Page 2 :
www.tntextbooks.in, , Government of TamilNadu, First Edition, , -, , 2018, , Revised Edition -, , 2019, , (Published under new syllabus), , NOT FOR SALE, , Content Creation, , The wise, possess all, , State Council of Educational, Research and Training, © SCERT 2018, , Printing & Publishing, , Tamil NaduTextbook and Educational, Services Corporation, www.textbooksonline.tn.nic.in, (ii), , PreliminaryT-Combine.indd 2, , 26-12-2019 11:55:59
Page 3 :
www.tntextbooks.in, , CONTENTS, Chapter, No., 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 3, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 4, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, , CHAPTER, SET LANGUAGE, Introduction, Set, Representation of a Set, Types of Sets, Set Operations, Properties of Set Operations, De Morgan’s Laws, Application on Cardinality of Sets:, REAL NUMBERS, Introduction, Rational Numbers, Irrational Numbers, Real Numbers, Radical Notation, Surds, Rationalisation of Surds, Scientific Notation, ALGEBRA, Introduction, Polynomials, Remainder Theorem, Algebraic Identities, Factorisation, Division of Polynomials, Greatest Common Divisor (GCD), Linear Equation in Two Variables, GEOMETRY, Introduction, Types of Angles, Quadrilaterals, Parts of a Circle, Properties of Chords of a Circle, Cyclic Quadrilaterals, Practical Geometry, , Page, No., 1, 1, 2, 3, 6, 12, 19, 24, 29, 41, 41, 42, 45, 55, 57, 60, 68, 70, 78, 78, 81, 92, 97, 102, 108, 114, 116, 138, 138, 139, 144, 159, 162, 170, 174, , MONTH, , June, , July, , August, , September, , July, August, October, November, , (iii), , PreliminaryT-Combine.indd 3, , 26-12-2019 11:55:59
Page 4 :
www.tntextbooks.in, , 5, , COORDINATE GEOMETRY, , 185, , 5.1, , Mapping the Plane, , 185, , 5.2, , Devising a Coordinate System, , 187, , 5.3, , Distance between any Two Points, , 193, , 5.4, , The Mid-point of a Line Segment, , 203, , 5.5, , Points of Trisection of a Line Segment, , 208, , 5.6, , Section Formula, , 210, , 5.7, , The Coordinates of the Centroid, , 214, , TRIGONOMETRY, , 222, , 6.1, , Introduction, , 222, , 6.2, , Trigonometric Ratios of Some Special Angles, , 228, , 6.3, , Trigonometric Ratios for Complementary Angles, , 233, , 6.4, , Method of using Trigonometric Table, , 235, , MENSURATION, , 248, , 7.1, , Introduction, , 248, , 7.2, , 249, , 7.4, , Heron’s Formula, Application of Heron’s Formula in Finding Areas, of Quadrilaterals, Surface Area of Cuboid and Cube, , 7.5, , Volume of Cuboid and Cube, , 258, , 8, , STATISTICS, , 266, , 8.1, , Introduction, , 266, , 8.2, , Collection of Data, , 267, , 8.3, , Measures of Central Tendency, , 269, , 8.4, , Arithmetic Mean, , 269, , 8.5, , Median, , 278, , 8.6, , Mode, , 284, , PROBABILITY, , 291, , 9.1, , Introduction, , 291, , 9.2, , Basic Ideas, , 292, , 9.3, , Classical Approach, , 295, , 9.4, , Empirical Approach, , 296, , 9.5, , Types of Events, , 299, , ANSWERS, , 305, , 6, , 7, , 7.3, , 9, , 251, , October, , November, , December, , January, , 253, , February, , March, , (iv), , PreliminaryT-Combine.indd 4, , 26-12-2019 11:55:59
Page 5 :
www.tntextbooks.in, , Captions, used in this, Textbook, , எண்ெணன்ப ஏைன எழுத்ெதன்ப இவ்விரண்டும், கண்ெணன்ப வாழும் உயிர்க்கு, - குறள் 392, Numbers and letters, they are known as, eyes to humans, they are., Kural 392, , Learning Outcomes, To transform the classroom processes into, learning centric with a set of bench marks, Note, To provide additional inputs for students in the, content, Activity / Project, To encourage students to involve in activities to, learn mathematics, ICT Corner, To encourage learner’s understanding of content through, application of technology, Thinking Corner, To kindle the inquisitiveness of students in learning mathematics. To, make the students to have a diverse thinking, Points to Remember, To recall the points learnt in the topic, Multiple Choice Questions, To provide additional assessment items on the content, Progress Check, Self evaluation of the learner’s progress, Exercise, To evaluate the learners’ in, understanding the content, , “The essence of mathematics is not to make simple things complicated, but to make complicated things simple” -S. Gudder, Let's use the QR code in the text books! How?, • Download the QR code scanner from the Google PlayStore/ Apple App Store into your smartphone, • Open the QR code scanner application, • Once the scanner button in the application is clicked, camera opens and then bring it closer to the QR code in the text, book., • Once the camera detects the QR code, a url appears in the screen.Click the url and go to the content page., , (v), , PreliminaryT-Combine.indd 5, , 26-12-2019 11:56:00
Page 6 :
www.tntextbooks.in, , SYMBOLS, , =, , equal to, , |||ly, , similarly, , !, , not equal to, , T, , symmetric difference, , 1, , less than, , N, , natural numbers, , #, , less than or equal to, , W, , whole numbers, , 2, , greater than, , Z, , integers, , $, , greater than or equal to, , R, , real numbers, , ., , equivalent to, , 3, , triangle, , j, , union, , +, , angle, , k, , intersection, , =, , perpendicular to, , U, , universal Set, , ||, , parallel to, , d, , belongs to, , (, , implies, , z, , does not belong to, , `, , therefore, , 1, , proper subset of, , a, , since (or) because, , 3, , subset of or is contained in, , Y, 1, , not a proper subset of, , -, , approximately equal to, , M, , not a subset of or is not contained in, , | (or) :, , such that, , Al (or) A c, , complement of A, , / (or) ,, , congruent, , Q (or) { }, , empty set or null set or void set, , /, , identically equal to, , n(A), , number of elements in the set A, , p, , pi, , P(A), , power set of A, , !, , plus or minus, , ∑, , summation, , P(E), , probability of an event E, , E-book, , absolute value, , Evaluation, , DIGI Links, , (vi), , PreliminaryT-Combine.indd 6, , 26-12-2019 11:56:01
Page 7 :
www.tntextbooks.in, , Z, W, , 1, , B, , A, x, , a, , r, z, , N, , b, y, , c, , C, , SET LANGUAGE, , A set is a many that allows itself to thought of as a one, -Georg Cantor, , The theory of sets was developed by German, mathematician Georg Cantor. Today it is used, in almost every branch of Mathematics., , In, , Mathematics, sets are convenient because all, mathematical structures can be regarded as sets., Georg Cantor, (AD (CE) 1845 - 1918), , Learning Outcomes, To describe and represent a set in different forms., To identify different types of sets., To understand and perform set operations and apply this in Venn diagram., To know the commutative, associative and distributive properties among sets., To understand and verify De Morgan’s laws., To use set language in solving life oriented word problems., , 1.1 Introduction, In our daily life, we often deal with collection of objects like books, stamps, coins,, etc. Set language is a mathematical way of representing a collection of objects., Let us look at the following pictures. What do they represent?, Here, Fig.1.1 represents a collection of fruits and Fig. 1.2 represents a collection of, house- hold items., Set Language, , 1-SET LANGUAGE.indd 1, , 1, , 26-12-2019 13:31:59
Page 8 :
www.tntextbooks.in, , We observe in these cases, our, attention turns from one individual, object to a collection of objects based on, their characteristics. Any such collection, is called a set., , 1.2 Set, A set is a well-defined collection, of objects., , Fig. 1.1, , Fig. 1.2, , Here “well-defined collection of objects” means that given a specific object it must be, possible for us to decide whether the object is an element of the given collection or not., The objects of a set are called its members or elements., , For example,, 1., 2., 3., , The collection of all books in a District Central Library., The collection of all colours in a rainbow., The collection of prime numbers., , We see that in the adjacent box,, statements (1), (2), and (4) are well defined Which of the following are sets ?, and therefore they are sets. Whereas (3) 1. Collection of Natural numbers., and (5) are not well defined because the 2. Collection of English alphabets., words good and beautiful are difficult to 3. Collection of good students in a class., agree on. I might consider a student to be 4. Collection of States in our country., good and you may not. I might consider 5. Collection of beautiful flowers in a garden., the Jasmine is the beautiful flower but you, may not. So we will consider only those collections to be sets where there is no such ambiguity., Therefore (3) and (5) are not sets., Activity - 1, Discuss and give as many examples of collections from your daily life situations, which, are sets and which are not sets., Note, z Elements of a set are listed only once., , T, z The order of listing the elements of the set does not change the set., , For example, the collection 1,2,3,4,5,6,7,8, … as well as the collection 1, 3, 2, 4, 5, 7, 6, 8, …, are the same though listed in different order. Since it is necessary to know whether an object, is an element in the set or not, we do not want to list that element many times., 2, , 1-SET LANGUAGE.indd 2, , 9th Standard Mathematics, , 26-12-2019 13:32:00
Page 9 :
www.tntextbooks.in, , Notation, A set is usually denoted by capital letters of the English Alphabets A, B, P, Q, X, Y, etc., The elements of a set is written within curly brackets “{ }”, If x is an element of a set A or x belongs to A, we write x ! A., If x is not an element of a set A or x does not belongs to A, we write x g A., , For example,, Consider the set A = {2,3,5,7} then, 2 is an element of A; we write 2 ! A, 5 is an element of A; we write 5 ! A, 6 is not an element of A; we write 6 g A, Example 1.1, Consider the set A = {Ashwin, Murali Vijay, Vijay Shankar, Badrinath }., Fill in the blanks with the appropriate symbol ! or g ., (i) Murali Vijay ______ A. (ii) Ashwin ______ A., (iv) Ganguly ______ A., , (iii) Badrinath ______A., , (v) Tendulkar ______ A, , Solution, (i) Murali Vijay ! A. (ii) Ashwin ! A, (iv) Ganguly g A., , (iii) Badrinath ! A, , (v) Tendulkar g A., , 1.3 Representation of a Set, The collection of odd numbers can be described in many ways:, (1) “The set of odd numbers” is a fine description, we understand it well., (2) It can be written as {1, 3, 5, …}, (3) Also, it can be said as the collection of all numbers x where x is an odd number., All of them are equivalent and useful. For instance,the two descriptions “The collection, of all solutions to the equation x–5 = 3” and {8} refer to the same set., A set can be represented in any one of the following three ways or forms:, , 1.3.1 Descriptive Form, In descriptive form, a set is described in words., , For example,, (i) The set of all vowels in English alphabets., (ii) The set of whole numbers., Set Language, , 1-SET LANGUAGE.indd 3, , 3, , 26-12-2019 13:32:01
Page 10 :
www.tntextbooks.in, , 1.3.2 Set Builder Form or Rule Form, In set builder form, all the elements are described by a rule., , For example,, , Note, , A = {x : x is a vowel in English alphabets}, , (i), , The symbol ‘:’ or ‘|’ stands for, “such that”., , (ii) B = {x|x is a whole number}, , 1.3.3 Roster Form or Tabular Form, A set can be described by listing all the elements of the set., Note, , For example,, A = {a, e, i, o, u} , , (i), , (ii) B = {0,1,2,3,…}, Can this form of representation be possible always?, , Three dots (…) in the example (ii), is called ellipsis. It indicates that, the pattern of the listed elements, continues in the same manner., , Activity-2, Write the following sets in respective forms., S.No., , Descriptive Form, , Set Builder Form, , 1, , The set of all natural, numbers less than 10, , Roster Form, , {x : x is a multiple of 3,, x ! N}, , 2, 3, , {2,4,6,8,10}, , 4, , The set of all days in a week., , 5, , {…-3,-2,-1,0,1,2,3…}, , Example 1.2, Write the set of letters of the following words in Roster form, (i) ASSESSMENT, , (ii) PRINCIPAL, , Solution, , 4, , 1-SET LANGUAGE.indd 4, , (i) ASSESSMENT, , (ii) PRINCIPAL, , X= {A, S, E, M, N, T}, , Y={P, R, I, N, C, A, L}, , 9th Standard Mathematics, , 26-12-2019 13:32:01
Page 11 :
www.tntextbooks.in, , Exercise 1.1, 1. Which of the following are sets?, (i), , The collection of prime numbers upto 100., , (ii) The collection of rich people in India., (iii) The collection of all rivers in India., (iv) The collection of good Hockey players., 2., , 3., , List the set of letters of the following words in Roster form., (i) INDIA , , (ii) PARALLELOGRAM, , (iii) MISSISSIPPI , , (iv) CZECHOSLOVAKIA, , Consider the following sets A = {0, 3, 5, 8}, B = {2, 4, 6, 10} and C = {12, 14,18, 20}., (a) State whether True or False:, , , , (i) 18 ! C, , (ii) 6 g A, , , , (v) 5 ! B, , (vi) 0 ! B, , (iii) 14 g C, , (iv) 10 ! B, , (iii) 18 ____ B, , (iv) 4 _____ B, , (b) Fill in the blanks:, , 4., , (i) 3 ! ____, , (ii) 14 ! _____, , Represent the following sets in Roster form., (i), , A = The set of all even natural numbers less than 20., , 1, (ii) B = {y : y = 2n , n ! N , n ≤ 5}, (iii) C = {x : x is perfect cube, 27 < x < 216}, (iv) D = {x : x ! Z , –5 < x ≤ 2}, 5., , Represent the following sets in set builder form., (i), , B = The set of all Cricket players in India who scored double centuries in One, Day Internationals., , (ii) C = $ 12 , 23 , 43 , ... ., (iii) D = The set of all tamil months in a year., (iv) E = The set of odd Whole numbers less than 9., Set Language, , 1-SET LANGUAGE.indd 5, , 5, , 26-12-2019 13:32:01
Page 12 :
www.tntextbooks.in, , 6., , Represent the following sets in descriptive form., (i), , P ={ January, June, July}, , (ii) Q = {7,11,13,17,19,23,29}, (iii) R = {x : x ! N , x < 5}, (iv) S = {x : x is a consonant in English alphabets}, , 1.4 Types of Sets, There is a very special set of great interest: the empty collection ! Why should, one care about the empty collection? Consider the set of solutions to the equation, x2+1 = 0. It has no elements at all in the set of Real Numbers. Also consider all rectangles, with one angle greater than 90 degrees. There is no such rectangle and hence this describes, an empty set., So, the empty set is important, interesting and deserves a special symbol too., , 1.4.1 Empty Set or Null Set, A set consisting of no element is called the empty, set or null set or void set., It is denoted by Q or { }., , Thinking Corner, Are the sets {0} and {∅} empty, sets?, , For example,, (i), , A={x : x is an odd integer and divisible by 2}, , ` A={ } or Q, (ii) The set of all integers between 1 and 2., , 1.4.2.Singleton Set, A set which has only one element is called a singleton set., , For example,, , (i), , A = {x : 3 < x < 5, x ! N }, , (ii) The set of all even prime numbers., , 1.4.3 Finite Set, A set with finite number of elements is called a finite set., , 6, , 1-SET LANGUAGE.indd 6, , Note, An empty set has no, elements, so Q is a, finite set., , 9th Standard Mathematics, , 26-12-2019 13:32:02
Page 13 :
www.tntextbooks.in, , For example,, 1. The set of family members., 2. The set of indoor/outdoor games you play., 3. The set of curricular subjects you learn in school., 4., A = {x : x is a factor of 36}, , 1.4.4 Infinite Set, A set which is not finite is called an infinite set., , Thinking Corner, , Is the set of natural numbers, a finite set?, , For example,, (i), , {5,10,15,...}, , (ii) The set of all points on a line., , To discuss further about the types of sets, we need to know the cardinality of sets., Cardinal number of a set : When a set is finite, it is very useful to know how many, elements it has. The number of elements in a set is called the Cardinal number of the set., The cardinal number of a set A is denoted by n(A), Example 1.3, , If A = {1,2,3,4,5,7,9,11}, find n(A)., , Solution, A = {1,2,3,4,5,7,9,11}, Since set A contains 8 elements, n(A) = 8., , Thinking Corner, If A = {1, b, b, {4, 2},, {x, y, z}, d, {d}},, then n(A) is____, , 1.4.5 Equivalent Sets, Two finite sets A and B are said to be equivalent if they contain the same number of, elements. It is written as A ≈ B., If A and B are equivalent sets, then n(A) = n(B), , For example,, Consider A = { ball, bat} and, , B = {history, geography}., Here A is equivalent to B because n(A) = n(B) = 2., , Thinking Corner, Let A={x : x is a colour in, national flag of India} and, B={Red, Blue, Green}. Are, these two sets equivalent?, , Example 1.4, , Are P = { x : –3 ≤ x ≤ 0, x ! Z } and Q = The set of all prime factors, of 210, equivalent sets?, Solution, P = {–3, –2, –1, 0}, The prime factors of 210 are 2,3,5,and 7 and so, Q = {2, 3, 5, 7}, n(P) =4 and n(Q) = 4. Therefore P and Q are equivalent sets. , Set Language, , 1-SET LANGUAGE.indd 7, , 7, , 26-12-2019 13:32:02
Page 14 :
www.tntextbooks.in, , 1.4.6 Equal Sets, Two sets are said to be equal if they contain, exactly the same elements, otherwise they are said to, be unequal., , Thinking Corner, Are the sets ∅, {0}, {∅}, equal or equivalent?, , In other words, two sets A and B are said to be, equal, if, (i), , every element of A is also an element of B, , (ii) every element of B is also an element of A, , For example,, Consider the sets A = {1, 2, 3, 4} and B = {4, 2, 3, 1}, Since A and B contain exactly the same elements, A and B are equal sets., , Note, zz, , If A and B are equal sets, we write A = B., , zz, , If A and B are unequal sets, we write A≠B., , A set does not change, if one or more elements of the set are repeated., , For example, if we are given, A={a, b, c} and B={a, a, b, b, b, c} then, we write B = { a, b, c, }. Since, every element of A is, also an element of B and every element of B is also an element of A, the sets A and B are equal., , Example 1.5, Are A = {x : x ! N , 4 ≤ x ≤ 8} and, B = { 4, 5, 6, 7, 8} equal sets?, , Solution, A = { 4, 5, 6, 7, 8}, B = { 4, 5, 6, 7, 8}, A and B are equal sets., , Note, Equal sets are equivalent sets but equivalent, sets need not be equal sets. For example,, if A = { p,q,r,s,t} and B= { 4,5,6,7,8}. Here, n(A)=n(B), so A and B are equivalent but, not equal., , 1.4.7 Universal Set, A Universal set is a set which contains all the elements of all the sets under consideration, and is usually denoted by U., , For example,, (i) , If we discuss about elements in Natural numbers, then the universal set U is the, set of all Natural numbers. U={x : x ∈ N }., (ii) If A={earth, mars, jupiter}, then the universal set U is the planets of solar system., 8, , 1-SET LANGUAGE.indd 8, , 9th Standard Mathematics, , 26-12-2019 13:32:02
Page 15 :
www.tntextbooks.in, , 1.4.8 Subset, Let A and B be two sets. If every element of A is also an element of B, then A is called a, subset of B. We write A 3 B., A 3 B is read as “A is a subset of B”, Thus A 3 B, if a ! A implies a ! B., , Thinking Corner, , Z, W, , If A is not a subset of B, we write A M B, Clearly, if A is a subset of B, then n(A) ≤ n(B)., , N, , Since every element of A is also an element of B,, the set B must have at least as many elements as A, thus, n(A) ≤ n(B)., , Is W subset of N or Z ?, , The other way is also true. Suppose that n(A) > n(B), then A has more elements than, B, and hence there is at least one element in A that cannot be in B, so A is not a subset of B., , For example,, (i) {1} 3 {1,2,3}, , (ii) {2,4} M {1,2,3}, , Example 1.6, , Insert the appropriate symbol 3 or M in each blank to make a true, statement. (i) {10, 20, 30} ____ {10, 20, 30, 40}, (ii) {p, q, r} _____ {w, x, y, z}, Solution, (i), , , {10, 20, 30} ____ {10, 20, 30, 40}, Since every element of {10, 20, 30} is also an element of, {10, 20, 30, 40}, we get {10, 20, 30} 3 {10, 20, 30, 40}., , (ii) {p, q, r} _____ {w, x, y, z}, , , Since the element p belongs to {p, q, r} but does not belong to, {w, x, y, z}, shows that {p, q, r} M {w, x, y, z}., Activity-3, , Discuss with your friends and give examples of subsets of sets from your daily life, situation., Example 1.7, Write all the subsets of A = {a, b}., Solution, A= {a,b}, , , Subsets of A are Q ,{a}, {b} and {a, b}., Set Language, , 1-SET LANGUAGE.indd 9, , 9, , 26-12-2019 13:32:03
Page 16 :
www.tntextbooks.in, , Note, zz If A 3 B and B 3 A, then A=B., , In fact this is how we defined equality of sets., zz Empty set is a subset of every set., , This is not easy to see ! Let A be any set. The only way for the empty set to be not a, subset of A would be to have an element x in it but with x not in A. But how can x be, in the empty set ? That is impossible. So this only way being impossible, the empty, set must be a subset of A. (Is your head spinning? Think calmly, explain it to a friend,, and you will agree it is alright !), zz Every set is a subset of itself. (Try and argue why?), , 1.4.9. Proper Subset, Let A and B be two sets. If A is a subset of B and A≠B, then A is called a proper subset, of B and we write A 1 B., , For example,, If A={1,2,5} and B={1,2,3,4,5} then A is a proper subset of B ie. A 1 B., , 1.4.10 Disjoint Sets, Two sets A and B are said to be disjoint if they do, not have common elements., In other words, if A∩B=∅, then A and B are said, to be disjoint sets., , Fig. 1.3, , Example 1.8, Verify whether A={20, 22, 23, 24} and B={25, 30, 40, 45} are disjoint sets., Note, , Solution, A = {20,22, 23, 24} , B={25, 30, 40, 45}, A∩B = {20,22, 23, 24} ∩ {25, 30, 40, 45}, , , ={ }, , Since A∩B = ∅, A and B are disjoint sets., , 10, , If A∩B≠ ∅, then A and B, are said to be overlapping, sets .Thus if two sets, have atleast one common, element, they are called, overlapping sets., , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 10, , 26-12-2019 13:32:03
Page 17 :
www.tntextbooks.in, , 1.4.11 Power Set, The set of all subsets of a set A is called the power set of ‘A’. It is denoted by P(A)., , For example,, (i), , If A={2, 3}, then find the power set of A., The subsets of A are ∅ , {2},{3},{2,3}., , , , The power set of A,, , P(A) = {∅ ,{2},{3},{2,3}}, (ii) If A = {∅ , {∅}}, then the power set of A is { ∅ , {∅ , {∅}}, {∅} , {{∅}} }., An important property., We already noted that n(A) # n[P(A)]. But how big is P(A) ? Think about this a bit, and, see whether you come to the following conclusion:, (i) If n(A) = m, then n[P(A)] = 2m, (ii) The number of proper subsets of a set A is n[P(A)]–1 = 2m–1., Example 1.9, , Find the number of subsets and the number of proper subsets of a set, , X={a, b, c, x, y, z}., Solution Given X={a, b, c, x, y, z}.Then, n(X) =6, The number of subsets = n[P(X)] = 26 , The number of proper subsets, , , Thinking Corner, = 64, , = n[P(X)]-1 = 2 –1, 6, , = 64 – 1, , = 63, , Exercise 1.2, 1., , Every set has only, one improper subset., Verify this fact using, any set., , Find the cardinal number of the following sets., (i), , M = {p, q, r, s, t, u}, , (ii) P = {x : x = 3n+2, n∈ W and x< 15}, (iii) Q = {y : y = 34n , n ∈ N and 2 < n ≤5}, , (iv) R = {x : x is an integers, x∈ Z and –5 ≤ x <5}, (v) S = The set of all leap years between 1882 and 1906., 2., , Identify the following sets as finite or infinite., (i) X = The set of all districts in Tamilnadu., (ii) Y = The set of all straight lines passing through a point., (iii) A = { x : x ∈ Z and x <5}, (iv) B = { x : x2–5x+6 = 0, x ∈ N }, Set Language, , 1-SET LANGUAGE.indd 11, , 11, , 26-12-2019 13:32:03
Page 18 :
www.tntextbooks.in, , 3., , Which of the following sets are equivalent or unequal or equal sets?, (i), , A = The set of vowels in the English alphabets., , B = The set of all letters in the word “VOWEL”, (ii) C = {2,3,4,5} D = { x : x ∈ W , 1< x<5}, , 4., , (iii) X = { x : x is a letter in the word “LIFE”} , , Y = { F, I, L, E}, , (iv) G = { x : x is a prime number and 3 < x < 23}, , H = { x : x is a divisor of 18}, , Identify the following sets as null set or singleton set., (i), , A = {x : x ∈ N , 1 < x < 2}, , (ii) B = The set of all even natural numbers which are not divisible by 2, (iii) C = {0}., (iv) D = The set of all triangles having four sides., 5., , State which pairs of sets are disjoint or overlapping?, (i), , A = {f, i, a, s} and B={a, n, f, h, s}, , (ii) C = {x : x is a prime number, x >2} and D ={x:x is an even prime number}, (iii) E = {x : x is a factor of 24} and F={x : x is a multiple of 3, x < 30}, 6., , If S = {square, rectangle, circle, rhombus, triangle}, list the elements of the following, subset of S., (i) The set of shapes which have 4 equal sides., (ii) The set of shapes which have radius., (iii) The set of shapes in which the sum of all interior angles is 180o., (iv) The set of shapes which have 5 sides., , 7., , If A = {a, {a, b}}, write all the subsets of A., , 8., , Write down the power set of the following sets:, (i) A = {a, b}, , 9., , (ii) B = {1, 2, 3} (iii) D = {p, q, r, s}, , (iv) E = ∅, , Find the number of subsets and the number of proper subsets of the following sets., (i) W = {red, blue, yellow}, , 10. (i), , (ii) X = { x2 : x ∈ N , x2 ≤ 100}., , If n(A) = 4, find n[P(A)]. , , (ii) If n(A)=0, find n[P(A)]., , (iii) If n[P(A)] = 256, find n(A)., , 1.5 Set Operations, We started with numbers and very soon we learned arithmetical operations on them., In algebra we learnt expressions and soon started adding and multiplying them as well,, writing (x2+2),(x-3) etc. Now that we know sets, the natural question is, what can we do with, sets, what are natural operations on them ?, 12, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 12, , 26-12-2019 13:32:03
