., Find the areas of the minor and the major segments of the circle., , SOLUTION, , Let AB be the chord of a circle of centre O and radius = 14 cm, such that ∠AOB = 90 ° ., ∴, , πr 2θ, cm 2, 360, 22, 90 ⎞, ⎟ cm 2, = ⎛⎜, × 14 × 14 ×, ⎝ 7, 360 ⎠, , area of the sector OACBO =, , = 154 cm 2 ., 1 2, r sin θ, 2, 1, = ⎛⎜ × 14 × 14 × sin 90 ° ⎞⎟ cm 2, ⎝2, ⎠, , Area of COAB =, , = 98 cm 2 ., Area of the minor segment ACBA, = (area of the sector OACBO) – (area of the C OAB), = (154 − 98) cm 2 = 56 cm 2 ., Area of the major segment BDAB, = (area of the circle) – (area of the minor segment), ⎡ 22, ⎤, = ⎢ ⎛⎜, × 14 × 14 ⎞⎟ − 56 ⎥ cm 2 = ( 616 − 56) cm 2 = 560 cm 2 ., ⎝, ⎠, 7, ⎣, ⎦, EXAMPLE 17, , The perimeter of a sector of a circle of radius 14 cm is 68 cm. Find the, area of the sector., , SOLUTION, , Let O be the centre of a circle of radius 14 cm., Let OACBO be the sector whose perimeter is, 68 cm., Then, OA + OB + arc ACB = 68 cm, ⇒ 14 cm + 14 cm + arc ACB = 68 cm, ⇒ arc ACB = ( 68 − 28) cm = 40 cm., 1, ∴ ar(sector OACBO) = ⎛⎜ × radius × arc length ⎞⎟, ⎝2, ⎠, 1, = ⎛⎜ × 14 × 40 ⎞⎟ cm 2, ⎝2, ⎠, = 280 cm 2 .
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688, , Secondary School Mathematics for Class 10, , EXAMPLE 18, , The minute hand of a clock is 12 cm long. Find the area of the face of, the clock described by the minute hand in 35 minutes., , SOLUTION, , Angle described by the minute hand in 60 minutes = 360°., Angle described by the minute hand in 35 minutes, °, 360, = ⎛⎜, × 35 ⎞⎟ = 210 ° ., ⎝ 60, ⎠, ∴, , θ = 210 ° and r = 12 cm., , Area swept by the minute hand in 35 minutes, ⎛ πr 2θ ⎞ ⎛ 22, 210 ⎞, ⎟ cm 2 = 264 cm 2 ., =⎜, × 12 × 12 ×, ⎟ = ⎜⎝, 7, 360 ⎠, ⎝ 360 ⎠, EXAMPLE 19, , A car has wheels which are 80 cm in diameter. How many complete, revolutions does each wheel make in 10 minutes, when the car is, travelling at a speed of 66 km an hour?, , SOLUTION, , Distance covered by wheel in 1 minute, 66 × 1000 × 100 ⎞, ⎟ cm = 110000 cm., = ⎛⎜, ⎝, ⎠, 60, 22, Circumference of a wheel = 2 πR = ⎛⎜ 2 ×, × 40 ⎞⎟ cm, ⎝, ⎠, 7, 1760, =, cm., 7, Number of revolutions made by a wheel in 1 minute, 7 ⎞, ⎟ cm., = ⎛⎜ 110000 ×, ⎝, 1760 ⎠, Number of revolutions made by a wheel in 10 minutes, 7, = ⎛⎜ 110000 ×, × 10 ⎞⎟ = 4375., ⎝, ⎠, 1760, , EXAMPLE 20, , A car has two wipers which do not overlap. Each wiper has a blade of, length 21 cm, sweeping through an angle of 120°. Find the total area, cleaned at each sweep of the blades., , SOLUTION, , Clearly, each wiper sweeps a sector of a circle of radius 21 cm,, making an angle of 120° at the centre of the circle., So, the required area swept by two wipers, ⎛, πr 2θ ⎞ ⎛, 22, 120 ⎞, ⎟ cm 2, = ⎜2×, × 21 × 21 ×, ⎟ = ⎜2×, 360 ⎠ ⎝, 7, 360 ⎠, ⎝, = 924 cm 2 .
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Area of Circle, Sector and Segment, , 689, , EXAMPLE 21, , To warn ships for underwater rocks, a lighthouse spreads a redcoloured light over a sector of angle 72° to a distance of 15 km. Find, the area of the sea over which the ships are warned. [Use π = 3.14.], , SOLUTION, , Required area = area of the sector in which, r = 15 km, and θ = 72 °, ⎛ πr 2θ ⎞ ⎛, 72 ⎞, ⎟ km 2, =⎜, ⎟ = ⎜⎝ 3.14 × 15 × 15 ×, 360 ⎠, ⎝ 360 ⎠, 314 × 45 ⎞, 1413, ⎟ km 2 =, = ⎛⎜, km 2, ⎝ 100 ⎠, 10, = 141. 3 km 2 ., , EXAMPLE 22, , An umbrella has 8 ribs which are equally spaced (as shown in the, figure). Assuming the umbrella to be a flat circle of radius 42 cm,, find the area between the two consecutive ribs of the umbrella., , SOLUTION, , Since the ribs are equally spaced, so the angle made by two, 360 ⎞ °, ⎟ = 45 ° ., consecutive ribs at the centre = ⎛⎜, ⎝ 8 ⎠, Area between two consecutive ribs, = area of a sector of a circle with r = 42 cm and θ = 45 °, 45 22, = ⎛⎜, ×, × 42 × 42 ⎞⎟ cm 2, ⎝ 360 7, ⎠, = 693 cm 2 ., , EXAMPLE 23, , A round table cover has six equal designs, as shown in the figure. If, the radius of the cover is 28 cm, find the cost of making the designs at, the rate of ` 0. 35 per cm 2 . [Use 3 = 1 .7 .], [CBSE 2009]
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690, SOLUTION, , Secondary School Mathematics for Class 10, , Let O be the centre of the table cover and let it be divided into, six equal designs, each being a segment. Let one of these, 360 ⎞ °, ⎟ = 60 ° ., segments be ACBA. Clearly, ∠AOB = ⎛⎜, ⎝ 6 ⎠, ar(segment ACBA) = ar(sector OACBO) – ar(equilateral SOAB), ⎛ πr 2θ, 3 2⎞, =⎜, −, a ⎟, 4, ⎝ 360, ⎠, ⎡ 22, ⎞⎤, 60 ⎞ ⎛ 3, ⎟ −⎜, = ⎢ ⎛⎜, × 28 × 28 ×, × 28 × 28 ⎟ ⎥ cm 2, ⎝, 360 ⎠ ⎝ 4, ⎠⎦, ⎣ 7, [Q a = OA = 28 cm], 1232 17, = ⎛⎜, −, × 7 × 28 ⎞⎟ cm 2, ⎝ 3, ⎠, 10, 1232 1666 ⎞, ⎟ cm 2, = ⎛⎜, −, ⎝ 3, 5 ⎠, ( 6160 − 4998), cm 2, 15, 1162, =, cm 2 ., 15, 1162, 2324, ar(all the six segments) = ⎛⎜, × 6 ⎞⎟ cm 2 =, cm 2 ., ⎝ 15, ⎠, 5, =, , 2324 35 ⎞, ⎟ = ` 162.68., Cost of designs = ` ⎛⎜, ×, ⎝ 5, 100 ⎠, EXAMPLE 24, , SOLUTION, , A broach is made with silver wire in the form of a, circle with diameter 35 mm.The wire is also used, in making 5 diameters which divide the circle into, ten equal sectors, as shown in the figure., Find (i) the total length of the silver wire, required,, (ii) the area of each sector of the broach., (i) Total length of the wire required, = (circumference of given circle), + (length of five diameters), 22 35 ⎞, ⎟ mm + (5 × 35) mm = (110 + 175) mm, = ⎛⎜ 2 ×, ×, ⎝, 7, 2 ⎠, = 285 mm = 28.5 cm., (ii) The given circle has been divided into 10 equal sectors.
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Area of Circle, Sector and Segment, , 691, , 1, × area of the given circle, 10, 1 22 35 35 ⎞, ⎟ mm 2, = ⎛⎜ ×, ×, ×, ⎝ 10 7, 2, 2 ⎠, 385, =, mm 2 = 96.25 mm 2 ., 4, , Area of each sector =, , EXAMPLE 25, , A horse is tied to a peg at one corner of a square-shaped grass field of, side 15 m by means of a 5-m-long rope (as shown in the figure)., Find (i) the area of that part of the field in which the horse can graze,, (ii) the increase in the grazing area if the rope were 10 m long, instead of 5 m. [Use π = 3.14.], , SOLUTION, , (i) Clearly, the required area is the area of a quadrant of a, circle of radius 5 m., 1, 1, ∴ required area = × πr 2 = ⎛⎜ × 3.14 × 5 2 ⎞⎟ m 2 = 19.625 m 2 ., ⎝, ⎠, 4, 4, (ii) Let the length of the rope be 10 m. Then,, 1, 1, area grazed = × πR 2 = ⎛⎜ × 3.14 × 10 2 ⎞⎟ m 2 = 78 .50 m 2 ., ⎝, ⎠, 4, 4, Increase in grazing area = (78 .50 − 19 . 625) m 2, = 58.875 m 2 ., , AREAS OF COMBINATIONS OF PLANE FIGURES, SOLVED, EXAMPLE 1, , EXAMPLES, , In the given figure, ABCD is a square of, side 10 cm and semicircles are drawn, with each side of the square as diameter., Find the area of the shaded region., [Use π = 3.14.]
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692, SOLUTION, , Secondary School Mathematics for Class 10, , Let us mark the unshaded regions as I, II, III and IV as shown, in the given figure., Let these regions meet at a common point O. Then,, (area of I) + (area of III), = ar(sq ABCD) – {ar(semicircle AOD), + ar(semicircle BOC)}, 1, 1, ⎧, ⎫, = ⎨(10 × 10) − ⎛⎜ × 3.14 × 5 2 + × 3.14 × 5 2 ⎞⎟ ⎬ cm 2, ⎝, ⎠, 2, 2, ⎩, ⎭, = (100 − 78 .5) cm 2 = 21 .5 cm 2 ., Similarly, (area of II) + (area of IV) = 21.5 cm 2 ., Area of shaded region = ar(sq ABCD) – ar(I + II + III + IV), = {(10 × 10) − ( 2 × 21 .5)} cm 2, = (100 − 43) cm 2 = 57 cm 2 ., , EXAMPLE 2, , In the given figure, two circular flower beds have been shown on two, sides of a square lawn ABCD of side AB = 42 m. If the centre of each, circular flower bed is the point of intersection O of the diagonals of, the square lawn, find, (i) the sum of the areas of the lawn and the flower beds,, (ii) the sum of the areas of two flower beds., [CBSE 2015], , SOLUTION, , Area of the square lawn ABCD = ( 42 × 42) m 2 ., Let OA = OB = r metres. Then, ∠AOB = 90 ° ., ∴, , OA 2 + OB2 = AB2 ⇒ r 2 + r 2 = ( 42) 2, ⇒ 2r 2 = 1764 ⇒ r 2 = 882., , (i) Sum of the areas of the lawn and the flower beds, = ar(sector OAEBO) + ar(sector OCFDO), + ar(AOAD) + ar(AOBC), 22, 90, 22, 90 ⎞ 2, ⎞⎟ m 2 + ⎛⎜, ⎟ m, = ⎛⎜, × 882 ×, × 882 ×, ⎝ 7, ⎠, ⎝, 360, 7, 360 ⎠, 1, 1, + ⎛⎜ × r × r ⎞⎟ m 2 + ⎛⎜ × r × r ⎞⎟ m 2, ⎝2, ⎠, ⎝2, ⎠
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694, , Secondary School Mathematics for Class 10, , EXAMPLE 4, , Find the area of the shaded region in the given, figure, where a circular arc of radius 6 cm has, been drawn with vertex O of an equilateral, triangle OAB of side 12 cm as centre., [Take 3 = 1 .73 and π = 3.14.], , SOLUTION, , Since SOAB is equilateral, we have ∠AOB = 60 ° ., Required area, = {(area of equilateral SOAB) + (area of circle with r = 6 cm)}, − (area of sector of a circle with r = 6 cm and θ = 60 °), ⎧⎛ 3, ⎞, 60 ⎞ ⎫, ⎟ ⎬ cm 2, = ⎨⎜, × 12 × 12 ⎟ + ( π × 6 × 6) − ⎛⎜ π × 6 × 6 ×, ⎝, ⎠, 4, 360, ⎝, ⎠, ⎭, ⎩, = {(1 .73 × 36) + 36 π − 6 π} cm 2 = {(1 .73 × 36) + ( 30 π)} cm 2, 314 ⎞ ⎫, 1557 471 ⎞, ⎧ 173, ⎟ ⎬ cm 2 = ⎛⎜, ⎟ cm 2, = ⎨ ⎛⎜, × 36 ⎞⎟ + ⎛⎜ 30 ×, +, ⎝, ⎠, ⎝, ⎠, ⎝, 100 ⎭, 25, 5 ⎠, ⎩ 100, 1557 + 2355 ⎞, 3912, ⎟ cm 2 =, = ⎛⎜, cm 2 = 156.48 cm 2 ., ⎝, ⎠, 25, 25, Hence, the area of the shaded region is 156.48 cm 2 ., , EXAMPLE 5, , In the given figure, AB and CD are the diameters of a circle with, centre O, perpendicular to each other. OA is the diameter of the, smaller circle. If OB = 7 cm, find the area of the shaded region., [CBSE 2010, ’13], , SOLUTION, , Clearly, the diameter of the larger circle is 14 cm and the, diameter of the smaller circle is 7 cm., So, the radius of the larger circle is 7 cm and that of the smaller, circle is 3.5 cm., Area of the shaded region, = {(area of smaller circle) + (area of larger semicircle)}, − (area of SCBD)
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Area of Circle, Sector and Segment, , 695, , 2, ⎤, ⎡⎧, 7 ⎫, 1, 1, = ⎢ ⎨ π × ⎛⎜ ⎞⎟ ⎬ + ⎧⎨ × π × 7 × 7 ⎫⎬ − ⎧⎨ × CD × OB ⎫⎬ ⎥ cm 2, ⎝ 2⎠, ⎭ ⎩2, ⎭ ⎥⎦, ⎭ ⎩2, ⎣⎢ ⎩, , 1, ⎧ 22 49 ⎞ ⎛ 1 22, ⎫, = ⎨ ⎛⎜, × ⎟ +⎜ ×, × 49 ⎞⎟ − ⎛⎜ × 14 × 7 ⎞⎟ ⎬ cm 2, ⎝, ⎠, ⎝, ⎠, ⎝, ⎠, 7, 4, 2, 7, 2, ⎩, ⎭, 77, = ⎛⎜, + 77 − 49 ⎞⎟ cm 2 = ( 38 .5 + 28) cm 2 = 66 .5 cm 2 ., ⎝ 2, ⎠, Hence, the area of the shaded region is 66.5 cm 2 ., EXAMPLE 6, , Calculate the area of the designed region in the given figure, common, between two quadrants of circles of radius 7 cm each., , SOLUTION, , Area of the designed region, = 2[ar(quadrant ABCE) – ar(SABC)], 1, 1, = 2 ⎡ π × (7 ) 2 − × 7 × 7 ⎤ cm 2, 2, ⎣⎢ 4, ⎦⎥, 49 ⎤, ⎡ 1 22, = 2 ⎢ ⎛⎜ ×, × 49 ⎞⎟ − ⎛⎜ ⎞⎟ ⎥ cm 2, ⎝, ⎠, ⎝, 2 ⎠⎦, ⎣ 4 7, 77 49 ⎞, 28, = 2 ⎛⎜, − ⎟ cm 2 = ⎛⎜ 2 × ⎞⎟ cm 2 = 28 cm 2 ., ⎝ 2, ⎠, ⎝, 2, 2⎠, Hence, the area of the designed region is 28 cm 2 ., , EXAMPLE 7, , In a circular table cover of radius 42 cm, a design is formed, leaving, an equilateral triangle ABC in the middle, as shown in the figure., Find the area of the design. [Use 3 = 1.73.], [CBSE 2013C], , SOLUTION, , Let O be the centre of the circle. Join BO and CO.
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Area of Circle, Sector and Segment, , 697, , 1, = æç ´ 6 ´ 8 ö÷ cm 2 = 24 cm 2 ., è2, ø, Let the radius of the incircle be r. Then,, ar(SABC ) = ar( SIAB) + ar( SIBC ) + ( SICA), 1, 1, 1, Þ 24 cm 2 = æç ´ AB ´ r ö÷ + æç ´ BC ´ r ö÷ + æç ´ CA ´ r ö÷, è2, ø è2, ø è2, ø, 1, 1, = r( AB + BC + CA) = r( 6 cm + 10 cm + 8 cm) = (12 cm)r, 2, 2, Þ r = 2 cm., , \ area of the shaded region, = ar(SABC) – (area of incircle with r = 2 cm), = ( 24 - pr 2 ) cm 2 = ( 24 - 3.14 ´ 2 ´ 2) cm 2, = ( 24 - 12 .56) cm 2 = 11.44 cm 2 ., Hence, the required area is 11.44 cm 2 ., EXAMPLE 9, , In an equilateral triangle of side 12 cm, a circle is inscribed touching, its sides. Find the area of the portion of the triangle not included in, the circle. [Take 3 = 1.73 and p = 3.14.], , SOLUTION, , Let ABC be an equilateral triangle of side 12 cm., Let AD ^ BC . Then, D is the midpoint of BC., \, , BD = DC = 6 cm and AB = 12 cm., , \, , AD = AB2 - BD 2 [by Pythagoras’ theorem], = (12) 2 - 6 2 cm = (144 - 36) cm, = 108 cm = 6 3 cm., , Let O be the centre of the inscribed circle., Then, O is the centroid of CABC., 1, 1, \ AO : OD = 2 : 1 and OD = AD = ´ 6 3 cm = 2 3 cm., 3, 3, \, , r = OD = 2 3 cm.
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698, , Secondary School Mathematics for Class 10, , Required area = area of the shaded region, = ar(SABC) − (area of the incircle), ⎡⎛ 3, ⎤, ⎞, = ⎢⎜, × 12 × 12 ⎟ − { 3.14 × ( 2 3 ) 2} ⎥ cm 2, 4, ⎝, ⎠, ⎣, ⎦, = [ 36 3 − ( 3.14 × 12)] cm 2, = ( 36 × 1 .73 − 37 . 68) cm 2, = ( 62.28 − 37 .68) cm 2 = 24.6 cm 2 ., Hence, the required area is 24.6 cm 2 ., EXAMPLE 10, , The area of an equilateral triangle is, 100 3 cm 2 . Taking each vertex as, centre, a circle is described with radius, equal to half the length of the side of the, triangle, as shown in the figure. Find, the area of that part of the triangle, which is not included in the circles., [Take π = 3.14 and 3 = 1.732.], , SOLUTION, , Let each side of the triangle be a cm. Then,, ⎛ 3 2⎞, area of the triangle = ⎜, a ⎟ cm 2 ., ⎝ 4, ⎠, 3 2, a = 100 3 ⇒ a2 = 400 ⇒ a = 20., 4, Thus, the length of each side of CABC is 20 cm., ∴ radius of each circle = 10 cm., Required area, = (area of CABC) – 3(area of a sector with r = 10 cm, θ = 60 °), 60 ⎞, ⎟ cm 2, = ⎛⎜ 100 3 − 3 × 3.14 × 10 × 10 ×, ⎝, 360 ⎠, = {(100 × 1.732) − 157} cm 2 = (173.2 − 157 ) cm 2 = 16.2 cm 2 ., , ∴, , Hence, the area of the required part is 16.2 cm 2 ., EXAMPLE 11, , In the given figure, ABCD is a, trapezium with AB||CD and, ∠BCD = 60 ° . If BFEC is a sector, of a circle with centre C and, AB = BC = 7 cm and DE = 4 cm,, then find the area of the shaded, region.[Use π = 22 7 and 3 = 1.73.], [CBSE 2010]
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Area of Circle, Sector and Segment, SOLUTION, , 699, , Draw BL ^ CD., Clearly, CE = CB = 7 cm., And, CD = CE + DE = 7 cm + 4 cm = 11 cm., From right SCLB, we have, BL, 3, BL, 3, = sin 60 °=, Þ, =, BC, 2, 7 cm, 2, Þ BL =, , 7 ´ 1 .73, 12 .11, 7 3, cm =, cm =, cm = 6.05 cm., 2, 2, 2, , Required shaded area, = ar(trapezium ABCD) – ar(sector BCEFB), ì pr 2 q ü, 1, = ìí ´ ( AB + CD) ´ BL üý - í, ý , where r = CB = 7 cm, þ î 360 þ, î2, 22, 60 ü ù, é 1, 2, = êìí (7 + 11) ´ 6.05 üý - ìí ´ 7 ´ 7 ´, ý cm, 360 þ úû, þ î7, ëî 2, = (54.45 - 25.66) cm 2 = 28 .79 cm 2 ., Hence, the area of the shaded region is 28.79 cm 2 ., EXAMPLE 12, , In the given figure, ABCD is a trapezium in which AB||DC,, AB = 18 cm, DC = 32 cm and the distance between AB and DC is, 14 cm. If arcs of equal radii 7 cm have been drawn with centres A, B, C, and D then find the area of the shaded region., [CBSE 2015, ’17], , SOLUTION, , ar(trap. ABCD) =, , 1, (sum of parallel sides), 2, ´ (distance between them), 1, = ìí (18 + 32) ´ 14 üý cm 2 = 350 cm 2 ., þ, î2, , Sum of the areas of the 4 sectors, = area of a circle of radius 7 cm, 22, = æç ´ 7 ´ 7 ö÷ cm 2 = 154 cm 2 ., è 7, ø, Area of the shaded region = ( 350 - 154) cm 2 = 196 cm 2 ., Hence, the area of the shaded region is 196 cm 2 .
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700, , Secondary School Mathematics for Class 10, , EXAMPLE 13, , Find the area of the shaded region in the figure, given below., [CBSE 2015], [Take π = 3.14.], , SOLUTION, , We have, ar(sq ABCD) = (14 × 14) cm 2 = 196 cm 2 ., Let the radius of each semicircle be a cm., Then, ( a + 2 a + a) = {14 − ( 3 + 3)} cm = 8 cm., ∴ 4 a = 8 ⇒ a = 2., So, the radius of each semicircle = 2 cm., 1, Area of 4 semicircles = ⎛⎜ 4 × × 3.14 × 2 × 2 ⎞⎟ cm 2 = 25.12 cm 2 ., ⎝, ⎠, 2, Length of each side of the smaller square, = 2 a cm = ( 2 × 2) cm = 4 cm., Area of smaller square = ( 4 × 4) cm 2 = 16 cm 2 ., Area of unshaded region = ( 25.12 + 16) cm 2 = 41.12 cm 2, Area of shaded region = (196 − 41.12) cm 2 = 154.88 cm 2, Hence, the required area is 154.88 cm 2 ., , EXAMPLE 14, , Three horses are tethered with 7-metre-long ropes at the three corners, of a triangular field having sides 20 m, 34 m and 42 m. Find the, area of the plot which can be grazed by the horses. Also, find the area, of the plot which remains ungrazed., [CBSE 2009], , SOLUTION, , Let ∠A = θ1° , ∠B = θ 2° and ∠C = θ 3° ., Area which can be grazed by the horses, = sum of the areas of three sectors, with central angles θ1° , θ 2° and θ 3° ,, and each with r = 7 m
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Area of Circle, Sector and Segment, , 701, , ⎛ πr 2θ1 πr 2θ 2 πr 2θ 3 ⎞ 2, =⎜, +, +, ⎟ m , where r = 7 m, 360, 360 ⎠, ⎝ 360, =, , ⎛ πr 2 × 180 ⎞ 2, πr 2, (θ1 + θ 2 + θ 3 ) m 2 = ⎜, ⎟ m, 360, ⎝ 360 ⎠, [B θ1 + θ 2 + θ 3 = ∠A + ∠B + ∠C = 180 ° ], , 22, 1, = ⎛⎜, × 7 × 7 × ⎞⎟ m 2 = 77 m 2 ., ⎝ 7, 2⎠, Sides of the plot are a = 20 m, b = 34 m, c = 42 m., 1, ∴ s = ( a + b + c) = 48 m , ( s − a) = 28 m , ( s − b) = 14 m and, 2, ( s − c) = 6 m., ∴ area of the whole plot = area of CABC, = s( s − a)( s − b)( s − c) sq units, = 48 × 28 × 14 × 6 m 2 = 336 m 2 ., Area ungrazed = ( 336 − 77 ) m 2 = 259 m 2 ., EXAMPLE 15, , The inside perimeter of a running track, as shown in the figure, is, 340 m. The length of each straight portion is 60 m, and the curved, portions are semicircles. If the track is, 7 m wide, find the area of the track., Also, find the outer perimeter of the, track., [CBSE 2008C], , SOLUTION, , Length of inner curved portion = ( 340 − 2 × 60) m = 220 m., ∴ the length of each inner curved part = 110 m., Let the radius of each inner curved part be r., 22, Then, πr = 110 m ⇒, × r = 110 m, 7, 7 ⎞, ⎟ m = 35 m., ⇒ r = ⎛⎜ 110 ×, ⎝, 22 ⎠, ∴ inner radius = 35 m, outer radius = ( 35 + 7 ) m = 42 m., ∴ area of the track, = (area of 2 rectangles, each 60 m × 7 m), + (area of the circular ring with R = 42 m, r = 35 m), 22, ⎡, ⎤, = ⎢( 2 × 60 × 7 ) +, × {( 42) 2 − ( 35) 2} ⎥ m 2, 7, ⎣, ⎦, 22, ⎡, ⎤, = 840 +, × ( 42 + 35)( 42 − 35) m 2, 7, ⎣⎢, ⎦⎥
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702, , Secondary School Mathematics for Class 10, , = ( 840 + 1694) m 2 = 2534 m 2 ., Length of the outer boundary of the track, 22, = æç 2 ´ 60 + 2 ´, ´ 42 ö÷ m = 384 m., 7, è, ø, Hence, the outer perimeter of the track is 384 m., EXAMPLE 16, , In the given figure, SABC is a, right-angled triangle with ÐB = 90 °,, AB = 48 cm and BC = 14 cm. With AC as, diameter a semicircle is drawn and with, BC as radius, a quadrant of a circle is, drawn. Find the area of the shaded region., [Use p = 227 .], , SOLUTION, , [CBSE 2009C], , In right SABC , we have, AC 2 = AB2 + BC 2 = [( 48) 2 + (14) 2 ] cm 2, = ( 2304 + 196) cm 2 = 2500 cm 2, 50, cm = 25 cm., Þ AC = 2500 cm = 50 cm Þ R =, 2, Area of the shaded region, = ar( SABC ) + (area of semicircle on AC ), - ar(quadrant BCD), 1, 1, 1, æ, ö, = ç ´ 14 ´ 48 ÷ + p ´ 25 ´ 25 - p ´ BC 2, 4, è2, ø 2, 1 22, 1 22, é, = 336 + ´, ´ 625 - ´, ´ 14 ´ 14 ù cm 2, êë, úû, 2 7, 4 7, 6875, = æç 336 +, - 154 ö÷ cm 2 = (182 + 982.14) cm 2, 7, è, ø, = 1164.14 cm 2 ., Hence, the area of the shaded region is 1164.14 cm 2 ., , EXAMPLE 17, , Find the area of the shaded region of the figure given below., , [CBSE 2011]
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704, EXAMPLE 19, , Secondary School Mathematics for Class 10, , In the given figure, two concentric circles with centre O, have radii, 21 cm and 42 cm. If ∠AOB = 60 ° find the area of the shaded region., [CBSE 2014], , SOLUTION, , ar(region ACDB) = ar(sector AOB) – ar(sector COD), 60 ⎞ ⎛ 22, 60 ⎞ ⎫, ⎧ 22, ⎟ −⎜, ⎟ ⎬ cm 2, = ⎨ ⎛⎜, × 42 × 42 ×, × 21 × 21 ×, ⎝, ⎠, ⎝, 360, 7, 360 ⎠ ⎭, ⎩ 7, = ( 924 − 231) cm 2 = 693 cm 2 ., 22, ⎧ 22, ⎫, ar(circular ring) = ⎨ ⎛⎜, × 42 × 42 ⎞⎟ − ⎛⎜, × 21 × 21 ⎞⎟ ⎬ cm 2, ⎠ ⎝ 7, ⎠⎭, ⎩⎝ 7, = (5544 − 1386) cm 2 = 4158 cm 2 ., Required area = ar(circular ring) – ar(region ACDB), = ( 4158 − 693) cm 2 = 3465 cm 2 ., Hence, the area of the shaded region is 3465 cm 2 ., , EXAMPLE 20, , In the given figure, SABC is right-angled at A with AB = 6 cm and, AC = 8 cm. A circle with centre O has been inscribed inside the, triangle. Find the value of r, the radius of the inscribed circle., , SOLUTION, , BC = AB2 + AC 2 = 6 2 + 8 2, =, , 36 + 64 = 100 = 10 cm., , Let O be the centre of the inscribed circle and r be its radius., Join OA , OB and OC.
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Area of Circle, Sector and Segment, , 705, , ar (C OAC ) + ar (C OAB ) + ar (C BOC ) = ar (C ABC ), 1, 1, 1, 1, ⇒ ⎛⎜ × 8 × r ⎞⎟ + ⎛⎜ × 6 × r ⎞⎟ + ⎛⎜ × 10 × r ⎞⎟ = × 6 × 8 ., ⎝2, ⎠ ⎝2, ⎠ ⎝2, ⎠ 2, ⇒ ( 4r + 3r + 5r) = 24 ⇒ 12r = 24 ⇒ r = 2 cm., Hence, r = 2 cm., , EXERCISE 16A, Use π =, , 22, , unless stated otherwise., 7, , 1. The circumference of a circle is 39.6 cm. Find its area., 2. The area of a circle is 98.56 cm 2 . Find its circumference., 3. The circumference of a circle exceeds its diameter by 45 cm. Find the, circumference of the circle., 4. A copper wire when bent in the form of a square encloses an area of, 484 cm 2 . The same wire is now bent in the form of a circle. Find the area, enclosed by the circle., 5. A wire when bent in the form of an equilateral triangle encloses an area, of 121 3 cm 2 . The same wire is bent to form a circle. Find the area, enclosed by the circle., 6. The length of a chain used as the boundary of a semicircular park is, 108 m. Find the area of the park., 7. The sum of the radii of two circles is 7 cm, and the difference of their, circumferences is 8 cm. Find the circumferences of the circles., 8. Find the area of a ring whose outer and inner radii are respectively, 23 cm and 12 cm., 9., , (i) A path of 8 m width runs around the outside of a circular park, whose radius is 17 m. Find the area of the path., (ii) A park is of the shape of a circle of diameter 7 m. It is surrounded, by a path of width of 0.7 m. Find the expenditure of cementing the, path, if its cost is ` 110 per sq m., [CBSE 2017], , 10. A racetrack is in the form of a ring whose inner circumference is 352 m, and outer circumference is 396 m. Find the width and the area of, the track., 11. A sector is cut from a circle of radius 21 cm. The angle of the sector is, 150°. Find the length of the arc and the area of the sector.
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706, , Secondary School Mathematics for Class 10, , 12. A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the, centre of the circle. Find the area of major and minor segments of, [CBSE 2017], the circle., 13. The length of an arc of a circle, subtending an angle of 54° at the centre, is 16.5 cm. Calculate the radius, circumference and area of the circle., 14. The radius of a circle with centre O is 7 cm. Two radii OA and OB are, drawn at right angles to each other. Find the areas of minor and major, segments., 15. Find the lengths of the arcs cut off from a circle of radius 12 cm by a, chord 12 cm long. Also, find the area of the minor segment., [Take π = 3.14 and, , 3 = 1 .73.], , 16. A chord 10 cm long is drawn in a circle whose radius is 5 2 cm. Find the, areas of both the segments. [Take π = 3.14.], 17. Find the area of both the segments of a circle of radius 42 cm with, 3, and 3 = 1 .73.], central angle 120°. [Given, sin 120 ° =, 2, 18. A chord of a circle of radius 30 cm makes an angle of 60° at the centre of, the circle. Find the areas of the minor and major segments. [Take, π = 3.14 and 3 = 1 .732.], 19. In a circle of radius 10.5 cm, the minor arc is one-fifth of the major arc., Find the area of the sector corresponding to the major arc., 20. The short and long hands of a clock are 4 cm and 6 cm long respectively., Find the sum of distances travelled by their tips in 2 days., [Take π = 3.14.], 21. Find the area of a quadrant of a circle whose circumference is 88 cm., 22. A rope by which a cow is tethered is increased from 16 m to 23 m. How, much additional ground does it have now to graze?, 23. A horse is placed for grazing inside a rectangular field 70 m by 52 m., It is tethered to one corner by a rope 21 m long. On how much area can it, graze? How much area is left ungrazed?, 24. A horse is tethered to one corner of a field which is in the shape of an, equilateral triangle of side 12 m. If the length of the rope is 7 m, find the, area of the field which the horse cannot graze. Take 3 = 1.732. Write, the answer correct to 2 places of decimal., 25. Four cows are tethered at the four corners of a square field of side 50 m, such that each can graze the maximum unshared area. What area will, be left ungrazed? [Take π = 3.14.]
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Area of Circle, Sector and Segment, , 707, , 26. In the given figure, OPQR is a rhombus, three of, whose vertices lie on a circle with centre O. If the, area of the rhombus is 32 3 cm 2 , find the radius, of the circle., 27. The side of a square is 10 cm. Find (i) the area of the inscribed circle, and, (ii) the area of the circumscribed circle. [Take π = 3.14.], 28. If a square is inscribed in a circle, find the ratio of the areas of the circle, and the square., 29. The area of a circle inscribed in an equilateral triangle is 154 cm 2 ., Find the perimeter of the triangle. [Take 3 = 1.73.], 30. The radius of the wheel of a vehicle is 42 cm. How many revolutions, will it complete in a 19.8-km-long journey?, 31. The wheels of the locomotive of a train are 2.1 m in radius. They make, 75 revolutions in one minute. Find the speed of the train in km per hour., 32. The wheels of a car make 2500 revolutions in covering a distance of, 4.95 km. Find the diameter of a wheel., 33. A boy is cycling in such a way that the wheels of his bicycle are making, 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate, the speed (in km/hr) at which the boy is cycling., 34. The wheel of a motorcycle is of radius 35 cm. How many revolutions, per minute must the wheel make so as to keep a speed of 66 km/hr?, 35. The diameters of the front and rear wheels of a tractor are 80 cm and 2 m, respectively. Find the number of revolutions that a rear wheel makes to, cover the distance which the front wheel covers in 800 revolutions., 36. Four equal circles are described about the four, corners of a square so that each touches two of, the others, as shown in the figure. Find the area, of the shaded region, if each side of the square, [CBSE 2007], measures 14 cm., 37. Four equal circles, each of radius 5 cm,, touch each other, as shown in the figure., Find the area included between them., [Take π = 3.14.], , 38. Four equal circles, each of radius a units, touch each other. Show that, 6, the area between them is ⎛⎜ a2 ⎞⎟ sq units., ⎝7 ⎠
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708, , Secondary School Mathematics for Class 10, , 39. Three equal circles, each of radius 6 cm,, touch one another as shown in the figure., Find the area enclosed between them., [Take π = 3.14 and 3 = 1.732.], 40. If three circles of radius a each, are drawn such that each touches the, 4 2, other two, prove that the area included between them is equal to, a ., 25, [Take 3 = 1.73 and π = 3.14.], 41. In the given figure, ABCD is a trapezium of area 24.5 cm 2 . If AD|| BC ,, ∠DAB = 90 ° , AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region., [CBSE 2014], D, , E, , C, , A, , B, , 42. ABCD is a field in the shape of a trapezium,, AD|| BC , ∠ABC = 90 ° and ∠ADC = 60 °., Four sectors are formed with centres A, B, C, and D, as shown in the figure. The radius of, each sector is 14 m. Find the following:, (i) total area of the four sectors,, (ii) area of the remaining portion, given that AD = 55 m, BC = 45 m, [CBSE 2013C], and AB = 30 m., 43. Find the area of the shaded region in the given, figure, where a circular arc of radius 6 cm has, been drawn with vertex of an equilateral, triangle of side 12 cm as centre and a sector, of circle of radius 6 cm with centre B is, made. [Use 3 = 1 .73 and π = 3 .14.], [CBSE 2014], , 44. In the given figure, ABCD is a rectangle with, AB = 80 cm and BC = 70 cm, ∠AED = 90 °, and DE = 42 cm. A semicircle is drawn,, taking BC as diameter. Find the area of the, [CBSE 2014], shaded region.
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Area of Circle, Sector and Segment, , 709, , 45. In the given figure, from a rectangular region, ABCD with AB = 20 cm, a right triangle AED, with AE = 9 cm and DE = 12 cm, is cut off., On the other end, taking BC as diameter,, a semicircle is added on outside the, region. Find the area of the shaded region., [Use π = 3 .14.], [CBSE 2014], 46. In the given figure, O is the centre of the, circle with AC = 24 cm, AB = 7 cm and, ∠BOD = 90 ° . Find the area of shaded region., [Use π = 3 .14.], [CBSE 2012, ’17], , 47. In the given figure, a circle is inscribed in an, equilateral triangle ABC of side 12 cm. Find the, radius of inscribed circle and the area of the, shaded region. [Use 3 = 1 .73 and π = 3 .14.], [CBSE 2014], , 48. On a circular table cover of radius 42 cm, a, design is formed by a girl leaving an equilateral, triangle ABC in the middle, as shown in the, figure. Find the covered area of the design., 22, [Use 3 = 1 .73 and π =, [CBSE 2013C], ⋅], 7, 49. The perimeter of the quadrant of a circle is 25 cm. Find its area., [CBSE 2012], , 50. A chord of a circle of radius 10 cm subtends a right angle at the centre., Find the area of the minor segment. [Use π = 3.14.], [CBSE 2012], 51. The radius of a circular garden is 100 m. There is a road 10 m wide,, running all around it. Find the area of the road and the cost of levelling, it at ` 20 per m 2 . [Use π = 3.14.], [CBSE 2011], 52. The area of an equilateral triangle is 49 3 cm 2 . Taking each angular, point as centre, circles are drawn with radius equal to half the length of, the side of the triangle. Find the area of the triangle not included in the, circles. [Take 3 = 1.73.], [CBSE 2009]
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710, , Secondary School Mathematics for Class 10, , 53. A child draws the figure of an aeroplane as, shown. Here, the wings ABCD and FGHI are, parallelograms, the tail DEF is an isosceles, triangle, the cockpit CKI is a semicircle and, CDFI is a square. In the given figure, BP ⊥ CD ,, and EL ⊥ DF. If CD = 8 cm,, HQ ⊥ FI, BP = HQ = 4 cm and DE = EF = 5 cm, find the, area of the whole figure. [Take π = 3.14.], 54. A circular disc of radius 6 cm is divided into three sectors with central, angles 90° , 120° and 150°. What part of the whole circle is the sector with, central angle150°? Also, calculate the ratio of the areas of the three sectors., 55. A round table cover has six equal designs, as shown in the given figure. If the radius, of the cover is 35 cm then find the total, area of the design. [Use 3 = 1.732 and, [CBSE 2009, ’14], π = 3.14.], 56. In the given figure, PQ = 24 cm, PR = 7 cm, and O is the centre of the circle. Find the, area of the shaded region. [Take π = 3.14.], [CBSE 2009], , 57. In the given figure, SABC is right-angled at A. Find the area of the, shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the, [CBSE 2009], incircle of SABC. [Take π = 3.14.], , 58. In the given figure, SABC is right-angled, at A. Semicircles are drawn on AB, AC and, BC as diameters. It is given that AB = 3 cm, and AC = 4 cm. Find the area of the, [CBSE 2017], shaded region.
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Area of Circle, Sector and Segment, , 711, , 59. PQRS is a diameter of a circle of radius, 6 cm. The lengths PQ, QR and RS are, equal. Semicircles are drawn with PQ and, QS as diameters, as shown in the given, figure. If PS = 12 cm, find the perimeter, and area of the shaded region., [Take π = 3.14.], 60. The inside perimeter of a running track shown in the figure is 400 m., The length of each of the straight portions is, 90 m, and the ends are semicircles. If the, track is 14 m wide everywhere, find the, area of the track. Also, find the length of the, outer boundary of the track., 61. In the given figure, ABCD is a rectangle of, dimensions 21 cm × 14 cm. A semicircle is, drawn with BC as diameter. Find the area, and the perimeter of the shaded region in, [CBSE 2017], the figure., , 21 cm, , A, , B, , 14 cm, , D, , C, , 62. In the given figure, two concentric circles, with centre O have radii 21 cm and 42 cm., If ∠AOB = 60 °, find the area of the shaded, 22, [CBSE 2017], region. [Use π =, ⋅], 7, 63. Three semicircles each of diameter 3 cm,, a circle of diameter 4.5 cm and a, semicircle of radius 4.5 cm are drawn in, the given figure. Find the area of the, [CBSE 2017], shaded region., 3 cm, , 64. In the given figure, the side of square is, 28 cm and radius of each circle is half of, the length of the side of the square, where, O and O′ are centres of the circles. Find, [CBSE 2017], the area of shaded region., , O, , 3 cm, , 3 cm
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712, , Secondary School Mathematics for Class 10, , 65. With the vertices A, B and C of a triangle ABC, as centres, arcs are drawn with radii 5 cm each, as shown in the given figure. If AB = 14 cm,, BC = 48 cm and CA = 50 cm then find the, area of the shaded region. [Use π = 3.14.], , 66. If the diameters of the concentric circles, shown in the figure below are in the ratio 1 : 2 :, 3 then find the ratio of the areas of three, regions., , ANSWERS (EXERCISE 16A), , 1. 124.74 cm 2, , 2. 35.2 cm, , 3. 66 cm, , 5. 346.5 cm 2, , 6. 693 m 2, , 7. 26 cm, 18 cm 8. 1210 cm 2, , 9. (i) 1056 m (ii) ` 1863.40, 2, , 11. 55 cm, 577.5 cm 2, , 4. 616 cm 2, , 10. 7 m, 2618 m 2, , 12. 9.08 cm 2 , 305.2 cm 2, , 13. 17.5 cm, 110 cm, 962.5 cm 2, , 14. 14 cm 2 , 140 cm 2, , 15. (12.56 cm, 62.8 cm), 13.08 cm, , 16. 14.25 cm 2 , 142.75 cm 2, , 17. 1085.07 cm 2 , 4458.93 cm 2, , 18. 81.3 cm 2 , 2744.7 cm 2, , 19. 288.75 cm 2, , 20. 1909.12 cm 2 21. 154 cm 2, , 23. 346.5 m 2 , 3293.5 m 2, 2, , 26. 8 cm 27. (i) 78.5 cm (ii) 157 cm, 30. 7500, 35. 320, , 2, , 31. 59.4 km/hr 32. 63 cm, 36. 42 cm, , 2, , 46. 283.96 cm, , 25. 537.5 m 2, , 28. π : 2, , 29. 72.66 cm, , 33. 15.84 km/hr, , 37. 21.5 cm 39. 5.76 cm, 43. 137.64 cm 2, , 2, , 34. 500, , 41. 14.875 cm 2, , 44. 2499 cm 2, , 45. 334.31 cm 2, , 47. r = 2 3 cm, ar(shaded region) = 24.6 cm 2, , 2, , 48. 3255.21 cm 2, , 49. 38.5 cm 2, , 51. 6594 m 2 , ` 131880, , 52. 7.77 cm 2, , 55. 663.95 cm 2, , 50. 28.5 cm 2, 53. 65.12 cm 2, , 56. 161.31 cm 2 57. 11.44 cm 2, , 59. 37.68 cm, 37.68 cm, 62. 3465 cm 2, , 24. 36.69 m 2, , 2, , 42. (i) 616 m 2 (ii) 884 m 2, , 22. 858 m 2, , 63. 12, , 2, , 2, , 60. 6216 m , 488 m, , 3, cm 2, 8, , 64. 1708 cm 2, , 54., , 5, , 3 : 4 :5, 12, , 58. 6 cm 2, , 61. 217 cm 2 , 78 cm, 65. 296.75 cm 2, , 66. 1 : 3 : 5
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Area of Circle, Sector and Segment, , 713, , HINTS TO SOME SELECTED QUESTIONS, 22, 3. 2 πR − 2 R = 45 ⇒ 2 R( π − 1) = 45 ⇒ 2 R ⎛⎜, − 1⎞⎟ = 45 ⇒ R = 10 . 5 cm., ⎝7, ⎠, 4. a2 = 484 ⇒ a = 484 = 22 cm., Length of the wire = 4 a = ( 4 × 22 ) cm = 88 cm., 2 πR = 88., 5., , 3 2, a = 121 3 ⇒ a2 = 484 ⇒ a = 484 = 22 cm., 4, Length of the wire = 3 a = ( 3 × 22 ) cm = 66 cm., 2 πR = 66., , 22, 6. πR + 2 R = 168 ⇒ ⎛⎜, + 2 ⎞⎟ R = 168 ⇒ R = 21 cm., ⎝7, ⎠, Area of the park =, , 1, πR 2 ., 2, , 7. Let the radii of the circles be x cm and (7 − x ) cm., Then, 2 πx − 2 π(7 − x ) = 8 ⇒ 2 πx = 26., Substitute this value of 2πx and find 2 π(7 − x )., 9. (i) R 1 = 17 m and R 2 = ( 17 + 8 ) m = 25 m., Area of the path = π{( 25 ) 2 − ( 17 ) 2 } m 2 ., 22, 54, 13. 2 ×, ×r ×, = 16.5. Find r., 7, 360, 22, 90, 1, 14. Area of the minor segment = ⎧⎨ × 7 × 7 ×, − × 7 × 7 × sin 90 ° ⎫⎬ cm 2 = 14 cm 2 ., 360 2, ⎩7, ⎭, 22, Area of the major segment = ⎧⎨ ⎛⎜ × 7 × 7 ⎞⎟ − 14 ⎫⎬ cm 2 = 140 cm 2 ., ⎠, ⎩⎝ 7, ⎭, 15. SOAB is equilateral. So, ∠AOB = 60 ° ., 2 πRθ ⎞ ⎛ 2 π × 12 × 60 ⎞, arc ACB = ⎛⎜, ⎟=⎜, ⎟ cm, ⎝ 360 ⎠ ⎝, ⎠, 360, = 4 π cm = ( 4 × 3.14 ) cm = 12.56 cm., arc BDA = ( 2πr − arc ACB), = ( 2 × 3.14 × 12 − 12.56 ) cm = 62.8 cm., ar(minor segment ACBA) = ar(sector OACBO) – ar(SOAB), ⎛ πr 2θ, 3 2⎞, =⎜, −, a ⎟, 360, 4, ⎝, ⎠, ⎧ 3.14 × 12 × 12 × 60, =⎨, −, 360, ⎩, , 3 × 12 × 12 ⎫, 2, ⎬ cm, 4, ⎭, , = (75. 36 − 62 . 28 ) cm 2 = 13.08 cm 2 ., 16. Let OA = 5 2 cm,OB = 5 2 cm and AB = 10 cm., Then, OA 2 + OB 2 = AB 2 ⇒ ∠AOB = 90 °.
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714, , Secondary School Mathematics for Class 10, ar(minor segment ACBA), = ar(sector OACBO) – ar(SOAB), ⎧ 3.14 × (5 2 ) 2 × 90 1, ⎫, =⎨, − × 5 2 × 5 2 ⎬ cm 2, 360, 2, ⎩, ⎭, 57, 2, 2, =, cm = 14.25 cm ., 4, ar(major segment) = { 3.14 × (5 2 ) 2 − 14.25} cm 2 = ( 157 − 14.25 ) cm 2, = 142 . 75 cm 2 ., , 17. ar(minor segment ACBA), = ar(sector OACBO) – ar(SOAB), 2, ⎛ πr θ 1 2, ⎞, =⎜, − r sin θ⎟ = ( 1848 − 762 . 93 ) cm 2, ⎝ 360 2, ⎠, , = 1085.07 cm 2 ., ⎤, ⎡ 22, ar(major segment) = ⎢ ⎛⎜ × 42 × 42 ⎞⎟ − 1085.07 ⎥ cm 2 ., ⎠, ⎦, ⎣⎝ 7, 18. ar(minor segment ACBA), 2, ⎛ πr θ 1 2, ⎞, = ar(sector OACBO) – ar(SOAB) = ⎜, − r sin θ⎟, ⎝ 360 2, ⎠, , = ⎛⎜, ⎝, , 3.14 × 30 × 30 × 60 1, − × 30 × 30 × sin 60 ° ⎞⎟, ⎠, 360, 2, , = ( 471 − 389 . 7 ) = 81. 3 cm 2 ., ar(major segment BDAB) = ( πR 2 − 81. 3 ) cm 2 = ( 2826 − 81. 3 ) = 2744.7 cm 2 ., 19. Let the length of the major arc be x cm. Then,, x, the length of the minor arc = cm., 5, x, 6x, Circumference of the circle = ⎛⎜ x + ⎞⎟ cm =, cm., ⎝, 5⎠, 5, ∴, , 6x ⎛, 22 21 ⎞, = ⎜2×, × ⎟ ⇒ x = 55., 5 ⎝, 7, 2⎠, , Length of the major arc = 55 cm., 1, 1 21, Area of the sector corresponding to major arc = ⎛⎜ rl ⎞⎟ = ⎛⎜ ×, × 55 ⎞⎟ cm 2 ., ⎝2 ⎠ ⎝2 2, ⎠, 20. In 2 days, the short hand will complete 4 rounds., ∴, , length moved by it = 4(circumference of the circle with r = 4 cm), = ( 4 × 2 π × 4 ) cm = 32 π cm ., , In 2 days, the long hand will complete 48 rounds., ∴, , length moved by it = 48(circumference of the circle with r = 6 cm), , ∴, , sum of the lengths moved = ( 32 π + 576 π ) cm = ( 608 π ) cm., , = ( 48 × 2 π × 6 ) cm = 576 π cm.
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Area of Circle, Sector and Segment, 22, × R = 88 ⇒ R = 14 cm., 7, 1, Required area = ⎛⎜ πR 2 ⎞⎟ cm 2 = 154 cm 2 ., ⎝4, ⎠, , 21. 2 πR = 88 ⇒ 2 ×, , 22. Additional ground = π × [( 23 ) 2 − ( 16 ) 2 ] m 2 ., 23. Area which can be grazed, = area of the quadrant of radius 21 m, 1 22, = ⎛⎜ ×, × 21 × 21⎞⎟ m 2 = 346.5 m 2 ., ⎝4 7, ⎠, Area ungrazed = [(70 × 52 ) − 346.5] m 2 ., 24. Each angle of an equilateral triangle is 60°., Area which cannot be grazed, = (area of the equilateral C ABC), – (area of the sector with r = 7 m,θ = 60 °), ⎡ 3, 60 ⎤ 2 ( 108 3 − 77 ) 2, 22, =⎢, × ( 12 ) 2 −, × (7 ) 2 ×, m, ⎥m =, 360, 3, 7, 4, ⎣, ⎦, = 36.69 m 2 ., 25. Ungrazed area = shaded area, ⎡, π × ( 25 ) 2 × 90 ⎤ 2, = ⎢(50 × 50 ) − 4 ×, ⎥m ., 360, ⎣, ⎦, = [2500 − ( 625 × 3.14 )] m 2, = ( 2500 − 1962 . 5 ) m 2 = 537 . 5 m 2 ., 26. OP = OR = OQ = r ., Let OQ and PR intersect at S., Since diagonals of a rhombus bisect each other at right angles, we have, 1, OS = r and ∠OSR = 90 ° ., 2, ∴, , SR = OR 2 − OS2 = r 2 −, , ∴, , PR = 2 × SR =, , 3r, ⋅, 2, , 3 r., , Area of the rhombus =, ∴, , r2, =, 4, , 1, 1, × OQ × PR = × r ×, 2, 2, , 3r =, , 3 2, r ., 2, , 3 2, r = 32 3 ⇒ r = 8 cm., 2, , 27. Diameter of the inscribed circle = side of the square = 10 cm., ∴ radius of the inscribed circle = 5 cm., Diameter of the circumscribed circle = diagonal of the square, = ( 2 × 10 ) cm., ∴, , radius of the circumscribed circle = 5 2 cm., , 715
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716, , Secondary School Mathematics for Class 10, , 28. Let the radius of the circle be r cm., Then, diagonal of the square = diameter of the circle = 2r cm., Area of the circle = ( πr 2 ) cm 2 ., 1, 1, Area of the square = × (diagonal) 2 = × 4 r 2 = ( 2 r 2 ) cm 2 ., 2, 2, 29. Let the radius of the incircle be r cm., 7, Then, πr 2 = 154 ⇒ r 2 = ⎛⎜ 154 × ⎞⎟ ⇒ r = 7 cm., ⎝, 22 ⎠, Let each side of the triangle be a cm and its height be h cm., h, Then, r = ⇒ h = 3r = 21 cm., 3, h = a2 −, , a2, =, 4, , 3a, ⇒, 2, , 3a, = 21., 2, , 3, = 14 3., 3, 22, 30. Circumference = ⎛⎜ 2 ×, × 42 ⎞⎟ cm = 264 cm., ⎝, ⎠, 7, ∴, , a=, , 42, ×, 3, , Required number of revolutions = ⎛⎜, ⎝, , 19.8 × 1000 × 100 ⎞, ⎟ = 7500., ⎠, 264, , 22, 66, 31. Circumference = ⎛⎜ 2 ×, × 2 . 1⎞⎟ m =, m., ⎝, ⎠, 5, 7, 66, Distance covered in 60 minutes = ⎛⎜ × 75 × 60 ⎞⎟ m,, ⎝ 5, ⎠, 66, 1 ⎞, speed in km/hr = ⎛⎜ × 75 × 60 ×, ⎟ km/hr., ⎝ 5, 1000 ⎠, 32. Distance covered in 1 revolution = ⎛⎜, ⎝, 2 πR = 198 ⇒ 2 R ×, , 4.95 × 1000 × 100 ⎞, ⎟ cm = 198 cm., ⎠, 2500, , 22, 7, = 198 ⇒ 2 R = ⎛⎜ 198 × ⎞⎟ = 63 cm., ⎝, 22 ⎠, 7, , 33. Number of revolutions per hour = ( 140 × 60 ) = 8400., 22, Distance covered in 1 revolution = 2 πR = ⎛⎜ × 60 ⎞⎟ cm., ⎝7, ⎠, Distance covered in 1 hour = distance covered in 8400 revolutions, 22, 1, 1 ⎞, = ⎛⎜ × 60 × 8400 ⎞⎟ cm = ⎛⎜ 22 × 72000 ×, ×, ⎟ km, ⎝7, ⎠, ⎝, 100 1000 ⎠, =, , 1584, km = 15.84 km = speed per hr., 100, , 35. Radius of the front wheel = 40 cm =, , 2, m., 5, , Circumference of the front wheel = ⎛⎜ 2 π ×, ⎝, , 2⎞, 4π, m., ⎟ m=, 5⎠, 5
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Area of Circle, Sector and Segment, , 719, , Required area = (area of the circle) – {ar(SBAC) + ar(quadrant COD)}, 25 25 ⎞ ⎫, 1, 25 25 ⎞ ⎧ ⎛ 1, = ⎛⎜ 3.14 ×, × ⎟ − ⎨ ⎜ × 24 × 7 ⎞⎟ + ⎜⎛ × 3.14 ×, × ⎟⎬, ⎝, ⎠ ⎝4, 2, 2 ⎠⎭, 2, 2 ⎠ ⎩⎝ 2, 25 25 ⎞, 3, = ⎧⎨ ⎛⎜ × 3.14 ×, × ⎟ − 84 ⎫⎬ cm 2 = ( 367 .96 − 84 ) cm 2 = 283.96 cm 2 ., 2, 2⎠, ⎩⎝ 4, ⎭, 47. Draw AOD⊥ BC . Then, D is the midpoint of BC., ∴, , AC = 12 cm and DC = 6 cm., AD 2 = AC 2 − DC 2 = ( 12 ) 2 − 6 2 = 108 ⇒ AD = 108 = 6 3 cm., , ∴, , h = 6 3 cm. Now, 3r = h = 6 3 ⇒ r = 2 3 cm., , So, the radius of inscribed circle is 2 3 cm., ⎤, ⎡⎛ 3, ⎞, ar(shaded region) = ⎢ ⎜, × 12 × 12 ⎟ − { 3.14 × ( 2 3 ) 2 }⎥ cm 2, 4, ⎝, ⎠, ⎦, ⎣, = [( 36 × 1. 73 ) − ( 12 × 3.14 )] cm 2, = ( 62.28 − 37 .68 ) cm 2, = 24.6 cm 2 ., Note h = 3r in case of incircle., 48. Let O be the centre of the circumcircle. Join OB and draw OD ⊥ BC . Then, OB = 42 cm, and ∠OBD = 30 ° ., OD, 1, OD 1, = sin 30 ° = ⇒, = ⇒ OD = 21 cm., OB, 2, 42 2, BD 2 = OB 2 − OD 2 = ( 42 ) 2 − ( 21) 2, = ( 42 + 21)( 42 − 21) = 63 × 21., ∴, , BD = 63 × 21 = 21 3 cm ⇒ BC = 2 × 21 3 = 42 3 cm., , Area of the shaded region, = (area of a circle with r = 42 cm) – ar(equilateral SABC with a = 42 3 cm), 173, 22, ., = ⎧⎨ ⎛⎜ × 42 × 42 ⎞⎟ −, × 42 3 × 42 3 ⎫⎬ cm 2, ⎠, 4, ⎩⎝ 7, ⎭, = (5544 × 2288 . 79 ) cm 2 = 3255.21 cm 2 ., Note : a = r 3 in case of circumcircle., 2 πr × 90 ⎞, 25r, 49. Perimeter of the quadrant = ⎛⎜ 2r +, cm., ⎟ cm =, ⎝, 360 ⎠, 7, 25r, = 25 ⇒ r = 7 ., ∴, 7, 1, 1 22, 77, Area of the quadrant = πr 2 = ⎛⎜ ×, × 7 × 7 ⎞⎟ cm 2 =, cm 2 = 38 . 5 cm 2 ., ⎝4 7, ⎠, 4, 2, 50. ar(minor segment), = ar(sector OACBO) – ar( SOAB), 90 ⎞ ⎛ 1, 2, ⎞⎫, = ⎧⎨ ⎛⎜ 3.14 × 10 × 10 ×, ⎟ − ⎜ × 10 × 10 ⎟ ⎬ cm, ⎠⎭, 360 ⎠ ⎝ 2, ⎩⎝, = (78.5 − 50 ) cm 2 = 28 . 5 cm 2 .