Page 19 :
www.tntextbooks.in, , When two or more sets combine together to form one set under the given conditions,, then operations on sets can be carried out. We can visualize the relationship between sets, and set operations using Venn diagram., John Venn was an English mathematician. He invented Venn diagrams which pictorially, represent the relations between sets.Venn diagrams are used in the field of Set Theory,, Probability, Statistics, Logic and Computer Science., , 1.5.1 Complement of a Set, The Complement of a set A is the set of all elements of U (the universal set) that are not in A., It is denoted by A′ or Ac. In symbols A′= {x : x∈U, x∉A}, , Venn diagram for complement of a set, , A (shaded region), , A′ (shaded region), , Fig. 1.4, , Fig. 1.5, , For example,, If U = {all boys in a class} and A= {boys who play Cricket}, then complement of the, set A is A′= {boys who do not play Cricket}., Example 1.10, If U = {c, d, e, f, g, h, i, j} and A = { c, d, g, j} , find A′., Solution, U = {c, d, e, f, g, h, i, j}, A = {c, d, g, j}, A′ ={e, f, h, i}, , 1.5.2 Union of Two Sets, , Note, zz, , (A′)′ = A, , zz, , U′ = ∅, , zz, , ∅′ = U, , The union of two sets A and B is the set of all elements which are either in A or in B or, in both. It is denoted by A∪B and read as A union B., In symbol, A∪B = {x : x ∈A or x∈B}, Set Language, , 1-SET LANGUAGE.indd 13, , 13, , 26-12-2019 13:32:03
Page 20 :
www.tntextbooks.in, , The union of two sets can be represented by Venn diagram as given below, , Sets A and B have common elements, , Sets A and B are disjoint, , Fig. 1.6, , Fig. 1.7, , For example,, If P ={Asia,Africa, Antarctica, Australia} and Q = {Europe, North America, South, America}, then the union set of P and Q is P∪Q = {Asia, Africa, Antartica, Australia, Europe,, North America, South America}., Note, zz A∪A = A, , zz A ⊆ A∪B and B ⊆ A∪B, , zz A∪∅ = A, , zz A∪B = B∪A (union of two sets is, , zz A∪U = U where A is any subset of, , commutative), , universal set U, , Example 1.11, , If P={m, n} and Q= {m, i, j}, then, represent, P and Q in Venn diagram and hence find P∪Q., Solution, Given P={m, n} and Q= {m, i, j}, From the venn diagram,, P∪Q={n, m, i, j}., , Fig. 1.8, , 1.5.3 Intersection of Two Sets, The intersection of two sets A and B is the set of all elements common to both A, and B. It is denoted by A∩B and read as A intersection B., In symbol , A∩B={x : x∈A and x∈B}, , Intersection of two sets can be represented by a, Venn diagram as given below, For example,, If A = {1, 2, 6}; B = {2, 3, 4}, then A∩B = {2}, because 2 is common element of the sets A and B., 14, , Fig. 1.9, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 14, , 26-12-2019 13:32:04
Page 21 :
www.tntextbooks.in, , Example 1.12, , Let A = {x : x is an even natural number and 1< x ≤ 12} and, , B = { x : x is a multiple of 3, x ∈ N and x≤12} be two sets. Find A∩B., Solution, Here A = {2, 4, 6, 8, 10, 12} and B = {3, 6, 9, 12}, A∩B = {6, 12}, Note, , Example 1.13, If A = {2, 3} and C = { }, find A∩C., , A∩A = A, , zz, , ∩U = A where A is any subset, A, of universal set U, , zz, , A∩B ⊆ A and A∩B ⊆ B, , zz, , ∩B = B∩A (Intersection of two, A, sets is commutative), , Solution, There is no common element and hence, A∩C ={ }, , A∩∅ = ∅, , zz, , (ii), , Note, zz When B ⊂ A, the union and intersection of A and B are represented in Venn, , diagram as follows, , B⊂A, , shaded region is, A∪B = A, , Fig. 1.10, , Fig. 1.11, , shaded region is, A∩B =B, Fig. 1.12, , zz If A and B are any two non empty sets such that A∪B = A∩B, then A = B, zz Let n(A) = p and n(B) = q then, , (a) Minimum of n(A∪B) = max{p, q}, (b) Maximum of n(A∪B) = p + q, (c) Minimum of n(A∩B) = 0, (d) Maximum of n(A∩B) = min{p, q}, , Set Language, , 1-SET LANGUAGE.indd 15, , 15, , 26-12-2019 13:32:04
Page 22 :
www.tntextbooks.in, , 1.5.4 Difference of Two Sets, Let A and B be two sets, the difference of sets A and B is the set of all elements which, are in A, but not in B. It is denoted by A–B or A\B and read as A difference B., In symbol, A–B = { x : x ∈ A and x ∉ B}, B–A = { y : y ∈ B and y ∉ A}., , Venn diagram for set difference, , A–B, , B–A, , A–B, , Fig. 1.13, , Fig. 1.14, , Fig. 1.15, , Example 1.14, If A={–3, –2, 1, 4} and B= {0, 1, 2, 4}, find (i) A–B (ii) B–A., Solution, A–B ={–3, -2, 1, 4} – {0, 1, 2, 4} = { -3, -2}, B–A = {0, 1, 2, 4} –{–3, -2, 1, 4} = { 0, 2}, , 1.5.5 Symmetric Difference of Sets, , Note, � A′ = U – A, zz A–B = A∩B′, , The symmetric difference of two sets A and B zz A–A = ∅, is the set (A–B)∪(B–A). It is denoted by AΔB. zz A–∅ = A, AΔB={ x : x ∈ A–B or x ∈ B–A}, Example 1.15, , zz A–B = B–A + A=B, zz A–B = A and B–A= B if A∩B =∅, , If A = {6, 7, 8, 9} and B={8, 10, 12}, find AΔB., Solution, A–B = {6, 7, 9}, B–A = {10, 12}, AΔB = (A–B)∪(B–A) = {6, 7, 9}∪{10,12}, , Thinking Corner, What is (A − B ) ∩ (B − A) ?, , AΔB = {6, 7, 9, 10, 12}., 16, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 16, , 26-12-2019 13:32:04
Page 23 :
www.tntextbooks.in, , Example 1.16, Represent AΔB through Venn diagram., Solution, AΔB= (A–B) ∪ (B–A), , A–B, , B–A, , (A–B) ∪ (B–A), , Fig. 1.16, , Fig. 1.17, , Fig. 1.18, , Note, , , �, , AΔA=∅, , �, , AΔB={x : x∈A∪B and x∉A∩B}, , �, , AΔB = BΔA, , �, , AΔB= (A∪B) – (A∩B), , Example 1.17, From the given Venn diagram, write the elements of, (i) A, (ii) B, (iii) A–B, (iv) B–A, (v) A′, , (vi) B′, , (vii) U, , Solution, , Fig. 1.19, , (i) A = {a, e, i, o, u} (ii) B = {b, c, e, o}, (iii) A–B = {a, i, u} (iv) B–A = {b, c}, (v) A′ = {b, c, d, g} (vi) B′ = {a, d, g, i, u}, (vii) U = {a, b, c, d, e, g, i, o, u}, Example 1.18, Draw Venn diagram and shade the region representing the following sets, (i) A′, , (ii) (A–B)′, , (iii) (A∪B)′, Set Language, , 1-SET LANGUAGE.indd 17, , 17, , 26-12-2019 13:32:04
Page 24 :
www.tntextbooks.in, , Solution, (i), , A′ (ii), , (A–B)′, , A′, , A–B, , Fig. 1.20, , Fig. 1.21, , (A–B)′, , Fig. 1.22, , (iii) (AUB)′, , A∪B, Fig. 1.23, , (A∪B)′, Fig. 1.24, , Exercise 1.3, 1., , Using the given Venn diagram, write the elements, of, (i) A, , (ii) B, , (iii) A∪B (iv) A∩B, , (v) A–B (vi) B–A (vii) A′, , (viii) B′, , (ix) U, 2., , Fig. 1.25, , Find A∪B, A∩B, A–B and B–A for the following sets., (i), , A = {2, 6, 10, 14} and B={2, 5, 14, 16}, , (ii) A = {a, b, c, e, u} and B={a, e, i, o, u}, (iii) A = {x : x ∈N, x ≤ 10} and B={x : x ∈W, x < 6}, (iv) A = Set of all letters in the word “mathematics” and, B = Set of all letters in the word “geometry”, 18, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 18, , 26-12-2019 13:32:05
Page 25 :
www.tntextbooks.in, , If U={a, b, c, d, e, f, g, h}, A={b, d, f, h} and B={a, d, e, h}, find the following sets., , 3., , (i) A′, , (ii) B′ (iii) A′∪B′, , (vi) (A∩B)′, , (iv) A′∩B′, , (v) (A∪B)′, , (vii) (A′)′ (viii) (B′)′, , Let U={0, 1, 2, 3, 4, 5, 6, 7}, A={1, 3, 5, 7} and B={0, 2, 3, 5, 7}, find the following sets., , 4., , (i) A′, , (ii) B′ (iii) A′∪B′, , (vi) (A∩B)′, 5., , (vii) (A′)′, , (iv)A′∩B′, , ( v)(A∪B)′, , (viii) (B′)′, , Find the symmetric difference between the following sets., (i), , P = {2, 3, 5, 7, 11} and Q={1, 3, 5, 11}, , (ii) R = {l, m, n, o, p} and S = {j, l, n, q}, (iii) X = {5, 6, 7} and Y={5, 7, 9, 10}, 6., , Using the set symbols, write down the expressions for the shaded region in the following, X, , Y, , X, , Fig. 1.26, , 7., , Y, , X, , Fig. 1.27, , Y, , Fig. 1.28, , Let A and B be two overlapping sets and the universal set be U. Draw appropriate Venn, diagram for each of the following,, (i), , A∪B, , (ii) A∩B, , (iii) (A∩B)′ (iv) (B–A)′, , (v) A′∪B′, , (vi) A′∩B′, , (vii) What do you observe from the Venn diagram (iii) and (v)?, , 1.6 Properties of Set Operations, It is an interesting investigation to find out if operations among sets (like union, intersection,, etc) follow mathematical properties such as Commutativity, Associativity, etc., We have seen, numbers having many of these properties; whether sets also possess these, is to be explored., We first take up the properties of set operations on union and intersection., Set Language, , 1-SET LANGUAGE.indd 19, , 19, , 26-12-2019 13:32:05
Page 26 :
www.tntextbooks.in, , 1.6.1 Commutative Property, In set language, commutative situations can be seen, when we perform operations. For example, we can look, into the Union (and Intersection) of sets to find out if the, operation is commutative., Let A = {2, 3, 8, 10} and B = {1, 3, 10, 13} be two sets., , Note, For any set A,, zz A ∪ A = A and A ∩ A = A, , [Idempotent Laws]., zz A ∪ f = A and A ∩ U = A, , [Identity Laws]., , Then, A ∪ B = {1, 2, 3, 8, 10, 13} and, B ∪ A = {1, 2, 3, 8, 10, 13}, , , , From the above, we see that A ∪ B = B ∪ A., This is called Commutative property of union of sets., Now, A ∩ B = {3, 10} and B ∩ A = {3, 10} . Then, we see that A ∩ B = B ∩ A ., This is called Commutative property of intersection of sets., Commutative property: For any two sets A and B, (i) A ∪ B = B ∪ A (ii) A ∩ B = B ∩ A, Example 1.19, , If A = {b, e, f , g } and B = {c, e, g, h } , then verify the commutative, , property of (i) union of sets (ii) intersection of sets., Solution, Given, A = {b, e, f , g } and B = {c, e, g, h }, (i), , , , A È B = {b, c, e, f , g, h }, , ... (1), , B È A = {b, c, e, f , g, h }, , ... (2), , Thinking Corner, Given, P l ,n , p and, P Q j ,l ,m,n ,o, p . If P and Q, , are disjoint sets, then what will be, Q and P ∩ Q ?, , From (1) and (2) we have A ∪ B = B ∪ A, It is verified that union of sets is commutative., , (ii), , A Ç B = {e, g } ... (3), , , B Ç A = {e, g } ... (4), , , From (3) and (4) we get, A ∩ B = B ∩ A, , , , It is verified that intersection of sets is commutative., , 20, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 20, , 26-12-2019 13:32:07
Page 27 :
www.tntextbooks.in, , Note, Recall that subtraction on numbers is not commutative. Is set difference, commutative? We expect that the set difference is not commutative as, well. For instance, consider A = {a, b, c }, B = {b, c, d }. A − B = {a },, B − A = {d } ; we see that A − B ≠ B − A., , 1.6.2 Associative Property, Now, we perform operations on union and intersection for three sets., Let A = {−1, 0, 1, 2}, B = {−3, 0, 2, 3} and C = {0, 1, 3, 4} be three sets., Now,, , B È C = {-3, 0, 1, 2, 3, 4}, , A È (B È C ) = {−1, 0, 1, 2} ∪ {−3, 0, 1, 2, 3, 4}, = {-3, -1, 0, 1, 2, 3, 4} ... (1), Then,, , A È B = {-3, -1, 0, 1, 2, 3}, , (A È B ) È C = {−3, −1, 0, 1, 2, 3} ∪ {0, 1, 3, 4}, = {-3, -1, 0, 1, 2, 3, 4} ... (2), From (1) and (2), A ∪ (B ∪ C ) = (A ∪ B ) ∪ C ., This is associative property of union among sets A, B, and C., Now,, B Ç C = {0, 3}, A Ç (B Ç C ) = {−1, 0, 1, 2} ∩ {0, 3}, = {0} , Then,, , ... (3), , A Ç B = {0, 2}, , (A Ç B ) Ç C = {0, 2} Ç {0, 1, 3, 4}, = {0} , , ... (4), , From (3) and (4), A ∩ (B ∩ C ) = (A ∩ B ) ∩ C ., This is associative property of intersection among sets A, B and C., Associative property: For any three sets A, B and C, (i), , (ii) A ∩ (B ∩ C ) = (A ∩ B ) ∩ C, , A ∪ (B ∪ C ) = (A ∪ B ) ∪ C, , Example 1.20, , 1 1 3 , 1 3, 1 1, 5 , 5 , If A = − , 0, , , 2, B = 0, , , 2, and C = − , , 1, 2, ,, 2, , 4 4, , , , , , 4 4, , 2 , , 2 4, , 2 , , then verify A ∩ (B ∩ C ) = (A ∩ B ) ∩ C ., Set Language, , 1-SET LANGUAGE.indd 21, , 21, , 26-12-2019 13:32:10
Page 28 :
www.tntextbooks.in, , Solution, Now,, , (B Ç C ) =, , A Ç (B Ç C ) =, Then,, , AÇB =, , (A Ç B ) Ç C =, , 1, , 2, 5 , 4, 2 , 1 , , 2 ... (1), 4 , 1 3 , 0, , , 2, 4 4 , 1 , , 2 ... (2), 4 , , Note, The set difference in general is, not associative, that is, (A–B)–C ¹ A–(B–C)., But, if the sets A, B and C are, mutually disjoint then the set, difference is associative, that is, (A–B)–C=A–(B–C)., , From (1) and (2), it is verified that, (A Ç B ) Ç C = A Ç (B Ç C ), , Exercise 1.4, 1. If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} , then find, (i), , (P È Q ) È R (ii) (P Ç Q ) Ç S , , (iii) (Q Ç S ) Ç R, , 2. Test for the commutative property of union and intersection of the sets, P = { x : x is a real number between 2 and 7} and, Q = { x : x is a rational number between 2 and 7}., 3. If A = {p,q,r,s}, B = {m,n,q,s,t} and C = {m,n,p,q,s}, then verify the associative property, of union of sets., , {, , }, , 4. Verify the associative property of intersection of sets for A = −11, 2, 5, 7 ,, B=, , {, , }, , 3, 5, 6, 13 and C =, , {, , }, , 2, 3, 5, 9 ., , 5. If A={ x : x = 2n , n ∈ W and n<4}, B = {x : x = 2n, n ∈ and n £ 4} and, C = {0, 1, 2, 5, 6} , then verify the associative property of intersection of sets., , 1.6.3 Distributive Property, In lower classes, we have studied distributive property of multiplication over addition, on numbers. That is, a × (b + c) = (a × b) + (a × c). In the same way we can define distributive, properties on sets., 22, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 22, , 26-12-2019 13:32:12
Page 29 :
www.tntextbooks.in, , Distributive property: For any three sets A, B and C, , (i), , A ∩ (B ∪ C ) = (A ∩ B ) ∪ (A ∩ C ) [Intersection over union], , (ii), , A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ) [Union over intersection], , Example 1.21, , If A = {0, 2, 4, 6, 8}, B = {x : x is a prime number and x < 11} and, , C = {x : x Î N and 5 ≤ x < 9 } then verify A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ) ., Solution, , A = {0, 2, 4, 6, 8} , B = {2, 3, 5, 7} and C = {5, 6, 7, 8}, , Given, First, we find, , B ÇC, , {, , }, , = {5, 7} , A ∪ (B ∩ C ) = 0, 2, 4, 5, 6, 7, 8, , ... (1), , A È B = {0, 2, 3, 4, 5, 6, 7, 8} , A È C = {0, 2, 4, 5, 6, 7, 8}, , Next,, , Then, (A ∪ B ) ∩ (A ∪ C ) = {0, 2, 4, 5, 6, 7, 8} ... (2), From (1) and (2), it is verified that A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C )., Example 1.22, Solution, , Verify A ∩ (B ∪ C ) = (A ∩ B ) ∪ (A ∩ C ) using Venn diagrams., , .... (1), , B ÈC, , , , A ∩ (B ∪ C ), , , .... (2), , A Ç B , , A ÇC, , , , (A ∩ B ) ∪ (A ∩ C ), , Fig.1.29, , From (1) and (2), A ∩ (B ∪ C ) = (A ∩ B ) ∪ (A ∩ C ) is verified., Set Language, , 1-SET LANGUAGE.indd 23, , 23, , 26-12-2019 13:32:14
Page 30 :
www.tntextbooks.in, , 1.7 De Morgan’s Laws, Augustus De Morgan (1806 – 1871) was a British mathematician., He was born on 27 th June 1806 in Madurai, Tamilnadu, India. His, father was posted in India by the East India Company. When he was, seven months old, his family moved back to England. De Morgan was, educated at Trinity College, Cambridge, London. He formulated laws for, set difference and complementation. These are called De Morgan’s laws., , 1.7.1 De Morgan’s Laws for Set Difference, These laws relate the set operations union, intersection and set difference., Let us consider three sets A, B and C as A = {−5, −2, 1, 3}, B = {−3, −2, 0, 3, 5} and, , C = {−2, −1, 0, 4, 5} ., , Now,, , Then,, , B ÈC, , = {-3, -2, -1, 0, 3, 4, 5}, , A − (B ∪ C ) = {−5, 1} , , ... (1), , A - B = {-5, 1} and A − C = {−5, 1, 3}, , (A − B ) ∪ (A − C ) = {−5, 1, 3} ... (2), (A − B ) ∩ (A − C ) = {−5, 1} , , ... (3), , From (1) and (2), we see that, A − (B ∪ C ) ≠ (A − B ) ∪ (A − C ), But note that from (1) and (3), we see that, A − (B ∪ C ) = (A − B ) ∩ (A − C ), Now,, , Thinking Corner, (A − B ) ∪ (A − C ) ∪ (A ∩ B ) =____, , B Ç C = {−2, 0, 5}, , A − (B ∩ C ) = {−5, 1, 3}, , ... (4), , From (3) and (4) we see that, A − (B ∩ C ) ≠ (A − B ) ∩ (A − C ), But note that from (2) and (4), we get A − (B ∩ C ) = (A − B ) ∪ (A − C ), De Morgan’s laws for set difference : For any three sets A, B and C, (i) A − (B ∪ C ) = (A − B ) ∩ (A − C ), 24, , (ii) A − (B ∩ C ) = (A − B ) ∪ (A − C ), , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 24, , 26-12-2019 13:32:16
Page 31 :
www.tntextbooks.in, , Example 1.23, Solution, , Verify A − (B ∪ C ) = (A − B ) ∩ (A − C ) using Venn diagrams., , , .... (1), , B ÈC, , A − (B ∪ C ), , .... (2), , A-B, , , , A -C, , , , (A − B ) ∩ (A − C ), , Fig.1.30, , From (1) and (2), we get A − (B ∪ C ) = (A − B ) ∩ (A − C ) . Hence it is verified., Example 1.24, , If P = {x : x Î W and 0 < x < 10}, Q = {x : x = 2n+1, n Î W and n<5}, , and R = {2, 3, 5, 7, 11, 13} , then verify P − (Q ∩ R) = (P − Q ) ∪ (P − R), Solution The roster form of sets P, Q and R are, P = {1, 2, 3, 4, 5, 6, 7, 8, 9}, Q = {1, 3, 5, 7, 9}, , and R = {2, 3, 5, 7, 11, 13}, First, we find (Q Ç R) = {3, 5, 7}, Then, P − (Q ∩ R) = {1, 2, 4, 6, 8, 9} ... (1), Next, P–Q = {2, 4, 6, 8} and, P–R = {1, 4, 6, 8, 9}, and so, (P − Q ) ∪ (P − R) = {1, 2, 4, 6, 8, 9} ... (2), , Finding the elements of set Q, Given, x = 2n + 1, n, n, n, n, n, , = 0 → x = 2(0) + 1 = 0 + 1 = 1, = 1 → x = 2(1) + 1 = 2 + 1 = 3, = 2 → x = 2(2) + 1 = 4 + 1 = 5, = 3 → x = 2(3) + 1 = 6 + 1 = 7, = 4 → x = 2(4) + 1 = 8 + 1 = 9, , Therefore, x takes values such as, 1, 3, 5, 7 and 9., , Hence from (1) and (2), it is verified that P − (Q ∩ R) = (P − Q ) ∪ (P − R)., Set Language, , 1-SET LANGUAGE.indd 25, , 25, , 26-12-2019 13:32:18
Page 32 :
www.tntextbooks.in, , 1.7.2 De Morgan’s Laws for Complementation, These laws relate the set operations on union, intersection and complementation., Let us consider universal set U={0,1,2,3,4,5,6}, A={1,3,5} and B={0,3,4,5}., Now,, , Thinking Corner, , A È B = {0, 1, 3, 4, 5} , , Then,, , (A ∪ B )′ = {2, 6} .....(1), , Next,, , A¢ = {0, 2, 4, 6} and B ¢ = {1, 2, 6}, , Then,, , A′ ∩ B ′ = {2, 6} .....(2), , Check whether, , A − B = A ∩ B′, , From (1) and (2), we get (A ∪ B )′ = A′ ∩ B ′, , Also, A ∩ B = {3, 5},, , Thinking Corner, , (A ∩ B )′ = {0, 1, 2, 4, 6} .....(3), , (A − B ) ∪ (B − A′) =____, , A′ = {0, 2, 4, 6} and B ′ = {1, 2, 6}, A′ ∪ B ′ = {0, 1, 2, 4, 6} .....(4), , From (3) and (4), we get (A ∩ B )′ = A′ ∪ B ′, De Morgan’s laws for complementation : Let ‘U’ be the universal set containing, finite sets A and B. Then (i) (A ∪ B )′ = A′ ∩ B ′ (ii) (A ∩ B )′ = A′ ∪ B ′, , Example 1.25, , Verify (A ∪ B )′ = A′ ∩ B ′ using Venn diagrams., , Solution, , .... (1), , 26, , AÈB, , , , (A ∪ B )′, , , , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 26, , 26-12-2019 13:32:20
Page 33 :
www.tntextbooks.in, , .... (2), , , A¢, , , , B ¢ , , A′ ∩ B ′ , , Fig.1.31, , From (1) and (2), it is verified that (A ∪ B )′ = A′ ∩ B ′, Example 1.26, , {, , If U = {x : x ∈ , − 2 ≤ x ≤ 10} ,, , }, , {, , }, , A = x : x = 2p + 1, p ∈ , − 1 ≤ p ≤ 4 , B = x : x = 3q + 1, q ∈ , − 1 ≤ q < 4 ,, , verify De Morgan’s laws for complementation., Solution, , Thinking Corner, , Given U = {−2, −1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ,, , A ( A B ) =____, , A = {−1, 1, 3, 5, 7, 9} and B = {−2, 1, 4, 7, 10}, , Law (i) (A ∪ B )′ = A′ ∩ B ′, Now,, , Then,, , , A È B = {-2, -1, 1, 3, 4, 5, 7, 9, 10}, (A ∪ B )′ = {0, 2, 6, 8} ..... (1), A¢ = {-2, 0, 2, 4, 6, 8, 10} and B ¢ = {-1, 0, 2, 3, 5, 6, 8, 9}, A′ ∩ B ′ = {0, 2, 6, 8} ..... (2), , , From (1) and (2), it is verified that (A ∪ B )′ = A′ ∩ B ′, Law (ii) (A ∩ B )′ = A′ ∪ B ′, Now,, , Thinking Corner, (A ∪ B )′ ∪ (A′ ∩ B ) =___, , A Ç B = {1, 7}, , , , (A ∩ B )′ = {-2, -1, 0, 2, 3, 4, 5, 6, 8, 9, 10}, , ..... (3), , Then,, , A′ ∪ B ′ = {-2, -1, 0, 2, 3, 4, 5, 6, 8, 9, 10}, , ..... (4), , From (3) and (4), it is verified that (A ∩ B )′ = A′ ∪ B ′, , Set Language, , 1-SET LANGUAGE.indd 27, , 27, , 26-12-2019 13:32:24
Page 34 :
www.tntextbooks.in, , Exercise 1.5, 1. Using the adjacent Venn diagram, find the following sets:, (i) A - B, , (ii) B - C, , (iii) A′ ∪ B ′, , (iv) A′ ∩ B ′, , (v) (B ∪ C )′, , (vi) A − (B ∪ C ), , (vii) A − (B ∩ C ), , 2. If K = {a, b, d, e, f }, L = {b, c, d, g } and M = {a, b, c, d, h } , then find the following:, (i) K ∪ (L ∩ M ) , , (ii) K ∩ (L ∪ M ), , (iii) (K ∪ L) ∩ (K ∪ M ) , , (iv) (K∩ L) ∪ (K ∩ M ), , and verify distributive laws., 3. If A = {x : x ∈ , −2 < x ≤ 4}, B={x : x ∈ W , x ≤ 5}, C = {−4, −1, 0, 2, 3, 4} , then, verify A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ) ., , 4. Verify A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ) using Venn diagrams., 5. If A = {b, c, e, g, h } , B = {a, c, d, g, i} and C = {a, d, e, g, h } , then show that, A − (B ∩ C ) = (A − B ) ∪ (A − C ) ., 6. If A = {x : x = 6n, n Î W and n<6}, B = {x : x = 2n, n Î N and 2<n £ 9} and, C = {x : x = 3n, n Î N and 4 £ n<10}, then show that A − (B ∩ C ) = (A − B ) ∪ (A − C ), 7. If A = {–2,0,1,3,5}, B = {–1,0,2,5,6} and C = {–1,2,5,6,7}, then show that A − (B ∪ C ), = (A − B ) ∩ (A − C ) ., a +1, 2n - 1, , a Î W and a £ 5}, B = {y : y =, , n Î W and n < 5} and, 2, 2, , 1 3 , C = −1, − , 1, , 2 , then show that A − (B ∪ C ) = (A − B ) ∩ (A − C ) ., , 2 2 , , 8. If A = {y : y =, , 9. Verify A − (B ∩ C ) = (A − B ) ∪ (A − C ) using Venn diagrams., 10. If U = {4,7,8,10,11,12,15,16}, A={7,8,11,12} and B = {4,8,12,15} then verify, De Morgan’s Laws for complementation., 11. Verify (A ∩ B )′ = A′ ∪ B ′ using Venn diagrams., 28, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 28, , 26-12-2019 13:32:27
Page 35 :
www.tntextbooks.in, , 1.8 Application on Cardinality of Sets:, We have learnt about the union, intersection, complement and difference of sets., Now we will go through some practical problems on sets related to everyday life., , Results :, If A and B are two finite sets, then, (i), , n(A∪B) = n(A)+n(B)– n(A∩B), , (ii) n(A–B) = n(A) – n(A∩B), (iii) n(B–A) = n(B)– n(A∩B), (iv) n(A′) = n(U) – n(A), , Fig. 1.32, , Note, From the above results we may get,, �, , n(A∩B) = n(A)+n(B) – n(A∪B), , zz, , If A and B are disjoint sets then, n(A∪B) = n(A)+n(B)., , �, , n(U) = n(A)+n(A′), , Example 1.27, , From the Venn diagram, verify that, n(A∪B) = n(A)+n(B) – n(A∩B), Solution From the venn diagram,, , , A = {5, 10, 15, 20}, , , , B = {10, 20, 30, 40, 50,}, A∪B = {5, 10, 15, 20, 30, 40, 50}, , Then, , A∩B = {10, 20}, n(A) = 4, n(B) = 5, n(A∪B) = 7, n(A∩B) = 2, , , n(A∪B) = 7 " (1), , Fig. 1.33, , n(A)+n(B)–n(A∩B) = 4+5–2, , =7 " (2), From (1) and (2), n(A∪B) = n(A)+n(B)–n(A∩B)., Example 1.28, , If n(A) = 36, n(B) = 10, n(A∪B)=40, and n(A′)=27 find n(U) and n(A∩B)., , Solution, n(A) = 36, n(B) =10, n(A∪B)=40, n(A′)=27, (i), n(U) = n(A)+n(A′) = 36+27 = 63, (ii) n(A∩B) = n(A)+n(B)–n(A∪B) = 36+10-40 = 46-40 = 6, Set Language, , 1-SET LANGUAGE.indd 29, , 29, , 26-12-2019 13:32:27
Page 36 :
www.tntextbooks.in, , Activity-4, Fill in the blanks with appropriate cardinal numbers., S.No., , n(A), , n(B), , n(A∪B), , 1, , 30, , 45, , 65, , 2, , 20, , 3, , 50, , 65, , 4, , 30, , 43, , Example 1.29, , n(A∩B), , 55, , n(A–B), , n(B–A), , 10, 25, , 70, , Let A={b, d, e, g, h} and B = {a, e, c, h}. Verify that n(A–B) = n(A)–n(A∩B)., , Solution, A = {b, d, e, g, h}, B = {a, e, c, h}, A – B = {b, d, g}, n(A–B) = 3 , , , , ... (1), , A ∩ B = {e, h}, , , , , n(A ∩ B) = 2 ,, , n(A) = 5, , n(A) – n(A∩B) = 5-2, = 3 , , ... (2), , Form (1) and (2) we get n(A–B) = n(A)–n(A∩B)., Example 1.30, In a school, all students play either Hockey or Cricket or both. 300 play Hockey, 250, play Cricket and 110 play both games. Find, (i) the number of students who play only Hockey., (ii) the number of students who play only Cricket., (iii) the total number of students in the School., Solution:, Let H be the set of all students who play Hockey and C be the set of all students who, play Cricket., Then n(H) = 300, n(C) = 250 and n(H ∩ C) = 110, Using Venn diagram,, From the Venn diagram,, (i), 30, , The number of students who play only Hockey = 190, , Fig. 1.34, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 30, , 26-12-2019 13:32:27
Page 37 :
www.tntextbooks.in, , (ii) The number of students who play only Cricket = 140, (iii) The total number of students in the school = 190+110+140 =440, Aliter, (i), , The number of students who play only Hockey, , n(H–C ) = n(H) – n(H∩C), , , = 300 –110 = 190, , (ii) The number of students who play only Cricket, n(C–H ) = n(C) – n(H∩C), , , = 250 – 110 = 140, , (iii) The total number of students in the school, n(HUC) = n(H) + n(C) – n(H∩C), , , = 300+250 – 110 = 440, , Example 1.31, , In a party of 60 people, 35 had Vanilla ice cream, 30 had Chocolate ice, , cream. All the people had at least one ice cream. Then how many of them had,, (i) both Vanilla and Chocolate ice cream., (ii) only Vanilla ice cream., (iii) only Chocolate ice cream., Solution :, Let V be the set of people who had Vanilla ice cream and C be the set of people who, had Chocolate ice cream., Then n(V) = 35, n(C) = 30, n(V , C) = 60,, Let x be the number of people who had both ice creams., From the Venn diagram, , , 35 – x + x +30 – x = 60, , , 65 – x = 60, x= 5, Hence 5 people had both ice creams., (i) Number of people who had only Vanilla ice, cream = 35 – x, , = 35– 5 = 30, , Fig. 1.35, , (ii) Number of people who had only Chocolate ice cream = 30 – x, , , = 30 – 5 = 25, Set Language, , 1-SET LANGUAGE.indd 31, , 31, , 26-12-2019 13:32:27
Page 38 :
www.tntextbooks.in, , We have learnt to solve problems involving two sets using the formula, n(A ∪ B ) = n(A) + n(B ) − n(A ∩ B ). Suppose we have three sets, we can apply this formula, to get a similar formula for three sets., For any three finite sets A, B and C, n(A È B È C ) = n(A) + n(B ) + n(C ) −n(A ∩ B ) − n(B ∩ C ) − n(A ∩ C ) + n(A ∩ B ∩ C ), , Note, Let us consider the following results which will be useful in solving problems, using Venn diagram. Let three sets A, B and C represent the students. From, the Venn diagram,, Number of students in only set A = a, only set B = b, only set C = c ., zz Total number of students in only one set = (a + b + c) A, zz Total number of students in only two sets = (x + y + z ), zz Number of students exactly in three sets = r, , (a + b + c + x + y + z + r ), , in, , 3, , r, z, , zz Total number of students in atleast two sets (two or, , more sets) = x + y +z + r, zz Total, number of students, , B, x, , a, , sets, , =, , b, y, , c, , Fig.1.36, , C, , Example 1.32, , In a college, 240 students play cricket, 180 students play football, 164, students play hockey, 42 play both cricket and football, 38 play both football and hockey,, 40 play both cricket and hockey and 16 play all the three games. If each student participate, in atleast one game, then find (i) the number of students in the college (ii) the number of, students who play only one game., Solution Let C, F and H represent sets of students who play Cricket, Football and Hockey, respectively., Then , n(C ) = 240, n(F ) = 180, n(H ) = 164, n(C ∩ F ) = 42,, n(F ∩ H ) = 38, n(C ∩ H ) = 40, n(C ∩ F ∩ H ) = 16., , Let us represent the given data in a Venn diagram., (i) The number of students in the college, = 174+26+116+22+102+24+16 = 480, (ii) The number of students who play only one game, = 174+116+102 = 392, 32, , Fig.1.37, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 32, , 26-12-2019 13:32:29
Page 39 :
www.tntextbooks.in, , Example 1.33, , In a residential area with 600 families, , 3, 1, owned scooter, owned car,, 5, 3, , 1, owned bicycle, 120 families owned scooter and car, 86 owned car and bicylce while 90, 4, 2, families owned scooter and bicylce. If, of families owned all the three types of vehicles,, 15, , then find (i) the number of families owned atleast two types of vehicle. (ii) the number of, families owned no vehicle., Solution Let S, C and B represent sets of families who owned Scooter, Car and Bicycle, U(600), respectively., 3, × 600 = 360, 5, 1, 1, n(C ) = × 600 = 200 n(B ) = × 600 = 150, 3, 4, 2, n(S ∩ C ∩ B ) = × 600 = 80, 15, , Given,, , n(U) = 600 n(S ) =, , From Venn diagram,, (i) The number of families owned atleast two, types of vehicles = 40+6+10+80 = 136, (ii) The number of families owned no vehicle, , = 600 – (owned atleast one vehicle), , = 600 − (230 + 40 + 74 + 6 + 54 + 10 + 80), , = 600 − 494 = 106, , Fig.1.38, , Example 1.34, , In a group of 100 students, 85 students speak Tamil, 40 students speak, English, 20 students speak French, 32 speak Tamil and English, 13 speak English and French, and 10 speak Tamil and French. If each student knows atleast any one of these languages,, then find the number of students who speak all these three languages., Solution Let A, B and C represent sets of students who speak Tamil, English and French, respectively., Given,, n(A È B È C ) = 100, n(A) = 85, n(B) = 40, n(C) = 20,, n(A∩ B ) = 32, n(B∩ C ) = 13, n(A∩ C ) = 10 ., We know that,, n(A È B È C ) = n(A) + n(B ) + n(C) − n(A ∩ B ) − n(B ∩ C ) − n(A ∩ C ) + n(A ∩ B ∩ C ), , 100 = 85 + 40 + 20 − 32 − 13 − 10 + n(A ∩ B ∩ C ), Then, n(A ∩ B ∩ C ) = 100 − 90 = 10, Therefore, 10 students speak all the three languages., Set Language, , 1-SET LANGUAGE.indd 33, , 33, , 26-12-2019 13:32:31
Page 40 :
www.tntextbooks.in, , Example 1.35, , A survey was conducted among 200 magazine subscribers of three, different magazines A, B and C. It was found that 75 members do not subscribe magazine A,, 100 members do not subscribe magazine B, 50 members do not subscribe magazine C and 125, subscribe atleast two of the three magazines. Find, (i) Number of members who subscribe exactly two magazines., (ii) Number of members who subscribe only one magazine., 5), , Solution, Total number of subscribers = 200, Magazine Do not subscribe Subscribe, A, , 75, , 125, , B, , 100, , 100, , C, , 50, , 150, , B(10, , 12, A(, , 0), , b, , x, , a, , r, , z, , y, , c, , From the Venn diagram,, Fig.1.39, Number of members who subscribe only one magazine = a + b + c, Number of members who subscribe exactly two magazines = x + y + z, and 125 members subscribe atleast two magazines., That is, x + y + z + r = 125 ... (1), , ), , C(150, , Now, n(A ∪ B ∪ C ) = 200 , n(A) = 125, n(B) = 100, n(C) = 150, n(A Ç B) = x + r, , n(B Ç C) = y + r,, n(A Ç C) = z + r,, n(A Ç B Ç C) = r, We know that,, n(A È B È C) = n(A)+ n(B)+ n(C)– n(A Ç B) – n(B Ç C)– n(A Ç C)+ n(A Ç B Ç C), , , 200 = 125+100+150–x–r–y–r–z–r+r, , , , = 375–(x+y+z+r)–r, , , , = 375–125–r x + y + z + r = 125, , , , , , , , , , 200 = 250–r, Þ r = 50, From (1) x + y + z + 50 = 125, We get, x + y + z = 75, , Therefore, number of members who subscribe exactly two magazines = 75., From Venn diagram,, , (a + b + c) + (x + y + z + r ) = 200 ... (2), substitute (1) in (2),, , a + b + c + 125 = 200, , a + b + c = 75, Therefore, number of members who subscribe only one magazine = 75., 34, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 34, , 26-12-2019 13:32:33
Page 41 :
www.tntextbooks.in, , Exercise 1.6, 1., , (i) If n(A) = 25, n(B) = 40, n(A∪B) = 50 and n(B′) = 25 , find n(A∩B) and n(U)., (ii) If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B′) = 350, find n(B) and n(U)., , 2., , If U = {x : x ∈ N , x ≤ 10}, A = {2,3,4,8,10} and B = {1,2,5,8,10}, then verify that, n(A ∪ B) = n(A) + n(B) – n(A ∩ B), , 3. Verify n(A ∪ B ∪ C ) = n(A) + n(B ) + n(C ) − n(A ∩ B ) − n(B ∩ C ) − n(A ∩ C ), +n(A ∩ B ∩ C ) for the following sets., (i) , A = {a, c, e, f , h } , B = {c, d, e, f } and C = {a, b, c, f }, (ii) , A = {1, 3, 5} , B = {2, 3, 5, 6} and C = {1, 5, 6, 7}, 4., , In a class, all students take part in either music or drama or both. 25 students take part, in music, 30 students take part in drama and 8 students take part in both music and, drama. Find, (i), , The number of students who take part in only music., , (ii) The number of students who take part in only drama., (iii) The total number of students in the class., 5., , In a party of 45 people, each one likes tea or coffee or both. 35 people like tea and, 20 people like coffee. Find the number of people who, (i) like both tea and coffee., , (ii) do not like Tea. , , (iii) do not like coffee., , 6., , In an examination 50% of the students passed in Mathematics and 70% of students, passed in Science while 10% students failed in both subjects. 300 students passed in, both the subjects. Find the total number of students who appeared in the examination,, if they took examination in only two subjects., , 7., , A and B are two sets such that n(A–B) = 32 + x, n(B–A) = 5x and n(A∩B) = x. Illustrate, the information by means of a Venn diagram. Given that n(A) = n(B), calculate the, value of x., , 8., , Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned, both A and B cars. Is this data correct?, , 9., , In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper,, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17, families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers, and 3 families buy all the three newspapers. If each family buy atleast one of these, newspapers then find, Set Language, , 1-SET LANGUAGE.indd 35, , 35, , 26-12-2019 13:32:33
Page 42 :
www.tntextbooks.in, , (i), , Number of families buy only one newspaper, , (ii), , Number of families buy atleast two newspapers, , (iii), , Total number of families in the colony., , 10., , A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120, grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer, grew atleast any one of the above three, then find the number of farmers who grew all, the three., , 11., , In the adjacent diagram, if n(U) = 125 , y is two times of x, and z is 10 more than x, then find the value of x , y and z., , 12., , Each student in a class of 35 plays atleast one, game among chess, carrom and table tennis. 22, play chess, 21 play carrom, 15 play table tennis,, 10 play chess and table tennis, 8 play carrom and, table tennis and 6 play all the three games. Find the number of students who, play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom, (Hint: Use Venn diagram), , 13., , In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by, bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students, come to school exactly by two modes of transport?, , Exercise 1.7, Multiple Choice Questions, 1., , 2., 3., 4., 5., , 36, , Which of the following is correct?, (1) {7} ∈ {1,2,3,4,5,6,7,8,9,10}, , (2) 7 ∈ {1,2,3,4,5,6,7,8,9,10}, , (3) 7 ∉ {1,2,3,4,5,6,7,8,9,10}, , (4) {7} M {1,2,3,4,5,6,7,8,9,10}, , The set P = {x | x ∈ Z , –1< x < 1} is a, (1) Singleton set, (2) Power set, (3) Null set, , (4) Subset, , If U ={x | x ∈ N , x < 10} and A = {x | x ∈ N , 2 ≤ x < 6} then (A′)′ is, (1) {1, 6, 7, 8, 9}, (2) {1, 2, 3, 4} (3) {2, 3, 4, 5}, (4) { }, If B⊆ A then n(A∩B) is, (1) n(A–B), (2) n(B), , (3) n(B – A), , (4) n(A), , If A = {x, y, z} then the number of non- empty subsets of A is, (1) 8, (2) 5, (3) 6, (4) 7, 9th Standard Mathematics, , 1-SET LANGUAGE.indd 36, , 26-12-2019 13:32:34
Page 43 :
www.tntextbooks.in, , 6., 7., 8., 9., , Which of the following is correct?, (1) ∅ ⊆ {a, b}, (2) ∅ ∈ {a, b}, , (3) {a} ∈ {a, b}, , (4) a ⊆ {a, b}, , If A∪B = A∩B, then, (1) A≠B, (2) A = B, , (3) A ⊂ B, , (4) B ⊂ A, , If B – A is B, then A∩B is, (1) A, (2) B, , (3) U, , (4) ∅, , From the adjacent diagram n[P(A∆B)] is, (1) 8, (2) 16, (3) 32, (4) 64, , Fig. 1.40, , 10. If n(A) = 10 and n(B) = 15, then the minimum and maximum number of elements, in A ∩ B is, (1) 10,15, (2) 15,10 (3) 10,0, (4) 0,10, 11. Let A = {∅} and B = P(A), then A∩B is, (1) { ∅, {∅} }, (2) {∅}, (3) ∅, , (4) {0}, , 12. In a class of 50 boys, 35 boys play Carrom and 20 boys play Chess then the number, of boys play both games is, (1) 5, (2) 30, (3) 15, (4) 10., If U = {x : x ∈ and x < 10} , A = {1, 2, 3, 5, 8} and B = {2, 5, 6, 7, 9} , then n (A ∪ B )′ is, , 13., , , , (1) 1, , (2) 2, , (3) 4, , , , (4) 8, , 14. For any three sets P, Q and R, P − (Q ∩ R) is, (1) P − (Q ∪ R) , (2) (P ∩ Q ) − R, (3) (P − Q ) ∪ (P − R) , , (4) (P − Q ) ∩ (P − R), , 15. Which of the following is true?, (1) A − B = A ∩ B , , (2) A − B = B − A, , (3) (A ∪ B )′ = A′ ∪ B ′ , , (4) (A ∩ B )′ = A′ ∪ B ′, , 16. If n(A ∪ B ∪ C ) = 100, n(A) = 4x , n(B ) = 6x , n(C ) = 5x , n(A ∩ B ) = 20,, n(B ∩ C ) = 15, n(A ∩ C ) = 25 and n(A ∩ B ∩ C ) = 10 , then the value of x is, (1) 10, (2) 15, (3) 25, (4) 30, 17. For any three sets A, B and C, (A − B ) ∩ (B − C ) is equal to, (1) A only, (2) B only, (3) C only, (4) f, , Set Language, , 1-SET LANGUAGE.indd 37, , 37, , 26-12-2019 13:32:37
Page 44 :
www.tntextbooks.in, , 18. If J = Set of three sided shapes, K = Set of shapes with two equal sides and L = Set of, shapes with right angle, then J Ç K Ç L is, (1) Set of isoceles triangles, (2) Set of equilateral triangles, (3) Set of isoceles right triangles, (4) Set of right angled triangles, 19. The shaded region in the Venn diagram is, (1) Z − (X ∪ Y ) (2) (X ∪ Y ) ∩ Z, (3) Z − (X ∩ Y ) (4) Z ∪ (X ∩ Y ), 20. In a city, 40% people like only one fruit, 35% people like only two, fruits, 20% people like all the three fruits. How many percentage of people do not like, any one of the above three fruits?, (1) 5, , (2) 8, , (3) 10, , (4) 15, , Points to Remember, A set is a well defined collection of objects., Sets are represented in three forms (i) Descriptive form, , (ii) Set – builder form, , (iii) Roster form., If every element of A is also an element of B, then A is called a subset of B., If A 3 B and A≠B, then A is a proper subset of B., The power set of the set A is the set of all the subsets of A and it is denoted by P(A)., The number of subsets of a set with m elements is 2m., The number of proper subsets of a set with m elements is 2m -1., If A∩B = Q then A and B are disjoint sets. If A∩B ≠Q then A and B are overlapping., The difference of two sets A and B is the set of all elements in A but not in B., The symmetric difference of two sets A and B is the union of A-B and -A., Commutative Property, , For any two sets A and B,, A∪B = B ∪A ;, , 38, , A∩B = B ∩A, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 38, , 26-12-2019 13:32:38
Page 45 :
www.tntextbooks.in, , Associative Property, , For any three sets A, B and C, A ∪ (B ∪ C ) = (A ∪ B ) ∪ C, , ;, , A ∩ (B ∩ C ) = (A ∩ B ) ∩ C, , Distributive Property, , For any three sets A, B and C, A ∩ (B ∪ C ) = (A ∩ B ) ∪ (A ∩ C ), A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ), , Intersection over union , , , , , Union over intersection, , De Morgan’s Laws for Set Difference, , For any three sets A, B and C, A − (B ∪ C ) = (A − B ) ∩ (A − C ), A − (B ∩ C ) = (A − B ) ∪ (A − C ), De Morgan’s Laws for Complementation, , Consider an Universal set and A, B are two subsets, then, (A ∪ B )′ = A′ ∩ B ′ ;, , (A ∩ B )′ = A′ ∪ B ′, , Cardinality of Sets, , If A and B are any two sets, then n(A ∪ B ) = n(A) + n(B ) − n(A ∩ B ), If A, B and C are three sets, then, n(A ∪ B ∪ C ) = n(A) + n(B ) + n(C ) − n(A ∩ B ) − n(B ∩ C ) − n(A ∩ C ) + n(A ∩ B ∩ C ), , Set Language, , 1-SET LANGUAGE.indd 39, , 39, , 26-12-2019 13:32:39
Page 46 :
www.tntextbooks.in, , ICT Corner 1, Expected Result is shown in this picture, , Step – 1 : Open the Browser, type the URL Link given, below (or) Scan the QR Code. GeoGebra work sheet, named “Set Language” will open. In the work sheet there are two activities. 1. Venn Diagram for, two sets and 2. Venn Diagram for three sets. In the first activity Click on the boxes on the right, side to see the respective shading and analyse., Step - 2 : Do the same for the second activity for three sets., Browse in the link, , Set Language: https://ggbm.at/BrG952dw or Scan the QR Code., , ICT Corner 2, Expected Result is shown in this picture, Step – 1, Open the Browser and copy and paste the Link given below (or) by, typing the URL given (or) Scan the QR Code., Step - 2, GeoGebra worksheet “Union of Sets” will appear. You can create new problems by clicking on the box, “NEW PROBLEM”, Step-3, Enter your answer by typing the correct numbers in the Question Box and then hit enter. If you have any, doubt, you can hit the “HINT” button, Step-4, If your answer is correct “GREAT JOB” menu will appear. And if your answer is Wrong “Try Again!”, menu will appear., Keep on working new problems until you get 5 consecutive trials as correct., , Browse in the link, , Union of Sets: https://www.geogebra.org/m/ufxdh47G, 40, , 9th Standard Mathematics, , 1-SET LANGUAGE.indd 40, , 26-12-2019 13:32:39
Page 47 :
2, , cm, 9.3, , www.tntextbooks.in, 1, cm, 9.3, , –4, cm, , –3, , 1, , - 2, , –2, , –1, , 0, , Y, C, , 3, , B, , 2.5, , 2, , 2, , 2, , 1, , 3, , 18, , 1.5, 1, , 2, , 3, , 4, , 0.5, , -0.5, , O, , 0.5, , 1, , -0.5, , 1.5, , A, 2, , 2.5, , 3, , X, , 3, , REAL NUMBERS, , “When I consider what people generally want in, calculating, I found that it always is a number”, - Al-Khwarizmi, Al-Khwarizmi, a Persian scholar, is, credited with identification of surds as, something noticeable in mathematics., He referred to the irrational numbers, as ‘inaudible’, which was later translated to the, Latin word ‘surdus’ (meaning ‘deaf ’ or ‘mute’). In, , Al-Khwarizmi, (A D (CE) 780 - 850), , mathematics, a surd came to mean a root (radical) that, cannot be expressed (spoken) as a rational number., , Learning Outcomes, Â To know that there exists infinitely many rational numbers between two given, rational numbers., Â To represent rational and irrational numbers on number line and express them in, decimal form., Â To visualize the real numbers on the number line., Â To identify surds., Â To carry out basic operations of addition, subtraction, multiplication and division, using surds., Â To rationalise denominators of surds., Â To understand the scientific notation., , 2.1 Introduction, Numbers, numbers, everywhere!, Â Do you have a phone at home? How many digits does its dial have?, Â What is the Pin code of your locality? How is it useful?, Â When you park a vehicle, do you get a ‘token’? What is its purpose?, Â Have you handled 24 ‘carat’ gold? How do you decide its purity?, Â How high is the ‘power’ of your spectacles?, Real Numbers, , 2-Real Numbers.indd 41, , 41, , 26-12-2019 13:34:41
Page 48 :
www.tntextbooks.in, , How much water does the overhead tank in, your house can hold?, Does your friend have fever? What is his, body temperature?, , Â, Â, , You have learnt about many types of numbers so far., Now is the time to extend the ideas further., , 2.2, , Rational Numbers, , When you want to count the number of, books, , in, , your, , cupboard,, , you, , start, , with, , 1, 2, 3, … and so on. These counting numbers 1,, 2, 3, … , are called Natural numbers. You know to, show these numbers on a line (see Fig. 2.1)., 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , Fig. 2.1, , We use N to denote the set of all Natural numbers., N = { 1, 2, 3, … }, , If the cupboard is empty (since no books are there). To denote such a situation we use the, symbol 0. Including zero as a digit you can now consider the numbers 0, 1, 2, 3, … and call, them Whole numbers. With this additional entity, the number line will look as shown below, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , Fig. 2.2, , We use W to denote the set of all Whole numbers., W = { 0, 1, 2, 3, … }, Certain conventions lead to more varieties of numbers. Let us agree that certain, conventions may be thought of as “positive” denoted by a ‘+’ sign. A thing that is ‘up’ or, ‘forward’ or ‘more’ or ‘increasing’ is positive; and anything that is ‘down’ or ‘backward’ or, ‘less’ or ‘decreasing’ is “negative” denoted by a ‘–’ sign., You can treat natural numbers as positive numbers and rename them as positive, integers; thereby you have enabled the entry of negative integers –1, –2, –3, … ., Note that –2 is “more negative” than –1. Therefore, among –1 and –2, you find that –2 is, smaller and –1 is bigger. Are –2 and –1 smaller or greater than –3? Think about it., The number line at this stage may be given as follows:, –4, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , Fig. 2.3, 42, , 2-Real Numbers.indd 42, , 9th Standard Mathematics, , 26-12-2019 13:34:42
Page 49 :
www.tntextbooks.in, , We use Z to denote the set of all Integers., , , Z = { …, –3, –2, –1, 0, 1, 2, 3, … }., , When you look at the figures (Fig. 2.2 and 2.3) above, you are sure to get amused by the, gap between any pair of consecutive integers. Could there be some numbers in between?, You have come across fractions already. How will you mark the point that shows 12 on, , 1 1 1, 3, ... etc., Z ? It is just midway between 0 and 1. In the same way, you can plot , , , 2, 4, 3 4 5, These are all fractions of the form ba where a and b are integers with one restriction that, , b ≠ 0. (Why?) If a fraction is in decimal form, even then the setting is same., , Because of the connection between fractions and ratios of lengths, we name them as, Rational numbers. Here is a rough picture of the situation:, –7, , –6, , –5 –4.3 –4, , –3, , –2, , –1 –0.5 0 0.5 1, , 2, , 3, , 4, , 4.8 5, , 6, , 7, , Fig. 2.4, , A rational number is a fraction indicating the quotient of two integers, excluding division, by zero., Since a fraction can have many equivalent fractions , there are many possible forms for, , 8, the same rational number. Thus 13 , 62 , 24, all these denote the same rational number., , 2.2.1 Denseness Property of Rational Numbers, +, Consider a, b where a > b and their AM(Arithmetic Mean) given by a 2 b . Is this AM, a rational number? Let us see., p, If a = q (p, q integers and q ! 0 ); b = rs (r, s integers and s ! 0 ), then, p, r, a + b = q + s = ps + qr, which is a rational number., 2, 2, 2qs, , We have to show that this rational number lies between a and b., a-`, , a + b = 2a - a - b = a - b, which is > 0 since a>b., 2 j, 2, 2, , +, Therefore, a >` a 2 b j ... (1), a+ b, a + b - 2b = a - b, which is > 0 since a>b., ` 2 j- b=, 2, 2, +, Therefore, ` a 2 b j > b , ... (2), +, From (1) and (2) we see that a >` a 2 b j >b, which can be visualized as follows:, Real Numbers, , 2-Real Numbers.indd 43, , 43, , 26-12-2019 13:34:44
Page 50 :
www.tntextbooks.in, , a+ b, 2, , b, , a, , Fig. 2.5, , Thus, for any two rational numbers, their average/mid point is rational. Proceeding, similarly, we can generate infinitely many rational numbers., Example 2.1, , Find any two rational numbers between 12 and 32 ., , Solution 1, +, 7, A rational number between 12 and 32 = 12 ` 12 + 32 j = 12 ` 3 6 4 j = 12 ` 76 j = 12, 7, 7 = 1 6 + 7 = 1 13 = 13, A rational number between 12 and 12, = 12 ` 12 + 12, j 2 ` 12 j 2 ` 12 j 24, 7, Hence two rational numbers between 12 and 32 are 12, and 13, (of course, there are, 24, many more!), There is an interesting result that could help you to write instantly rational numbers, between any two given rational numbers., , Result, p, , p, , p+ r, , is a rational number,, If q and rs are any two rational numbers such that q < rs , then, q+ s, p p+ r, such that q < + < rs ., q, , s, , Let us take the same example: Find any two rational numbers between 12 and 32, Solution 2, , +, +, +, 1 2, < gives 12 < 1 + 2 < 32 or 12 < 53 < 32 gives 12 < 1 + 3 < 53 < 3 + 2 < 32 or 12 < 74 < 53 < 58 1 32, 2 3, 2 3, 2 5, 5 3, , Solution 3, Any more new methods to solve? Yes, if decimals are your favourites, then the above, example can be given an alternate solution as follows:, 1, = 0.5 and 32 = 0.66..., 2, Hence rational numbers between 12 and 32 can be listed as 0.51, 0.57,0.58,…, , , Solution 4, , There is one more way to solve some problems. For example, to find four rational, , numbers between 94 and 35 , note that the LCM of 9 and 5 is 45; so we can write 94 = 20, and, 45, 3 27, =, ., 5 45, 22 23 24, Therefore, four rational numbers between 94 and 35 are 21, ,, ,, ,, , ..., 45 45 45 45, , 44, , 2-Real Numbers.indd 44, , 9th Standard Mathematics, , 26-12-2019 13:34:49
Page 51 :