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Area of Circle, Sector and Segment, radius of the circle (r ) =, , ∴, , Area of the shaded region, , 721, , 25, cm., 2, , 25, = ar ⎛⎜ semicircle with r = ⎞⎟ − ar( SPQR ), ⎝, 2⎠, 25, 25, 1, 1, = ⎧⎨ ⎛⎜ × 3.14 ×, × ⎞⎟ − ⎛⎜ × 7 × 24 ⎞⎟ ⎫⎬ cm 2, ⎠⎭, 2, 2 ⎠ ⎝2, ⎩⎝ 2, = ( 245. 31 − 84 ) cm 2 = 161. 31 cm 2 ., 57. AC = BC 2 − AB 2 = 100 − 36 = 64 ⇒ AC = 8 cm., 2, , Draw OD ⊥ AB, OE ⊥ AC and OF ⊥ CB., Let OD = OE = OF = r cm., Then, AEOD is a square such that AD = AE = r ., [Q tangents to a circle from a point are equal.], CF = CE = ( 8 − r ) cm and BF = BD = ( 6 − r ) cm., BF + CF = 10 ⇒ ( 6 − r ) + ( 8 − r ) = 10 ⇒ r = 2., ar(shaded region) = ar(SABC ) – (area of the incircle), , 1, 1, = ⎧⎨ × AB × AC − π × 2 2 ⎫⎬ cm 2 = ⎛⎜ × 6 × 8 − 3.14 × 4 ⎞⎟ cm 2 = 11. 44 cm 2 ., ⎝2, ⎠, ⎩2, ⎭, 58. Area of the shaded region, = {ar ( SABC ) + ar(semicircle APB ) + ar(semicircle AQC )} – ar(semicircle BAC), 1, 5 5 ⎤, 1, 3 3, 1, ⎡ 1, = ⎢ ⎛⎜ × 3 × 4 ⎞⎟ + ⎛⎜ π × × ⎞⎟ + ⎛⎜ π × 2 × 2 ⎞⎟ − ⎛⎜ π × × ⎞⎟ ⎥ cm 2, ⎠ ⎝2, ⎠ ⎝2, 2 2⎠ ⎦, 2 2⎠ ⎝ 2, ⎣⎝ 2, 1, 9, 25, = ⎧⎨ 6 + π ⎛⎜ + 4 − ⎞⎟ ⎫⎬ cm 2 = ( 6 + 0 ) cm 2 = 6 cm 2 ., 2 ⎝4, 4 ⎠⎭, ⎩, 59. PQ = QR = RS = 4 cm,QS = 8 cm., Perimeter = arc PTS + arc PBQ + arc QES, = ( π × 6 + π × 2 + π × 4 ) cm = 12 π cm., Area = (area of the semicircle PBQ) + (area of the semicircle PTS), – (area of the semicircle QES), 1, 1, 1, ⎡, 2, 2, =, π × ( 2 ) + π × ( 6 ) − π × ( 4 ) 2 ⎤ cm 2 = ( 12 π ) cm 2 ., ⎢⎣ 2, ⎥⎦, 2, 2, 22, 60. 90 + 90 + 2 πr = 400 ⇒ 2 ×, × r = 220 ⇒ r = 35 m., 7, So, R = ( 35 + 14 ) m = 49 m., 22, Area of the track = ⎡( 90 × 14 ) +, × {( 49 ) 2 − ( 35 ) 2 }⎤ m 2 = 6216 m 2 ., ⎢⎣, ⎥⎦, 7
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722, , Secondary School Mathematics for Class 10, Length of the outer boundary = ( 90 + 90 + 2πR ) m , where R = 49 m, = 488 m., , ................................................................, , EXERCISE 16B, Very-Short-Answer Questions, 1. The difference between the circumference and radius of a circle is, 22, [CBSE 2013], 37 cm. Using π =, , find the circumference of the circle., 7, 2. The circumference of a circle is 22 cm. Find the area of its quadrant., [CBSE 2012], , 3. What is the diameter of a circle whose area is equal to the sum of the, areas of two circles of diameters 10 cm and 24 cm?, [CBSE 2012], 4. If the area of a circle is numerically equal to twice its circumference,, then what is the diameter of the circle?, [CBSE 2011], 5. What is the perimeter of a square which circumscribes a circle of radius, a cm?, [CBSE 2011], 6. Find the length of the arc of a circle of diameter 42 cm which subtends, an angle of 60° at the centre., [CBSE 2012], 7. Find the diameter of the circle whose area is equal to the sum of the, areas of two circles having radii 4 cm and 3 cm., [CBSE 2011], 8. Find the area of a circle whose circumference is 8π., , [CBSE 2014], , 9. Find the perimeter of a semicircular protractor whose diameter is, 14 cm., [CBSE 2009], 10. Find the radius of a circle whose perimeter and area are numerically, equal., 11. The radii of two circles are 19 cm and 9 cm. Find the radius of the circle, which has circumference equal to the sum of the circumferences of the, two circles., 12. The radii of two circles are 8 cm and 6 cm. Find the radius of the circle, having area equal to the sum of the areas of the two circles., 13. Find the area of the sector of a circle having radius 6 cm and of angle 30°., [Take π = 3.14.], 14. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre., Find the length of the arc., 15. The circumferences of two circles are in the ratio 2 : 3. What is the ratio, between their areas?
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Area of Circle, Sector and Segment, , 723, , 16. The areas of two circles are in the ratio 4 : 9. What is the ratio between, their circumferences?, Short-Answer Questions, 17. A square is inscribed in a circle. Find the ratio of the areas of the circle, and the square., 18. The circumference of a circle is 8 cm. Find the area of the sector whose, central angle is 72°., 19. A pendulum swings through an angle of 30° and describes an arc 8.8 cm, in length. Find the length of the pendulum., 20. The minute hand of a clock is 15 cm long. Calculate the area swept by it, in 20 minutes. [Take π = 3.14.], 21. A sector of 56° , cut out from a circle, contains 17 .6 cm 2 . Find the radius, of the circle., 22. The area of the sector of a circle of radius 10.5 cm is 69.3 cm 2 . Find the, central angle of the sector., 23. The perimeter of a certain sector of a circle of radius 6.5 cm is 31 cm., Find the area of the sector., 24. The radius of a circle is 17.5 cm. Find the area of the sector enclosed by, two radii and an arc 44 cm in length., 25. Two circular pieces of equal radii and maximum area, touching each, other are cut out from a rectangular cardboard of dimensions, [CBSE 2013], 14 cm × 7 cm. Find the area of the remaining cardboard., 26. In the given figure, ABCD is a square of, side 4 cm. A quadrant of a circle of radius, 1 cm is drawn at each vertex of the square, and a circle of diameter 2 cm is also, drawn. Find the area of the shaded region., [CBSE 2012], [Use π = 3.14.], 27. From a rectangular sheet of paper ABCD with AB = 40 cm and, AD = 28 cm, a semicircular portion with BC as diameter is cut off. Find, [CBSE 2012], the area of the remaining paper., 28. In the given figure, OABC is a square of, side 7 cm. If COPB is a quadrant of a, circle with centre C find the area of the, [CBSE 2012], shaded region.
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724, , Secondary School Mathematics for Class 10, , 29. In the given figure, three sectors of a, circle of radius 7 cm, making angles of, 60°, 80° and 40° at the centre are, shaded. Find the area of the shaded, [CBSE 2012], region., , 30. In the given figure, PQ and AB are, respectively the arcs of two concentric, circles of radii 7 cm and 3.5 cm with, centre O. If ∠POQ = 30 ° , find the area, [CBSE 2012], of the shaded region., 31. In the given figure, find the area of the, shaded region, if ABCD is a square of, side 14 cm and APD and BPC are, [CBSE 2012], semicircles., , 32. In the given figure, the shape of the top, of a table is that of a sector of a circle, with centre O and ∠AOB = 90 ° . If, AO = OB = 42 cm then find the, perimeter of the top of the table., [CBSE 2012], , 33. In the given figure, ABCD is a square, of side 7 cm, DPBA and DQBC are, quadrants of circles each of the radius, 7 cm. Find the area of shaded region., , 34. In the given figure, OABC is a quadrant, of a circle with centre O and radius, 3.5 cm. If OD = 2 cm, find the area of, [CBSE 2017], the shaded region.
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Area of Circle, Sector and Segment, , 725, , 35. Find the perimeter of the shaded, region in the figure, if ABCD is a square, of side 14 cm and APB and CPD are, [CBSE 2011], semicircles., , 36. In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of, [CBSE 2011], the circle which is outside the square., 37. In the given figure, APB and CQD are semicircles of diameter 7 cm each,, while ARC an BSD are semicircles of diameter 14 cm each. Find the, [CBSE 2011], (i) perimeter, (ii) area of the shaded region., , 38. In the given figure, PSR, RTQ and PAQ are three semicircles of, diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of, [CBSE 2014], shaded region. [Use π = 3.14.], , 39. In the given figure, a square OABC is, inscribed in a quadrant OPBQ of a, circle. If OA = 20 cm, find the area of, the shaded region. [Use π = 3.14.], [CBSE 2014]
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726, , Secondary School Mathematics for Class 10, , 40. In the given figure, APB and AQO are semicircles and AO = OB. If the, perimeter of the figure is 40 cm, find the area of the shaded region., [CBSE 2015], , 41. Find the area of a quadrant of a circle whose circumference is 44 cm., [CBSE 2011], , 42. In the given figure, find the area of the, shaded region, where ABCD is a square, of side 14 cm and all circles are of the, [CBSE 2014], same diameter., , 43. Find the area of the shaded region in, the given figure, if ABCD is a rectangle, with sides 8 cm and 6 cm and O is the, [CBSE 2014], centre of the circle., , 44. A wire is bent to form a square enclosing an area of 484 m 2 . Using the, [CBSE 2014], same wire, a circle is formed. Find the area of the circle., 45. A square ABCD is inscribed in a circle of radius r. Find the area of the, square., 46. The cost of fencing a circular field at the rate of ` 25 per metre is ` 5500., The field is to be ploughed at the rate of 50 paise per m 2 . Find the cost of, 22, ploughing the field. [Take π =, ⋅], 7, 47. A park is in the form of a rectangle 120 m by 90 m., At the centre of the park, there is a circular lawn as, shown in the figure. The area of the park excluding, the lawn is 2950 m 2 . Find the radius of the circular, lawn. [Given, π = 3.14.]
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Area of Circle, Sector and Segment, , 727, , 48. In the given figure, PQSR represents, a flower bed. If OP = 21 m and, OR = 14 m, find the area of the, flower bed., , 49. In the given figure, O is the centre of the, bigger circle, and AC is its diameter., Another circle with AB as diameter is, drawn. If AC = 54 cm and BC = 10 cm,, find the area of the shaded region., , 50. From a thin metallic piece in the shape, of a trapezium ABCD in which AB||CD, and ∠BCD = 90 ° , a quarter circle BFEC, is removed. Given, AB = BC = 3 .5 cm, and DE = 2 cm, calculate the area of, remaining (shaded) part of metal sheet., [CBSE 2011], , 51. Find the area of the major segment APB, of a circle of radius 35 cm and, ∠AOB = 90 °, as shown in the given, figure., ANSWERS (EXERCISE 16B), , 1. 44 cm, 6. 22 cm, , 77, cm 2, 8, 7. 10 cm, 2., , 3. 26 cm, 8. 16π, 2, , 11. 28 cm, , 12. 10 cm, , 16. 2 : 3, , 17. π : 2, , 18. 123.2 cm, , 21. 6 cm, , 22. 72°, , 26. 9.72 cm 2, , 4. 8 cm, , 5. 8a cm, , 9. 36 cm, , 10. 2 units, , 14. 22 cm, , 15. 4 : 9, , 19. 16.8 cm, , 20. 235.5 cm 2, , 23. 58.5 cm 2, , 24. 385 cm 2, , 25. 21 cm 2, , 27. 812 cm 2, , 28. 10.5 cm 2, , 29. 77 cm 2, , 30., , 31. 42 cm 2, , 32. 282 cm, , 33. 28 cm 2, , 34. 6.125 cm 2, , 77, cm 2, 8, 35. 72 cm, , 36. 63 cm 2, , 37. (i) 66 cm (ii) 115.5 cm 2, , 38. 31.4 cm, , 39. 228 cm 2, , 40. 96.25 cm 2 41. 38.5 cm 2, , 13. 9.42 cm, , 42. 42 cm 2, , 2, , 43. 30.57 cm 2 44. 616 m 2
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728, , Secondary School Mathematics for Class 10, , 46. ` 1925, , 45. ( 2r 2 ) sq units, 49. 770 cm 2, , 47. 50 m, , 50. 6.125 cm 2 51. 3500 cm 2, HINTS TO SOME SELECTED QUESTIONS, , 1. ( 2 πR − R ) = 37 ⇒ R( 2 π − 1) = 37, 22, ⇒ R ⎛⎜ 2 ×, − 1⎞⎟ = 37 ⇒ R = 7 cm., ⎝, ⎠, 7, 22, circumference = 2 πR = ⎛⎜ 2 ×, × 7 ⎞⎟ cm = 44 cm., ⎝, ⎠, 7, , ∴, , 22, 7, × R = 22 ⇒ R = cm., 2, 7, 1, 1 22 7 7 ⎞, 77, 2, ⎛, Area of the quadrant = × πR = ⎜ ×, × × ⎟ cm 2 =, cm 2 ., ⎝4 7, 4, 2 2⎠, 8, , 2. 2 πR = 22 ⇒ 2 ×, , 3. πR 2 = {π × 5 2 + π × ( 12 ) 2 } = ( 25 π + 144 π ) = 169 π, ⇒ R 2 = 169 = ( 13 ) 2 ⇒ R = 13 cm ⇒ 2 R = 26 cm., 4. πR 2 = 4 πR ⇒ R = 4 ⇒ diameter = 8 cm., 5. Side of the square, = 2 × radius of circle = 2a., Perimeter of the square, = ( 4 × 2 a) = 8 a cm., 6. Length of the arc =, , 2 πRθ ⎛, 22, 60 ⎞, = ⎜2×, × 21 ×, ⎟ cm = 22 cm., ⎝, 7, 360 ⎠, 360, , 7. πR 2 = π × ( 4 ) 2 + π × ( 3 ) 2 = 25 π ⇒ R 2 = 25 ⇒ R = 5 ⇒ 2 R = 10 cm., 9. Perimeter = ( πR + 2 R ) = R( π + 2 ), 22, = 7 ⎛⎜, + 2 ⎞⎟ cm = 36 cm., ⎝7, ⎠, 11. 2 πR = ( 2 π × 19 ) + ( 2 π × 9 ) ⇒ 2 πR = 2 π × ( 19 + 9 ) = 2 π × 28, ⇒ R = 28 cm., 12. πR = ( π × 8 ) + ( π × 6 ) = π( 64 + 36 ) = 100 π, 2, , 2, , 2, , ⇒ R 2 = 100 ⇒ R = 10 cm., 13. Area of the sector =, 14. Length of the arc =, 15., , 30 ⎞, πR 2θ ⎛, 2, 2, = ⎜ 3.14 × 36 ×, ⎟ cm = 9.42 cm ., 360 ⎝, 360 ⎠, 2 πRθ ⎛, 22, 60 ⎞, = ⎜2×, × 21 ×, ⎟ cm = 22 cm., ⎝, 360, 7, 360 ⎠, , 2 πR 1 2, R, 2, = ⇒ 1 = ⋅, 2 πR 2 3, R2 3, ∴, , πR 1, , 2, , πR 2, , 2, , =, , R1, , 2, , R2, , 2, , 2, , 2, , 2, 4, ⎛R ⎞, = ⎜ 1 ⎟ = ⎛⎜ ⎞⎟ = ⋅, ⎝ 3⎠, 9, ⎝ R2 ⎠, , Required ratio = 4 : 9., , 48. 192.5 m 2
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Area of Circle, Sector and Segment, , 16., , πR 1, , 2, , πR 2, , 2, , 2, , 2, , =, , 2, , 4, R, 4, 2, R, 2, ⎛R ⎞, ⇒ 1 2 = ⇒ ⎜ 1 ⎟ = ⎛⎜ ⎞⎟ ⇒ 1 = ⋅, ⎝ 3⎠, 9, 9, R2 3, ⎝ R2 ⎠, R2, , 2 πR 1 R 1 2, =, = ⋅, 2 πR 2 R 2 3, Required ratio = 2 : 3., 17. Let the radius of the circle be r cm., Then, diagonal of the square = diameter of the circle = 2r cm., Area of the circle = ( πr 2 ) cm 2 ., 1, 1, Area of the square = × (diagonal) 2 = × 4 r 2 = ( 2 r 2 ) cm 2 ., 2, 2, ∴, , required ratio = πr 2 : 2r 2 = π : 2., 22, 18. 2 πR = 88 ⇒ 2 ×, × R = 88 ⇒ R = 14 cm., 7, Area of the sector =, , 72 ⎞, πR 2θ ⎛ 22, 2, 2, = ⎜ × 14 × 14 ×, ⎟ cm = 123.2 cm ., 360 ⎝ 7, 360 ⎠, , 19. Length of the pendulum = radius of the sector = r cm., 22, 30, 168, Arc length = 8.8 ⇒ 2 ×, ×r ×, = 8.8 ⇒ r =, = 16.8 cm., 10, 7, 360, °, 360, 20. Angle described by the minute hand in 20 minutes = ⎛⎜, × 20 ⎞⎟ = 120 °., ⎝ 60, ⎠, Now, θ = 120 ° and r = 15 cm., 2, ⎛ πr θ ⎞, 2, area swept by minute hand in 20 min = ⎜, ⎟ cm, ⎝ 360 ⎠, , ∴, , 120 ⎞, 2, = ⎛⎜ 3.14 × 15 × 15 ×, ⎟ cm, ⎝, 360 ⎠, = 235 . 5 cm 2 ., 21., , 22, 56, πr θ, = 17 .6 ⇒, ×r2 ×, = 17 .6 ⇒ r 2 = 36 ⇒ r = 6 cm., 360, 7, 360, , 22., , 22 21 21, 693, πr 2θ, = 69. 3 ⇒, ×, ×, ×θ=, × 360 ⇒ θ = 72 ° ., 360, 7, 2, 2, 10, , 2, , 2 π × 6 .5 × θ ⎞, 2 πrθ ⎞ ⎛, 23. Perimeter of the sector = ⎛⎜ 2r +, ⎟ = ⎜ 2 × 6 .5 +, ⎟ cm, ⎝, ⎠, 360 ⎠ ⎝, 360, 0, , ∴, , 13 +, , Area =, 24., , 18 × 360 ⎞, 13 πθ, 13 πθ, = 31 ⇒, = 18 ⇒ θ = ⎛⎜, ⎟ ⋅, ⎝ 13 π ⎠, 360, 360, , 13 13 18 × 360, 1 ⎞, πr 2θ ⎛, 2, 2, = ⎜π×, ×, ×, ×, ⎟ cm = 58 . 5 cm ., 360 ⎝, 2, 2, 13 π, 360 ⎠, , 2 πrθ, 35, θ, 44 × 72, = 44 ⇒ 2 π ×, ×, = 44 ⇒ θ =, ⋅, 360, 2, 360, 7π, Area =, , 35 35 44 × 72, 1 ⎞, πr 2θ ⎛, 2, 2, = ⎜π×, ×, ×, ×, ⎟ cm = 385 cm ., 360 ⎝, 2, 2, 7π, 360 ⎠, , 729
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730, , Secondary School Mathematics for Class 10, , 25. Clearly, the diameter of each circle is 7 cm., 22 7 7 ⎞ ⎫, Required area = ⎧⎨( 14 × 7 ) − ⎛⎜ 2 ×, × × ⎟ ⎬ cm 2, ⎝, 7 2 2⎠ ⎭, ⎩, = ( 98 − 77 ) cm 2 = 21 cm 2 ., 1, 26. Required area = ⎧⎨( 4 × 4 ) − 4 × π × ( 1) 2 + π × ( 1) 2 ⎫⎬ cm 2, 4, ⎩, ⎭, = ( 16 − 2 × 3.14 × 1) cm 2 = 9 . 72 cm 2 ., 1 22, 27. Required area = ⎧⎨( 40 × 28 ) − ⎛⎜ ×, × 14 × 14 ⎞⎟ ⎫⎬ cm 2, ⎝2 7, ⎠⎭, ⎩, = ( 1120 − 308 ) cm 2 = 812 cm 2 ., , 1 22, 28. Required area = ⎧⎨(7 × 7 ) − ⎛⎜ ×, × 7 × 7 ⎞⎟ ⎫⎬ cm 2, ⎝, ⎠⎭, 4, 7, ⎩, 77, 21, = ⎛⎜ 49 − ⎞⎟ cm 2 =, cm 2 = 10 . 5 cm 2 ., ⎝, 2⎠, 2, ( 60 °+ 80 ° × 40 ° ) ⎫, 2, 29. Required area = ⎧⎨ π × 7 2 ×, ⎬ cm, 360 °, ⎩, ⎭, 22, 180 ⎞, 2, = ⎛⎜ × 49 ×, ⎟ = 77 cm ., ⎝7, 360 ⎠, 30. Required area = (area of sector OQP) – (area of sector OAB), ⎡ ⎛ π × 7 2 × 30 ⎞ ⎛, 7 7, 30 ⎞ ⎤, 2, = ⎢⎜, ⎟ ⎥ cm, ⎟ − ⎜π× × ×, ⎝, ⎠, 360, 2, 2, 360, ⎝, ⎠, ⎣, ⎦, 22 49 ⎤, 77 77 ⎞, 77, ⎡ 22 49, − ⎟ cm 2 =, cm 2 ., = ⎢ ⎛⎜ × ⎞⎟ − ⎛⎜ × ⎞⎟ ⎥ cm 2 = ⎜⎛, ⎝ 6 24 ⎠, 8, ⎣ ⎝ 7 12 ⎠ ⎝ 7 48 ⎠ ⎦, 31. Required area = (area of sq ABCD) – (area of 2 semicircles), 22, ⎤, ⎡, = ⎢( 14 × 14 ) − ⎛⎜ × 7 × 7 ⎞⎟ ⎥ cm 2, ⎝7, ⎠⎦, ⎣, = ( 196 − 154 ) cm 2 = 42 cm 2 ., 32. Required perimeter = (circumference – arc AB) + (OA + OB), 90 ⎞, 22, 22, ⎫, = ⎧⎨ ⎛⎜ 2 ×, × 42 ⎞⎟ − ⎛⎜ 2 ×, × 42 ×, ⎟ + ( 42 + 42 )⎬ cm, ⎠ ⎝, 360 ⎠, 7, 7, ⎩⎝, ⎭, = ( 264 − 66 + 83 ) cm = 282 cm., 33. Required area = (area of quadrant ABPD + area of quadrant CDQB), – ar(area of square ABCD), 1 22, 1 22, = ⎧⎨ ⎛⎜ ×, × 7 × 7 ⎞⎟ + ⎛⎜ ×, × 7 × 7 ⎞⎟ − (7 × 7 )⎫⎬ cm 2, ⎠ ⎝4 7, ⎠, ⎩⎝ 4 7, ⎭, 77 77, = ⎛⎜, +, − 49 ⎞⎟ cm 2 = (77 − 49 ) cm 2 = 28 cm 2 ., ⎝ 2, ⎠, 2
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732, , Secondary School Mathematics for Class 10, 1 22 7 7 ⎞ ⎤, ⎡ 1 22, Area of the shaded region = ⎢ ⎛⎜ ×, × 7 × 7 ⎞⎟ + ⎛⎜ ×, × × ⎟ cm 2, ⎠ ⎝ 2 7 2 2 ⎠ ⎥⎦, ⎣⎝ 2 7, 77 ⎞, 385, 2, = ⎛⎜ 77 +, cm 2 = 96.25 cm 2 ., ⎟ cm =, ⎝, 4⎠, 4, 22, × R = 44 ⇒ R = 7 cm., 7, 1 22, Area of the quadrant = ⎛⎜ ×, × 7 × 7 ⎞⎟ cm 2, ⎝4 7, ⎠, , 41. 2 πR = 44 ⇒ 2 ×, , =, 42. Radius of each circle =, , 77, cm 2 = 38 . 5 cm 2 ., 2, 7, cm = 3.5 cm., 2, , Shaded area = ar(square ABCD) – 4ar(circle with r = 3.5 cm), 22 7 7 ⎞ ⎫, = ⎧⎨( 14 × 14 ) − ⎛⎜ 4 ×, × × ⎟ ⎬ cm 2, ⎝, 7, 2 2⎠ ⎭, ⎩, = ( 196 − 154 ) cm 2 = 42 cm 2 ., 43. AC 2 = AB 2 + BC 2 = ( 8 2 + 6 2 ) = ( 64 + 36 ) = 100 ⇒ AC = 100 = 10 cm., ∴, , r = 5 cm., , 550, 22, Shaded area = ⎧⎨ ⎛⎜ × 5 × 5 ⎞⎟ − ( 8 × 6 )⎫⎬ cm 2 = ⎛⎜, − 48 ⎞⎟ cm 2, ⎝, ⎠, ⎝, ⎠, 7, 7, ⎩, ⎭, (550 − 336 ), 214, =, cm 2 =, cm 2, 7, 7, = 30 . 57 cm 2 ., 44. Side of the square = 484 m = 22 m., Circumference of the circle = perimeter of square = ( 22 × 4 ) m = 88 m, 22, 7, ⇒ 2×, × R = 88 ⇒ R = ⎛⎜ 88 × ⎞⎟ = 14 m., ⎝, 44 ⎠, 7, 22, Area of the circle = πR 2 = ⎛⎜ × 14 × 14 ⎞⎟ m 2 = 616 m 2 ., ⎝7, ⎠, 45. Let O be the centre of the given circle and let ABCD be the, square inscribed in it., So, diag. AC = 2r ., ar(square ABCD) =, , 1, 1, × (diagonal) 2 = ⎧⎨ × ( 2r ) 2 ⎫⎬ sq units = ( 2r 2 ) sq units., 2, ⎩2, ⎭, , 46. Rate of fencing = ` 25 per metre. Cost of fencing = ` 5500., total cost, 5500 ⎞, Perimeter =, = ⎛⎜, ⎟ m = 220 m., rate/metre ⎝ 25 ⎠, Let R be the radius of the field. Then,, 22, 7, 2 πR = 220 ⇒ 2 ×, × R = 220 ⇒ R = ⎛⎜ 220 × ⎞⎟ m = 35 m., ⎝, 7, 44 ⎠
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Area of Circle, Sector and Segment, ∴, , 733, , 22, area of the field = πR 2 = ⎛⎜ × 35 × 35 ⎞⎟ = 3850 m 2 ., ⎝7, ⎠, , 50 ⎞, Cost of ploughing at 50 P per m 2 = ` ⎛⎜ 3850 ×, ⎟ = ` 1925., ⎝, 100 ⎠, 47. Area of the lawn = [( 120 × 90 ) − 2950] m 2 = 7850 m 2 ., ∴, , 3.14 × r 2 = 7850. Find r., , 48. Area of flower bed = ar(quadrant OPQ) – ar(quadrant ORS), 1 22, ⎫, ⎧ 1 22, = ⎨ ⎛⎜ ×, × 21 × 21⎞⎟ − ⎛⎜ ×, × 14 × 14 ⎞⎟ ⎬ m 2 = 192.5 m 2 ., ⎠ ⎝4 7, ⎠⎭, ⎩⎝ 4 7, 49. Diameter of the larger circle = AC = 54 cm., Diameter of the smaller circle = AB = ( AC − BC ) = 44 cm., Area of the shaded region = {π × ( 27 ) 2 − π × ( 22 ) 2 } cm 2 ., = {π × ( 27 + 22 )( 27 − 22 )} cm 2, 22, = ⎛⎜ × 49 × 5 ⎞⎟ cm 2 ., ⎝7, ⎠, 50. Clearly, AB = BC = CE = 3 cm and DE = 2 cm., Area of the shaded part = ar(trapezium ABCD) – ar(quadrant BCE), 1 22 7 7 ⎫ ⎤, ⎡ 1, = ⎢ ⎧⎨ ( AB + CD ) × BC ⎫⎬ − ⎧⎨ ×, × × ⎬ cm 2, 2, 4, 7 2 2 ⎭ ⎥⎦, ⎩, ⎭, ⎩, ⎣, 77 ⎤ 49, ⎡ 1, = ⎢ ⎧⎨ ( 3 . 5 + 5 . 5 ) × 3 . 5 ⎫⎬ − ⎥ =, = 6.125 cm 2 ., ⎭ 8⎦ 8, ⎣⎩ 2, 51. Area of the minor segment ACBA = ar(sector OACBO) – ar(SOAB), 90 ⎞ ⎛ 1, ⎡ 22, ⎞⎤, 2, 2, = ⎢ ⎛⎜ × 35 × 35 ×, ⎟ − ⎜ × 35 × 35 ⎟ ⎥ cm = 350 cm ., ⎠⎦, 360 ⎠ ⎝ 2, ⎣⎝ 7, ⎫, ⎧ 22, Area of the major segment = ⎨ ⎛⎜ × 35 × 35 ⎞⎟ − 350 ⎬ cm 2, ⎠, ⎭, ⎩⎝ 7, = ( 3850 − 350 ) cm 2 = 3500 cm 2 ., , ................................................................, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. The area of a circle is 38.5 cm 2 . The circumference of the circle is, (a) 6.2 cm, , (b) 12.1 cm, , (c) 11 cm, , (d) 22 cm, , 2. The area of a circle is 49 π cm . Its circumference is, 2, , (a) 7 π cm, , (b) 14π cm, , (c) 21π cm, , (d) 28π cm, , 3. The difference between the circumference and radius of a circle is, 37 cm. The area of the circle is, (a) 111 cm 2, , (b) 184 cm 2, , (c) 154 cm 2, , (d) 259 cm 2
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734, , Secondary School Mathematics for Class 10, , 4. The perimeter of a circular field is 242 m. The area of the field is, (a) 9317 m 2, , (b) 18634 m 2, , (c) 4658.5 m 2, , (d) none of these, , 5. On increasing the diameter of a circle by 40%, its area will be increased by, (a) 40%, , (b) 80%, , (c) 96%, , (d) 82%, , 6. On decreasing the radius of a circle by 30%, its area is decreased by, (a) 30%, , (b) 60%, , (c) 45%, , (d) none of these, , 7. The area of a circle is the same as the area of a square. Their perimeters, are in the ratio, (b) 2 : π, , (a) 1 : 1, , (c) π : 2, , (d) π : 2, , 8. The circumference of a circle is equal to the sum of the circumferences, of two circles having diameters 36 cm and 20 cm. The radius of the new, circle is, (a) 16 cm, , (b) 28 cm, , (c) 42 cm, , (d) 56 cm, , 9. The area of a circle is equal to the sum of the areas of two circles of radii, 24 cm and 7 cm. The diameter of the new circle is, (a) 25 cm, , (b) 31 cm, , (c) 50 cm, , (d) 62 cm, , 10. If the perimeter of a square is equal to the circumference of a circle then, the ratio of their areas is, (a) 4 : π, , (b) π : 4, , (c) π : 7, , (d) 7 : π, , 11. If the sum of the areas of two circles with radii R1 and R 2 is equal to the, area of a circle of radius R then, (a) R1 + R 2 = R, , (b) R1 + R 2 < R, , (c) R12 + R 22 < R 2 (d) R12 + R 22 = R 2, , 12. If the sum of the circumferences of two circles with radii R1 and R 2 is, equal to the circumference of a circle of radius R then, (a) R1 + R 2 = R, , (b) R1 + R 2 > R, , (c) R1 + R 2 < R, , (d) none of these, , 13. If the circumference of a circle and the perimeter of a square are, equal then, (a) area of the circle = area of the square, (b) (area of the circle) > (area of the square), (c) (area of the circle) < (area of the square), (d) none of these, 14. The radii of two concentric circles are 19 cm and 16 cm respectively. The, area of the ring enclosed by these circles is, (a) 320 cm 2, , (b) 330 cm 2, , 2, , (d) 340 cm 2, , (c) 332 cm
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Area of Circle, Sector and Segment, , 735, , 15. The areas of two concentric circles are 1386 cm 2 and 962.5 cm 2 . The, width of the ring is, (a) 2.8 cm, , (b) 3.5 cm, , (c) 4.2 cm, , (d) 3.8 cm, , 16. The circumferences of two circles are in the ratio 3 : 4. The ratio of their, areas is, (a) 3 : 4, , (b) 4 : 3, , (c) 9 : 16, , (d) 16 : 9, , 17. The areas of two circles are in the ratio 9 : 4. The ratio of their, circumferences is, (a) 3 : 2, , (b) 4 : 9, , (c) 2 : 3, , (d) 81 : 16, , 18. The radius of a wheel is 0.25 m. How many revolutions will it make in, covering 11 km?, (a) 2800, , (b) 4000, , (c) 5500, , (d) 7000, , 19. The diameter of a wheel is 40 cm. How many revolutions will it make in, covering 176 m?, (a) 140, , (b) 150, , (c) 160, , (b) 166, , 20. In making 1000 revolutions, a wheel covers 88 km. The diameter of the, wheel is, (a) 14 m, , (b) 24 m, , (c) 28 m, , (d) 40 m, , 21. The area of a sector of angle θ° of a circle with radius R is, (a), , 2 πRθ, 180, , (b), , πR 2θ, 180, , (c), , 2 πRθ, 360, , (d), , πR 2θ, 360, , 22. The length of an arc of a sector of angle θ° of a circle with radius R is, (a), , 2 πRθ, 180, , (b), , 2 πRθ, 360, , (c), , πR 2θ, 180, , (d), , πR 2θ, 360, , 23. The length of the minute hand of a clock is 21 cm. The area swept by the, [CBSE 2012], minute hand in 10 minutes is, (a) 231 cm 2, , (b) 210 cm 2, , (c) 126 cm 2, , (d) 252 cm 2, , 24. A chord of a circle of radius 10 cm subtends a right angle at the centre., The area of the minor segments (given, π = 3.14) is, (a) 32.5 cm 2, , (b) 34.5 cm 2, , (c) 28.5 cm 2, , (d) 30.5 cm 2, , 25. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre., The length of the arc is, (a) 21 cm, , (b) 22 cm, , (c) 18.16 cm, , (d) 23.5 cm, , 26. In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre., If 3 = 1 .73 then the area of the segment of the circle is, (a) 120.56 cm 2, , (b) 124.63 cm 2, , (c) 118.24 cm 2, , (d) 130.57 cm 2
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736, , Secondary School Mathematics for Class 10, ANSWERS (MCQ), , 3. (c), , 4. (c), , 5. (c), , 6. (d), , 7. (d), , 8. (b), , 9. (c), , 10. (b), , 1. (d), , 11. (d) 12. (a), , 2. (b), , 13. (b), , 14. (b), , 15. (b), , 16. (c), , 17. (a), , 18. (d), , 19. (a), , 20. (c), , 21. (d) 22. (b), , 23. (a), , 24. (c), , 25. (b), , 26. (a), , HINTS TO SOME SELECTED QUESTIONS, 22, 385, 7, ⎛ 385 7 ⎞ 49, × R2 =, ⇒ R2 = ⎜, × ⎟=, ⇒ R = cm., 4, 7, 10, 2, ⎝ 10 22 ⎠, 22 7 ⎞, ⎛, Circumference = 2 πR = ⎜ 2 ×, × ⎟ cm = 22 cm., ⎝, 7, 2⎠, , 1. πR 2 = 38.5 ⇒, , 2. πR 2 = 49 π ⇒ R 2 = 49 ⇒ R = 7 cm., Circumference = 2 πR = (2 π × 7 ) cm = 14 π cm., 22, 7 ⎞, 3. 2πR − R = 37 ⇒ R ⎛⎜ 2 ×, − 1⎞⎟ = 37 ⇒ R = ⎛⎜ 37 ×, ⎟ = 7., ⎝, ⎠, ⎝, 7, 37 ⎠, 22, Area = πR 2 = ⎛⎜ × 7 × 7 ⎞⎟ cm 2 = 154 cm 2 ., ⎝7, ⎠, 4. 2πR = 242 ⇒ 2 ×, , 22, 7, 77, × R = 242 ⇒ R = ⎛⎜ 242 × ⎞⎟ =, ⋅, ⎝, 7, 44 ⎠, 2, , ⎛ 22 77 77 ⎞ 2, Area = πR 2 = ⎜ ×, × ⎟ m = 4658.5 m 2 ., 2, 2⎠, ⎝ 7, 5. Let original diameter be 100 units. Then, original radius = 50., ∴, , original area = π × (50 ) 2 = 2500 π., , New diameter = 140. New radius = 70., New area = π × (70 ) 2 = 4900 π., 2400, Increase % = ⎛⎜, × 100 ⎞⎟ % = 96%., ⎝ 2500, ⎠, 6. Let original radius be 100 units., Then, original area = π × ( 100 ) 2 = 10000 π., New radius = 70 units., New area = π × (70 ) 2 = 4900 π, 5100, Decrease % = ⎛⎜, × 100 ⎞⎟ % = 51%., ⎝ 10000, ⎠, 7. a2 = πR 2 ⇒, , R2 1, R, 1, = ⇒ =, ., a2 π, a, π, , Ratio of their perimeters =, , 2 πR π ⎛ R ⎞, π, 1, π, = ×⎜ ⎟= ×, =, = π : 2., 4a, 2 ⎝ a⎠, 2, 2, π, , 8. Circumference of new circle = ( 2 π × 18 + 2 π × 10 ) = 2 π × ( 18 + 10 ) = ( 2 π × 28 ) cm., 2 πR = 2 π × 28 cm ⇒ R = 28 cm., ∴, , radius of the new circle is 28 cm.
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Area of Circle, Sector and Segment, 9. Area of the new circle = {π × ( 24 ) 2 + π × (7 ) 2 } cm 2, = {π × (576 + 49 )} cm 2 = ( 625 π ) cm 2 ., ∴, , πR = 625 π ⇒ R = 625 = ( 25 ) 2 ⇒ R = 25 cm., , ∴, , radius of the new circle = 25 cm., , 2, , 2, , 10. 4 a = 2 πr ⇒, , a π, = ⋅, r 2, 2, , Area of square, a2, 1, π, π2 π, = 2 = × ⎛⎜ ⎞⎟ =, = = π : 4., Area of circle, πr, π ⎝ 2⎠, 4π 4, 11. πR 12 + πR 22 = πR 2 ⇒ R 12 + R 22 = R 2 ., 12. 2 πR 1 + 2 πR 2 = 2 πR ⇒ R 1 + R 2 = R., R 2, R2, 4, = ⇒ 2 = 2, a π, a, π, , 13. 2 πR = 4 a ⇒, ⇒, , πR 2 4, = > 1 [Q π = 3.14 < 4], a2, π, , ⇒ πR 2 > a2, ⇒ area of circle > area of square., 14. Required area = π{( 19 ) 2 − ( 16 ) 2 } cm 2, 22, = ⎛⎜ × 35 × 3 ⎞⎟ cm 2 = 330 cm 2 ., ⎝7, ⎠, 7, 15. πR 2 = 1386 ⇒ R 2 = ⎛⎜ 1386 × ⎞⎟ = 441 = ( 21) 2 ⇒ R = 21 cm., ⎝, 22 ⎠, 9625 7 ⎞ ( 49 × 25 ), 7 × 5⎞, 35, cm., × ⎟=, ⇒ r = ⎛⎜, πr 2 = 962.5 ⇒ r 2 = ⎛⎜, ⎟ cm =, ⎝ 10, ⎝ 2 ⎠, 22 ⎠, 4, 2, 35 ⎞, 7, Width of the ring = ( R − r ) = ⎛⎜ 21 −, ⎟ cm = cm = 3.5 cm., ⎝, 2 ⎠, 2, 2, , 16., , 2 πR 1 3, R, 3, 3, ⎛R ⎞, = ⇒ 1 = ⇒ ⎜ 1 ⎟ = ⎛⎜ ⎞⎟, ⎝ 4⎠, 2 πR 2 4, R2 4, ⎝ R2 ⎠, ⇒, , 17., , πR 1, , 2, , πR 2, , 2, , R1, , 2, , R2, , 2, , 2, , =, , 2, , =, , 9, πR 1, 9, ⇒, =, ⋅, 2, 16, 16, πR 2, 2, , 9, R, 9, 3, ⎛R ⎞, ⇒ 1 2 = ⇒ ⎜ 1 ⎟ = ⎛⎜ ⎞⎟, ⎝ 2⎠, 4, 4, ⎝ R2 ⎠, R2, ⇒, , 2, , 2, , 2 πR 1 3, R1 3, = ⇒, = ⋅, 2 πR 2 2, R2 2, , 22 25 ⎞, 11, 18. Distance moved in 1 revolution = 2 πR = ⎛⎜ 2 ×, ×, m., ⎟m=, ⎝, 7 100 ⎠, 7, Total distance covered = 11 km = 11000 m., 7, Number of revolutions = ⎛⎜ 11000 × ⎞⎟ = 7000., ⎝, 11 ⎠, , 737
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738, , Secondary School Mathematics for Class 10, , 22 40 ⎞, 44, 19. Distance moved in 1 revolution = πd = ⎛⎜ ×, m., ⎟m=, ⎝ 7 100 ⎠, 35, Total distance covered = 176 m., 35 ⎞, Number of revolutions = ⎛⎜ 176 ×, ⎟ = 140., ⎝, 44 ⎠, 20. Distance moved in 1 revolution =, πd = 88 ⇒, , 88000, m = 88 m., 1000, , 22, × d = 88, 7, , 7, ⇒ d = ⎛⎜ 88 × ⎞⎟ = 28 m., ⎝, 22 ⎠, 23. Area swept by minute hand in 60 minutes = πR 2 ., Area swept by it in 10 minutes, 2, 22, ⎛ πR, ⎞, =⎜, × 10 ⎟ cm 2 = ⎛⎜ × 21 × 21 ×, ⎝7, ⎝ 60, ⎠, , 1⎞, 2, ⎟ cm, 6⎠, , = 231 cm 2 ., 24. ar(minor segment ACBA) = ar(sector OACBO) – ar(SOAB), 2, ⎛ πr θ 1, ⎞, =⎜, − ×r ×r⎟, ⎝ 360 2, ⎠, , = ⎛⎜, ⎝, , 3.14 × 10 × 10 × 90 1, − × 10 × 10 ⎞⎟ cm 2, ⎠, 360, 2, , = (78 . 5 − 50 ) cm 2 = 28 . 5 cm 2 ., 25. Arc length =, , 2 πrθ ⎛, 22, 60 ⎞, = ⎜2×, × 21 ×, ⎟ cm = 22 cm., 7, 360 ⎠, 360 ⎝, , 2, θ, θ⎞, ⎛ πr θ, 26. ar(segment) = ⎜, − r 2 sin cos ⎟, 2, 2⎠, ⎝ 360, , 22, 120 ⎞, = ⎛⎜ × 14 × 14 ×, ⎟ − ( 14 × 14 × sin 60 ° cos 60 ° ), ⎝7, 360 ⎠, ⎛ 616, ⎞, 3 1, =⎜, −, × × 14 × 14 ⎟ cm 2, 2, 2, ⎝ 3, ⎠, = ( 205. 33 − 49 × 1. 73 ) cm 2, = ( 205. 33 − 84 . 77 ) cm 2, = 120 . 56 cm 2 ., , ................................................................
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Area of Circle, Sector and Segment, , 739, , TEST YOURSELF, MCQ, , 1. In the given figure, a square OABC has been, inscribed in the quadrant OPBQ. If OA = 20 cm, then the area of the shaded region is [take, π = 3.14], (a) 214 cm 2, , (b) 228 cm 2, , 2, , (d) 248 cm 2, , (c) 242 cm, , 2. The diameter of a wheel is 84 cm. How many revolutions will it make to, cover 792 m?, (a) 200, , (b) 250, , (c) 300, , (d) 350, , 3. The area of a sector of a circle with radius r, making an angle of x° at the, centre is, x, x, x, x, (b), (d), (c), (a), × 2 πr, × 2 πr, × πr 2, × πr 2, 180, 180, 360, 360, 4. In the given figure, ABCD is a rectangle, inscribed in a circle having length 8 cm and, breadth 6 cm. If π = 3.14 then the area of the, shaded region is, (b) 266 cm 2, (a) 264 cm 2, (c) 272 cm 2, , (d) 254 cm 2, , Short-Answer Questions, 5. The circumference of a circle is 22 cm. Find its area. [Take π =, , 22, ⋅], 7, , 6. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre., Find the length of the arc., 7. The minute hand of a clock is 12 cm long. Find the area swept by it in, 35 minutes., 8. The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the, area of the sector., 9. A chord of a circle of radius 14 cm makes a right angle at the centre., Find the area of the sector., 10. In the give figure, the sectors of two concentric, circles of radii 7 cm and 3.5 cm are shown. Find, the area of the shaded region.
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740, , Secondary School Mathematics for Class 10, , 11. A wire when bent in the form of an equilateral triangle encloses an area, of 121 3 cm 2 . If the same wire is bent into the form of a circle, what, 22, will be the area of the circle? [Take π =, ⋅], 7, 12. The wheel of a cart is making 5 revolutions per second. If the diameter, 22, of the wheel is 84 cm, find its speed in km per hour. [Take π =, ⋅], 7, 13. OACB is a quadrant of a circle with centre O and, its radius is 3.5 cm. If OD = 2 cm, find the area of, (i) the quadrant OACB (ii) the shaded region., 22, [Take π =, ⋅], 7, , 14. In the given figure, ABCD is a square each of, whose sides measures 28 cm. Find the area of, 22, the shaded region. [Take π =, ⋅], 7, , 15. In the given figure, an equilateral triangle has, been inscribed in a circle of radius 4 cm. Find the, area of the shaded region. [Take π = 3.14 and, 3 = 1.73.], , 16. The minute hand of a clock is 7.5 cm long. Find the area of the face of the, clock described by the minute hand in 56 minutes., Long-Answer Questions, 17. A racetrack is in the form of a ring whose inner circumference is 352 m, and outer circumference is 396 m. Find the width and the area of, the track., 18. A chord of a circle of radius 30 cm makes an angle of 60° at the centre of, the circle. Find the areas of the minor and major segments. [Take, π = 3.14 and 3 = 1.732.], 19. Four cows are tethered at the four corners of a square field of side 50 m, such that each can graze the maximum unshared area. What area will, be left ungrazed? [Take π = 3.14.]
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Area of Circle, Sector and Segment, , 741, , 20. A square tank has an area of 1600 m 2 . There are four semicircular, plots around it. Find the cost of turfing the plots at ` 12.50 per m 2 . [Take, π = 3.14.], ANSWERS (TEST YOURSELF), , 1. (b), 5. 38.5 cm, , 2. (c), 2, , 6. 22 cm, , 9. 154 cm 2, 13. (i) 9.625 cm, , 3. (d), 7. 264 cm, , 10. 9.625 cm 2, 2, , (ii) 6.125 cm, , 4. (b), 2, , 11. 346.5 cm 2, , 2, , 14. 168 cm, , 16. 165 cm 2, , 17. 7 m, 2618 m 2, , 19. 537.5 m 2, , 20. ` 31400, _, , 2, , 8. 44.8 cm 2, 12. 47.52 km/hr, 15. 29.48 cm 2, , 18. 81.27 cm 2 , 2744.73 cm 2
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SOLIDS, , The objects having definite shape and size are called solids., A solid occupies a definite space., , CUBOID Solids like matchbox, chalkbox, a tile, a book, an almirah, a room,, etc., are in the shape of a cuboid., , FORMULAE, , For a cuboid of length = l, breadth = b and height = h, we have, (i) Volume = ( l × b × h) cubic units., (ii) Total surface area = 2( lb + bh + lh) sq units., (iii) Lateral surface area = [2( l + b) × h] sq units., (iv) Diagonal = l 2 + b 2 + h 2 units., , CUBE, , Solids like ice cubes, sugar cubes, dice, etc., are in the shape of a cube., For a cube having each edge = a units, we have, (i) Volume = a3 cubic units., (ii) Total surface area = 6 a2 sq units., (iii) Lateral surface area = 4 a2 sq units., (iv) Diagonal = 3a units., , FORMULAE, , CYLINDER Solids like measuring jars, circular pillars,, circular pencils, circular pipes, road rollers, gas, cylinders, etc., are said to have a cylindrical shape., FORMULAE For a cylinder of base radius = r and height, (or length) = h, we have, (i) Volume = ( πr 2 h) cubic units., (ii) Curved surface area = 2πrh sq units., (iii) Total surface area = ( 2 πrh + 2 πr 2 ) sq units, = 2πr( h + r) sq units., 742
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Volume and Surface Areas of Solids, , 743, , Solids like iron pipes, rubber tubes, are in the shape of hollow cylinders., , HOLLOW CYLINDERS, , FORMULAE, , Consider a hollow cylinder having, external radius = R, internal radius = r, and height = h., , Then, we have, (i) Volume of material = (external volume) - (internal volume), = ( pR 2 h - pr 2 h) cubic units, = ph( R 2 - r 2 ) cubic units., (ii) Curved surface area of hollow cylinder, = (external surface area) + (internal surface area), = ( 2 pRh + 2 prh) sq units, = 2ph( R + r) sq units., (iii) Total surface area of hollow cylinder, = (curved surface area) + (area of the base rings), = {( 2 pRh + 2 prh) + 2( pR 2 - pr 2 )} sq units, = {2 ph( R + r) + 2 p( R 2 - r 2 )} sq units, = {2 ph( R + r) + 2 p( R + r)( R - r)} sq units, = 2p( R + r)( h + R - r) sq units., Solids like ice-cream cones, conical tents,, funnels, etc., are having the shape of a cone., , CONE, , Consider a cone in which, base radius = r, height = h and slant height, l = h 2 + r 2 ., , FORMULAE, , Then, we have, , 1 2, pr h cubic units., 3, (ii) Curved surface area of the cone = prl = pr r 2 + h 2 sq units., (i) Volume of the cone =, , (iii) Total surface area of the cone = (curved surface area) + (area of the base), = ( prl + pr 2 ) = pr( l + r) sq units., SPHERE Objects like a football, a cricket ball, etc., are, said to have the shape of a sphere., FORMULAE For a sphere of radius r, we have, 4, (i) Volume of the sphere = æç pr 3 ö÷ cubic units., è3, ø, (ii) Surface area of the sphere = ( 4 pr 2 ) sq units.
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744, , Secondary School Mathematics for Class 10, , A plane through the centre of a sphere cuts it into two equal, parts. Each part is called a hemisphere., , HEMISPHERE, , FORMULAE, , For a hemisphere of radius r, we have, , 2 3, πr cubic units., 3, (ii) Curved surface area of the hemisphere = ( 2 πr 2 ) sq units., (i) Volume of the hemisphere =, , (iii) Total surface area of the hemisphere = ( 3 πr 2 ) sq units., SPHERICAL SHELL Consider a spherical shell having external radius = R and, internal radius = r. Then, we have, , Volume of material = (external volume) – (internal volume), 4, 4, = ⎛⎜ πR 3 − πr 3 ⎞⎟ cubic units, ⎝3, ⎠, 3, =, , 4, π( R 3 − r 3 ) cubic units., 3, , VOLUME AND SURFACE AREA OF A COMBINATION OF SOLIDS, SOLVED, , EXAMPLES, , EXAMPLE 1, , Three cubes each of volume 216 cm 3 are joined end to end to form a, cuboid. Find the total surface area of the resulting cuboid., , SOLUTION, , Edge of each cube = 216 cm = 6 cm., , [CBSE 2012], 3, , Length of resulting cuboid, l = 18 cm., Breadth of resulting cuboid, b = 6 cm., Height of resulting cuboid, h = 6 cm., , ∴ total surface area of the cuboid, = 2( lb + bh + lh) = 2(18 × 6 + 6 × 6 + 18 × 6) cm 2, = 2(108 + 36 + 108) cm 2, EXAMPLE 2, , = ( 2 × 252) cm 2 = 504 cm 2 ., A sphere and a cube have equal surface areas. Show that the ratio of, the volume of sphere to that of the cube is 6 : π., [CBSE 2011]
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Volume and Surface Areas of Solids, SOLUTION, , 745, , Let the radius of the sphere be r and the edge of the cube be a., Then,, surface area of the sphere = surface area of the cube, 2, ⇒ 4 πr 2 = 6 a2 ⇒ a2 = πr 2, 3, ⇒ a=r, , 2π, ⋅, 3, , ∴ required ratio = volume of sphere : volume of cube, =, , 4 3 3 4 3 2 3 2π, πr : a = πr : πr ⋅, 3, 3, 3, 3, , 2π, π, = 2:, = 6 : π., 3, 3, A vessel is in the form of a hemispherical bowl mounted by a hollow, cylinder. The diameter of the hemisphere is 14 cm and the total, height of the vessel is 13 cm. Find, = 2:, , EXAMPLE 3, , (i) the capacity of the vessel,, SOLUTION, , [CBSE 2006C], , (ii) the inner surface area of the vessel., Radius of the hemisphere, r = 7 cm., , [CBSE 2013], , Total height of the vessel, H = 13 cm., Height of the cylinder, h = (13 – 7) cm = 6 cm., Radius of the cylinder, = radius of the hemisphere = 7 cm., (i) Capacity of the vessel, = volume of the hemisphere + volume of the cylinder, 2, = πr 3 + πr 2 h, 3, 22, ⎡ 2 22, ⎤, = ⎢ ⎛⎜ ×, × 7 × 7 × 7 ⎞⎟ + ⎛⎜, × 7 × 7 × 6 ⎞⎟ ⎥ cm 3, ⎝, ⎠, ⎝, ⎠, 3, 7, 7, ⎣, ⎦, 2156, = ⎛⎜, + 924 ⎞⎟ cm 3 = (718.67 + 924) cm 3, ⎝ 3, ⎠, = 1642 . 67 cm 3 ., (ii) Inner surface area of the vessel, = curved surface area of the hemisphere, + curved surface area of the cylinder, = 2 πr 2 + 2 πrh
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746, , Secondary School Mathematics for Class 10, , 22, 22, ⎡, ⎤, = ⎢ ⎛⎜ 2 ×, × 7 × 7 ⎞⎟ + ⎛⎜ 2 ×, × 7 × 6 ⎞⎟ ⎥ cm 2, ⎝, ⎠, ⎝, ⎠, 7, 7, ⎣, ⎦, EXAMPLE 4, , SOLUTION, , = ( 308 + 264) cm 2 = 572 cm 2 ., Due to some floods, some welfare associations jointly requested the, government to get 100 tents fixed immediately and offered to, contribute 50% of the cost. If the lower part of each tent is of the form, of a cylinder of diameter 4.2 m and height 4 m with the conical upper, part of same diameter but of height 2.8 m, and the canvas to be used, costs ` 100 per square metre, find the amount the associations will, have to pay. What values are shown by these associations?, Radius of the cylinder, r = 2.1 m., Height of the cylinder, h = 4 m., Radius of the cone = radius of the cylinder = 2.1 m., Height of the cone, H = 2.8 m., Slant height of the cone, l = r 2 + H 2, = ( 2.1) 2 + ( 2.8) 2 m, = 4.41 + 7 .84 m, = 12.25 m = 3 .5 m., Area of the canvas required for each tent, = curved surface area of the cylinder, + curved surface area of the cone, 22, 22, ⎡, ⎤, = 2 πrh + πrl = ⎢ ⎛⎜ 2 ×, × 2.1 × 4 ⎞⎟ + ⎛⎜, × 2.1 × 3 .5 ⎞⎟ ⎥ m 2, ⎝, ⎠, ⎝, ⎠, 7, 7, ⎣, ⎦, = (52.8 + 23.1) m 2 = 75 . 9 m 2 ., Total area of the canvas required for 100 tents, = (75.9 × 100) m 2 = 7590 m 2 ., Total cost of 100 tents = ` (7590 × 100) = ` 759000., Amount to be paid by the associations, 50, = 50% of ` 759000 = ` ⎛⎜, × 759000 ⎞⎟ = ` 379500., ⎝ 100, ⎠, Hence, the associations will have to pay ` 379500., Values shown: We must, , (a) help the people in need, (b) donate a part of our income to charity, (c) join hands together for a noble cause like community, service.
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Volume and Surface Areas of Solids, EXAMPLE 5, , SOLUTION, , 747, , A cylindrical tub of radius 5 cm and length 9.8 cm is full of water., A solid in the form of a right circular cone mounted on a hemisphere, is immersed into the tub. If the radius of the hemisphere is 3.5 cm and, the total height of the solid is 8.5 cm, find the volume of water left in, the tub., Radius of the cylinder, r = 5 cm., Height of the cylinder, h = 9.8 cm., Radius of the hemisphere, R = 3.5 cm., Radius of the cone, = radius of the hemisphere, 7, = 3 .5 cm = cm., 2, Height of the cone, H = (8.5 – 3.5) cm = 5 cm., Volume of the water in the cylindrical tub, 22, = πr 2 h = ⎛⎜, × 5 × 5 × 9.8 ⎞⎟ cm 3 = 770 cm 3 ., ⎝ 7, ⎠, Volume of the solid immersed in the tub, = volume of the hemisphere + volume of the cone, 2, 1, = πR 3 + πR 2 H, 3, 3, ⎡ 2 22 7 7 7 ⎞ ⎛ 1 22 7 7, ⎤, = ⎢ ⎛⎜ ×, × × × ⎟ +⎜ ×, × × × 5 ⎞⎟ ⎥ cm 3, ⎠⎦, ⎣⎝ 3 7 2 2 2 ⎠ ⎝ 3 7 2 2, 539 385 ⎞, 924 ⎞, ⎟ cm 3 = ⎛⎜, ⎟ cm 3 = 154 cm 3 ., = ⎛⎜, +, ⎝ 6, ⎝ 6 ⎠, 6 ⎠, ∴ volume of water left in the tub = (770 − 154) cm 3 = 616 cm 3 ., , EXAMPLE 6, , A solid wooden toy is in the form of a hemisphere surmounted by a, cone of same radius. The radius of the hemisphere is 3.5 cm and the, total wood used in the making of toy is 166 5 cm 3 . Find the height of, 6, , the toy. Also, find the cost of painting the hemispherical part of the, [CBSE 2015], toy at the rate of ` 10 per cm 2 ., SOLUTION, , Let the height of the conical part be h., Radius of the hemisphere, r, = radius of the cone = 3.5 cm =, , 7, cm., 2, , Volume of the wood used in making the toy, = volume of the hemisphere + volume of the cone
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748, , Secondary School Mathematics for Class 10, , 2 3 1 2, 1, πr + πr h = πr 2 ( 2r + h), 3, 3, 3, 1 22 7 7 ⎛, 7, 77, = ×, × × × ⎜ 2 × + h ⎞⎟ =, (7 + h)., ⎝, ⎠, 3 7 2 2, 2, 6, 5, 1001, But, the volume of the wood = 166 cm 3 =, cm 3 ., 6, 6, 77, 1001, 1001 6, ∴, (7 + h) =, ⇒7 +h =, ×, = 13, 6, 6, 6, 77, =, , ⇒ h = (13 − 7 ) cm = 6 cm., So, the height of the toy = h + r = ( 6 + 3 .5) cm = 9 . 5 cm., Area to be painted = curved surface area of the hemisphere, 22 7 7 ⎞, = 2 πr 2 = ⎛⎜ 2 ×, × × ⎟ cm 2, ⎝, 7 2 2⎠, = 77 cm 2 ., EXAMPLE 7, , SOLUTION, , Hence, cost of painting = ` (77 × 10) = ` 770., A solid is in the shape of a cone mounted on a hemisphere of same base, radius. If the curved surface areas of the hemispherical part and the, conical part are equal then find the ratio of the radius and the height, of the conical part., [CBSE 2012], Let the radius of each of the cone and the hemisphere be r. Let, the height of the cone be h and its slant height be l., Then, curved surface area of the hemisphere, = curved surface area of the cone, ⇒ 2 πr = πrl ⇒ 2 πr 2 = πr r 2 + h 2, 2, , ⇒ 2r = r 2 + h 2 ⇒ 4r 2 = r 2 + h 2, [squaring both sides], ⇒ 3r 2 = h 2 ⇒, , 2, , r, 1, r, = ⇒ =, 2, h, 3, h, , 1, 1, =, ⋅, 3, 3, , Hence, the required ratio is 1 : 3 ., EXAMPLE 8, , A solid cylinder of diameter 12 cm and height 15 cm is melted and, recast into 12 toys in the shape of a right circular cone mounted on a, hemisphere. Find the radius of the hemisphere and total height of the, toy, if the height of the cone is 3 times the radius., , SOLUTION, , Radius of the cylinder, r = 6 cm and, height of the cylinder, h = 15 cm.