www.tntextbooks.in, , Exercise 2.1, 1., , Which arrow best shows the position of 11, on the number line?, 3, A, , –5, , –4, , B, , –3, , –2, , –1, , 0, , 1, , 2, , C D, , 3, , 4, , 5, , 2, 2. Find any three rational numbers between 117 and 11, ., -, , 3., , Find any five rational numbers between (i) 14 and 15 (ii) 0.1 and 0.11 (iii) –1 and –2, , 2.3 Irrational Numbers, , You know that each rational number is assigned to a point on the, number line and learnt about the denseness property of the rational numbers., Does it mean that the line is entirely filled with the rational numbers and, there are no more numbers on the number line? Let us explore., , 2, , 1, , 1, Consider an isosceles right-angled triangle whose base and height, Fig.2.6, are each 1 unit long. Using Pythagoras theorem, the hypotenuse can, be seen having a length 12 + 12 = 2 (see Fig. 2.6 ). Greeks found that this 2 is neither, a whole number nor an ordinary fraction. The belief of relationship between points on the, number line and all numbers was shattered! 2 was called an irrational number., , An irrational number is a number that cannot be expressed as an ordinary ratio of two integers., , Examples, , GOLDEN RATIO (1:1.6), The Golden Ratio has, a, b, been heralded as the, most beautiful ratio in, a+b, art and architecture., a+b : a = a : b, Take a line segment and, divide it into two smaller segments such that the, ratio of the whole line segment (a+b) to segment, a is the same as the ratio of segment a to the, segment b., , 1., , Apart from, 2 , one can, produce a number of examples, for such irrational numbers., Here are a few: 5 , 7 , 2 3 , f, , 2., , π , the ratio of the circumference, of a circle to the diameter of that, same circle, is another example, for an irrational number., , 3., , e, also known as Euler’s, number, is another common, irrational number., , 4., , The golden ratio, also known as golden mean, or golden section, is a number often, stumbled upon when taking the ratios of distances in simple geometric figures such, as the pentagon, the pentagram, decagon and dodecahedron, etc., it is an irrational, number., , +, This gives the proportion a a b = ab, Notice that ‘a’ is the geometric mean of a+b and b., , Real Numbers, , 2-Real Numbers.indd 45, , 45, , 26-12-2019 13:34:50
Page 52 :
www.tntextbooks.in, , 2.3.1 Irrational Numbers on the Number Line, Where are the points on the number line that correspond to the irrational numbers?, As an example, let us locate, Remember that, , 2 on the number line. This is easy., , 2 is the length of the diagonal of the square whose side is 1 unit, , (How?)Simply construct a square and transfer the length of one of its diagonals to our, number line. (see Fig.2.7)., 1, 1, , - 2, , –4, , –3, , –2, , –1, , 2, , 2, , 0, , 1, , 2, , 3, , 4, , Fig.2.7, , You started with Natural numbers, and extended it to rational numbers, and then irrational numbers. You may, wonder if further extension on the, number line waits for us. Fortunately, it stops and you can learn about it in, higher classes., , Squares on grid sheets can be used to, produce irrational lengths., Here are a few examples :, 5, , 10, , 5, , 10, , 8 or, 2 2, 8, , 2, , 2, 2, , 2, 2, 2, , 2, 2, , 2, , 18, , We draw a circle with centre at 0, on the number line,with a radius equal, to that of diagonal of the square. This, circle cuts the number line in two points,, locating 2 on the right of 0 and – 2, on its left. (You wanted to locate 2 ;, you have also got a bonus in – 2 ), , 3, , 2, , =, , Representation of a Rational, number as terminating and non, terminating decimal, helps us to, understand irrational numbers. Let us see the decimal expansion of rational numbers., , 2.3.2 Decimal Representation of a Rational Number, If you have a rational number written as a fraction, you get the decimal representation, by long division. Study the following examples where the remainder is always zero., , 46, , 2-Real Numbers.indd 46, , 9th Standard Mathematics, , 26-12-2019 13:34:51
Page 53 :
www.tntextbooks.in, , Consider the examples,, 2.71875, 32 87.00000, 64, , 230, 224, 60, 32, 280, 256, 21 =, 0.84 , 240, 25, 224, 160, 160, 0, , 0.875, 8 7.000, 64, 60, 56, 40, 40, 0, , 0.84, 25 21.00, 200, 100, 100, 0, , 7 =, 0.875 , 8, , - 87, = - 2.71875, 32, , Note, These show that the process could lead to a decimal with finite, number of decimal places. They are called terminating decimals., , Can the decimal representation of a rational number lead to forms of decimals that do not, terminate? The following examples (with non-zero remainder) throw some light on this point., Example 2.2, decimal form (i) 114, Solution, -, , 0.3636…., 11 4.0000, 33, 70, 66, 40, 33, 70, 66, 4, h, , Represent the following as The reciprocals of Natural Numbers are, (ii), , 11, 75, , Rational numbers. It is interesting to note, their decimal forms. See the first ten., S.No., , 0.1466…, 75 11.0000, 75, 350, 300, 500, 450, 500, 450, 50, h, , = 0.146, Thus we see that, 114 = - 0.36 11, 75, , Reciprocal, , Decimal, Representation, , 1, , 1 =, 1.0, 1, , Terminating, , 2, , 1 =, 0.5, 2, , Terminating, , 3, , 1 =, 0.3, 3, , Non-terminating, Recurring, , 4, , 1 =, 0.25, 4, 1 =, 0.2, 5, , Terminating, , 1 =, 0.1 6, 6, , Non-terminating, Recurring, , 5, 6, , Terminating, , Real Numbers, , 2-Real Numbers.indd 47, , 47, , 26-12-2019 13:34:53
Page 54 :
www.tntextbooks.in, , A rational number can be expressed by, (i), , either a terminating, , (ii) or a non-terminating and recurring, (repeating) decimal expansion., The converse of this statement is also true., That is, if the decimal expansion of a number, is terminating or non-terminating and recurring,, then the number is a rational number., , 7, 8, , Non-terminating, 1 =, 0.142857, Recurring, 7, 1 =, 0.125, 8, , Terminating, , 9, , 1 =, 0.1, 9, , 10, , Non-terminating, Recurring, , 1 =, 0.1, 10, , Terminating, , Note, In this case the decimal expansion does not terminate!, The remainders repeat again and again! We get non-terminating but recurring block of digits., , 2.3.3 Period of Decimal, In the decimal expansion of the rational numbers, the number of repeating decimals is, called the length of the period of decimals., , For example,, 25 =, 3.571428 has the length of the period of decimal = 6, 7, 27 =, (ii) 110, 0.2 45 has the length of the period of decimal = 2, , (i), , Example 2.3, , 1, Express the rational number 27, in recurring decimal form by using, , in recurring decimal form., the recurring decimal expansion of 13 . Hence write 59, 27, Solution, We know that 13 = 0.3, Therefore,, Also,, , , 1 = 1 #1 = 1 #, 0.333... = 0.037037... = 0.037, 9 3 9, 27, 59 = 5 = + 5, 2, 2, 27, 27, 27, = 2 + `5 # 1 j, 27, , = 2 + (5 # 0. 037) = 2 + (5 # 0.037037037...) = 2 + 0.185185... = 2.185185... = 2. 185, , 2.3.4 Conversion of Terminating Decimals into Rational Numbers, Let us now try to convert a terminating decimal, say 2.945 as rational number in the, fraction form., , 2.945, = 2 + 0.945, 48, , 2-Real Numbers.indd 48, , 9th Standard Mathematics, , 26-12-2019 13:34:55
Page 55 :
www.tntextbooks.in, , , , 2.945, , 9 + 4 + 5, = 2 + 10, 100 1000, , , , 900 + 40 + 5, = 2 + 1000, (making denominators common), 1000 1000, , , , 945, = 2 + 1000, , , , 2945, 589, = 1000, which is required, or, 200, , 2945, (In the above, is it possible to write directly 2.945 = 1000, ?), , Example 2.4, , p, , Convert the following decimal numbers in the form of q , where p, and q are integers and q ! 0 : (i) 0.35 (ii) 2.176 (iii) – 0.0028, Solution, (i) , , 35, 7, 0.35 = 100, = 20, , 2176, 272, (ii) 2.176 = 1000, = 125, , 28, 7, (iii) –0.0028 = 10000, = 2500, -, , -, , 2.3.5 Conversion of Non-terminating and recurring decimals into Rational Numbers, It was very easy to handle a terminating decimal. When we come across a decimal, such as 2.4, we get rid of the decimal point, by just divide it by 10., 24, Thus 2.4 = 10, , which is simplified as 12, . But, when we have a decimal such as 2.4 ,, 5, , the problem is that we have infinite number of 4s and hence will need infinite number of 0s, in the denominator. For example,, , , 4, 2.4 = 2 + 10, , , , 4 + 4, 2.44 = 2 + 10, 100, , , , 4 + 4 + 4, 2.444 = 2 + 10, 100 1000, , How tough it is to have infinite 4’s and work with them. We need to get rid of the, infinite sequence in some way. The good thing about the infinite sequence is that even if we, pull away one , two or more 4 out of it, the sequence still remains infinite., Let, , x = 2. 4 , , ...(1), , Then, 10x = 24. 4 , ...(2) [When you multiply by 10, the decimal moves one, place to the right but you still have infinite 4s left over)., Subtract the first equation from the second to get,, 9x = 24. 4 – 2. 4 = 22 (Infinite 4s subtract out the infinite 4s and the left out is, 24 – 2 = 22), Real Numbers, , 2-Real Numbers.indd 49, , 49, , 26-12-2019 13:34:57
Page 56 :
www.tntextbooks.in, , , , x = 22, , the required value., 9, , We use the same exact logic to convert any number with a non terminating repeating, part into a fraction., Example 2.5, , Convert the following decimal numbers in the form of, , p, ^ p, q ! Z and q ! 0h ., q, , (i) 0.3 (ii) 2.124, , (iii) 0.45, , (iv) 0.568, , Solution, (i) Let x = 0.3 = 0.3333… , , , , , (Here period of decimal is 1, multiply equation (1) by 10), 10 x = 3.3333..., (2) – (1):, , , , 9 x = 3 or, , x = 13, , (ii) Let x = 2.124 = 2.124124124…, , , (2)–(1): 999 x = 2122, , , , (1), (2), , x = 2122, 999, , (iii) Let x = 0.4 5 = 0.45555… , , , (2), , (Here period of decimal is 3, multiply equation (1) by 1000.), 1000 x = 2124.124124124… , , , , (1), , (1), , (Here the repeating decimal digit is 5, which is the second digit after the decimal, point, multiply equation (1) by 10), 10 x = 4.5555… , , (2), , , , (Now period of decimal is 1, multiply equation (2) by 10), , , , 100 x = 45.5555… , , , , (3) – (2):, , 41, 90 x = 41 or x = 90, , (iv) Let x = 0.568 = 0.5686868… , , , , , 10 x = 5.686868… , , , , (2), , (Now period of decimal is 2, multiply equation (2) by 100), , 1000 x = 568.686868… , , 2-Real Numbers.indd 50, , (1), , (Here the repeating decimal digit is 68, which is the second digit after the decimal, point, so multiply equation (1) by 10), , , , 50, , (3), , (3), , (3) – (2): 990 x = 563 or x = 563, ., 990, , 9th Standard Mathematics, , 26-12-2019 13:35:00
Page 57 :
www.tntextbooks.in, , Note, To determine whether the decimal form of a rational number will terminate or non - terminate,, we can make use of the following rule, p, , p, , If a rational number q , q ! 0 can be expressed in the form m # n , where p ! Z and, 2, 5, m, n ! W , then rational number will have a terminating decimal expansion. Otherwise,, the rational number will have a non- terminating and recurring decimal expansion, Example 2.6, , Without actual division, classify the decimal expansion of the, following numbers as terminating or non – terminating and recurring., (i) 13, , 64, Solution, (a), (b), (c), (d), , 43, 31, 71, (ii) 125, (iii) 375, (iv) 400, -, , 13, = 136 ,, So 13, has a terminating decimal expansion., 64, 64, 2, - 71, - 71, - 71, =, ,, So, has a terminating decimal expansion., 3, 125, 125, 5, 43, 43, = 1 43 3 , So 375, has a non – terminating recurring decimal expansion., 375, 3 #5, 31, 31, = 4 31 2 , So 400, has a terminating decimal expansion., 400, #, 2 5, , Example 2.7, , Verify that 1 = 0.9, , Solution, Let x = 0.9 = 0.99999… , , (1), (2), , Subtract (1) from (2), 9x = 9 or x =1, Thus,, , 7 = 6.9999…, 3.7 = 3.6999…, , (Multiply equation (1) by 10), 10 x = 9.99999… , , 1 = 0.9999…, , 0.9 = 1, , The pattern suggests that, any terminating decimal, can be represented as a nonterminating and recurring, decimal expansion with an, endless block of 9’s., , Exercise 2.2, 1., , 2., , Express the following rational numbers into decimal and state the kind of decimal expansion, 2, 3, 22, 327, (i) (ii) - 5 , (iii), (iv), 7, 3, 200, 11, 1, Express 13, in decimal form. Find the length of the period of decimals., Real Numbers, , 2-Real Numbers.indd 51, , 51, , 26-12-2019 13:35:05
Page 58 :
www.tntextbooks.in, , 3., , 1, Express the rational number 33, in recurring decimal form by using the recurring, 1, in recurring decimal form., decimal expansion of 11 . Hence write 71, 33, , 4., , Express the following decimal expression into rational numbers., , 5., , (i) 0.24, , (ii) 2.327, , (iii) –5.132, , (iv) 3.17, , (v) 17.2 15, , (vi) - 21.2137, , Without actual division, find which of the following rational numbers have terminating, decimal expansion., (i), , 7, 128, , (ii), , 21, 15, , (iii) 4, , 9, 35, , (iv), , 219, 2200, , 2.3.6 Decimal Representation to Identify Irrational Numbers, It can be shown that irrational numbers, when expressed as decimal numbers, do not, terminate, nor do they repeat. For example, the decimal representation of the number π, starts with 3.14159265358979..., but no finite number of digits can represent π exactly, nor, does it repeat., Consider the following decimal expansions:, (i) 0.1011001110001111… , , (ii) 3.012012120121212…, , (iii) 12.230223300222333000… , , (in), , 2 = 1.4142135624…, , Are the above numbers terminating (or) recurring and non- terminating? No… They, are neither terminating, nor non–terminating and recurring. Hence they are not rational, p, numbers. They cannot be written in the form of q ,where p, q , ! Z and q ! 0 . They are, irrational numbers., A number having non- terminating and non- recurring decimal expansion is an irrational number., , Example 2.8, Solution, , Find the decimal expansion of, , 3 ., , We often write, 2 = 1.414,, 1.7320508…, 1 3.00,00,00,00,00,…, 3 = 1.732, π = 3.14 etc. These, , 1, are only approximate values and, 27 200, not exact values. In the case of, 189, the irrational number π, we take, 343 1100, , (which gives the, frequently 22, 1029, 7, 3462, 7100, value 3.142857) to be its correct, 6924, value but in reality these are, 346405, 1760000, only approximations., This is, 1732025, because, the decimal expansion, 34641008, 279750000, of an irrational number is non , 277128064, terminating and non-recurring., 2621936, , , None of them gives an exact value!, , 52, , 2-Real Numbers.indd 52, , 9th Standard Mathematics, , 26-12-2019 13:35:07
Page 59 :
www.tntextbooks.in, , Thus, by division method,, , 3 = 1.7320508…, , It is found that the square root of every positive non perfect square number is an, irrational number., , 2 , 3 , 5 , 6 , 7 , … are all irrational numbers., , Example 2.9, (i) 10 , Solution, (i), (ii), , Classify the numbers as rational or irrational:, (ii), , 49, , (iii) 0.025, , (iv) 0.76, , (v) 2.505500555..., , 2, , (vi) 2, , 10 is an irrational number ( since 10 is not a perfect square number)., 49 = 7 =, , 7, , a rational number(since 49 is a perfect square number). , 1, , (iii) 0.025 is a rational number (since it is a terminating decimal). , (iv) 0.7 6 = 0.7666…. is a rational number ( since it is a non – terminating and, recurring decimal expansion)., (v) 2.505500555…. is an irrational number ( since it is a non – terminating and, non–recurring decimal). , (vi), , 2, =, 2, , number)., , 2, =, 2# 2, , 1, is an irrational number ( since 2 is not a perfect square, 2, , Note, The above example(vi) it is not to be misunderstood as, be integers and not an irrational number., , Example 2.10, , p, form, because both p and q must, q, , Find any 3 irrational numbers between 0.12 and 0.13 ., , Solution, Three irrational numbers between 0.12 and 0.13, , are, , 0.12010010001…, 0.12040040004…, 0.12070070007…, Note, We state (without proof) an important result worth remembering., If ‘a’ is a rational number and b is an irrational number then each one of the, following is an irrational number:, b, (i) a + b ; (ii) a – b ;, (iii) a b ;, (iv) a ;, (v) a ., b, , For example, when you consider the rational number 4 and the irrational number 5 ,, then, , 5, 4 + 5 , 4 – 5 , 4 5 , 4 , 4 ... are all irrational numbers., 5, Real Numbers, , 2-Real Numbers.indd 53, , 53, , 26-12-2019 13:35:09
Page 60 :
www.tntextbooks.in, , Example 2.11, , Give any two rational numbers lying between 0.5151151115…. and, , 0.5353353335…, Solution Two rational numbers between the given two irrational numbers are 0.5152 and 0.5352, Example 2.12, , Find whether x and y are rational or irrational in the following., , (i) a = 2 + 3 , b = 2 - 3 ; x = a + b, y = a - b, (ii) a =, , 2 + 7, b =, , 2 - 7 ; x = a + b, y = a - b, Note, (iii) a = 75 , b = 3 ; x = ab, y = ba, From these, (iv) a = 18 , b = 3 ; x = ab, y = ba, examples, it is clear that the sum ,, , difference, product, quotient of any two, irrational numbers could be rational or, irrational., , Solution, (i) Given that a = 2 + 3 , b = 2 - 3, x = a + b = (2 + 3 ) +(2 – 3 ) = 4 ,, a rational number. , y = a – b = (2 +, (ii) Given that a =, , 3 ) – (2– 3 ) = 2 3 , an irrational number., 2 +7 , b =, , 2 –7, , x = a + b = ( 2 +7 ) +( 2 –7) = 2 2 , an irrational number., y = a – b = ( 2 +7 )–( 2 –7) = 14 , a rational number. , (iii) Given that a = 75 , b =, x = ab =, , 75 # 3 =, , y = ba =, , 75, =, 3, , 75 =, 3, , (iv) Given that a = 18 ,, , 3, , 75 # 3 =, , 5 # 5 # 3 # 3 = 5 # 3 = 15 , a rational number., , 25 = 5 , rational number., , b=, , 3, , x = ab = 18 # 3 = 18 # 3 = 6 # 3 # 3 = 3 6 , an irrational number., 18, =, y = ba =, 3, , Example 2.13, , 18 =, 3, , 6 , an irrational number., , Represent 9.3 on a number line., , Solution, ÂÂ Draw a line and mark a point A on it., ÂÂMark a point B such that AB = 9.3 cm., ÂÂMark a point C on this line such that BC = 1 cm., ÂÂFind the midpoint of AC by drawing perpendicular bisector of AC and let it be O, ÂÂ With O as center and OC = OA as radius, draw a semicircle., 54, , 2-Real Numbers.indd 54, , 9th Standard Mathematics, , 26-12-2019 13:35:12
Page 61 :
www.tntextbooks.in, , 9.3 cm, , ÂÂDraw a line BD, which is perpendicular to AB at B., ÂÂNow BD =, 9.3 , which can be marked in the number line as the value of, BE = BD = 9.3, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , cm, 0 cm, 9, 10, , 8, , cm, , Fig.2.8, , 9.3 cm, , Exercise 2.3, 1., , Represent the following irrational numbers on the number line., (i), , 2., , (ii), , 4.7, , (iii), , 6.5, , Find any two irrational numbers between, , 6, 12, (iii) 2 and, and, 7, 13, Find any two rational numbers between 2.2360679….. and 2.236505500…., (i), , 3., , 3, , 0.3010011000111…. and 0.3020020002…., , (ii), , 3, , 2.4 Real Numbers, , The real numbers consist of all the rational numbers and all the irrational numbers., Real numbers can be thought of as points on an infinitely long number line called the, real line, where the points corresponding to integers are equally spaced., Real Numbers, , Rational, Numbers, , Terminating, , Irrational, Numbers, , Non-terminating, and Recurring, , Non-terminating, and non-recurring, , Real Numbers, , 2-Real Numbers.indd 55, , 55, , 26-12-2019 13:35:13
Page 62 :
www.tntextbooks.in, , Any real number can be determined by a possibly infinite decimal representation, (as we, have already seen decimal representation of the rational numbers and the irrational numbers)., , 2.4.1, , The Real Number Line, , Visualisation through Successive Magnification., We can visualise the representation of numbers on the number line, as if we glimpse, through a magnifying glass., Example 2.14, , Represent 4.863 on the number line., Solution , 4.863 lies between 4 and 5(see Fig. 2.9), (i), , Divide the distance between 4 and 5 into 10 equal intervals., , (ii) Mark the point 4.8 which is second from the left of 5 and eighth from the right of 4, (iii) 4.86 lies between 4.8 and 4.9. Divide the distance into 10 equal intervals., (iv) Mark the point 4.86 which is fourth from the left of 4.9 and sixth from the right of 4.8, (v) 4.863 lies between 4.86 and 4.87. Divide the distance into 10 equal intervals., (vi) Mark point 4.863 which is seventh from the left of 4.87 and third from the right of, 4.86., -3, , -2, , -1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 4.0, , 4.1, , 4.2, , 4.3, , 4.4, , 4.5, , 4.6, , 4.7, , 4.8, , 4.9, , 5.0, , 4.80, , 4.81, , 4.82, , 4.83, , 4.84, , 4.85, , 4.86, , 4.87, , 4.88, , 4.89, , 4.90, , 4.860 4.861, , Example 2.15, , 4.862 4.863 4.864 4.865, , 4.866 4.867 4.868 4.869 4.870, , Fig. 2.9, , Represent 3.45 on the number line upto 4 decimal places., , Solution , 3.45 = 3.45454545….., 56, , 2-Real Numbers.indd 56, , 9th Standard Mathematics, , 26-12-2019 13:35:13
Page 63 :
www.tntextbooks.in, , , , = 3.4545 ( correct to 4 decimal places)., , The number lies between 3 and 4, , -4, , -3, , -2, , -1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 3.0, , 3.1, , 3.2, , 3.3, , 3.4, , 3.5, , 3.6, , 3.7, , 3.8, , 3.9, , 4.0, , 3.40, , 3.41, , 3.42, , 3.43, , 3.44, , 3.45, , 3.46, , 3.47, , 3.48, , 3.49, , 3.50, , 3.450 3.451, , 3.452, , 3.453, , 3.454, , 3.455, , 3.456, , 3.457, , 3.458, , 3.459, , 3.460, , 3.4540 3.4541 3.4542 3.4543 3.4544 3.4545 3.4546 3.4547 3.4548 3.4549, , 3.4550, , Fig. 2.10, , Exercise 2.4, 1., , Represent the following numbers on the number line., (i) 5.348 (ii) 6.4 upto 3 decimal places., , (iii) 4.73 upto 4 decimal places., , 2.5 Radical Notation, Let n be a positive integer and r be a real number. If, rn = x, then r is called the nth root of x and we write, n, , x =r, The symbol n (read as nth root) is called a radical;, n is the index of the radical (hitherto we named it as, exponent); and x is called the radicand., , Note, It is worth spending some, time on the concepts of the, ‘square root’ and the ‘cube, root’, for better understanding, of surds., Real Numbers, , 2-Real Numbers.indd 57, , 57, , 26-12-2019 13:35:14
Page 64 :
www.tntextbooks.in, , What happens when n = 2? Then we get r 2 = x, so that r is, , 2, , x , our good old friend,, , the square root of x. Thus 2 16 is written as 16 , and when n =3, we get the cube root of x,, namely 3 x . For example,, , 3, , 8 is cube root of 8, giving 2. (Is not 8 = 23?), , How many square roots are there for 4? Since (+2)×(+2), = 4 and also (–2)×(–2) = 4, we can say that both +2 and –2 are Thinking Corner, square roots of 4. But it is incorrect to write that 4 = ± 2 ., Which one of the, This is because, when n is even, it is an accepted convention to, following is false?, n, th, reserve the symbol x for the positive n root and to denote, (1) The square root of, the negative nth root by – n x . Therefore we need to write, 9 is 3 or –3., 4 = 2 and − 4 =− 2., (2) 9 = 3, When n is odd, for any value of x, there is exactly one real, (3) − 9 = − 3, nth root. For example, 3 8 = 2 and 5 −32 =− 2., (4), 9 = ±3, , 2.5.1 Fractional Index, , Consider again results of the form r =, , n, , x., , In the adjacent notation, the index of the radical (namely Root Index Radical Symbol, n which is 3 here) tells you how many times the answer (that, 3, 64 = 4, is 4) must be multiplied with itself to yield the radicand., To express the powers and roots, there is one more way, Radicand, Number, of representation. It involves the use of fractional indices., We write, , n, , 1, n, , x as x ., , With this notation, for example, 3, , 1, 3, , 1, 2, , 64 is 64 and 25 is 25 ., Observe in the following table just some representative patterns arising out of this new acquaintance:, Power, Radical Notation, Index Notation, Read as, 26 = 64, , 2 = 64, , 25 = 32, , 2 = 5 32, , 6, , 24 = 16, , 4, , 2 = 16, , 23 = 8, 2 =4, 2, , 3, , 2= 8, , 58, , 2-Real Numbers.indd 58, , 2 is the 6th root of 64, , 1, 5, , 2 is the 5th root of 32, , 1, 4, , 2 is the 4th root of 16, , 1, 3, , 2 is the cube root of 8, meaning 2 is the 3rd root of 8, , 1, 2, , 2 is the square root of 4, meaning 2 is the 2nd root of 4, , 2 = 64, 2 = 32, 2 = 16, 2=8, , 2 = 2 4 or simply, 2= 4, , 1, 6, , 2=4, , 9th Standard Mathematics, , 26-12-2019 13:35:16
Page 65 :
www.tntextbooks.in, , Example 2.16, , Express the following in the form 2n :, 1, (ii) 32, (iii), (iv) 2, (v), 4, , (i) 8, Solution, , (i) 8 = 2 × 2 × 2 ;, , therefore 8 = 2, , (ii) 32 = 2 × 2 × 2 × 2 × 2 = 2, , 3, , 5, , (iii), , 1/2, , (iv) 2 = 2, (v), , m, x n , (where m and n, , We interpret, , m, xn, , m, In=, symbols, x n, , Example 2.17, , (ii) 64, , −2, 3, , 1, 1, m n, n, (=, or ( )m, , x ), , x, , n, , x m or (n x )m, , Find the value of (i) 81, , Solution, (i) 81 =, , are Positive Integers), , either as the nth root of the mth power of x or as the mth power of, , the nth root of x., , 5, 4, , 1, 1, 1, −2, =, = 2 =2, 4, 2×2, 2, , 3, 1 3, 2 , 2, 2 × 2 × 2 = 2 which may be written as 2, , , , 8 =, , Meaning of, , 8., , ( 81), , 5, , 4, , =, , 1, 64, , 2, 3, , 5, 4, , (ii) 64, , -2, 3, , 4 4 5, 5, = 3 = 3 = 3 × 3 × 3 × 3 × 3 = 243, , , 1, 1, 1, =, =, (How?), =, 2, 2, 16, 3, 4, 64, , ( ), , Exercise 2.5, 1. Write the following in the form of 5n:, (i) 625, , (ii), , 1, , (iii), , 5, , (iv), , 5, , 125, , 2. Write the following in the form of 4n:, (i) 16, , (ii) 8, , (iii) 32, , 3. Find the value of, 1, 2, , ( ), , (i) 49, , 2, 5, , ( ), , (ii) 243, , (iii) 9, , 4. Use a fractional index to write:, (i), , 5, , (ii), , 2, , 7, , (iii), , −2, , -3, 2, , 64 3, , (vi) , 125 , , ( 49 ), 3, , 5, , 1 7, , (iv) , 3, 100 , Real Numbers, , 2-Real Numbers.indd 59, , 59, , 26-12-2019 13:35:18
Page 66 :
www.tntextbooks.in, , 5. Find the 5th root of, (i) 32, , (ii) 243, , (iii) 100000, , (iv), , 2.6 Surds, , 1024, 3125, , Having familiarized with the concept of Real numbers, representing them on the, number line and manipulating them, we now learn about surds, a distinctive way of, representing certain approximate values., Can you simplify, number 2. How about, , 4 and remove the, , symbol? Yes; one can replace, , 1, ? It is easy; without, 9, , symbol, the answer is, , 4 by the, , 1, . What about, 3, , 0.01 ?. This is also easy and the solution is 0.1, In the cases of, that the symbol, , 4,, , 1, and, 9, , 0.01 , you can resolve to get a solution and make sure, , is not seen in your solution. Is this possible at all times?, , Consider 18 . Can you evaluate it and also remove the radical symbol? Surds are, unresolved radicals, such as square root of 2, cube root of 5, etc., They are irrational roots of equations with rational coefficients., A surd is an irrational root of a rational number., is rational., , Examples :, , n, , a is a surd, provided n ,, , n 1 , ‘a’, , 2 is a surd. It is an irrational root of the equation x2 = 2. (Note that, , x2 – 2 = 0 is an equation with rational coefficients., , 2 is irrational and may be shown as, , 1.4142135… a non-recurring, non-terminating decimal)., 3, , 1, 3, , 3 (which is same as 3 ) is a surd since it is an irrational root of the equation, , x3 – 3 = 0. ( 3 is irrational and may be shown as 1.7320508… a non-recurring,, non-terminating decimal)., You will learn solving (quadratic) equations like x2 – 6x + 7 = 0 in your next class. This, is an equation with rational coefficients and one of its roots is 3 + 2 , which is a surd., 1, 1, a surd? No; it can be simplified and written as rational number . How about, 5, 25, 16, 2, ? It is not a surd because it can be simplified as ., 3, 81, Is, , 4, , The famous irrational number, , p is not a surd!, , expressed as a rational number under the, , Though it is irrational, it cannot be, , symbol. (In other words, it is not a root of any, , equation with rational co-efficients)., 60, , 2-Real Numbers.indd 60, , 9th Standard Mathematics, , 26-12-2019 13:35:19
Page 67 :
www.tntextbooks.in, , Why surds are important? For calculation purposes we assume approximate value as, 2 = 1.414 , 3 = 1.732 and so on., , ( ), , , , 2, , 2, , ( ), , 2, , = (1.414) = 1.99936 ≠ 2 ;, , 3, , 2, , 2, , = (1.732) = 3.999824 ≠ 3, , Hence, we observe that 2 and 3 represent the more accurate and precise values than, their assumed values. Engineers and scientists need more accurate values while constructing, the bridges and for architectural works. Thus it becomes essential to learn surds., Progress Check, 1. Which is the odd one out? Justify your answer., (i), , 50, ,, 98, , 36 ,, , 1,, , 5, , 1.44 ,, , 32 ,, , 120, , 1, 10, 2. Are all surds irrational numbers? - Discuss with your answer., (ii), , 7,, , 3, , 48,, , 5 + 3,, , 36,, , 1.21,, , 3. Are all irrational numbers surds? Verify with some examples., , 2.6.1 Order of a Surd, The order of a surd is the index of the root to be extracted. The order of the surd n a, is n. What is the order of, , 5, , 99 ? It is 5., , Surds can be classified in different ways:, (i) Surds of same order : Surds of same order are surds for which the index of the root to, be extracted is same. (They are also called equiradical surds)., 3, 2, , For example,, , , , x, a ,, 3, , 5,, 3,, , 3, , 3, , 2, , m are all 2nd order (called quadratic) surds ., 1, 3, , (x - 2), (ab) are all 3rd order (called cubic) surds., , 10 ,, , 4, , 6 and 8, , 2, 5, , are surds of different order., , (ii) Simplest form of a surd : A surd is said to be in simplest form, when it is expressed, as the product of a rational factor and an irrational factor. In this form the surd has, , (a) the smallest possible index of the radical sign., , , (b) no fraction under the radical sign., , , , (c) no factor is of the form an, where a is a positive integer under index n., Real Numbers, , 2-Real Numbers.indd 61, , 61, , 26-12-2019 13:35:20
Page 68 :
www.tntextbooks.in, , Example 2.18, (i), , Can you reduce the following numbers to surds of same order :, 4, , (ii), , 3, , (iii) 3 3, , 3, , Solution, (i) 3 = 3, , 1, 2, , =3, , (ii), , 4, , 3 =3, , 6, 12, , 1, 4, , =3, , (iii), , 3, , 3, 12, , 3, , =3, =3, , 1, 3, , 4, 12, , =, , 12, , 3, , 6, , =, , 12 3, , 3, , =, , 12, , 3, , 4, , =, , 12, , 729, , =, , 12, , 27, , =, , 12, , 81, , The last row has surds of same order., Example 2.19, 1., , Express the surds in the simplest form: i), , 2., , Show that, , 3, , (i), , =, , 8, , ii) 3 192, , 7 > 45., , Solution, 1., , (ii), 2., , 3, , 8, 3, , 7 =, , 4 ×2, 3, , 192 =, 12, , =, , 2 2, , 4×4×4×3 = 43 3, , 4, , 7 = 12 2401, , 1, 4, , 3, 12, , 4, , 5=5 =5, , 12, , 2401 > 12 125, , Therefore,, , 3, , =, , 12, , 3, , 5 = 12 125, , 7 > 45., , (iii) Pure and Mixed Surds : A surd is called a pure surd if its coefficient in its simplest form, is 1. For example,, , 3,, , 3, , 6,, , 4, , 7,, , 5, , 49 are pure surds. A surd is called a mixed surd if, , its co-efficient in its simplest form is other than 1. For example, 5 3 , 2 4 5 , 3, mixed surds., , 4, , 6 are, , (iv) Simple and Compound Surds : A surd with a single term is said to be a simple surd., For example,, , 3 , 2 5 are simple surds. The algebraic sum of two (or more) surds is, , called a compound surd. For example,, compound surds., 62, , 2-Real Numbers.indd 62, , 5 + 3 2, 3 − 2 7 , 5 − 7 2 + 6 3 are, , 9th Standard Mathematics, , 26-12-2019 13:35:22
Page 69 :
www.tntextbooks.in, , (v), , Binomial Surd : A binomial surd is an algebraic sum (or difference) of 2 terms both of, which could be surds or one could be a rational number and another a surd. For, 1, − 19, 5 + 3 2, 3 − 2 7 are binomial surds., 2, , example,, , Example 2.20, , 3, , Arrange in ascending order:, , 2, , 2,, , 4,, , 4, , 3, , Solution, The order of the surds, , 3, , 2,, , 2, , 4 and, , 4, , 3 are 3, 2, 4., , L.C.M. of 3, 2, 4 = 12., 1 4 , 1 6 , 3 12 12 4 12, , , 12 6, 2, 2 = 2 = 2 = 2 = 16 ;, 4 = 4 2 = 4 12 = 4 = 12 4096, , , , , 1 3 , , , 12 3, 4, 3 = 3 4 = 312 = 3 = 12 27, , , The ascending order of the surds 3 2, 4 3, 2 4 is 12 16 < 12 27 < 12 4096, 3, , that is,, , 3, , 2, 4 3, 2 4 ., , 2.6.2 Laws of Radicals, For positive integers m, n and positive rational numbers a and b, it is worth remembering, the following properties of radicals:, S.No., , Radical Notation, , 1., , ( ), n, , a, , 2., , n, , 3., , m n, , 4., , n, , n, , =a = a, , (a ), , n, , 1, n, , n, , a, , n, , b, , =n, , a, b, , n, , 1, n, , a × n b = n ab, a = mn a =, , Index Notation, = a = (a, 1, n, , 1, n n, , ), , 1, n, , a ×b = (ab ), n m, , a, , 1, 1 m, n, , (a ), a, b, , 1, n, 1, n, , =a, , 1, mn, , 1, 1 n, m, , ( ), , = a, , 1, , a n, = , b , , We shall now discuss certain problems which require the laws of radicals for simplifying., Example 2.21, (ii), , 3, , −2, , (1024), , Express each of the following surds in its simplest form (i) 3 108, , and find its order, radicand and coefficient., Real Numbers, , 2-Real Numbers.indd 63, , 63, , 26-12-2019 13:35:23
Page 70 :
www.tntextbooks.in, , Solution, 3, , (i), , 108 = 3 27 × 4, 3, , 3, , 3, , 3, , 2, 2, 3, 3, 3, , , = 3 ×4, , = 3 × 3 4 (Laws of radicals - ii), , = 3× 3 4, , ( Laws of radicals- i), , 108, 54, 27, 9, 3, 1, , order = 3; radicand = 4; coefficient = 3, , 3, 3, 3, (ii) (1024) = 3 2 × 2 × 2 × 2, , , 3, 3, 3, , = 3 2 ×2 ×2 ×2, , , −2 , , 2 1024, 2 512, 2 256, −2, 2 128, [Laws of radicals - (i)], , 2, 64, 3 3 3 3 3 3 3 −2, 32, = 2 × 2 × 2 × 2 [Laws of radicals – (ii)] 2, , , , 2, 16, −2, 2, 8, , = 2 × 2 × 2 × 3 2 [Laws of radicals – (i)], , , 2, 4, 2, 2, −2, , , , , 1, 1, 2, 2, , = 8 × 3 2 = × , 3, , , , , , , 1, 8 2 , 3, , −2, , (, , ), , (, , ), , 1, , =, 64, , , , , , 1, 4, 1, 1, order = 3 ; radicand =, ; coefficient =, 4, 64, (These results can also be obtained using index notation)., 3, , , Note, Consider the numbers 5 and 6. As 5 = 25 and 6 = 36, Therefore, 26, 27,, between 5 and 6., , 28,, , 29,, , 30,, , 2, , 31,, , 32,, , 33,, , 34, and, , 35 are surds, , 2, , Consider 3 2 = 3 × 2 = 18 , 2 3 = 2 × 3 = 12, Therefore, 17,, , 15, 14,, , 13 are surds between 2 3 and 3 2 ., , 2.6.3 Four Basic Operations on Surds, , (i) Addition and subtraction of surds : Like surds can be added and subtracted, using the following rules:, , a n b ± c n b = (a ± c)n b , where b > 0., , 64, , 2-Real Numbers.indd 64, , 9th Standard Mathematics, , 26-12-2019 13:35:26
Page 71 :
www.tntextbooks.in, , Example 2.22, , (i) Add 3 7 and 5 7 . Check whether the sum is rational or, , irrational. (ii) Subtract 4 5 from 7 5 . Is the answer rational or irrational?, Solution, (i) 3 7 + 5 7 = (3 + 5) 7 = 8 7 . The answer is irrational., (ii) 7 5 - 4 5 = (7 − 4) 5 = 3 5 . The answer is irrational., Example 2.23, , Simplify the following:, (ii) 2 3 40 + 3 3 625 − 4 3 320, , (i) 63 − 175 + 28, Solution, (i), , 63 − 175 + 28 = 9 × 7 − 25 × 7 + 4 × 7, = 3 7 −5 7 +2 7, , (, , ), , = 3 7 +2 7 −5 7, = 5 7 −5 7, =0, (ii) 2 3 40 + 3 3 625 − 4 3 320, = 2 3 8 × 5 + 3 3 125 × 5 − 4 3 64 × 5, 3, , 3, , 3, , 3, , 3, , 3, , = 2 2 ×5 + 3 5 ×5 − 4 4 ×5, = 2 × 2 3 5 +3 × 5 3 5 −4 × 4 3 5, = 4 3 5 + 15 3 5 − 16 3 5, = (4 + 15 − 16) 3 5 = 3, , 3, , 5, , (ii) Multiplication and division of surds, Like surds can be multiplied or divided by using the following rules:, Multiplication property of surds, (i), , n, , a × n b = n ab, n, , n, , Division property of surds, (iii), , n, , (ii) a b ×c d = ac bd, where b, d > 0, , (iv), , n, n, , a, b, n, , a b, cn d, , =n, =, , a, b, anb, where b, d > 0, c d, Real Numbers, , 2-Real Numbers.indd 65, , 65, , 26-12-2019 13:35:28
Page 72 :
www.tntextbooks.in, , Example 2.24, Multiply, , 3, , 40 and, , 3, , 16 ., , Solution, 3, , 40 × 3 16 =, , (, , 3, , =, , ) × ( 2×2×2×2), 5 ) × (2 × 2 ) = 4 × ( 2 × 5 ), 3, , 2×2×2×5, , (2 ×, , 3, , 3, , 3, , 3, , =, , 4 × 3 2 ×55, , = 4 3 10, Example 2.25, Compute and give the answer in the simplest form: 2 72 ´ 5 32 ´ 3 50, Solution, , (, , ) (, , ) (, , 2 72 × 5 32 × 3 50 = 2 × 6 2 × 5× 4 2 × 3× 5 2, , ), , = 2×5×3×6×4×5× 2× 2× 2, = 3600×2 2, , Let us simplify:, 72 = 36×2 = 6 2, 32 = 16×2 = 4 2, 50 = 25×2 = 5 2, , = 7200 2, Example 2.26, Divide, , 9, , 6, , 8 by, , 6., , Solution, 9, , 8, , 6, , 6, , =, , 8, 6, , =, , 1, 9, 1, 6, , 8, , 2, 18, 3, 18, , (Note that 18 is the LCM of 6 and 9), , (How?), , 6, 1, 1, 82 18, , , 18, 8×8 , , , = 3 (How ?) = , 6 × 6 × 6 , 6 , 1, 3 18, , 1, , 8 18, 2 , , = = , 27 , 3 , , , 66, , 2-Real Numbers.indd 66, , , , , , , 1, , 2 6, = =, 3 , , 6, , 2, 3, , Activity - 1, Is it interesting to see this pattern ?, 4, , 4, 4, =4, and, 15, 15, , 5, , 5, 5, =5, 24, 24, , Verify it. Can you frame 4 such new, surds?, , 9th Standard Mathematics, , 26-12-2019 13:35:29
Page 73 :
www.tntextbooks.in, , Activity - 2, Take a graph sheet and mark O, A, B, C as follows., Y, , In the square OABC,, OA = AB = BC = OC = 1unit, , 2, 1.5, 1, , Consider right angled DOAC, , B, , A, , 2, , 0.5, , = 2 unit [By Pythagoras theorem], , C, O, Xʹ -0.5 , , 0.5, , 2, , AC = 1 + 1, , 2, 1, , 1.5, , 2, , X, , -0.5, , The length of the diagonal (hypotenuse), AC = 2 , which is a surd., , Yʹ, , Consider the following graphs:, Y, C, , 3, , 3, , B, , 2, , 2.5, , 2, , 2, , 1.5, , 3, D, , 1, , 2, , 0.5, , A, , A, 0.5, , 1, , 1.5, , -0.5, , Yʹ, , 2, , 18, , 1.5, 1, , 0.5, O, Xʹ -0.5 , , B, , 2.5, , E, , 2, , Y, C, , 2.5, , 3, , X, , O, Xʹ -0.5 , , 0.5, , 1, , -0.5, , 1.5, , 2, , 2.5, , 3, , 2, , 2, , X, , 3, , Yʹ, , Let us try to find the length of AC in two different ways :, AC = AD + DE + EC, (diagonals of units squares), =, , 2+ 2+ 2, , AC = 3 2 units, , 2, , 2, , AC = OA + OC = 3 + 3, =, , 9+9, , AC = 18 units, , Are they equal? Discuss. Can you verify the same by taking different squares of different, lengths?, Real Numbers, , 2-Real Numbers.indd 67, , 67, , 26-12-2019 13:35:30
Page 74 :
www.tntextbooks.in, , Exercise 2.6, 1. Simplify the following using addition and subtraction properties of surds:, (i) 5 3 + 18 3 − 2 3 (ii) 4 3 5 + 2 3 5 − 3 3 5, (iv) 5 3 40 + 2 3 625 − 3 3 320, , (iii) 3 75 + 5 48 − 243 , , 2. Simplify the following using multiplication and division properties of surds:, 3 ´, , (i), , 5 ´, , (, , (iv) 7 a − 5 b, , (ii) 35 ¸, , 2, , )(, , 7 a +5 b, , 225, (v) , −, , ), , 729, , (iii), , 7, , 3, , 27 ´ 3 8 ´, , 3, , 125, , 25 , 16, ÷, , 144 , 81, , , 3. If 2 = 1.414, 3 = 1.732, 5 = 2.236, 10 = 3.162 , then find the values of the, following correct to 3 places of decimals., (i), , 40 - 20, , (ii), , 300 + 90 − 8, , 4. Arrange surds in descending order : (i), 5. Can you get a pure surd when you find, , 3, , (i) the sum of two surds , (iii) the product of two surds , , 5, 9 4, 6 3 (ii), , 2 3, , 5, 3 4 7 ,, , 3, , (ii) the difference of two surds, (iv) the quotient of two surds, , Justify each answer with an example., 6. Can you get a rational number when you compute, (i) the sum of two surds , (iii) the product of two surds , , (ii) the difference of two surds, (iv) the quotient of two surds, , Justify each answer with an example., , 2.7 Rationalisation of Surds, , Rationalising factor is a term with which a term is multiplied or divided to make the, whole term rational., Examples:, (i), (ii), , 3 is a rationalising factor of, 7, , 4, , 5 is a rationalising factor of, , 3 (since 3 × 3 = the rational number 3), 7, , 7, , 3, , 7, , 5 (since their product = 5 = 5 , a rational), , Thinking Corner, 1. In the example (i) above, can 12 also be a rationalising factor? Can you think of, any other number as a rationalising factor for 3 ?, 7, , 3, , 2. Can you think of any other number as a rationalising factor for 5 in example (ii) ?, 3. If there can be many rationalising factors for an expression containing a surd, is there, any advantage in choosing the smallest among them for manipulation?, 68, , 2-Real Numbers.indd 68, , 9th Standard Mathematics, , 26-12-2019 13:35:31
Page 75 :
www.tntextbooks.in, , Progress Check, Identify a rationalising factor for each one of the following surds and verify the, same in each case:, (i), , (ii) 5 12, , 18, , (iii), , 3, , (iv), , 49, , 1, 8, , 2.7.1 Conjugate Surds, Can you guess a rationalising factor for 3 + 2 ? This surd has one rational part and, one radical part. In such cases, the rationalising factor has an interesting form., A rationalising factor for 3 + 2 is 3 - 2 . You can very easily check this., , (, , )(, , ), , ( ), , 2, , 3+ 2 3− 2 = 3 −, , 2, , 2, , = 9 −2, = 7, a rational., What is the rationalising factor for a + b where a and b are rational numbers? Is it, a- b ?, , Check it. What could be the rationalising factor for a + b where a and b are, , rational numbers? Is it a - b ? Or, is it − a + b ? Investigate., Surds like a + b and a - b are called conjugate surds. What is the conjugate of, b + a ? It is − b + a . You would have perhaps noted by now that a conjugate is usually, obtained by changing the sign in front of the surd!, Example 2.27, , 7, , Rationalise the denominator of (i), , 14, , Solution, , (ii), , 5+ 3, 5− 3, , (i) Multiply both numerator and denominator by the rationalising factor 14 ., 7, 14, (ii), , =, , 5+ 3, 5− 3, , 7, 14, , ×, , 14, 14, , =, , 7 14, 14, =, 14, 2, , (, )×(, )=(, ), (5 − 3 ) (5 + 3 ) 5 − ( 3 ), 5 + ( 3) + 2× 5× 3, =, =, , 5+ 3, , 5+ 3, , 5+ 3, , 2, , 2, , 2, , 2, , 2, , 25 − 3, , =, , 2 ×[14 + 5 3 ], 25 + 3 + 10 3, 28 + 10 3, =, =, 22, 22, 22, , =, , 14 + 5 3, 11, Real Numbers, , 2-Real Numbers.indd 69, , 69, , 26-12-2019 13:35:33
Page 76 :
www.tntextbooks.in, , Exercise 2.7, 1. Rationalise the denominator, 1, 5, (i), , (ii), , 50, , 75, , (iii), , 3 5, , 18, , , , (iv), , 3 5, 6, , 2. Rationalise the denominator and simplify, 48 + 32, , (i), (iii), , (ii), , 27 − 18, 2 6- 5, , (iv), , 3 5 -2 6, , 3. Find the value of a and b if, , 7 −2, 7 +2, , 2, , 2 = 1.414 , find the value of, , 3+ 2, 5, 6 +2, , −, , 5, 6 −2, , = a 7 +b, , 4. If x = 5 + 2, then find the value of x +, 5. Given, , 5 3+ 2, , 1, x, , 2, , 8 -5 2, 3 -2 2, , (to 3 places of decimals)., , 2.8 Scientific Notation, Suppose you are told that the diameter, of the Sun is 13,92,000 km and that of the, Earth is 12,740 km, it would seem to be a, daunting task to compare them. In contrast, if, 13,92,000 is written as 1.392 ×106 and 12,740, as 1.274×104, one will feel comfortable. This, sort of representation is known as scientific notation., Since, , 1.392 ⋅ 10, , 6, , 4, , 14, 2, ⋅ 10, 13, , 108., , 1.274 ⋅ 10, You can imagine 108 Earths could line up across the face of the sun., Scientific notation is a way of representing numbers that are too large or too small, to be, conveniently written in decimal form. It allows the numbers to be easily recorded and handled., , 2.8.1 Writing a Decimal Number in Scientific Notation, Here are steps to help you to represent a number in scientific form:, (i) Move the decimal point so that there is only one non-zero digit to its left., (ii) Count the number of digits between the old and new decimal point. This gives, ‘n’, the power of 10., 70, , 2-Real Numbers.indd 70, , 9th Standard Mathematics, , 26-12-2019 13:35:35
Page 77 :
www.tntextbooks.in, , (iii) If the decimal is shifted to the left, the exponent n is positive. If the decimal is, shifted to the right, the exponent n is negative., n, , Expressing a number N in the form of N = a ×10 where, 1 ≤ a < 10 and ‘n’ is an, integer is called as Scientific Notation., The following table of base 10 examples may make things clearer:, Decimal notation, , Scientific, notation, , 100, , Decimal notation, , 0.01, , 1 × 102, , 1,000, 10,000, , 1 × 10-5, , 0.000001, , 1 × 106, , 1,00,00,000, , 1 × 10-4, , 0.00001, , 1 × 105, , 10,00,000, , 1 × 10-3, , 0.0001, , 1 × 104, , 1,00,000, , 1 × 10-2, , 0.001, , 1 × 103, , 1 × 10-6, , 0.0000001, , 1 × 107, , Scientific, notation, , 1 × 10-7, , Let us look into few more examples., Example 2.28, , Express in scientific notation (i) 9768854 (ii) 0.04567891, , (iii) 72006865.48, Solution, (i), , 9, , 7, , 6, , 8, , 8, , 5, , 4, , 6, , 5, , 4, , 3, , 2, , 1, , . 0, , = 9.768854 ´ 10, , 6, , The decimal point is to be moved six places to the left. Therefore n = 6., (ii), , 0, , ., , 0, , 4, , 1, , 5, , 6, , 7, , 8, , 9, , −2, , 1 = 4.567891 × 10, , 2, , The decimal point is to be moved two places to the right. Therefore n = −2 ., (iii), , 7, , 2, 7, , 0, 6, , 0, , 6, , 8, , 6, , 5 . 48, , 5, , 4, , 3, , 2, , 1, , = 7.200686548 × 10, , 7, , The decimal point is to be moved seven places to the left. Therefore n = 7 ., , 2.8.2 Converting Scientific Notation to Decimal Form, The reverse process of converting a number in scientific notation to the decimal form, is easily done when the following steps are followed:, (i) Write the decimal number., Real Numbers, , 2-Real Numbers.indd 71, , 71, , 26-12-2019 13:35:35
Page 78 :
www.tntextbooks.in, , (ii) Move the decimal point by the number of places specified by the power of 10, to, the right if positive, or to the left if negative. Add zeros if necessary., (iii) Rewrite the number in decimal form., Example 2.29, (i) 6.34 ´ 10, , Write the following numbers in decimal form:, , 4, , −5, , (ii) 2.00367 × 10, , Solution, (i)V 6.34 ´ 10, Þ, , 4, , 6 ., , 3, , 4, , 0, , 0, , 1, , 2, , 3, , 4, , = 63400, , −5, , (ii), , 2.00367 × 10, 0, , 0, , 0, , 0, , 0, , 2, , 5, , 4, , 3, , 2, , 1, , .00367, , = 0.0000200367, , 2.8.3 Arithmetic of Numbers in Scientific Notation, (i) If the indices in the scientific notation of two numbers are the same, addition (or, subtraction) is easily performed., Example 2.30, , The mass of the Earth is 5.97×1024 kg and that of the Moon is, , 0.073 ×1024 kg. What is their total mass?, Solution, Total mass = 5.97×1024 kg + 0.073 ×1024 kg, = (5.97 + 0.073) ×1024 kg, = 6.043 ×1024 kg, (ii) The product or quotient of numbers in scientific notation can be easily done if, we make use of the laws of radicals appropriately., Example 2.31, (i), (iii), 72, , 2-Real Numbers.indd 72, , Write the following in scientific notation :, 4, , 3, , (50000000), 3, , (ii) (0.00000005), 4, , (300000) ×(2000), , 3, , 4, , (iv) (4000000) ÷ (0.00002), , 9th Standard Mathematics, , 26-12-2019 13:35:36
Page 79 :
www.tntextbooks.in, , Solution, , (, , 4, , (i) (50000000) = 5.0 × 10, , 7, , ), , 4, , ( ), , 4, , = (5.0) × 10, , 4, , 7, , (, , −22, , 30, , = 1.25 × 10, , 4, , 3, , × 2.0 × 10, , 5, , 3, , 3, , 4, , 4, , 4, , 3, , 15, , 12, , 1, , 15, , 1, , 1, , 15, , 1, , 12, , 1+15 +1+12, , (, , ) (, ), = (4.0) × (10 ) ÷ (2.0) × (10 ), = 4.0 × 10, , 3, , 6, , 3, , −5, , ÷ 2.0 × 10, 3, , 6, , −5, , 4, , 18, , =, , −20, , 12, 29, , = 4.0 × 10, , = 4.32 × 10, , 4, , 4, , 64.0 × 10, , 16.0 × 10, 18, +20, = 4 × 10 × 10, , = 2.7 × 1.6 × 10 × 10 × 10 × 10, = 4.32 × 10, , 4, , (iv) (4000000) ÷ (0.00002), , ) (, ) , = (3.0) × (10 ) × (2.0) × (10 ), = (27.0)× (10 )× (16.0)× (10 ), = (2.7 × 10 )× (10 )× (1.6 × 10 )× (10 ), 3, , −24, , 2, , = 1.25 × 10 × 10, , 28, , (iii) (300000) × (2000), 3, , 3, , −24, , 2, , 5, , −8, , = (125.0)× (10), , = 6.25 × 10 ×10, , = 3.0 × 10, , ), , 3, , ( ), , 3, , 28, , 3, , −8, , = (5.0) × 10, , = 625.0 × 10, , = 6.25 × 10, , (, , 3, , (ii) (0.00000005) = 5.0 × 10, , 38, , Thinking Corner, 1. Write two numbers in scientific notation whose product is 2. 83104., 2. Write two numbers in scientific notation whose quotient is 2. 83104., , Exercise 2.8, 1. Represent the following numbers in the scientific notation:, (i) 569430000000, (iii) 0.0000006000, , (ii) 2000.57, (iv) 0.0009000002, , 2. Write the following numbers in decimal form:, 6, 4, (i) 3.459 ´ 10, (ii) 5.678 ´ 10, −5, −7, (iii) 1.00005 × 10, (iv) 2.530009 × 10, 3. Represent the following numbers in scientific notation:, 2, , 4, , 11, , 3, , (ii) (0.000001) ÷ (0.005), (300000) ×(20000), 6, 4, 3, 2, (iii) {(0.00003) × (0.00005) } ÷ {(0.009) × (0.05) }, (i), , Real Numbers, , 2-Real Numbers.indd 73, , 73, , 26-12-2019 13:35:39
Page 80 :
www.tntextbooks.in, , 4. Represent the following information in scientific notation:, (i) The world population is nearly 7000,000,000., (ii) One light year means the distance 9460528400000000 km., (iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg., , 5. Simplify:, (i) (2.75 × 107) + (1.23 × 108), (iii) (1.02 × 1010) × (1.20 × 10–3), , (ii) (1.598×1017) – (4.58 ×1015), (iv) (8.41 × 104) ' (4.3 × 105), , Activity - 3, The following list shows the mean distance of the planets of the solar system from the, Sun. Complete the following table. Then arrange in order of magnitude starting with the, distance of the planet closest to the Sun., Decimal form, (in Km), , Planet, , Jupiter, Mercury, , 58000000, , Uranus, Venus, Neptune, , 2870000000, 108000000, 4500000000, , Mars, , Earth, , Scientific Notation, (in Km), , 7.78´108, 2.28´108, , 1.5´108, 1.43´108, , Saturn, , Exercise 2.9, Multiple Choice Questions, 1., , 2., , If n is a natural number then n is, (1) always a natural number., , (2) always an irrational number., , (3) always a rational number, , (4) may be rational or irrational, , Which of the following is not true?., (1) Every rational number is a real number., , (2) Every integer is a rational number., , (3) Every real number is an irrational number. (4) Every natural number is a whole number., , 3., , 4., 74, , 2-Real Numbers.indd 74, , Which one of the following, regarding sum of two irrational numbers, is true?, (1) always an irrational number., , (2) may be a rational or irrational number., , (3) always a rational number., , (4) always an integer., , Which one of the following has a terminating decimal expansion?., 9th Standard Mathematics, , 26-12-2019 13:35:39
Page 81 :