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Volume and Surface Areas of Solids, , 749, , ∴ volume of the cylinder = πr 2 h, = ( π × 6 × 6 × 15) cm 3, = (540 π) cm 3 ., Volume of 12 toys = (540 π) cm 3 ., 540 π ⎞, ⎟ cm 3 = ( 45 π) cm 3 ., ∴ volume of 1 toy = ⎛⎜, ⎝ 12 ⎠, Let the radius of each of the hemisphere and cone be R cm., Then, height of the cone, H = (3R) cm., Volume of 1 toy = volume of the hemisphere, + volume of the cone, 2, 1, = πR 3 + πR 2 H, 3, 3, 2, 1, ⎛, 3, = ⎜ πR + πR 2 × 3 R ⎞⎟ cm 3, ⎝3, ⎠, 3, ⎛ 5 πR 3 ⎞, 3, =⎜, ⎟ cm ., ⎝ 3 ⎠, ∴, , 3, 5 πR 3, = 45 π ⇒ R 3 = ⎛⎜ 45 × ⎞⎟ = 27 = 3 3 ⇒ R = 3., ⎝, 5⎠, 3, , Total height of the toy = ( R + 3 R ) cm = 4 R cm, EXAMPLE 9, , SOLUTION, , = ( 4 × 3) cm = 12 cm., An iron pillar has some part in the form of a right circular cylinder, and the remaining in the form of a right circular cone. The radius of, the base of each of the cone and the cylinder is 8 cm. The cylindrical, part is 240 cm high and the conical part is 36 cm high. Find the, weight of the pillar if 1 cubic centimetre of iron weighs 7.5 g., Radius of the cylinder, r = 8 cm., Radius of the cone, r = 8 cm., Height of the cylinder, h = 240 cm., Height of the cone, H = 36 cm., Total volume of the iron, = volume of the cylinder, + volume of the cone, 1, 1, = πr 2 h + πr 2 H = πr 2 ⎛⎜ h + H ⎞⎟, ⎝, 3, 3 ⎠, 1, ⎡ 22, ⎤, = ⎢ × 8 × 8 × ⎛⎜ 240 + × 36 ⎞⎟ ⎥ cm 3, ⎝, ⎠⎦, 3, ⎣7, = 50688 cm 3 .
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750, , Secondary School Mathematics for Class 10, , ∴ weight of the pillar = volume in cm 3 × weight per cm 3, ⎛ 50688 × 7 .5 ⎞, =⎜, ⎟ kg = 380.16 kg., ⎝, 1000 ⎠, Hence, the weight of the pillar is 380.16 kg., EXAMPLE 10, , A solid is composed of a cylinder with hemispherical ends. If the, whole length of the solid is 104 cm and the radius of each of its, hemispherical ends is 7 cm, find the cost of polishing its surface at the, rate of ` 10 per dm 2 ., [CBSE 2006C], , SOLUTION, , Radius of each hemispherical end = 7 cm., Height of each hemispherical part, = its radius = 7 cm., Height of the cylindrical part, = (104 − 2 × 7 ) cm = 90 cm., Area of surface to be polished, = 2 (curved surface area of the hemisphere), + (curved surface area of the cylinder), = [2( 2 πr ) + 2 πrh] sq units, 2, , 22, 22, ⎡, ⎤, = ⎢ ⎛⎜ 4 ×, × 7 × 7 ⎞⎟ + ⎛⎜ 2 ×, × 7 × 90 ⎞⎟ ⎥ cm 2, ⎝, ⎠, ⎝, ⎠, 7, 7, ⎣, ⎦, = ( 616 + 3960) cm 2 = 4576 cm 2, ⎛ 4576 ⎞, 2, 2, =⎜, ⎟ dm = 45 .76 dm, ⎝ 10 × 10 ⎠, , [Q 10 cm = 1 dm]., , ∴ cost of polishing the surface of the solid, = ` (45.76 × 10) = ` 457.60., EXAMPLE 11, , SOLUTION, , A gulabjamun contains sugar syrup up to about 30% of its volume., Find approximately how much syrup would be found in 45 gulabjamuns, each shaped like a cylinder with two hemispherical ends of, length 5 cm and diameter 2.8 cm., [CBSE 2008], 2.8 ⎞, ⎟ cm = 1.4 cm., Radius of each hemispherical part, r = ⎛⎜, ⎝ 2 ⎠
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Volume and Surface Areas of Solids, , 751, , Radius of the cylindrical part, r = 1.4 cm., Length of the cylindrical part, h = (5 − 2 × 1.4) cm = 2.2 cm., Volume of one gulabjamun, = volume of 2 hemispherical parts, + volume of the cylindrical part, 2, 4, = ⎛⎜ 2 × πr 3 + πr 2 h ⎞⎟ = ⎛⎜ πr 3 + πr 2 h ⎞⎟, ⎝, ⎠, ⎝, ⎠, 3, 3, 4, 4, ⎡ 22, ⎤, = πr 2 ⎛⎜ r + h ⎞⎟ = ⎢ × 1.4 × 1.4 × ⎛⎜ × 1.4 + 2.2 ⎞⎟ ⎥ cm 3, ⎝3, ⎠ ⎣7, ⎝3, ⎠⎦, 12.2 ⎞, 75.152 ⎞, ⎟ cm 3 = ⎛⎜, ⎟ cm 3 ., = ⎛⎜ 6.16 ×, ⎝, ⎝ 3 ⎠, 3 ⎠, Volume of 45 gulabjamuns, 75.152, = ⎛⎜, × 45 ⎞⎟ cm 3 = 1127 .28 cm 3 ., ⎝ 3, ⎠, Volume of the syrup = (30% of 1127.28) cm 3, 30, = ⎛⎜, × 1127 .28 ⎞⎟ cm 3 = 338.184 cm 3 ., ⎝ 100, ⎠, Hence, 45 gulabjamuns contain approximately 338 cm 3 of, syrup., EXAMPLE 12, , SOLUTION, , Rahul, an engineering student, prepared a model shaped like a, cylinder with two cones attached at its two ends. The diameter of the, model is 3 cm and its length is 12 cm. If each conical part has a height, of 2 cm, find the (i) volume of air contained in Rahul’s model, (assume the outer and inner dimensions of the model to be nearly the, same), (ii) cost of painting the outer surface of the model at ` 12.50, per cm 2 ., 3, Radius of each conical part, r = cm., 2, 3, Radius of cylindrical part, r = cm., 2
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752, , Secondary School Mathematics for Class 10, , Height of each conical part, h = 2 cm., Length of the cylindrical part, H = (12 – 2 × 2) cm = 8 cm., Slant height of each conical part,, 2, , 3, l = r 2 + h 2 = ⎛⎜ ⎞⎟ + 2 2 cm =, ⎝ 2⎠, =, , 9, + 4 cm, 4, , 25, 5, cm = cm., 4, 2, , (i) Volume of air contained in the model, = volume of 2 conical parts, + volume of the cylindrical part, = 2×, , 1 2, 2, πr h + πr 2 H = πr 2 ⎛⎜ h + H ⎞⎟, ⎝3, ⎠, 3, , 2, ⎡ 22 3 3, ⎤, = ⎢ × × × ⎛⎜ × 2 + 8 ⎞⎟ ⎥ cm 3, ⎠⎦, 2 2 ⎝3, ⎣7, 22 3 3 28 ⎞, = ⎛⎜, × × × ⎟ cm 3 = 66 cm 3 ., ⎝ 7, 2 2 3⎠, (ii) Total surface area of the model, = curved surface area of 2 conical parts, + curved surface area of the cylindrical part, = 2 × πrl + 2 πrH = 2 πr( l + H), 22 3, = ⎡2 ×, × × ( 2 .5 + 8) ⎤ cm 2, ⎢⎣, ⎥⎦, 7, 2, 22 3, = ⎛⎜ 2 ×, × × 10 .5 ⎞⎟ cm 2 = 99 cm 2 ., ⎝, ⎠, 7, 2, ∴ cost of painting the model = ` ( 99 × 12 .50), = ` 1237 .50., EXAMPLE 13, , SOLUTION, , A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in, diameter; the diameter of the spherical part is 8.5 cm. By measuring, the amount of water it holds, Radhika finds its volume to be 345 cm 3 ., Check mathematically whether she is correct, taking the above as the, inside measurements and π = 3.14., ⎛ 8 .5 ⎞, Radius of the spherical part, R = ⎜, ⎟ cm = 4.25 cm., ⎝ 2 ⎠, Radius of the cylindrical neck, r = 1 cm., Height of the cylindrical neck, h = 8 cm.
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Volume and Surface Areas of Solids, , Volume of the vessel, = volume of the spherical part, + volume of the cylindrical part, 4, 4, = πR 3 + πr 2 h = π ⎛⎜ R 3 + r 2 h ⎞⎟, ⎝, ⎠, 3, 3, 4, ⎡, ⎤, = ⎢ 3.14 × ⎛⎜ × 4.25 × 4.25 × 4.25 + 1 × 1 × 8 ⎞⎟ ⎥ cm 3, ⎝, ⎠, 3, ⎣, ⎦, = [ 3.14 × (102. 354 + 8)] cm 3, = ( 3.14 × 110. 354) cm 3, = 346 .51156 cm 3 ≈ 346 .51 cm 3 ., EXAMPLE 14, , SOLUTION, , Hence, the volume found by Radhika is almost correct., A juice seller was serving his customers using glasses, as shown in the figure. The inner diameter of the, cylindrical glass was 5 cm, but the bottom of the glass, had a hemispherical raised portion which reduced the, capacity of the glass. If the height of a glass was, 10 cm, find the apparent capacity of the glass and its, actual capacity. [Use π = 3.14.], [CBSE 2009], 5, cm., 2, Height of the cylindrical glass, h = 10 cm., 5, Radius of the hemispherical part, r = cm., 2, Radius of the cylindrical glass, r =, , Apparent capacity of the glass, = volume of the cylinder, = πr 2 h, 5 5, = ⎛⎜ 3.14 × × × 10 ⎞⎟ cm 3, ⎝, ⎠, 2 2, = 196.25 cm 3 ., Volume of the hemispherical part, 2, 2, 5 5 5, = πr 3 = ⎛⎜ × 3.14 × × × ⎞⎟ cm 3, ⎝, 3, 3, 2 2 2⎠, ⎛ 196 . 25 ⎞, =⎜, ⎟ cm 3 = 32 .708 cm 3, ⎝ 6 ⎠, ≈ 32 .71 cm 3 ., , 753
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754, , Secondary School Mathematics for Class 10, , ∴ actual capacity of the glass, = apparent capacity of the glass, − volume of the hemispherical part, EXAMPLE 15, , SOLUTION, , = (196.25 − 32.71) cm 3 = 163.54 cm 3 ., A pen stand made of wood is in the shape of a cuboid with four conical, depressions to hold pens. The dimensions of the cuboid are 15 cm by, 10 cm by 3.5 cm. The diameter of each of the depressions is 1 cm and, the depth is 1.4 cm. Find the volume of wood in the entire stand., , 1, cm., 2, Depth of each conical depression, h = 1.4 cm., Radius of each conical depression, r =, , Volume of the wood in the stand, , EXAMPLE 16, , SOLUTION, , = (volume of the cuboid, – volume of 4 conical depressions), 1, ⎡, ⎤, ⎛, = ⎢(15 × 10 × 3 .5) − ⎜ 4 × πr 2 h ⎞⎟ ⎥ cm 3, ⎝, ⎠, 3, ⎣, ⎦, 4 22 1 1, ⎡, ⎤, ⎛, ⎞, = ⎢525 − ⎜ ×, × × × 1.4 ⎟ ⎥ cm 3, ⎝3 7 2 2, ⎠⎦, ⎣, 4.4 ⎞, ⎟ cm 3 = (525 − 1.47 ) cm 3, = ⎛⎜ 525 −, ⎝, 3 ⎠, = 523.53 cm 3 ., A solid toy is in the form of a right circular cylinder with a, hemispherical shape at one end and a cone at the other end. Their, common diameter is 4.2 cm, and the heights of the cylindrical and, conical portions are 12 cm and 7 cm respectively. Find the volume of, the given toy. [Take π = 22 7 .], The shape of the toy is given below.
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Volume and Surface Areas of Solids, , 755, , 4.2 ⎞, ⎟ cm = 2.1 cm., Radius of the hemispherical part, r = ⎛⎜, ⎝ 2 ⎠, Radius of the cylindrical part, r = 2.1 cm., Radius of the conical part, r = 2.1 cm., Height of the cylindrical part, h = 12 cm., Height of the conical part, H = 7 cm., Volume of the toy, = volume of the hemispherical part, + volume of the cylindrical part, + volume of the conical part, 2 3, 1, 2, 1, = πr + πr 2 h + πr 2 H = πr 2 ⎛⎜ r + h + H ⎞⎟, ⎝, 3, 3, 3, 3 ⎠, 7 ⎤, 2, ⎡ 22, = ⎢ × 2.1 × 2.1 × ⎛⎜ × 2.1 + 12 + ⎞⎟ ⎥ cm 3, ⎝, 3⎠⎦, 3, 7, ⎣, 47 .2 ⎞, ⎟ cm 3 = 218.064 cm 3 ., = ⎛⎜ 13.86 ×, ⎝, 3 ⎠, Hence, the volume of the given toy is 218.064 cm 3 ., EXAMPLE 17, , SOLUTION, , The largest possible sphere is carved out from a solid wooden cube of, side 7 cm. Find, (i) the volume of the sphere,, [CBSE 2011], (ii) the percentage of wood wasted in the process., Clearly, diameter of the largest possible sphere = 7 cm., 7, Radius of the sphere, r = cm., 2, (i) Volume of the sphere, 4, 4 22 7 7 7 ⎞, = πr 3 = ⎛⎜ ×, × × × ⎟ cm 3, ⎝ 3 7 2 2 2⎠, 3, 539 ⎞, ⎟ cm 3 = 179.67 cm 3 ., = ⎛⎜, ⎝ 3 ⎠, (ii) Volume of the cube = (7 × 7 × 7 ) cm 3 = 343 cm 3 ., Volume of wood wasted, = volume of cube – volume of sphere, 539 ⎞, 490 ⎞, ⎟ cm 3 = ⎛⎜, ⎟ cm 3 ., = ⎛⎜ 343 −, ⎝, ⎝ 3 ⎠, 3 ⎠, ∴ percentage of wood wasted, 490, 1, 1000 ⎞, 13, ⎟ % = 47 %., = ⎛⎜, ×, × 100 ⎞⎟ % = ⎛⎜, ⎝ 3, ⎠, ⎝ 21 ⎠, 343, 21
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756, EXAMPLE 18, , SOLUTION, , Secondary School Mathematics for Class 10, , From a solid right circular cylinder with height 12 cm and radius of, the base 5 cm, a right circular cone of the same height and the same, base radius is removed. Find the volume and total surface area of the, remaining solid. [Use π = 3.14.], [CBSE 2014], Radius of each of the cylinder and cone, r = 5 cm., Height of each of the cylinder and cone, h = 12 cm., Slant height of the cone,, l = r 2 + h 2 = 5 2 + (12) 2 cm, = 169 cm = 13 cm., Volume of the remaining solid, = volume of the cylinder – volume of the cone., 1, 2, = πr 2 h − πr 2 h = πr 2 h, 3, 3, 2, = ⎛⎜ × 3.14 × 5 × 5 × 12 ⎞⎟ cm 3 = 628 cm 3 ., ⎝3, ⎠, Total surface area of the remaining solid, = curved surface area of the cylinder, + curved surface area of the cone, + area of the upper circular face of the cylinder, = 2 πrh + πrl + πr 2 = πr( 2 h + l + r), = [ 3.14 × 5 × ( 2 × 12 + 13 + 5)] cm 2, = ( 3.14 × 5 × 42) cm 2 = 659.4 cm 2 ., , EXAMPLE 19, , A hemispherical depression is cut out from one face of a cubical block of, side 7 cm, such that the diameter of the hemisphere is equal to the edge, of the cube. Find the surface area of the remaining solid. [CBSE 2014], , SOLUTION, , Edge of the cube, a = 7 cm., 7, cm., 2, Surface area of remaining solid, = total surface area of the cube, – area of the top of hemispherical part, + curved surface area of the hemisphere, , Radius of the hemisphere, r =, , = 6 a 2 − πr 2 + 2 πr 2 = 6 a 2 + πr 2, 22 7 7 ⎞, = ⎛⎜ 6 × 7 × 7 +, × × ⎟ cm 2, ⎝, 7 2 2⎠, = ( 294 + 38 .5) cm 2 = 332 .5 cm 2 .
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Volume and Surface Areas of Solids, EXAMPLE 20, , SOLUTION, , 757, , A wooden article was made by scooping out a, hemisphere from each end of a solid cylinder, as, shown in the figure. If the height of the cylinder is, 12 cm and its base is of radius 4.2 cm, find the total, surface area of the article. Also, find the volume of, the wood left in the article., [CBSE 2009C], Radius of the hemisphere, r = 4.2 cm., Radius of the cylinder, r = 4.2 cm, Height of the cylinder, h = 12 cm, Total surface area of the article, = curved surface area of cylinder, + curved surface area of 2 hemispheres, = 2 πrh + 2 × 2 πr 2 = 2 πrh + 4 πr 2, 22, = 2 πr( h + 2r) = ⎡ 2 ×, × 4. 2 × (12 + 2 × 4 . 2) ⎤ cm 2, 7, ⎣⎢, ⎦⎥, = ( 26.4 × 20.4) cm 2 = 538.56 cm 2 ., Total volume of the wood left in the article, = volume of the cylinder – volume of 2 hemispheres, 2, 4, = πr 2 h − 2 × πr 3 = πr 2 ⎛⎜ h − r ⎞⎟, ⎝, 3, 3 ⎠, 4, ⎡ 22, ⎤, = ⎢ × 4 . 2 × 4 . 2 × ⎛⎜ 12 − × 4 . 2 ⎞⎟ ⎥ cm 3, ⎝, ⎠⎦, 3, ⎣7, , EXAMPLE 21, , = (55.44 × 6.4) cm 3 = 354.816 cm 3 ., In the given figure, from a cuboidal solid metallic block of dimensions, 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled, out. Find the surface area of the remaining block. [HOTS] [CBSE 2015], 7 cm, , 5 cm, , 15 cm, , SOLUTION, , Length of the cuboid, l = 15 cm., Breadth of the cuboid, b = 10 cm., Height of the cuboid, h = 5 cm., , 10, , cm
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758, , Secondary School Mathematics for Class 10, , Radius of the cylinder, r =, , 7, cm., 2, , Height of the cylinder, h = 5 cm., Surface area of the remaining block, = total surface area of the cuboid, – area of two circular faces of the cylinder, + curved surface area of the cylinder, = 2( lb + bh + lh) − 2 × πr 2 + 2 πrh, , EXAMPLE 22, , SOLUTION, , 22 7 7, 22 7, ⎡, ⎤, = ⎢2(15 × 10 + 10 × 5 + 15 × 5) − 2 ×, × × + 2×, × × 5 ⎥ cm 2, 7 2 2, 7 2, ⎣, ⎦, = (550 − 77 + 110) cm 2 = 583 cm 2 ., A wooden toy rocket is in the shape of a cone mounted on a cylinder., The height of the entire rocket is 26 cm, while the height of the conical, part is 6 cm. The base of the conical portion has a diameter of 5 cm,, while the base diameter of the cylindrical portion is 3 cm. If the, conical portion is to be painted orange and the cylindrical portion, yellow, find the area of the rocket painted with each of these colours., [Take π = 3.14.], [HOTS], , 5, cm., 2, Height of the conical part, h = 6 cm., 3, Radius of the cylindrical part, R = cm., 2, Radius of the conical part, r =, , Height of cylindrical part, H = (26 − 6) cm, = 20 cm., Slant height of the conical part,, 2, , 5, l = r 2 + h 2 = ⎛⎜ ⎞⎟ + 6 2 cm, ⎝ 2⎠, =, , 25, 169, 13, + 36 cm =, cm =, cm., 4, 4, 2, , Area to be painted orange, = curved surface area of the cone, + base area of the cone – base area of the cylinder, = πrl + πr 2 − πR 2 = π (rl + r 2 − R 2 ), 5 13 5 5 3 3 ⎞ ⎤, ⎡, = ⎢ 3.14 × ⎛⎜ ×, + × − × ⎟ cm 2, ⎝, 2 2 2 2 2 2 ⎠ ⎥⎦, ⎣
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Volume and Surface Areas of Solids, , 759, , 81, 65 25 9 ⎞ ⎤, ⎡, = ⎢ 3.14 × ⎛⎜, +, − ⎟ ⎥ cm 2 = ⎛⎜ 3.14 × ⎞⎟ cm 2, ⎝, ⎠, ⎝, 4⎠, 4, 4, 4, ⎣, ⎦, = ( 3.14 × 20.25) cm 2 = 63 .585 cm 2 ., Area to be painted yellow, = curved surface area of the cylinder, + base area of the cylinder, = 2 πRH + πR 2 = πR ( 2 H + R ), 3 ⎤, 3, ⎡, = ⎢ 3.14 × × ⎛⎜ 2 × 20 + ⎞⎟ ⎥ cm 2, ⎝, 2⎠⎦, 2, ⎣, 3 83, 781.86 ⎞, ⎟ cm 2, = ⎛⎜ 3.14 × × ⎞⎟ cm 2 = ⎛⎜, ⎝, ⎝ 4 ⎠, 2 2⎠, EXAMPLE 23, , = 195.465 cm 2 ., Deepa runs an industry in a shed which is, in the shape of a cuboid surmounted by a, half-cylinder. The base of the shed is of, dimensions 7 m × 15 m and the height of, the cuboidal portion is 8 m., (i) Find the volume of air that the shed can hold., (ii) Suppose the machinery in the shed occupies a total space of, 300 m 3 and there are 20 workers, each of whom occupy about, 0.08 m 3 space on an average. Then, how much air is in, the shed?, (iii) Find the internal surface area of the shed, excluding the floor., [HOTS], , SOLUTION, , The shed consists of a cuboid with dimensions l = 15 m,, 7, b = 7 m, h = 8 m and a half-cylinder of radius, r = m and, 2, height, H = 15 m., (i) Volume of air that the shed can hold, 1, = volume of the cuboid + × volume of the cylinder, 2, 1, 1 22 7 7, = lbh + πr 2 H = ⎛⎜ 15 × 7 × 8 + ×, × × × 15 ⎞⎟ m 3, ⎝, ⎠, 2, 2 7 2 2, 1155 ⎞ 3 ⎛ 4515 ⎞ 3, ⎟ m =⎜, ⎟ m, = ⎛⎜ 840 +, ⎝, ⎝ 4 ⎠, 4 ⎠, = 1128 .75 m 3 .
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760, , Secondary School Mathematics for Class 10, , (ii) Total space occupied by machinery and 20 workers, = ( 300 + 0.08 × 20) m 3 = 301.6 m 3 ., Volume of air in the shed when there are machinery and, workers inside it = (1128.75 – 301.6) m 3 = 827.15 m 3, (iii) Total internal surface area of the shed (excluding floor), = (surface area of walls) + (surface area of ceiling), = (area of two walls each measuring 15 m × 8 m), + (area of two walls each measuring 7 m × 8 m), + (area of two semicircles each of radius 3.5 m), + (curved surface area πrh of half-cylinder), 1 22 7 7 ⎞, ⎡, = ⎢( 2 × 15 × 8) + ( 2 × 7 × 8) + ⎛⎜ 2 × ×, × × ⎟, ⎝, 2 7, 2 2⎠, ⎣, 22 7, ⎤, + ⎛⎜, × × 15 ⎞⎟ ⎥ m 2, ⎝ 7, ⎠, 2, ⎦, EXAMPLE 24, , SOLUTION, , = ( 240 + 112 + 38.5 + 165) m 2 = 555.5 m 2 ., A solid is made up of a cube and a, hemisphere attached on its top, as shown in, the figure. Each edge of the cube measures, 5 cm and the hemisphere has a diameter of, 4.2 cm. Find the total area to be painted., [Take π = 227 .], [CBSE 2009], Edge of the cube, a = 5 cm., 4.2 ⎞, ⎟ cm = 2.1 cm., Radius of the hemisphere, r = ⎛⎜, ⎝ 2 ⎠, Total area to be painted, = total surface area of the cube, − base area of the hemisphere, + curved surface area of the hemisphere, = 6 a 2 − πr 2 + 2 πr 2 = 6 a 2 + πr 2, 22, = ⎛⎜ 6 × 5 × 5 +, × 2.1 × 2.1 ⎞⎟ cm 2, ⎝, ⎠, 7, , EXAMPLE 25, , = (150 + 13.86) cm 2 = 163.86 cm 2 ., A solid toy is in the form of a hemisphere surmounted by a right, circular cone of height 2 cm and diameter of base 4 cm. If a right, circular cylinder circumscribes the toy, find how much more space, than the toy it will cover. [Use π = 3.14.]
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Volume and Surface Areas of Solids, SOLUTION, , 761, , Let OAB be the cone and ACB be the, hemisphere, having the same base, AB. Let the right circular cylinder, DEFG circumscribe the given solid., Radius of each of the cone, cylinder, and hemisphere, r = 2 cm., Height of the cone, h = 2 cm., Height of the cylinder,, H = h + r = ( 2 + 2) cm = 4 cm., Volume of the toy, = (volume of the hemisphere) + (volume of the cone), 2, 1, = πr 3 + πr 2 h, 3, 3, 2, 1, = ⎛⎜ π × 2 3 + π × 2 2 × 2 ⎞⎟ cm 3, ⎝3, ⎠, 3, 16 π 8 π ⎞, ⎟ cm 3 = 8 π cm 3 ., = ⎛⎜, +, ⎝ 3, 3 ⎠, Volume of the cylinder = πr 2 H, = ( π × 2 2 × 4) cm 3 = (16 π) cm 3 ., Required volume = volume of the cylinder – volume of the toy, = (16 π − 8 π) cm 3 = 8 π cm 3, = ( 8 × 3.14) cm 3 = 25.12 cm 3 ., , EXERCISE 17A, 1. Two cubes each of volume 27 cm 3 are joined end to end to form a solid., [CBSE 2011, ’14], Find the surface area of the resulting cuboid., 2. The volume of a hemisphere is 2425 1 cm 3 . Find its curved surface area., 2, , [CBSE 2012], , 2, , 3. If the total surface area of a solid hemisphere is 462 cm , find its, [CBSE 2014], volume., 4., , (i) A 5-m-wide cloth is used to make a conical tent of base diameter, 14 m and height 24 m. Find the cost of cloth used at the rate of ` 25, [CBSE 2014], per metre., (ii) The radius and height of a solid right-circular cone are in the, ratio of 5 : 12. If its volume is 314 cm 3 , find its total surface area., [CBSE 2017], [Take π = 3.14.]
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762, , Secondary School Mathematics for Class 10, , 5. If the volumes of two cones are in the ratio of 1 : 4 and their diameters, are in the ratio of 4 : 5, find the ratio of their heights., 6. The slant height of a conical mountain is 2.5 km and the area of its base, is 1.54 km 2 . Find the height of the mountain., 7. The sum of the radius of the base and the height of a solid cylinder is, 37 metres. If the total surface area of the cylinder be 1628 sq metres, find, its volume., 8. The surface area of a sphere is 2464 cm 2 . If its radius be doubled, what, will be the surface area of the new sphere?, 9. A military tent of height 8.25 m is in the form of a right circular cylinder, of base diameter 30 m and height 5.5 m surmounted by a right circular, cone of same base radius. Find the length of canvas used in making the, tent, if the breadth of the canvas is 1.5 m., [CBSE 2012], 10. A tent is in the shape of a right circular cylinder up to a height of 3 m, and conical above it. The total height of the tent is 13.5 m and the radius, of its base is 14 m. Find the cost of cloth required to make the tent at the, [CBSE 2005], rate of ` 80 per square metre. [Take π = 227 .], 11. A circus tent is cylindrical to a height of 3 m and conical above it. If its base, radius is 52.5 m and the slant height of the conical portion is 53 m, find the, [CBSE 2004], area of canvas needed to make the tent. [Take π = 227 .], 12. A rocket is in the form of a circular cylinder closed at the lower end and, a cone of the same radius is attached to the top. The radius of the, cylinder is 2.5 m, its height is 21 m and the slant height of the cone is, 8 m. Calculate the total surface area of the rocket., 13. A solid is in the shape of a cone surmounted on a hemisphere, the, radius of each of them being 3.5 cm and the total height of the solid is, [CBSE 2012], 9.5 cm. Find the volume of the solid., 14. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere, of same radius on its circular face. The total height of the toy is 15.5 cm., Find the total surface area of the toy., [CBSE 2017], 15. A toy is in the shape of a cone mounted on a hemisphere of same base, radius. If the volume of the toy is 231 cm 3 and its diameter is 7 cm, find, the height of the toy., [CBSE 2012], 16. A cylindrical container of radius 6 cm and height 15 cm is filled with, ice cream. The whole ice cream has to be distributed to 10 children in, equal cones with hemispherical tops. If the height of the conical portion, is 4 times the radius of its base, find the radius of the ice cream cone., 17. A vessel is in the form of a hemispherical bowl surmounted by a hollow, cylinder. The diameter of the hemisphere is 21 cm and the total height, of the vessel is 14.5 cm. Find its capacity.
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Volume and Surface Areas of Solids, , 763, , 18. A toy is in the form of a cylinder with hemispherical ends. If the whole, length of the toy is 90 cm and its diameter is 42 cm, find the cost of, [CBSE 2014], painting the toy at the rate of 70 paise per sq cm., 19. A medicine capsule is in the shape of a cylinder with two hemispheres, stuck to each of its ends. The length of the entire capsule is 14 mm and the, diameter of the capsule is 5 mm. Find its surface area., 20. A wooden article was made by scooping out a, hemisphere from each end of a cylinder, as shown in, the figure. If the height of the cylinder is 20 cm and its, base is of diameter 7 cm, find the total surface area of, the article when it is ready., [CBSE 2008C], 21. A solid is in the form of a right circular cone mounted on a hemisphere., The radius of the hemisphere is 2.1 cm and the height of the cone is, 4 cm. The solid is placed in a cylindrical tub full of water in such a, way that the whole solid is submerged in water. If the radius of the, cylinder is 5 cm and its height is 9.8 cm, find the volume of the water, [CBSE 1996C, 2000C], left in the tub., 22. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical, cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the, volume of the remaining solid. Also, find the total surface area of the, [HOTS] [CBSE 2009], remaining solid. [Take π = 3.14.], 23. From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical, cavity of the same height and same diameter is hollowed out. Find the, [CBSE 2014], total surface area of the remaining solid., 24. From a solid cylinder of height 14 cm and base diameter 7 cm, two equal, conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the, [CBSE 2011], volume of the remaining solid., 25. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its, weight, a conical hole is drilled in the cylinder. The conical hole has a, radius of 3 2 cm and its depth is 8 9 cm. Calculate the ratio of the volume, of metal left in the cylinder to the volume of metal taken out in conical, [CBSE 2015], shape., 26. A spherical glass vessel has a cylindrical neck 7 cm long and 4 cm in, diameter. The diameter of the spherical part is 21 cm. Find the quantity, [CBSE 2009C], of water it can hold. [Use π = 227 .], 27. The given figure represents a solid consisting of a cylinder surmounted, by a cone at one end and a hemisphere at the other. Find the volume of, the solid.
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764, , Secondary School Mathematics for Class 10, , 28. From a cubical piece of wood of side 21 cm, a hemisphere is carved out, in such a way that the diameter of the hemisphere is equal to the side of, the cubical piece. Find the surface area and volume of the remaining, piece., [CBSE 2014], 29., , (i) A hemisphere of maximum possible diameter is placed over a, cuboidal block of side 7 cm. Find the surface area of the solid so, [CBSE 2017], formed., (ii) A cubical block of side 10 cm is surmounted by a hemisphere., What is the largest diameter that the hemisphere can have? Find, the cost of painting the total surface area of the solid so formed, at, [CBSE 2015], the rate of ` 5 per 100 sq cm. [Use π = 3.14.], , 30. A toy is in the shape of a right circular cylinder with a hemisphere on one, end and a cone on the other. The radius and height of the cylindrical part, are 5 cm and 13 cm respectively. The radii of the hemispherical and the, conical parts are the same as that of the cylindrical part. Find the surface, area of the toy, if the total height of the toy is 30 cm., 31. The inner diameter of a glass is 7 cm and it, has a raised portion in the bottom in the, shape of a hemisphere, as shown in the, figure. If the height of the glass is 16 cm,, find the apparent capacity and the actual, capacity of the glass., 32. A wooden toy is in the shape of a cone, mounted on a cylinder, as shown in the, figure. The total height of the toy is 26 cm,, while the height of the conical part is 6 cm., The diameter of the base of the conical, part is 5 cm and that of the cylindrical, part is 4 cm. The conical part and the, cylindrical part are respectively painted, red and white. Find the area to be painted, by each of these colours. [Take π = 227 .]
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Volume and Surface Areas of Solids, , 765, , ANSWERS (EXERCISE 17A), , 1. 90 cm 2, , 2. 693 cm 2, , 3. 718.67 cm 2, , 4. (i) ` 2750 (ii) 282.6 cm 2, , 5. 25 : 64, , 6. 2.4 km, , 7. 4620 m 3, , 8. 9856 cm 2, , 9. 825 m, , 10. ` 82720, , 11. 9735 m 2, , 12. 412.5 m 2, , 13. 166.83 cm 3 14. 214.5 cm 2 15. 18 cm, 17. 3811.5 cm, , 3, , 18. ` 8316, , 19. 220 mm, , 16. 3 cm, 2, , 21. 732.116 cm 3 22. 602.88 cm 2 23. 73.92 cm 2, 25. 1463 : 22, , 26. 4939 cm 3, , 27. 376.016 m 3, , 29. (i) 332.5 cm 2 (ii) 10 cm, ` 53.93, , 20. 594 cm 2, 24. 502.04 cm 3, 28. 6835.5 cm 3 , 2992.5 cm 2, , 30. 770 cm 2 31. 526.17 cm 3 , 616 cm 3, , 32. Area painted red = 58.143 cm 2 , Area painted white = 264 cm 2, HINTS TO SOME SELECTED QUESTIONS, 4. (i) Area of the cloth required = curved surface area of the tent = πrl = πr r 2 + h 2 ., Length of the cloth required =, , area, ⋅, width, , 5. Let the radii of the cones be 4r and 5r and their heights be h1 and h2 respectively. Then,, 1, π × ( 4r ) 2 × h1, 1, 16 h1 1, 1 25 25, h, 3, =, ⋅, = ⇒, = ⇒ 1 = ×, 1, 2, 4, 25, 4, 4, 16 64, h, h, 2, 2, π × (5r ) × h2, 3, 6. πr 2 = 1. 54. Find r., Calculate l, using the formula, l = r 2 + h 2 ., 7. h + r = 37 m and 2 πr ( h + r ) = 1628. Calculate r. Then, find h and calculate volume., 8. 4 πr 2 = 2464. Find r., Now, calculate surface area using 2r as radius., 9. Total area of the canvas required, = curved surface area of the cylinder + curved surface area of the cone., area, Length of the canvas =, ⋅, width, 13. Height of the conical part = (9.5 – 3.5) cm = 6 cm., Volume of the solid = volume of the cone + volume of the hemisphere., 18. Length of the cylindrical part = ( 90 − 2 × 21) cm = 48 cm., Total area to be painted, = 2 × curved surface area of a hemisphere + curved surface area of the cylinder., 20. Total surface area, = curved surface area of the cylinder + 2× curved surface area of each hemisphere.
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766, , Secondary School Mathematics for Class 10, , 22. Volume of the remaining solid = volume of the cylinder – volume of the cone., Surface area of the remaining solid, = curved surface area of the cylinder + curved surface area of the cone, + area of upper circular face of the cylinder., 28. Surface area of the remaining piece, = total surface area of the cube – area of the top of the hemispherical part, + curved surface area of the hemisphere., Volume of the remaining piece, = volume of the cube – volume of the hemisphere., 29. (ii) Largest possible diameter = edge of the cube., Total surface area of the solid, = surface area of the cube – area of the top of the hemispherical part, + curved surface area of the hemisphere., , ................................................................, , CONVERSION OF SOLID FROM ONE SHAPE, TO ANOTHER AND MIXED PROBLEMS, EXAMPLE 1, , The dimensions of a metallic cuboid are 100 cm × 80 cm × 64 cm. It, is melted and recast into a cube. Find the surface area of the cube., [ CBSE 2011], , SOLUTION, , Let each edge of the cube be a., Then, volume of the cube = a3 ., Volume of the cuboid = (100 × 80 × 64) cm 3, = 512000 cm 3 ., Volume of the cube = volume of the cuboid, ⇒ a3 = 512000 cm 3, ⇒ a = 3 512 × 1000 cm = ( 8 × 10) cm = 80 cm., Surface area of the cube = 6 a2, = (6 × 80 × 80) cm 2 = 38400 cm 2 ., , EXAMPLE 2, , How many spherical solid bullets can be made out of a solid cube of, lead whose edge measures 44 cm, each bullet being 4 cm in diameter?, , SOLUTION, , Edge of the cube, a = 44 cm., , [CBSE 2014], , Volume of the cube = a3 = ( 44 × 44 × 44) cm 3 ., Radius of each spherical bullet, r = 2 cm., 4, Volume of each spherical bullet = πr 3, 3
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Volume and Surface Areas of Solids, , 767, , 4 22, 704 ⎞, ⎟ cm 3, = ⎛⎜ ×, × 2 × 2 × 2 ⎞⎟ cm 3 = ⎛⎜, ⎝3 7, ⎠, ⎝ 21 ⎠, ∴ required number of bullets, , EXAMPLE 3, , =, , volume of the cube, volume of each spherical bullet, , =, , 44 × 44 × 44, 21, = 44 × 44 × 44 ×, = 2541., 704 / 21, 704, , How many silver coins, 1.75 cm in diameter and of thickness 2 mm,, must be melted to form a cuboid of dimensions, 5.5 cm × 10 cm × 3.5 cm?, , SOLUTION, , 175 ⎞, 7, ⎛ 1 .75 ⎞, ⎟ cm = cm., Radius of each coin, r = ⎜, ⎟ cm = ⎛⎜, ⎝, ⎠, ⎝ 2 ⎠, 200, 8, 2, 1, Thickness of each coin, h = 2 mm = ⎛⎜ ⎞⎟ cm = cm., ⎝ 10 ⎠, 5, Volume of each coin = πr 2 h, 22 7 7 1 ⎞, 77 ⎞, ⎟ cm 3 ., = ⎛⎜, × × × ⎟ cm 3 = ⎛⎜, ⎝ 7 8 8 5⎠, ⎝ 160 ⎠, Volume of the cuboid = (5 .5 × 10 × 3 .5) cm 3, , EXAMPLE 4, , SOLUTION, , = 192 .5 cm 3 ., volume of the cuboid, ∴ number of coins =, volume of each coin, 160, = 192 .5 ×, = 400., 77, A metallic sphere of radius 10.5 cm is melted and then recast into, smaller cones, each of radius 3.5 cm and height 3 cm. How many, cones are obtained?, [CBSE 2004, ’12, ’17], 21, cm., Radius of the sphere, R =, 2, 4, 4, 21 21 21 ⎞, Volume of the sphere = πR 3 = ⎛⎜ π ×, ×, × ⎟ cm 3, ⎝3, 3, 2, 2, 2⎠, 3087 π ⎞, ⎟ cm 3 ., = ⎛⎜, ⎝ 2 ⎠, 7, Radius of each cone, r = cm., 2, Height of each cone, h = 3 cm.
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768, , Secondary School Mathematics for Class 10, , Volume of each cone =, , 1 2, 1, 7 7, πr h = ⎛⎜ π × × × 3 ⎞⎟ cm 3, ⎝3, ⎠, 3, 2 2, , 49 π ⎞, ⎟ cm 3 ., = ⎛⎜, ⎝ 4 ⎠, volume of the sphere, volume of each cone, 3087 π, 4 ⎞, ⎟ = 126., = ⎛⎜, ×, ⎝ 2, 49 π ⎠, , ∴ required number of cones =, , EXAMPLE 5, , The internal and external radii of a hollow sphere are 3 cm and 5 cm, respectively. The sphere is melted to form a solid cylinder of height, 2 2 cm. Find the diameter and the curved surface area of the, 3, , cylinder., SOLUTION, , External radius of the sphere, r1 = 5 cm., Internal radius of the sphere, r2 = 3 cm., Volume of metal in the hollow sphere, 4, 4, = π (r13 − r23 ) = ⎡ π × (5 3 − 3 3 ) ⎤ cm 3, ⎢, ⎥⎦, 3, ⎣3, 392 π ⎞, ⎟ cm 3 ., = ⎛⎜, ⎝ 3 ⎠, Let the radius of the solid cylinder be R cm., 8, Height of the solid cylinder, h = cm., 3, Volume of the solid cylinder, ⎛ 8 πR 2 ⎞, 8, 3, = πR 2 h = ⎛⎜ πR 2 × ⎞⎟ cm 3 = ⎜, ⎟ cm ., ⎝, 3⎠, ⎝ 3 ⎠, Volume of cylinder = volume of metal in the hollow sphere, ⇒, , 8πR 2, 392 π, =, ⇒ R 2 = 49 ⇒ R = 7 ., 3, 3, , ∴, , diameter of the cylinder formed = (2 × 7) cm = 14 cm., , Curved surface area of the cylinder, 22, 8, = 2 πrh = ⎛⎜ 2 ×, × 7 × ⎞⎟ cm 2, ⎝, 7, 3⎠, =, , 352, 1, cm 2 = 117 cm 2 ., 3, 3
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Volume and Surface Areas of Solids, EXAMPLE 6, , SOLUTION, , 769, , A hollow sphere of internal and external diameters 4 cm and 8 cm is, melted to form a cone of base diameter 8 cm. Find the height and the, slant height of the cone., [CBSE 2011], External radius of the sphere, r1 = 4 cm., Internal radius of the sphere, r2 = 2 cm., Volume of metal in the hollow sphere, 4, 4, 224 π ⎞, ⎟ cm 3 ., = π (r13 − r23 ) = ⎡ π × ( 4 3 − 2 3 ) ⎤ cm 3 = ⎛⎜, ⎢⎣ 3, ⎥⎦, ⎝ 3 ⎠, 3, Radius of the cone formed, R = 4 cm., Let the height of the cone be h., Volume of the cone, =, , 1, πR 2 h =, 3, , ⎛⎜ 1 π × 4 × 4 × h ⎞⎟ cm 3 = ⎛ 16 πh ⎞ cm 3 ., ⎜, ⎟, ⎝3, ⎠, ⎝ 3 ⎠, , Volume of the cone = volume of metal in the sphere, 16 πh 224 π, =, ⇒ h = 14 cm., ⇒, 3, 3, Slant height, l = R 2 + h 2 = 4 2 + 14 2 cm, = 212 cm = 2 53 cm., , EXAMPLE 7, , SOLUTION, , Hence, the height of the cone is 14 cm and its slant height is, 2 53 cm., A girl empties a cylindrical bucket full of sand, of base radius 18 cm, and height 32 cm, on the floor to form a conical heap of sand. If the, height of this conical heap is 24 cm then find its slant height correct, to one place of decimal., [CBSE 2014], Radius of the cylindrical bucket, R = 18 cm., Height of the cylindrical bucket, H = 32 cm., Volume of sand in the bucket = πR 2 H = ( π × 18 × 18 × 32) cm 3 ., Let the radius of the conical heap be r., Height of the conical heap, h = 24 cm., 1, 1, Volume of the conical heap = πr 2 h = ⎛⎜ π × r 2 × 24 ⎞⎟ cm 3, ⎝3, ⎠, 3, 2, 3, = ( 8 πr ) cm ., Volume of the conical heap = volume of sand in the bucket, 18 × 18 × 32, ⇒ 8 πr 2 = π × 18 × 18 × 32 ⇒ r 2 =, = 1296, 8, ⇒ r = 1296 = 36 cm.
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770, , Secondary School Mathematics for Class 10, , Slant height, l = r 2 + h 2, = ( 36) 2 + ( 24) 2 cm = 1296 + 576 cm, = 1872 cm = 43.26 cm., EXAMPLE 8, , SOLUTION, , ≈ 43. 3 cm (correct to one decimal place)., 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and, recast into a metallic sphere. Find the diameter of the sphere and, hence find its surface area., [CBSE 2015], 35 ⎞, 7, ⎛ 3 .5 ⎞, ⎛, Radius of each cone, r = ⎜, ⎟ cm = ⎜ ⎟ cm = cm., ⎝ 20 ⎠, ⎝ 2 ⎠, 4, Height of each cone, h = 3 cm., 1, Volume of each cone = πr 2 h, 3, 1, 7 7, 49 π ⎞, ⎟ cm 3 ., = ⎛⎜ × π × × × 3 ⎞⎟ cm 3 = ⎛⎜, ⎝3, ⎠, ⎝ 16 ⎠, 4 4, Total volume of 504 cones, 49 π, 49 π × 63 ⎞, ⎟ cm 3 ., = ⎛⎜, × 504 ⎞⎟ cm 3 = ⎛⎜, ⎝ 16, ⎠, ⎝, ⎠, 2, Let the radius of the sphere be R., 4, Then, volume of the sphere = πR 3 ., 3, Volume of the sphere = total volume of 504 cones, 4, 49 π × 63, 49 × 63 3 ⎞, πR 3 =, ⇒ R 3 = ⎛⎜, × ⎟, ⇒, ⎝ 2, 3, 2, 4⎠, 7 × 7 ×7 × 9 × 3⎞, 7 × 3⎞, 21, ⎟ ⇒ R = ⎛⎜, ⎟ cm =, cm., ⇒ R 3 = ⎛⎜, ⎝, ⎠, ⎝ 2 ⎠, 2×4, 2, 21, Diameter of the sphere = ⎛⎜, × 2 ⎞⎟ cm = 21 cm., ⎝ 2, ⎠, 22 21 21 ⎞, Surface area of the sphere = 4 πR 2 = ⎛⎜ 4 ×, ×, × ⎟ cm 2, ⎝, 7, 2, 2⎠, , EXAMPLE 9, , SOLUTION, , = 1386 cm 2 ., Two spheres of same metal weigh 1 kg and 7 kg. The radius of the, smaller sphere is 3 cm. The two spheres are melted to form a single, big sphere. Find the diameter of the new sphere. [HOTS] [CBSE 2015], Since both the spheres are made of same metal, their weights, are directly proportional to their volumes, or the ratio of their, volumes is equal to the ratio of their weights.
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Volume and Surface Areas of Solids, , 771, , Radius of first small sphere, r1 = 3 cm., Let the radius of second small sphere be r2 cm., ∴, , volume of first small sphere, 1, =, volume of second small sphere 7, , 4 3, πr1, 1, 33 1, ⇒ 3, =, ⇒ 3 = ⇒ r23 = 27 × 7 = 189., 4 3 7, 7, r2, πr, 3 2, Sum of volumes of two small spheres, 4, 4, = π (r13 + r23 ) = ⎡ π ( 3 3 + 189) ⎤ cm 3, 3, ⎣⎢ 3, ⎦⎥, 4, = ⎛⎜ π × 216 ⎞⎟ cm 3 ., ⎝3, ⎠, Let the radius of the new sphere be R., Volume of new sphere = sum of volumes of two small spheres, 4, 4, ⇒, πR 3 = π × 216 ⇒ R 3 = 216, 3, 3, ⇒ R = 3 216 = 6 cm ⇒ diameter = 2 R = 12 cm., EXAMPLE 10, , The diameter of a copper sphere is 6 cm. The sphere is melted and, drawn into a long wire of uniform circular cross section. If the length, of the wire is 36 m, find its thickness., , SOLUTION, , Radius of the sphere, R = 3 cm., 4, Volume of the sphere = πR 3, 3, 4, = ⎛⎜ π × 3 × 3 × 3 ⎞⎟ cm 3, ⎝3, ⎠, = ( 36 π) cm 3 ., Length of the wire, h = 36 m = 3600 cm., Let the radius of the wire be r., Volume of the wire = πr 2 h = ( πr 2 × 3600) cm 3 ., But, volume of the wire = volume of the sphere, 1, ⇒ 3600 πr 2 = 36 π ⇒ r 2 =, 100, 1, 1, cm =, cm = 1 mm., ⇒ r=, 100, 10, Hence, thickness of the wire = its diameter = 2 mm.
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772, , Secondary School Mathematics for Class 10, , EXAMPLE 11, , A solid cylinder of diameter 12 cm and height 15 cm is melted and, recast into 12 toys in the shape of a right circular cone mounted on a, hemisphere. Find the radius of the hemisphere and the total height of, the toy, if the height of the conical part is thrice its radius., , SOLUTION, , Radius of the cylinder, R = 6 cm., , [CBSE 2005C], , Height of the cylinder, H = 15 cm, Volume of the cylinder = πR 2 H, = ( π × 6 × 6 × 15) cm 3, = (540 π) cm 3 ., Let the radius of each of the conical and, hemispherical parts be r., Then, height of the conical part, h = 3r., Volume of each toy, = volume of the hemisphere + volume of the cone, 2, 1, 2, 1, = πr 3 + πr 2 h = ⎛⎜ πr 3 + πr 2 × 3r ⎞⎟ cm 3, ⎝3, ⎠, 3, 3, 3, ⎛ 5 πr 3 ⎞, 2, 3, = ⎛⎜ πr 3 + πr 3 ⎞⎟ cm 3 = ⎜, ⎟ cm ., ⎝3, ⎠, ⎝ 3 ⎠, Now, total volume of 12 toys = volume of the cylinder, ⇒ 12 ×, , 540, 5 πr 3, = 540 π ⇒ r 3 =, = 27 ⇒ r =, 20, 3, , 3, , 27 = 3 cm., , Hence, radius of the hemisphere = 3 cm., Total height of the toy = (r + h) = (r + 3r) = 4r, EXAMPLE 12, , SOLUTION, , = ( 4 × 3) cm = 12 cm., Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be, melted to form a right circular cylinder of height 10 cm and diameter, 4.5 cm., 1 .5, 15, 3, Radius of each cylindrical coin, r =, cm =, cm = cm., 2, 20, 4, 2, 1, Thickness of each cylindrical coin, h = 2 mm =, cm = cm., 10, 5, 3, 3, 1, Volume of each coin = πr 2 h = ⎛⎜ π × × × ⎞⎟ cm 3, ⎝, 4 4 5⎠, 9π ⎞, ⎟ cm 3 ., = ⎛⎜, ⎝ 80 ⎠
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Volume and Surface Areas of Solids, , 773, , ⎛ 4 .5 ⎞, Radius of the required cylinder, R = ⎜, ⎟ cm = 2.25 cm., ⎝ 2 ⎠, Height of the required cylinder, H = 10 cm., Volume of the required cylinder, = πR 2 H = ( π × 2.25 × 2.25 × 10) cm 3, 225 225, 405 π ⎞, ⎟ cm 3 ., = ⎛⎜ π ×, ×, × 10 ⎞⎟ cm 3 = ⎛⎜, ⎝, ⎠, ⎝, 100 100, 8 ⎠, Required number of coins, volume of the required cylinder ⎛ 405 π 80 ⎞, ⎟ = 450., =, =⎜, ×, ⎝ 8, volume of each coin, 9π ⎠, EXAMPLE 13, , A solid iron rectangular block of dimensions 4.4 m , 2.6 m and 1 m, is cast into a hollow cylindrical pipe of internal radius 30 cm and, thickness 5 cm. Find the length of the pipe., [CBSE 2017], , SOLUTION, , Volume of iron = ( 440 × 260 × 100) cm 3 ., Internal radius of the pipe = 30 cm., External radius of the pipe = ( 30 + 5) cm = 35 cm., Let the length of the pipe be h cm., Volume of iron in the pipe, = (external volume) − (internal volume), = [π ( 35) 2 h − π ( 30) 2 h] cm 3 = πh[( 35) 2 − ( 30) 2 ] cm 3, = ( 65 × 5) πh cm 3 = ( 325 πh) cm 3 ., ∴, , 325 πh = 440 × 260 × 100, , ⎛ 440 × 260 × 100 × 7 ⎞, ⇒ length = h cm = ⎜, ⎟ cm, 325 × 22, ⎝, ⎠, = 11200 cm = 112 m., EXAMPLE 14, , SOLUTION, , Hence, the length of the pipe is 112 m., A hemispherical bowl of internal radius 9 cm is full of liquid. The, liquid is to be filled into cylindrical-shaped small bottles, each of, diameter 3 cm and height 4 cm. How many bottles are needed to, empty the bowl?, [CBSE 2005, ’12], Radius of the hemispherical bowl, R = 9 cm., 2, Volume of liquid in the bowl = πR 3, 3, 2, = ⎛⎜ π × 9 × 9 × 9 ⎞⎟ cm 3, ⎝3, ⎠, = ( 486 π) cm 3 .
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774, , Secondary School Mathematics for Class 10, , Radius of each cylindrical bottle, r =, , 3, cm., 2, , Height of each cylindrical bottle, h = 4 cm., Volume of each cylindrical bottle, 3 3, = πr 2 h = ⎛⎜ π × × × 4 ⎞⎟ cm 3 = ( 9 π) cm 3 ., ⎝, ⎠, 2 2, Required number of bottles, volume of liquid in the bowl 486 π, =, =, = 54., volume of each bottle, 9π, EXAMPLE 15, , A hemispherical bowl of internal diameter 36 cm contains liquid., This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find, the height of each bottle if 10% liquid is wasted in this transfer., , SOLUTION, , Radius of the hemispherical bowl, R = 18 cm., , [CBSE 2015], , Volume of liquid in the bowl,, 2, 2, r = πR 3 = ⎛⎜ π × 18 × 18 × 18 ⎞⎟ cm 3, ⎝, ⎠, 3, 3, = ( 3888π) cm 3 ., Volume of liquid filled into bottles, = ( 90% of 3888 π) cm 3, 90, = ⎛⎜, × 3888 π ⎞⎟ cm 3 ., ⎝ 100, ⎠, Let the height of each cylindrical bottle be h., Radius of each cylindrical bottle, r = 3 cm., Volume of each cylindrical bottle, = πr 2 h = ( π × 3 2 × h) cm 3 = ( 9 πh) cm 3 ., Now, volume of 72 cylindrical bottles, , EXAMPLE 16, , = volume of liquid filled into bottles, 90, ⇒ 72 × 9 πh =, × 3888 π, 100, 90 × 3888 ⎞, 27, ⎟ cm =, ⇒ h = ⎛⎜, cm = 5.4 cm., ⎝ 100 × 72 × 9 ⎠, 5, A conical vessel whose internal radius is 5 cm and height 24 cm, is, full of water. The water is emptied into a cylindrical vessel with, internal radius 10 cm. Find the height to which the water rises in the, cylindrical vessel.