www.tntextbooks.in, , (1), , 5., , 5, 64, 25, , (3), , 14, 15, , (4), , 1, 12, , 9, 4, , (2), , (3), , 7, 11, , (4) π, , An irrational number between 2 and 2.5 is, (1) 11, , 7., , 8, 9, , Which one of the following is an irrational number, (1), , 6., , (2), , (2), , (3) 2.5, (4) 8, The smallest rational number by which 13 should be multiplied so that its decimal, 5, , expansion terminates with one place of decimal is, , 8., , 1, 3, (1), (2), 10, 10, If 17 = 0.142857 then the value of 57 is, (1) 0.142857, , 9., , (4) 30, , (3) 0.571428, , (4) 0.714285, , (3), , (4), , Find the odd one out of the following., (1), , 10., , (2) 0.714285, , (3) 3, , 32 # 2, , 27, 3, , (2), , 72 # 8, , 54, 18, , 0.34 + 0.34 =, (1) 0.6 87, , (2) 0.68, , (3) 0.68, , (4) 0.687, , 11. Which of the following statement is false?, (1) The square root of 25 is 5 or −5 , , (3), , 25 = 5, , (2) − 25 = − 5 , , (4), , 25 = ± 5, , 12. Which one of the following is not a rational number?, (1), , 13., , 8, 18, , 7, 3, , (3), , 0.01, , (4), , 13, , 27 + 12 =, (1), , 14. If, , 39, , (2) 5 6, , (3) 5 3, , (4) 3 5, , (2) 4, , (3) 8, , (4) 16, , (2) 8 21, , (3) 8 10, , (4) 6 21, , 80 = k 5, then k =, , (1) 2 , , 15., , (2), , 4 7× 2 3 =, (1) 6 10, , 16. When written with a rational denominator, the expression, (1), , 2, 3, , (2), , 3, 2, , (3), , 6, 3, , 2 3, 3 2, , can be simplified as, , (4), , 2, 3, Real Numbers, , 2-Real Numbers.indd 75, , 75, , 26-12-2019 13:35:44
Page 82 :
www.tntextbooks.in, , (, , ), , 2, , 17. When 2 5 − 2 is simplified, we get, (1) 4 5 + 2 2, , 18., , (2) 22 - 4 10, , −3, , (4) 2 10 - 2, , −3, , (0.000729) 4 × (0.09) 4 = ______, 3, 5, 10, 10, , (1), , (3) 8 - 4 10, , (2), , 3, , 5, , 2, , (3), , 10, , 10, , (4), , 2, , 6, , 6, , 3, 3, 19. If 9x = 3 92 ,, then x = ______, (1) 2, (2) 4, (3) 1, (4) 5, 3, 3, 3, 3, 20. The length and breadth of a rectangular plot are 5×10 5 and 4×10 4 metres, respectively. Its area is ______., 3, , (1) 9×101 m2, , (2) 9×109 m2, , 3, , (3) 2×1010 m2, , (4) 20×1020 m2, , Points to Remember, p, , When the decimal expansion of q , q ! 0 terminates that is, comes to an end, the, decimal is called a terminating decimal., p, In the decimal expansion of q , q ! 0 when the remainder is not zero, we have a repeating, (recurring) block of digits in the quotient. In this case, the decimal expansion is called, non-terminating and recurring., p, p, If a rational number q , q ! 0 can be expressed in the form m # n , where p ! Z and, 2, 5, m, n ! W , then the rational number will have a terminating decimals. Otherwise, the, rational number will have a non-terminating repeating (recurring) decimal., A rational number can be expressed either a terminating or a non- terminating, recurring decimal., An irrational number is a non-terminating and non-recurring decimal, i.e. it cannot be, p, written in form q , where p and q are both integers and q ! 0., The union of all rational numbers and all irrational numbers is called the set of real, numbers., Every real number is either a rational number or an irratonal number., If a real number is not rational number, then it must be an irrational number., If ‘a’ is a positive rational number, ‘n’ is a positive integer and if, , n, , a is an irrational, , n, , number, then a is called as a surd., If ‘m’, ‘n’ are positive integers and a, b are positive rational numbers, then, , ( ), , n, , n, , a, n, b, b, The process of multiplying a surd by another surd to get a rational number is called, Rationalisation., , (i), , n, , a, , n, , =a = a, , n, , (ii) n a × n b = n ab (iii), , m n, , a = mn a =, , n m, , a, , (iv), , a, , =n, , n, , Expressing a number N in the form of N = a ×10 where, 1 ≤ a < 10 and ‘n’ is an, integer is called as Scientific Notation., 76, , 2-Real Numbers.indd 76, , 9th Standard Mathematics, , 26-12-2019 13:35:46
Page 83 :
www.tntextbooks.in, , ICT Corner-1, Expected Result is shown in this picture, , Step – 1, Open the Browser and copy and paste the Link given below, (or) by typing the URL given (or) Scan the QR Code., Step - 2, GeoGebra workbook named “Real Numbers” will open. There are several worksheets in the, workbook. Open the worksheet named “Square root spiral – 1st part”, Step-3, Drag the slider named “Steps “. The construction of Square root of numbers 2,3,4,5,…. will, appear step by step., Step-4, By dragging the slider named “Unit segment” you can enlarge the diagram for more clarity., Now you can draw the same in a paper and measure the values obtained, , Browse in the link, , Union of Sets: https://www.ggbm.at/m6GQc6mQ, , ICT Corner-2, Expected Result is, shown in this picture, Step – 1, Open the Browser type the URL Link given below, (or) Scan the QR Code. GeoGebra work sheet, named “Real Numbers” will open. In the work sheet there are two activities. 1. Rationalising, the denominator for surds and, 2. Law of exponents., In the first activity procedure for rationalising the denominator is given. Also, example is given, under. To change the values of a and b enter the value in the input box given., Step - 2, In the second activity law of exponents is given. Also, example is given on right side. To change, the value of m and n move the sliders and check the answers., Browse in the link, , Real Numbers: https://ggbm.at/BYEWDpHU or Scan the QR Code., , Real Numbers, , 2-Real Numbers.indd 77, , 77, , 26-12-2019 13:35:46
Page 84 :
www.tntextbooks.in, 0, c1 =, +, b 1y, 0, a 1x +, c2 =, +, y, b2, a 2x +, , 3, , ALGEBRA, In real life, I assure you, there is no such things as Algebra., - Fran Lebowitz, Diophantus of Alexandria an Alexandrian Hellenistic, mathematician who lived for about 84 years, was born, between A.D(C.E) 201 and A.D(C.E) 215. Diophantus, was the author of a series of books called Arithmetica., His texts deal with solving algebraic equations. He is, , Diophantus, , also called as “the father of algebra”., , Learning Outcomes, zz To understand the classification of polynomials and to perform basic operations., zz To evaluate the value of a polynomial and understand the zeros of polynomial., zz To understand the remainder and factor theorems., zz To use Algebraic Identities in factorisation., zz To factorise a quadratic and a cubic polynomial., zz To use synthetic division to factorise a polynomial., zz To find GCD of polynomials., zz Able to draw graph for a given linear equation., zz To solve simultaneous linear equations in two variables by Graphical method and, Algebraic method, zz To understand consistency and inconsistency of linear equations in two variables., , 3.1 Introduction, Why study polynomials?, This chapter is going to be all about polynomial expressions in algebra. These are, your friends, you have already met, without being properly introduced! We will properly, introduce them to you, and they are going to be your friends in whatever mathematical, journey you undertake from here on., 78, , 3 Algebra_Term1.indd 78, , 9th Standard Mathematics, , 26-12-2019 13:39:24
Page 85 :
www.tntextbooks.in, , (a+1) 2 = a2 + 2a + 1, Now that’s a polynomial. That does not look very special, does it? We have seen a lots, of algebraic expressions already, so why to bother about these? There are many reasons why, polynomials are interesting and important in mathematics., For now, we will just take one example showing their use. Remember, we studied lots, of arithmetic and then came to algebra, thinking of variables as unknown numbers. Actually, we can now get back to numbers and try to write them in the language of algebra., Consider a number like 5418. It is actually 5 thousand 4 hundred and eighteen., Write it as:, 5 # 1000 + 4 # 100 + 1 # 10 + 8, which again can be written as:, 5 # 103 + 4 # 102 + 1 # 101 + 8, , Height, , Now it should be clear what this is about. This is of the form 5x3 + 4x2 + x + 8, which is, a polynomial. How does writing in this form help? We always write numbers in decimal, system, and hence always x = 10. Then what is the fun? Remember divisibility rules? Recall, that a number is divisible by 3 only if the sum of, Y, its digits is divisible by 3. Now notice that if x, divided by 3 gives 1 as remainder, then it is the, same for x2, x3, etc. They all give remainder 1, y, s, when divided by 3. So you get each digit, multiplied by 1, added together, which is the, θ, sum of digits. If that is divisible by 3, so is the, x, X, whole number. You can check that the rule for, Fig. 3.1, divisibility by 9, or even divisibility by 2 or 5,, can be proved similarly with great ease. Our purpose is not to prove divisibility rules but to, show that representing numbers as polynomials, h1, shows us many new number patterns. In fact,, h0, h2, many objects of study, not just numbers, can, be represented as polynomials and then we, can learn many things about them., t0, t2, t1, Time, In algebra we think of x2, 5x2–3, 2x+7 etc, Fig. 3.2, as functions of x. We draw pictures to see how, the function varies as x varies, and this is very helpful to understand the function. And now,, it turns out that a good number of functions that we encounter in science, engineering,, business studies, economics, and of course in mathematics, all can be approximated by, polynomials, if not actually be represented as polynomials. In fact, approximating functions, using polynomials is a fundamental theme in all of higher mathematics and a large number, of people make a living, simply by working on this idea., Algebra, , 3 Algebra_Term1.indd 79, , 79, , 26-12-2019 13:39:25
Page 86 :
www.tntextbooks.in, , Polynomials are extensively used in, biology, computer science, communication, systems ... the list goes on. The given pictures, (Fig. 3.1, 3.2 & 3.3) may be repsresented as a, quadratic polynomial. We will not only learn, what polynomials are but also how we can use, them like in numbers, we add them, multiply, them, divide one by another, etc.,, Observe the given figures., , Fig. 3.3, , y, y, , 2x, 2x, , y, , 2x, Fig. 3.4, , Fig. 3.5, , Fig. 3.6, , The total area of the above 3 figures is 4x2 + 2xy + y2 , we call this expression as an, algebraic expression. Here for different values of x and y we get different values of areas., Since the sides x and y can have different values, they are called variables. Thus, a variable is, a symbol which can have various numerical values., Variables are usually denoted by letters such as x, y, z, etc. In the above algebraic, expression the numbers 4, 2 are called constants. Hence the constant is a symbol, which has, a fixed numeric value., , Constants, Any real number is a constant. We can form numerical expressions using constants, and the four arithmetical operations., Examples of constant are 1, 5, –32, 37 , - 2 , 8.432, 1000000 and so on., , Variables, The use of variables and constants together in expressions give us ways of representing, a range of numbers, one for each value of the variable. For instance, we know the expression, 2pr, it stands for the circumference of a circle of radius r. As we vary r, say, 1cm, 4cm, 9cm, etc, we get larger and larger circles of circumference 2p, 8p, 18p etc.,, The single expression 2pr is a short and compact description for the circumference of, all these circles. We can use arithmetical operations to combine algebraic expressions and, get a rich language of functions and numbers. Letters used for representing unknown real, numbers called variables are x, y, a, b and so on., 80, , 3 Algebra_Term1.indd 80, , 9th Standard Mathematics, , 26-12-2019 13:39:25
Page 87 :
www.tntextbooks.in, , Algebraic Expression, An algebraic expression is a combination of constants and variables combined together, with the help of the four fundamental signs., Examples of algebraic expression are, x3 - 4x2 + 8x - 1, 4xy2 + 3x2 y - 54 xy + 9, 5x2 - 7x + 6, , Coefficients, Any part of a term that is multiplied by the variable of the term is called the coefficient, of the remaining term., , For example,, x2 + 5x - 24 is an algebraic expression containing three terms. The variable of this, , expression is x, coefficient of x 2 is 1, the coefficient of x is 5 and the constant is, –24 (not 24)., Activity-1, Write the Variable, Coefficient and Constant in the given algebraic expression, Expression, , x+7, , Variable, , x, , Coefficient, , 1, , Constant, , 7, , 3y – 2, , 5x2, , 2xy +11, , - 1 p +7, 2, , -8 +3a, , x, y, -1, 2, , –8, , 3.2 Polynomials, A polynomial is an arithmetic expression consisting of variables and constants that involves, four fundamental arithmetic operations and non-negative integer exponents of variables., , Polynomial in One Variable, An algebraic expression of the form p^ xh = an x n + an - 1 x n - 1 + ... + a2 x2 + a1 x + a0 is called, Polynomial in one variable x of degree ‘n’ where a0, a1, a2, ...an are constants (an ! 0) and n is, a whole number., In general polynomials are denoted by f^ xh, g^ xh, p^ t h, q^ z h and r(x) and so on., , Note, The coefficient of variables in the algebraic expression may have any real numbers,, T, where as the powers of variables in polynomial must have only non-negative integral, powers that is, only whole numbers. Recall that a0 = 1 for all a., Algebra, , 3 Algebra_Term1.indd 81, , 81, , 26-12-2019 13:39:26
Page 88 :
www.tntextbooks.in, , For example,, S.No, , Polynomial /, not a polynomial, , Given expression, , Reason, , 1, , 4y3 + 2y2 + 3y + 6, , Polynomial, , Non- negative integral power, , 2, , 4x- 4 + 5x4, , Not a polynomial, , One of the powers is negative (–4), , 3, , m2 +, , Polynomial, , Non- negative integral power, , 4, , 5 y2, , Polynomial, , Non- negative integral power, , Not a polynomial, , 1, One of the power is negative ` = r- 1j, r, , Not a polynomial, , 1, power of q is fraction ^ q = q 2 h, , Polynomial, , Non- negative integral power, , Not a polynomial, , One of the power of n is a fraction, , 4 +, m 8, 5, , 5, , 2r2 + 3r - 1 +, , 6, , 8+ q, , 7, 8, , 1, r, , 8 p2 + 5p - 7, 4, , 5n 5 + 6n - 1, , Standard Form of a, Polynomial, , Activity-2, , We can write, a, polynomial p(x) in the, decreasing or increasing, order of the powers of, x. This way of writing, the polynomial is called, the standard form of a, polynomial., , For example:, , (i), , 4, 5, , Write the following polynomials in standard form., Sl.No., , Polynomial, , Standard Form, , 1, , 5m4 - 3m + 7m2 + 8, , 2, , 2 + 3y 8y 12 + 5 y2, 3, , 3, , 12p2 - 8p5 - 10p4 - 7, , 8x4 + 4x3 - 7x2 - 9x + 6 , , (ii) 5 – 3y + 6y2 + 4y3 – y4, , Degree of the Polynomial, , In a polynomial of one variable, the highest power of the variable is called the degree, of the polynomial., In case of a polynomial of more than one variable, the sum of the powers of the variables in, each term is considered and the highest sum so obtained is called the degree of the polynomial., This is intended as the most significant power of the polynomial. Obviously when, we write x2+5x the value of x2 becomes much larger than 5x for large values of x. So we, could think of x2 + 5x being almost the same as x2 for large values of x. So the higher, the power, the more it dominates. That is why we use the highest power as important, information about the polynomial and give it a name., 82, , 3 Algebra_Term1.indd 82, , 9th Standard Mathematics, , 26-12-2019 13:39:26
Page 89 :
www.tntextbooks.in, , Example 3.1, , Find the degree of each term for the following polynomial and, , also find the degree of the polynomial, , 6ab8 + 5a2 b3 c2 - 7ab + 4b2 c + 2, , Solution, Given polynomial is 6ab8 + 5a2 b3 c2 - 7ab + 4b2 c + 2, Degree of each of the terms is given below., 6ab8 has degree, , (1+8), , = 9, , 5a2 b3 c2 has degree (2+3+2) = 7, , 7ab has degree, , (1+1), , = 2, , 4b2 c has degree, , (2+1), , = 3, , The constant term 2 is always regarded as having degree Zero., The degree of the polynomial 6ab8 + 5a2 b3 c2 - 7ab + 4b2 c + 2 ., = the largest exponent in the polynomial, = 9, , A very Special Polynomial, , We have said that coefficients can be any real numbers. What if the, coefficient is zero? Well that term becomes zero, so we won’t write it. What if, all the coefficients are zero? We acknowledge that it exists and give it a name., It is the polynomial having all its coefficients to be zero., g (t) = 0t4 + 0t2 - 0t ,, , h (p) = 0p2 - 0p + 0, , From the above example we see that we cannot talk of the degree of the zero polynomial,, since the above two have different degrees but both are zero polynomial. So we say that the, degree of the zero polynomial is not defined., The degree of the zero polynomial is not defined, , Types of Polynomials, (i), , Polynomial on the basis of number of terms, , MONOMIAL, BINOMIAL, TRINOMIAL, , A polynomial having one term is called a monomial, Examples :, , 5, 6m, 12ab, , A polynomial having two terms is called a Binomial, Examples : 5x + 3, 4a - 2, 10p + 1, A polynomial having three terms is called a Trinomial, Example : 4x2 + 8x - 12, 3a2 + 4a + 10, Algebra, , 3 Algebra_Term1.indd 83, , 83, , 26-12-2019 13:39:27
Page 90 :
www.tntextbooks.in, , (ii) Polynomial based on degree, A polynomial of degree zero is called constant polynomial, Examples : 5, - 7, 23 , 5, , CONSTANT, , A polynomial of degree one is called linear polynomial, , LINEAR, , Examples : 410x - 7, A polynomial of degree two is called quadratic, polynomial, , QUADRATIC, , Example : 2 5 x2 + 8x - 4, A polynomial of degree three is called cubic polynomial, , CUBIC, , Example : 12y3,, , Example 3.2, , 6m3 - 7m + 4, , Classify the following polynomials based on number of terms., , S.No. Polynomial, , Type of polynomial, based of terms, , No of Terms, , (i), , 5t3 + 6t + 8t2 3 Terms, , Trinomial, , (ii), , y–7, , 2 Terms, , Binomial, , (iii), , 2 4, r, 3, , 1 Term, , Monomial, , (iv), , 6y5 + 3y - 7, , 3 Terms, , Trinomial, , (v), , 8m 2 + 7m 2, , Like Terms. So, it is 15m2 which is 1 term only, , Monomial, , Example 3.3, S.No., 3, , (ii), , z3 - z2 + 3, , (iv), , 3 Algebra_Term1.indd 84, , Polynomial, , (i), , (iii), , 84, , Classify the following polynomials based on their degree., , 4 z+ 7, , 7, y2 - 8, , Degree, , Type, , Degree one, , Linear polynomial, , Degree three, , Cubic polynomial, , Degree zero, , Constant polynomial, , Degree two, , Quadratic polynomial, , 9th Standard Mathematics, , 26-12-2019 13:39:28
Page 91 :
www.tntextbooks.in, , 3.2.1 Arithmetic of Polynomials, , We now have a rich language of polynomials, and we have seen that they can be, classified in many ways as well. Now, what can we do with polynomials? Consider a, polynomial on x., We can evaluate the polynomial at a particular value of x. We can ask how the, function given by the polynomial changes as x varies. Write the polynomial equation, p(x) = 0 and solve for x. All this is interesting, and we will be doing plenty of all this as we, go along. But there is something else we can do with polynomials, and that is to treat them, like numbers! We already have a clue to this at the beginning of the chapter when we saw, that every positive integer could be represented as a polynomial., Following arithmetic, we can try to add polynomials, subtract one from another,, multiply polynomials, divide one by another. As it turns out, the analogy between numbers, and polynomials runs deep, with many interesting properties relating them. For now, it is, fun to simply try and define these operations on polynomials and work with them., , Addition of Polynomials, The addition of two polynomials is also a polynomial., , Note, Only like terms can be added. 3x2 + 5x2 gives 8x2 but unlike terms such as 3x2 and 5x3, when added gives 3x2 + 5x3, a new polynomial., Example 3.4, , If p^ xh = 4x2 - 3x + 2x3 + 5 and q^ xh = x2 + 2x + 4 , then find, p^ x h + q^ x h, , Solution, Given Polynomial, , Standard form, , p^ xh = 4x2 - 3x + 2x3 + 5, , 2x3 + 4x2 - 3x + 5, , q ^ x h = x 2 + 2x + 4, , x 2 + 2x + 4, , p ^ x h + q ^ x h = 2 x 3 + 5x 2 - x + 9, , We see that p(x) + q(x) is also a polynomial. Hence the sum of any two polynomials, is also a polynomial., , Subtraction of Polynomials, The subtraction of two polynomials is also a polynomial., Note, , Only like terms can be subtracted. 8x2 – 5x2 gives 3x2 but when 5x3 is, subtracted from 3x2 we get, 3x2 –5x3, a new polynomial., Algebra, , 3 Algebra_Term1.indd 85, , 85, , 26-12-2019 13:39:28
Page 94 :
www.tntextbooks.in, , (ii), (iii), , f^ yh = 6y2 - 7y + 2, h^ z h = z5 - 6z4 + z, , g^ yh = 7y + y3, f^ z h = 6z2 + 10z - 7, , 7., , What should be added to 2x3 + 6x2 - 5x + 8 to get 3x3 - 2x2 + 6x + 15 ?, , 8., , What must be subtracted from 2x4 + 4x2 - 3x + 7 to get 3x3 - x2 + 2x + 1 ?, , 9., , Multiply the following polynomials and find the degree of the resultant polynomial:, (i) p^ xh = x2 - 9, q^ xh = 6x2 + 7x - 2, (ii) f^ xh = 7x + 2, g (x) = 15x - 9, 2, (iii) h^ xh = 6x - 7x + 1, f^ xh = 5x - 7, , 10. The cost of a chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total, amount paid by him in terms of x and y. If x =10, y =5 find the amount paid by him., 11. The length of a rectangle is (3x+2) units and it’s breadth is (3x–2) units. Find its area in, terms of x. What will be the area if x = 20 units., 12., , p^ xh is a polynomial of degree 1 and q^ xh is a polynomial of degree 2. What kind of, , the polynomial p(x) × q(x) is ?, , 3.2.2 Value and Zeros of a Polynomial, , Consider the two graphs given below. The first is linear, the second is quadratic. The, first intersects the X axis at one point (x = –3) and the second at two points (x = -1 and, x = 2). They both intersect the Y axis only at one point. In general, every polynomial has a, graph and the graph is shown as a picture (since we all like pictures more than formulas,, don’t we?). But also, the graph contains a lot of useful information like whether it is a straight, line, what is the shape of the curve, how many places it cuts the x-axis, etc., Y, , Y, 6, , y = x+3, , 3, , 5, , 2, , 4, , 1, , 3, , Xʹ, , 2, , -3, , -2, , Xʹ, , -3, , -2, , 0, , -1, , 1, , 2, , 3, , 4, , Yʹ, , 88, , 3 Algebra_Term1.indd 88, , 1, , 2, , 3, , X, , X, , y = x2–x–2, , -2, -1, , -1, , , , 0, , -3, , 1, -4, , -1, , Fig. 3.7 , , Yʹ, , Fig. 3.8, , 9th Standard Mathematics, , 26-12-2019 13:39:31
Page 95 :
www.tntextbooks.in, , In general, the value of a polynomial p(x) at x=a, denoted p(a), is obtained by replacing, x by a, where a is any real number., Notice that the value of p(x) can be zero for many possible values of x as in the second, graph 3.8. So it is interesting to ask, for how many values of x, does p(x) become zero, and, for which values ? We call these values of x, the zeros of the polynomial p(x)., Once we see that the values of the polynomial are what we plot in the graph of the, polynomial, it is also easy to notice that the polynomial becomes zero exactly when the, graph intersects the X-axis., The number of zeros depends on the line or curves intersecting x-axis., For Fig. 3.7, Number of zeros is equal to 1, For Fig. 3.8, Number of zeros is equal to 2, , Value of a Polynomial, , Value of a polynomial p^ xh at x = a is p(a) obtained on, replacing x by a ^a ! Rh, , Note, Number of zeros of a, polynomial < the degree, of the polynomial, , For example,, , , , Consider, The value of, , , , f^ xh = x2 + 3x - 1 ., , f(x) at x = 2 is, f(2) = 22+3(2)–1 = 4+6-1 = 9., , Zeros of Polynomial, (i), , , Consider the polynomial, , p(x) = 4x3 –6x2 + 3x –14, , The value of p(x) at x = 1 is p(1) = 4(1)3 – 6(1)2 + 3(1) – 14, , , , = 4 – 6 + 3 – 14, , = –13, , , Then, we say that the value of p(x) at x = 1 is – 13., , , , If we replace x by 0, we get, , p(0) = 4(0)3 – 6(0)2 + 3(0) –14, , = 0 – 0 + 0 – 14, = –14, , , we say that the value of p(x) at x = 0 is – 14., , , , The value of p(x) at x = 2 is p(2) = 4(2)3 – 6(2)2 + 3(2) – 14, = 32 – 24 + 6 – 14, , = 0, Since the value of p(x) at x = 2 is zero, we can say that 2 is one of the zeros of p(x), where p(x) = 4x3 –6x2 + 3x –14., Algebra, , 3 Algebra_Term1.indd 89, , 89, , 26-12-2019 13:39:32
Page 96 :