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Volume and Surface Areas of Solids, SOLUTION, , 775, , Radius of the conical vessel, r = 5 cm., Height of the conical vessel, h = 24 cm., Volume of the conical vessel, 1, 1, = πr 2 h = ⎛⎜ π × 5 × 5 × 24 ⎞⎟ cm 3 = ( 200 π) cm 3 ., ⎝, ⎠, 3, 3, Radius of the cylindrical vessel, R = 10 cm., Let the height to which water rises in the vessel be H. Then,, volume of the water in the cylindrical vessel, = πR 2 H = ( π × 10 × 10 × H) cm 3 = (100 πH) cm 3 ., Volume of the water in cylindrical vessel, = volume of the water in the conical vessel, 200, ⇒ 100 πH = 200 π ⇒ H =, = 2 cm., 200, Hence, the required height is 2 cm., , EXAMPLE 17, , A sphere of diameter 6 cm is dropped into a right circular cylindrical, vessel, partly filled with water. The diameter of the cylindrical vessel, is 12 cm. If the sphere is completely submerged in water, by how, much will the level of water rise in the cylindrical vessel?, , SOLUTION, , Radius of the sphere, R = 3 cm., 4, 4, Volume of the sphere = πR 3 = ⎛⎜ π × 3 3 ⎞⎟ cm 3 = ( 36 π) cm 3 ., ⎝3, ⎠, 3, Radius of the cylindrical vessel, r = 6 cm., Let the rise in water level be h., Increase in volume of water when the sphere is submerged, = πr 2 h = ( π × 6 × 6 × h) cm 3 = ( 36 πh) cm 3 ., But this volume must be equal to the volume of the sphere., ∴, , 36 πh = 36 π ⇒ h = 1 cm., , Hence, rise in the water level is 1 cm., EXAMPLE 18, , Sushant has a vessel of the form of an inverted cone, open at the top,, of height 11 cm and radius of the top as 2.5 cm and is full of water., Metallic spherical balls each of diameter 0.5 cm are put in the vessel, due to which two-fifths of the water in the vessel flows out. Find how, many balls were put in the vessel. Sushant made the arrangement so, that the water that flows out irrigates the flower beds. What value has, been shown by Sushant?, [CBSE 2014]
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776, SOLUTION, , Secondary School Mathematics for Class 10, , Height of the conical vessel, h = 11 cm., Radius of the conical vessel, r = 2.5 cm., Volume of the conical vessel, 1, 1, = πr 2 h = ⎛⎜ π × 2 .5 × 2 .5 × 11 ⎞⎟ cm 3, ⎝, ⎠, 3, 3, 1, 5 5, 275 π ⎞, ⎟ cm 3 ., = ⎛⎜ π × × × 11 ⎞⎟ cm 3 = ⎛⎜, ⎝3, ⎠, ⎝, 2 2, 12 ⎠, Volume of water flown out after putting spherical balls, 2, 275 π ⎞, 55 π ⎞, ⎟ cm 3 = ⎛⎜, ⎟ cm 3 ., = ⎛⎜ of, ⎝5, ⎠, ⎝, 12, 6 ⎠, Radius of each spherical ball,, 0 .5, 25 ⎞, 1, ⎟ cm = cm., R=, cm = 0.25 cm = ⎛⎜, ⎝ 100 ⎠, 2, 4, Volume of each spherical ball, 4, 4, 1 1 1, π, = πR 3 = ⎛⎜ π × × × ⎞⎟ cm 3 = ⎛⎜ ⎞⎟ cm 3 ., ⎝3, ⎝ 48 ⎠, 3, 4 4 4⎠, ∴ required number of balls, volume of water flown out, 55 π ⎞ 48, ⎟×, =, = ⎛⎜, = 440., volume of each spherical ball ⎝ 6 ⎠ π, , EXAMPLE 19, , SOLUTION, , Value shown: We must not waste water and try to put every, drop of it to optimum use., A vessel full of water is in the form of an inverted cone of height 8 cm, and the radius of its top, which is open, is 5 cm. 100 spherical lead, balls are dropped into the vessel. One-fourth of the water flows out of, the vessel. Find the radius of a spherical ball., [CBSE 2015], Height of the conical vessel, h = 8 cm., Radius of the conical vessel, r = 5 cm., Volume of the conical vessel, 1, 1, 200 π ⎞, ⎟ cm 3 ., = πr 2 h = ⎛⎜ π × 5 × 5 × 8 ⎞⎟ cm 3 = ⎛⎜, ⎝3, ⎠, ⎝ 3 ⎠, 3, Volume of water flown out on dropping spherical balls, 1, 200 π ⎞, 50 π ⎞, ⎟ cm 3 = ⎛⎜, ⎟ cm 3 ., = ⎛⎜ of, ⎝4, ⎝ 3 ⎠, 3 ⎠, Let the radius of each spherical ball be R., 4, Volume of each spherical ball = πR 3 ., 3
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Volume and Surface Areas of Solids, , 777, , Now, volume of 100 balls = volume of water flow out, 50 3, 1 ⎞ 1, 4, 50 π, ⎟ =, ⇒ 100 × πR 3 =, ⇒ R 3 = ⎛⎜ × ×, ⎝ 3 4 100 ⎠ 8, 3, 3, 1 1, = cm = 0 .5 cm., 8 2, A housing society used to collect rain water from the roof of its, building 22 m × 20 m to a cylindrical vessel having diameter of base, 2 m and height 3.5 m and then pump this water into the main water, tank so that all members can use it. On a particular day the rain, water collected from the roof just filled the cylindrical vessel. Then,, find the rainfall in centimetre., [CBSE 2013C], Length of the roof = 22 m. Breadth of the roof = 20 m., ⇒ R=, , EXAMPLE 20, , SOLUTION, , 3, , Let the rainfall be x cm., Volume of water on the roof, x ⎞ 3 ⎛ 22 x ⎞ 3, ⎟ m =⎜, ⎟ m ., = ⎛⎜ 22 × 20 ×, ⎝, ⎝ 5 ⎠, 100 ⎠, Radius of the cylindrical vessel, r = 1 m., Height of the cylindrical vessel, h = 3.5 m., Volume of water in the cylindrical vessel when just full, 22, 7, = πr 2 h = ⎛⎜, × 1 × 1 × ⎞⎟ m 3 = 11 m 3 ., ⎝ 7, 2⎠, Now, volume of water on the roof, = volume of water in the cylindrical vessel, 22 x, 11 × 5 ⎞, ⎟ = 2.5., ⇒, = 11 ⇒ x = ⎛⎜, ⎝ 22 ⎠, 5, Hence, the rainfall is 2.5 cm., EXAMPLE 21, , SOLUTION, , Water flows through a circular pipe whose internal diameter is 2 cm,, at the rate of 0.7 m per second into a cylindrical tank, the radius of, whose base is 40 cm. By how much will the level of water rise in the, tank in half an hour?, [CBSE 2006C], Internal radius of the pipe, r = 1 cm., Length of water flowing in 1 second, h = 0.7 m = 70 cm., Volume of water flowing in 1 second, = πr 2 h = ( π × 1 × 1 × 70) cm 3 = (70 π) cm 3 ., Volume of water flowing in 30 minutes, = (70 π × 60 × 30) cm 3 = (126000 π) cm 3 .
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778, , Secondary School Mathematics for Class 10, , Radius of cylindrical tank, R = 40 cm., Let the rise in level of water be H cm., Volume of water in the tank, = πR 2 H = ( π × 40 × 40 × H) cm 3, = (1600 πH) cm 3 ., Volume of water in the tank, = volume of water flown through the pipe, 126000 315, ⇒ 1600 πH = 126000 π ⇒ H =, =, = 78 .75., 1600, 4, Hence, rise in level = 78.75 cm., EXAMPLE 22, , SOLUTION, , Water is flowing at the rate of 15 km per hour through a pipe of, diameter 14 cm into a rectangular tank which is 50 m long and 44 m, wide. Find the time in which the level of water in the tank will rise by, 21 cm., [CBSE 2006, ’11], Length of the tank = 50 m., Width of the tank = 44 m., 21, Desired rise in level = 21 cm =, m., 100, Desired volume of water in the tank, 21 ⎞ 3, ⎟ m = 462 m 3 ., = ⎛⎜ 50 × 44 ×, ⎝, 100 ⎠, 7, Radius of the pipe, r = 7 cm =, m., 100, Length of water flowing in 1 hour, h = 15 km = 15000 m., Volume of water flowing from the pipe in 1 hour, 22, 7, 7, = πr 2 h = ⎛⎜, ×, ×, × 15000 ⎞⎟ m 3 = 231 m 3 ., ⎝ 7 100 100, ⎠, desired volume, ∴ required time =, volume of water flown in 1 hour, 462 ⎞, ⎟ hours = 2 hours., = ⎛⎜, ⎝ 231 ⎠, , EXAMPLE 23, , SOLUTION, , A farmer connects a pipe of internal diameter 20 cm from a canal into, a cylindrical tank in his field, which is 10 m in diameter and 2 m, deep. If water flows through the pipe at the rate of 3 km/hr, in how, much time will the tank be filled?, [CBSE 2008C], 10, 1, ⎞⎟ m =, Radius of the pipe, r = 10 cm = ⎛⎜, m., ⎝ 100 ⎠, 10
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Volume and Surface Areas of Solids, , 779, , Length of water flowing through the pipe in 1 hour,, h = 3 km = 3000 m., Volume of water that flows through the pipe in 1 hour, 1, 1, = πr 2 h = ⎛⎜ π ×, ×, × 3000 ⎞⎟ m 3 = ( 30 π) m 3 ., ⎝, ⎠, 10 10, Radius of the cylindrical tank, R = 5 m., Depth of the tank, H = 2 m., Volume of the tank = πR 2 H = ( π × 5 × 5 × 2) m 3, = (50 π) m 3 ., Time taken to fill the tank, volume of the tank, =, volume of water flown in 1 hour, 50 π ⎞, 5, ⎟ hours = hours = 1 hr 40 min., = ⎛⎜, ⎝ 30 π ⎠, 3, Hence, the required time is 1 hr 40 min., EXAMPLE 24, , SOLUTION, , Water is flowing at the rate of 2.52 km/hr through a cylindrical pipe, into a cylindrical tank, the radius of whose base is 40 cm. If the, increase in the level of water in the tank, in half an hour is 3.15 m,, find the internal diameter of the pipe., [CBSE 2015], Let the internal radius of the pipe be r., Length of water flowing rhrough the pipe in 1 hour,, h = 2.52 km = 2520 m., Volume of water flowing through the pipe in 1 hour, = πr 2 h = ( π × r 2 × 2520) m 3 ., Volume of water flowing through the pipe in half an hour, 1, = ⎛⎜ × πr 2 × 2520 ⎞⎟ m 3 = (1260 πr 2 ) m 3 ., ⎝2, ⎠, Radius of the cylindrical tank, R = 40 cm =, , 40, 2, m = m., 100, 5, , Increase in level of water, H = 3.15 m., Volume of water filled in the tank, 2 2 315 ⎞ 3 ⎛ 63 π ⎞ 3, ⎟ m =⎜, ⎟ m ., = πR 2 H = ⎛⎜ π × × ×, ⎝, ⎝ 125 ⎠, 5 5 100 ⎠, Now, volume of water flown in half an hour, = volume of water filled in the tank
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780, , Secondary School Mathematics for Class 10, , ⇒ 1260 πr 2 =, ⇒ r=, , 63 π, 63, 1, ⇒ r2 =, =, 125, 125 × 1260 2500, , 1, 1, 1, =, m = ⎛⎜ × 100 ⎞⎟ cm = 2 cm., ⎝ 50, ⎠, 2500 50, , Hence, the internal diameter of the pipe = (2 × 2) cm = 4 cm., EXAMPLE 25, , A 21-m-deep well with diameter 6 m is dug and the earth from, digging is evenly spread to form a platform 27 m × 11 m. Find the, height of the platform., [CBSE 2015], , SOLUTION, , Depth of the well, h = 21 m., Radius of the well, r = 3 m., Volume of the earth taken out, 22, = πr 2 h = ⎛⎜, × 3 × 3 × 21 ⎞⎟ m 3 = 594 m 3 ., ⎝ 7, ⎠, Area of the platform = ( 27 × 11) m 2 = 297 m 2 ., Height of the platform, volume of the earth ⎛ 594 ⎞, ⎟ m = 2 m., =, =⎜, area of the platform ⎝ 297 ⎠, , EXAMPLE 26, , A well of diameter 4 m is dug 14 m deep. The earth taken out of it is, spread evenly all around the well to form a 40-cm-high embankment., Find the width of the embankment., [HOTS] [CBSE 2015], , SOLUTION, , Radius of the well, r = 2 m., Depth of the well, h = 14 m., Volume of the earth taken out, 22, = πr 2 H = ⎛⎜, × 2 × 2 × 14 ⎞⎟ m 3 = 176 m 3 ., ⎝ 7, ⎠, Let the width of the embankment be x metres., Internal radius of the embankment, r = 2 m., External radius of the embankment, R = (2 + x) m., 40 ⎞, ⎟ m., Height of the embankment, H = 40 cm = ⎛⎜, ⎝ 100 ⎠, Volume of the embankment, 40 π, = πH( R 2 − r 2 ) = ⎧⎨, [( 2 + x) 2 − 2 2 ] ⎫⎬ m 3, ⎩ 100, ⎭, π, 2, = ⎡ ( x 2 + 4 x) ⎤ m 3 ., ⎢⎣ 5, ⎥⎦
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Volume and Surface Areas of Solids, , 781, , Now, volume of the embankment = volume of the earth dug out, 2 22, 176 × 5 × 7, ×, × ( x 2 + 4 x) = 176 ⇒ x 2 + 4 x =, = 140, ⇒, 5 7, 2 × 22, ⇒ x 2 + 4 x − 140 = 0 ⇒ x 2 + 14 x − 10 x − 140 = 0, ⇒ x( x + 14) − 10( x + 14) = 0 ⇒ ( x + 14)( x − 10) = 0, ⇒ x = 10 [Q x ≠ −14]., Hence, width of the embankment = 10 m., EXAMPLE 27, , From each end of a solid metal cylinder, metal was scooped out in, hemispherical form of same diameter. The height of the cylinder is, 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is, melted and converted into a cylindrical wire of 1.4 cm thickness., Find the length of the wire., [HOTS] [CBSE 2015], , SOLUTION, , Height of the cylinder, h = 10 cm., Radius of each of the cylinder and hemisphere, r = 4.2 cm., Volume of the solid, = volume of the cylinder – volume of 2 hemispheres, 2, 4, = πr 2 h − 2 × πr 3 = πr 2 ⎛⎜ h − r ⎞⎟, ⎝, 3, 3 ⎠, 4, ⎡, ⎤, = ⎢ π × 4.2 × 4.2 × ⎛⎜ 10 − × 4.2 ⎞⎟ ⎥ cm 3, ⎝, ⎠⎦, 3, ⎣, = ( π × 4.2 × 4.2 × 4.4) cm 3 ., Let the length of the wire be H., 1.4 ⎞, 7, ⎟ cm = 0 .7 cm =, Radius of the wire, R = ⎛⎜, cm., ⎝ 2 ⎠, 10, 7, 7, Volume of the wire = πR 2 H = ⎛⎜ π ×, ×, × H ⎞⎟ cm 3 ., ⎝, ⎠, 10 10, Now, volume of the wire = volume of the solid, 7, 7, ⇒ π×, ×, × H = π × 4.2 × 4.2 × 4.4, 10 10, 4.2 × 4.2 × 4.4 × 10 × 10 ⎞, 42 × 42 × 4.4 ⎞, ⎟ cm = ⎛⎜, ⎟ cm, ⇒ H = ⎛⎜, ⎝, ⎠, ⎝, ⎠, 7 ×7, 7 ×7, = 158.4 cm., , EXAMPLE 28, , 50 circular plates each of radius 7 cm and thickness 5 mm are placed, one above another to form a solid right circular cylinder. Find the, total surface area of the cylinder so formed., [HOTS] [CBSE 2013C]
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782, SOLUTION, , Secondary School Mathematics for Class 10, , Clearly, we have, radius of the cylinder so formed, r = 7 cm, and, height of the cylinder so formed, h = (50 ´ 5) mm, = 250 mm = 25 cm., \, , EXAMPLE 29, , SOLUTION, , total surface area of the cylinder so formed, 22, = 2 pr( h + r) = é2 ´, ´ 7 ´ ( 25 + 7 ) ù cm 2, êë, úû, 7, 2, 2, = ( 2 ´ 22 ´ 32) cm = 1408 cm ., A copper wire, 3 mm in diameter, is wound about a cylinder whose, length is 12 cm and diameter 10 cm, so as to cover the curved surface, of the cylinder. Find the length and mass of the wire, assuming the, density of copper to be 8.88 g per cm 3 . [Use p = 3.14.], [HOTS], Diameter of the wire = 3 mm = 0.3 cm., Length of the cylinder = 12 cm., Radius of the cylinder = 5 cm., length of the cylinder 12, Number of turns =, =, = 40., diameter of the wire, 0. 3, Length of one turn = circumference of the base of the cylinder, = 2pr = (2 ´ 3.14 ´ 5) cm = 31.4 cm., Length of 40 turns = (31.4 ´ 40) cm = 1256 cm., , Total length of the wire wrapped, l = 1256 cm., 0. 3 ö, Radius of the wire, r = æç, ÷ cm = 0.15 cm., è 2 ø, Volume of the wire = pr 2 l = ( 3.14 ´ 0.15 ´ 0.15 ´ 1256) cm 3, = 88 .7364 cm 3 » 88 .74 cm 3 ., \ mass of the wire = volume ´ density = (88.74 ´ 8.88) g, = 788.01 g » 788 g., EXAMPLE 30, , A cistern, internally measuring 150 cm ´ 120 cm ´ 110 cm, has, 129600 cm 3 of water in it. Porous bricks are placed in the water until, the cistern is full to the brim. Each brick absorbs one-seventeenth of, its own volume of water. How many bricks can be put in without, overflowing the water, each brick being 22.5 cm ´ 7.5 cm ´ 6.5 cm?, , SOLUTION, , Total volume of the cistern = (150 ´ 120 ´ 110) cm 3, = 1980000 cm 3 ., Volume of water in the cistern = 129600 cm 3 ., Volume of empty space in the cistern, = (1980000 – 129600) cm 3 = 1850400 cm 3 .
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Volume and Surface Areas of Solids, , 783, , Total volume of each brick = (22.5 ´ 7.5 ´ 6.5) cm 3, = 1096.875 cm 3 ., 1, of its own volume,, Since each brick absorbs water equal to, 17, 1, so the effective volume of each brick may be taken as æç 1 - ö÷ ,, 17, è, ø, 16, of its actual volume., i.e.,, 17, 16, So, the effective volume of each brick = æç ´ 1096.875 ö÷ cm 3 ., ø, è 17, , EXAMPLE 31, , SOLUTION, , \ required number of bricks, volume of empty space in the cistern, =, effective volume of each brick, 1850400, 17 ´ 1850400, =, =, æ 16 ´ 1096.875 ö 16 ´ 1096.875, ç, ÷, è 17, ø, 17 ´ 1850400, =, = 1792.41 » 1792., 17550, A right triangle whose sides are 3 cm and 4 cm (other than, hypotenuse) is made to revolve about its hypotenuse. Find the, volume and surface area of the double cone so formed. (Choose the, value of p as found appropriate)., [HOTS], Let SAOB be the given right triangle in which ÐAOB = 90 °., When SAOB is rotated about the hypotenuse AB, the two, cones generated are AO¢O and BOO¢., Clearly, OA = 3 cm, OB = 4 cm., In right-angled SAOB, we have, AB = (OA) 2 + (OB) 2 = 3 2 + 4 2 cm, = 25 cm = 5 cm., Let OP = x cm. Then,, 1, 1, ´ OA ´ OB = ´ AB ´ OP, 2, 2, 1, 1, 3´ 4 ö, Þ, ´ 3 ´ 4 = ´ 5 ´ OP Þ OP = æç, ÷ cm = 2.4 cm., 2, 2, è 5 ø, In right-angled SAPO, we have, AP = (OA) 2 - (OP) 2 = 3 2 - ( 2.4) 2 cm, = 9 - 5 .76 cm = 3.24 cm = 1.8 cm.
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784, , Secondary School Mathematics for Class 10, , BP = AB − AP = (5 − 1.8) cm = 3.2 cm., ∴ radius of each of the cones AO′O and BOO′, r = OP = 2.4 cm., Height of the cone AO′ O , h = AP = 1.8 cm., Height of the cone BOO′ , H = BP = 3.2 cm., Volume of the double cone, = volume of the the cone AO′O + volume of the cone BOO′, 1, 1, 1, = πr 2 h + πr 2 H = πr 2 ( h + H), 3, 3, 3, 1, = ⎛⎜ × 3.14 × 2.4 × 2.4 × 5 ⎞⎟ cm 3 = 30.144 cm 3 ., ⎝3, ⎠, Slant height of the cone AO′O , l1 = OA = 3 cm., Slant height of the cone BO′O , l2 = OB = 4 cm., Surface area of the double cone, = curved surface area of the cone AO′O, + curved surface area of the cone BOO′, 22, = πrl1 + πrl 2 = πr( l 1 + l 2) = ⎡ × 2 . 4 × ( 3 + 4) ⎤ cm 2, ⎢⎣ 7, ⎥⎦, 22, = ⎛⎜, × 2.4 × 7 ⎞⎟ cm 2 = 52.8 cm 2 ., ⎝ 7, ⎠, , EXERCISE 17B, 1. A solid metallic cuboid of dimensions 9 m × 8 m × 2 m is melted and, recast into solid cubes of edge 2 m. Find the number of cubes so formed., [CBSE 2017], , 2. A cone of height 20 cm and radius of base 5 cm is made up of modelling, clay. A child reshapes it in the form of a sphere. Find the diameter of the, [CBSE 2011], sphere., 3. Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to, form a single solid sphere. Find the radius of the resulting sphere., [CBSE 2012], , 4. A solid metal cone with radius of base 12 cm and height 24 cm is melted, to form solid spherical balls of diameter 6 cm each. Find the number of, [CBSE 2005C], balls thus formed., 5. The radii of internal and external surfaces of a hollow spherical shell are, 3 cm and 5 cm respctively. It is melted and recast into a solid cylinder of, [CBSE 2012], diameter 14 cm. Find the height of the cylinder.
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Volume and Surface Areas of Solids, , 785, , 6. The internal and external diameters of a hollow hemispherical shell are, 6 cm and 10 cm respectively. It is melted and recast into a solid cone of, base diameter 14 cm. Find the height of the cone so formed. [CBSE 2005C], 7. A copper rod of diameter 2 cm and length 10 cm is drawn into a wire of, uniform thickness and length 10 m. Find the thickness of the wire., [CBSE 2012], , 8. A hemispherical bowl of internal diameter 30 cm contains some liquid., This liquid is to be filled into cylindrical-shaped bottles each of, diameter 5 cm and height 6 cm. Find the number of bottles necessary to, [CBSE 2004, ’11], empty the bowl., 9. A solid metallic sphere of diameter 21 cm is melted and recast into a, number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find, [CBSE 2004], the number of cones so formed., 10. A spherical cannon ball 28 cm in diameter is melted and recast into a, right circular conical mould, base of which is 35 cm in diameter. Find, [CBSE 2001C], the height of the cone., 11. A spherical ball of radius 3 cm is melted and recast into three spherical, balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius, of the third ball., 12. A spherical shell of lead whose external and internal diameters are, respectively 24 cm and 18 cm, is melted and recast into a right circular, cylinder 37 cm high. Find the diameter of the base of the cylinder., 13. A hemisphere of lead of radius 9 cm is cast into a right circular cone of, height 72 cm. Find the radius of the base of the cone., 14. A spherical ball of diameter 21 cm is melted and recast into cubes, each, of side 1 cm. Find the number of cubes so formed., 15. How many lead balls, each of radius 1 cm, can be made from a sphere of, radius 8 cm?, 16. A solid sphere of radius 3 cm is melted and then cast into small, spherical balls, each of diameter 0.6 cm. Find the number of small balls, so obtained., 17. The diameter of a sphere is 42 cm. It is melted and drawn into a, cylindrical wire of diameter 2.8 cm. Find the length of the wire., 18. The diameter of a copper sphere is 18 cm. It is melted and drawn into a, long wire of uniform cross section. If the length of the wire is 108 m,, find its diameter.
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786, , Secondary School Mathematics for Class 10, , 19. A hemispherical bowl of internal radius 9 cm is full of water. Its, contents are emptied into a cylindrical vessel of internal radius 6 cm., [CBSE 2012], Find the height of water in the cylindrical vessel., 20. A hemispherical tank, full of water, is emptied by a pipe at the rate of, 25 litres per second. How much time will it take to empty half the tank, 7, [CBSE 2012], if the diameter of the base of the tank is 3 m?, 21. The rain water from a roof of 44 m x 20 m drains into a cylindrical tank, having diameter of base 4 m and height 3.5 m. If the tank is just full, find, [CBSE 2014], the rainfall in cm., 22. The rain water from a 22 m × 20 m roof drains into a cylindrical vessel, of diameter 2 m and height 3.5 m. If the rain water collected from the, roof fills 4 th of the cylindrical vessel then find the rainfall in centimetre., 5, , [CBSE 2015, ’17], , 23. A solid right circular cone of height 60 cm and radius 30 cm is dropped, in a right circular cylinder full of water, of height 180 cm and radius, 60 cm. Find the volume of water left in the cylinder, in cubic metres., [CBSE 2015], , 24. Water is flowing through a cylindrical pipe of internal diameter 2 cm,, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per, second. Determine the rise in level of water in the tank in half an hour., [CBSE 2013], , 25. Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm, into a rectangular tank which is 60 m long and 22 m wide. Determine the, time in which the level of water in the tank will rise by 7 cm. [CBSE 2011], 26. Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed, of 25 km/hr. How much area can it irrigate in 40 minutes, if 10 cm of, [CBSE 2017], standing water is required for irrigation?, 27. A farmer connects a pipe of internal diameter 25 cm from a canal into a, cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep., If water flows through the pipe at the rate of 3.6 km/hr, in how much, time will the tank be filled? Also, find the cost of water if the canal, [CBSE 2009C], department charges at the rate of ` 0.07 per m 3 ., 28. Water running in a cylindrical pipe of inner diameter 7 cm, is collected, in a container at the rate of 192.5 litres per minute. Find the rate of flow, [CBSE 2013], of water in the pipe in km/hr., 29. 150 spherical marbles, each of diameter 14 cm, are dropped in a, cylindrical vessel of diameter 7 cm containing some water, which are, completely immersed in water. Find the rise in the level of water in, [CBSE 2014], the vessel.
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Volume and Surface Areas of Solids, , 787, , 30. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of, diameter 7 cm, containing some water. Find the number of marbles that, should be dropped into the beaker so that the water level rises by 5.6 cm., 31. In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth, taken out of it is spread all around to a width of 5 m to form an, embankment. Find the height of the embankment. What value of the, [CBSE 2014], villagers is reflected here?, 32. In a corner of a rectangular field with dimensions 35 m × 22 m, a well, with 14 m inside diameter is dug 8 m deep. The earth dug out is spread, evenly over the remaining part of the field. Find the rise in the level of, [HOTS], the field., 33. A copper wire of diameter 6 mm is evenly wrapped on a cylinder of, length 18 cm and diameter 49 cm to cover its whole surface. Find the, length and the volume of the wire. If the density of copper be 8.8 g per, [HOTS], cu-cm, find the weight of the wire., 34. A right triangle whose sides are 15 cm and 20 cm (other than, hypotenuse), is made to revolve about its hypotenuse. Find the volume, and surface area of the double cone so formed. (Choose value of π as, [HOTS], found appropriate), 35. In a hospital, used water is collected in a cylindrical tank of diameter, 2 m and height 5 m. After recycling, this water is used to irrigate a, park of hospital whose length is 25 m and breadth is 20 m. If the tank is, filled completely then what will be the height of standing water used, for irrigating the park? Write your views on recycling of water., [CBSE 2017], , ANSWERS (EXERCISE 17B), , 1. 18, , 2. 10 cm, , 3. 12 cm, , 4. 32, , 2, cm, 3, 10. 35.84 cm, 15. 512, 5. 2, , 6. 4 cm, 7. 2 mm, 8. 60, 9. 504, 11. 2.5 cm, 12. 12 cm, 13. 4.5 cm, 14. 4851, 16. 1000, 17. 63 m, 18. 0.6 cm, 19. 13.5 cm, 20. 16 minutes 30 seconds, 21. 5 cm, 22. 2 cm, 23. 1.98 m 3, 2, 24. 180 cm, 25. 1 hour, 26. 1620000 m 27. 1 hr 36 min, ` 19.80, 28. 3 km/hr 29. 56 m, 30. 150, 31. 2.8 m;, Value: We must labour hard to make maximum use of the available, resources., 32. 2 m, 33. 46.2 m, 1306.8 cm 3 , 11.5 kg 34. 3768 cm 3 , 1320 cm 2
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788, , Secondary School Mathematics for Class 10, , VOLUME AND SURFACE AREA OF A FRUSTUM OF A CONE, FRUSTUM OF A CONE, , When a cone is cut by a plane parallel to the base of the cone then the portion, between the plane and the base is called the frustum of the cone., FORMULAE, , Let R and r be the radii of the base and the top of the frustum of, , a cone., Let h be its height and l be its slant height., Then,, (i) Volume of the frustum of the cone, πh 2, =, [R + r 2 + Rr] cubic units., 3, (ii) Lateral surface area of the frustum of the cone, = πl( R + r), where l 2 = h 2 + ( R − r) 2 sq units., (iii) Total surface area of the frustum of the cone, = (area of the base) + (area of the top) + (lateral surface area), = [πR 2 + πr 2 + πl( R + r)] = π[R 2 + r 2 + l( R + r)] sq units., DERIVATION OF THE ABOVE FORMULAE, , Let H, R and l 1 be the height, radius of, the base and slant height respectively, of a cone OAB. Suppose it is cut by a, plane PQ parallel to its base at a height, h from the base to form the frustum, ABQP., Let the radius of the base of cone, OPQ be r and l be the slant height of, the frustum so formed., Clearly, we have OC = H , DC = h , OD = OC − DC = ( H − h), DQ = r ,, CB = R , OB = l 1 , BQ = l , OQ = OB – BQ = ( l 1 − l)., Clearly, SODQ ~ SOCB., l1 − l, OD DQ OQ, H−h r, ∴, =, =, ⇒, =, =, ⋅, OC, CB, OB, H, R, l1, H−h r, =, ⇒ R ( H − h) = Hr ⇒ HR − hR = Hr, H, R, hR, ⇒ H( R − r) = hR ⇒ H =, ⋅, R −r, l1 − l r, And,, =, ⇒ ( l1 − l) R = l1r ⇒ l1 R − lR = l1r, l1, R, , ... (i), , ... (ii)
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790, , Secondary School Mathematics for Class 10, , Volume of milk that the bucket can hold, 15400 ⎞, ⎟ litres = 15.4 litres., = ⎛⎜, ⎝ 1000 ⎠, ∴ cost of milk = ` (15.4 × 30) = ` 462., EXAMPLE 2, , SOLUTION, , A bucket open at the top, and made up of a metal sheet is in the form, of the frustum of a cone. The depth of the bucket is 24 cm and the, diameters of its upper and lower circular ends are 30 cm and 10 cm, respectively. Find the cost of metal sheet used in it at the rate of ` 10, per 100 m 2 . [Use π = 3.14.], [CBSE 2013], Radius of the upper end, R = 15 cm., Radius of the lower end, r = 5 cm., Depth of the bucket, h = 24 cm., Slant height of the bucket, l = h 2 + ( R − r) 2, = ( 24) 2 + (15 − 5) 2 cm, , EXAMPLE 3, , = 676 cm = 26 cm., Area of metal sheet used, = curved surface area + area of the bottom, = πl( R + r) + πr 2, = [ 3.14 × 26 × (15 + 5) + 3.14 × 5 × 5] cm 2, = ( 3.14 × 26 × 20 + 3.14 × 25) cm 2, = [ 3.14 × (520 + 25)] cm 2 = ( 3.14 × 545) cm 2 = 1711. 3 cm 2 ., 1711. 3, ∴ cost of metal sheet used = ` ⎛⎜, × 10 ⎞⎟ = ` 171.13., ⎝ 100, ⎠, A bucket is in the form of a frustum of a cone with a capacity of, 12308.8 cm 3 of water. The radii of the top and bottom circular ends, are 20 cm and 12 cm respectively. Find the height of the bucket and, the area of the metal sheet used in its making. [Use π = 3.14 .], [CBSE 2006C], , SOLUTION, , Here, R = 20 cm , r = 12 cm, and volume = 12308.8 cm 3, Let the height of the bucket be h cm., Volume of the bucket, = volume of frustum of the cone., 1, πh( R 2 + r 2 + Rr) = 12308.8, ∴, 3, 1, ⇒, × 3.14 × h[( 20) 2 + (12) 2 + 20 × 12] = 12308.8, 3, 12308.8 × 3, ⇒ 784 h =, 3.14
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Volume and Surface Areas of Solids, , 791, , æ 12308.8 ´ 3 ö, ÷ = 15., Þ h = çç, ÷, è 3.14 ´ 784 ø, Slant height of the bucket, l = h 2 + ( R - r) 2 units, = (15) 2 + ( 20 - 12) 2 cm = (15) 2 + 8 2 cm, = 225 + 64 cm = 289 cm = 17 cm., , EXAMPLE 4, , SOLUTION, , Area of the metal sheet used, = (curved surface area) + (area of the bottom), = [pl( R + r) + pr 2 ], = [ 3.14 ´ 17 ´ ( 20 + 12) + 3.14 ´ 12 ´ 12] cm 2, = [ 3.14 ´ (17 ´ 32) + 3.14 ´ 144] cm 2, = [ 3.14 ´ (544 + 144)] cm 2, = ( 3.14 ´ 688) cm 2 = 2160. 32 cm 2 ., In the figure, from the top of a solid cone of height 12 cm and base, radius 6 cm, a cone of height 4 cm is removed by a plane parallel, to the base. Find the total surface area of the remaining solid., [Use p = 227 and 5 = 2.236.], [CBSE 2015], , Clearly, the remaining solid is the frustum CDBA of cone OPD., Radius of lower end of the frustum, R = CD = 6 cm., Height of the frustum,, h = AC = OC - OA, = (12 - 4) cm = 8 cm., Let the radius of upper face be r cm., Now, SOAB ~ SOCD., OA AB, 4 r, So,, =, Þ, =, OC CD, 12 6, 4, Þ r = æç ´ 6 ö÷ cm = 2 cm., è 12, ø
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792, , Secondary School Mathematics for Class 10, , Slant height of the frustum,, l = h 2 + ( R − r) 2 = 8 2 + ( 6 − 2) 2 = 80 cm = 4 5 cm., Total surface area of the frustum, = area of base + area of top + curved surface area, = πR 2 + πr 2 + πl( R + r) = π [R 2 + r 2 + l( R + r)], 22, =, × [6 2 + 2 2 + 4 × 2.236 × ( 6 + 2)] cm 2, 7, 22, 22, = ⎡ × ( 36 + 4 + 71 .552) ⎤ cm 2 = ⎛⎜, × 11 .552 ⎞⎟ cm 2, ⎝ 7, ⎠, ⎣⎢ 7, ⎦⎥, = ( 22 × 15 . 936) cm 2 = 350 .592 cm 2 ., EXAMPLE 5, , An open metallic bucket is in the shape of a, frustum of a cone mounted on hollow, cylindrical base made of metallic sheet. If, the diameters of the two circular ends of the, bucket are 45 cm and 25 cm, the total, vertical height of the bucket is 30 cm and, that of the cylindrical portion is 6 cm, find, the area of the metallic sheet used to make, the bucket. Also, find the volume of water it, can hold. [Take π = 227 .], , SOLUTION, , Here R =, , 45, 25, cm , r =, cm = 12.5 cm., 2, 2, Height of the frustum of the cone, h = ( 30 − 6) cm = 24 cm., , ∴, , h = 24 cm., , Slant height of the frustum of the cone,, l = h 2 + ( R − r) 2 units = ( 24) 2 + ( 22.5 − 12.5) 2 cm, = ( 24) 2 + (10) 2 cm = 576 + 100 cm, = 676 cm = 26 cm., Area of metallic sheet used, = (curved surface area of the frustum of the cone), + (area of the base) + (curved surface area of the cylinder), = πl( R + r) + πr 2 + 2 πrH , where H = 6 cm, = π ⋅ {l( R + r) + r 2 + 2rH} sq units, 22, =, ⋅ {26(22.5 + 12.5) + (12.5) 2 + 2 × 12.5 × 6} cm 2, 7, 22, =, ⋅ {( 26 × 35) + (12.5 × 12.5) + 150} cm 2, 7
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Volume and Surface Areas of Solids, , 793, , 22, 22, ⋅ {910 + 156.25 + 150} cm 2 = ⎛⎜, × 1216.25 ⎞⎟ cm 2, ⎝ 7, ⎠, 7, 2, 2, = ( 22 × 173.75) cm = 3822.5 cm ., =, , EXAMPLE 6, , SOLUTION, , EXAMPLE 7, , Volume of water which the bucket can hold, 1, = πh[R 2 + r 2 + Rr] cm 3, 3, 2, ⎡ 45 2, 1 22, 25, 45 25 ⎞ ⎤, ⎟ ⎥ cm 3, = ×, × 24 × ⎢ ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ + ⎛⎜, ×, ⎝ 2⎠, ⎝ 2⎠, ⎝ 2, ⎠, 3 7, 2, ⎣, ⎦, 176 ⎛ 2025 625 1125 ⎞, 176, 3775, ⎞⎟ cm 3, ⎟ cm 3 = ⎛⎜, =, ×⎜, +, +, ×, ⎝ 4, ⎝ 7, 7, 4, 4 ⎠, 4 ⎠, 166100, =, cm 3 = 23728.57 cm 3 = 23.73 litres., 7, A shuttlecock used for playing, badminton has the shape of a frustum of, a cone mounted on a hemisphere. The, external diameters of the frustum are, 5 cm and 2 cm, and the height of the, entire shuttlecock is 7 cm. Find its, external surface area., 5, Here, R = cm, r = 1 cm and, 2, h = (7 − 1) cm = 6 cm., Let l be the slant height of the frustum of the cone., 2, ⎧, ⎫, 5, ∴ l 2 = h 2 + ( R − r) 2 = ⎨ 6 2 + ⎛⎜ − 1 ⎞⎟ ⎬ cm 2, ⎝2, ⎠, ⎩, ⎭, 9⎞, 153, ⎛, 2, 2, = ⎜ 36 + ⎟ cm =, cm, ⎝, 4⎠, 4, 153, 12. 36, ⇒ l=, cm =, cm = 6.18 cm., 2, 2, External surface area of the shuttlecock, = (its lateral surface area + area of the base), = [πl( R + r) + 2 πr 2 ] sq units, 22, 5, ⎧ 22, ⎫, =⎨, × 6.18 × ⎛⎜ + 1 ⎞⎟ + 2 ×, × 1 × 1 ⎬ cm 2, ⎝, ⎠, 7, 2, ⎩7, ⎭, 44 ⎞, ⎛, 2, ⎟ cm = ( 67 .98 + 6.28) cm 2 = 74.26 cm 2 ., = ⎜ 11 × 6.18 +, ⎝, 7 ⎠, The slant height of the frustum of a cone is 4 cm and the perimeters of, its circular ends are 18 cm and 6 cm. Find the curved surface area of, the frustum., [CBSE 2017]
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794, SOLUTION, , Secondary School Mathematics for Class 10, , Let the radii of its circular ends be R cm and r cm, and its slant, height be l cm. Then,, 2 πR = 18 ⇒ πR = 9., 2 πr = 6 ⇒ πr = 3., Also, l = 4 cm (given)., Curved surface area of the frustum, = πl( R + r) sq units, , EXAMPLE 8, , SOLUTION, , = l × ( πR + πr) = 4 × ( 9 + 3) cm 2 = 48 cm 2 ., The height of a cone is 10 cm. The cone is divided into two parts, using a plane parallel to its base at the middle of its height. Find the, ratio of the volumes of the two parts., [CBSE 2017], Let OAB be the cone and OQ be its axis, and let P be the, midpoint of OQ., Let OQ = h cm., Then, OP = PQ =, , h, cm and QB = 10 cm., 2, , Also, C OPD ~ C OQB., ( h/2) PD, OP PD, =, ⇒, =, ⇒ PD = 5 cm., ∴, OQ QB, h, 10, The plane CD divides the cone into two parts, namely, (i) a smaller cone of radius = 5 cm and height (h/2) cm,, (ii) frustum of a cone in which, ⎛ h⎞, R = 10 cm , r = 5 cm and height = ⎜ ⎟ cm., ⎝ 2⎠, Volume of the smaller cone, h⎞, ⎛1, ⎛ 25 πh ⎞, 3, = ⎜ × π × 5 × 5 × ⎟ cm 3 = ⎜, ⎟ cm ., ⎝3, ⎝ 6 ⎠, 2⎠, Volume of the frustum of the cone, =, , 1, h, ⎛ 175 πh ⎞, 3, π × ⋅ [(10) 2 + (5) 2 + 10 × 5] cm 3 = ⎜, ⎟ cm ., ⎝ 6 ⎠, 3, 2, , 25 πh 175 πh, :, = 25 : 175 = 1: 7 ., 6, 6, The height of a cone is 30 cm. A small cone is cut off at the top by a, plane parallel to the base. If its volume be 1/27 of the volume of the, given cone, at what height above the base is the section made?, , Ratio of the required volumes =, EXAMPLE 9, , [HOTS] [CBSE 2005C]
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Volume and Surface Areas of Solids, SOLUTION, , 795, , Height of the given cone = 30 cm., Let the radius of its base be R cm., 1, Volume of the given cone = ⎛⎜ πR 2 × 30 ⎞⎟ cm 3 = (10 πR 2 ) cm 3 ., ⎝3, ⎠, Let the radius and height of the smaller cone be r cm and h cm, respectively., 1, Then, volume of the smaller cone = ⎛⎜ πr 2 h ⎞⎟ cm 3 ., ⎝3, ⎠, 1 2, 1, [given], ∴, πr h =, ⋅(10 πR 2 ), 3, 27, 2, , R, 9h, ⋅, ⇒ ⎛⎜ ⎞⎟ =, ⎝r⎠, 10, Now, C OAB ~ C OCD., AB OA, R 30, ∴, =, ⇒, =, ⋅, CD OC, r, h, , ... (i), , ... (ii), , From (i) and (ii), we get, ⎛⎜ 30 ⎞⎟ = 9 h ⇒ 30 × 30 = 9 h, ⎝ h ⎠, h×h, 10, 10, 2, , ⇒ h3 =, ∴, , 30 × 30 × 10, = 1000 ⇒ h 3 = (10) 3 ⇒ h = 10., 9, , height of the smaller cone = 10 cm., , Height of the section from the base = ( 30 − 10) cm = 20 cm., EXAMPLE 10, , A hollow cone is cut by a plane parallel to the base and the upper, portion is removed. If the curved surface of the remainder is 8/9 of, the curved surface of the whole cone, find the ratio of the line, segments into which the altitude of the cone is divided by the plane., , SOLUTION, , Let OAB be the given hollow cone cut by the plane CD parallel, to base AB and let cone OCD be removed. Then, the remainder, is the frustum CABD of the given cone., , [CBSE 2004, ’04C], , Let OE = h units, OF = H units,, OD = l units , OB = L units,, ED = r units, FB = R units., In COED and COFB, we have, ∠EOD = ∠FOB (common),, ∠OED = ∠OFB = 90 °.
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796, , Secondary School Mathematics for Class 10, , ∴, , C OED ~ C OFB, OE OD ED, h, l, r, =, =, ⇒ = = ⋅ ... (i) [by Thales’ theorem], ⇒, OF OB, FB, H L R, , Now, (curved surface area of the frustum CABD), 8, = (curved surface area of the cone OAB), 9, ⇒ (curved surface area of the cone OCD), = (curved surface of the cone OAB), − (curved surface of the frustum CABD), = (curved surface of the cone OAB), 8, − (curved surface of the cone OAB), 9, 1, = (curved surface of the cone OAB), 9, 1, ⇒ πrl = πRL, 9, r ⎛ l⎞ 1, ⇒ ⎛⎜ ⎞⎟ ⎜ ⎟ =, ⎝ R ⎠ ⎝ L⎠ 9, , h 1, ⎛ h h⎞ 1, ⇒ ⎜ × ⎟ = ⇒, =, ⎝ H H⎠ 9, H 3, , ⇒ H = 3 h., , [using (i)], ... (ii), , Now, EF = (OF − OE) = ( H − h) = ( 3 h − h) = 2 h., OE, h, 1, ∴, =, = ⋅, EF 2 h 2, EXAMPLE 11, , SOLUTION, , Hence, OE : EF = 1 : 2., A metallic right circular cone is 20 cm high and has a vertical angle, of 60°. This is cut into two parts at the middle of its height by a plane, parallel to the base. If the frustum so obtained is drawn into a wire of, diameter 1 16 cm, find the length of the wire., [CBSE 2017], Let OAB be the cone in which ∠AOB = 60 ° ., Clearly ∠DOE = 30 ° , OE = 10 cm , OF = 20 cm., Let ED = r cm and FB = R cm., ED, ED, 1, = tan 30 ° ⇒, =, ∴, OE, 10, 3, 1 ⎞, 10, ⎟ cm ⇒ r =, ⇒ ED = ⎛⎜ 10 ×, cm, ⎝, 3⎠, 3, and,, , FB, FB, 1, = tan 30 ° ⇒, =, OF, 20, 3
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Volume and Surface Areas of Solids, , 797, , 1 ⎞, 20, ⎟ cm ⇒ R =, ⇒ FB = ⎛⎜ 20 ×, cm., ⎝, 3⎠, 3, Also, EF = 10 cm., Thus, ABCD is the frustum of a cone in which, 20, 10, R=, cm, r =, cm and h = 10 cm., 3, 3, 1, Volume of this frustum = πh( R 2 + r 2 + Rr), 3, 1, 400 100 200 ⎫, 3, = × π × 10 ⎧⎨, +, +, ⎬ cm, 3, 3, 3 ⎭, ⎩ 3, = ⎛⎜, ⎝, , π × 10 700 ⎞, 7000 π ⎞, ⎟ cm 3 = ⎛⎜, ⎟ cm 3 ., ×, ⎝ 9 ⎠, 3, 3 ⎠, , Let the length of the wire be l., 1, cm., Radius of the wire, r1 =, 32, 2, , 1, Volume of the wire = πr12 l = π × ⎛⎜ ⎞⎟ × l., ⎝ 32 ⎠, ∴, , 7000 π, πl, 7000 × 32 × 32 ⎞, ⎟ cm, =, ⇒ l = ⎛⎜, ⎝, ⎠, 9, 32 × 32, 9, , 7000 × 32 × 32 ⎞, 71680 ⎞, ⎟ m = ⎛⎜, ⎟ m, ⇒ l = ⎛⎜, ⎝, ⎠, ⎝ 9 ⎠, 9 × 100, = 7964.44 m., , EXERCISE 17C, 1. A drinking glass is in the shape of a frustum of a cone of height 14 cm., The diameters of its two circular ends are 16 cm and 12 cm. Find the, [CBSE 2012], capacity of the glass., 2. The radii of the circular ends of a solid frustum of a cone are 18 cm and, 12 cm and its height is 8 cm. Find its total surface area. [Use π = 3.14.], [CBSE 2011], , 3. A metallic bucket, open at the top, of height 24 cm is in the form of the, frustum of a cone, the radii of whose lower and upper circular ends are, 7 cm and 14 cm respectively. Find, (i) the volume of water which can completely fill the bucket;, (ii) the area of the metal sheet used to make the bucket., , [CBSE 2014]
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798, , Secondary School Mathematics for Class 10, , 4. A container, open at the top, is in the form of a frustum of a cone of, height 24 cm with radii of its lower and upper circular ends as 8 cm and, 20 cm respectively. Find the cost of milk which can completely fill the, container at the rate of ` 21 per litre., [CBSE 2014], 5. A container made of a metal sheet open at the top is of the form of, frustum of cone, whose height is 16 cm and the radii of its lower and, upper circular edges are 8 cm and 20 cm respectively. Find, (i) the cost of metal sheet used to make the container if it costs ` 10 per, [CBSE 2013], 100 cm 2, (ii) the cost of milk at the rate of ` 35 per litre which can fill it, [CBSE 2017], completely., 6. The radii of the circular ends of a solid frustum of a cone are 33 cm and, 27 cm, and its slant height is 10 cm. Find its capacity and total surface, [CBSE 2005], area. [Take π = 22 7 .], 7. A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the, diameters of the top and the bottom are 56 cm and 42 cm respectively., Find how many litres of water the bucket can hold. [Take π = 22 7 .], , [CBSE 2005C], , 8. A bucket made up of a metal sheet is in the form of a frustum of a cone, of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the bucket if the cost of metal sheet used is, [CBSE 2006, ’08, ’08C], ` 15 per 100 cm 2 . [Use π = 3.14.], 9. A bucket made up of a metal sheet is in the form of frustum of a cone. Its, depth is 24 cm and the diameters of the top and bottom are 30 cm and, 10 cm respectively. Find the cost of milk which can completely fill the, bucket at the rate of ` 20 per litre and the cost of metal sheet used if it, costs ` 10 per 100 cm 2 . [Use π = 3.14.], [CBSE 2006, ’09C], 10. A container in the shape of a frustum of a cone having diameters of its, two circular faces as 35 cm and 30 cm and vertical height 14 cm, is, completely filled with oil. If each cm 3 of oil has mass 1.2 g then find the, cost of oil in the container if it costs ` 40 per kg., [CBSE 2009C], 11. A bucket is in the form of a frustum of a cone and it can hold 28.49 litres, of water. If the radii of its circular ends are 28 cm and 21 cm, find the, height of the bucket., [CBSE 2012], 12. The radii of the circular ends of a bucket of height 15 cm are 14 cm and, r cm (r < 14). If the volume of bucket is 5390 cm 3 , find the value of r., [CBSE 2011], , 13. The radii of the circular ends of a solid frustum of a cone are 33 cm, and 27 cm and its slant height is 10 cm. Find its total surface area., [CBSE 2012], [Use π = 3.14.]
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Volume and Surface Areas of Solids, , 799, , 14. A tent is made in the form of a frustum of a cone surmounted by, another cone. The diameters of the base and the top of the frustum are, 20 m and 6 m respectively, and the height is 24 m. If the height of the, tent is 28 m and the radius of the conical part is equal to the radius of the, top of the frustum, find the quantity of canvas required. [Take π = 22 7 .], 15. A tent consists of a frustum of a cone, surmounted by a cone. If the, diameters of the upper and lower circular ends of the frustum be 14 m, and 26 m respectively, the height of the frustum be 8 m and the slant, height of the surmounted conical portion be 12 m, find the area of the, canvas required to make the tent. (Assume that the radii of the upper, circular end of the frustum and the base of the surmounted conical, [CBSE 2008], portion are equal.), 16. The perimeters of the two circular ends of a frustum of a cone are 48 cm, and 36 cm. If the height of the frustum is 11 cm, find its volume and, [CBSE 2014], curved surface area., 17. A solid cone of base radius 10 cm is cut into two parts through the, midpoint of its height, by a plane parallel to its base. Find the ratio of, [CBSE 2013], the volumes of the two parts of the cone., 18. The height of a right circular cone is 20 cm. A small cone is cut off at the, top by a plane parallel to the base. If its volume be 1 8 of the volume of, the given cone, at what height above the base is the section made?, [HOTS] [CBSE 2014], , 19. A solid metallic right circular cone 20 cm high and whose vertical angle, is 60°, is cut into two parts at the middle of its height by a plane parallel, to its base. If the frustum so obtained be drawn into a wire of diameter, 1 cm, find the length of the wire., [HOTS] [CBSE 2014], 12, 20. A fez, the cap used by the Turks, is shaped like, the frustum of a cone. If its radius on the open, side is 10 cm, radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of, material used for making it. [Use π = 22 7 ⋅], 21. An oil funnel made of tin sheet consists of a, 10 cm long cylindrical portion attached to a, frustum of a cone. If the total height is 22 cm,, diameter of the cylindrical portion is 8 cm and, the diameter of the top of the funnel is 18 cm,, find the area of the tin sheet required to make, the funnel.
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800, , Secondary School Mathematics for Class 10, , 22. A right circular cone is divided into three parts by trisecting its height, by two planes drawn parallel to the base. Show that the volumes of the, three portions starting from the top are in the ratio 1 : 7 : 19. [CBSE 2017], ANSWERS (EXERCISE 17C), , 1. 2170.67 cm, 2. 2411.52 cm 2 3. (i) 8624 cm 3 (ii) 1804 cm 2 4. ` 329.49, 5. (i) ` 145.83 (ii) ` 366.08, 6. 22704 cm 3 , 7599.43 cm 2 7. 28.49 litres, 8. ` 293.90, 9. ` 163.28, ` 171.13 10. ` 558.80, 11. 15 cm 12. r = 7, 13. 7592.52 cm 2 14. 1068.57 m 2 15. 892.6 m 2 16. 1554 cm 3 , 468.91 cm 2, 17. 1 : 7, 18. 10 cm, 19. 4480 m 20. 710.28 cm 2 21. 782.57 cm 2, 3, , HINTS TO SOME SELECTED QUESTIONS, 10. r = 15 cm, R = 17.5 cm, h = 14 cm., 1, 3175 × 11 ⎞, 3, Volume of the container = π( R 2 + r 2 + Rr )h = ⎛⎜, ⎟ cm ., ⎝, ⎠, 3, 3, 3175 × 11, Mass of oil in the container = ⎛⎜, × 1.2 ⎞⎟ g = 13.97 kg., ⎝, ⎠, 3, Cost of oil = ` (13.97 × 40) = ` 558.80., 11. r = 21 cm, R = 28 cm., Volume of the bucket = (28.49 x 1000) cm 3 ., 1, π[( 28 ) 2 + ( 21) 2 + 28 × 21]h = 28490. Solve for h., ∴, 3, 14. R = 10 m, r = 3 m and h = 24 m., Let l be the slant height of the frustum. Then,, l = h 2 + ( R − r ) 2 = ( 24 ) 2 + 7 2 m, = 576 + 49 m = 625 m = 25 m., Let L be the slant height of the conical part., For this conical part, r = 3 m and H = 4 m., ∴, , L=, , 3 2 + 4 2 m = 25 m = 5 m., , Quantity of canvas, = (lateral surface area of the frustum), + (lateral surface area of the cone), 22, = {πl( R + r ) + πrL} m 2 =, × {( 25 × 13 ) + ( 3 × 5 )} m 2 = 1068.57 m 2 ., 7, 15. R = 13 m, r = 7 m and h = 8 m., ∴, , l = h 2 + ( R − r ) 2 = 10 m., , Required area, 22, 22, = ⎧⎨ × 10 × ( 13 + 7 ) +, × 7 × 12 ⎫⎬ m 2 ., 7, 7, ⎩, ⎭
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Volume and Surface Areas of Solids, , 801, , 16. 2 πr = 36 and 2 πR = 48. Find r and R., 20. Area of material = curved surface area + area of closed, small circular face, = πl(r + R ) + πr 2 ., 21. Height of the cylindrical portion, h = 10 cm., Height of the frustum, H = (22 – 10) cm = 12 cm., Radius of the cylindrical portion, r = 4 cm., Radius of smaller end of the frustum, r = 4 cm., Radius of bigger end of the frustum, R = 9 cm., Area of tin sheet required, = curved surface area of the frustum + curved surface area of the cylindrical portion, = πl(r + R ) + 2 πrh., , ................................................................, , EXERCISE 17D, Very-Short-Answer Questions, 1. A river 1.5 m deep and 36 m wide is flowing at the rate of 3.5 km/hr., Find the amount of water (in cubic metres) that runs into the sea per, minute., 2. The volume of a cube is 729 cm 3 . Find its surface area., 3. How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?, 4. Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively, are melted and formed into a single cube. Find the edge of the new cube, formed., 5. Five identical cubes, each of edge 5 cm, are placed adjacent to each, other. Find the volume of the resulting cuboid., 6. The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their, surface areas., 7. The volume of a right circular cylinder with its height equal to the, 1, radius is 25 cm 3 . Find the height of the cylinder., 7, 8. The ratio between the radius of the base and the height of a cylinder is, 2 : 3. If the volume of the cylinder is 12936 cm 3 , find the radius of the, base of the cylinder., 9. The radii of two cylinders are in the ratio of 2 : 3 and their heights are in, the ratio of 5 : 3. Find the ratio of their volumes., 10. 66 cubic cm of silver is drawn into a wire 1 mm in diameter. Calculate, the length of the wire in metres.
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802, , Secondary School Mathematics for Class 10, , 11. If the area of the base of a right circular cone is 3850 cm 2 and its height is, 84 cm, find the slant height of the cone., 12. A cylinder with base radius 8 cm and height 2 cm is melted to form a, cone of height 6 cm. Calculate the radius of the base of the cone., 13. A right cylinderical vessel is full of water. How many right cones, having the same radius and height as those of the right cylinder will be, needed to store that water?, 14. The volume of a sphere is 4851 cm 3 . Find its curved surface area., 15. The curved surface area of a sphere is 5544 cm 3 . Find its volume., 16. The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio, of their volumes., 17. A solid metallic sphere of radius 8 cm is melted and recast into, spherical balls each of radius 2 cm. Find the number of spherical balls, obtained., 18. How many lead shots each 3 mm in diameter can be made from a, cuboid of dimensions 9 cm × 11 cm × 12 cm?, 19. A metallic cone of radius 12 cm and height 24 cm is melted and made, into spheres of radius 2 cm each. How many spheres are formed?, 20. A hemisphere of lead of radius 6 cm is cast into a right circular cone of, height 75 cm. Find the radius of the base of the cone., 21. A copper sphere of diameter 18 cm is drawn into a wire of diameter, 4 mm. Find the length of the wire., 22. The radii of the circular ends of a frustum of height 6 cm are 14 cm and, 6 cm respectively. Find the slant height of the frustum., 23. Find the ratio of the volume of a cube to that of a sphere which will fit, inside it., 24. Find the ratio of the volumes of a cylinder, a cone and a sphere, if each, has the same diameter and same height?, Short-Answer Questions, 25. Two cubes each of volume 125 cm 3 are joined end to end to form a solid., [CBSE 2013C], Find the surface area of the resulting cuboid., 26. Three metallic cubes whose edges are 3 cm, 4 cm and 5 cm, are melted, and recast into a single large cube. Find the edge of the new cube, [CBSE 2011], formed., 27. A solid metallic sphere of diameter 8 cm is melted and drawn into a, cylindrical wire of uniform width. If the length of the wire is 12 m, find, [CBSE 2013], its width.
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Volume and Surface Areas of Solids, , 803, , 28. A 5-m-wide cloth is used to make a conical tent of base diameter 14 m, and height 24 m. Find the cost of cloth used, at the rate of ` 25 per metre., [CBSE 2014], , 29. A wooden toy was made by scooping out a hemisphere of same radius, from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy., [CBSE 2013], , 30. Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted, and converted into a single cube whose diagonal is 12 3 cm. Find the, [CBSE 2013C], edges of the three cubes., HINT, , 3 × edge = 12 3 ⇒ edge = 12 cm., ∴ ( 3 x ) 3 + ( 4 x ) 3 + (5 x ) 3 = ( 12 ) 3 . Solve to find x., , 31. A hollow sphere of external and internal diameters 8 cm and 4 cm, respectively is melted into a solid cone of base diameter 8 cm. Find the, [CBSE 2013C], height of the cone., 32. A bucket of height 24 cm is in the form of frustum of a cone whose, circular ends are of diameter 28 cm and 42 cm. Find the cost of milk at, [CBSE 2013C], the rate of ` 30 per litre, which the bucket can hold., 33. The interior of a building is in the form of a right circular cylinder of, diameter 4.2 m and height 4 m surmounted by a cone of same diameter., The height of the cone is 2.8 m. Find the outer surface area of the, [CBSE 2014], building., 34. A metallic solid right circular cone is of height 84 cm and the radius of, its base is 21 cm. It is melted and recast into a solid sphere. Find the, [CBSE 2014], diameter of the sphere., 35. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere, of same radius. The total height of the toy is 15.5 cm. Find the total, [CBSE 2012], surface area of the toy., 36. If the radii of the circular ends of a bucket 28 cm high, are 28 cm and, [CBSE 2011], 7 cm, find its capacity and total surface area., 37. A bucket is in the form of a frustum of a cone with a capacity of, 12308.8 cm 3 of water. The radii of the top and bottom circular ends are, 20 cm and 12 cm respectively. Find the height of the bucket., [CBSE 2006C], (Use π = 3.14.), 38. A milk container is made of metal sheet in the shape of frustum of a cone, whose volume is 10459 3 cm 3 . The radii of its lower and upper circular, 7, ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used, [CBSE 2010], in making the container at the rate of ` 1.40 per cm 2 .