www.tntextbooks.in, , Roots of a Polynomial Equation, In general, if p(a) = 0 we say that a is zero of polynomial p(x) or a is the root of, polynomial equation p(x) = 0, Example 3.7, , If f^ xh = x2 - 4x + 3 , then find the values of f^1 h, f^- 1h, f^2h, f^3 h ., Also find the zeros of the polynomial f(x)., Solution, f ^ x h = x 2 - 4x + 3, , x2 –4x + 3, Value at, x=1, , x = –1, , x=2, , x=3, , 12 –4(1) + 3, , (–1)2 –4(–1)+3, , 22 –4(2) + 3, , 32 –4(3) + 3, , 1–4+3, , 1+4+3, , 4–8+3, , 9–12+3, , 8, , –1, , 0, , Value of P(x), 0, , Fig. 3.9, , Since the value of the polynomial f(x) at x = 1 and x = 3 is zero, as the zeros of polynomial, f(x) are 1 and 3., Example 3.8, (i) f(x) = 2x + 1, Solution, (i) Given that, , Find the Zeros of the following polynomials., (ii) f(x) = 3x – 5, f(x) = 2x + 1 = 2` x + 12 j, = 2` x - (- 12 )j, , (ii), , 90, , 3 Algebra_Term1.indd 90, , 1, 1, 1, f`- j = 2 8- - `- jB = 2(0) = 0, 2, 2, 2, 1, 1, Since f`- 2 j = 0, x = - 2 is the zero of f(x), Given that, f(x) = 3x – 5 = 3` x - 53 j, 5, 5 5, f` j = 3` - j = 3(0) = 0, 3, 3 3, 5, 5, x = 3 is the zero of f(x), Since f` 3 j =0,, , 9th Standard Mathematics, , 26-12-2019 13:39:32
Page 97 :
www.tntextbooks.in, , Example 3.9, , Find the roots of the following polynomial equations., , (i) 5x – 3 = 0, , (ii) –7 –4x = 0, , Note, , Solution, , (i) A zero of a polynomial can be, , (i) 5x – 3 = 0, , any real number not necessarily, , (or), , 5x = 3, , Then,, (ii) –7 – 4x = 0, (or), , zero., , x = 53, , (ii) A non zero constant polynomial, has no zero., , 4x = – 7, , (iii) By convention, every real, , x = -7 = - 47, 4, , Then,, Example 3.10, , number is zero of the zero, polynomial, , Check whether –3 and 3 are zeros of the polynomial x2 – 9, , Solution, f(x) = x2 – 9, , Let, Then,, , f(–3) = (–3)2 – 9 = 9 – 9 = 0, f(+3) = 32 – 9 = 9 – 9 = 0, , ` -3 and 3 are zeros of the polynomial x2 – 9, , Exercise 3.2, 1., , Find the value of the polynomial f^ yh = 6y - 3y2 + 3 at, (i) y = 1, , (ii) y = –1, , (iii) y = 0, , 2., , If p(x) = x2– 2 2x +1, find p(2 2) ., , 3., , Find the zeros of the polynomial in each of the following :, (i) p(x) = x – 3, , (ii) p(x) = 2x + 5, , (iii) q(y) = 2y – 3, , (iv) f(z) = 8z, , (v) p(x) = ax when a ! 0, , (vi) h^ xh = ax + b, a ! 0, a, b ! R, , 4., , Find the roots of the polynomial equations ., (i) 5x – 6 = 0, (ii) x +3 = 0, (iii) 10x + 9 = 0, (iv) 9x – 4 = 0, , 5., , Verify whether the following are zeros of the polynomial indicated against them, or not., (i) p^ xh = 2x - 1, x = 12, -, , (iii) p^ xh = ax + b, x = ab, , (ii) p^ xh = x3 - 1, x = 1, (iv) p^ xh = ^ x + 3h ^ x - 4h, x = 4 , x = –3, Algebra, , 3 Algebra_Term1.indd 91, , 91, , 26-12-2019 13:39:33
Page 98 :
www.tntextbooks.in, , 6., , Find the number of zeros of the following polynomials represented by their graphs., (i), , (ii), , y=0, , , , Fig. 3.10 , , (iii), , (iv), , Fig. 3.11, , (v), , y, , x, , xl, , yl, , , , Fig. 3.12 , , Fig. 3.13 , , Fig. 3.14, , 3.3 Remainder Theorem, , In this section , we shall study a simple and an elegant method of finding the remainder., In the case of divisibility of a polynomial by a linear polynomial we use a well known, theorem called Remainder Theorem., If a polynomial p(x) of degree greater than or equal to one is divided by a linear polynomial, (x–a) then the remainder is p(a), where a is any real number., Significance of Remainder theorem : It enables us to find the remainder without, actually following the cumbersome process of long division., , Note, (i) If p(x) is divided by (x+a), then the remainder is p(– a), (ii) If p(x) is divided by (ax–b), then the remainder is p( ba ), , (iii) If p(x) is divided by (ax+b), then the remainder is p(– ba ), 92, , 3 Algebra_Term1.indd 92, , 9th Standard Mathematics, , 26-12-2019 13:39:34
Page 99 :
www.tntextbooks.in, , Example 3.11, S.No., 1, , Question, Find the remainder when, f(x) = x + 3x + 3x + 1 is, divided by x+1., 3, , 2, , 2, , Check whether f(x) = x3– x + 1, is a multiple of g(x) = 2 – 3x, , 3, , Find the remainder when, f(x) = x3–ax2 + 6x – a, is divided by (x – a), , Solution, , Hint, , f(x) = x3 + 3x2 + 3x + 1, , g(x) = x+1, , f(–1) = (–1)3+3(–1)2+3(–1)+1, , g(x) = 0, , = –1+3–3+1 = 0, , x +1 = 0, , Hence, the remainder is 0, 2 3 2, 2, ` f` j = ` j - + 1, 3, 3, 3, 8 -2+, =, 1, 27 3, 8 - 18 + 27 = 17, =, !0, 27, 27, & f(x) is not multiple of g(x), , x = –1, g(x) = 2–3x, =0, gives x =, , 2, 3, , Let, , We have, f(x) = x3–ax2 + 6x –a, , g(x) = x–a, , f(a) = a - a^ah + 6a - a, 2, , 3, , = a - a + 5a, 3, , 3, , g(x) = 0, x–a=0, , = 5a, , x=a, , Hence the required remainder is 5a, 4, , For what value of k is the Let, Let, polynominal, f(x) =2x4 + 3x3 + 2kx2 + 3x + 6 g(x)=x+2, 2x4 + 3x3 + 2kx2 + 3x + 6, , exactly divisible by (x + 2) ?, , If f(x) is exactly divisible by g(x)=0, (x+2), then the remainder must be, x+2 = 0, zero, i.e ., f(–2), , x = –2, , =0, , i.e ., 2(–2)4 + 3(–2)3 + 2k(–2)2 +, 3(–2)+6=0, 2(16) + 3(–8) + 2k(4) –6 + 6=0, 32–24 + 8k = 0, 8k = –8 ,, , k = –1, , Hence f(x) is exactly divisible by, (x–2) when k = –1, Algebra, , 3 Algebra_Term1.indd 93, , 93, , 26-12-2019 13:39:34
Page 100 :
www.tntextbooks.in, , Example 3.12, , Without actual division , prove that f(x) = 2x4 - 6x3 + 3x2 + 3x - 2, is exactly divisible by x2 –3x + 2, Solution :, Let, , f(x) = 2x4 - 6x3 + 3x2 + 3x - 2, g(x) = x2 - 3x + 2, = x 2 - 2x - x + 2, = x^ x - 2h - 1^ x - 2h, = ^ x - 2h^ x - 1h, , we show that f(x) is exactly divisible by (x–1) and (x–2) using remainder theorem, f(1) = 2^1 h4 - 6^1 h3 + 3^1 h2 + 3^1 h - 2, = 2–6+3+3–2, =0, f(2) = 2^2h4 - 6^2h3 + 3^2h2 + 3^2h - 2, = 32 – 48 + 12 + 6 – 2, = 0, f(x) is exactly divisible by (x – 1) (x – 2), i.e., f(x) is exactly divisible by x2 –3x + 2, , `, , If p(x )is divided by (x - a ) with the remainder p(a ) = 0 , then (x - a ) is a factor of, p(x ). Remainder Theorem leads to Factor Theorem., , 3.3.1 Factor Theorem, If p(x )is a polynomial of degree n ³ 1 and ‘a’ is any real number then, (i) p(a ) = 0 implies (x - a ) is a factor of p(x )., (ii) (x - a ) is a factor of p(x ) implies p(a ) = 0 ., Proof, If p(x ) is the dividend and (x - a ) is a divisor, then by division, algorithm we write, p(x ) = (x − a )q(x ) + p(a ) where q(x ) is the quotient, and p(a ) is the remainder., (i), , If p(a ) = 0, we get p(x ) = (x − a )q(x ) which shows that (x - a ) is a factor of p(x )., , (ii), , Since (x - a ) is a factor of p(x ) , p(x ) = (x − a )g(x ) for some polynomial g(x )., In this case, p(a ) = (a − a ) g(a ), Thinking Corner, = 0 × g(a ), For any two integers a (a ¹ 0), =0, and b, a divides b if, Hence, p(a ) = 0, when (x - a ) is a factor of p(x )., b = ax, for some integer x., , 94, , 3 Algebra_Term1.indd 94, , 9th Standard Mathematics, , 26-12-2019 13:39:37
Page 101 :
www.tntextbooks.in, , Note, z (x - a ) is a factor of p(x ), if p(a) = 0, , ( x–a = 0, x = a), , z (x + a) is a factor of p(x ), if p(–a) = 0, , ( x+a = 0, x = –a), , , b, ax + b = 0, ax = −b, x = − b , z (ax+b) is a factor of p(x ), if p − = 0, , a , a , , , b , ax − b = 0, ax = b, x = b , z (ax–b) is a factor of p(x ), if p = 0, , a , a , , , x − a = 0 or x − b = 0, z (x–a) (x–b) is a factor of p(x ) , if p(a) = 0 and p(b) = 0 , , x = b , x = a or, , , Example 3.13, , 3, , 2, , Show that (x + 2) is a factor of x − 4x − 2x + 20, , Let p(x ) = x − 4x − 2x + 20, , To find the zero, of x+2;, , By factor theorem, (x + 2) is factor of p(x ), if p(−2) = 0, , put x + 2 = 0, , Solution, 3, , 2, , 3, , 2, , p(−2) = (−2) − 4(−2) − 2(−2) + 20, , we get, , x = –2, , = −8 − 4(4) + 4 + 20, p(−2) = 0, 3, , 2, , Therefore, (x + 2) is a factor of x − 4x − 2x + 20, Example 3.14, , 3, , 2, , To find the zero of, 3x–2;, , Is (3x - 2) a factor of 3x + x − 20x + 12 ?, , put 3x – 2 = 0, , Solution, , 3x = 2, 3, , 2, , Let p(x ) = 3x + x − 20x +12, , we get, , x=, , 2, By factor theorem, (3x - 2) is a factor, if p = 0, 3 , 3, 2, 2 2, 2, 2, p = 3 + − 20 + 12, 3 3 , 3 , 3 , Progress Check, 8 4, 2, = 3 + − 20 + 12, 27 9 , 3 , =, , 8 4 120 108, + −, +, 9 9, 9, 9, , 2, 3, , 1. (x+3) is a factor of p(x), if p(__) = 0, 2. (3–x) is a factor of p(x), if p(__) = 0, 3. (y–3), , is a factor of p(y), if p(__) = 0, , 4. (–x–b) is a factor of p(x), if p(__) = 0, 5. (–x+b) is a factor of p(x), if p(__) = 0, Algebra, , 3 Algebra_Term1.indd 95, , 95, , 26-12-2019 13:39:39
Page 102 :
www.tntextbooks.in, , 2, (120 − 120), p =, 3 , 9, =0, Therefore,(3x - 2) is a factor of, 3, , 2, , 3x + x − 20x + 12 , Example 3.15, 3, , Find the value of m, if (x - 2) is a factor of the polynomial, , 2, , 2x − 6x + mx + 4 ., Solution, 3, , 2, , Let p(x ) = 2x − 6x + mx, By factor theorem, (x - 2), p(2), 3, 2, 2(2) − 6(2) + m(2) + 4, 2(8) − 6(4) + 2m + 4, −4 + 2m, m, , +4, is a factor of p(x ), if p(2) = 0, =0, =0, =0, =0, =2, , To find the zero, of x–2;, put, , x–2=0, , we get, , x=2, , Exercise 3.3, 1., , Check whether p(x) is a multiple of g(x) or not ., p(x) = x3 - 5x2 + 4x - 3 ; g(x) = x – 2, , 2., , By remainder theorem, find the remainder when, p(x) is divided by g(x) where,, (i) p(x) = x3 - 2x2 - 4x - 1 ; g^ xh = x + 1, (ii), , p(x) = 4x3 - 12x2 + 14x - 3; g^ xh = 2x - 1, , (iii), , p(x) = x3 - 3x2 + 4x + 50 ; g^ xh = x - 3, , 3., , Find the remainder when 3x3 - 4x2 + 7x - 5 is divided by (x+3), , 4., , What is the remainder when x2018 +2018 is divided by x–1, , 5., , For what value of k is the polynomial, p^ xh = 2x3 - kx2 + 3x + 10 exactly divisible by (x–2), , 6., , If two polynomials 2x3 + ax2 + 4x – 12 and x3 + x2 –2x+ a leave the same, , 7., , remainder when divided by (x – 3), find the value of a and also find the remainder., Determine whether (x - 1) is a factor of the following polynomials:, 3, , 2, , i ) x + 5x – 10x + 4, 96, , 3 Algebra_Term1.indd 96, , 4, , 2, , ii ) x + 5x – 5x + 1, , 9th Standard Mathematics, , 26-12-2019 13:39:41
Page 103 :
www.tntextbooks.in, , 8., , Using factor theorem, show that (x - 5) is a factor of the polynomial, 3, , 2, , 2x − 5x − 28x + 15, 3, , 9., , 2, , Determine the value of m , if (x + 3) is a factor of x − 3x − mx + 24 ., , , 1, 2, 10. If both (x - 2) and x − are the factors of ax + 5x + b , then show that a = b., , , 2, 3, , 2, , 11. If (x - 1) divides the polynomial kx − 2x + 25x − 26 without remainder, then find, the value of k ., 2, , 12. Check if (x + 2) and (x - 4) are the sides of a rectangle whose area is x – 2x – 8 by, using factor theorem., , 3.4 Algebraic Identities, An identity is an equality that remains true regardless of the values chosen for its, variables., We have already learnt about the following identities:, 2, , 2, , (1) (a + b) ≡ a + 2ab + b, 2, , (3) (a + b)(a − b) ≡ a − b, , 2, , 2, , 2, , 2, , (2) (a − b) ≡ a − 2ab + b, , 2, , 2, , (4) (x + a )(x + b) ≡ x + (a + b)x + ab, , Note, 2, , 2, , 2, , 2, , 2, , 2, , (i) a + b = (a + b) − 2ab, , (ii) a + b = (a − b) + 2ab, , 2, , 1 , , (iii) a + 2 = a + − 2, , a, a, , 2, , 1 , , (iv) a + 2 = a − + 2, , a, a, , 2, , 1, , Example 3.16, 2, , (ii) (2a - 3b), , 2, , 1, , 2, , Expand the following using identities: (i) (3x + 4y ), (iii) (5x + 4y )(5x − 4y ), , (iv), , (m + 5)(m − 8), , Solution, (i), , 2, , (3x + 4y ), 2, , 2, , 2, , (3x + 4y ) = (3x ) + 2(3x )(4y ) + (4y ), 2, , = 9x + 24xy + 16y, (ii), , 2, , 2, , (2a - 3b), , 2, , 2, , 2, , (2a - 3b) = (2a ) − 2(2a )(3b) + (3b), 2, , = 4a − 12ab + 9b, , 2, 2, 2, , we have (a + b) = a + 2ab + b , put [a = 3x , b = 4y ], , we have (a − b)2 = a 2 − 2ab + b 2 , , , put [a = 2a, b = 3b ], , 2, , Algebra, , 3 Algebra_Term1.indd 97, , 97, , 26-12-2019 13:39:44
Page 106 :
www.tntextbooks.in, , 3, , 3, , 3.4.3 Expansion of (x + y ) and (x - y ), 3, , 2, , (x + a )(x + b)(x + c) ≡ x + (a + b + c)x + (ab + bc + ca )x + abc, substituting a = b = c = y in the identity, 3, , 2, , we get, (x + y )(x + y )(x + y ) = x + (y + y + y )x + (yy + yy + yy )x + yyy, 3, , 2, , 2, , = x + (3y )x + (3y )x + y, 3, , 2, , 3, , 2, , 3, , (or) (x + y )3 ≡ x + y + 3xy(x + y ), , 2, , 3, , (or) (x − y ) ≡ x − y − 3xy(x − y ), , Thus, (x + y )3 ≡ x + 3x y + 3xy + y, , 3, , 3, , 3, , 3, , by replacing y by -y, we get, 3, , 3, , 2, , (x − y ) ≡ x − 3x y + 3xy − y, Example 3.21, , 3, , 3, , Expand (5a - 3b), , Solution, We know that,, , 3, , = x − 3x y + 3xy − y, , 3, , = (5a ) − 3(5a ) (3b) + 3(5a )(3b) – (3b), , (x - y ), , 3, , 2, , 3, , (5a - 3b), , 2, , 3, , 2, , 3, , 2, , 2, , 3, , 2, , 3, , = 125a − 3(25a )(3b) + 3(5a )(9b ) – (3b), 3, , 2, , 2, , = 125a − 225a b + 135ab − 27b, The following identity is also used:, 3, , 3, , 3, , 2, , 2, , 3, , 2, , x + y + z − 3xyz ≡ (x + y + z )(x + y + z − xy − yz − zx ), We can check this by performing the multiplication on the right hand side., Note, (i) If (x + y + z ) = 0 then x 3 + y 3 + z 3 = 3xyz, Some identities involving sum, difference and product are stated without proof, (i) x 3 + y 3 ≡ (x + y )3 − 3xy(x + y ), (ii) x 3 − y 3 ≡ (x − y )3 + 3xy(x − y ), Example 3.22, , Find the product of, 2, , 2, , 2, , (2x + 3y + 4z )(4x + 9y + 16z − 6xy − 12yz − 8zx ), Solution, 2, , 2, , 2, , 3, , 3, , 3, , We know that, (a + b + c)(a + b + c − ab − bc − ca ) = a + b + c − 3abc, , 100, , 3 Algebra_Term1.indd 100, , 9th Standard Mathematics, , 26-12-2019 13:39:54
Page 107 :
www.tntextbooks.in, , 2, , 2, , 2, , (2x + 3y + 4z )(4x + 9y + 16z − 6xy − 12yz − 8zx ), 3, , 3, , 3, , = (2x ) + (3y ) + (4z ) − 3(2x )(3y )(4z ), 3, , Example 3.23, , 3, , 3, , = 8x + 27y + 64z − 72xyz, 3, , 3, , Evaluate 10 − 15 + 5, , 3, , Solution, 3, , 3, , 3, , We know that, if a + b + c = 0 , then a + b + c = 3abc, Here, a + b + c = 10 − 15 + 5 = 0, 3, , 3, , 3, , = 3(10)(−15)(5), , 3, , 3, , = −2250, , Therefore, 10 + (−15) + 5, 3, , Replace, a by 10, b by –15,, c by 5, , 10 − 15 + 5, , Exercise 3.4, 1., , Expand the following:, 2, , 2, , (ii) (−p + 2q + 3r ), (iv) (3a + 1)(3a − 2)(3a + 4), , (i) (x + 2y + 3z ), (iii) (2p + 3)(2p − 4)(2p − 5), , 2, , 2., , Using algebraic identity, find the coefficients of x , x and constant term without, actual expansion., (i) (x + 5)(x + 6)(x + 7), (ii), (2x + 3)(2x − 5)(2x − 6), , 3., , If (x + a )(x + b)(x + c) = x + 14x + 59x + 70 , find the value of, 1 1 1, 2, 2, 2, (i) a + b + c, (ii) + +, (iii) a + b + c, (iv), a b c, Expand:, , 3, 1, 3, (ii) x + , (i) (3a - 4b), , y , Evaluate the following by using identities:, , 4., , 5., , 3, , (i), , 98, , 3, , (ii), , 2, , a, b, c, +, +, bc ac ab, , 3, , 1001, , 2, , 2, , 2, , 6., , If (x + y + z ) = 9 and (xy + yz + zx ) = 26, then find the value of x + y + z ., , 7., , Find 27a + 64b , if 3a + 4b = 10 and ab = 2 ., , 8., , Find x - y , if x − y = 5 and xy = 14 ., , 3, , 3, , 3, , 3, , 1, 1, 3, = 6 , then find the value of a + 3 ., a, a, 1, 1, 1, 2, 3, 10. If x + 2 = 23 , then find the value of x + and x + 3 ., x, x, x, 3, , 1, 1, 3, 11. If y − = 27 , then find the value of y - 3 ., , y , y, 9., , 3 Algebra_Term1.indd 101, , If a +, , Algebra 101, , 26-12-2019 13:39:58
Page 108 :
www.tntextbooks.in, , 12. Simplify:, , 2, , 2, , 2, , (i) (2a + 3b + 4c)(4a + 9b + 16c − 6ab − 12bc − 8ca ), 2, , 2, , 2, , (ii) (x − 2y + 3z )(x + 4y + 9z + 2xy + 6yz − 3xz ), 13. By using identity evaluate the following:, 1 27, 3, 3, 3, (i) 7 − 10 + 3 , (ii) 1 + −, 8, 8, 3, 3, 3, 14. If 2x − 3y − 4z = 0 , then find 8x - 27y - 64z ., , 3.5 Factorisation, Factorisation is the reverse of multiplication., For Example :, , Multiply 3 and 5; we get product 15., , Factorise 15; we get factors 3 and 5., For Example :, , 2, , Multiply (x + 2) and (x + 3); we get product x + 5x + 6 ., 2, , Factorise x + 5x + 6 ; we get factors (x + 2) and (x + 3)., (x + 3)(x + 2), , Factorise, , Multiply, , (x + 3)(x + 2), x (x + 2) + 3(x + 2), 2, , x + 2x + 3x + 6, , x (x + 2) + 3(x + 2), 2, , x + 2x + 3x + 6, , 2, , 2, , x + 5x + 6, , x + 5x + 6, , Thus, the process of converting the given higher degree polynomial as the product, of factors of its lower degree, which cannot be further factorised is called factorisation., Two important ways of factorisation are :, (i), , By taking common factor, , (ii) By grouping them, , ab + ac, a × b +a × c, a(b + c) factored form, , a + b − pa − pb, (a + b) − p(a + b) group in pairs, (a + b)(1 − p) factored form, , When a polynomial is factored, we “factored out” the common factor., Example 3.24, , Factorise the following:, 3, , 2, , (i) am + bm + cm (ii) a - a b, , (iii) 5a − 10b − 4bc + 2ac (iv) x + y − 1 − xy, , Solutions, (i), , 102, , 3 Algebra_Term1.indd 102, , am + bm + cm, , (ii), , 3, , 2, , a -a b, 2, , 2, , am + bm + cm, , a ⋅ a − a ⋅ b group in pairs, , m(a + b + c) factored form, , a × (a − b) factored form, , 2, , 9th Standard Mathematics, , 26-12-2019 13:40:00
Page 109 :
www.tntextbooks.in, , (iii) 5a − 10b − 4bc + 2ac, 5a − 10b + 2ac − 4bc, 5(a − 2b) + 2c(a − 2b), (a − 2b)(5 + 2c), , (iv), , x + y − 1 − xy, x − 1 + y − xy, (x − 1) + y(1 − x ), (x - 1) - y(x - 1), (x - 1)(1 - y ), , (a–b)=–(b–a), , 3.5.1 Factorisation using Identity, 2, , 2, , 2, , 2, , 2, , 2, , 2, , (iii) a − b ≡ (a + b)(a − b), 3, , 3, , 3, , 3, , 2, , 2, , (ii) a − 2ab + b ≡ (a − b), , (i) a + 2ab + b ≡ (a + b), , 2, , 2, , 2, , (iv)a + b + c + 2ab + 2bc + 2ca ≡ (a + b + c), , 2, , 2, , 3, , 3, , 2, , 2, , (v) a + b ≡ (a + b)(a − ab + b ) (vi) a − b ≡ (a − b)(a + ab + b ), 3, , 2, , 2, , 2, , (vii) a + b + c − 3abc ≡ (a + b + c)(a + b + c − ab − bc − ca ), Note, 2, , 2, , 2, , 2, , 2, , 2, , (a + b) + (a − b) = 2(a + b );, );, (a + b) − (a − b) = 4ab, , 4, , 4, , 2, , 2, , a − b = ((a, a + b )(a + b)(a − b), 6, , 6, , 2, , 2, , 2, , 2, , a − b = (a + b)(a − b)(a − a, abb + b )(a + a, abb + b ), , ;, , Progress Check, 2, , 2, , 2, , 2, , , , , 1, 1, 1, Prove: (i) a + + a − = 2 a 2 + 2 , , , , a, a, a , , Example 3.25, , Factorise the following:, , 2, , (i) 9x + 12xy + 4y, 3, , 2, , , 1 , 1, (ii) a + − a − = 4, , a , a, , 2, , 2, , 2, , (ii) 25a − 10a + 1, , 4, , 2, , 2, , (iii) 36m − 49 n, , 2, , 2, , (v) x - 16 (vi) x + 4y + 9z − 4xy + 12yz − 6xz, , (iv) x - x, Solution, (i), , 2, , 9x + 12xy + 4y, , 2, , 2, , 2, , 2, , 2, , 2, , = (3x ) + 2(3x )(2y ) + (2y ) [ a + 2ab + b = (a + b) ], 2, , = (3x + 2y ), (ii), , 2, , 2, , 2, , 25a − 10a + 1 = (5a ) − 2(5a )(1) + 1, 2, , = (5a − 1), (iii), , 2, , 2, , 2, , 2, , 36m - 49n . = (6m ) - (7n ), , = (6m + 7n )(6m − 7n ), , a 2 − 2ab + b 2 = (a − b)2 , , , a 2 − b 2 = (a + b)(a − b), , , Algebra 103, , 3 Algebra_Term1.indd 103, , 26-12-2019 13:40:05
Page 111 :
www.tntextbooks.in, , Exercise 3.5, 1., , Factorise the following expressions:, 2, , 2, , 2, , (ii) ab − ac − mb + mc, , (i) 2a + 4a b + 8a c, 2., , Factorise the following:, 2, , 2, , (ii) 3a − 24ab + 48b, , (i) x + 4x + 4, 1, 2, (iv) m + 2 − 23, m, 3., , (v) 6 - 216x, , 2, , 5, , (iii) x - 16x, 1, 2, (vi) a + 2 − 18, a, , 2, , Factorise the following:, 2, , 2, , 2, , (i) 4x + 9y + 25z + 12xy + 30yz + 20xz, 2, , 2, , 2, , 4., , (ii) 25x + 4y + 9z − 20xy + 12yz − 30xz, Factorise the following:, , 5., , (ii) 27x - 8y, (i) 8x + 125y, Factorise the following:, , 3, , 3, , 3, , 3, , 3, , 3, , 6, , (iii) a - 64, 3, , 3, , 3, , (ii) l - 8m - 27n - 18lmn, , (i) x + 8y + 6xy − 1, , 3.5.2 Factorising the Quadratic Polynomial (Trinomial) of the type, 2, , ax + bx + c, a ≠ 0, 2, , The linear factors of ax + bx + c will be in the form(kx + m ) and (lx + n ), 2, , 2, , Thus, ax + bx + c = (kx + m )(lx + n ) = klx + (lm + kn )x + mn, 2, , Comparing coefficients of x , x and constant term c on both sides., We have, a = kl , b = (lm + kn ) and c = mn, where ac is the product of kl and mn, that is, equal to the product of lm and kn which are the coefficient of x. Therefore, (kl × mn ) = (lm × kn ) ., 2, , Steps to be followed to factorise ax + bx + c :, 2, , Step 1 : Multiply the coefficient of x and constant term, that is ac ., Step 2 : Split ac into two numbers whose sum and product is equal to b and ac, respectively., Step 3 : The terms are grouped into two pairs and factorise., Product of Sum of Product of Sum of, numbers numbers numbers numbers, Example 3.27, 2, Factorise 2x + 15x + 27, ac = 54, b = 15, ac = 54, b = 15, Solution, 1 × 54, 55, –1 × –54, –55, 2, Compare with ax + bx + c, 2 × 27, 29, –2 × –27, –29, we get, a = 2, b = 15, c = 27, 3 × 18, 21, –3 × –18, –21, 6×9, 15, –6 × –9, –15, product ac = 2 × 27 = 54 and sum b = 15, The required factors are 6 and 9, We find the pair 6, 9 only satisfies “b = 15”, Algebra 105, , 3 Algebra_Term1.indd 105, , 26-12-2019 13:40:12
Page 112 :
www.tntextbooks.in, , and also “ac = 54”., \ we split the middle term as 6x and 9x, 2, , 2, , 2x + 15x + 27 = 2x + 6x + 9x + 27, = 2x (x + 3) + 9(x + 3), = (x + 3)(2x + 9), 2, , Therefore, (x + 3) and (2x + 9) are the factors of 2x + 15x + 27 ., Example 3.28, , 2, , Factorise 2x − 15x + 27 Product of Sum of Product of Sum of, Solution, numbers numbers numbers numbers, 2, Compare with ax + bx + c, ac = 54, b = –15, ac = 54, b =–15, a = 2, b = −15, c = 27, 1 × 54, 55, –1 × –54, –55, 2 × 27, 29, –2 × –27, –29, product ac = 2×27 = 54, sum b=–15, 3 × 18, 21, –3 × –18, –21, \ we split the middle term as –6x and –9x, 6×9, 15, –6 × –9, –15, 2, 2, 2x − 15x + 27 = 2x − 6x − 9x + 27, The required factors are –6 and –9, = 2x (x − 3) − 9(x − 3), = (x − 3)(2x − 9), 2, , Therefore, (x - 3) and (2x - 9) are the factors of 2x − 15x + 27., Example 3.29, , 2, , Factorise 2x + 15x − 27 Product of Sum of Product, Sum of, Solution, numbers numbers of numbers numbers, 2, Compare with ax + bx + c, ac =–54, b = 15, ac =–54, b = 15, Here, a = 2, b = 15, c = −27, –1 × 54, 53, 1 × –54, –53, –2 × 27, 25, 2 × –27, –25, product ac = 2×–27 = –54, sum b=15, –3 × 18, 15, 3 × –18, –15, \ we split the middle term as 18x and –3x, –6 × 9, 3, 6 × –9, –3, 2, 2, 2x + 15x − 27 = 2x + 18x − 3x − 27, The required factors are –3 and 18, = 2x (x + 9) − 3(x + 9), = (x + 9)(2x − 3), 2, , Therefore, (x + 9) and (2x − 3) are the factors of 2x + 15x − 27., Example 3.30, , 2, , Factorise 2x - 15x - 27, , Solution, 2, , Compare with ax + bx + c, Here,a = 2, b = −15, c = −27, product ac = 2×–27=–54, sum b=–15, \ we split the middle term as –18x and 3x, 106, , 3 Algebra_Term1.indd 106, , Product of Sum of Product of Sum of, numbers numbers numbers numbers, ac =–54, b =–15, ac =–54, b =–15, –1 × 54, 53, 1 × –54, –53, –2 × 27, 25, 2 × –27, –25, –3 × 18, 15, 3 × –18, –15, –6 × 9, 3, 6 × –9, –3, The required factors are 3 and –18, , 9th Standard Mathematics, , 26-12-2019 13:40:15
Page 113 :
www.tntextbooks.in, , 2, , 2, , 2x - 15x - 27 = 2x − 18x + 3x − 27, = 2x (x − 9) + 3(x − 9), = (x − 9)(2x + 3), 2, , Therefore, (x − 9) and (2x + 3) are the factors of 2x - 15x - 27, Example 3.31, , 2, , Factorise (x + y ) + 9(x + y ) + 20, , Solution, , Product of Sum of Product of Sum of, numbers numbers numbers numbers, ac = 20, b=9, ac = 20, b=9, 1 × 20, 21, –1 × –20, –21, 2 × 10, 12, –2 × –10, –12, 4×5, 9, –4 × –5, –9, The required factors are 4 and 5, , 2, , Let x + y = p , we get p + 9p + 20, 2, , Compare with ax + bx + c,, We get a = 1, b = 9, c = 20, product ac = 1×20 = 20, sum b=9, \ we split the middle term as 4p and 5p, 2, , 2, , p + 9p + 20 = p + 4 p + 5p + 20, = p(p + 4) + 5(p + 4), = (p + 4)(p + 5), 2, , Put, p = x + y we get, (x + y ) + 9(x + y ) + 20 = (x + y + 4)(x + y + 5), , Exercise 3.6, 1., , Factorise the following:, 2, , (ii) z + 4z − 12, , 2, , (v) y - 16y - 80, , (i) x + 10x + 24, (iv) t + 72 − 17t, 2., , 2, , (iii), , 2, , 2, , (vi) a + 10a − 600, , p - 6p - 16, 2, , Factorise the following:, 2, , 2, , (ii) 5x - 29xy - 42y, , (i) 2a + 9a + 10, 2, , (iv) 6x + 16xy + 8y, , 2, , 2, , 2, , 2, , (iii) 9 − 18x + 8x, , 2, , 2 2, , (v) 12x + 36x y + 27y x, , 2, , (vi) (a + b ) + 9 (a + b ) + 18, 3., , Factorise the following:, 2, , (i) (p - q ) - 6(p - q ) - 16, 4, , 2, , (iv) a − 3a + 2, , 2, , (ii) m + 2mn − 24n, 3, , 2, , 2, , (v) 8m - 2m n - 15mn, , (iii), 2, , (vi), , 2, , 5a + 2a − 3 5, 1, x, , 2, , +, , 1, y, , 2, , +, , 2, xy, , Algebra 107, , 3 Algebra_Term1.indd 107, , 26-12-2019 13:40:18
Page 114 :
www.tntextbooks.in, , 3.6 Division of Polynomials, , Let us consider the numbers 13 and 5. When 13 is divided by 5 what is the quotient, and remainder.?, Yes, of course, the quotient is 2 and the remainder is 3. We write 13 = (5×2)+3, Let us try., Divide, , Expressed as, , Remainder, , Divisor, , 11 by 4, , (4×2)+3, , 3, , 4, , 22 by 11, , (11×2)+0, , 0, , 11, , Dividend = ( Divisor × Quotient ) + Remainder., , From the above examples, we observe that the remainder is less than the divisor., , 3.6.1 Division Algorithm for Polynomials, Let p(x) and g(x) be two polynomials such that degree of p(x)> degree of g(x) and, g(x) ! 0. Then there exists unique polynomials q(x) and r(x) such that, , p(x) = g (x) # q (x) + r (x) , … (1), where, r(x) = 0 or degree of r(x) < degree of g(x)., The polynomial p(x) is the Dividend, g(x) is the Divisor, q(x) is the Quotient and, r(x) is the Remainder. Now (1) can be written as, , Dividend = ( Divisor × Quotient ) + Remainder., If r(x) is zero, then we say p(x) is a multiple of g(x). In other, words, g(x) divides p(x)., , If it looks complicated, don’t worry! it is important to know how to divide, polynomials, and that comes easily with practice. The examples below will help you., Example 3.32, , Divide x3 - 4x2 + 6x by x, where , x ! 0, , Solution, We have 3, x - 4x2 + 6x, x3 - 4x2 + 6x, , =, ,x ! 0, x, x, x, x, = x2 - 4x + 6, Example 3.33, , Find the quotient and the remainder when (5x2 - 7x + 2) ' (x - 1), , Solution, , (5x2 - 7x + 2) ' (x - 1), , 108, , 3 Algebra_Term1.indd 108, , 9th Standard Mathematics, , 26-12-2019 13:40:23