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804, , Secondary School Mathematics for Class 10, , 39. A solid metallic sphere of diameter 28 cm is melted and recast into a, number of smaller cones, each of diameter 4 2 cm and height 3 cm. Find, 3, , the number of cones so formed., 40. A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is, full of water. A solid cone of base diameter 7 cm and height 6 cm is, completely immersed in water. Find the volume of water (i) displaced, [CBSE 2009], out of the cylinder (ii) left in the cylinder., ANSWERS (EXERCISE 17D), , 1. 3150 m 3, , 2. 486 cm 2, , 3. 1000, , 4. 12 cm, , 6. 4 : 9, , 7. 2 cm, , 8. 14 cm, , 9. 20 : 27, , 11. 91 cm, , 12. 8 cm, , 13. 3, , 14. 1386 cm 2, , 15. 38808 cm 3, , 16. 8 : 125, , 17. 64, , 18. 84000, , 19. 108, , 20. 2.4 cm, , 21. 243 m, , 22. 10 cm, , 23. 6 : π, , 24. 3 : 1 : 2, , 25. 250 cm 2, , 26. 6 cm, , 27., , 28. ` 2750, , 29. 205.33 cm 3, , 31. 14 cm, , 32. ` 702.24, , 8, cm, 15, , 30. 6 cm, 8 cm, 10 cm, 34. 42 cm, , 35. 231 cm 2, , 36. 30184 cm 3 , 6468 cm 2, , 5. 625 cm 3, 10. 84 m, , 33. 75.9 cm 2, 37. 15 cm, , 38. ` 4224, 39. 126, 40. 77 cm , 748 cm, ................................................................, 3, , 3, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following:, 1. A cylindrical pencil sharpened at one edge is the combination of, (a) a cylinder and a cone, (b) a cylinder and frustum of a cone, (c) a cylinder and a hemisphere, (d) two cylinders, 2. A shuttlecock used for playing badminton is the, combination of, (a) cylinder and a hemisphere, (b) frustum of a cone and a hemisphere, (c) a cone and a hemisphere, (d) a cylinder and a sphere, , Shuttlecock
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Volume and Surface Areas of Solids, , 805, , 3. A funnel is the combination of, (a) a cylinder and a cone, (b) a cylinder and a hemisphere, (c) a cylinder and frustum of a cone, (d) a cone and a hemisphere, , Funnel, , 4. A surahi is a combination of, (a) a sphere and a cylinder, (b) a hemisphere and a cylinder, (c) a cylinder and a cone, (d) two hemispheres, , Surahi, , 5. The shape of a glass (tumbler) is usually in the form of, (a) a cylinder, (b) frustum of a cone, (c) a cone, (d) a sphere, , Glass, , 6. The shape of a gilli in the gilli-danda game is a combination of, (a) a cone and a cylinder, (b) two cylinders, , Gilli, , (c) two cones and a cylinder, (d) two cylinders and a cone, 7. A plumbline (sahul) is the combination of, (a) a hemisphere and a cone, (b) a cylinder and a cone, (c) a cylinder and frustum of a cone, (d) a cylinder and a sphere, , Plumbline, , 8. A cone is cut by a plane parallel to its base and the upper, part is removed. The part that is left over is called, (a) a cone, , (b) a sphere, , (c) a cylinder, (d) frustum of a cone, 9. During conversion of a solid from one shape to another, the volume of, the new shape will, (a) decrease, (b) increase, (c) remain unaltered, (d) be doubled
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806, , Secondary School Mathematics for Class 10, , 10. In a right circular cone, the cross section made by a plane parallel to the, base is a, (a) sphere, (b) hemisphere (c) circle, (d) a semicircle, 11. A solid piece of iron in the form of a cuboid of dimensions, (49 cm × 33 cm × 24 cm) is moulded to form a solid sphere. The radius of, the sphere is, (a) 19 cm, , (b) 21 cm, , (c) 23 cm, , (d) 25 cm, , 12. The radius (in cm) of the largest right circular cone that can be cut out, [CBSE 2011], from a cube of edge 4.2 cm is, (a) 2.1, , (b) 4.2, , (c) 8.4, , (d) 1.05, , 13. A metallic solid sphere of radius 9 cm is melted to form a solid cylinder, of radius 9 cm. The height of the cylinder is, [CBSE 2014], (a) 12 cm, , (b) 18 cm, , (c) 36 cm, , (d) 96 cm, , 14. A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow, cylinder of height 40 cm. The radius of the cylinder (in cm) is [CBSE 2014], 80, (d) 5, (a) 3.5, (b) 7, (c), 7, 15. The number of solid spheres, each of diameter 6 cm, that can be made, by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is, [CBSE 2014], , (a) 2, , (b) 4, , (c) 5, , (d) 6, , 16. The surface areas of two spheres are in the ratio 16 : 9. The ratio of their, volumes is, [CBSE 2013C], (a) 64 : 27, , (b) 16 : 9, , (c) 4 : 3, , (d) 16 3 : 9 3, , 17. If the surface area of a sphere is 616 cm 2 , its diameter (in cm) is, [CBSE 2012], , (a) 7, (b) 14, (c) 28, (d) 56, 18. If the radius of a sphere becomes 3 times then its volume will become, [CBSE 2011], , (a) 3 times, , (b) 6 times, , (c) 9 times, , (d) 27 times, , 19. If the height of a bucket in the shape of frustum of a cone is 16 cm and, the diameters of its two circular ends are 40 cm and 16 cm then its slant, [CBSE 2013C], height is, (a) 20 cm, , (b) 12 5 cm, , (c) 8 13 cm, , (d) 16 cm, , 20. A sphere of diameter 18 cm is dropped into a cylindrical vessel of, diameter 36 cm, partly filled with water. If the sphere is completely, [CBSE 2011], submerged then the water level rises by, (a) 3 cm, , (b) 4 cm, , (c) 5 cm, , (d) 6 cm
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Volume and Surface Areas of Solids, , 807, , 21. A solid right circular cone is cut into two parts at the middle of its, height by a plane parallel to its base. The ratio of the volume of the, [CBSE 2012], smaller cone to the whole cone is, (a) 1 : 2, , (b) 1 : 4, , (c) 1 : 6, , (d) 1 : 8, , 22. The radii of the circular ends of a bucket of height 40 cm are 24 cm and, [CBSE 2012], 15 cm. The slant height (in cm) of the bucket is, (a) 41, , (b) 43, , (c) 49, , (d) 51, , 23. A solid is hemispherical at the bottom and conical (of same radius), above it. If the surface areas of the two parts are equal then the ratio of, [CBSE 2011], its radius and the slant height of the conical part is, (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 4, , (d) 4 : 1, , 24. If the radius of the base of a right circular cylinder is halved, keeping, the height the same, then the ratio of the volume of the cylinder thus, [CBSE 2012], obtained to the volume of original cylinder is, (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 4, , (d) 4 : 1, , 25. A cubical ice-cream brick of edge 22 cm is to be distributed among some, children by filling ice-cream cones of radius 2 cm and height 7 cm up to, its brim. How many children will get the ice-cream cones?, (a) 163, , (b) 263, , (c) 363, , (d) 463, , 26. A mason constructs a wall of dimensions (270 cm × 300 cm × 350 cm), with bricks, each of size (22.5 cm × 11.25 cm × 8.75 cm) and it is assumed, 1, space is covered by the mortar. Number of bricks used to, that, 8, , construct the wall is, (a) 11000, , (b) 11100, , (c) 11200, , (d) 11300, , 27. Twelve solid spheres of the same size are made by melting a solid, metallic cylinder of base diameter 2 cm and height 16 cm. The diameter, of each sphere is, (a) 2 cm, , (b) 3 cm, , (c) 4 cm, , (d) 6 cm, , 28. The diameters of two circular ends of a bucket are 44 cm and 24 cm, and, the height of the bucket is 35 cm. The capacity of the bucket is, (a) 31.7 litres, , (b) 32.7 litres, , (c) 33.7 litres, , (d) 34.7 litres, , 29. The slant height of a bucket is 45 cm and the radii of its top and bottom, are 28 cm and 7 cm respectively. The curved surface area of the, bucket is, (a) 4953 cm 2, , (b) 4952 cm 2, , (c) 4951 cm 2, , (d) 4950 cm 2
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808, , Secondary School Mathematics for Class 10, , 30. The volumes of two spheres are in the ratio 64 : 27. The ratio of their, surface areas is, (a) 9 : 16, , (b) 16 : 9, , (c) 3 : 4, , (d) 4 : 3, , 31. A hollow cube of internal edge 22 cm is filled with spherical marbles of, 1, diameter 0.5 cm and space of the cube remains unfilled. Number of, 8, , marbles required is, (a) 142296, , (b) 142396, , (c) 142496, , (d) 142596, , 32. A metallic spherical shell of internal and external diameters 4 cm and, 8 cm respectively, is melted and recast into the form of a cone of base, diameter 8 cm. The height of the cone is, (a) 12 cm, , (b) 14 cm, , (c) 15 cm, , (d) 8 cm, , 33. A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with, two hemispheres stuck to each of its ends. The length of the entire, capsule is 2 cm. The capacity of the capsule is, (a) 0.33 cm 3, , (b) 0.34 cm 3, , (c) 0.35 cm 3, , (d) 0.36 cm 3, , 34. The length of the longest pole that can be kept in a room, (12 m × 9 m × 8 m) is, (a) 29 m, , (b) 21 m, , (c) 19 m, , (d) 17 m, , 35. The length of the diagonal of a cube is 6 3 cm. Its total surface area is, (a) 144 cm 2, , (b) 216 cm 2, , (c) 180 cm 2, , (d) 108 cm 2, , 36. The volume of a cube is 2744 cm 3 . Its surface area is, (a) 196 cm 2, , (b) 1176 cm 2, , (c) 784 cm 2, , (d) 588 cm 2, , 37. The total surface area of a cube is 864 cm 2 . Its volume is, (a) 3456 cm 3, , (b) 432 cm 3, , (c) 1728 cm 3, , (d) 3456 cm 3, , 38. How many bricks each measuring ( 25 cm × 11.25 cm × 6 cm) will be, required to construct a wall ( 8 m × 6 m × 22.5 cm)?, (a) 8000, , (b) 6400, , (c) 4800, , (d) 7200, , 39. The area of the base of a rectangular tank is 6500 cm 2 and the volume of, water contained in it is 2.6 m 3 . The depth of water in the tank is, (a) 3.5 m, , (b) 4 m, , (c) 5 m, , (d) 8 m, , 40. The volume of a wall, 5 times as high as it is broad and 8 times as long as, it is high, is 12.8 m 3 . The breadth of the wall is, (a) 30 cm, , (b) 40 cm, , (c) 22.5 cm, , (d) 25 cm
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Volume and Surface Areas of Solids, , 809, , 41. If the areas of three adjacent faces of a cuboid are x, y, z respectively, then the volume of the cuboid is, (a) xyz, , (b) 2xyz, , (c) xyz, , (d) 3 xyz, , 42. The sum of length, breadth and height of a cuboid is 19 cm and its, diagonal is 5 5 cm. Its surface area is, (a) 361 cm 2, , (b) 125 cm 2, , (c) 236 cm 2, , (d) 486 cm 2, , 43. If each edge of a cube is increased by 50%, the percentage increase in the, surface area is, (a) 50%, , (b) 75%, , (c) 100%, , (d) 125%, , 44. How many bags of grain can be stored in a cuboidal granary, (8 m × 6 m × 3 m), if each bag occupies a space of 0.64 m 3 ?, (a) 8256, , (b) 90, , (c) 212, , (d) 225, , 45. A cube of side 6 cm is cut into a number of cubes each of side 2 cm. The, number of cubes formed is, (a) 6, , (b) 9, , (c) 12, , (d) 27, , 46. In a shower, 5 cm of rain falls. The volume of the water that falls on, 2 hectares of ground, is, (a) 100 m 3, , (b) 10 m 3, , (c) 1000 m 3, , (d) 10000 m 3, , 47. Two cubes have their volumes in the ratio 1 : 27. The ratio of their, surface areas is, (a) 1 : 3, , (b) 1 : 8, , (c) 1 : 9, , (d) 1 : 18, , 48. The diameter of the base of a cylinder is 4 cm and its height is 14 cm. The, volume of the cylinder is, (a) 176 cm 3, , (b) 196 cm 3, , (c) 276 cm 3, , (d) 352 cm 3, , 49. The diameter of a cylinder is 28 cm and its height is 20 cm. The total, surface area of the cylinder is, (a) 2993 cm 2, , (b) 2992 cm 2, , (c) 2292 cm 2, , (d) 2229 cm 2, , 50. The height of a cylinder is 14 cm and its curved surface area is 264 cm 2 ., The volume of the cylinder is, (a) 308 cm 3, , (b) 396 cm 3, , (c) 1232 cm 3, , (d) 1848 cm 3, , 51. The curved surface area of a cylinder is 1760 cm 2 and its base radius is, 14 cm. The height of the cylinder is, (a) 10 cm, , (b) 15 cm, , (c) 20 cm, , (d) 40 cm, , 52. The ratio of the total surface area to the lateral surface area of a cylinder, with base radius 80 cm and height 20 cm is, (a) 2 : 1, , (b) 3 : 1, , (c) 4 : 1, , (d) 5 : 1
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810, , Secondary School Mathematics for Class 10, , 53. The curved surface area of a cylindrical pillar is 264 m 2 and its volume, is 924 m 3 . The height of the pillar is, (a) 4 m, , (b) 5 m, , (c) 6 m, , (d) 7 m, , 54. The ratio between the radius of the base and the height of the cylinder is, 2 : 3. If its volume is 1617 cm 3 , the total surface area of the cylinder is, (a) 308 cm 2, , (b) 462 cm 2, , (c) 540 cm 2, , (d) 770 cm 2, , 55. The radii of two cylinders are in the ratio 2 : 3 and their heights are in, the ratio 5 : 3. The ratio of their volumes is, (a) 27 : 20, , (b) 20 : 27, , (c) 4 : 9, , (d) 9 : 4, , 56. Two circular cylinders of equal volume have their heights in the ratio, 1 : 2. The ratio of their radii is, (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 2, , (d) 1 : 4, , 57. The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved, surface area is, (a) 60 π cm 2, , (b) 65 π cm 2, , (c) 30 π cm 2, , (d) none of these, , 58. The diameter of the base of a cone is 42 cm and its volume is 12936 cm 3 ., Its height is, (a) 28 cm, , (b) 21 cm, , (c) 35 cm, , (d) 14 cm, , 59. The area of the base of a right circular cone is 154 cm 2 and its height is, 14 cm. Its curved surface area is, (a) 154 5 cm 2, , (b) 154 7 cm 2, , (c) 77 7 cm 2, , (d) 77 5 cm 2, , 60. On increasing each of the radius of the base and the height of a cone by, 20% its volume will be increased by, (a) 20%, , (b) 40%, , (c) 60%, , (d) 72.8%, , 61. The radii of the base of a cylinder and a cone are in the ratio 3 : 4. If they, have their heights in the ratio 2 : 3, the ratio between their volumes is, (a) 9 : 8, , (b) 3 : 4, , (c) 8 : 9, , (d) 4 : 3, , 62. A metallic cylinder of radius 8 cm and height 2 cm is melted and, converted into a right circular cone of height 6 cm. The radius of the, base of this cone is, (a) 4 cm, , (b) 5 cm, , (c) 6 cm, , (d) 8 cm, , 63. The height of a conical tent is 14 m and its floor area is 346.5 m 2 . How, much canvas, 1.1 m wide, will be required for it ?, (a) 490 m, , (b) 525 m, , (c) 665 m, , (d) 860 m
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Volume and Surface Areas of Solids, , 64. The diameter of a sphere is 14 cm. Its volume is, 1, (b) 1439 cm 3, (c) 1437 cm 3, (a) 1428 cm 3, 3, , 811, , (d) 1440 cm 3, , 65. The ratio between the volumes of two spheres is 8 : 27. What is the ratio, between their surface areas?, (a) 2 : 3, , (b) 4 : 5, , (c) 5 : 6, , (d) 4 : 9, , 66. A hollow metallic sphere with external diameter 8 cm and internal, diameter 4 cm is melted and moulded into a cone having base radius, 4 cm. The height of the cone is, (a) 12 cm, , (b) 14 cm, , (c) 15 cm, , (d) 18 cm, , 67. A metallic cone having base radius 2.1 cm and height 8.4 cm is melted, and moulded into a sphere. The radius of the sphere is, (a) 2.1 cm, , (b) 1.05 cm, , (c) 1.5 cm, , (d) 2 cm, , 3, , 68. The volume of a hemisphere is 19404 cm . The total surface area of the, hemisphere is, (a) 4158 cm 2, , (b) 16632 cm 2, , (c) 8316 cm 2, , (d) 3696 cm 2, , 69. The surface area of a sphere is 154 cm 2 . The volume of the sphere is, 2, 1, 1, (a) 179 cm 3, (b) 359 cm 3, (c) 1437 cm 3 (d) none of these, 3, 3, 3, 70. The total surface area of a hemisphere of radius 7 cm is, (a) (588 π) cm 2, , (b) ( 392 π) cm 2, , (c) (147π) cm 2, , (d) ( 98 π) cm 2, , 71. The circular ends of a bucket are of radii 35 cm and 14 cm and the height, of the bucket is 40 cm. Its volume is, (a) 60060 cm 3, , (b) 80080 cm 3, , (c) 70040 cm 3, , (d) 80160 cm 3, , 72. If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is, (a) 1815. 3 cm 2, , (b) 1711.3 cm 2, , (c) 2025.3 cm 2, , (d) 2360 cm 2, , 73. A circus tent is cylindrical to a height of 4 m and conical above it. If its, diameter is 105 m and its slant height is 40 m, the total area of canvas, required is, (a) 1760 m 2, , (b) 2640 m 2, , 2, , (d) 7920 m 2, , (c) 3960 m
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812, , Secondary School Mathematics for Class 10, , Matching of columns, , 74. Match the following columns:, Column I, , Column II, , (a) A solid metallic sphere of (p) 18, radius 8 cm is melted and the, material is used to make solid, right cones with height 4 cm, and radius of the base 8 cm., How many cones are formed?, (b) A 20-m-deep well with (q) 8, diameter 14 m is dug up and, the earth from digging is, evenly spread out to form a, platform 44 m by 14 m. The, height of the platform is ...... m., (c) A sphere of radius 6 cm is, melted and recast into the, shape of a cylinder of radius, 4 cm. Then, the height of the, cylinder is ...... cm., , (r) 16 : 9, , (d) The volumes of two spheres are, in the ratio 64 : 27. The ratio of, their surface areas is ...... ., , (s) 5, , The correct answer is, (a)–......,, , (b)–......,, , (c)–......,, , (d)–...... ., , 75. Match the following columns:, Column I, , Column II, , (a) The radii of the circular ends of (p) 2418π, a bucket in the form of frustum, of a cone of height 30 cm are, 20 cm and 10 cm respectively., The capacity of the bucket is, ...... cm 3 . [Take π =, , 22, 7, , ⋅]
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Volume and Surface Areas of Solids, , 813, , (b) The radii of the circular ends (q) 22000, of a conical bucket of height, 15 cm are 20 cm and 12 cm, respectively. The slant height, of the bucket is ...... cm., (c) The radii of the circular ends, of a solid frustum of a cone are, 33 cm and 27 cm and its slant, height is 10 cm. The total, surface area of the bucket is, ...... cm 2 ., , (r) 12, , (d) Three solid metallic spheres of, radii 3 cm, 4 cm and 5 cm are, melted to form a single solid, sphere. The diameter of the, resulting sphere is ...... cm., , (s) 17, , The correct answer is, (a)–......,, , (b)–......,, , (c)–......,, , (d)–...... ., , Assertion-and-Reason Type, Each question consists of two statements, namely, Assertion (A) and, Reason (R). For selecting the correct answer, use the following code:, (a) Both Assertion (A) and Reason (R) are true and Reason (R) is a, correct explanation of Assertion (A)., (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a, correct explanation of Assertion (A)., (c) Assertion (A) is true and Reason (R) is false., (d) Assertion (A) is false and Reason (R) is true., 76., , Assertion (A), , Reason (R), , If the radii of the circular ends of a, bucket 24 cm high are 15 cm and, 5 cm respectively, then the surface, area of the bucket is 545π cm 2 ., , If the radii of the circular ends, of the frustum of a cone are R and r, respectively and its height is h,, then its surface area is, π{R 2 + r 2 + l( R − r)},, where l 2 = h2 + ( R + r) 2 ., , The correct answer is (a)/(b)/(c)/(d).
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814, , Secondary School Mathematics for Class 10, , Assertion (A), , 77., , Reason (R), , A hemisphere of radius 7 cm is to The total surface volume of a, 2, be painted outside on the surface. hemisphere is 3 πr ., The total cost of painting at ` 5 per, cm 2 is ` 2300., The correct answer is (a)/(b)/(c)/(d)., Assertion (A), , 78., , Reason (R), , The number of coins 1.75 cm in Volume of a cylinder of base, diameter and 2 mm thick from a radius r and height h is given by, V = ( πr 2 h) cubic units., melted cuboid (10 cm × 5.5 cm ×, And,, volume of a cuboid, 3.5 cm) is 400., = ( l × b × h) cubic units., , The correct answer is (a)/(b)/(c)/(d)., Assertion (A), , 79., , Reason (R), , If the volumes of two spheres are Volume of a sphere = 4 πR 3 ., 3, in the ratio 27 : 8 then their surface, Surface area of a sphere = 4πR 2 ., areas are in the ratio 3 : 2., The correct answer is (a)/(b)/(c)/(d)., Assertion (A), , 80., , Reason (R), , The curved surface volume of a Volume of a cone = πr 2 h., cone of base radius 3 cm and, height 4 cm is (15π) cm 2 ., The correct answer is (a)/(b)/(c)/(d)., ANSWERS (MCQ), , 1. (a), 10. (c), 19. (a), 28. (b), 37. (c), 46. (c), 55. (b), 64. (c), 73. (d), 76. (d), , 2. (b), 3. (c), 4. (a), 5. (b), 11. (b) 12. (a) 13. (a) 14. (a), 20. (a) 21. (d) 22. (a) 23. (a), 29. (d) 30. (b) 31. (a) 32. (b), 38. (b) 39. (b) 40. (b) 41. (c), 47. (c) 48. (a) 49. (b) 50. (b), 56. (b) 57. (b) 58. (a) 59. (a), 65. (d) 66. (b) 67. (a) 68. (a), 74. (a)–(q), (b)–(s), (c)–(p), (d)–(r), 77. (d) 78. (a) 79. (d) 80. (c), , 6. (c), 7. (a), 8. (d), 9. (c), 15. (c) 16. (a) 17. (b) 18. (d), 24. (c) 25. (c) 26. (c) 27. (a), 33. (d) 34. (d) 35. (b) 36. (b), 42. (c) 43. (d) 44. (d) 45. (d), 51. (c) 52. (d) 53. (c) 54. (d), 60. (d) 61. (a) 62. (d) 63. (b), 69. (a) 70. (c) 71. (b) 72. (b), 75. (a)–(q), (b)–(s), (c)–(p), (d)–(r)
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Volume and Surface Areas of Solids, , 815, , HINTS TO SOME SELECTED QUESTIONS, 9. During conversion of a solid from one shape to another, the volume remains the same., 10. In a right circular cone, the cross section made by a plane parallel to the base is a, 11., , 4 3, 4 22, πr = 49 × 33 × 24 ⇒, ×, × r 3 = 49 × 33 × 24, 3, 3 7, 21, = 21 × 21 × 21 ⇒ r = 21 cm., ⇒ r 3 = 49 × 33 3 × 24 3 ×, 88 8 1, , 12. Diameter of the cone = edge of the cube = 4.2 cm., ∴ radius = 2.1 cm., 4, 4, π × 9 × 9 × 9 = π × 9 × 9 × h ⇒ h = ⎛⎜ × 9 ⎞⎟ cm = 12 cm., ⎝3, ⎠, 3, 1, 7, 14. 2 πr = 22 ⇒ r = ⎛⎜ 22 × × ⎞⎟ cm = 3.5 cm., ⎝, 2 22 ⎠, , 13., , volume of the cylinder, volume of each sphere, π × 2 × 2 × 45, =, = 5., 4, × π× 3× 3× 3, 3, , 15. Number of spheres =, , 4 πR 2 16, R 2 16, R, =, ⇒ 2 =, ⇒, =, 2, 4 πr, 9, r, 9, r, , 16 4, =, 9, 3, 4, 3, πR, 64, R 3 4 3 64, ⇒ 3 = 3 =, ⇒ 3, =, ⋅, 4 3 27, 3, 27, r, πr, 3, 1 7, 2, 2, 17. 4 πr = 616 ⇒ r = 616 × ×, = 49 ⇒ r = 7 cm., 4 22, 4, 18. Volume of a sphere of radius r, V = πr 3 ., 3, 4, 4, Volume of a sphere of radius 3r = π × ( 3r ) 3 = 27 × πr 3 = 27 V., 3, 3, , 16., , 19. l = h 2 + ( R − r ) 2 = ( 16 ) 2 + ( 20 − 8 ) 2 cm, = 256 + 144 cm =, , 400 cm = 20 cm., , 20. Increase in volume of water = volume of the sphere, ⇒, , π × 18 × 18 × h =, , 4, 4 9× 9× 9⎞, π × 9 × 9 × 9 ⇒ h = ⎛⎜ ×, ⎟ cm = 3 cm., ⎝ 3 18 × 18 ⎠, 3, 2, , r, h, 1, π × ⎛⎜ ⎞⎟ × ⎛⎜ ⎞⎟, ⎝ 2⎠, ⎝ 2⎠ 1, 3, 21. Ratio of volumes =, = ⋅, 1, 8, π×r2 × h, 3, 22. l = h 2 + ( R − r ) 2 = ( 40 ) 2 + ( 24 − 15 ) 2 cm, = 1600 + 81 cm = 1681 cm = 41 cm.
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816, , Secondary School Mathematics for Class 10, , 23. 2 πr 2 = πrl ⇒, , r 1, = ⋅, l 2, 2, , r, π × ⎛⎜ ⎞⎟ × h, ⎝ 2⎠, 1, 24. Ratio of volumes =, = ⋅, π×r2 × h, 4, 25. Let the number of cones be n. Then,, 1, 1 22, n × πr 2 h = 22 × 22 × 22 ⇒ n × ×, × 2 × 2 × 7 = 22 × 22 × 22, 3, 3 7, 3, ∴ n = ⎛⎜ 22 × 22 × 22 × ⎞⎟ = 363., ⎝, 88 ⎠, 7, 26. Volume of the wall filled with bricks = ⎛⎜ × 270 × 300 × 350 ⎞⎟ cm 3 ., ⎝8, ⎠, 9, 1125 225 875 35 ⎞ ⎛ 9 × 225 × 35 ⎞, ⎛ 225, 3, Volume of each brick = ⎜, ×, ×, ⎟ cm ., ⎟=⎜, ⎠, 100 4, 100 4 ⎠ ⎝, 32, ⎝ 10 2, , 7, 32, ⎫ = 11200., Number of bricks = ⎧⎨ × 270 × 300 × 350 ×, ⎬, 9 × 225 × 35 ⎭, ⎩8, 27. Let the diameter of each sphere be d cm. Then,, 3, , 12 ×, ⇒, , 4, 4 3, d, πr = πR 2 h ⇒ 12 × π × ⎛⎜ ⎞⎟ = π × 12 × 16, ⎝ 2⎠, 3, 3, , 2 d 3 = 16 ⇒ d 3 = 8 = 2 3 ⇒ d = 2 cm., , 28. Here, R = 22 cm, r = 12 cm and h = 35 cm., 1, Capacity of the bucket = πh( R 2 + r 2 + Rr ) cm 3, 3, 1 22, = ×, × 35 × [( 22 ) 2 + ( 12 ) 2 + 22 × 12] cm 3, 3 7, 110, 110 × 892 ⎞, = ⎛⎜, × 892 ⎞⎟ cm 3 = ⎛⎜, ⎟ litres = 32.7 litres., ⎝ 3, ⎠, ⎝ 3 × 1000 ⎠, 29. Here, l = 45 cm, R = 28 cm and r = 7 cm., 22, × 45 × ( 28 + 7 ) cm 2 = 4950 cm 2 ., ∴ curved surface area = πl( R + r ) =, 7, 4, 3, 3, πR 3, 64, 4, R 3 64, R, R 4, 30. 3, =, ⇒ 3 =, ⇒ ⎛⎜ ⎞⎟ = ⎛⎜ ⎞⎟ ⇒, = ⋅, 4 3 27, ⎝r ⎠, ⎝ 3⎠, 27, 3, r, r, πr, 3, 2, , Ratio of their surface areas =, , 2, , 4 πR 2 R 2 ⎛ R ⎞, 4, 16, =, = ⎜ ⎟ = ⎛⎜ ⎞⎟ =, = 16 : 9., ⎝ 3⎠, 4 πr 2 r 2 ⎝ r ⎠, 9, , 7, 31. Space filled in the cube = ⎛⎜ × 22 × 22 × 22 ⎞⎟ cm 3 = (7 × 1331) cm 3 ., ⎝8, ⎠, Radius of each marble =, , 0.5, 5, 1, cm =, cm = cm., 2, 20, 4
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Volume and Surface Areas of Solids, 4 3 ⎛ 4 22 1 1, πr = ⎜ ×, × × ×, ⎝3 7 4 4, 3, 7 × 1331 × 24 × 7 ⎞, Number of marbles = ⎛⎜, ⎟ = 142296., ⎝, ⎠, 11, , Volume of each marble =, , 1⎞, 3, ⎛ 11 ⎞ cm 3 ., ⎟ cm = ⎜, ⎟, ⎝ 24 × 7 ⎠, 4⎠, , 32. Here, R = 4 cm and r = 2 cm., 4, 4, Volume of spherical shell = π{(4) 3 − ( 2 3 )} cm 3 = ⎛⎜ π × 56 ⎞⎟ cm 3 ., ⎝3, ⎠, 3, Let the height of the cone be h cm., 1, 4, π × 4 × 4 × h = π × 56 ⇒ h = 14 cm., ∴, 3, 3, 33. Radius of the capsule =, , 0.5, cm = 0.25 cm., 2, , Let the length of cylindrical part be x cm., Then, 0.25 + x + 0.25 = 2 ⇒ x + 0.5 = 2 ⇒ x = 1.5 cm., 2, Capacity of the capsule = ⎛⎜ πr 3 × 2 ⎞⎟ + πr 2 h, ⎝3, ⎠, 4 22, 22, = ×, × (0.25) 3 +, × (0.25) 2 × 1.5, 3 7, 7, 4 22 1 1 1 ⎞ ⎛ 22 11 1 1 15 3 ⎞, = ⎛⎜ ×, × × × ⎟+⎜, × × ×, ⎟, ⎝3 7, 4 4 4⎠ ⎝ 7, 4 4 10 2 ⎠, 11, 33 ⎛ 22 + 99 ⎞ 121, =, +, =⎜, = 0.36 cm 3 ., ⎟=, 168 112 ⎝ 336 ⎠ 336, 34. Length of the longest pole = length of diagonal of the room, = l 2 + b 2 + h2, = ( 12 ) 2 + 9 2 + 8 2 m = 289 m = 17 m., 3 a = 6 3 ⇒ a = 6., , 35., , Total surface area = 6 a2 cm 2 = ( 6 × 6 × 6 ) cm 2 = 216 cm 2 ., 36. a3 = 2744 = ( 2 3 × 7 3 ) ⇒ a = ( 2 × 7 ) = 14 cm., surface area = 6 a2 cm 2 = ( 6 × 14 × 14 ) cm 2 = 1176 cm 2 ., , ∴, , 37. 6 a = 864 ⇒ a2 = 144 ⇒ a = 12 cm., 2, , ∴, , volume = ( 12 × 12 × 12 ) cm 3 = 1728 cm 3 ., , 38. Volume of the wall = ( 800 × 600 × 22.5 ) cm 3 ., volume of the wall, Number of bricks =, volume of 1 brick, ⎛ 800 × 600 × 22.5 ⎞, =⎜, ⎟ = 6400., ⎝ 25 × 11 . 25 × 6 ⎠, 39. Area of the base =, , 6500, 13 2, m2 =, m ., 100 × 100, 20, , Let the depth of water be d metres. Then,, 13, 26 20 ⎞, × d = 2.6 ⇒ d = ⎛⎜ ×, ⎟ = 4., ⎝ 10 13 ⎠, 20, , 817
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818, , Secondary School Mathematics for Class 10, , 40. Let breadth = x cm. Then, height = 5x cm and length = 40x cm, ∴ 40 x × x × 5 x = 12.8 × 100 × 100 × 100, ⇒ x 3 = 64000 = ( 40 × 40 × 40 ) ⇒ x = 40., 41. Let length = 2, breadth = b and height = h. Then, lb = x , bh = y, and lh = z, ⇒, , lb × bh × lh = xyz ⇒ lbh = xyz, , ⇒, , volume = xyz., , 42. l + b + h = 19 ⇒ ( l + b + h ) 2 = ( 19 ) 2 = 361, ⇒ ( l 2 + b 2 + h 2 ) + 2( lb + bh + lh ) = 361, ⇒ (5 5 ) 2 + 2( lb + bh + lh ) = 361, ⇒ 2( lb + bh + lh ) = 361 − 125 = 236 cm 2 ., 43. Let original edge be a. Original surface area = 6 a2 ., 150 a 3 a, New edge = 150% of a =, =, ⋅, 100, 2, 2, , 3a, 27 a2, New surface area = 6 × ⎛⎜ ⎞⎟ =, ⋅, ⎝ 2⎠, 2, 2, 2, ⎛ 27 a, ⎞ 15 a, Increase in area = ⎜, ⋅, − 6 a2 ⎟ =, 2, ⎝ 2, ⎠, , 1, ⎛ 15 a, ⎞, Increase % = ⎜, × 2 × 100 ⎟ = 125% ., 6a, ⎝ 2, ⎠, 2, , ( 8 × 6 × 3), = 225., 0.64, volume of given cube, 45. Number of cubes formed =, volume of each small cube, ( 6 × 6 × 6), = 27 ., =, ( 2 × 2 × 2), 44. Number of bags =, , 46. Required volume = ( area × depth), 5 ⎞ 3, 3, = ⎛⎜ 2 × 10000 ×, ⎟ m = 1000 m ., ⎝, 100 ⎠, 3, , 47., , 3, , a3, 1, a, 1, a 1, =, ⇒ ⎛⎜ ⎞⎟ = ⎛⎜ ⎞⎟ ⇒ = ⋅, ⎝ b⎠, ⎝ 3⎠, b 3 27, b 3, 2, , S1 6 a2 a2 ⎛ 1 ⎞, 1, =, =, =⎜ ⎟ = ⋅, S2 6 b 2 b 2 ⎝ 3 ⎠, 9, Required ratio = 1 : 9., 48. Here, r = 2 cm and h = 14 cm., ∴, , ⎛ 22, ⎞, volume = πr 2 h = ⎜ × 2 × 2 × 14 ⎟ cm 3 = 176 cm 3 ., ⎝7, ⎠, , 49. Here, r = 14 cm and h = 20 cm., Total surface area = 2 πrh + 2 πr 2, = 2πr ( h + r ), 22, = 2×, × 14 × ( 20 + 14 ) cm 2 = 2992 cm 2 ., 7
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Volume and Surface Areas of Solids, , 819, , 22, 264, × r × 14 = 264 ⇒ r =, = 3., 7, 88, 22, Volume = ⎛⎜ × 3 × 3 × 14 ⎞⎟ cm 3 = 396 cm 3 ., ⎝7, ⎠, , 50. 2 ×, , 22, × 14 × h = 1760, 7, 1760, = 20 cm., ⇒ h=, 88, , 51. 2 πrh = 1760 ⇒ 2 ×, , 52., , Total surface area, 2 πr (h + r ) ( h + r ), =, =, Lateral surface area, 2 πrh, h, =, , ( 20 + 80 ) 100 5, =, = ⋅, 20, 20 1, , 53. 2 πrh = 264 and πr 2 h = 924, 2 × 924, πr 2 h 924, =, ⇒ r=, ⇒, = 7 m., 2 πrh 264, 264, 22, 264, × 7 × h = 264 ⇒ h =, = 6 cm., ∴ 2×, 7, 44, 54. Let radius = 2x and height = 3x. Then,, 22, × ( 2 x ) 2 × 3 x = 1617, πr 2 h = 1617 ⇒, 7, 7, 1, 7 7 7, 7, × ⎞⎟ = ⎛⎜ × × ⎞⎟ = ⎛⎜ ⎞⎟, ⇒ x 3 = ⎛⎜ 1617 ×, ⎝, 22 12 ⎠ ⎝ 2 2 2 ⎠ ⎝ 2 ⎠, ⇒ x=, ∴, , r = 7 cm and h =, , 3, , 7, cm., 2, , 21, cm., 2, , Total surface area = 2πr ( h + r ), 22, 21, = 2×, × 7 × ⎛⎜ + 7 ⎞⎟ cm 2 = 770 cm 2 ., ⎝ 2, ⎠, 7, 55. Let the radii be 2r , 3r and heights be 5h and 3h. Then, ratio of their volumes, π × ( 2r ) 2 × 5 h 20, =, ⋅, =, π × ( 3r ) 2 × 3 h 27, 2, , 56. π × r 2 × h = π × R 2 × 2 h ⇒, ∴, , r2 2, r, r, = ⇒ ⎛⎜ ⎞⎟ = 2 ⇒, = 2., ⎝ R⎠, R2 1, R, , required ratio = 2 : 1., , 57. Here, r = 5 cm and h = 12 cm., ∴ l 2 = (r 2 + h 2 ) = (5 ) 2 + ( 12 ) 2 = 169 ⇒ l = 169 = 13 cm., Curved surface area = πrl = ( π × 5 × 13 ) cm 2 = ( 65 π ) cm 2 ., 58., , 1 2, 1 22, ⎛ 12936 ⎞, × 21 × 21 × h = 12936 ⇒ h = ⎜, πr h = 12936 ⇒ ×, ⎟ cm = 28 cm., 3, 3 7, ⎝ 22 × 21 ⎠
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820, , Secondary School Mathematics for Class 10, , 59. πr 2 = 154 ⇒, , 22, 7, × r 2 = 154 ⇒ r 2 = ⎛⎜ 154 × ⎞⎟ = 49 ⇒ r = 7., ⎝, 7, 22 ⎠, , Now, r = 7 cm and h = 14 cm., ∴, , l = r 2 + h 2 = 49 + 196 = 245 = 7 5 cm., , Curved surface area = πrl, 22, = ⎛⎜ × 7 × 7 5 ⎞⎟ cm 2 = 154 5 cm 2 ., ⎝7, ⎠, 60. Let the original radius be r and height be h., 1, Then, original volume = πr 2 h = V., 3, 120 r ⎞ 6r, New radius = 120% of r = ⎛⎜, ⋅, ⎟=, ⎝ 100 ⎠ 5, 120 h ⎞ 6 h, New height = 120% of h = ⎛⎜, ⋅, ⎟=, ⎝ 100 ⎠, 5, 2, , New volume =, , 1, 6r, 6 h 216 ⎛ 1 2 ⎞ 216, π × ⎛⎜ ⎞⎟ ×, =, V., ⎜ πr h ⎟ =, ⎝ 5⎠, ⎠ 125, 3, 5, 125 ⎝ 3, , 216, 91V, Increase = ⎛⎜, V − V ⎞⎟ =, ⋅, ⎝ 125, ⎠ 125, 91V 100 ⎞, Increase % = ⎛⎜, ×, ⎟ % = 72.8%., ⎝ 125, V ⎠, 61. Let the radii of the cylinder and cone be 3r and 4r respectively and their heights be 2h, and 3h respectively. Then,, π( 3r ) 2 × ( 2 h ), 54 9, V1, =, =, = ⋅, 1, 2, V2, π ( 4r ) × ( 3 h ) 48 8, 3, 62. π × 8 × 8 × 2 =, , 1, πR 2 × 6 ⇒ R 2 = 64 ⇒ R = 8 cm., 3, 2, , 3465 7 ⎞ 441 ⎛ 21 ⎞, 21, 63. πR 2 = 346.5 ⇒ R 2 = ⎛⎜, × ⎟=, =⎜ ⎟ ⇒ R=, m., ⎝ 10, ⎝ 2⎠, 22 ⎠, 4, 2, l = R 2 + h2 =, , 441, + ( 14 ) 2 =, 4, , 1225 35, m., =, 4, 2, , 22 21 35 ⎞ 2, 2, Area of canvas = πRl = ⎛⎜ ×, ×, ⎟ m = 577.5 m ., ⎝7, 2, 2 ⎠, Length of canvas =, , area, 577.50, =, m = 525 m., width, 1.1, , 4, 4 22, 1, 64. Volume = ⎛⎜ πr 3 ⎞⎟ = ⎛⎜ ×, × 7 × 7 × 7 ⎞⎟ cm 3 = 1437 cm 3 ., ⎝3, ⎠ ⎝3 7, ⎠, 3, 4, 3, 3, πR 3, 8, R3, 8, R, 2, R 2, 3, 65., ⇒ ⎛⎜ ⎞⎟ = ⎛⎜ ⎞⎟ ⇒, = ⋅, =, ⇒ 3 =, 4 3 27, ⎝r ⎠, ⎝ 3⎠, r, 27, r, 3, πr, 3
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Volume and Surface Areas of Solids, 2, , Ratio between their surface areas =, 66., 67., , 68., , 821, 2, , 4 πR 2 R 2 ⎛ R ⎞, 2, 4, =, = ⎜ ⎟ = ⎛⎜ ⎞⎟ = ⋅, ⎝ 3⎠, 4 πr 2 r 2 ⎝ r ⎠, 9, , 4, 4, 1, π × ( 4 ) 3 − π × ( 2 ) 3 = π × ( 4 ) 2 × h ⇒ 16 h = 224 ⇒ h = 14 cm., 3, 3, 3, 1, 4, π × ( 2.1) 2 × 8.4 = π × r 3, 3, 3, 1, 3, 3, ⎛, ⇒ r = ⎜ × 4.41 × 8.4 × ⎞⎟ = ( 2.1) 3 ⇒ r = 2.1 cm., ⎝3, 4⎠, 2 22, 21, ×, × R 3 = 19404 ⇒ R 3 = ⎛⎜ 19404 × ⎞⎟, ⎝, 3 7, 44 ⎠, ⇒, , R 3 = ( 21) 3 ⇒ R = 21 cm., 22, Total surface area = 3 πR 2 = ⎛⎜ 3 ×, × 21 × 21⎞⎟ cm 2 = 4158 cm 2 ., ⎝, ⎠, 7, 22, × R 2 = 154, 7, 7, 7, ⇒ R 2 = ⎛⎜ 154 × ⎞⎟ ⇒ R = ⋅, ⎝, 88 ⎠, 2, , 69. 4 πR 2 = 154 ⇒ 4 ×, , 4, πR 3, 3, 4 22 7 7 7 ⎞, 2, = ⎛⎜ ×, × × × ⎟ cm 3 = 179 cm 3 ., ⎝3 7, 3, 2 2 2⎠, , Volume of the sphere =, , 70. Total surface area = 3 πR 2 = ( 3 π × 7 × 7 ) cm 2 = ( 147 π ) cm 2 ., πh 2, 71. Volume of the bucket =, ( R + r 2 + Rr ), 3, 22 1, =, × × 40 × [( 35 ) 2 + ( 14 ) 2 + ( 35 × 14 )] cm 3, 7, 3, = 80080 cm 3 ., 72. l = h 2 + ( R − r ) 2 = ( 24 ) 2 + ( 15 − 5 ) 2 = 26 cm., Surface area of the bucket = π[l( R + r ) + r 2 ], = 3.14 × [26 × 20 + 25] cm 2 = 1711.3 cm 2 ., 22 105, 22 105, 73. Area of canvas = ⎛⎜ 2 ×, ×, ×4+, ×, × 40 ⎞⎟ m 2, ⎝, ⎠, 7, 2, 7, 2, = ( 1320 + 6600 ) m 2 = 7920 m 2 ., 74. (a) Let the number of cones be n. Then,, 4, 1, πR 3 = n × πr 2 h ⇒ 4 R 3 = nr 2 h ⇒ 4 × 8 3 = n × 8 2 × 4 ⇒ n = 8., 3, 3, 22, (b) Volume of the earth dug out = πr 2 h = ⎛⎜ × 7 × 7 × 20 ⎞⎟ m 3 = ( 22 × 140 ) m 3 ., ⎝7, ⎠, Let the height of the platform be h metres., Volume of the earth on platform = ( 44 × 14 × h ) m 3 ., 22 × 140, m = 5 m., ∴ 44 × 14 × h = 22 × 140 ⇒ h =, 44 × 14
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822, , Secondary School Mathematics for Class 10, (c) Volume of the sphere =, , 4 3, πr ., 3, , Volume of the cylinder = πR 2 h., 4 3, 4, πr = πR 2 h ⇒ r 3 = R 2 h, ∴, 3, 3, 4, 8× 6× 6, × 6× 6× 6 = 4× 4× h ⇒ h =, = 18 cm., ∴, 3, 4× 4, 4, 3, 3, πR 3, 64, 4, R 3 64, R, R 4, 3, (d), =, ⇒ 3 =, ⇒ ⎛⎜ ⎞⎟ = ⎛⎜ ⎞⎟ ⇒, = ⋅, 4 3 27, ⎝r ⎠, ⎝ 3⎠, 27, 3, r, r, πr, 3, 2, 2, 4 πR 2 R 2 ⎛ R ⎞, ⎛ 4 ⎞ = 16 = 16 : 9., Ratio of their surface areas =, =, =, =, ⎜, ⎟, ⎜, ⎟, ⎝ 3⎠, 4 πr 2 r 2 ⎝ r ⎠, 9, The correct answer is (a)–(q), (b)–(s), (c)–(p), (d)–(r)., 75. (a) R = 20 cm, r = 10 cm and h = 30 cm., 1, Capacity = πh( R 2 + r 2 + Rr ), 3, 1 22, = ⎧⎨ ×, × 15 × ( 400 + 100 + 200 )⎫⎬ cm 3 = 11000 cm 3 ., ⎩3 7, ⎭, (b) R = 20 cm, r = 12 cm and h = 15 cm., l = h 2 + ( R − r ) 2 = ( 15 ) 2 + ( 20 − 12 ) 2 cm = ( 15 ) 2 + 8 2 cm, = 225 + 64 cm = 289 cm = 17 cm., (c) R = 33 cm, r = 27 cm and l = 10 cm., Total surface area = π[R 2 + r 2 + l( R + r )], = π[( 33 ) 2 + ( 27 ) 2 + 10 × ( 33 + 27 )] cm 2, = π × ( 2418 ) cm 2 = ( 2418 π ) cm 2 ., (d), , 4, 4, πR 3 = π × [( 3 ) 3 + ( 4 ) 3 + (5 ) 3 ] ⇒ R 3 = 216 = ( 6 ) 3 ⇒ R = 6 ., 3, 3, ∴, , diameter = 12 cm., , The correct answer is (a)–(q), (b)–(s), (c)–(p), (d)–(r)., 76. The Reason (R) is clearly true., Now, h = 24 cm, R = 15 cm and r = 5 cm., ∴, , l 2 = h 2 + ( R − r ) 2 = {( 24 ) 2 + ( 15 − 5 ) 2 } cm 2 = (576 + 100 ) cm 2 = 676 cm 2, , ⇒, , l = 676 cm = 26 cm., , ∴, , surface area = π{R 2 + r 2 + l( R + r )}, = π{( 15 ) 2 + 5 2 + 26 × 10} cm 2 = π{225 + 25 + 260} cm 2, = (510π ) cm 2 , which is true., , ∴, , Assertion (A) is true and Reason (R) gives Assertion (A)., , Hence, the correct answer is (a)., 77. The Reason (R) is clearly true., 22, Total surface area = 3 πr 2 = ⎛⎜ 3 ×, × 7 × 7 ⎞⎟ cm 2 = 462 cm 2 ., ⎝, ⎠, 7
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Volume and Surface Areas of Solids, , 823, , Cost of painting = ` ( 462 × 5 ) = ` 2310., ∴, , Assertion (A) is false and Reason (R) is true., , So, the correct answer is (d)., 78. The Reason (R) is clearly true., 1.75, 2, 1, Radius of each coin =, cm, thickness =, cm = cm., 2, 10, 5, 22 1.75 1.75 1 ⎞, ×, × ⎟ cm 3, ∴ volume of each coin = πr 2 h = ⎛⎜ ×, ⎝7, 2, 2, 5⎠, 11 175 175 ⎞, 77, 3, = ⎛⎜ ×, ×, cm 3 ., ⎟ cm =, ⎝ 70 100 100 ⎠, 160, 55 35 ⎞, 385, 3, Volume of cuboid = ⎛⎜ 10 ×, ×, cm 3 ., ⎟ cm =, ⎝, 10 10 ⎠, 2, 385 160 ⎞, Number of coins = ⎛⎜, ×, ⎟ = 400, which is true., ⎝ 2, 77 ⎠, Clearly, Reason (R) is true and Assertion (A) gives Reason (R)., So, the correct answer is (a)., 79. We know that the volume of a sphere is, , 4, πR 3 and its surface area is 4 πR 2 . So, the, 3, , Reason (R) is clearly true., Now, let R and r be the radii of two given spheres. Then,, 4, 3, 3, πR 3, 27, 3, V1, R 3 27, R, R 3, 3, =, =, ⇒ 3 =, ⇒ ⎛⎜ ⎞⎟ = ⎛⎜ ⎞⎟ ⇒, = ⋅, 4 3, ⎝r ⎠, ⎝ 2⎠, 8, 8, r, 2, V2, r, πr, 3, 2, , ∴, , ratio of their surface areas =, , 2, , 4 πR 2 R 2 ⎛ R ⎞, 3, 9, =, = ⎜ ⎟ = ⎛⎜ ⎞⎟ = = 9 : 4., ⎝ 2⎠, 4 πr 2 r 2 ⎝ r ⎠, 4, , So, the given result is false., Thus, Assertion (A) is false and Reason (R) is true., So, the correct answer is (d)., 80. We know that the volume of a cone is, , 1 2, πr h., 3, , So, the given result, i.e., Reason (R), is false., Now, r = 3 cm and h = 4 cm., ∴, , curved surface area of the cone, = πr × r 2 + h 2 = π × 3 ×, , 32 + 42, , = 3 π 25 cm = ( 15 π ) cm , which is true., 2, , 2, , Thus, Assertion (A) is true and Reason (R) is false., Hence, the correct answer is (c)., , ................................................................
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824, , Secondary School Mathematics for Class 10, , TEST YOURSELF, Very-Short-Answer Questions, 1. Find the number of solid spheres, each of diameter 6 cm, that could be, moulded to form a solid metallic cylinder of height 45 cm and diameter, 4 cm., 2. Two right circular cylinders of equal volumes have their heights in the, ratio 1 : 2. What is the ratio of their radii?, 3. A circus tent is cylindrical to a height of 4 m and conical above it. If its, diameter is 105 m and its slant height is 40 m, find the total area of the, canvas required., 4. The radii of the top and bottom of a bucket of slant height 45 cm are, 28 cm and 7 cm respectively. Find the curved surface area of the bucket., Short-Answer Questions, 5. A solid metal cone with radius of base 12 cm and height 24 cm is melted, to form solid spherical balls of diameter 6 cm each. Find the number of, balls formed., 6. A hemispherical bowl of internal diameter 30 cm is full of a liquid. This, liquid is filled into cylindrical-shaped bottles each of diameter 5 cm and, height 6 cm. How many bottles are required?, 7. A solid metallic sphere of diameter 21 cm is melted and recast into, small cones, each of diameter 3.5 cm and height 3 cm. Find the number, of cones so formed., 8. The diameter of a sphere is 42 cm. It is melted and drawn into a, cylindrical wire of diameter 2.8 cm. Find the length of the wire., 9. A drinking glass is in the shape of frustum of a cone of height 21 cm, with 6 cm and 4 cm as the diameters of its two circular ends. Find the, capacity of the glass., 10. Two cubes, each of volume 64 cm 3 , are joined end to end. Find the total, surface area of the resulting cuboid., 11. The radius of the base and the height of a solid right circular cylinder, are in the ratio 2 : 3 and its volume is 1617 cm 3 . Find the total surface, area of the cylinder. [Take π =, , 22, 7, , ⋅], , 12. A toy is in the form of a cone mounted on a hemisphere of common base, radius 7 cm. The total height of the toy is 31 cm. Find the total surface, area of the toy.
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Volume and Surface Areas of Solids, , 825, , 13. A hemispherical bowl of internal radius 9 cm is full of water. This water, is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm., Find the number of bottles needed to fill the whole water of the bowl., 14. The surface areas of a sphere and a cube are equal. Find the ratio of their, 22, , volumes. [Take π =, , 7, , ⋅], , 15. The slant height of the frustum of a cone is 4 cm and the perimeters, (i.e., circumferences) of its circular ends are 18 cm and 6 cm. Find the, curved surface area of the frustum., 16. A solid is composed of a cylinder with hemispherical ends. If the whole, length of the solid is 104 cm and the radius of each hemispherical end is, 7 cm, find the surface area of the solid., Long-Answer Questions, 17. From a solid cylinder whose height is 15 cm and diameter 16 cm, a, conical cavity of the same height and same diameter is hollowed out., Find the total surface area of the remaining solid. [Use π = 3.14.], 18. A solid rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into, a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm., Find the length of the pipe., 19. An open metal bucket is in the shape of a frustum of a cone, mounted on, a hollow cylindrical base made of the same metallic sheet. The, diameters of the two circular ends of the bucket are 45 cm and 25 cm,, the total vertical height of the bucket is 40 cm and that of the cylindrical, base is 6 cm. Find the area of the metallic sheet used to make the bucket., Also, find the volume of water the bucket can hold, in litres., 20. A farmer connects a pipe of internal diameter 20 cm from a canal into a, cylindrical tank which is 10 m in diameter and 2 m deep. If the water, flows through the pipe at the rate of 4 km/hr, in how much time will, the tank be filled completely?, ANSWERS (TEST YOURSELF), , 1. 5, 6. 60, , 7. 504, , 11. 770 cm 2, 16. 4576 cm, , 3. 7920 m 2, , 2. 2 : 1, , 8. 63 m, , 12. 858 cm 2, 2, , 13. 54, , 4. 4950 cm 2, 9. 418 cm, , 5. 32, , 3, , 10. 160 cm 2, 15. 710.3 cm 2, , 14. 231 : 121, , 2, , 17. 443.14 cm 18. 112 m, , 20. 1 hour 15 minutes, _, , 2, , 19. 4860.9 cm , 33.62 litres
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826, , Secondary School Mathematics for Class 10, , Mean, Median,, Mode of Grouped Data,, Cumulative Frequency, Graph and Ogive, , 18, INTRODUCTION, , Suppose we want to compare the wage distribution of workers in two, factories and determine which factory pays more to its workers. If we, compare on the basis of individual workers, we cannot conclude anything., However, if for the given data, we get a representative value that signifies, the characteristics of the data, the comparison becomes easy., A certain value representative of the whole data and signifying its, characteristics is called an average of the data., An average tends to lie centrally with the values of the variable, arranged in ascending order of magnitude. So, we call an average a measure, of central tendency of the data., Three measures of central tendency are useful for analysing the, data, namely, (a) Mean, , (b) Median, , (c) Mode., , MEAN OF GROUPED DATA, We already know that, mean , , sum of observations ·, number of observations, , Thus, if x1, x2, …, xn are n observations with respective frequencies, f1, f2, …, fn then the mean is given by, n, , x, , fi xi, , f1 x1 f2 x2 … fn xn i 1, , ,, fi, f1 f2 … fn, , where the Greek letter (read as, sigma) stands for summation of the terms., There are three methods for computing the mean of grouped data., In this method, the midpoint or class mark of each class, interval is taken to represent the observations falling in the class. We, , I. DIRECT METHOD, , 826
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 827, , proceed stepwise as follows:, Step 1., , For each class interval, find the class mark xi as, upper class limit lower class limit, ·, xi , 2, , Step 2., , Calculate f i xi for each i., , Step 3., , Calculate the mean using the formula, f i xi, ·, mean , f i, , EXAMPLE 1, , Find the mean of the following data:, Class interval, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 8, , 12, , 10, , 11, , 9, [CBSE 2007], , SOLUTION, , We may prepare the table as shown:, Class interval, , Frequency, fi, , Class mark, xi, , 0–10, , 8, , 5, , 40, , 10–20, , 12, , 15, , 180, , 20–30, , 10, , 25, , 250, , 30–40, , 11, , 35, , 385, , 40–50, , 9, , 45, , 405, , f i 50, , , EXAMPLE 2, , SOLUTION, , ( f i # xi), , mean , , ( f i # xi) 1260, , ( f i # xi) 1260, , 25.2., 50, f i, , The arithmetic mean of the following frequency distribution is 25., Determine the value of p., [CBSE 2006], Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 5, , 18, , 15, , p, , 6, , We have, Class, interval, , Frequency, fi, , Midvalue, xi, , ( f i # xi), , 0–10, , 5, , 5, , 25, , 10–20, , 18, , 15, , 270
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828, , Secondary School Mathematics for Class 10, , 20–30, , 15, , 25, , 375, , 30–40, , p, , 35, , 35p, , 6, , 45, , 40–50, , 270, , f i (44 p), , , mean, x , , ( f i # xi) (940 35p), , (940 35p), ( f i # xi), 25, &, (44 p), f i, , & (940 35p) 25 (44 p), & (35p 25p) (1100 940), & 10p 160 & p 16., Hence, p 16., EXAMPLE 3, , If the mean of the following frequency distribution is 65.6, find the, missing frequencies f 1 and f 2 ., [CBSE 2010], Class, , 10–30 30–50 50–70 70–90 90–110 110–130 Total, , Frequency, SOLUTION, , 5, , 8, , 20, , f1, , 2, , f2, , 50, , We have, 5 8 f 1 20 f 2 2 50 & f 2 (15 f 1) ., Now, we may prepare the table given below:, Class, interval, , Frequency, fi, , Class mark, xi, , (f i # xi), , 10–30, , 5, , 20, , 100, , 30–50, , 320, , 8, , 40, , 50–70, , f1, , 60, , 60f 1, , 70–90, , 20, , 80, , 1600, , 90–110, , 15 f 1, , 100, , 1500 100f 1, , 110–130, , 2, , 120, , f i 50, , , , mean , , 240, ( f i # xi) (3760 40f 1), , ( f i # xi) (3760 40f 1), ·, , 50, f i, , But, mean = 65.6 (given)., (3760 40f 1), 65.6 & 3760 40f 1 3280, , 50, & 40f1 480 & f1 12., Thus, f 1 12 and f 2 (15 12) 3.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 829, II. ASSUMED-MEAN METHOD This method is usually used when the numerical, values of xi and f i are large due to which calculations become tedious and, time-consuming., , We proceed stepwise as follows:, Step 1., , For each class interval, calculate the class mark xi by using the, formula,, upper class limit lower class limit, ·, xi , 2, , Step 2., , Choose a suitable value of xi in the middle as the assumed, mean and denote it by A., , Step 3., , Calculate the deviations di (xi A) for each i., , Step 4., , Calculate the product (f i di) for each i., , Step 5., , Find n f i ., , Step 6., , Calculate the mean, x , by using the formula,, f i di, x A n ·, , MATHEMATICAL DERIVATION OF ASSUMED MEAN FORMULA, , We have, mean of the deviations, d , , f i di, ·, f i, , But, di (xi A), where A is the assumed mean., , , d, , f i (xi A) f i xi f i A, f i, , , xA, xA, f i, f i, f i, f i, , x Ad, f i di, ·, x A, f i, EXAMPLE 4, , Data regarding the weights of students of Class X of a school is, given below., Weight, (in kg), Number of, students, , 50–52 52–54 54–56 56–58 58–60 60–62 62–64, 18, , 21, , 17, , 28, , Compute the mean weight of the students., , 16, , 35, , 15, , [CBSE 2014]
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830, , Secondary School Mathematics for Class 10, , SOLUTION, , Let A 57 be the assumed mean. Then, we have, Class, interval, , Frequency, fi, , Class, mark, xi, , Deviation, di (xi A), (xi 57), , ( f i # di), , 50–52, , 18, , 51, , –6, , –108, , 52–54, , 21, , 53, , –4, , –84, , 54–56, , 17, , 55, , –2, , –34, , 56–58, , 28, , 57 = A, , 0, , 0, , 58–60, , 16, , 59, , 2, , 32, , 60–62, , 35, , 61, , 4, , 140, , 15, , 63, , 6, , 62–64, , f i 150, , , x A, , 90, ( f i # di) 36, , ( f i # di), a57 36 k 57 0.24 57.24., 150, f i, , Hence, mean weight = 57.24 kg., This method is also used to simplify calculations, in the process of computing the mean. It is particularly used when the, values of (xi A) are large and divisible by the class size (= upper limit –, lower limit)., III. STEP-DEVIATION METHOD, , We proceed stepwise as shown below., Step 1., , For each class interval, calculate the class mark xi, where, , upper class limit lower class limit, ·, 2, Step 2. Choose a suitable value of xi in the middle as the assumed, mean and denote it by A., xi , , Step 3., , Calculate h (upper limit lower limit), which is the same for, all the classes., , Step 4., , Calculate ui , , Step 5., , Calculate f i ui for each class., , Step 6., , Calculate f i and f i ui ., , Step 7., , Calculate the mean by using the formula, f i ui, 3·, x A )h #, f i, , (xi A), for each class., h
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 831, MATHEMATICAL DERIVATION OF STEP-DEVIATION FORMULA, , We have u , , ( f i ui), ·, f i, , xi A, $, h, (xi A), f i, ( f i xi Af i), h, G, 1=, u, h, f i, f i, , But, ui , , , , (f i xi), f i, A G, 1=, h f i, f i, 1 ( x A), h, hu x A, x A hu, ( f i ui), o·, x A he, f i, EXAMPLE 5, , SOLUTION, , Calculate the arithmetic mean of the following frequency distribution,, using the step-deviation method:, [CBSE 2001C], Class interval, , Frequency, , 0–50, , 17, , 50–100, , 35, , 100–150, , 43, , 150–200, , 40, , 200–250, , 21, , 250–300, , 24, , Here, h 50. Let the assumed mean be A 125., For calculating the mean, we prepare the table given as, follows:, Class Frequency Midvalue, (xi A), ui , interval, fi, xi, h, 0–50, , 17, , 50–100, 100–150, , (f i # ui), , 25, , –2, , –34, , 35, , 75, , –1, , –35, , 43, , 125 = A, , 0, , 0
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832, , Secondary School Mathematics for Class 10, , 150–200, , 40, , 175, , 1, , 40, , 200–250, , 21, , 225, , 2, , 42, , 24, , 275, , 3, , 250–300, , 72, , ( f i ui) 154 69, 85, , f i 180, Thus, we have, , A 125, h 50, f i 180 and ( f i ui) 85., Mean, x A )h #, , ( f i ui), 3, f i, , 125 &50 # 85 0 125 23.61 148.61., 180, Hence, the mean of the given frequency is 148.61., EXAMPLE 6, , SOLUTION, , The following table gives the distribution of expenditures of different, families on education. Find the mean expenditure on education of a, family., [CBSE 2004C], Expenditure (in `), , Number of families, , 1000–1500, , 24, , 1500–2000, , 10, , 2000–2500, , 33, , 2500–3000, , 28, , 3000–3500, , 30, , 3500–4000, , 22, , 4000–4500, , 16, , 4500–5000, , 7, , Here, h 500., For calculating the mean, we prepare the table given as, follows:, Class, interval, , Frequency Midvalue, (xi A), ui , fi, xi, h, , ( f i # ui), , 1000–1500, , 24, , 1250, , –3, , –72, , 1500–2000, , 10, , 1750, , –2, , –20, , 2000–2500, , 33, , 2250, , –1, , –33, , 28, , 2750 A, , 0, , 0, , 2500–3000
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 833, , 3000–3500, , 30, , 3250, , 1, , 30, , 3500–4000, , 22, , 3750, , 2, , 44, , 4000–4500, , 16, , 4250, , 3, , 48, , 4500–5000, , 7, , 4750, , 4, , 28, , f i 170, , (f i # ui) 25, , Thus, A 2750, h 500, f i 170 and (f i # ui) 25., , , mean A )h #, , ( f i # ui), 3, f i, , 2750 a 500 # 25 k 2750 73.53 2823.53., 170, Hence, the required mean expenditure = ` 2823.53., EXAMPLE 7, , SOLUTION, , Compute the arithmetic mean for the following data:, Marks obtained, , Number of students, , Less than 10, , 14, , Less than 20, , 22, , Less than 30, , 37, , Less than 40, , 58, , Less than 50, , 67, , Less than 60, , 75, , The above data may be written as:, Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Frequency, , 14, , 8, , 15, , 21, , 9, , 8, , Here, h 10., Let assumed mean = midpoint of (30–40) = 35., Now, we form the table as under., (xi A), h, (xi 35), , 10, , ui , , Class, , Frequency, fi, , Midvalue, xi, , 0–10, , 14, , 5, , –3, , –42, , 10–20, , 8, , 15, , –2, , –16, , 20–30, , 15, , 25, , –1, , –15, , ( f i # ui)
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834, , Secondary School Mathematics for Class 10, , 30–40, , 21, , 35 = A, , 0, , 0, , 40–50, , 9, , 45, , 1, , 9, , 8, , 55, , 2, , 50–60, , 16, , ( f i # ui) 48, , f i 75, , , , h #(f i # ui), 10 #(48), m1, 3 '35 c, 75, f i, 35 6.4 28.6., , mean, x )A , , MEAN FOR AN INCLUSIVE SERIES, EXAMPLE 8, , Find the arithmetic mean of the following frequency distribution:, Class, Frequency, , 25–29 30–34 35–39 40–44 45–49 50–54 55–59, 14, , 22, , 16, , 6, , 5, , 3, , 4, , [CBSE 2006C], SOLUTION, , The given series is an inclusive series. Making it an exclusive, series, we get, , Class, , Frequency Midvalue, fi, xi, , (xi A), h, (xi 42), , 5, , ui , , ( f i # ui), , 24.5–29.5, , 14, , 27, , –3, , –42, , 29.5–34.5, , 22, , 32, , –2, , –44, , 34.5–39.5, , 16, , 37, , –1, , –16, , 39.5–44.5, , 6, , 42 = A, , 0, , 0, , 44.5–49.5, , 5, , 47, , 1, , 5, , 49.5–54.5, , 3, , 52, , 2, , 6, , 54.5–59.5, , 4, , 57, , 3, , 12, , f i 70, , ( f i # ui) 79, , Thus, A 42, h 5, f i 70 and (f i # ui) 79., (f i # ui), ( 79), 1, 3 42 '5 #, mean, x A )h #, 70, f i, 42 5.64 36.36., Hence, the required arithmetic mean is 36.36., REMEMBER The value of mean obtained by all the above three methods is, , the same.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 835, , f, , EXERCISE 18A, , 1. If the mean of 5 observations x, x 2, x 4, x 6 and x 8 is 11, find the, value of x., [CBSE 2014], 2. If the mean of 25 observations is 27 and each observation is decreased, by 7, what will be the new mean?, [CBSE 2014], 3. Compute the mean of the following data:, Class, , 1–3, , 3–5, , 5–7, , 7–9, , Frequency, , 12, , 22, , 27, , 19, , [CBSE 2013], , 4. Find the mean of the following data, using direct method:, Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Frequency, , 7, , 5, , 6, , 12, , 8, , 2, , 5. Calculate the mean of the following data, using direct method:, Class, , 25–35, , 35–45, , 45–55, , 55–65, , 65–75, , Frequency, , 6, , 10, , 8, , 12, , 4, , 6. Compute the mean of the following data, using direct method:, Class, , 0–100, , 100–200, , 200–300, , 300–400, , 400–500, , Frequency, , 6, , 9, , 15, , 12, , 8, [CBSE 2005C], , 7. Using an appropriate method, find the mean of the following frequency, distribution:, Class, interval, , 84–90, , 90–96, , Frequency, , 8, , 10, , 96–102 102–108 108–114 114–120, 16, , 23, , 12, , 11, , Which method did you use, and why?, 8. If the mean of the following frequency distribution is 24, find the value, of p., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 3, , 4, , p, , 3, , 2, [CBSE 2013]
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836, , Secondary School Mathematics for Class 10, , 9. The following distribution shows the daily pocket allowance of, children of a locality. If the mean pocket allowance is ` 18, find the, missing frequency f., Daily pocket, allowance (in `), , 11–13 13–15 15–17 17–19 19–21 21–23 23–25, , Number of, children, , 7, , 6, , 9, , f, , 13, , 5, , 4, , 10. If the mean of the following frequency distribution is 54, find the value, of p., Class, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , Frequency, , 7, , p, , 10, , 9, , 13, [CBSE 2006C], , 11. The mean of the following data is 42. Find the missing frequencies x, and y if the sum of frequencies is 100., Class interval, , 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80, , Frequency, , 7, , x, , 10, , y, , 13, , 10, , 14, , 9, , [CBSE 2014], , 12. The daily expenditure of 100 families are given below. Calculate f 1 and, f 2 if the mean daily expenditure is ` 188., Expenditure, (in `), , 140–160, , 160–180, , 180–200, , 200–220, , 220–240, , Number of, families, , 5, , 25, , f1, , f2, , 5, [CBSE 2014], , 13. The mean of the following frequency distribution is 57.6 and the total, number of observations is 50., Class, , 0–20, , 20–40, , 40–60, , 60–80, , Frequency, , 7, , f1, , 12, , f2, , 80–100 100–120, 8, , 5, , Find f1 and f 2 ., 14. During a medical check-up, the number of heartbeats per minute of 30, patients were recorded and summarised as follows:, Number of heartbeats per minute, Number of patients, , 65–68 68–71 71–74 74–77 77–80 80–83 83–86, 2, , 4, , 3, , 8, , 7, , 4, , 2
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 837, , Find the mean heartbeats per minute for these patients, choosing a, suitable method., 15. Find the mean marks per student, using assumed-mean method:, Marks, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Number of, students, , 12, , 18, , 27, , 20, , 17, , 6, , 16. Find the mean of the following frequency distribution, using the, assumed-mean method:, Class, , 100–120, , 120–140, , 140–160, , 160–180, , 180–200, , Frequency, , 10, , 20, , 30, , 15, , 5, , 17. Find the mean of the following data, using the assumed-mean method:, Class, , 0–20, , 20–40, , 40–60, , 60–80, , Frequency, , 20, , 35, , 52, , 44, , 80–100 100–120, 38, , 31, , 18. The following table gives the literacy rate (in percentage) in 40 cities., Find the mean literacy rate, choosing a suitable method., Literacy rate (%), , 45–55, , 55–65, , 65–75, , 75–85, , 85–95, , Number of cities, , 4, , 11, , 12, , 9, , 4, [CBSE 2014], , 19. Find the mean of the following frequency distribution using stepdeviation method., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 7, , 10, , 15, , 8, , 10, [CBSE 2014], , 20. Find the mean of the following data, using step-deviation method:, Class, , 5–15, , 15–25, , 25–35, , 35–45, , 45–55, , 55–65, , 65–75, , Frequency, , 6, , 10, , 16, , 15, , 24, , 8, , 7, [CBSE 2013], , 21. The weights of tea in 70 packets are shown in the following table:, Weight, (in grams), Number of, packets, , 200–201 201–202 202–203 203–204 204–205 205–206, 13, , 27, , 18, , 10, , 1, , 1, , Find the mean weight of packets using step-deviation method. [CBSE 2013]
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838, , Secondary School Mathematics for Class 10, , 22. Find the mean of the following frequency distribution using a suitable, method:, Class, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Frequency, , 25, , 40, , 42, , 33, , 10, [CBSE 2013], , 23. In an annual examination, marks (out of 90) obtained by students of, Class X in mathematics are given below:, Marks, obtained, , 0–15, , 15–30, , 30–45, , 45–60, , 60–75, , 75–90, , Number of, students, , 2, , 4, , 5, , 20, , 9, , 10, , Find the mean marks., , [CBSE 2014], , 24. Find the arithmetic mean of the following frequency distribution using, step-deviation method:, Age (in years), , 18–24, , 24–30, , 30–36, , 36–42, , 42–48, , 48–54, , Number of workers, , 6, , 8, , 12, , 8, , 4, , 2, , 25. Find the mean of the following data using step-deviation method:, Class, , 500–520 520–540 540–560 560–580 580–600 600–620, , Frequency, , 14, , 9, , 5, , 4, , 3, , 5, , 26. Find the mean age from the following frequency distribution:, Age (in years), , 25–29 30–34 35–39 40–44 45–49 50–54 55–59, , Number of persons, HINT, , 4, , 14, , 22, , 16, , 6, , 5, , 3, , Change the given series to the exclusive series., , 27. The following table shows the age distribution of patients of malaria in, a village during a particular month:, Age (in years), , 5–14, , 15–24, , 25–34, , 35–44, , 45–54, , 55–64, , Number of cases, , 6, , 11, , 21, , 23, , 14, , 5, , Find the average age of the patients., 28. Weight of 60 eggs were recorded as given below:, Weight (in grams) 75–79 80–84 85–89 90–94 95–99 100–104 105–109, Number of eggs, , 4, , 9, , 13, , 17, , Calculate their mean weight to the nearest gram., , 12, , 3, , 2
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 839, , 29. The following table shows the marks scored by 80 students in an, examination:, Less, Less, Less, Less, Less, Less, Less, Less, than 5 than 10 than 15 than 20 than 25 than 30 than 35 than 40, , Marks, Number of, students, , 3, , 10, , 25, , 49, , 65, , 73, , 78, , 80, , Calculate the mean marks correct to 2 decimal places., ANSWERS (EXERCISE 18A), , 1. x 7, , 2. 20, , 3. 5.325, , 4. 28.75, , 5. 49.5, , 6. 264, , 7. 103.05, , 8. p 18, , 9. f 20, , 10. p 11, , 11. x 12, y 25 12. f 1 50, f 2 15, , 13. f 1 8, f 2 10, , 14. 75.9, , 15. 28, , 16. 145, , 17. 62.55, , 18. 69.5%, , 19. 25.8, , 20. 40.81, , 21. 201.96 g, , 22. 42.53, , 23. 55.5, , 24. 33.3 years, , 25. 544, , 26. 39.36 years 27. 34.87 years 28. 90 g, , 29. 18.56, , MEDIAN FOR GROUPED DATA, The median is a measure of central tendency which gives the value of the, middlemost observation in the data., As we have already studied, for an ungrouped data comprising n, observations:, ]Z] n 1, ]]a 2 kth observation, if n is odd, ]], Median = [] n, ]]` jth observation `n 1jth observation, if n is even, 2, ]] 2, ], 2, \, METHOD FOR FINDING THE MEDIAN FOR GROUPED DATA, , We proceed stepwise as follows:, Step 1. For the given frequency distribution and obtain N f i ., N·, 2, N, Step 3. Look at the cumulative frequency just greater than, and find, 2, the corresponding class, known as median class (as the middlemost observation lies in this class)., Step 2., , Find
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840, , Secondary School Mathematics for Class 10, Step 4., , Compute the median using the formula:, Z], _b, N, ]]], a cf kbbb, 2, b,, Median, Me l ]][h #, b`b, f, ]], bb, ]], b, \, a, where l lower limit of median class;, h width of median class;, f frequency of median class;, cf cumulative frequency of the class preceding the, median class;, N f i., , SOLVED EXAMPLES, EXAMPLE 1, , Find the median of the following data:, Marks, , 20–30 30–40 40–50 50–60 60–70 70–80 80–90, , Number of, students, , 5, , 15, , 25, , 20, , 7, , 8, , 10, [CBSE 2013], , SOLUTION, , We prepare the cumulative frequency table, as given below., Class, , Frequency (f i), , Cumulative, frequency, , 20–30, , 5, , 5, , 30–40, , 15, , 20, , 40–50, , 25, , 45, , 50–60, , 20, , 65, , 60–70, , 7, , 72, , 70–80, , 8, , 80, , 80–90, , 10, , 90, , N f i 90, N, 45., 2, The cumulative frequency just greater than 45 is 65 and the, corresponding class is 50–60., Now, N 90 &, , Thus, the median class is 50–60.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 841, , , , l 50, h 10, f 20, cf c.f. of preceding class = 45,, and, , N, 45., 2, , Z], N _bb, ]], cf kbb, a, ], 2, b 50 '10 # (45 45) 1, Median, Me l ]][h #, b`b, 20, f, ]], bb, ]], b, \, a, 50 0 50., Hence, the median mark is 50., EXAMPLE 2, , Find the median wage from the following data:, , Wages (in `) 800–820 820–840 840–860 860–880 880–900 900–920 920–940, Number of, workers, , 7, , 14, , 19, , 25, , 20, , 10, , 5, [CBSE 2013], , SOLUTION, , We prepare the cumulative frequency table, as given below., Cumulative, frequency, , Class, , Frequency (f i), , 800–820, , 7, , 7, , 820–840, , 14, , 21, , 840–860, , 19, , 40, , 860–880, , 25, , 65, , 880–900, , 20, , 85, , 900–920, , 10, , 95, , 920–940, , 5, , 100, , N f i 100, , N, 50., 2, The cumulative frequency just greater than 50 is 65 and the, corresponding class is 860–880., Now, N 100 &, , Thus, the median class is 860–880., , , l 860, h 20, f 25, cf c.f. of preceding class = 40,, and, , N, 50., 2
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842, , Secondary School Mathematics for Class 10, , Z], N _bb, ]], cf kbb, a, ], 2, b 860 '20 # (50 40) 1, Median, Me l ][]h #, `b, 25, f, ]], bb, ]], bb, \, a, 860 20 # 10 860 8 868., 25, Hence, the median wage is ` 868., MEDIAN FOR INCLUSIVE SERIES, EXAMPLE 3, , Find the median for the following frequency distribution:, Height (in cm) 160–162 163–165 166–168 169–171 172–174, Frequency, , SOLUTION, , 15, , 117, , 136, , 118, , 14, , The given series is in inclusive form. Converting it to exclusive, form and preparing the cumulative frequency table, we get, Class, , Frequency (f i), , Cumulative, frequency, , 159.5–162.5, , 15, , 15, , 162.5–165.5, , 117, , 132, , 165.5–168.5, , 136, , 268, , 168.5–171.5, , 118, , 386, , 171.5–174.5, , 14, , 400, , N f i 400, , N, 200., 2, The cumulative frequency just greater than 200 is 268 and the, corresponding class is 165.5–168.5., Now, N 400 &, , Thus, the median class is 165.5–168.5., , , l 165.5, h 3, f 136, cf c.f. of preceding class = 132,, , N, 200., 2, Z], _b, N, ]], a cf kbbb, ]], 2, b, Median, Me l ][h #, b`b, f, ]], bb, ]], b, \, a, and
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 843, , (200 132), 1 165.5 a 3 # 68 k, 136, 136, 165.5 1.5 167., 165.5 '3 #, , Hence, median height is 167 cm., EXAMPLE 4, , Given below is the distribution of IQ of 100 students. Find the, median IQ., , IQ, , 75–84, , 85–94, , Frequency, , 8, , 11, , SOLUTION, , 95–104 105–114 115–124 125–134 135–144, 26, , 31, , 18, , 4, , 2, , The given series is in inclusive form. Converting it to, exclusive form and preparing the cumulative frequency table,, we get, Cumulative, frequency, , Class, , Frequency (f i), , 74.5–84.5, , 8, , 8, , 84.5–94.5, , 11, , 19, , 94.5–104.5, , 26, , 45, , 104.5–114.5, , 31, , 76, , 114.5–124.5, , 18, , 94, , 124.5–134.5, , 4, , 98, , 134.5–144.5, , 2, , 100, , N f i 100, , N, 50., 2, The cumulative frequency just greater than 50 is 76 and the, corresponding class interval is 104.5–114.5., Now, N 100 &, , Thus, the median class is 104.5–114.5., l 104.5, h 10, f 31, cf c.f. of preceding class = 45, N, and, 50., 2, Z], _b, N, ]], a cf kbbb, ]], 2, b, Median, Me l []h #, b`b, f, ]], bb, ]], b, \, a,
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844, , Secondary School Mathematics for Class 10, , (50 45), 1 104.5 50, 31, 31, 104.5 1.6 106.1., Hence, the median IQ is 106.1., 104.5 '10 #, , EXAMPLE 5, , SOLUTION, , Calculate the median for the following data:, Marks obtained, , No. of students, , Below 10, , 6, , Below 20, , 15, , Below 30, , 29, , Below 40, , 41, , Below 50, , 60, , Below 60, , 70, , From the given table, we may get back frequencies and, cumulative frequencies as shown below., Class interval, , Frequency (f i), , Cumulative, frequency, , 0–10, , 6, , 6, , 10–20, , 9, , 15, , 20–30, , 14, , 29, , 30–40, , 12, , 41, , 40–50, , 19, , 60, , 10, , 70, , 50–60, , N f i 70, , N, 35., 2, The cumulative frequency just greater than 35 is 41 and the, corresponding class is 30–40., Thus, the median class is 30–40., l 30, h 10, f 12, cf c.f. of preceding class = 29,, N, and, 35., 2, ]Z], b_, N, ]], a cf kbbb, 2, b 30 '10 # (35 29) 1 35., Median, Me l ][]h #, `b, 12, f, ]]], bbb, ], b, \, a, Hence, the required median is 35., Now, N 70 &
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 845, EXAMPLE 6, , Find the missing frequencies in the following frequency distribution, table, if N = 100 and median is 32., , Marks, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Total, , Number of, students, , 10, , ?, , 25, , 30, , ?, , 10, , 100, [CBSE 2013], , SOLUTION, , Let f 1 and f 2 be the missing frequencies of class intervals 10–20, and 40–50 respectively. Then,, 10 f 1 25 30 f 2 10 100 & f 1 f 2 25., Median is 32, which lies in 30–40. So, the median class is 30–40., l 30, h 10, f 30, N 100 and cf 10 f 1 25 f 1 35., ]Z], b_, N, ]], a cf kbbb, 2, b, Now, median, Me l ][]h #, `b, f, ]], bb, ]], bb, \, a, {50 (f 1 35)}, F 32, , 30 <10 #, 30, (15 f 1), 32 & (15 f 1) b & f 1 9., , 30 , 3, , f 1 9 and f 2 25 9 16., Hence, f 1 9 and f 2 16., , f, , EXERCISE 18B, , 1. In a hospital, the ages of diabetic patients were recorded as follows., Find the median age., Age (in years), , 0–15, , 15–30, , 30–45, , 45–60, , Number of patients, , 5, , 20, , 40, , 50, , 60–75, 25, [CBSE 2014], , 2. Compute the median from the following data:, Marks, , 0–7, , 7–14, , 14–21, , 21–28, , 28–35, , 35–42, , 42–49, , Number of, students, , 3, , 4, , 7, , 11, , 0, , 16, , 9, , 3. The following table shows the daily wages of workers in a factory:, Daily wages (in `), , 0–100, , 100–200, , 200–300, , 300–400, , 400–500, , Number of workers, , 40, , 32, , 48, , 22, , 8, , Find the median daily wage income of the workers.
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846, , Secondary School Mathematics for Class 10, , 4. Calculate the median from the following frequency distribution:, Class, , 5–10, , Frequency, , 5, , 10–15 15–20 20–25 25–30 30–35 35–40 40–45, 6, , 15, , 10, , 5, , 4, , 2, , 2, , 5. Given below is the number of units of electricity consumed in a week in, a certain locality:, Consumption, (in units), , 65–85 85–105 105–125 125–145 145–165 165–185 185–205, , Number of, consumers, , 4, , 5, , 13, , 20, , 14, , 7, , 4, , Calculate the median., 6. Calculate the median from the following data:, Height, 135–140 140–145145–150150–155155–160160–165165–170170–175, (in cm), No. of, boys, , 6, , 10, , 18, , 22, , 20, , 15, , 6, , 3, , 7. Calculate the missing frequency from the following distribution, it, being given that the median of the distribution is 24., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 5, , 25, , ?, , 18, , 7, , 8. The median of the following data is 16. Find the missing frequencies a, and b if the total of frequencies is 70., Class, , 0–5, , 5–10, , Frequency, , 12, , a, , 10–15 15–20 20–25 25–30 30–35 35–40, 12, , 15, , b, , 6, , 6, , 4, [CBSE 2013], , 9. In the following data the median of the runs scored by 60 top batsmen, of the world in one-day international cricket matches is 5000. Find the, missing frequencies x and y., Runs scored 2500–3500 3500–4500 4500–5500 5500–6500 6500–7500 7500–8500, Number of, batsmen, , x, , 5, , y, , 12, , 6, , 2, , 10. If the median of the following frequency distribution is 32.5, find the, values of f 1 and f 2 ., Class, interval, , 0–10, , Frequency, , f1, , 10–20 20–30 30–40 40–50 50–60 60–70 Total, 5, , 9, , 12, , f2, , 3, , 2, , 40
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 847, , 11. Calculate the median for the following data:, Age (in years), , 19–25, , 26–32, , 33–39, , 40–46, , 47–53, , 54–60, , Frequency, , 35, , 96, , 68, , 102, , 35, , 4, , HINT, , Convert it to exclusive form., , 12. Find the median wages for the following frequency distribution:, Wages per day, (in `), , 61–70, , 71–80, , 81–90, , No. of women, workers, , 5, , 15, , 20, , HINT, , 91–100 101–110 111–120, 30, , 20, , 8, , Convert it to exclusive form., , 13. Find the median from the following data:, Class, , 1–5, , Frequency, , 7, , HINT, , 6–10 11–15 16–20 21–25 26–30 31–35 36–40 41–45, 10, , 16, , 32, , 24, , 16, , 11, , 5, , 2, , Convert it to exclusive form., , 14. Find the median from the following data:, Marks, , No. of students, , Below 10, , 12, , Below 20, , 32, , Below 30, , 57, , Below 40, , 80, , Below 50, , 92, , Below 60, , 116, , Below 70, , 164, , Below 80, , 200, , ANSWERS (EXERCISE 18B), , 1. 46.5 years, 6. 153.64 cm, , 2. 28, , 3. ` 206.25, , 7. 25, , 8. a 8, b 7, , 10. f 1 3, f 2 6 11. 36 years 12. ` 93.50, , 4. 19.5, , 5. 136.5 units, , , 9. x 25, y 10, 13. 19.95, , 14. 53.33, , MODE OF A GROUPED DATA, Mode is that value of a variate which occurs most often, i.e., the value of the, observation having the maximum frequency.
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848, , Secondary School Mathematics for Class 10, , More precisely, mode is that value of the variable at which the, concentration of the data is maximum., In a frequency distribution, the class having maximum, frequency is called the modal class., , MODAL CLASS, , Formula for Calculating Mode:, , We have, , (f k f k 1), 3,, mode, Mo xk h · ), (2f k f k 1 f k 1), , where, , xk lower limit of the modal class interval;, f k frequency of the modal class;, f k 1 frequency of the class preceding the modal class;, f k 1 frequency of the class succeeding the modal class;, h width of the class interval., , SOLVED EXAMPLES, EXAMPLE 1, , Find the mode of the following data:, Class, , 0–20 20–40 40–60 60–80 80–100 100–120 120–140, , Frequency, , 6, , 8, , 10, , 12, , 6, , 5, , 3, [CBSE 2013], , SOLUTION, , Clearly, the modal class is 60–80, as it has the maximum, frequency., x k 60, h 20, fk 12, fk 1 10, fk 1 6., (fk fk 1), 3, Mode, Mo xk )h #, (2fk fk 1 fk 1), (12 10), 2, (2 #12 10 6), 2, 2, 60 '20 #, 1 60 &20 # 0 60 5 65., 8, (24 16), 60 (20 #, , Hence, mode = 65., EXAMPLE 2, , The distribution of sale of shirts sold in a month in a departmental, store is as under., Size, (in cm), Number of, shirts sold, , 80–85 85–90 90–95 95–100 100–105 105–110 110–115, 33, , 27, , 85, , 155, , Calculate the modal size of shirts sold., , 110, , 45, , 15, [CBSE 2014]
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 849, , Clearly, the modal class is 95–100 as it has the maximum, frequency., , SOLUTION, , xk 95, h 5, fk 155, fk 1 85, fk 1 110., , , , (fk fk 1), 3, (2fk fk 1 fk 1), (155 85), 2, 95 (5 #, (2 #155 85 110), , Mode, Mo xk )h #, , 95 '5 #, , 95 70 95 3.04 98.04., 23, modal size = 98.04 cm., , , EXAMPLE 3, , 70, 70, 0, 1 95 &5 #, 115, (310 195), , Given below is the frequency distribution of the heights of players in, a school., Height (in cm), , 160–162 163–165 166–168 169–171 172–174, , Number of, students, , 15, , 118, , 142, , 127, , 18, , Find the modal height and interpret it., SOLUTION, , The given data is an inclusive series. So, we convert it into an, exclusive form, as given below., Class, , Frequency, , 159.5–162.5 162.5–165.5 165.5–168.5 168.5–171.5 171.5–174.5, 15, , 118, , 142, , 127, , 18, , Clearly, the class 165.5–168.5 has maximum frequency, so it is, the modal class., , , xk 165.5, fk 142, fk 1 118, fk 1 127 and h 3., , Mode, Mo xk )h #, , (fk fk 1), 3, (2fk fk 1 fk 1), , (142 118), 2, (2 #142 118 127), 165.5 & 3 # 24 0 165.5 24, 39, 13, 165.5 1.85 167.35., 165.5 (3 #, , Thus, modal height = 167.35 cm., This means that the height of maximum number of players in, the school is 167.35 cm (approx.).
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850, EXAMPLE 4, , Secondary School Mathematics for Class 10, , The following table shows the ages of the patients admitted in a, hospital during a month:, Age (in years), , 6–15, , 16–25, , 26–35, , 36–45, , 46–55, , 56–65, , Number of, patients, , 6, , 11, , 21, , 23, , 14, , 5, , Find the mode and the mean of the data given above. Compare and, interpret the two measures of central tendency., SOLUTION, , The given data is an inclusive series. Making it an exclusive, series, we get, Frequency Class mark, fi, xi, , Class, , (xi A), h, (xi 40.5), , 10, , f i ui, , ui , , 5.5–15.5, , 6, , 10.5, , –3, , –18, , 15.5–25.5, , 11, , 20.5, , –2, , –22, , 25.5–35.5, , 21, , 30.5, , –1, , –21, , 35.5–45.5, , 23, , 40.5 = A, , 0, , 0, , 45.5–55.5, , 14, , 50.5, , 1, , 14, , 5, , 60.5, , 2, , 55.5–65.5, , f i 80, , 10, f i ui 37, , Thus, A 40.5, h 10, f i 80, f i ui 37., f i ui, 3, f i, (37), 1 40.5 37, 40.5 '10 #, 80, 8, 40.5 4.63 35.87., , Mean, x A )h #, , Also, the modal class of the given data is 35.5–45.5, as it has, the maximum frequency., , , xk 35.5, fk 23, fk 1 21, fk 1 14 and h 10., (fk fk 1), 3, (2fk fk 1 fk 1), (23 21), 2 35.5 a10 # 2 k, 35.5 (10 #, 11, (2 # 23 21 14), 35.5 1.82 37.32., , Mode, Mo xk )h #, , Clearly, mode > mean.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 851, Interpretation The modal age is 37.32 years. This means that, the maximum number of patients admitted in the hospital, during the given month are of the age 37.32 years (approx.)., , The mean age is 35.87 years. This means that on an average the, age of a patient admitted to the hospital is 35.87 years., EXAMPLE 5, , The mode of the following series is 36. Find the missing frequency in it., Class interval, Frequency, , SOLUTION, , 0–10 10–20 20–30 30–40 40–50 50–60 60–70, 8, , 10, , …, , 16, , 12, , 6, , 7, , Since the mode of the given series is 36 and maximum, frequency 16 lies in the class 30–40, so the modal class is 30–40., Let the missing frequency be x. Then,, , , xk 30, fk 16, fk 1 x, fk 1 12 and h 10., , Also, mo 36., Using the formula, Mo xk )h #, 30 (10 #, , (fk fk 1), 3, we get, (2fk fk 1 fk 1), , (16 x), 2 36, (32 x 12), , , , 10 #(16 x), 6 & 160 10x 120 6x, (20 x), , , , 4x 40 & x 10., , Hence, the missing frequency is 10., EXAMPLE 6, , SOLUTION, , Compare the modal ages of two groups of students appearing for an, entrance examination., Age (in years), , 16–18, , 18–20, , 20–22, , 22–24, , 24–26, , Group A, , 50, , 78, , 46, , 28, , 23, , Group B, , 54, , 89, , 40, , 25, , 17, , Case I., , Computation of Modal Age of Group A:, , In group A, class 18–20 has the maximum frequency., So, 18–20 is the modal class and its frequency is 78., , , xk 18, fk 78, fk 1 50, fk 1 46 and h 2., , , , mode, Mo xk )h #, , (fk fk 1), 3, (2fk fk 1 fk 1)
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852, , Secondary School Mathematics for Class 10, , 18 (2 #, , Case II., , (78 50), 2 18 & 2 # 28 0, 60, (2 #78 50 46), , 18 14 18 0.93 18.93., 15, Computation of Modal Age of Group B:, , In group B, class 18–20 has the maximum frequency., So, 18–20 is the modal class and its frequency is 89., xk 18, fk 89, fk 1 54, fk 1 40 and h 2., , , , (fk fk 1), 3, (2fk fk 1 fk 1), (89 54), (2 # 35), 2 18 , 18 (2 #, (2 # 89 54 40), (178 94), 18 70 18 5 18 0.83 18.83., 6, 84, Hence, the modal ages of students in groups A and B are, 18.93 years and 18.83 years respectively., , EXAMPLE 7, , Marks, , , , mode, Mo xk )h #, , , , (modal age in group A) > (modal age in group B)., , The following table shows the marks obtained by 100 students of, Class X in a school during a particular academic session. Find the, mode of this distribution., Less, Less, Less, Less, Less, Less, Less, Less, than 10 than 20 than 30 than 40 than 50 than 60 than 70 than 80, , Number of, students, , 7, , 21, , 34, , 46, , 66, , 77, , 92, , 100, [CBSE 2013], , SOLUTION, , Clearly, the above data can be written as:, , Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , 70–80, , Frequency, , 7, , 14, , 13, , 12, , 20, , 11, , 15, , 8, , Clearly, the modal class is 40–50, as it has the maximum, frequency., , , xk 40, fk 20, fk 1 12, fk 1 11, h 10., (fk fk 1), 3, (2fk fk 1 fk 1), (20 12), 2, 40 (10 #, (2 # 20 12 11), , Mode, Mo xk )h #
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 853, , 40 '10 #, , 8, 8, 1 40 &10 # 0, 17, (40 23), , 40 80 40 4.71 44.71., 17, Hence, mode = 44.71., f, , EXERCISE 18C, , 1. Find the mode of the following frequency distribution:, Marks, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Frequency, , 12, , 35, , 45, , 25, , 13, [CBSE 2014], , 2. Compute the mode of the following data:, Class, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , Frequency, , 25, , 16, , 28, , 20, , 5, [CBSE 2013], , 3. Heights of students of Class X are given in the following frequency, distribution:, Height (in cm), , 150–155, , 155–160, , 160–165, , 165–170, , 170–175, , Number of, students, , 15, , 8, , 20, , 12, , 5, , Find the modal height., , [CBSE 2014], , Also, find the mean height. Compare and interpret the two measures of, central tendency., 4. Find the mode of the following distribution:, Class interval 10–14 14–18 18–22 22–26 26–30 30–34 34–38 38–42, Frequency, , 8, , 6, , 11, , 20, , 25, , 22, , 10, , 4, , 5. Given below is the distribution of total household expenditure of 200, manual workers in a city., Expenditure (in `), , No. of manual workers, , 1000–1500, , 24, , 1500–2000, , 40, , 2000–2500, , 31
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854, , Secondary School Mathematics for Class 10, , 2500–3000, , 28, , 3000–3500, , 32, , 3500–4000, , 23, , 4000–4500, , 17, , 4500–5000, , 5, , Find the expenditure done by maximum number of manual workers., 6. Calculate the mode from the following data:, Monthly salary (in `), , No. of employees, , 0–5000, , 90, , 5000–10000, , 150, , 10000–15000, , 100, , 15000–20000, , 80, , 20000–25000, , 70, , 25000–30000, , 10, , 7. Compute the mode from the following data:, Age (in years), , 0–5, , 5–10, , Number of patients, , 6, , 11, , 10–15 15–20 20–25 25–30 30–35, 18, , 24, , 17, , 13, , 5, , 8. Compute the mode from the following series:, Size, , 45–55, , 55–65, , 65–75, , 75–85, , Frequency, , 7, , 12, , 17, , 30, , 85–95 95–105 105–115, 32, , 6, , 10, , 9. Compute the mode of the following data:, Class, interval, Frequency, , 1–5 6–10 11–15 16–20 21–25 26–30 31–35 36–40 41–45 46–50, 3, , 8, , 13, , 18, , 28, , 20, , 13, , 8, , 6, , 4, , 10. The agewise participation of students in the Annual Function of a, school is shown in the following distribution., [CBSE 2014], Age (in years), , 5–7, , 7–9, , 9–11, , 11–13, , 13–15, , 15–17, , 17–19, , Number of, students, , x, , 15, , 18, , 30, , 50, , 48, , x, , Find the missing frequencies when the sum of frequencies is 181. Also,, find the mode of the data.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 855, ANSWERS (EXERCISE 18C), , 1. 33.33, 2. 52, 3. Modal height = 163 cm. This means that the height of maximum number, , of students is 163 cm. Mean height = 161.17 cm. This means that on an, average the height of a student of the class is 161.17 cm., 4. 28.5, 5. ` 1820, 6. ` 7727.27, 7. 17.3 years, 8. 81.5, 9. 23.28, 10. Each of two missing frequencies is 10; 11.18 years, COMPARATIVE STUDY OF THE THREE MEASURES OF CENTRAL TENDENCY, , The mean is the most frequently used measure of central tendency because, it takes into account all the observations and lies between the largest and, the smallest observations of the entire data. It also enables us to compare, two or more distributions, e.g., by comparing the mean marks of students, of different classes in a particular examination, we can conclude which, class has a better performance. However, the mean is useful when there are, no extreme values in the data set, i.e., if the data are normally distributed., However, the median is considered a better measure of central, tendency when there are a few extreme values that could greatly influence, the mean and distort what might be considered typical, as in finding the, typical productivity rate of workers, average wage in a country, etc., e.g.,, in a group of employees, if the salary of one is 10 times the mean salary of, others then the mean salary of the group will be high. In such a case, the, median may better represent the typical salary level of the group., The mode is particularly useful for dealing with categorical data,, where most popular item is to be ascertained, e.g., to find the most popular, TV programme being watched, the consumer item in greatest demand, the, colour of the vehicle used by most of the people, etc. It is particularly useful, for the manufacturers. Thus, a shirt manufacturer is more interested in the, modal size of shirts sold, so as to produce it in large numbers., EMPIRICAL RELATIONSHIP BETWEEN THE THREE MEASURES OF CENTRAL, TENDENCY, , 3(Median) = Mode + 2(Mean), , MISCELLANEOUS QUESTIONS ON MEAN, MODE AND MEDIAN, EXAMPLE, , 100 surnames were randomly picked up from a telephone directory, and the distribution of the number of letters of the English alphabet, in the surnames are obtained as follows:
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856, , Secondary School Mathematics for Class 10, , Number of, letters, , 1–4, , 4–7, , 7–10, , 10–13, , 13–16, , 16–19, , Number of, surnames, , 6, , 30, , 40, , 16, , 4, , 4, , Determine the median and mean number of letters in the surnames., Also, find the modal size of surnames., [CBSE 2014], SOLUTION, , Class, , We have, Class mark Frequency Cumulative, frequency, xi, fi, , (xi A), h, (xi 11.5), , 3, , ( f i ui), , ui , , 1–4, , 2.5, , 6, , 6, , –3, , –18, , 4–7, , 5.5, , 30, , 36, , –2, , –60, , 7–10, , 8.5, , 40, , 76, , –1, , –40, , 10–13, , 11.5 = A, , 16, , 92, , 0, , 0, , 13–16, , 14.5, , 4, , 96, , 1, , 4, , 16–19, , 17.5, , 4, , 100, , 2, , 8, , f i 100, , ( f i ui) 106, , Clearly, A 11.5, h 3, f i 100 and ( f i ui) 106., ( f i ui), (106), 1, 3 11.5 '3 #, 100, f i, 11.5 3.18 8.32., N, Here, N 100 &, 50., 2, Cumulative frequency just greater than 50 is 76 and the, corresponding class is 7–10. Thus, the median class is 7–10., N, l 7, h 3, f 40, cf c.f. of preceding class = 36,, 50., 2, ]Z], b_, N, ]], a cf kbbb, 2, b, Median, Me l ][]h #, `b, f, ]]], bbb, ], b, \, a, (50 36), 1 7 a3 # 14 k, 7 '3 #, 40, 40, Mean, x A )h #, , 7 21 7 1.05 8.05., 20
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 857, , Mode 3(median) 2(mean), 3 # 8.05 2 # 8.32 24.15 16.64 7.51., f, , EXERCISE 18D, , 1. Find the mean, mode and median of the following frequency, distribution:, Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Frequency, , 4, , 4, , 7, , 10, , 12, , 8, , 5, [CBSE 2010], , 2. Find the mean, median and mode of the following data:, Class, , 0–20, , 20–40, , 40–60, , Frequency, , 6, , 8, , 10, , 60–80 80–100 100–120 120–140, 12, , 6, , 5, , 3, [CBSE 2008], , 3. Find the mean, median and mode of the following data:, Class, , 0–50 50–100 100–150 150–200 200–250 250–300 300–350, , Frequency, , 2, , 3, , 5, , 6, , 5, , 3, , 1, [CBSE 2013], , 4. Find the mode, median and mean for the following data:, Marks obtained, , 25–35, , 35–45, , 45–55, , 55–65, , 65–75, , 75–85, , Number of, students, , 7, , 31, , 33, , 17, , 11, , 1, [CBSE 2009], , 5. A survey regarding the heights (in cm) of 50 girls of a class was, conducted and the following data was obtained:, Height (in cm), , 120–130 130–140 140–150 150–160 160–170, , Number of girls, , 2, , 8, , 12, , 20, , 8, , Find the mean, median and mode of the above data., , Total, 50, , [CBSE 2008], , 6. The following table gives the daily income of 50 workers of a factory:, Daily income (in `), , 100–120, , 120–140, , 140–160, , 160–180, , 180–200, , Number of workers, , 12, , 14, , 8, , 6, , 10, , Find the mean, mode and median of the above data., , [CBSE 2009]
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858, , Secondary School Mathematics for Class 10, , 7. The table below shows the daily expenditure on food of 30 households, in a locality:, Daily expenditure (in `), , Number of households, , 100–150, , 6, , 150–200, , 7, , 200–250, , 12, , 250–300, , 3, , 300–350, , 2, , Find the mean and median daily expenditure on food., , [CBSE 2009C], , ANSWERS (EXERCISE 18D), , 1. Mean = 38.2, median = 40, mode = 43.6, 2. Mean = 62.4, median = 61.67, mode = 60.21, 3. Mean = 169, median = 170.83, mode = 174.49, 4. Mean = 49.70, median = 48.64, mode = 46.52, 5. Mean = 149.8 cm, median = 152.5 cm, mode = 157.9 cm, , 6. Mean = ` 145.2, median = ` 137.5, mode = ` 122.1, 7. Mean = ` 205, median = ` 208.33, mode = ` 214.99, , CUMULATIVE FREQUENCY CURVE (OR OGIVE), An ogive is a freehand graph showing the curve of a cumulative frequency, distribution. The term ’ogive‘ is derived from the word ’ogee‘ meaning a, shape consisting of a concave arc flowing into a convex arc, so forming an, S-shaped curve with vertical ends., Let a grouped frequency distribution be given to us., FOR A ‘LESS THAN’ SERIES:, , On a graph paper, we mark the upper class limits along the x-axis and the, corresponding cumulative frequencies along the y-axis., (i) On joining these points successively by line segments, we get a, polygon, called cumulative frequency polygon., (ii) On joining these points successively by smooth curves, we get a, curve, known as cumulative frequency curve, or an ogive.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 859, FOR A ‘GREATER THAN’ SERIES, , On a graph paper, we mark the lower class limits along the x-axis and the, corresponding cumulative frequencies along the y-axis., (i) On joining these points successively by line segments, we get a, polygon, called cumulative frequency polygon., (ii) On joining these points successively by smooth curves, we get a, curve, known as cumulative frequency curve, or an ogive., HOW TO OBTAIN MEDIAN FROM CUMULATIVE FREQUENCY CURVE?, METHOD 1, , We proceed stepwise as follows:, , Draw either the less than type or the more than type cumulative, frequency curve for the given frequency distribution., N, Step 2. Locate, on the y-axis. Let it be P., 2, Step 3. From point P, draw a line PQ parallel to the x-axis cutting the, curve at a point Q., Step 1., , Step 4., , From Q, draw QM perpendicular to the x-axis cutting the, x-axis at M., , The point of intersection of this perpendicular with the x-axis, i.e., the, abscissa or x-coordinate of M gives the median of the data., METHOD 2, , We proceed stepwise as follows:, , Step 1., , Draw both ogives (i.e., of the ‘less than’ type and of the, ‘more than’ type) for the given frequency distribution on the, same axis., , Step 2., , The two ogives intersect each other at a point, say P., , Step 3., , From P, draw PQ perpendicular to the x-axis meeting the, x-axis at Q., , Clearly, the abscissa or x-coordinate of point Q gives the median of, the data., , SOLVED EXAMPLES, EXAMPLE 1, , Following is the age distribution of a group of students. Draw the, cumulative frequency curve of ‘less than’ type and hence obtain the, median value.
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860, , SOLUTION, , Secondary School Mathematics for Class 10, , Age (in years), , Frequency, , 4–5, , 36, , 5–6, , 42, , 6–7, , 52, , 7–8, , 60, , 8–9, , 68, , 9–10, , 84, , 10–11, , 96, , 11–12, , 82, , 12–13, , 66, , 13–14, , 48, , 14–15, , 50, , 15–16, , 16, , From the given table, we may prepare the (‘less than’ type), cumulative frequency table as shown below:, Age (in years), , cf, , Less than 5, , 36, , Less than 6, , 78, , Less than 7, , 130, , Less than 8, , 190, , Less than 9, , 258, , Less than 10, , 342, , Less than 11, , 438, , Less than 12, , 520, , Less than 13, , 586, , Less than 14, , 634, , Less than 15, , 684, , Less than 16, , 700, , On a graph paper, we take the scale.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 861, , , Scale: *Along the x-axis, 5 small div. 1., , Along the y-axis, 1 small div. 10., And, plot the points A(5, 36), B(6, 78), C(7, 130), D(8, 190),, E(9, 258), F(10, 342), G(11, 438), H(12, 520), I(13, 586), J(14, 634),, K(15, 684) and L(16, 700)., We join freehand these points successively a to get the, cumulative frequency curve, or an ogive., Here, N 700 &, , N, 350., 2, , Take a point P(0, 350) on the y-axis and draw PQ x-axis,, meeting the curve at Q. Draw QM = x-axis, intersecting the, x-axis at M whose coordinates are (10, 0)., Hence, median = 10 years., EXAMPLE 2, , For the following frequency distribution, draw a cumulative frequency, curve of ‘more than’ type and hence obtain the median value., Class interval, Frequency, , 0–10 10–20 20–30 30–40 40–50 50–60 60–70, 5, , 15, , 20, , 23, , 17, , 11, , 9
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862, SOLUTION, , Secondary School Mathematics for Class 10, , From the given table, we may prepare the ‘more than’ series as, shown below:, More than 60, , 9, , More than 50, , 20, , More than 40, , 37, , More than 30, , 60, , More than 20, , 80, , More than 10, , 95, , More than 5, , 100, , , Scale: )Along the x-axis, 10 small div. 5., Along the y-axis, 1 small div. 1., Now plot the points A(5, 100), B(10, 95), C(20, 80), D(30, 60),, E(40, 37), F(50, 20) and G(60, 9) on a graph paper., Join AB, BC, CD, DE, EF and FG freehand to get the required, curve, as shown below., N, Here, N 100 &, 50., 2, , From P(0, 50) draw PQ x-axis, meeting the curve at Q. Draw, QM = OX, meeting x-axis at M whose coordinates are (35, 0)., Hence, median = 35.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 863, EXAMPLE 3, , The following table gives production yield per hectare of wheat of, 100 farms of a village:, Production yield, (kg/ha), Number of farms, , 40–45 45–50 50–55 55–60 60–65 65–70, 4, , 6, , 16, , 20, , 30, , 24, , Change the distribution to a ‘more than’ type distribution and draw, its ogive., [CBSE 2009C, ’13], SOLUTION, , We may prepare the ‘more than’ series as shown below:, More than 65, , 24, , More than 60, , 54, , More than 55, , 74, , More than 50, , 90, , More than 45, , 96, , More than 40, , 100, , , Scale: )Along the x-axis, 1 small div. 1., Along the y-axis, 1 small div. 1., On a graph paper, we plot the points A(40, 100), B(45, 96),, C(50, 90), D(55, 74), E(60, 54) and F(65, 24)., Join AB, BC, CD, DE and EF freehand to get a ‘More Than Ogive’.
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864, EXAMPLE 4, , Secondary School Mathematics for Class 10, , During the medical check-up of 35 students of a class their weights, were recorded as follows:, Weight (in kg) 38–40 40–42 42–44 44–46 46–48 48–50 50–52, No. of students, , 3, , 2, , 4, , 5, , 14, , 4, , 3, , Draw a less than type and a more than type ogive from the given, data. Hence, obtain the median weight from the graph., SOLUTION, , [CBSE 2009], , (i) Less Than Series:, We may prepare the less than series as under., Weight (in kg), , Number of students, , Less than 40, , 3, , Less than 42, , 5, , Less than 44, , 9, , Less than 46, , 14, , Less than 48, , 28, , Less than 50, , 32, , Less than 52, , 35, , , Scale: )Along the x-axis, 5 small div. 1 kg., Along the y-axis, 10 small div. 5 students., Now, plot the points A(40, 3), B(42, 5), C(44, 9), D(46, 14),, E(48, 28), F(50, 32) and G(52, 35) on a graph paper., Join AB, BC, CD, DE, EF and FG freehand to get the curve, representing ’Less Than Series‘., (ii) More Than Series:, We may prepare the more than series as under., Weight (in kg), , Number of students, , More than 38, , 35, , More than 40, , 32, , More than 42, , 30, , More than 44, , 26
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 865, , More than 46, , 21, , More than 48, , 7, , More than 50, , 3, , On the same graph paper as above, we plot the points, P(38, 35), Q(40, 32), R(42, 30), S(44, 26), T(46, 21), U(48, 7), and V(50, 3)., Join PQ, QR, RS, ST, TU and UV freehand to get the curve, representing ‘More Than Series’., , The two curves intersect at the point L. Draw LM = OX., , , Score, , 250–300, , 300–350, , 350–400, , 400–450, , 450–500, , 500–550, , 550–600, , The table given below shows the frequency distribution of the scores, obtained by 200 candidates in a BCA entrance examination., 200–250, , EXAMPLE 5, , median weight = OM = 46.5 kg., , No. of candidates, , 30, , 15, , 45, , 20, , 25, , 40, , 10, , 15, , Draw cumulative frequency curves by using (i) ‘less than’ series and, (ii) ‘more than’ series., Hence, find the median.
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866, SOLUTION, , Secondary School Mathematics for Class 10, , (i) Less Than Series:, We may prepare the less than series as under., Score, , Number of candidates, , Less than 250, , 30, , Less than 300, , 45, , Less than 350, , 90, , Less than 400, , 110, , Less than 450, , 135, , Less than 500, , 175, , Less than 550, , 185, , Less than 600, , 200, , , Scale: )Along the x-axis, 1 small div. 5., Along the y-axis, 1 small div. 2., We plot the points A(250, 30), B(300, 45), C(350, 90),, D(400, 110), E(450, 135), F(500, 175), G(550, 185) and, H(600, 200) on a graph paper., Join AB, BC, CD, DE, EF, FG and GH freehand to get the, curve representing ‘Less Than Series’., (ii) More Than Series:, We may prepare the more than series as under., Score, , Number of candidates, , Less than 200, , 200, , More than 250, , 170, , More than 300, , 155, , More than 350, , 110, , More than 400, , 90, , More than 450, , 65, , More than 500, , 25, , More than 550, , 15, , Now, on the same graph paper as above, we plot the, points P(200, 200), Q(250, 170), R(300, 155), S(350, 110),, T(400, 90), U(450, 65), V(500, 25), W(550, 15)., Join PQ, QR, RS, ST, TU, UV and VW freehand to get the, curve representing ‘More Than Series’.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 867, , The two curves intersect at the point L. Draw LM = OX,, cutting the x-axis at M., , Clearly, M represents 375., Hence, median = 375., f, , EXERCISE 18E, , 1. Find the median of the following data by making a ’less than ogive‘., Marks, Number of, students, , 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100, 5, , 3, , 4, , 3, , 3, , 4, , 7, , 9, , 7, , 8, [CBSE 2014], , 2. The given distribution shows the number of wickets taken by the, bowlers in one-day international cricket matches:, Number of Less, Less, Less, Less, Less, Less, Less, Less, wickets than 15 than 30 than 45 than 60 than 75 than 90 than 105 than 120, Number of, bowlers, , 2, , 5, , 9, , 17, , 39, , 54, , 70, , 80, , Draw a ’less than type‘ ogive from the above data. Find the median., [CBSE 2014]
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868, , Secondary School Mathematics for Class 10, , 3. Draw a ’more than‘ ogive for the data given below which gives the, marks of 100 students., Marks, , 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80, , Number of, students, , 4, , 6, , 10, , 10, , 25, , 22, , 18, , 5, , [CBSE 2013], , 4. The heights of 50 girls of Class X of a school are recorded as follows:, Height (in cm) 135–140 140–145 145–150 150–155 155–160 160–165, Number of girls, , 5, , 8, , 9, , 12, , 14, , Draw a ’more than type‘ ogive for the above data., , 2, [CBSE 2014], , 5. The monthly consumption of electricity (in units) of some families of a, locality is given in the following frequency distribution:, Monthly, consumption 140–160 160–180 180–200 200–220 220–240 240–260 260–280, (in units), Number of, families, , 3, , 8, , 15, , 40, , 50, , 30, , 10, , Prepare a ‘more than type‘ ogive for the given frequency distribution., [CBSE 2014], , 6. The following table gives the production yield per hectare of wheat of, 100 farms of a village., Production, yield (kg/ha), , 50–55, , 55–60, , 60–65, , 65–70, , 70–75, , 75–80, , Number of, farms, , 2, , 8, , 12, , 24, , 38, , 16, , Change the distribution to a ’more than type‘ distribution and draw its, ogive. Using ogive, find the median of the given data., [CBSE 2013], 7. The table given below shows the weekly expenditures on food of some, households in a locality., Weekly expenditure (in `), , Number of households, , 100–200, , 5, , 200–300, , 6, , 300–400, , 11, , 400–500, , 13, , 500–600, , 5
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 869, , 600–700, , 4, , 700–800, , 3, , 800–900, , 2, , Draw a ‘less than type ogive’ and a ‘more than type ogive’ for this, distribution., 8. From the following frequency distribution, prepare the ’more than‘, ogive., Score, , Number of candidates, , 400–450, , 20, , 450–500, , 35, , 500–550, , 40, , 550–600, , 32, , 600–650, , 24, , 650–700, , 27, , 700–750, , 18, , 750–800, , 34, , Total, , 230, , Also, find the median., 9. The marks obtained by 100 students of a class in an examination are, given below:, Marks, , Number of students, , 0–5, , 2, , 5–10, , 5, , 10–15, , 6, , 15–20, , 8, , 20–25, , 10, , 25–30, , 25, , 30–35, , 20, , 35–40, , 18, , 40–45, , 4, , 45–50, , 2
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870, , Secondary School Mathematics for Class 10, , Draw cumulative frequency curves by using (i) ’less than‘ series and, (ii) ’more than‘ series., Hence, find the median., 10. From the following data, draw the two types of cumulative frequency, curves and determine the median., Height (in cm), , Frequency, , 140–144, , 3, , 144–148, , 9, , 148–152, , 24, , 152–156, , 31, , 156–160, , 42, , 160–164, , 64, , 164–168, , 75, , 168–172, , 82, , 172–176, , 86, , 176–180, , 34, , SUMMARY OF THE RESULTS AND FORMULAE, 1. MEAN OF THE GROUPED DATA, (i) DIRECT METHOD, , Mean ( x ) , , fi xi, 1, , where xi (lower limit + upper limit) of ith class interval, 2, fi, , and f i is its frequency., (ii), , f i (xi x ) 0., , (iii) ASSUMED-MEAN METHOD, , Mean ( x ) A , , each i., , fi di, , where A is the assumed mean and di (xi A) for, fi
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 871, (iv) STEP-DEVIATION METHOD, , Mean ( x ) A )h #, , ui , , (xi A), ·, h, , ( f i # ui), 3, where A assumed mean, h class size and, f i, , 2. (i) MEDIAN CLASS, , N, Let N f i . Then, the class whose cumulative frequency is just greater than a k, 2, is the median class., (ii) MEDIAN FOR GROUPED DATA, , Z], N _bb, ]], a, cf kbb, ], 2, b, where l lower limit of the median class;, Median (Me) l ]][h #, b`b, f, ]], bb, ]], b, \, a, h width of the median class; f frequency of the median class; N f i ;, cf cumulative frequency of the class preceding the median class., 3., , Mode = 3(median) – 2(mean)., , 4. (i) MODAL CLASS, , The class having maximum frequency is called the modal class., (ii), , (fk fk 1), 2, where xk lower limit of the modal, Mode (Mo) xk h · (, (2fk fk 1 fk 1, , class interval; fk frequency of the modal class; fk 1 frequency of the, class preceding the modal class; fk 1 frequency of the class succeeding the, modal class; h width of the class interval., 5. OGIVES (OR CUMULATIVE FREQUENCY CURVES), (i) FOR ‘LESS THAN’ SERIES, , On a graph paper, we mark the upper class limits along x-axis and the, corresponding cumulative frequencies along y-axis. Join these points, successively to obtain a curve, called ogive for less than series., (ii) FOR ‘GREATER THAN’ SERIES, , On the same graph paper as above, we mark the lower class limits, along x-axis and the corresponding cumulative frequencies along y-axis., Join these points successively to obtain a curve, called ogive for greater, than series., (iii), , The abscissa of the point of intersection of the two ogives gives us the median.
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872, , Secondary School Mathematics for Class 10, , f, , EXERCISE 18F, , Very-Short-Answer Questions, 1. Write the median class of the following distribution:, Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Frequency, , 4, , 4, , 8, , 10, , 12, , 8, , 4, [CBSE 2009], , 2. What is the lower limit of the modal class of the following frequency, distribution?, Age (in years), , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Number of, patients, , 16, , 13, , 6, , 11, , 27, , 18, [CBSE 2009], , 3. The monthly pocket money of 50 students of a class are given in the, following distribution:, Monthly pocket, money (in `), , 0–50, , Number of, students, , 2, , 50–100 100–150 150–200 200–250 250–300, 7, , 8, , 30, , 12, , 1, , Find the modal class and also give class mark of the modal class., [CBSE 2014], , 4. A data has 25 observations arranged in a descending order. Which, [CBSE 2014], observation represents the median?, 5. For a certain distribution, mode and median were found to be 1000 and, 1250 respectively. Find mean for this distribution using an empirical, [CBSE 2014], relation., 6. In a class test, 50 students obtained marks as follows:, Marks, obtained, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , Number of, students, , 4, , 6, , 25, , 10, , 5, , Find the modal class and the median class., 7. Find the class marks of classes 10–25 and 35–55., , [CBSE 2014], [CBSE 2008]
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 873, , 8. While calculating the mean of a given data by the assumed-mean, method, the following values were obtained:, A 25, f i di 110, f i 50., Find the mean., 9. The distributions X and Y with total number of observations 36 and 64,, and mean 4 and 3 respectively are combined. What is the mean of the, resulting distribution X + Y?, 10. In a frequency distribution table with 12 classes, the class-width is, 2.5 and the lowest class boundary is 8.1, then what is the upper class, boundary of the highest class?, 11. The observations 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 are arranged in, ascending order. What is the value of x if the median of the data is 63?, 12. The median of 19 observations is 30. Two more observations are, made and the values of these are 8 and 32. Find the median of the 21, observations taken together., HINT, , Since 8 is less than 30 and 32 is more than 30, so the value of median, (middle value) remains unchanged., , x x x, x, 13. If the median of , , , x and , where x 0, is 8, find the value of x., 5 4 2, 3, HINT, , x x x x, Arranging the observations in ascending order, we have , , , , x., 5 4 3 2, x, Median 8., 3, , 14. What is the cumulative frequency of the modal class of the following, distribution?, Class, , 3–6, , 6–9, , 9–12, , 12–15, , 15–18, , 18–21, , 21–24, , Frequency, , 7, , 13, , 10, , 23, , 4, , 21, , 16, , Short-Answer Questions, 15. Find the mode of the given data:, Class interval, , 0–20, , 20–40, , 40–60, , Frequency, , 15, , 6, , 18, , 60–80, 10, [CBSE 2015], , 16. The following are the ages of 300 patients getting medical treatment in, a hospital on a particular day:, Age (in years), , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Number of, patients, , 60, , 42, , 55, , 70, , 53, , 20, , Form a ’less than type‘ cumulative frequency distribution., , [CBSE 2013]
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874, , Secondary School Mathematics for Class 10, , 17. In the following data, find the values of p and q. Also, find the median, class and modal class., Class, , Frequency (f), , Cumulative frequency (cf), , 100–200, , 11, , 11, , 200–300, , 12, , p, , 300–400, , 10, , 33, , 400–500, , q, , 46, , 500–600, , 20, , 66, , 600–700, , 14, , 80, [CBSE 2013], , 18. The following frequency distribution gives the monthly consumption, of electricity of 64 consumers of a locality., Monthly, consumption, (in units), , 65–85, , Number of, consumers, , 4, , 85–105 105–125 125–145 145–165 165–185, 5, , 13, , 20, , 14, , 8, , Form a ’more than type‘ cumulative frequency distribution., 19. The following table gives the life-time (in days) of 100 electric bulbs of, a certain brand., Life-time Less than Less than Less than Less than Less than Less than, (in days), 50, 100, 150, 200, 250, 300, Number of, bulbs, , 7, , 21, , 52, , 79, , 91, , 100, , From this table, construct the frequency distribution table., 20. The following table gives the frequency distribution of the percentage, of marks obtained by 2300 students in a competitive examination., Marks obtained, 11–20, (in per cent), Number of, students, , 141, , 21–30, , 31–40, , 41–50, , 51–60, , 61–70, , 71–80, , 221, , 439, , 529, , 495, , 322, , 153, , (a) Convert the given frequency distribution into the continuous form., (b) Find the median class and write its class mark., (c) Find the modal class and write its cumulative frequency.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 875, , 21. If the mean of the following distribution is 27, find the value of p., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 8, , p, , 12, , 13, , 10, , 22. Calculate the missing frequency from the following distribution, it, being given that the median of the distribution is 24., Age (in years), , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Number of, persons, , 5, , 25, , ?, , 18, , 7, , ANSWERS (EXERCISE 18F), , 1. 30–40, , 2. 40, , 3. 150–200, 175 4. 13th, , 5. 1375, , 6. Modal class 40–60, median class 40–60, 9. 3.36, , 10. 38.1, , 11. 62, , 12. 30, , 13. 24, , 7. 17.5, 45, 14. 53, , 8. 27.2, , 15. 52, , 17. p 23, q 13; median class 400–500, modal class 500–600, 20. (b) 40.5–50.5, 45.5, , (c) 40.5–50.5, 1330, , 21. p 7, , 22. 25, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. Which of the following is not a measure of central tendency? [CBSE 2013], (a) Mean, , (b) Mode, , (c) Median, , (d) Range, , 2. Which of the following cannot be determined graphically?, (a) Mean, , (b) Median, , (c) Mode, , (d) None of these, , 3. Which of the following measures of central tendency is influenced by, extreme values?, (a) Mean, , (b) Median, , (c) Mode, , (d) None of these, , 4. The mode of a frequency distribution is obtained graphically from, (a) a frequency curve, , (b) a frequency polygon, , (c) a histogram, , (d) an ogive, , 5. The median of a frequency distribution is found graphically with the, help of, (a) a histogram, , (b) a frequency curve, , (c) a frequency polygon, , (d) ogives
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876, , Secondary School Mathematics for Class 10, , 6. The cumulative frequency table is useful in determining the, (a) mean, , (b) median, , (c) mode, , (d) all of these, , 7. The abscissa of the point of intersection of the Less Than Type and of, the More Than Type cumulative frequency curves of a grouped data, gives its, (a) mean, , (b) median, , (c) mode, , (d) none of these, , 8. If xi’s are the midpoints of the class intervals of a grouped data, f i’s are, the corresponding frequencies and x is the mean then f i (xi x ) ?, (a) 1, , (b) 0, , (c) –1, , (d) 2, , 9. For finding the mean by using the formula, x A h e, ui ?, (a), , (A xi), h, , (b), , (xi A), h, , (c), , (A xi), h, , f i ui, o, we have, f i, , (d) h(xi A), , f i di, 3 for finding the mean of the grouped data,, f i, the di’s are the deviations from A of, , 10. In the formula, x )A , , (a) lower limits of the classes, , (b) upper limits of the classes, , (c) midpoints of the classes, , (d) none of these, , 11. While computing the mean of the grouped data, we assume that the, frequencies are, (a) evenly distributed over the classes, (b) centred at the class marks of the classes, (c) centred at the lower limits of the classes, (d) centred at the upper limits of the classes, 12. The relation between mean, mode and median is, (a) mode = (3 # mean) – (2 # median), (b) mode = (3 # median) – (2 # mean), (c) median = (3 # mean) – (2 # mode), (d) mean = (3 # median) – (2 # mode), 13. If the ’less than type‘ ogive and ’more than type‘ ogive intersect each, other at (20.5, 15.5) then the median of the given data is, [CBSE 2013], (a) 5.5, , (b) 15.5, , (c) 20.5, , (d) 36.0
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 877, , 14. Consider the frequency distribution of the heights of 60 students of, a class:, Height (in cm), , No. of Students, , Cumulative Frequency, , 150–155, , 16, , 16, , 155–160, , 12, , 28, , 160–165, , 9, , 37, , 165–170, , 7, , 44, , 170–175, , 10, , 54, , 175–180, , 6, , 60, , The sum of the lower limit of the modal class and the upper limit of the, median class is, (a) 310, , (b) 315, , (c) 320, , (d) 330, , 15. Consider the following frequency distribution:, Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Frequency, , 3, , 9, , 15, , 30, , 18, , 5, , The modal class is, (a) 10–20, , (b) 20–30, , 16. Mode = ?, , (c) 30–40, , (d) 50–60, , (fk 1 fk), 3, (a) xk h · ), (2fk fk 1 fk 1), , (fk fk 1), 3, (b) xk h · ), (2fk fk 1 fk 1), , (fk fk 1), 3, (c) xk h · ), (fk 2fk 1 fk 1), , (fk fk 1), 3, (d) xk h · ), (fk fk 1 2fk 1), , 17. Median = ?, ]Z], ]Z], b_, N, N b_, ]], ]], a cf kbbb, acf kbbb, 2, 2 b, b, (b) l ][]h #, (a) l ][]h #, `b, `b, f, f, ]], ]], bb, bb, ]], ]], bb, bb, \Z, \, a_, a, ]], N bb, ]], cf kbb, a, 2, b, (c) l ]][h #, (d) none of these, b`b, f, ]], bb, ]], b, \, a, 18. If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be, (a) 7.2, , (b) 8.2, , (c) 9.2, , (d) 10.2
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878, , Secondary School Mathematics for Class 10, , 19. Look at the frequency distribution table given below:, Class interval, , 35–45, , 45–55, , 55–65, , 65–75, , Frequency, , 8, , 12, , 20, , 10, , The median of the above distribution is, (a) 56.5, , (b) 57.5, , (c) 58.5, , (d) 59, , 20. Consider the following table:, Class interval, , 10–14, , 14–18, , 18–22, , 22–26, , 26–30, , Frequency, , 5, , 11, , 16, , 25, , 19, , The mode of the above data is, (a) 23.5, , (b) 24, , (c) 24.4, , (d) 25, , 21. The mean and mode of a frequency distribution are 28 and 16, respectively. The median is, (a) 22, , (b) 23.5, , (c) 24, , (d) 24.5, , 22. The median and mode of a frequency distribution are 26 and 29, respectively. Then, the mean is, (a) 27.5, , (b) 24.5, , (c) 28.4, , (d) 25.8, , 23. For a symmetrical frequency distribution, we have, (a) mean < mode < median, , (b) mean > mode > median, 1, (c) mean = mode = median, (d) mode (mean + median), 2, 24. Look at the cumulative frequency distribution table given below:, Monthly income, , Number of families, , More than ` 10000, More than ` 14000, More than ` 18000, More than ` 20000, More than ` 25000, More than ` 30000, , 100, 85, 69, 50, 37, 15, , Number of families having income range 20000 to 25000 is, (a) 19, (b) 16, (c) 13, (d) 22, 25. The median of first 8 prime numbers is, (a) 7, (b) 9, (c) 11, , (d) 13, , 26. The mean of 20 numbers is zero. Of them, at the most, how many may, be greater than zero?, (a) 0, (b) 1, (c) 10, (d) 19
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 879, , 27. If the median of the data 4, 7, x 1, x 3, 16, 25, written in ascending, order, is 13 then x is equal to, (a) 13, , (b) 14, , (c) 15, , (d) 16, , 28. The mean of 2, 7, 6 and x is 15 and the mean of 18, 1, 6, x and y is 10., What is the value of y?, (a) 5, , (b) 10, , (c) 20, , (d) 30, , Matching of columns, , 29. Match the following columns:, Column I, , Column II, , (a) The most frequent value in a data is, known as …… ., , (p) standard deviation, , (b) Which of the following cannot, be determined graphically out of, mean, mode and median?, , (q) median, , (c) An ogive is used to determine …… ., , (r) mean, , (d) Out of mean, mode, median and, standard deviation, which is not a, measure of central tendency?, , (s) mode, , The correct answer is:, (a) –……,, , (b) –……,, , (c) –……,, , (d) –…… ., , Assertion-and-Reason Type, , Each question consists of two statements, namely, Assertion (A) and, Reason (R). For selecting the correct answer, use the following code:, (a) Both Assertion (A) and Reason (R) are true and Reason (R) is a, correct explanation of Assertion (A)., (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a, correct explanation of Assertion (A)., (c) Assertion (A) is true and Reason (R) is false., (d) Assertion (A) is false and Reason (R) is true., 30., , Assertion (A), , Reason (R), , If the median and mode of a Mean, median and mode of a, frequency distribution are 150 and frequency distribution are related as:, 154 respectively, then its mean is 148., mode = 3 median – 2 mean., , The correct answer is: (a)/(b)/(c)/(d).
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880, , Secondary School Mathematics for Class 10, , Assertion (A), , 31., , Reason (R), , Consider the following frequency The value of the variable which, distribution:, occurs most often is the mode., Class, interval, , 3– 6– 9– 12– 15– 18–, 6 9 12 15 18 21, , Frequency 2, , 5, , 21 23 10 12, , The mode of the above data is 12.4., , The correct answer is: (a)/(b)/(c)/(d)., ANSWERS (MCQ), , 1., 10., 19., 28., , 5. (d) 6. (b) 7. (b) 8. (b) 9. (b), (d) 2. (a) 3. (a) 4. (c), 11., 12., 13., 14., (c), (b), (b), (c), (b) 15. (c) 16. (b) 17. (a) 18. (c), (b) 20. (c) 21. (c) 22. (b) 23. (c) 24. (c) 25. (b) 26. (d) 27. (c), 30. (a) 31. (b), (c) 29. (a)–(s), (b)–(r), (c)–(q), (d)–(p), HINTS TO SOME SELECTED QUESTIONS, , 1. Range is not a measure of central tendency., 2. Mean cannot be determined graphically., 3. Only mean is affected by extreme values, while both median and mode remain, unaffected., 4. Mode of a frequency distribution can be obtained graphically from a histogram., 6. The cumulative frequency table is useful in determining the median., 8. Clearly, we have f i (xi x ) 0., 9. We have ui , , (xi A), ·, h, , 10. di’s are deviations from A of midpoints of the classes., 11. In computing the mean of the grouped data, we assume that the frequencies are, centred at the class marks of the classes., 13. Clearly, the abscissa of the point of intersection of both the ogives gives the median., 14. The class having maximum frequency is the modal class., So, the modal class is 150–155. Its lower limit is 150., N, 30 and the cumulative frequency just more than 30 is 37. Its class, 2, is 160–165, whose upper limit is 165., , Also, N 60 &, , Required sum = (150 + 165) = 315.
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 881, 15. The class 30–40 has maximum frequency. So, the modal class is 30–40., 18. Mode = (3 median) – (2 mean) (3 # 9 2 # 8.9) (27 17.8) 9.2., 19. We have:, , Class interval, , 35–45, , 45–55, , 55–65, , 65–75, , Frequency, , 8, , 12, , 20, , 10, , Cumulative, frequency, , 8, , 20, , 40, , 50, , N, 25, which lies in class interval 55–65., 2, b_, ]Z], N, ]], a cf kbbb, 2, b 55 (65 55) #(25 20) 57.5., ], Median l []h #, `b, 20, f, bb, ]], bb, ]], a, \, 20. The maximum frequency is 25 and the modal class is 22–26., Here, N 50 &, , , , xk 22, f k 25, f k 1 16, f k 1 19 and h 4., , , , (f k f k 1), 3, mode xk h · ), (2f k f k 1 f k 1), (22 4 #, , (25 16), 2 a22 4 # 9 k a22 12 k (22 2.4) 24.4., 5, 15, (50 16 19), , 21. 3 # median = (mode + 2 mean) (16 2 # 28) 72, , , median , , 72 , 24., 3, , 22. Mode = (3 median) – (2 mean), , , 2 mean = (3 median) – (mode) (3 # 26) 29 49, , , , mean , , 49 , 24.5., 2, , 23. We must have, mean = mode = median., 25. First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19., Required median = mean of fourth and fifth observations, , 7 11 9., 2, 26. Mean of 20 numbers = 0., , , sum of 20 numbers 0 # 20 0., , It is possible that 19 of these numbers may be positive and if their sum is a, the 20th, number is ( a) ., 27. Clearly, median of 6 observations, = mean of 3rd and 4th observations
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882, , Secondary School Mathematics for Class 10, , (x 1) (x 3) 2x 4, , x 2., 2, 2, x 2 13 or x 13 2 15., 276x , k 5 or 15 x 20 or x 5., 28. We have a, 4, 18 1 6 x y, m 10 or 25 5 y 50 or y 20., Also, c, 5, , , 30. Reason (R) is clearly true., Using the relation given in Reason (R), we have, 2 mean = (3 median) – (mode) (3 #150) (154) 450 154 296., , , mean = 148, which is true., , Thus, Assertion (A) and Reason (R) are both true and Reason (R) is the correct, explanation of Assertion (A)., Hence, the correct answer is (a)., 31. Reason (R) is clearly true., The maximum frequency is 23 and the modal class is 12–15., , , xk 12, f k 23, f k 1 21, f k 1 10 and h 3., , , , mode (12 3 #, , , , Assertion (A) is true., , (23 21), 2 a12 3 # 2 k 12.4., 15, (2 # 23 21 10), , But, Reason (R) is not a correct explanation of Assertion (A)., Hence, the correct answer is (b)., , TEST YOURSELF, MCQ, 1. Which one of the following measures is determined only after the, construction of cumulative frequency distribution?, (a) Mean, , (b) Median, , (c) Mode, , (d) None of these, , 2. If the mean of a data is 27 and its median is 33 then the mode is, (a) 30, , (b) 43, , (c) 45, , (d) 47, , 3. Consider the following distribution:, Class, , 0–5, , 5–10, , 10–15, , 15–20, , 20–25, , Frequency, , 10, , 15, , 12, , 20, , 9, , The sum of the lower limits of the median class and the modal class is, (a) 15, , (b) 25, , (c) 30, , (d) 35
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 883, , 4. Consider the following frequency distribution:, Class, , 0–5, , 6–11, , 12–17, , 18–23, , 24–29, , Frequency, , 13, , 10, , 15, , 8, , 11, , The upper limit of the median class is, (a) 16.5, , (b) 18.5, , (c) 18, , (d) 17.5, , Very-Short-Answer Questions, 5. If the mean and mode of a frequency distribution be 53.4 and 55.2, respectively, find the median., 6. In the table given below, the times taken by 120 athletes to run a 100-mhurdle race are given., Class, , 13.8–14, , Frequency, , 2, , 14–14.2 14.2–14.4 14.4–14.6 14.6–14.8 14.8–15, 4, , 15, , 54, , 25, , 20, , Find the number of athletes who completed the race in less than, 14.6 seconds., 7. Consider the following frequency distribution:, Class, , 0–5, , 6–11, , 12–17, , 18–23, , 24–29, , Frequency, , 13, , 10, , 15, , 8, , 11, , Find the upper limit of the median class., 8. The annual profits earned by 30 shops of a shopping complex in a, locality are recorded in the table shown below:, Profit (in lakhs `), , Number of shops, , More than or equal to 5, , 30, , More than or equal to 10, , 28, , More than or equal to 15, , 16, , More than or equal to 20, , 14, , More than or equal to 25, , 10, , More than or equal to 30, , 7, , More than or equal to 35, , 3, , If we draw the frequency distribution table for the above data, find the, frequency corresponding to the class 20–25.
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884, , Secondary School Mathematics for Class 10, , Short-Answers Questions, 9. Find the mean of the following frequency distribution:, Class, , 1–3, , 3–5, , 5–7, , 7–9, , Frequency, , 9, , 22, , 27, , 18, , 10. The maximum bowling speeds (in km/hr) of 33 players at a cricket, coaching centre are given below:, Speed in km/hr, , 85–100, , 100–115, , 115–130, , 130–145, , No. of players, , 10, , 4, , 7, , 9, , Calculate the median bowling speed., 11. The arithmetic mean of the following frequency distribution is 50., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Frequency, , 16, , p, , 30, , 32, , 14, , Find the value of p., 12. Find the median of the following frequency distribution:, Marks, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Number of students, , 6, , 16, , 30, , 9, , 4, , 13. Following is the distribution of marks of 70 students in a periodical test:, Less than Less than Less than Less than Less than, 10, 20, 30, 40, 50, , Marks, Number of, students, , 3, , 11, , 28, , 48, , 70, , Draw a cumulative frequency curve for the above data., 14. Find the median of the following data., Class interval, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Total, , Frequency, , 8, , 16, , 36, , 34, , 6, , 100, [CBSE 2014], , 15. For the following distribution draw a ’less than type‘ ogive and from, the curve find the median., Marks, obtained, , Less, than, 20, , Less, than, 30, , Less, than, 40, , Less, than, 50, , Less, than, 60, , Less, than, 70, , Less, than, 80, , Less, than, 90, , Less, than, 100, , Number of, students, , 2, , 7, , 17, , 40, , 60, , 82, , 85, , 90, , 100, , [CBSE 2014]
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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive 885, , 16. The median value for the following frequency distribution is 35 and the, sum of all the frequencies is 170. Using the formula for median, find the, missing frequencies., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Frequency, , 10, , 20, , ?, , 40, , ?, , 25, , 15, , 17. Find the missing frequencies f1 and f 2 in the table given below, it being, given that the mean of the given frequency distribution is 50., Class, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , Total, , Frequency, , 17, , f1, , 32, , f2, , 19, , 120, , 18. Find the mean of the following frequency distribution using stepdeviation method:, Class, , 84–90, , 90–96, , 96–102, , Frequency, , 15, , 22, , 20, , 102–108 108–114 114–120, 18, , 20, , 25, , Long-Answer Questions, 19. Find the mean, median and mode of the following data:, Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Frequency, , 6, , 8, , 10, , 15, , 5, , 4, , 2, , 20. Draw ’less than ogive‘ and ’more than ogive‘ on a single graph paper, and hence find the median of the following data:, Class interval, , 5–10, , 10–15, , 15–20, , 20–25, , 25–30, , 30–35, , 35–40, , Frequency, , 2, , 12, , 2, , 4, , 3, , 4, , 3, [CBSE 2014], , 21. The production yield per hectare of wheat of some farms of a village, are given in the following table:, Production, yield, 40–45 45–50 50–55 55–60 60–65 65–70 70–75 75–80 80–85, (in kg/ha), Number of, farms, , 1, , 9, , 15, , 18, , 40, , 26, , 16, , 14, , 10, , Draw a less than type ogive and a more than type ogive for this data., [CBSE 2014]
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886, , Secondary School Mathematics for Class 10, , 22. The following table gives the marks obtained by 50 students in a, class test:, Marks, , 11–15 16–20 21–25 26–30 31–35 36–40 41–45 46–50, , Number of, students, , 2, , 3, , 6, , 7, , 14, , 12, , 4, , 2, , Calculate the mean, median and mode for the above data., ANSWERS (TEST YOURSELF), , 1. (b), , 2. (c), , 3. (b), , 4. (d), , 6. 75, , 7. 17.5, , 8. 4, , 9. 5.42, , 10. 117.1 km/hr 11. 28, , 12. 24, , 14. 27.22, , 17. f 1 28, f 2 24, , 18. 102.75, , 19. Mean = 30, median = 30.67, mode = 32.01, 22. Mean = 32, median = 33, mode = 35, , , , 5. 54, 16. 35, 25
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Probability, , INTRODUCTION, , In class IX, we have studied the concept of empirical probability. Since, empirical probability is based on experiments, we also call it experimental, probability., Suppose we toss a coin 500 times and get a head, say, 240 times and, tail 260 times. Then, we would say that in a single throw of a coin, the, 240, 12, , i.e.,, probability of getting a head is, ⋅, 500, 25, Again, suppose we toss a coin 1000 times and get a head, say, 530 times, and tail 470 times. Then, we would say that in a single throw of a coin, the, 530, 53, probability of getting a head is, , i.e.,, ⋅, 1000, 100, Thus, in various experiments, we would get different probabilities for, the same event., However, theoretical probability overcomes the above problem. In this, chapter, by probability, we shall mean theoretical probability., PROBABILITY, , Probability is a concept which numerically measures the degree of certainty of the, occurrence of events., Before defining probability, we shall define certain concepts used, therein., EXPERIMENT An operation which can produce some well-defined outcomes is called, an experiment., RANDOM EXPERIMENT An experiment in which all possible outcomes are known,, and the exact outcome cannot be predicted in advance, is called a random, experiment., By a trial, we mean ‘performing a random experiment’., Examples (i) Tossing a fair coin, (ii) Rolling an unbiased die, (iii) Drawing a card from a pack of well-shuffled cards, (iv) Picking up a ball from a bag of balls of different colours, These are all examples of a random experiment., 887
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888, , Secondary School Mathematics for Class 10, , SOME DETAILS ABOUT THESE EXPERIMENTS, , I. Tossing a coin When we throw a coin, either a head (H) or a tail (T), appears on the upper face., II. Throwing a die A die is a solid cube, having 6 faces, marked 1, 2, 3,, 4, 5 and 6, or having 1, 2, 3, 4, 5 and 6 dots., In throwing a die, the outcome is the number or number of dots, appearing on the uppermost face., The plural of die is dice., III. Drawing a card from a well-shuffled deck of 52 cards., A deck of playing cards has in all 52 cards., (i) It has 13 cards each of four suits, namely, spades, clubs, hearts and diamonds., , (a) Cards of spades and clubs are black cards., (b) Cards of hearts and diamonds are red cards., , (ii) Kings, queens and jacks (or knaves) are known as face cards., Thus, there are in all 12 face cards., , LOOKING AT ALL POSSIBLE OUTCOMES IN VARIOUS EXPERIMENTS, , I. When we toss a coin, we get either a head (H) or a tail (T)., Thus, all possible outcomes are H, T., II. Suppose two coins are tossed simultaneously., Then, all possible outcomes are HH, HT, TH, TT., REMARKS, , HH means head on first coin and head on second coin., HT means head on first coin and tail on second coin, etc., , III. On rolling a die, the number on the upper face is the outcome., Thus, all possible outcomes are 1, 2, 3, 4, 5, 6.
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Probability, , 889, , IV. In drawing a card from a well-shuffled deck of 52 cards, total, number of possible outcomes is 52., EVENT, , The collection of all or some of the possible outcomes is called an event., , Examples, , (i) In throwing a coin, H is the event of getting a head., (ii) Suppose we throw two coins simultaneously and let, E be the event of getting at least one head. Then, E, contains HT, TH, HH., , EQUALLY LIKELY EVENTS A given number of events are said to be equally likely if, none of them is expected to occur in preference to the others., , For example, if we roll an unbiased die, each number is equally likely to, occur. If, however, a die is so formed that a particular face occurs most, often then the die is biased. In this case, the outcomes are not equally likely, to happen., PROBABILITY OF OCCURRENCE OF AN EVENT, , Probability of occurrence of an event E, denoted by P(E) is defined as:, P(E) =, , number of outcomes favourable to E, ⋅, total number of possible outcomes, , SURE EVENT, , It is evident that in a single toss of die, we will always get a number less, than 7., So, getting a number less than 7 is a sure event., 6, P(getting a number less than 7) = = 1., 6, Thus, the probability of a sure event is 1., IMPOSSIBLE EVENT, , In a single toss of a die, what is the probability of getting a number 8?, We know that in tossing a coin, 8 will never come up., So, getting 8 is an impossible event., P(getting 8 in a single throw of a die) =, , 0, = 0., 6, , Thus, the probability of an impossible event is zero.
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890, , Secondary School Mathematics for Class 10, , COMPLEMENTARY EVENT, , Let E be an event and (not E) be an event which occurs only when E does, not occur. We denote (not E) by E′, or E, called complement of event E., The event (not E) is called the complementary event of E., Clearly, P(E) + P(not E) = 1., ∴, , P(E) = 1 − P(not E)., , SUMMARY, , (i) For some event E, we have 0 ≤ P(E) ≤ 1., (ii) P(E) = 0 , when E is an impossible event., (iii) P(E) = 1, when E is a sure event., (iv) P(not E) = 1 − P(E). Thus, P(E )=1 − P(E)., , SOLVED, EXAMPLE 1, SOLUTION, , EXAMPLES, , A coin is tossed once. What is the probability of getting a head?, When a coin is tossed once, all possible outcomes are H and T., Total number of possible outcomes = 2., The favourable outcome is H., Number of favourable outcomes = 1., ∴, , EXAMPLE 2, SOLUTION, , P(getting a head), number of favourable outcomes, 1, = P ( H) =, = ⋅, total number of possible outcomes 2, , A die is thrown once. What is the probability of getting a prime, number?, In a single throw of a die, all possible outcomes are, 1, 2, 3, 4, 5, 6., Total number of possible outcomes = 6., Let E be the event of getting a prime number., Then, the favourable outcomes are 2, 3, 5., Number of favourable outcomes = 3., 3 1, = ⋅, 6 2, A die is thrown once. What is the probability that it shows (i) a ’3’,, (ii) a ‘5’, (iii) an odd number, (iv) a number greater than 4?, When a die is thrown, all possible outcomes are 1, 2, 3, 4, 5, 6., Total number of possible outcomes = 6., , ∴, EXAMPLE 3, SOLUTION, , P(getting a prime number) = P(E) =
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Probability, , 891, , (i) Let E1 be the event of getting a 3., Then, the number of favourable outcomes = 1., 1, P(getting a 3) = P(E1 ) = ⋅, ∴, 6, (ii) Let E2 be the event of getting a 5., Then, the number of favourable outcomes = 1., 1, P(getting a 5) = P(E2 ) = ⋅, ∴, 6, (iii) Let E3 be the event of getting an odd number., Then, the favourable outcomes are 1, 3, 5., Number of favourable outcomes = 3., 3 1, P(getting an odd number) = P(E3 ) = = ⋅, ∴, 6 2, , EXAMPLE 4, SOLUTION, , (iv) Let E4 be the event of getting a number greater than 4., Then, the favourable outcomes are 5, 6., Number of favourable outcomes = 2., 2 1, ∴ P(getting a number greater than 4) = P(E4 ) = = ⋅, 6 3, A die is thrown once. Find the probability of getting (i) an even prime, number, (ii) a multiple of 3., [CBSE 2012], When a die is thrown, all possible outcomes are 1, 2, 3, 4, 5, 6., Total number of possible outcomes = 6., (i) Let E1 be the event of getting an even prime number., Then, the favourable outcome is 2 only., Number of favourable outcomes = 1 ., 1, P(getting an even prime number) = P(E1 ) = ⋅, ∴, 6, , EXAMPLE 5, SOLUTION, , (ii) Let E2 be the event of getting a multiple of 3., Then, the favourable outcomes are 3 and 6., Number of favourable outcomes = 2., 2 1, P(getting a multiple of 3) = P(E2 ) = = ⋅, ∴, 6 3, Two coins are tossed simultaneously. What is the probability of, getting at least one head?, [CBSE 2014], When two coins are tossed simultaneously, all possible, outcomes are HH, HT, TH, TT., Total number of possible outcomes = 4., Let E be the event of getting at least one head.
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892, , EXAMPLE 6, , SOLUTION, , Secondary School Mathematics for Class 10, , Then, E is the event of getting 1 head or 2 heads., So, the favourable outcomes are HT, TH, HH., Number of favourable outcomes = 3., 3, ∴ P(getting at least one head) = P(E) = ⋅, 4, Three unbiased coins are tossed simultaneously. Find the probability, of getting (i) exactly 2 heads, (ii) at least 2 heads, (iii) at most, 2 heads., [CBSE 2015], When 3 coins are tossed simultaneously, all possible outcomes, are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT., Total number of possible outcomes = 8., , EXAMPLE 7, , SOLUTION, , (i) Let E1 be the event of getting exactly 2 heads., Then, the favourable outcomes are HHT, HTH, THH., Number of favourable outcomes = 3., 3, P( getting exactly 2 heads) = P(E1 ) = ⋅, ∴, 8, (ii) Let E2 be the event of getting at least 2 heads., Then, E2 is the event of getting 2 or 3 heads., So, the favourable outcomes are, HHT, HTH, THH, HHH., Number of favourable outcomes = 4., 4 1, P(getting at least 2 heads) = P(E2 ) = = ⋅, ∴, 8 2, (iii) Let E3 be the event of getting at most 2 heads., Then, E3 is the event of getting 0 or 1 head or 2 heads., So, the favourable outcomes are, TTT, HTT, THT, TTH, HHT, HTH, THH., Number of favourable outcomes = 7., 7, P(getting at most 2 heads) = P(E3 ) = ⋅, ∴, 8, Cards numbered 11 to 60 are kept in a box. If a card is drawn at, random from the box, find the probability that the number on the, drawn card is (i) an odd number, (ii) a perfect square number,, (iii) divisible by 5, (iv) a prime number less than 20., [CBSE 2014], All possible outcomes are 11, 12, 13, ... , 60., Number of all possible outcomes = ( 60 − 10) = 50., (i) Let E1 be the event that the number on the drawn card is, an odd number.
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Probability, , 893, , Then, the favourable outcomes are 11, 13, 15, ..., 59., Clearly, these numbers form an AP with a = 11 and d = 2., Let the number of these numbers be n. Then,, Tn = 59 ⇒ 11 + (n − 1) × 2 = 59 ⇒ (n − 1) × 2 = 48, ⇒ (n − 1) = 24 ⇒ n = 25., So, the number of favourable outcomes = 25., 25 1, P(getting an odd number) = P(E1 ) =, = ⋅, ∴, 50 2, (ii) Let E2 be the event that the number on the drawn card is a, perfect square number., Then, the favourable outcomes are 16, 25, 36, 49., So, the numbr of favourable outcomes = 4., ∴ P(getting a perfect square number) = P(E2 ) =, , 4, 2, =, ⋅, 50 25, , (iii) Let E3 be the event that the number on the drawn card is, divisible by 5., Then, the favourable outcomes are 15, 20, 25, ... , 60., Clearly, these numbers form an AP with a = 15 and d = 5., Let the number of these terms be m. Then,, Tm = 60 ⇒ 15 + (m − 1) × 5 = 60 ⇒ (m − 1) × 5 = 45, ⇒ m − 1 = 9 ⇒ m = 10., So, the number of favourable outcomes = 10., 10 1, ∴ P(getting a number divisible by 5) = P(E3 ) =, = ⋅, 50 5, (iv) Let E4 be the event that the number on the drawn card is a, prime number less than 20., Then, the favourable outcomes are 11, 13, 17, 19., So, the number of favourable outcomes = 4., 4, 2, =, ⋅, 50 25, A box contains 100 red balls, 200 yellow balls and 50 blue balls. If a, ball is drawn at random from the box, then find the probability that it, will be (i) a blue ball, (ii) not a yellow ball, (iii) neither yellow nor a, blue ball., [CBSE 2012], ∴ P(getting a prime number less than 20) = P(E4 ) =, , EXAMPLE 8, , SOLUTION, , Total number of all possible outcomes = total number of balls, = 100 + 200 + 50 = 350., (i) Number of blue balls = 50.
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894, , Secondary School Mathematics for Class 10, , ∴, , P(getting a blue ball) =, , 50, 1, = ⋅, 350 7, , (ii) Number of balls which are not yellow = 100 + 50 = 150., 150 3, P(getting a ball which is not yellow) =, ∴, = ⋅, 350 7, , EXAMPLE 9, , SOLUTION, , (iii) Number of balls which are neither yellow nor blue = 100., ∴ P(getting a ball which is neither yellow nor blue), 100 2, =, = ⋅, 350 7, A bag contains 5 red balls and some blue balls. If the probability of, drawing a blue ball from the bag is thrice that of a red ball, find the, number of blue balls in the bag., [CBSE 2007], Let the number of blue balls in the bag be x., Then, total number of balls = (5 + x)., Given, P(a blue ball) = 3 × P(a red ball), x, 5, ∴, = 3×, ⇒ x = 15., (5 + x), (5 + x), , EXAMPLE 10, , SOLUTION, , Hence, the number of blue balls in the bag is 15., A bag contains white, black and red balls only. A ball is drawn at, random from the bag. If the probability of getting a white ball is 3 10, and that of a black ball is 2 5 then find the probability of getting a red, ball. If the bag contains 20 black balls then find the total number of, balls in the bag., [CBSE 2015], Let E be the event of getting a red ball. Then,, P(getting a white ball) + P(getting a black ball) + P(E) = 1, 3 2, 7, ⇒, + + P(E) = 1 ⇒, + P(E) = 1, 10 5, 10, 7, 3, ⇒ P(E) = 1 −, =, 10 10, 3, ⇒ P(getting a red ball) =, ⋅, 10, Since P(getting a white ball) = P(getting a red ball), so the number, of white balls is equal to the number of red balls, say x., x, x, ∴ P(geting a red ball) =, =, ⋅, x + 20 + x 2 x + 20, x, 3, ∴, =, ⇒ 10 x = 6 x + 60 ⇒ 4 x = 60 ⇒ x = 15., 2 x + 20 10
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Probability, , 895, , Hence, the total number of balls in the bag, EXAMPLE 11, , SOLUTION, , = 2 x + 20 = 2 × 15 + 20 = 50., Two different dice are rolled together. Find the probability of getting, (i) the sum of numbers on two dice to be 5, (ii) even number on both, dice, (iii) a doublet., [CBSE 2015], When two dice are thrown simultaneously, all possible, outcomes are, (1, 1),, , (1, 2),, , (1, 3),, , (1, 4),, , (1, 5),, , (1, 6),, , (2, 1),, , (2, 2),, , (2, 3),, , (2, 4),, , (2, 5),, , (2, 6),, , (3, 1),, , (3, 2),, , (3, 3),, , (3, 4),, , (3, 5),, , (3, 6),, , (4, 1),, , (4, 2),, , (4, 3),, , (4, 4),, , (4, 5),, , (4, 6),, , (5, 1),, , (5, 2),, , (5, 3),, , (5, 4),, , (5, 5),, , (5, 6),, , (6, 1),, , (6, 2),, , (6, 3),, , (6, 4),, , (6, 5),, , (6, 6)., , Number of all possible outcomes = 36., (i) Let E1 be the event of getting two numbers whose sum, is 5., Then, the favourable outcomes are (1, 4) (2, 3), (3, 2), (4, 1)., Number of favourable outcomes = 4., 4, 1, = ⋅, ∴ P( getting two numbers whose sum is 5) = P(E1 ) =, 36 9, (ii) Let E2 be the event of getting a even numbers on both, dice., Then, the favourable outcomes are, (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)., Number of favourable outcomes = 9., 9, 1, P( getting even number on both dice) = P(E2 ) =, = ⋅, ∴, 36 4, , EXAMPLE 12, , (iii) Let E3 be the event of getting a doublet., Then, the favourable outcomes are, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)., Number of favourable outcomes = 6., 6, 1, P( getting a doublet) = P(E3 ) =, = ⋅, ∴, 36 6, Two dice are thrown at the same time. Find the probability that, the sum of the two numbers appearing on the top of the dice is more, than 9., [CBSE 2009C]
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896, SOLUTION, , Secondary School Mathematics for Class 10, , We know that when two dice are thrown the same time, then, the number of all possible outcomes is 36., Let E be the event that the sum of the numbers appearing on, the top of the two dice is more than 9., The favourable outcomes are, (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)., Number of favourable outcomes = 6., 6, 1, = ⋅, 36 6, A piggy bank contains hundred 50-p coins, fifty ` 1 coins, twenty, ` 2 and ten ` 5 coins. If it is equally likely that one of the coins will, fall out when the bank is turned upside down, find the probability the, coin falling out will be (i) a 50-p coin, (ii) of value more than ` 1,, (iii) of value less than ` 5 (iv) a ` 1 or ` 2 coin., [CBSE 2014], Total number of coins = (100 + 50 + 20 + 10) = 180., , ∴, EXAMPLE 13, , SOLUTION, , P(getting a sum more than 9) = P(E) =, , So, the number of all possible outcomes is 180., (i) Let E1 be the event of getting a 50-p coin., Then, the number of favourable outcomes = 100., 100 5, P(getting a 50-p coin) = P(E1 ) =, = ⋅, ∴, 180 9, (ii) LetE2 be the event of getting a coin of value more than ` 1., Then, it can be ` 2 or ` 5 coin., Number of all such coins = 20 + 10 = 30., P(getting a coin of value more than ` 1), ∴, 30 1, = P(E2 ) =, = ⋅, 180 6, (iii) Let E3 be the event of getting a coin of value less than ` 5., Then, it can be 50-p or ` 1 or ` 2 coin., Number of favourable outcomes, = number of all coins of 50-p, ` 1 and ` 2, = (100 + 50 + 20 = 170., P(getting a coin of value less than ` 5), ∴, 170 17, = P(E3 ) =, =, ⋅, 180 18, (iv) Let E4 be the event of getting a ` 1 or ` 2 coin., Number of all such coins = 50 + 20 = 70., Number of favourable outcomes = 70.
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Probability, , 70, 7, =, ⋅, 180 18, A game consists of tossing a one-rupee coin three times and noting, its outcome each time. Hanif wins if all the tosses give the same, result, i.e., three heads or three tails and loses otherwise. Calculate, the probability that Hanif will lose the game., [CBSE 2009C, ’11, ’17], In tossing a one-rupee coin three times, all possible outcomes, are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT., ∴, , EXAMPLE 14, , SOLUTION, , 897, , P(getting a ` 1 or ` 2 coin) = P(E4 ) =, , Total number of all possible outcomes = 8., Let E be the event of getting 3 heads or 3 tails., Then, E consists of HHH, TTT., Number of favourable outcomes of E = 2., P(that Hanif wins the game) = P(getting 3 heads or 3 tails), 2 1, = P(E) = = ⋅, 8 4, 1, 3, P(that Hanif loses the game) = 1 − P(E) = ⎛⎜ 1 − ⎞⎟ = ⋅, ⎝, ⎠, 4, 4, FACTS ABOUT PLAYING CARDS, , 1. A deck of playing cards has 52 cards., 2. There are 4 suits, namely (i) spades, (ii) clubs, (iii) hearts and, (iv) diamonds. There are 13 cards of each suit., I. Spades and clubs are black cards., II. Hearts and diamonds are red cards., 3. There are 12 face cards, namely 4 kings, 4 queens and 4 jacks., EXAMPLE 15, , SOLUTION, , One card is drawn at random from a well-shuffled deck of 52 cards., Find the probability that the card drawn is (i) a king, (ii) a red eight,, (iii) a spade, (iv) a red card, (v) the six of the clubs and (vi) a face card., Total number of all possible outcomes = 52., (i) There are 4 kings in all., ∴, , P(drawing a king) =, , 4, 1, =, ⋅, 52 13, , (ii) There are 2 red eights in all., ∴, , P(drawing a red eight) =, , 2, 1, =, ⋅, 52 26, , (iii) There are 13 cards of spades., 13 1, P(drawing a spade) =, ∴, = ⋅, 52 4
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898, , Secondary School Mathematics for Class 10, , (iv) There are 26 red cards., ∴, , P(drawing a red card) =, , 26 1, = ⋅, 52 2, , (v) There is one six of the clubs., ∴, , P(drawing the six of the clubs) =, , 1, ⋅, 52, , (vi) There are 12 face cards., 12, 3, =, ⋅, 52 13, One card is drawn from a well-shuffled deck of 52 cards. Find the, probability of getting (i) a king of red suit, (ii) a queen of black, suit, (iii) a jack of hearts, (iv) a red face card., [CBSE 2010], Total number of possible outcomes = 52., ∴, , EXAMPLE 16, , SOLUTION, , P(drawing a face card) =, , (i) Number of kings of red suit = 2., ∴, , P( getting a king of red suit) =, , 2, 1, =, ⋅, 52 26, , (ii) Number of queens of black suit = 2., ∴, , P( getting a queen of black suit) =, , 2, 1, =, ⋅, 52 26, , (iii) Number of jacks of hearts = 1., ∴, , P( getting a jack of hearts) =, , 1, ⋅, 52, , (iv) Red face cards are 2 kings, 2 queens, 2 jacks., Number of red face cards = 6., 6, 3, =, ⋅, 52 26, A card is drawn at random from a well-shuffled deck of 52 playing, cards. Find the probability that the card drawn is (i) a card of spades, or an ace, (ii) a black king, (iii) neither a jack nor a king, (iv) either a, king or a queen., [CBSE 2015], Total number of all possible outcomes = 52., ∴, , EXAMPLE 17, , SOLUTION, , P( getting a red face card) =, , (i) There are 13 cards of spades including one ace and there, are 3 more aces., So, the number of favourable cases = 13 + 3 = 16., 16, 4, P( getting a card of spades or an ace) =, ∴, =, ⋅, 52 13
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Probability, , 899, , (ii) There are 2 black kings., ∴, , P( getting a black king) =, , 2, 1, =, ⋅, 52 26, , (iii) There are 4 jacks and 4 kings., So, the number of cards which are neither jacks nor kings, = {52 − ( 4 + 4)} = 44., P( getting a card which is neither a jack nor a king), ∴, 44 11, =, =, ⋅, 52 13, , EXAMPLE 18, , SOLUTION, , (iv) There are 4 kings and 4 queens., So, the number of cards which are either kings or queens, = 4 + 4 = 8., P( getting a card which is either a king or a queen), ∴, 8, 2, =, =, ⋅, 52 13, One card is drawn at random from a well-shuffled deck of 52 playing, cards. Find the probability that the card drawn is (i) either a red card, or a king, (ii) neither a red card nor a queen., Total number of all possible outcomes = 52., (i) Let E1 be the event of getting a red card or a king., There are 26 red cards (including 2 kings) and there are 2, more kings., So, the number of favourable outcomes = 26 + 2 = 28., 28 7, P( getting a red card or a king) = P(E1 ) =, =, ⋅, ∴, 52 13, , EXAMPLE 19, , (ii) Let E1 be the event of getting a card which is neither a red, card nor a queen., There are 26 red cards (including 2 queens) and there are, 2 more queens., So, the number of non-favourable outcomes = 26 + 2 = 28., ∴ the number of favourable outcomes = 52 − 28 = 24., 24, 6, ∴ P( getting neither a red card nor a queen) =, =, ⋅, 52 13, From a pack of 52 playing cards jacks, queens, kings and aces of red, colour are removed. From the remaining, a card is drawn at random., Find the probability that the card drawn is, (i) a black queen (ii) a red card (iii) a ten, [CBSE 2006C], (iv) a picture card (jacks, queens and kings are picture cards).
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900, SOLUTION, , Secondary School Mathematics for Class 10, , Number of cards removed = 2 + 2 + 2 + 2 = 8., Total number of remaining cards = 52 − 8 = 44., Now, there are 2 jacks, 2 queens, 2 kings and 2 aces of black, colour only., (i) Number of black queens = 2., ∴, , P(getting a black queen) =, , 2, 1, =, ⋅, 44 22, , (ii) Remaining number of red cards = 26 − 8 = 18., 18, 9, P( getting a red card) =, ∴, =, ⋅, 44 22, (iii) Number of tens = 4., ∴, , EXAMPLE 20, , SOLUTION, , P( getting a ten) =, , 4, 1, =, ⋅, 44 11, , (iv) We know that jacks, queens and kings are picture cards., Out of 12 picture cards, it is given that 6 have been, removed. So, the remaining number of picture cards, = 12 − 6 = 6., 6, 3, P( getting a picture card) =, ∴, =, ⋅, 44 22, All the black face cards are removed from a pack of 52 playing cards., The remaining cards are well shuffled and then a card is drawn at, random. Find the probability of getting a (i) face card, (ii) red card,, (iii) black card, (iv) king., [CBSE 2014], Out of 52 playing cards; 2 black jacks, 2 black queens and, 2 black kings have been removed., Total number of remaining cards = (52 − 6) = 46., (i) Now, there are 6 face cards in the remaining cards., 6, 3, P( getting a face card) =, ∴, =, ⋅, 46 23, (ii) There are 26 red cards., ∴, , P( getting a red card) =, , 26 13, =, ⋅, 46 23, , (iii) Out of 46 cards, number of black cards = 26 − 6 = 20., 20 10, P( getting a black card) =, =, ⋅, ∴, 46 23, (iv) Now, these 46 cards have 2 kings., 2, 1, P( getting a king) =, ∴, =, ⋅, 46 23
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Probability, EXAMPLE 21, , SOLUTION, , 901, , Red queens and black jacks are removed from a pack of 52 playing, cards. A card is drawn at random from the remaining cards, after, reshuffling them. Find the probability that the drawn card is, (i) a king, (ii) of red colour, (iii) a face card, (iv) a queen. [CBSE 2014], After removing 2 red queens and 2 black jacks, the number of, remaining cards = 52 − ( 2 + 2) = 48., (i) Out of 48 cards, there are 4 kings., 4, 1, P( getting a king) =, ∴, =, ⋅, 48 12, (ii) Number of cards of red colour = 26 − 2 = 24., Total number of cards = 48., ∴, , P( getting a card of red colour) =, , 24 1, = ⋅, 48 2, , (iii) Number of face cards = 12 − ( 2 + 2) = 8., Total number of cards = 48., ∴, , P( getting a face card) =, , 8, 1, = ⋅, 48 6, , (iv) Number of queens in 48 cards = 4 − 2 = 2., 2, 1, P( getting a queen) =, ∴, =, ⋅, 48 24, , EXERCISE 19A, Very-Short-Answer Questions, 1. Fill in the blanks:, (i) The probability of an impossible event is ...... ., (ii) The probability of a sure event is ...... ., (iii) For any event E, P(E) + P(not E) = ...... ., (iv) The probability of a possible but not a sure event lies between ......, and ...... ., (v) The sum of probabilities of all the outcomes of an experiment is ...... ., 2. A coin is tossed once. What is the probability of getting a tail?, 3. Two coins are tossed simultaneously. Find the probability of getting, (i) exactly 1 head (ii) at most 1 head (iii) at least 1 head., 4. A die is thrown once. Find the probability of getting, (i) an even number, (ii) a number less than 5, (iii) a number greater than 2, (iv) a number between 3 and 6, (v) a number other than 3, (vi) the number 5.
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902, , Secondary School Mathematics for Class 10, , Short-Answer Questions, 5. A letter of English alphabet is chosen at random. Determine the, probability that the chosen letter is a consonant., [CBSE 2015], 6. A child has a die whose 6 faces show the letters given below:, A, , B, , C, , A, , D, , A, , The die is thrown once. What is the probability of getting (i) A, (ii) D?, 7. It is known that a box of 200 electric bulbs contains 16 defective bulbs., One bulb is taken out at random from the box. What is the probability, that the bulb drawn is (i) defective, (ii) nondefective?, 8. If the probability of winning a game is 0.7, what is the probability of, losing it?, 9. There are 35 students in a class of whom 20 are boys and 15 are girls., From these students one is chosen at random. What is the probability, that the chosen student is a (i) boy, (ii) girl?, 10. In a lottery there are 10 prizes and 25 blanks. What is the probability of, getting a prize?, 11. 250 lottery tickets were sold and there are 5 prizes on these tickets. If, Kunal has purchased one lottery ticket, what is the probability that he, wins a prize?, 12. 17 cards numbered 1, 2, 3, 4, …, 17 are put in a box and mixed, thoroughly. A card is drawn at random from the box. Find the, probability that the card drawn bears (i) an odd number (ii) a number, divisible by 5., [CBSE 2012], 13. A game of chance consists of spinning an arrow, which comes to rest, pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally, likely outcomes. Find the probability that the arrow will point at any, factor of 8., [CBSE 2015], 14. In a family of 3 children, find the probability of having at least one boy., [CBSE 2014], , 15. A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls., A ball is drawn at random from the bag. Find the probability that it is, (i) black, (ii) not green, (iii) red or white, (iv) neither red nor green., [CBSE 2012], , 16. A card is drawn at random from a well-shuffled pack of 52 cards., Find the probability of getting (i) a red king, (ii) a queen or a jack., [CBSE 2012]
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Probability, , 903, , 17. A card is drawn at random from a well-shuffled pack of 52 cards. Find, the probability that the drawn card is neither a king nor a queen., [CBSE 2013], , 18. A card is drawn from a well-shuffled pack of 52 cards. Find the, [CBSE 2013C], probability of getting (i) a red face card (ii) a black king., 19. Two different dice are tossed together. Find the probability that (i) the, number on each die is even, (ii) the sum of the numbers appearing on, [CBSE 2014], the two dice is 5., 20. Two different dice are rolled simultaneously. Find the probability that, [CBSE 2014], the sum of the numbers on the two dice is 10., 21. Two different dice are thrown together. Find the probability that, (i) the sum of the numbers appeared is less than 7., [CBSE 2011], (ii) the product of the numbers appeared is less than 18., , [CBSE 2017], , 22. Two dice are rolled together. Find the probability of getting such, [CBSE 2011], numbers on two dice whose product is a perfect square., 23. Two dice are rolled together. Find the probability of getting such, [CBSE 2013], numbers on the two dice whose product is 12., 24. Cards marked with numbers 5 to 50 are placed in a box and mixed, thoroughly. A card is drawn from the box at random. Find the, probability that the number on the taken out card is (i) a prime number, [CBSE 2008], less than 10 (ii) a number which is a perfect square., 25. A game of chance consists of spinning an arrow, which is equally likely to come to rest pointing, to one of the numbers 1, 2, 3, …, 12 as shown in, the figure. What is the probability that it will, point to, (i) 6?, (iii) a prime number?, , (ii) an even number?, (iv) a number which is a multiple of 5?, , 26. 12 defective pens are accidently mixed with 132 good ones. It is not, possible to just look at pen and tell whether or not it is defective. One, pen is taken out at random from this lot. Find the probability that the, pen taken out is good one., 27. A lot consists of 144 ballpoint pens of which 20 are defective and others, good. Tanvy will buy a pen if it is good, but will not buy it if it is, defective. The shopkeeper draws one pen at random and gives it to her., What is the probability that (i) she will buy it, (ii) she will not buy it?, 28. A box contains 90 discs which are numbered from 1 to 90. If one disc is, drawn at random from the box, find the probability that it bears (i) a
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904, , Secondary School Mathematics for Class 10, , two-digit number, (ii) a perfect square number, (iii) a number divisible, by 5., 29., , (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at, random from the lot. What is the probability that this bulb is, defective?, (ii) Suppose the bulb drawn in (i) is not defective and not replaced., Now, bulb is drawn at random from the rest. What is the, probability that this bulb is not defective?, , 30. A bag contains lemon-flavoured candies only. Hema takes out one, candy without looking into the bag. What is the probability that she, takes out (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?, 31. There are 40 students in a class of whom 25 are girls and 15 are boys., The class teacher has to select one student as a class representative., He writes the name of each student on a separate card, the cards being, identical. Then she puts cards in a bag and stirs them thoroughly., She then draws one card from the bag. What is the probability that the, name written on the card is the name of (i) a girl? (ii) a boy?, 32. One card is drawn from a well-shuffled deck of 52 cards. Find the, probability of drawing, (i) an ace, (iii) a ‘9’ of a black suit, , (ii) a ‘4’ of spades, (iv) a red king., , 33. A card is drawn at random from a well-shuffled deck of 52 cards. Find, [CBSE 2003], the probability of getting, (i) a queen, (iii) a king or an ace, , (ii) a diamond, (iv) a red ace., , 34. One card is drawn from a well-shuffled deck of 52 cards. Find the, probability of getting, (i) a king of red suit, (iii) a red face card, (v) a jack of hearts, , (ii) a face card, (iv) a queen of black suit, (vi) a spade., , 35. A card is drawn at random from a well-shuffled deck of playing cards., Find the probability that the card drawn is, [CBSE 2006], (i) a card of spades or an ace, (iii) either a king or a queen, , (ii) a red king, (iv) neither a king nor a queen., , 36. Two different dice are thrown together. Find the probability that the, numbers obtained have, (i) even sum, , (ii) even product., , [CBSE 2017]
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Probability, , 905, , 37. Two different dice are thrown together. Find the probability that the, numbers obtained, (i) have a sum less than 7, , (ii) have a product less than 16, , (iii) is a doublet of odd numbers., , [CBSE 2017], , 38. The king, the jack and the 10 of spades are lost from a pack of 52 cards, and a card is drawn from the remaining cards after shuffling. Find the, probability of getting a, (i) red card, , (ii) black jack, , (iii) red king, , (iv) 10 of hearts., , [CBSE 2017], , 39. Peter throws two different dice together and finds the product of the, two numbers obtained. Rina throws a die and squares the number, obtained. Who has the better chance to get the number 25? [CBSE 2017], ANSWERS (EXERCISE 19A), , 1, 3, 3, (ii) (iii), 2, 4, 4, 1, 2, 2, 1, 5, 1, 21, 1, 1, 5., 6. (i) (ii), 4. (i) (ii) (iii) (iv) (v) (vi), 2, 3, 3, 3, 6, 6, 26, 2, 6, 2, 23, 4, 3, 2, 1, (ii), 8. 0.3, 9. (i) (ii), 10., 11., 7. (i), 25, 25, 7, 7, 7, 50, 9, 3, 3, 7, 2, 11, 3, 2, (ii), 13., 14., 15. (i), (ii), (iii) (iv), 12. (i), 17, 17, 8, 8, 15, 15, 5, 5, 1, 2, 11, 3, 1, 1, 1, 1, 16. (i), (ii), 17., 18. (i), (ii), 19. (i) (ii), 20., 26, 13, 13, 26, 26, 4, 9, 12, 5, 13, 2, 1, 1, 5, (ii), 22., 23., 24. (i), (ii), 21. (i), 12, 18, 9, 9, 23, 46, 1, 1, 5, 1, 11, 31, 5, 25. (i), (ii) (iii), (iv), 26., 27. (i), (ii), 12, 2, 12, 6, 12, 36, 36, 9, 1, 1, 1, 15, (ii), (iii), 29. (i) (ii), 30. (i) 0 (ii) 1, 28. (i), 10, 10, 5, 5, 19, 5, 3, 1, 1, 1, 1, 31. (i) (ii), 32. (i), (ii), (iii), (iv), 8, 8, 13, 52, 26, 26, 1, 1, 2, 1, 1, 3, 3, 1, 1, 1, (ii) (iii), (iv), 34. (i), (ii), (iii), (iv), (v), (vi), 33. (i), 13, 4, 13, 26, 26, 13, 26, 26, 52, 4, 4, 1, 2, 11, 1, 3, 5, 25, 1, (ii), (iii), (iv), 36. (i) (ii), (ii), (iii), 35. (i), 37. (i), 13, 26, 13, 13, 2, 4, 12, 36, 4, 26, 1, 2, 1, 38. (i), (ii), (iii), (iv), 39. Rina, 49, 49, 49, 49, 1. (i) 0 (ii) 1 (iii) 1 (iv) 0, 1 (v) 1, , 2., , 1, 2, , 3. (i)
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906, , Secondary School Mathematics for Class 10, HINTS TO SOME SELECTED QUESTIONS, , 5. Out of 26 letters of English alphabet, there are 11 consonants., 21, ⋅, ∴ P(getting a consonant) =, 26, 6. There are 6 letters in all consisting of 3As, 1B, 1C and 1D., 3 1, 1, (ii) P(getting D) = ⋅, ∴ (i) P(getting A) = = ⋅, 6 2, 6, 7. Total number of bulbs = 200., Number of defective bulbs = 16., Number of non-defective bulbs = 200 − 16 = 184., 16, 2, (i) P(getting a defective bulb) =, =, ⋅, 200 25, 184 23, (ii) P(getting a non-defective bulb) =, =, ⋅, 200 25, 8. Let E be the event of winning the game. Then, P( E) = 0 . 7 ., Probability of losing the game = 1 − P( E) = ( 1 − 0 . 7 ) = 0. 3., 9. (i) P(choosing a boy) =, , 20 4, = ⋅, 35 7, , (ii) P(choosing a girl) =, , 15 3, = ⋅, 35 7, , 10. Total number of tickets = 10 + 25 = 35., Number of prizes = 10., 10 2, P(getting a prize) =, = ⋅, 35 7, 11. P(getting a prize) =, , 5, 1, =, ⋅, 250 50, , 12. Total number of cards = 17., (i) Let E1 be the event of choosing an odd number., These numbers are 1, 3, 5, ... , 17., Let their number be n. Then,, Tn = 17 ⇒ 1 + (n − 1) × 2 = 17 ⇒ n = 9., 9, ⋅, ∴ P( E1 ) =, 17, (ii) Let E2 be the event of choosing a number divisible by 5., Numbers divisible by 5 are 5, 10, 15. Their number is 3., 3, ⋅, ∴ P( E2 ) =, 17, 13. All possible outcomes are 1, 2, 3, 4, 5, 6, 7, 8., Number of all possible outcomes is 8., All factors of 8 are 2, 4, 8., Number of favourable outcomes = 3., Probability that the arrow will point at any factor of 8 =, , 3, ⋅, 8
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Probability, , 907, , 14. All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG., Number of all possible outcomes = 8., Let E be the event of having at least one boy., Then, E contains GGB, GBG, BGG, BBG, BGB, GBB, BBB., Number of cases favourable to E = 7., 7, ∴ required probability = P( E) = ⋅, 8, 15. Total number of balls = 4 + 5 + 2 + 4 = 15., (i) Number of black balls = 2., 2, P(getting a black ball) =, ⋅, 15, (ii) Number of balls which are not green = 4 + 5 + 2 = 11., 11, P(getting a ball which is not green) =, ⋅, 15, (iii) Number of balls which are red or white = 5 + 4 = 9., 9, 3, P(getting a ball which is red or white) =, = ⋅, 15 5, (iv) Number of balls which are neither red nor green = 4 + 2 = 6., 6 2, P(getting a ball which is neither red nor green) =, = ⋅, 15 5, 16. Total number of cards = 52., (i) Number of red kings = 2., ∴ P(getting a red king) =, , 2, 1, =, ⋅, 52 26, , (ii) There are 4 queens and 4 jacks., ∴ P(getting a queen or a jack) =, , 8, 2, =, ⋅, 52 13, , 17. Toal number of cards = 52., Total number of kings and queens = 4 + 4 = 8., Remaining number of cards = 52 − 8 = 44., ∴, , P(getting a card which is neither a king nor a queen) =, , 44 11, =, ⋅, 52 13, , 18. Total number of cards = 52., (i) 4 kings, 4 queens and 4 jacks are all face cards., Number of red face cards = 2 + 2 + 2 = 6., 6, 3, =, ⋅, ∴ P(getting a red face card) =, 52 26, (ii) Number of black kings = 2., ∴, , P(getting a black king) =, , 2, 1, =, ⋅, 52 26, , 19. When two different dice are thrown, then total number of outcomes = 36., (i) Let E1 be the event of getting an even number on each die., These numbers are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).
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908, , Secondary School Mathematics for Class 10, Number of favourable outcomes = 9., 9, 1, = ⋅, 36 4, (ii) Let E2 be the event of getting the sum 5. Then these numbers are, (1, 4), (2, 3), (4, 1), (3, 2)., ∴ P(getting an even number on both dice) = P( E1 ) =, , Number of favourable outcomes = 4., 4, 1, = ⋅, ∴ P( E2 ) =, 36 9, 20. Number of all possible outcomes is 36., Let E be the event of getting the sum 10 on the two dice., Then, the favourable outcomes are (4, 6), (6, 4), (5, 5)., Number of favouable outcomes = 3., 3, 1, =, ⋅, ∴ P( E) =, 36 12, 21., , (i) Number of all possible outcomes is 36., Let E be the event of getting the sum less than 7 on the two dice., Then, the favourable outcomes are, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3),, (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)., Number of favourable outcomes = 15., 15, 5, =, ⋅, ∴ P( E) =, 36 12, , 22. Number of all possible outcomes is 36., Let E be the event of getting the product of numbers on the two dice, as a perfect, square., Then, the favourable outcomes are, (1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)., Number of favourable outcomes = 8., 8, 2, = ⋅, ∴ P( E) =, 36 9, 23. Number of all possible outcomes is 36., Let E be the event of getting all those two numbers whose product is 12., Then, the favourable outcomes are (2, 6), (6, 2), (3, 4), (4, 3)., Number of favourable outcomes = 4., 4, 1, = ⋅, ∴ P( E) =, 36 9, 24. Total number of cards = 50 − 4 = 46., (i) Out of the given numbers, prime numbers less than 10 are 5 and 7., ∴ number of prime numbers less than 10 = 2., 2, 1, =, ⋅, ∴ P(getting a prime number less than 10) =, 46 23
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Probability, (ii) From given numbers, the perfect square numbers are 9, 16, 25, 36, 49., Their number is 5., ∴ P(getting a perfect square number) =, , 5, ⋅, 46, , 25. Number of all possible outcomes = 12., 1, (i) P(getting a 6) =, ⋅, 12, (ii) Out of the given numbers there are 6 even numbers., 6, 1, ∴ P(getting an even number) =, = ⋅, 12 2, (iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7, 11., Their number is 5., 5, ⋅, 12, (iv) Out of the given numbers, the multiples of 5 are 5, 10., ∴ P(getting a prime number) =, , Their number is 2., ∴ P(getting a multiple of 5) =, , 2, 1, = ⋅, 12 6, , 26. Total number of pens = 132 + 12 = 144., Number of good pens = 132., 132 11, =, ⋅, ∴ P(getting a good pen) =, 144 12, 27. Total number of pens = 144., Number of defective pens = 20., Number of non-defective pens = 144 − 20 = 124., 124 31, =, ⋅, 144 36, 20, 5, (ii) P(not buying the pen) = P(getting a defective pen) =, =, ⋅, 144 36, (i) P(buying the pen) = P(getting a non-defective pen) =, , 28. Total number of discs = 90., (i) Number of discs bearing 2-digit numbers, = number of numbers from 10 to 90 = ( 90 − 9 ) = 81., 81 9, P(getting a 2-digit number) =, =, ⋅, 90 10, (ii) Perfect square numbers are 12 , 2 2 , 3 2 , ... , 9 2 . Their number is 9., Number of discs bearing perfect square numbers = 9., 9, 1, =, ⋅, ∴ P(getting a perfect square number) =, 90 10, (iii) Numbers divisible by 5 are 5, 10, 15, ... , 90. They are 18 in number., 18 1, = ⋅, ∴ P(getting a number divisible by 5) =, 90 5, 29., , (i) Total number of bulbs = 20. Number of defective bulbs = 4., Number of non-defective bulbs = 20 − 4 = 16., , 909
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910, , Secondary School Mathematics for Class 10, 4 1, = ⋅, 20 5, (ii) After removing 1 non-defective bulb, we have, ∴ P(getting a defective bulb) =, , remaining number of bulbs = 20 − 1 = 19., Out of these, the number of non-defective bulbs = 16 − 1 = 15., 15, ⋅, ∴ P(getting a non-defective bulb) =, 19, 30. Suppose there are x candies in the bag., Then, number of orange-flavoured candies in the bag = 0., And, the number of lemon-flavoured candies in the bag = x., 0, ∴ (i) P(getting an orange-flavoured candy) = = 0., x, x, (ii) P(getting a lemon-flavoured candy) = = 1., x, 31. Total number of students in the class = 40., Number of girls = 25, number of boys = 15., 25 5, (i) P(selecting the name of a girl) =, = ⋅, 40 8, 15 3, (ii) P(selecting the name of a boy) =, = ⋅, 40 8, 32., , (i) There are 4 aces in all., ∴ P(getting an ace) =, , 4, 1, =, ⋅, 52 13, , (ii) There is one ‘4’ of spades., ∴ P(getting 4 of spades) =, , 1, ⋅, 52, , (iii) There are two nines of black suits., ∴ P(getting a ‘9’ of a black suit) =, , 2, 1, =, ⋅, 52 26, , (iv) There are two red kings., ∴ P(getting a red king) =, 33., , 2, 1, =, ⋅, 52 26, , (i) There are 4 queens., ∴ P(getting a queen) =, , 4, 1, =, ⋅, 52 13, , (ii) There are 13 diamonds., ∴ P(getting a diamond) =, , 13 1, = ⋅, 52 4, , (iii) There are 4 kings and 4 aces., ∴ P(getting a king or an ace) =, , 8, 2, =, ⋅, 52 13
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Probability, , 911, , (iv) There are 2 red aces., 2, 1, =, ⋅, 52 26, (i) There are 4 kings of red suits., ∴ P(getting a red ace) =, , 34., , ∴ P(getting a king of a red shuit) =, , 2, 1, =, ⋅, 52 26, , (ii) There are 12 face cards in all., 12, 3, =, ⋅, 52 13, (iii) There are 6 red face cards., ∴ P(getting a face card) =, , 6, 3, =, ⋅, 52 26, (iv) There are 2 queens of black suits., ∴ P(getting a red face card) =, , ∴ P(getting a queen of a black suit) =, , 2, 1, =, ⋅, 52 26, , (v) There is 1 jack of hearts., ∴ P(getting a jack of hearts) =, , 1, ⋅, 52, , (vi) There are 3 cards of spades., ∴ P(getting a spade) =, 35., , 13 1, = ⋅, 52 4, , (i) There are 13 cards of spades including 1 ace and 3 more aces are there., ∴ P(getting a card of spades or an ace) =, , 13 + 3 16 4, =, =, ⋅, 52, 52 13, , (ii) There are 2 red kings., ∴ P(getting a red king) =, , 2, 1, =, ⋅, 52 26, , (iii) There are 4 kings and 4 queens., ∴ P(getting either a king or a queen) =, , 4+ 4 8, 2, =, =, ⋅, 52, 52 13, , (iv) P(neither a king nor a queen) = 1 − P(either a king or a queen), 2, 11, = ⎛⎜ 1 − ⎞⎟ =, ⋅, ⎝, 13 ⎠ 13, , ................................................................, , EXERCISE 19B, Long-Answer Questions, 1. A box contains 25 cards numbered from 1 to 25. A card is drawn at, random from the bag. Find the probability that the number on the, drawn card is (i) divisible by 2 or 3, (ii) a prime number., [CBSE 2015]
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912, , Secondary School Mathematics for Class 10, , 2. A box contains cards numbered 3, 5, 7, 9,... , 35, 37. A card is drawn at, random from the box. Find the probability that the number on the card, is a prime number., [CBSE 2013], 3. Cards numbered 1 to 30 are put in a bag. A card is drawn at random, from the bag. Find the probability that the number on the drawn card is, (i) not divisible by 3, (ii) a prime number greater that 7, (iii)not a perfect, square number., [CBSE 2014], 4. Cards bearing numbers 1, 3, 5,... , 35 are kept in a bag. A card is drawn, at random from the bag. Find the probability of getting a card bearing, (i) a prime number less than 15, (ii) a number divisible by 3 and 5., [CBSE 2010], , 5. A box contains cards bearing numbers 6 to 70. If one card is drawn at, random from the box, find the probability that it bears (i) a one-digit, number, (ii) a number divisible by 5, (iii) an odd number less than 30,, (iv) a composite number between 50 and 70., [CBSE 2015], 6. Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and, mixed thoroughly. A card is drawn at random from the bag. Find the, probability that the number on the drawn card is (i) less than 19, (ii) a, prime number less than 20., [CBSE 2012], 7. Tickets numbered 2, 3, 4, 5, ..., 100, 101 are placed in a box and mixed, thoroughly. One ticket is drawn at random from the box. Find the, probability that the number on the ticket is, (i) an even number, (ii) a number less than 16, (iii) a number which is a perfect square, (iv) a prime number less than 40., 8., , (i) A box contains 80 discs, which are numbered from 1 to 80. If one, disc is drawn at random from the box, find the probability that it, bears a perfect square number., [CBSE 2011], (ii) A box contains 90 discs which are numbered 1 to 90. If one disc is, drawn at random from the box, find the probability that it bears, [CBSE 2017], (a) a two-digit number (b) a number divisible by 5, , 9. A piggy bank contains hundred 50-p coins, seventy ` 1 coin, fifty ` 2, coins and thirty ` 5 coins. If it is equally likely that one of the coins will, fall out when the bank is turned upside down, what is the probability, that the coin (i) will be a ` 1 coin? (ii) will not be a ` 5 coin (iii) will be, [CBSE 2013C], 50-p or a ` 2 coin?
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Probability, , 913, , 10. The probability of selecting a red ball at random from a jar that contains, 1, only red, blue and orange balls is ⋅ The probability of selecting a blue, 4, 1, ball at random from the same jar is ⋅ If the jar contains 10 orange balls,, 3, [CBSE 2015], find the total number of balls in the jar., 11. A bag contains 18 balls out of which x balls are red., (i) If one ball is drawn at random from the bag, what is the probability, that it is not red?, (ii) If two more red balls are put in the bag, the probability of drawing, 9, a red ball will be times the probability of drawing a red ball in the, 8, first case. Find the value of x., [CBSE 2015], 12. A jar contains 24 marbles. Some of these are green and others are blue. If, a marble is drawn at random from the jar, the probability that it is green, 2, is ⋅ Find the number of blue marbles in the jar., 3, 13. A jar contains 54 marbles, each of which some are blue, some are green, and some are white. The probability of selecting a blue marble at, 1, random is and the probability of selecting a green marble at random, 3, 4, is ⋅ How many white marbles does the jar contain?, 9, 14. A carton consists of 100 shirts of which 88 are good and 8 have minor, defects. Rohit, a trader, will only accept the shirts which are good. But,, Kamal, an another trader, will only reject the shirts which have major, defects. One shirt is drawn at random from the carton. What is the, probability that it is acceptable to (i) Rohit and (ii) Kamal?, 15. A group consists of 12 persons, of which 3 are extremely patient, other 6, are extremely honest and rest are extremely kind. A person from the, group is selected at random. Assuming that each person is equally, likely to be selected, find the probability of selecting a person who is, (i) extremely patient, (ii) extremely kind or honest. Which of the above, values you prefer more?, [CBSE 2013], 16. A die is rolled twice. Find the probability that, (i) 5 will not come up either time,, (ii) 5 will come up exactly one time,, (iii) 5 will come up both the times., , [CBSE 2014]
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914, , Secondary School Mathematics for Class 10, , 17. Two dice are rolled once. Find the probability of getting such numbers, on two dice whose product is a perfect square., [CBSE 2011], 18. A letter is chosen at random from the letters of the word ‘ASSOCIATION’., Find the probability that the chosen letter is a (i) vowel (ii) consonant, (iii) an S., 19. Five cards—the ten, jack, queen, king and ace of diamonds are well, shuffled with their faces downwards. One card is then picked up at, random. (a) What is the probability that the drawn card is the queen?, (b) If the queen is drawn and put aside and a second card is drawn, find, the probability that the second card is (i) an ace, (ii) a queen. [CBSE 2014], 20. A card is drawn at random from a well shuffled pack of 52 cards. Find, the probability that the card drawn is neither a red card nor a queen., [CBSE 2011], , 21. What is the probability that an ordinary year has 53 Mondays?, 22. All red face cards are removed from a pack of playing cards. The, remaining cards are well shuffled and then a card is drawn at random, from them. Find the probability that the drawn card is (i) a red card, (ii), a face card, (iii) a card of clubs., [CBSE 2015], 23. All kings, queens and aces are removed from a pack of 52 cards., The remaining cards are well-shuffled and then a card is drawn from, it. Find the probability that the drawn card is (i) a black face card,, (ii) a red card., [CBSE 2012], 24. A game consists of tossing a one-rupee coin three times, and noting its, outcome each time. Find the probability of getting (i) three heads, (ii) at, least 2 tails., [CBSE 2015], 25. Find the probability that a leap year selected at random will contain, 53 Sundays., 26. The probability of selecting a rotten apple randomly from a heap of, 900 apples is 0.18. What is the number of rotten apples in the heap?, [CBSE 2017], , 27. A bag contains 15 white and some black balls. If the probability of, drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag., [CBSE 2017], 28. Find the probability of getting the sum of two numbers, less than 3 or, more than 11, when a pair of distinct dice is thrown together. [CBSE 2017]
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Probability, , 915, , ANSWERS (EXERCISE 19B), , 16, 9, 11, 2, 1, 5, 5, 1, (ii), 2., 3. (i) (ii) (iii), 4. (i), (ii), 25, 25, 18, 3, 5, 6, 18, 9, 4, 1, 12, 3, 3, 7, (ii) (iii), (iv), 6. (i), (ii), 5. (i), 65, 5, 65, 13, 17, 51, 1, 7, 9, 3, 1, 9, 1, (iii), (iv), 8. (i), (ii) (a), (b), 7. (i) (ii), 2, 50, 100, 25, 10, 10, 5, (18 − x), 7, 22, 3, (ii), (iii), 10. 24, 11. (i), (ii) x = 8, 9. (i), 25, 25, 5, 18, 22, 24, 1, 3, 12. 8, 13. 12, 14. (i), (ii), 15. (i) (ii), 25, 25, 4, 4, 25, 5, 1, 2, 6, 5, 2, (ii), (ii), 17., 18. (i), (ii), (iii), 16. (i), 36, 18, 36, 9, 11, 11, 11, 1, 1, 6, 1, 10, 3, 6, 19. (a) (b) (i) (ii) 0, 20., 21., 22. (i) (ii) (iii), 5, 4, 13, 7, 23, 23, 23, 1, 1, 1, 1, 2, (ii), 24. (i) (ii), 25., 26. 162, 23. (i), 20, 2, 8, 2, 7, 1, 27. 45 28., 18, 1. (i), , HINTS TO SOME SELECTED QUESTIONS, 1. Number of all possible outcomes = 25., (i) Numbers divisible by 2 or 3 are 2, 4, 6, 8,10, 12, 14, 16, 18, 20, 22, 24, 3, 9, 15, 21., Their number is 16., ∴ P(getting a number divisible by 2 or 3) =, , 16, ⋅, 25, , (ii) Prime numbers from 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23., Their number is 9., ∴ P(getting a prime number) =, , 9, ⋅, 25, , 2. Given numbers 3, 5, 7, 9, ..., 35, 37 form an AP with a = 3 and d = 2., Let Tn = 37. Then,, 3 + (n − 1) × 2 = 37 ⇒ (n − 1) × 2 = 34 ⇒ n − 1 = 17 ⇒ n = 18., Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37., Their number is 11., ∴, , P(getting a prime number) =, , 11, ⋅, 18, , 3. Number of all possible outcomes = 30., (i) There are 10 numbers divisible by 3., Number of numbers not divisible by 3 = 30 − 10 = 20.
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916, , Secondary School Mathematics for Class 10, 20 2, = ⋅, 30 3, (ii) Prime numbers greater than 7 are 11, 13, 17, 19, 23, 29., ∴ P(getting a number not divisible by 3) =, , Number of these prime numbers = 6., 6, 1, = ⋅, 30 5, (iii) Perfect square numbers are 1, 4, 9, 16, 25. These are 5 in number., ∴ P(getting a prime number greater than 7) =, , Number of those numbers which are not perfect squares = 30 − 5 = 25., 25 5, = ⋅, ∴ P(getting non-perfect square numbers) =, 30 6, 4. The numbers 1, 3, 5, ..., 35 form an AP with a = 1 and d = 2., Let Tn = 35. Then,, 1 + (n − 1) × 2 = 35 ⇒ (n − 1) × 2 = 34 ⇒ n − 1 = 17 ⇒ n = 18., ∴, , number of all possible outcomes = 18., (i) Out of the given numbers, the prime numbers less than 15 are 3, 5, 7, 11, 13., Their number is 5., 5, ∴ P(getting a prime number) =, ⋅, 18, (ii) A number is divisible by 3 and 5 means, it must be divisible by 15., The numbers divisible by 15 are 15 and 30., Their number is 2., ∴ P(getting a number divisible by both 3 and 5) =, , 2, 1, = ⋅, 18 9, , 5. Given numbers 6, 7, 8, ..., 70 form an AP with a = 6 and d = 1., Let Tn = 70. Then, 6 + (n − 1) × 1 = 70 ⇒ n − 1 = 64 ⇒ n = 65., ∴, , total number of cards = 65., (i) Out of the given numbers, the one-digit numbers are 6, 7, 8, 9., Number of one-digit numbers = 4., ∴ P(getting a one-digit number) =, , 4, ⋅, 65, , (ii) Out of the given numbers, those divisible by 5 are 10, 15, 20, 25, ..., 70., Let Tn = 70. Then,, 10 + (n − 1) × 5 = 70 ⇒ (n − 1) × 5 = 60 ⇒ n − 1 = 12 ⇒ n = 13., 13 1, = ⋅, ∴ P(getting a number divisible by 5) =, 65 5, (iii) Out of the given numbers, odd numbers less than 30 are 7, 9, 11, 13, ... , 29., Let Tm = 29. Then,, 7 + (m − 1) × 2 = 29 ⇒ (m − 1) × 2 = 22 ⇒ m − 1 = 11 ⇒ m = 12., 12, ⋅, 65, , ∴ P(getting an odd number less than 30) =, , (iv) Number of numbers between 50 and 70 = numbers from 51 to 69., Their number = ( 69 − 51) + 1 = 19., Prime numbers between 50 and 70 = 53 , 59 , 61, 67 .
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Probability, , 917, , Number of prime numbers = 4., Number of composite numbers = 19 − 4 = 15., 15, 3, =, ⋅, ∴ P(getting a composite number) =, 65 13, 6. Given numbers 1, 3, 5, ..., 101 form an AP with a = 1 and d = 2., Let Tn = 101. Then,, 1 + (n − 1) × 2 = 101 ⇒ (n − 1) × 2 = 100 ⇒ n − 1 = 50 ⇒ n = 51., ∴, , total number of cards = 51., (i) Out of the given numbers, those less than 19 are 1, 3, 5, ..., 17., Let t m = 17., Then, 1 + (m − 1) × 2 = 17 ⇒ (m − 1) × 2 = 16 ⇒ m − 1 = 8 ⇒ m = 9., 9, 3, =, ⋅, ∴ P(getting a number less than 19) =, 51 17, (ii) Out of the given numbers, the prime numbers less than 20 are 3, 5, 7, 11, 17, 19., Their number is 7., , 7, ⋅, 51, 7. The tickets bear the numbers 2, 3, 4, ... , 100, 101. So, the number of tickets = 100., ∴ P(getting a prime number less than 20) =, , (i) Out of the given numbers, the even numbers are 2, 4, 6, 8, ..., 100., Their number is 50., ∴ P(getting an even number) =, , 50, 1, = ⋅, 100 2, , (ii) Out of the given numbers, the number of numbers less than 16 is 14., 14, 7, =, ⋅, ∴ P(getting number less than 16) =, 100 50, (iii) Out of the given numbers, the perfect square numbers are 2 2 , 3 2 , 4 2 , ..., 10 2 ., Their number is 9., ∴ P(getting a perfect square number) =, , 9, ⋅, 100, , (iv) Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37., Their number is 12., ∴ P(getting a prime number less than 40) =, 10. P(getting a red ball) =, , 12, 3, =, ⋅, 100 25, , 1, 1, , P(getting a blue ball) = ⋅, 4, 3, , Let P(getting an orange ball) be x., Since there are only thre types of balls in the jar, the sum of probabilities of drawing, these balls must be 1., 1 1, 7, 7, 5, ⋅, + + x = 1⇒, + x = 1 ⇒ x = ⎛⎜ 1 − ⎞⎟ =, ∴, ⎝, 4 3, 12, 12 ⎠ 12, ∴, , P(getting an orange ball) =, , 5, ⋅, 12, , Let the total number of balls in the jar be n., , ... (i)
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918, , Secondary School Mathematics for Class 10, Number of orange balls = 10., 10, ⋅, n, , ∴, , P(getting an orange ball) =, , ⇒, , 10 5, =, ⇒ 5n = 120 ⇒ n = 24 [using (i)]., n 12, , Hence, the total number of balls in the jar is 24., 11., , (i) Total number of balls = 18., Number of red balls = x., Number of balls which are not red = 18 − x., 18 − x, ∴ P(getting a ball which is not red) =, ⋅, 18, (ii) Now, total number of balls = 18 + 2 = 20., , Number of red balls now = x + 2., x+ 2, x, P(getting a red ball now) =, and P(getting a red ball in first case) =, 20, 18, x+ 2 9 x, = ×, ⇒ 144( x + 2 ) = 180 x, ⇒, 20, 8 18, 288, = 8., ⇒ 180 x − 144 x = 288 ⇒ 36 x = 288 ⇒ x =, 36, Hence, x = 8., 12. Total number of marbles in the jar = 24., Let the number of blue marbles be x., Then, the number of green marbles = 24 − x., 24 − x, P(getting a green marble) =, ⋅, 24, 2, But, P(getting a green marble) = (given)., 3, 24 − x 2, = ⇒ 72 − 3 x = 48 ⇒ 3 x = 24 ⇒ x = 8., ∴, 24, 3, Hence, the number of blue marbles in the jar is 8., 13. Total number of marbles in the jar = 54., 1, 4, P(getting a blue marble) = and P(getting a green marble) = ⋅, 3, 9, Let P(getting a white marble) be x., Since there are only three given types of marbles in the jar, we have, 1 4, 7, 7 2, + + x=1 ⇒, + x = 1 ⇒ x = 1− = ⋅, 3 9, 9, 9 9, 2, ∴ P(getting a white marble) = ⋅, 9, Let the number of white marbles be n., n, Then, P(getting a white marble) =, ⋅, 54, n 2, = ⇒ 9n = 108 ⇒ n = 12., ∴, 54 9, Hence, there are 12 white marbles in the jar.
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Probability, , 919, , 14. Total number of shirts = 100., Number of good shirts = 88., Number of shirts having minor defects = 8., Number of shirts having major defects = 100 − ( 88 + 8 ) = 4., 88 22, (i) P(that the drawn shirt is acceptable to Rohit) =, =, ⋅, ∴, 100 25, 88 + 8, 96 24, (ii) P(that the drawn shirt is acceptable to Kamal) =, =, =, ⋅, 100, 100 25, 15. Total number of persons in the group = 12., (i) Number of persons who are extremely patient = 3., 3 1, P(selecting a person who is extremely patient) =, = ⋅, 12 4, (ii) Number of persons who are extremely honest = 6., Number of persons who are extremely kind = 12 − ( 3 + 6 ) = 3., ∴, , P(selecting a person who is extremely kind or extremely honest) =, , 3+ 6, 9, 3, =, = ⋅, 12, 12 4, , From the given three values, we prefer honesty. Honesty can get rid of rampant, corruption which is a burning issue of the present society., 16. Number of all possible outcomes = 36., (i) All those cases where 5 comes up on at least one face are, (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)., Number of such cases = 11., Number of those cases where 5 will not come up any time = 36 − 11 = 25., 25, ⋅, ∴ P(that 5 will not come up either time) =, 36, (ii) All those case where 5 comes up exactly one time are, (1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)., Number of such cases = 10., ∴ P(that 5 will come up exactly one time) =, , 10, 5, =, ⋅, 36 18, , (iii) There is only one case, namely (5, 5) when 5 comes up both the times., 1, ⋅, ∴ P(that 5 will come up both the times) =, 36, 17. Number of all possible outcomes = 36., Let E be the event of getting 2 numbers on the two dice whose product is a perfect, square., Then, the favourable outcomes are (1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4) (5, 5), (6, 6)., 8, 2, Their number is 8. So, P( E) =, = ⋅, 36 9, 18. Total number of letters in the the given word = 11., (i) Number of vowels in the given word = 6., 6, ⋅, ∴ P(selecting a vowel) =, 11
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920, , Secondary School Mathematics for Class 10, (ii) Number of consonants in the given word = 5., 5, P(selecting a consonant) =, ⋅, 11, (iii) Number of S in the given word = 2., 2, ⋅, ∴ P(getting an S) =, 11, , 19. Total number of cards = 5., 1, (a) P(getting a queen) = ⋅, 5, (b) When the queen is set aside, remaining number of cards = 4., 1, (i) P(getting an ace now) = ⋅, 4, 0, (ii) P(getting a queen now) = = 0., 4, 20. There are 26 red cards containing 2 queens and 2 more queens are there., 28 7, P(getting a red card or a queen) =, =, ⋅, 52 13, 7, 6, =, ⋅, ∴ P(getting neither a red card nor a queen) = 1 −, 13 13, 21. An ordinary year has 365 days consisting of 52 weeks 1 day., This day can be any day of the week., 1, P(of this day to be Monday) = ⋅, 7, 22. There are 6 red face cards. These are removed., Remaining number of cards = 52 − 6 = 46., (i) Number of red cards now = 26 − 6 = 20., 20 10, =, ⋅, ∴ P(getting a red card) =, 46 23, (ii) Remaining number of face cards = 12 − 6 = 6., 6, 3, =, ⋅, ∴ P(getting a face card) =, 46 23, (iii) There are 12 cards of clubs., 12 6, =, ⋅, ∴ P(getting a card of clubs) =, 46 23, 23. 4 kings, 4 queens and 4 aces have been removed., Remaining number of cards = 52 − 12 = 40., (i) There are 2 black face cards in the remaining cards., These are 2 black jacks., 2, 1, ∴ P(getting 2 black face cards) =, =, ⋅, 40 20, (ii) Remaining number of red cards = 26 − ( 2 + 2 + 2 ) = 20., 20 1, = ⋅, ∴ P(getting a red card) =, 40 2, 24. When a coin is tossed 3 times, all possible outcomes are, HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
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Probability, , 921, , Number of all possible outcomes = 8., 1, (i) P(getting 3 heads) = ⋅, 8, (ii) At least 2 tails means 2 or 3 tails., All such cases are HTT, THT, TTH, TTT. Their number is 4., 4 1, ∴ P(getting at least 2 tails) = = ⋅, 8 2, 25. A leap year has 366 days = 52 weeks and 2 days., Now, 52 weeks contain 52 Sundays., The remaining 2 days can be:, (i) Sunday and Monday, (iii) Tuesday and Wednesday, (v) Thursday and Friday, (vii) Saturday and Sunday, , (ii) Monday and Tuesday, (iv) Wednesday and Thursday, (vi) Friday and Saturday, , Out of these 7 cases, their are 2 cases favouring it to be Sunday., 2, ∴ P(a leap year having 53 Sundays) = ⋅, 7, , ................................................................, , MULTIPLE-CHOICE QUESTIONS (MCQ), Choose the correct answer in each of the following questions:, 1. If P(E) denotes the probability of an event E then, (a) P(E) < 0, , (b) P(E) > 1, , (c) 0 ≤ P(E) ≤ 1, , [CBSE 2013C], , (d) −1 ≤ P(E) ≤ 1, , 2. If the probability of occurrence of an event is p then the probability of, non-happening of this event is, [CBSE 2013C], (a) ( p − 1), , (b) (1 − p), , (c) p, , 3. What is the probability of an impossible event?, 1, (b) 0, (c) 1, (a), 2, , ⎛, 1⎞, (d) ⎜ 1 − ⎟, p⎠, ⎝, (d) More than 1, , 4. What is the probability of a sure event?, 1, (c) 1, (d) Less than 1, (a) 0, (b), 2, 5. Which of the following cannot be the probability of an event? [CBSE 2011], 3, (c) 25%, (d) 0.3, (a) 1.5, (b), 5, 6. A number is selected at random from the numbers 1 to 30. What is the, probability that the selected number is a prime number?, [CBSE 2014], 2, 1, 1, 11, (b), (c), (d), (a), 3, 6, 3, 30
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922, , Secondary School Mathematics for Class 10, , 7. The probability that a number selected at random from the numbers, [CBSE 2014], 1, 2, 3, ..., 15 is a multiple of 4, is, 4, 2, 1, 1, (a), (b), (c), (d), 15, 15, 5, 3, 8. A box contains cards numbered 6 to 50. A card is drawn at random from, the box. The probability that the drawn card has a number which is a, [CBSE 2013], perfect square is, 1, 2, 4, 1, (b), (c), (d), (a), 45, 15, 45, 9, 9. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at, random from the box, the probability that it bears prime number less, [CBSE 2013], than 23 is, 7, 1, 4, 8, (b), (c), (d), (a), 90, 9, 45, 89, 10. Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at, random from the bag. The probability of getting a card with a prime, [CBSE 2012], number is, 1, 2, 3, 5, (a), (b), (c), (d), 2, 5, 10, 9, 11. One ticket is drawn at random from a bag containing tickets numbered, 1 to 40. The probability that the selected ticket has a number, which is a, [CBSE 2013C], multiple of 7, is, 1, 1, 1, 7, (a), (b), (c), (d), 7, 8, 5, 40, 12. Which of the following cannot be the probability of an event?, 1, 7, (a), (b) 0.3, (c) 33%, (d), 3, 6, 13. If the probability of winning a game is 0.4 then the probability of losing, it, is, 1, (c) 0.6, (d) none of these, (a) 0.96, (b), 0.4, 14. If an event cannot occur then its probability is, 1, 3, (a) 1, (b), (c), 2, 4, , (d) 0, , 15. There are 20 tickets numbered as 1, 2, 3, ..., 20 respectively. One ticket is, drawn at random. What is the probability that the number on the ticket, drawn is a multiple of 5?, 1, 1, 2, 3, (b), (c), (d), (a), 4, 5, 5, 10
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Probability, , 923, , 16. There are 25 tickets numbered as 1, 2, 3, 4, ..., 25 respectively. One ticket, is drawn at random. What is the probability that the number on the, ticket is a multiple of 3 or 5?, 2, 11, 12, 13, (b), (c), (d), (a), 5, 25, 25, 25, 17. Cards, each marked with one of the numbers 6, 7, 8, …, 15, are placed in, a box and mixed thoroughly. One card is drawn at random from the, box. What is the probability of getting a card with number less than 10?, [CBSE 2009C], , 3, (a), 5, , 1, (b), 3, , 1, (c), 2, , 2, (d), 5, , 18. A die is thrown once. The probability of getting an even number is, [CBSE 2013], , 1, (a), 2, , 1, (b), 3, , 1, (c), 6, , 5, (d), 6, , 19. The probability of throwing a number greater than 2 with a fair die is, [CBSE 2011], , 2, (a), 5, , 5, (b), 6, , 1, (c), 3, , 2, (d), 3, , 20. A die is thrown once. The probability of getting an odd number greater, than 3 is, [CBSE 2013C], 1, 1, 1, (b), (c), (d) 0, (a), 3, 6, 2, 21. A die is thrown once. The probability of getting a prime number is, [CBSE 2013], , 2, (a), 3, , 1, (b), 3, , 1, (c), 2, , 1, (d), 6, , 22. Two dice are thrown together. The probability of getting the same, number on both dice is, [CBSE 2012], 1, 1, 1, 1, (b), (c), (d), (a), 2, 3, 6, 12, 23. The probability of getting 2 heads, when two coins are tossed, is, [CBSE 2012], , (a) 1, , 3, (b), 4, , 1, (c), 2, , 1, (d), 4, , 24. Two dice are thrown together. The probability of getting a doublet is, [CBSE 2013], , 1, (a), 3, , 1, (b), 6, , 1, (c), 4, , 2, (d), 3
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924, , Secondary School Mathematics for Class 10, , 25. Two coins are tossed simultaneously. What is the probability of getting, at most one head?, 1, 1, 2, 3, (b), (c), (d), (a), 4, 2, 3, 4, 26. Three coins are tossed simultaneously. What is the probability of, getting exactly two heads?, 1, 1, 3, 3, (b), (c), (d), (a), 2, 4, 8, 4, 27. In a lottery, there are 8 prizes and 16 blanks. What is the probability of, getting a prize?, 1, 1, 2, (b), (c), (d) None of these, (a), 2, 3, 3, 28. In a lottery, there are 6 prizes and 24 blanks. What is the probability of, not getting a prize?, 3, 3, 4, (a), (b), (c), (d) None of these, 4, 5, 5, 29. A box contains 3 blue, 2 white and 4 red marbles. If a marble is drawn, at random from the box, what is the probability that it will not be a, [CBSE 2009C], white marble?, 1, 4, 7, 2, (b), (c), (d), (a), 3, 9, 9, 9, 30. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at, random. What is the probability of getting a black ball?, [CBSE 2008], 2, 3, 1, (b), (c), (d) None of these, (a), 5, 5, 10, 31. A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at, random. What is the probability that the ball drawn is not black?, 8, 2, 13, 1, (b), (c), (d), (a), 15, 15, 15, 3, 32. A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at, random. What is the probability that the ball drawn is neither black, nor white?, 1, 1, 1, 3, (b), (c), (d), (a), 4, 2, 3, 4, 33. A card is drawn at random from a well-shuffled deck of 52 cards. What, is the probability of getting a black king?, [CBSE 2009C], (a), , 1, 13, , (b), , 1, 26, , (c), , 2, 39, , (d) None of these
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Probability, , 925, , 34. From a well-shuffled deck of 52 cards, one card is drawn at random., What is the probability of getting a queen?, 1, 1, 4, (b), (c), (d) None of these, (a), 13, 26, 39, 35. One card is drawn at random from a well-shuffled deck of 52 cards., What is the probability of getting a face card?, 1, 3, 3, 4, (b), (c), (d), (a), 26, 26, 13, 13, 36. One card is drawn at random from a well-shuffled deck of 52 cards., What is the probability of getting a black face card?, 1, 3, 3, 3, (b), (c), (d), (a), 26, 26, 13, 14, 37. One card is drawn at random from a well-shuffled deck of 52 cards., What is the probability of getting a 6?, 3, 1, 1, (b), (c), (d) None of these, (a), 26, 52, 13, ANSWERS (MCQ), , 1. (c), , 2. (b), , 3. (b), , 4. (c), , 5. (a), , 6. (c), , 7. (c), , 8. (d), , 9. (c) 10. (a), , 11. (b) 12. (d) 13. (c) 14. (d) 15. (b) 16. (c) 17. (d) 18. (a) 19. (d) 20. (b), 21. (c) 22. (c) 23. (d) 24. (b) 25. (d) 26. (c) 27. (b) 28. (c) 29. (c) 30. (b), 31. (c) 32. (c) 33. (b) 34. (a) 35. (c) 36. (b) 37. (c), HINTS TO SOME SELECTED QUESTIONS, 2. P(occurrent of an event) = p, ⇒ P(non-occurence of this event) = ( 1 − p )., 5. The probability of an event cannot be greater than 1., 6. Prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29., Their number is 10., 10 1, = ⋅, 30 3, 7. Total number of given numbers = 15., ∴, , P(getting a prime number) =, , From given numbers, the multiples of 4 are 4, 8, 12., They are 3 in number., ∴, , P(getting a multiple of 4) =, , 3 1, = ⋅, 15 5, , 8. Given numbers are 6, 7, 8, 9, ..., 50., Number of these numbers = 50 − 5 = 45.
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926, , Secondary School Mathematics for Class 10, Perfect square numbers from these are 3 2 , 4 2 , 5 2 , 6 2 , 7 2 ., Their number is 5., , ∴, , P(getting a perfect square number) =, , 5, 1, = ⋅, 45 9, , 9. Total number of discs = 90., Prime number less than 23 are 2, 3, 5, 7, 11, 13, 17, 19., Their number is 8., P(getting a prime number less than 23) =, , 8, 4, =, ⋅, 90 45, , 10. Total number of cards = 10., Prime numbers from given numbers are 2, 3, 5, 7, 11., Their number is 5., ∴, , P(getting a prime number) =, , 5, 1, = ⋅, 10 2, , 11. Total number of tickets = 40., Tickets bearing the numbers as multiple of 7 bear the numbers 7, 14, 21, 28, 35., Their number is 5., 5, 1, = ⋅, 40 8, 13. P(losing the game) = 1 − P(winning the game) = ( 1 − 0.4 ) = 0.6., ∴, , P(getting a multiple of 7) =, , 14. If an event cannot occur then its probability is 0., 15. Total number of tickets = 20., Multiples of 5 are 5, 10, 15, 20., Their number is 4., ∴, , P(getting a multiple of 5) =, , 4 1, = ⋅, 20 5, , 16. Total number of tickets = 25., Multiples of 3 or 5 are 3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25., Number of these numbers = 12., ∴, , P(getting a multiple of 3 or 5) =, , 12, ⋅, 25, , 17. Total number of cards = 15 − 5 = 10., Number of cards with number less than 10 = 4., ∴, , P(getting a card with number less than 10) =, , 4 2, = ⋅, 10 5, , 18. Number of all possible outcomes = 6., Even numbers are 2, 4, 6. Their number is 3., 3 1, ∴ P(getting an even number) = = ⋅, 6 2, 19. Number of all possible outcomes = 6., Numbers greater than 2 are 3, 4, 5, 6. Their numbr is 4., 4 2, ∴ P(getting a number greater than 2) = = ⋅, 6 3
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Probability, , 927, , 20. Number of all possible outcomes = 6., Odd number greater than 3 is 5 only. Its number is 1., 1, ∴ P(getting an odd number greater than 3) = ⋅, 6, 21. Number of all possible outcomes = 6., Prime numbers are 2, 3, 5. Their number is 3., 3 1, ∴ P(getting a prime number) = = ⋅, 6 2, 22. Total number of all possible outcomes = 36., Getting same number on both dice means getting, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)., Their number is 6., ∴, , P(getting the same number on both dice) =, , 6, 1, = ⋅, 36 6, , 23. All possible outcomes are HH, HT, TH, TT. Their number is 4., Getting 2 heads, means getting HH. Its number is 1., 1, ∴ P(getting 2 heads) = ⋅, 4, 24. Number of all possible outcomes = 36., The doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)., Their number is 6., ∴, , P(getting a doublet) =, , 6, 1, = ⋅, 36 6, , 25. When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT., Their number is 4., All favourable outcomes are TT, TH, HT. Their number is 3., 3, ∴ P(getting at most 1 head) = ⋅, 4, 26. When 3 coins are tossed, all possible outcomes are HHH, HHT, HTH, THH, HTT, THT,, TTH, TTT., Their number is 8., Favourable outcomes are HHT, HTH, THH. Their number is 3., 3, ∴ P(getting exactly 2 heads) = ⋅, 8, 27. Total number of lottery tickets = 8 + 16 = 24., Number of prizes = 8., ∴, , P(getting a prize) =, , 8, 1, = ⋅, 24 3, , 28. Total number of tickets = 6 + 24 = 30., Number of blanks = 24., ∴, , P(not getting a prize) =, , 24 4, = ⋅, 30 5
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928, , Secondary School Mathematics for Class 10, , 29. Total number of marbles = 3 + 2 + 4 = 9., Number of non-white marbles = 3 + 4 = 7., 7, ∴ P(getting a non-white marble) = ⋅, 9, 30. Total number of balls in the bag = 4 + 6 = 10., Number of black balls = 6., ∴, , P(getting a black ball) =, , 6, 3, = ⋅, 10 5, , 31. Total number of balls in the bag = 8 + 2 + 5 = 15., Number of non-black balls = 8 + 5 = 13., 13, P(getting a non-black ball) =, ⋅, 15, 32. Total number of balls in the bag = 3 + 4 + 5 = 12., Number of non-black and non-white balls = 4., ∴, , P(getting a ball which is neither black nor white) =, , 33. Number of all possible outcomes = 52., Number of black kings = 2., ∴, , P(getting a black king) =, , 2, 1, =, ⋅, 52 26, , 34. Number of all possible outcomes = 52., Number of queens = 4., ∴, , P(getting a queen) =, , 4, 1, =, ⋅, 52 13, , 35. Total number of cards = 52., Number of face cards = 12, (4 kings + 4 queens + 4 jacks)., 12, 3, P(getting a face card) =, =, ⋅, 52 13, , ∴, , 36. Total number of cards = 52., Number of black face cards = 6, (2 kings +2 queens + 2 jacks)., 6, 3, =, ⋅, ∴ P(getting a face card) =, 52 26, 37. Total number of cards = 52., Number of 6 s = 4., ∴, , P(getting a 6) =, , 4, 1, =, ⋅, 52 13, , _, , 4, 1, = ⋅, 12 3
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Sample Paper I, , 929, , SAMPLE PAPER I, Time: 3 hours, , 80 Marks, , GENERAL INSTRUCTIONS, , 1. All questions are compulsory., 2. The paper consists of 30 questions divided into four sections:, A, B, C and D., 3. (i) Section A consists of 6 Very-Short-Answer (VSA) questions of, 1 mark each., (ii) Section B consists of 6 Short-Answer (SA) questions of, 2 marks each., (iii) Section C consists of 10 Short-Answer (SA) questions of, 3 marks each., (iv) Section D consists of 8 Long-Answer (LA) questions of, 4 marks each., 4. Use of calculator is not permitted., , SECTION A, 1. What is the distance between the points A(sin cos , 0) and, B(0, sin cos ) ?, 2. The angle of depression of a car standing on the ground from the top, of a 66 m tower, is 30. Find the distance of the car from the base of, the tower., 3. A rational number in its decimal expansion is 554.6023. What can you, say about the prime factors of q, when this number is expressed in the, p, form q ?, 4. If sin , , 1, then find the value of 2 cot 2 2., 3, , 5. In a family of 3 children, find the probability of having at least one boy., 929
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930, , Secondary School Mathematics for Class 10, , 6. If and are the zeros of the quadratic polynomial f (x) 5x 2 7x 1,, 1 1, find the value of ·, , , SECTION B, 7. The length of the minute hand of a clock is 5 cm. Find the area swept by, the minute hand during the time from 4:05 a.m. to 4:40 a.m., 8. Prove that three times the square of any side of an equilateral triangle, is equal to four times the square of the altitude., 9. In the given figure, O is the centre of the circle. PA and PB are tangent, segments. Show that the quadrilateral AOBP is cyclic., , A, O, , B, , P, , 10. Find the lengths of the medians of a 3ABC whose vertices are A(7, –3),, B(5, 3) and C(3, –1)., 11. If p, q are prime positive integers, prove that p q is an irrational, number., 12. Determine the value of k so that the following system of linear equations, has no solution:, (3k 1) x 3y 2 0,, (k 2 1) x (k 2) y 5 0., , SECTION C, 13. Show that the square of any positive integer cannot be of the form 5q 2, or 5q 3 for any integer q., 14. Prove that, , tan A cot A , 1 tan A cot A 1 sec A · cosec A., 1 cot A 1 tan A, , 15. A straight highway leads to the foot of a tower. A man standing at the, top of the tower observes a car at angle of depression of 30°, which is, approaching the foot of the tower with a uniform speed. Six seconds, later, the angle of depression of the car is found to be 60°. Find the, further time taken by the car to reach the foot of the tower.
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Sample Paper I, , 931, , 16. If the points A(1, –2), B(2, 3), C(–3, 2) and D(–4, –3) are the vertices of a, parallelogram ABCD then taking AB as the base find the height of the, parallelogram., 17. A plane left 30 minutes later than the scheduled time and in order to, reach its destination 1500 km away in time, it has to increase its speed, by 250 km/hr from its usual speed. Find its usual speed., 18. Obtain all the zeros of the polynomial f (x) 3x 4 6x 3 2x 2 10x 5 if, 5, 5, ·, two of its zeros are, and , 3, 3, 19. Find the mean, median and mode of the following data:, Classes, , 0–20, , 20–40, , 40–60, , 60–80, , Frequency, , 6, , 8, , 10, , 12, , 80–100 100–120 120–140, 6, , 5, , 3, , 20. Two customers are visiting a particular shop in the same week (Monday, to Saturday). Each is equally likely to visit the shop on any one day as, on another. What is the probability that both will visit the shop on, (i) the same day?, , (ii) different days?, , (iii) consecutive days?, 21. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at, P and Q intersect at a point T. Find the length TP., 22. Solve the following system of equations in x and y:, (a b) x (a b) y a 2 2ab b 2,, (a b)(x y) a 2 b 2., , SECTION D, 23. An open metal bucket is in the shape of a frustum of a cone, mounted, on a hollow cylindrical base made of the same metallic sheet. The, diameters of the two circular ends of the bucket are 45 cm and 25 cm,, the total vertical height of the bucket is 40 cm and that of the cylindrical, base is 6 cm. Find the area of the metallic sheet used to make the bucket., Also, find the volume of water the bucket can hold, in litres., 24. It can take 12 hours to fill a swimming pool using two pipes. If the, pipe of larger diameter is used for 4 hours and the pipe of smaller, diameter for 9 hours, only half of the pool can be filled. How long, would it take for each pipe to fill the pool separately?
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932, , Secondary School Mathematics for Class 10, , 25. If the angle of elevation of a cloud from a point h metres above a lake be, and the angle of depression of its reflection in the lake be , prove that, 2h sec , ·, the distance of the cloud from the point of observation is, tan tan , 26. The following table gives the yield per hectare of wheat of 100 farms of, a village:, Yield in kg/hectare, , 50–55, , 55–60, , 60–65, , 65–70, , 70–75, , 75–80, , Number of farms, , 2, , 8, , 12, , 24, , 38, , 16, , Change the above distribution to ‘more than type’ distribution and, draw its ogive., 27. Sides of a triangular field are 15 m, 16 m and 17 m. With the three, corners of the field a cow, a buffalo and a horse are tied separately with, ropes of length 7 m each to graze in the field. Find the area of the field, which cannot be grazed by the three animals., 28. The radius of incircle of a triangle is 4 cm and the segments into which, one side is divided by the point of contact are 6 cm and 8 cm. Determine, the other two sides of the triangle., 29. Two poles of height a metres and b metres are p metres apart. Prove that, the height of the point of intersection of the lines joining the top of each, ab, pole to the foot of the opposite pole is given by, metres., ab, 30. In the centre of a rectangular lawn of dimensions 50 m # 40 m, a, rectangular pond has to be constructed so that the area of the grass, surrounding the pond would be 1184 m 2. Find the length and breadth, of the pond., ANSWERS, , 1., , 2, , 2. 22 3 m, , 3. Since 554.6023 is a terminating decimal number so q must be of the, , form 2 m # 5 n, where m, n are natural numbers., 5, 7, 4. 18 5., 6. 7, 7. 45 cm 2, 10. 5 units; 5 units; 10 units, 6, 8, 24, 12. k 1, 15. 3 seconds 16., units 17. 750 km/hr, 26, 5 5 , 18., ,, , 1, 1 19. Mean = 61.66, median = 62.4, mode = 65, 3, 3
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Sample Paper I, , 20. (i), , 1, 6, , (ii), , 5, 6, , (iii), , 5, 36, , 21., , 20, cm, 3, , 933, , 22. x a b, y , , 23. 4860.9 cm 2, 33.62 litres, , 24. 20 hours, 30 hours, , 27. (24 21 77) m 2, , 28. 15 cm, 13 cm, , 30. Length = 34 m, Breadth = 24 m, , , , 2ab, ab
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934, , Secondary School Mathematics for Class 10, , SAMPLE PAPER II, Time: 3 hours, , 80 Marks, , GENERAL INSTRUCTIONS, , 1. All questions are compulsory., 2. The paper consists of 30 questions divided into four sections:, A, B, C and D., 3. (i) Section A consists of 6 Very-Short-Answer (VSA) questions of, 1 mark each., (ii) Section B consists of 6 Short-Answer (SA) questions of, 2 marks each., (iii) Section C consists of 10 Short-Answer (SA) questions of, 3 marks each., (iv) Section D consists of 8 Long-Answer (LA) questions of, 4 marks each., 4. Use of calculator is not permitted., , SECTION A, 1. Evaluate: sin 35° · sin 55° – cos 35° · cos 55°., 2. The decimal expansion of the rational number, after how many places of decimals?, , 83, will terminate, 23 # 54, , 3. If the point P(x, y) is equidistant from A(5, 1) and B(–1, 5) then find the, relation between x and y., 4. If and are the zeros of the polynomial f (x) x 2 5x k such that, 1, find the value of k., 5. The tops of two towers of heights x and y, standing on level ground,, subtend angles of 30° and 60° respectively at the centre of the line, joining their feet. Find x : y., 934
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Sample, Sample Paper, Paper II, I, , 935, , 6. A die is rolled twice. Find the probability that 4 will come up exactly, one time., , SECTION B, 7. Find the value of a and b for which the following system of equations, has infinitely many solutions:, x 2y 1,, (a b) x (a b) y a b 2., 8. Use Euclid’s division algorithm to find the HCF of 441, 567 and 693., 9. Find k so that the point (–4, 6) lies on the line segment joining A(k, 10), and B(3, –8). Also, find the ratio in which P divides AB., 10. Prove that the segment joining the points of contact of two parallel, tangents passes through the centre., 11. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find, the area of the triangle., 12. Find the area of the shaded region in the figure given below where, radii of the two concentric circles with centre O are 7 cm and 14 cm, respectively and +AOC 40c., , O, B 40º, D, , A, C, , SECTION C, 13. Three unbiased coins are tossed simultaneously. Find the probability of, getting, (i) exactly two heads;, , (ii) at least two heads;, , (iii) at most 2 heads., 14. Prove that the angle between two tangents drawn from an external, point to a circle is supplementary to the angle subtended by the line, segments joining the points of contact at the centre., 15. Solve the following system of equations in x and y:, ax by c,, bx ay 1 c.
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936, , Secondary School Mathematics for Class 10, , 16. Use Euclid’s division lemma to show that the cube of any positive, integer is either of the form 9m, 9m + 1 or 9m + 8 for some integer m., 1·, 2, 18. From the top of a building AB, 60 m high, the angles of depression of, the top and bottom of a vertical lamp-post CD are observed to be 30°, and 60° respectively. Find, 17. If 1 sin 2 3 sin · cos , prove that tan 1 or, , (i) the horizontal distance between AB and CD,, (ii) the height of the lamp-post., 19. Find the area of a parallelogram ABCD if three of its vertices are, A(2, 4), B(2 3 , 5) and C(2, 6)., 20. A train travelling at a uniform speed for 360 km would have taken, 48 minutes less to travel the same distance if its speed were 5 km/hr, more. Find the original speed of the train., 21. If two zeros of the polynomial f (x) x 4 6x 3 26x 2 138x 35 are, 2 ! 3 , find other zeros., 22. The following table shows the marks obtained by 100 students of a class, in a school during a particular academic session. Find the modal marks., Marks, Number of, students, , Less, Less, Less, Less, Less, Less, Less, Less, than 10 than 20 than 30 than 40 than 50 than 60 than 70 than 80, 7, , 21, , 34, , 46, , 66, , 77, , 92, , 100, , SECTION D, 23. There is a square field whose side is 44 m. A square flower bed is, prepared in its centre leaving a gravel path all around the flower bed., The total cost of laying the flower bed and gravelling the path at ` 2.75, and ` 1.50 per square metre respectively, is ` 4904. Find the width of the, gravel path., 24. Prove that the ratio of the areas of two similar triangles is equal to the, ratio of the squares of any two corresponding sides., 25. From an external point P, a tangent PT and a line segment PAB is, drawn to a circle with centre O. ON is perpendicular on the chord AB., Prove that, (i) PA · PB PN 2 AN 2, (ii) PN 2 AN 2 OP 2 OT 2, (iii) PA · PB PT 2
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Sample, Sample Paper, Paper III, , 937, , 26. ABCD is a trapezium with AB DC, AB 18 cm, DC 32 cm and the, distance between AB and DC is 14 cm. If arcs of equal radii 7 cm with, centres A, B, C and D are drawn find the area of the remaining region,, i.e., region of the trapezium excluding the circular arcs., 27. The table given below shows the frequency distribution of the scores, obtained by 200 candidates in an entrance examination:, Scores, , 200–, 250, , 250–, 300, , 300–, 350, , 350–, 400, , 400–, 450, , 450–, 500, , 500–, 550, , 550–, 600, , No. of candidates, , 30, , 15, , 45, , 20, , 25, , 40, , 10, , 15, , Draw cumulative frequency curve using ‘less than series’., 28. A window of a house is h metres above the ground. From the window,, the angles of elevation and depression of the top and bottom of another, house situated on the opposite side of the lane are found to be and, respectively. Prove that the height of the house is h(1 tan · tan ), metres., 29. 8 men and 12 boys can finish a piece of work in 10 days while 6 men, and 8 boys can finish it in 14 days. Find the time taken by one man, alone and that by one boy alone to finish the work., 30. From a solid cylinder whose height is 15 cm and diameter 16 cm, a, conical cavity of the same height and same diameter is hollowed out., Find the total surface area of the remaining solid. [Use 3.14.], ANSWERS, , 1. 0, , 2. 4, , 3. 2x 3y, , 4. 6, , 5, 6., 18, , 7. a 3, b 1, , 8. 63, , 9. k 6; 2 : 7, , 154, cm 2, 3, , 13. (i), , 11. 150 cm 2, 15. x , , 12., , 3, 8, , (ii), , c b , c a, y, a b a2 b2, a b a2 b2, , 19. 2 3 sq units, , 20. 45 km/hr, , 23. 2 metres, , 26. 196 cm, , 1, 2, , (iii), , (ii) 40 m, , 21. 7, –5, , 22. 44.71, , 29. Time taken by one man alone = 140 days and, , time taken by one boy alone = 280 days, , , , 7, 8, , 18. (i) 34.64 m, , 2, , 30. 443.14 cm 2, , 5. 1 : 3