Page 116 :
www.tntextbooks.in, , Solution, Step 1, , Arrange dividend and the divisor in standard form., 3, , 2, , 3x − 2x + 7x − 5, , (standard form of dividend), , x +3, , (standard form of divisor), , Write the coefficients of dividend in the first row. Put ‘0’ for missing term(s)., 3, Step 2, , −2, , 7, , −5, , (first row), , Find out the zero of the divisor., x + 3 = 0 implies x = −3, , Step 3, Write the zero of divisor in front of dividend in the first row. Put ‘0’ in the first, column of second row., –3, , 3, , –2, , 7, , –5, , 0, Step 4, , (first row), (second row), , Complete the second row and third row as shown below., –3, , 3, , –2, , 0, , –3×3, , –9, , 3, , 7, –3×–11, , –11, , 33, 40, , –5, –3×40, , (first row), , –120, , (second row), , –125, , (third row), , All the entries except the last one in the third row are the coefficients of the quotient., 2, , Then quotient is 3x − 11x + 40 and, , remainder is -125., , Example 3.36, , 3, , 2, , Find the quotient and remainder when (3x - 4x - 5) is divided, by (3x+1) using synthetic division., To find the, Solution, 3, , zero of 3x+1;, , 2, , Let p(x) = 3x - 4x - 5, d (x ) = (3x + 1), 3, , 2, , Standard form: p(x) = 3x − 4x + 0x − 5 and d (x ) = 3x + 1, -1, 3, , 110, , 3 Algebra_Term1.indd 110, , 3, , –4, , 0, , –1, , 3, , –5, , 0, , –5, , 5, 3, , -5, 9, , 5, 3, , -50, 9, , put 3x + 1 = 0, we get 3x = –1, x= -, , 1, 3, , (remainder), , , 1 2, 5 50, 3, 2, 3x - 4x - 5 = x + 3x − 5x + −, , 3 , 3 9, 9th Standard Mathematics, , 26-12-2019 13:40:26
Page 117 :
www.tntextbooks.in, , 3, , 2 5, (3x + 1), 5 50, × 3 x − x + −, , 3, 3, 9 9, , 2, , 3x - 4x - 5 =, , 2 5, 5 50 , = (3x + 1) x − x + − , , 3, 9 9 , , (since, p(x ) = d (x )q(x ) + r ), , 2 5, 5, -50, Hence the quotient is x − x + and remainder is, , 9, 3, 9 , Example 3.37, 3, , 4, , 3, , 2, , If the quotient on dividing x + 10x + 35x + 50x + 29 by (x + 4), , 2, , is x − ax + bx + 6 , then find the value of a, b and also remainder., Solution, Let, , 4, , 3, , 2, , 4, , 3, , 2, , To find the, zero of x+4;, , p(x ) = x + 10x + 35x + 50x + 29, , put x + 4 = 0, , Standard form = x + 10x + 35x + 50x + 29, Coefficient are, , 1, , –4, , 10, , 1, 0, 1, , 35, , 50, , 10, –4, 6, , 3, , 35, –24, 11, , x = –4, , we get, , 29, 50, –44, 6, , 29, –24, 5, , (remainder), , 2, , 3, , 2, , quotient x + 6x + 11x + 6 is compared with given quotient x − ax + bx + 6, 2, , coefficient of x is 6 = −a and coefficient of x is 11 = b, Therefore, a = −6 , b = 11 and remainder = 5 ., , Exercise 3.7, 1. Find the quotient and remainder of the following., (i) (4x3 + 6x2 – 23x +18) ' (x+3), , (ii) (8y3 – 16y2 + 16y –15) ' (2y–1), , (iii) (8x3 – 1) ' (2x–1), , (iv) (–18z + 14z2 + 24z3 +18) ' (3z+4), , 2. The area of a rectangle is x2 + 7x + 12. If its breadth is (x+3), then find its length., 3. The base of a parallelogram is (5x+4). Find its height, if the area is 25x2–16., 4. The sum of (x+5) observations is (x3+125). Find the mean of the observations., 5. Find the quotient and remainder for the following using synthetic division:, 3, , 2, , 3, , (i) (x + x − 7x − 3) ÷ (x − 3), 3, , 2, , (ii) (x + 2x − x − 4) ÷ (x + 2), , 2, , 4, , 2, , (iv) (8x − 2x + 6x + 5) ÷ (4x + 1), , (iii) (3x − 2x + 7x − 5) ÷ (x + 3), 4, , 2, , 6. If the quotient obtained on dividing (8x − 2x + 6x − 7) by (2x + 1) is, 3, , 2, , (4x + px − qx + 3) , then find p, q and also the remainder., Algebra 111, , 3 Algebra_Term1.indd 111, , 26-12-2019 13:40:28
Page 118 :
www.tntextbooks.in, , 3, , 2, , 2, , 7. If the quotient obtained on dividing 3x + 11x + 34x + 106 by x - 3 is 3x + ax + b ,, then find a, b and also the remainder., 3.6.3 Factorisation using Synthetic Division, In this section, we use the synthetic division method that helps to factorise a cubic, polynomial into linear factors. If we identify one linear factor of cubic polynomial p(x ), then using synthetic division we can get the quadratic factor of p(x ). Further if possible, one can factorise the quadratic factor into linear factors., Note, z For any non constant polynomial p(x), x = a is zero if and only if p(a) = 0, z x–a is a factor for p(x) if and only if p(a) = 0 (Factor theorem), , To identify (x – 1) and (x + 1) are the factors of a polynomial, z (x–1) is a factor of p(x) if and only if the sum of coefficients of p(x) is 0., z (x+1) is a factor of p(x) if and only if the sum of the coefficients of even power of x,, including constant is equal to the sum of the coefficients of odd powers of x, Example 3.38, , 3, , 2, , (i) Prove that (x - 1) is a factor of x − 7x + 13x − 7, 3, , 2, , (ii) Prove that (x + 1) is a factor of x + 7x + 13x + 7, Solution, (i), , 3, , 2, , 3, , 2, , Let p(x ) = x − 7x + 13x − 7, Sum of coefficients = 1 − 7 + 13 − 7 = 0, Thus (x - 1) is a factor of p(x ), , (ii) Let q(x ) = x + 7x + 13x + 7, Sum of coefficients of even powers of x and constant term = 7 + 7 = 14, Sum of coefficients of odd powers of x = 1 + 13 = 14, Hence, (x + 1) is a factor of q(x ), Example 3.39, , 3, , 2, , Factorise x + 13x + 32x + 20 into linear factors., , Solution, 3, , 2, , Let, p(x ) = x + 13x + 32x + 20, Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0, Hence, (x - 1) is not a factor., Sum of coefficients of even powers and constant term = 13 + 20 = 33, Sum of coefficients of odd powers = 1 + 32 = 33, Hence, (x + 1) is a factor of p(x ), Now we use synthetic division to find the other factors, 112, , 3 Algebra_Term1.indd 112, , 9th Standard Mathematics, , 26-12-2019 13:40:31
Page 119 :
www.tntextbooks.in, , Method I, –1, –2, , Method II, , 1, 0, , 13, –1, , 32 20, –12 –20, , 1, , 12, , 20, , 0, , –2, , –20, , 1, , 10, , 0, , –1, , 0 (remainder), , 1, , 13, , 32, , 20, , 0, , –1, , –12 –20, , 1, , 12, , 20, , 0 (remainder), 2, , Then p(x ) = (x + 1)(x + 12x + 20), , (remainder), , 2, , 2, , Now x + 12x + 20 = x + 10x + 2x + 20, , p(x ) = (x + 1)(x + 2)(x + 10), , =x (x + 10) + 2(x + 10), , Hence,, 3, , =(x + 2)(x + 10), , 2, , x + 13x + 32x + 20, = (x + 1)(x + 2)(x + 10), , Example 3.40, , 3, , 3, , 2, , Hence, x + 13x + 32x + 20, = (x + 1)(x + 2)(x + 10), , 2, , Factorise x − 5x − 2x + 24, , Solution, 3, , 2, , Let p(x ) = x − 5x − 2x + 24, When x = 1, p(1)= 1 − 5 − 2 + 24 = 18 ≠ 0;, (x - 1)is not a factor., When x = –1, p(-1)= −1 − 5 + 2 + 24 = 20 ≠ 0;, (x + 1) is not a factor., Therefore, we have to search for different values of x by trial and error method., When x = 2, 3, 2, p(2) =, = 2 − 5(2) − 2(2) + 24, = 8 − 20 − 4 + 24, =8¹0, Hence, (x–2) is not a factor, Note, When x = −2, 3, , p(-2)= (−2) − 5(−2) − 2(−2) + 24, 2, , = −8 − 20 + 4 + 24, p(-2) = 0, Hence, (x+2) is a factor, –2, 3, , 1, 0, 1, 0, 1, , –5, –2, –7, 3, –4, , Check whether 3 is a zero of, x 2 − 7x + 12 . If it is not, then, check for –3 or 4 or –4 and so, on., , –2, 24, +14 –24, 12, 0 (remainder), –12, 0 (remainder), , Thus, (x + 2)(x − 3)(x − 4) are the factors., 3, , 2, , 2, , Therefore, x − 5x − 2x + 24 = (x + 2)(x − 3)(x − 4), Algebra 113, , 3 Algebra_Term1.indd 113, , 26-12-2019 13:40:34
Page 120 :
www.tntextbooks.in, , Exercise 3.8, 1., , Factorise each of the following polynomials using synthetic division:, (ii) 2x 3 − 3x 2 − 3x + 2, (i) x 3 − 3x 2 − 10x + 24, (iii) −7x + 3 + 4x 3, (iv) x 3 + x 2 − 14x − 24, (vi) x 3 − 10x 2 − x + 10, (v) x 3 − 7x + 6, , 3.7 Greatest Common Divisor (GCD), The Greatest Common Divisor, abbreviated as GCD, of two or more polynomials is a, polynomial, of the highest common possible degree, that is a factor of the given two or more, polynomials. It is also known as the Highest Common Factor (HCF)., This concept is similar to the greatest common divisor of two integers., For example, Consider the expressions 14xy2 and 42xy. The common divisors of 14, and 42 are 2, 7 and 14. Their GCD is thus 14. The only common divisors of xy2 and xy are, x, y and xy; their GCD is thus xy., 14xy2 = 1 × 2 × 7 × x × y × y, 42xy = 1 × 2 × 3 × 7 × x × y, Therefore the requried GCD of 14xy2 and 42xy is 14xy., , To find the GCD by Factorisation, (i) Each expression is to be resolved into factors first., (ii) The product of factors having the highest common powers in those factors will be the GCD., (iii) If the expression have numerical coefficient, find their GCD separately and then prefix it as a, coefficient to the GCD for the given expressions., Example 3.41, , Find GCD of the following:, , 3 2, , 3, , (i) 16x y , 24xy z, 3, , 2, , (ii) (y + 1) and (y − 1), 2, , 2, , (iii) 2x - 18 and x - 2x - 3, 2, , 3, , 4, , (iv) (a - b) , (b - c) , (c - a ), Solutions, 3 2, , 3 2, , 4, , 3, , 2, , (i) 16x y = 2 × 2 × 2 × 2 × x y = 2 × x × y = 23 ´ 2 ´ x 2 ´ x ´ y 2, 114, , 3 Algebra_Term1.indd 114, , 9th Standard Mathematics, , 26-12-2019 13:40:35
Page 121 :
www.tntextbooks.in, , 3, , , , 3, , 3, , 3, , 24xy z = 2 × 2 × 2 × 3 × x × y × z = 2 × 3 × x × y × z = 23 ´ 3 ´ x ´ y ´ y 2 ´ z, 3, , , , Therefore, GCD = 2 xy, 3, , 3, , 2, 2, , 3, , (ii) y + 1 = y + 1 = (y + 1)(y − y + 1), 2, , , , 2, , 2, , y - 1 = y − 1 = (y + 1)(y − 1), , , , Therefore, GCD = (y + 1), 2, , 2, , (iii), , 2, , 2, , 2x - 18 = 2(x − 9) = 2(x − 3 ) = 2(x + 3)(x − 3), 2, , 2, , x - 2x - 3 = x − 3x + x − 3, , , = x (x − 3) + 1(x − 3), , , , = (x − 3)(x + 1), , , , Therefore, GCD = (x − 3), 2, , 3, , 4, , (iv) (a - b) , (b - c) , (c - a ), , , There is no common factor other than one., , Therefore, GCD = 1, , Exercise 3.9, 1., , Find the GCD for the following:, 5, , 11, , 9, , 3, , 3, , (i) p , p , p (ii) 4x , y , z, 2 2 3, , 3 2 4, , (iii) 9a b c , 15a b c , 2 3, , 2 3, , 3, , 2, , (v) ab c , a b c, a bc , 3, , 2, , (vii) 25ab c, 100a bc, 125ab, 2., , 8, , 3, , (iv) 64x , 240x, , 6, , 5 3 4, , 2, , 3, , 2 2, , (vi) 35x y z , 49x yz , 14xy z, (viii) 3abc, 5xyz, 7 pqr, , Find the GCD of the following:, (i) (2x + 5), (5x + 2) , 2, , 2, , (iii) 2a + a, 4a − 1 , 4, , 2, , (v) x - 1, x - 1 , , (ii) a, , m +1, , ,a, , 2, , m +2, , 3, , ,a, , (iv) 3a , 5b , 7c, 3, , 2, , m +3, , 4, 2, , (vi) a - 9ax , (a - 3x ), , Algebra 115, , 3 Algebra_Term1.indd 115, , 26-12-2019 13:40:39
Page 122 :
www.tntextbooks.in, , 3.8 Linear Equation in Two Variables, , A linear equation in two variables is of the form ax + by + c = 0 where a, b and c are, real numbers, both a and b are not zero (The two variables are denoted here by x and y and, c is a constant)., , Examples, Linear equation in two variables, , Not a linear equation in two variables., , 2x + y = 4, 1, –5x + = y, 2, , xy + 2x = 5 (Why?), x + y = 25 (Why?), x(x+1) = y (Why?), , 5x = 35y, , If an equation has two variables each of which is in first degree such that the variables are, not multiplied with each other, then it is a linear equation in two variables (If the degree of an, equation in two variables is 1, then it is called a linear equation in two variables)., An understanding of linear equation in two variables will be easy if it is done along, with a geometrical visualization (through graphs). We will make use of this resource., Why do we classify, for example, the equation 2x + y = 4 is a linear equation? You are, right; because its graph will be a line. Shall we check it up?, We try to draw its graph. To draw the graph of 2x + y = 4, we need some points on the, line so that we can join them. (These are the ordered pairs satisfying the equation)., To prepare table giving ordered pairs for 2x + y = 4. It is better, to take it as, , y = 4 – 2x. (Why? How?), When x = –4,, , y = 4 – 2(–4) = 4 + 8 = 12, , When x = –2,, , y = 4 – 2(–2) = 4 + 4 = 8, , When x = 0,, , y = 4 – 2(0) = 4 + 0 = 4, , When x = +1,, , y = 4 – 2(+1) = 4 – 2 = 2, , When x = +3,, , y = 4 – 2(+3) = 4 – 6 = −2, , Thus the values are tabulated as follows:, x-value, , –4, , –2, , 0, , 1, , 3, , y-value, , 12, , 8, , 4, , 2, , –2, , (To fix a line, do we need so many points? It is enough if we have two and probably one, more for verification.), 116, , 3 Algebra_Term1.indd 116, , 9th Standard Mathematics, , 26-12-2019 13:40:40
Page 123 :
www.tntextbooks.in, , (–4, 12), y=, 4, , Y, 12, , –, , 2x, , Scale, x axis 1 cm = 1 unit, y axis 1 cm = 2 units, , 10, (–2, 8), , 8, 6, 4 (0, 4), 2, , X′, , –4, , –3, , –2, , O, , –1, , (1, 2), , 1, , 2, , –2, , 3, , 4, (3, –2), , X, , –4, , Y′, Fig. 3.15, When you plot the points (−4,12), (−2,8), (0,4), (1,2) and (3,−2), you find that they all, lie on a line., This clearly shows that the equation 2x + y = 4 represents a line (and hence said to be, linear)., All the points on the line satisfy this equation and, hence the ordered pairs of all the points on the line are, the solutions of the equation., , Y, 4, 3, , Solution, (i), , Let us prepare a table to find the ordered pairs, of points for the line y = 3x − 1., , 1, , X′, , –2, , –1, , When x = 0 ,, , y = 3(0)–1 = –1, , When x = 1 ,, , y = 3(1)–1 = 2, , O, , 1, , –1, , (0, –1), , 2, , X, , –2, , We shall assume any value for x, for our, convenience let us take −1, 0 and 1., When x = −1, y = 3(–1)–1 = –4, , (1, 2), , 2, x–1, , (i) y = 3x − 1, , Draw the graph for the following:, 2, (ii) y = x + 3, 3, , y=3, , Example 3.42, , Scale, x axis 1 cm = 1 unit, y axis 1 cm = 1 unit, , –3, (–1, –4), , –4, , Y′, , Fig. 3.16, Algebra 117, , 3 Algebra_Term1.indd 117, , 26-12-2019 13:40:41
Page 124 :
www.tntextbooks.in, , , , x, , –1, , 0, , 1, , y, , –4, , –1, , 2, , The points (x,y) to be plotted :, , , , 6, , (−1, −4), (0, −1) and (1, 2)., , Y, , Scale, x axis 1 cm = 1 unit, y axis 1 cm = 1 unit, , 5, , (ii) Let us prepare a table to find the, ordered pairs of points, , , 4, , 2, for the line y = x + 3 ., 3, , 3, , (–3, 1), , X′, , –3, , 3.8.1 Simultaneous Linear Equations, , X, , Fig. 3.17, , Y, 8, , 6, , (1, 8), , Scale, x axis 1 cm = 1 unit, y axis 1 cm = 1 unit, , (2, 6), , 5, 4, 3, , 0, , 2, , (3, 4), , y=1, , What are simultaneous linear equations? These, consists of two or more linear equations with the same, variables., , 2, , 2x+, , With sufficient background of graphing an, equation, now we are set to study about system of, equations, particularly pairs of simultaneous equations., , O, Y′, , 5, , (−3,1), (0,3) and (3,5)., , –1, , –1, , The points (x,y) to be plotted :, , , , 3, , 1, , –2, , 7, , , , 1, , 2, , 2, When x = −3 , y = (−3) + 3 = 1, 3, 2, When x = 0 , y = (0) + 3 = 3, 3, 2, When x = 3 , y = (3) + 3 = 5, 3, x, –3, 0, 3, 1, , (3, 5), , 3 (0, 3), , , Let us assume −3, 0, 3 as x values., (why?), , y, , 2 x+3, =, y 3, , 1, , Why do we need them? A single equation like, 2x+y =10 has an unlimited number of solutions. The, 1, 2, 3, 4, 5 X, O, points (1,8), (2,6), (3,4) and many more lie on the, Fig. 3.18, graph of the equation, which means these are some of, its endless list of solutions. To be able to solve an equation like this, another equation needs to be, used alongside it; then it is possible to find a single ordered pair that solves both equations at the, same time., 118, , 3 Algebra_Term1.indd 118, , 9th Standard Mathematics, , 26-12-2019 13:40:43
Page 125 :
www.tntextbooks.in, , The equations we consider together in such settings make a meaningful situation and are, known as simultaneous linear equations., Real life Situation to understand the simultaneous linear equations, Consider the situation, Anitha bought two erasers and a pencil for ₹10. She does not, know the individual cost of each. We shall form an equation by considering the cost of, eraser as ‘x’ and that of pencil as ‘y’., That is 2x + y = 10, , ... (1), , Now, Anitha wants to know the individual cost of an eraser and a pencil. She tries to, solve the first equation, assuming various values of x and y., 2 × cost of eraser + 1 × cost of pencil = 10, , , 2(1)+8 =10, , , , 2(1.5)+7 =10, , , , 2(2)+6 =10, , , , 2(2.5)+5 =10, , , , 2(3)+4 =10, , , , , Points to be plotted :, x, , 1, , 1.5, , 2, , 2.5, , 3, , ..., , y, , 8, , 7, , 6, , 5, , 4, , ..., , She gets infinite number of answers. So she tries to, find the cost with the second equation., Again, Anitha needs some more pencils, and erasers. This time, she bought 3 erasers and, 4 pencils and the shopkeeper received ₹30 as the, total cost from her. We shall form an equation like, , Y, , (1, 8), , 8, , (1.5, 7), , 7, , (2, 6), , 6, , (2.5, 5), , 5, , the previous one., 4, , Even then she arrives at an infinite number, of answers., , 3, , 3(6)+4 (3) = 30, 3(8)+4 (1.5) = 30, , 1, , O, , 0, , 3(4)+4 (4.5) = 30, , =1, , 3(2)+4 (6) = 30, , 2, , +y, , 3 × cost of eraser + 4 × cost of pencil = 30, , (3, 4), , 2x, , The equation is 3x + 4y = 30 ...(2), , Scale, x axis 1 cm = 1 unit, y axis 1 cm = 1 unit, , 1, , 2, , 3, , 4, , 5, , X, , Fig. 3.19, , , Algebra 119, , 3 Algebra_Term1.indd 119, , 26-12-2019 13:40:44
Page 126 :
www.tntextbooks.in, , Points to be plotted :, x, , 2, , 4, , 6, , 8, , ..., , y, , 6, , 4.5, , 3, , 1.5, , ..., , While discussing this with her teacher, the teacher suggested that she can get a unique, answer if she solves both the equations together., By solving equations (1) and (2) we have the cost of an eraser as ₹2 and cost of a, pencil as ₹6. It can be visualised in, Y, (1, 8), the graph., 8, Scale, The equations we consider, together in such settings make, a meaningful situation and are, known as simultaneous linear, equations., , (1.5, 7), , 7, , (2, 6), , 6, , (2.5, 5), , 5, , (4, 4.5), , (3, 4), , 4, , (6, 3), , 3, , 3x, , 2x, , 2, , +y, , O, , y=, , 30, , 0, , 1, , +4, , (8, 1.5), , =1, , Thus a system of linear, equations consists of two or more, linear equations with the same, variables. Then such equations, are called Simultaneous linear, equations or System of linear, equations or a Pair of linear, equations., , x axis 1 cm = 1 unit, y axis 1 cm = 1 unit, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , X, , Fig. 3.20, , Example 3.43, , Check whether (5, −1) is a solution of the simultaneous equations, x – 2y = 7 and 2x + 3y = 7., Solution, Given, , x – 2y = 7, , …(1), , 2x + 3y = 7, , …(2), , When x = 5, y = −1 we get, From (1), , x – 2y = 5 – 2(−1) = 5 + 2 = 7 which is RHS of (1), , From (2), , 2x + 3y = 2(5) + 3(−1) = 10−3 = 7 which is RHS of (2), , Thus the values x = 5, y = −1 satisfy both (1) and (2) simultaneously. Therefore (5,−1), is a solution of the given equations., 120, , 3 Algebra_Term1.indd 120, , 9th Standard Mathematics, , 26-12-2019 13:40:44
Page 127 :
www.tntextbooks.in, , Progress Check, Examine if (3,3) will be a solution for the simultaneous linear equations, 2x – 5y – 2 = 0 and x + y – 6 = 0 by drawing a graph., , 3.8.2, , Methods of solving simultaneous linear equations, , There are different methods to find the solution of a pair of simultaneous linear, equations. It can be broadly classified as geometric way and algebraic ways., Geometric way, 1. Graphical method, , Algebraic ways, 1. Substitution method, 2. Elimination method, 3. Cross multiplication method, , Solving by Graphical Method, Already we have seen graphical representation of linear equation in two variables., Here we shall learn, how we are graphically representing a pair of linear equations in two, variables and find the solution of simultaneous linear equations., Example 3.44, , Use graphical method to solve the following system of equations:, , x + y = 5; 2x – y = 4., ...(1), , 2x – y = 4, , ...(2), , To draw the graph (1) is very easy. We can find the x and, y values and thus two of the points on the line (1)., , Y, 6, , 2, , To draw the graph of (2), we can adopt the same, procedure., When x = 0, (2) gives y = −4., , X′ O, –2, , (3, 2), Q (2, 0), 2, 4, , (5, 0), B 6, , X, , 2x, , Plot A and B ; join them to produce the line (1)., , 5, , Thus B(5,0) is another point on the line., , =, , Thus A(0,5) is a point on the line., , (0, 5), , +y, , 4, , A, , x, , When x = 0, (1) gives y = 5., When y = 0, (1) gives x = 5., , Scale, x axis 1 cm = 2 units, y axis 1 cm = 2 units, , 8, , =4, , x+y=5, , –y, , Solution, Given, , –4 P (0, –4), , Y′, , Fig. 3.21, , Thus P(0,−4) is a point on the line., Algebra 121, , 3 Algebra_Term1.indd 121, , 26-12-2019 13:40:45
Page 128 :
www.tntextbooks.in, , When y = 0, (2) gives x = 2., Thus Q(2,0) is another point on the line., Plot P and Q ; join them to produce the line (2)., The point of intersection (3, 2) of lines (1) and (2) is, a solution., , Note, It is always good to verify, if the answer obtained is, correct and satisfies both, the given equations., , The solution is the point that is common to both the, lines. Here we find it to be (3,2). We can give the solution, as x = 3 and y = 2., Example 3.45, , Use graphical method to solve the following system of equations:, 3x + 2y = 6; 6x + 4y = 8, Solution, Let us form table of values for each line and then fix the ordered pairs to be plotted., Graph of 3x + 2y = 6, Graph of 6x + 4y = 8, x, , –2, , 0, , 2, , x, , –2, , 0, , 2, , y, , 6, , 3, , 0, , y, , 5, , 2, , –1, , Points to be plotted :, , Points to be plotted :, , (−2,6), (0,3), (2,0), , (−2,5), (0,2), (2,−1), , When we draw the graphs of these two, equations, we find that they are parallel and, they fail to meet to give a point of intersection., As a result there is no ordered pair that can be, common to both the equations. In this case there, is no solution to the system., Example 3.46, , Use graphical method, to solve the following system of equations:, y = 2x + 1; −4x + 2y = 2, , Y, , (–2, 6), , Scale, x axis 1 cm = 2 units, y axis 1 cm = 2 units, , 6, , (–2, 5), 4, , (0, 3), , 6x +4y = 8, , 3x +2y = 6, , 2, (0, 2), , X′, , –2, , (2, 0), 2, 4, (2, –1), , O, , X, , –2, , Y′, Fig. 3.22, , Solution, Let us form table of values for each line and then fix the ordered pairs to be plotted., Graph of y = 2x + 1, Graph of −4x + 2y = 2, 2y = 4x + 2, y = 2x + 1, 122, , 3 Algebra_Term1.indd 122, , 9th Standard Mathematics, , 26-12-2019 13:40:45
Page 129 :
www.tntextbooks.in, , x, −2 −1 0, 1, 2, 2x, −4 −2 0, 2, 4, 1, 1, 1, 1, 1, 1, y = 2x+1 −3 −1 1, 3, 5, Points to be plotted :, (−2, −3), (−1, −1), (0, 1), (1, 3), (2, 5), Scale, Y, Here both the equations are identical; they were, x axis 1 cm = 2 units, y axis 1 cm = 2 units, 6, only represented in different forms. Since they are, (2, 5), identical, their solutions are same. All the points on one, 4, line are also on the other!, (1, 3), y=, , 2x+, , 1, , x, −2 −1 0, 1, 2, 2x, −4 −2 0, 2, 4, 1, 1, 1, 1, 1, 1, y = 2x+1 −3 −1 1, 3, 5, Points to be plotted :, (−2, −3), (−1, −1), (0,1), (1,3), (2, 5), , This means we have an infinite number of, solutions which are the ordered pairs of all the points, on the line., , 2, (0, 1), , X′, , Example 3.47, , l +b=, , 36, 2, , 2, , 4, , 6, , X, , 14, , X, , –4, , Y′, Fig. 3.23, , Y, , (8, 26), b+2, , 26, , l=3, , 24, 22, 20, 18, (2, 16), , 16, Length, , 2(l + b) = 36, , (–1, –1), –2, , (–2, –3), , The perimeter of a rectangle is 36, metres and the length is 2 metres more than three times, the width. Find the dimension of rectangle by using the, method of graph., Solution, Let us form equations for the given, statement., Let us consider l and b as the length and, breadth of the rectangle respectively., Now let us frame the equation for the first, statement, Perimeter of rectangle = 36, , O, , –2, , Scale, x axis 1 cm = 2 units, y axis 1 cm = 2 units, , (5, 17), (4, 14), (4, 14), (5, 13), , 14, 12, , (8, 10), , 10, , l = 18 − b, , ... (1), , 8, , b, , 2, , 4, , 5, , 8, , 6, , 18, , 18, , 18, , 18, , 18, , 4, , −b, , −2, , −4, , −5, , −8, , 2, , l = 18–b, , 16, , 14, , 13, , 10, , Points: (2,16), (4,14), (5,13), (8,10), , X′ O, , (2, 8), , Y′, , 2, , 4, , l=, , 6, , 8, 10, Breadth, , 18, , –b, , 12, , Fig. 3.24, Algebra 123, , 3 Algebra_Term1.indd 123, , 26-12-2019 13:40:46
Page 130 :
www.tntextbooks.in, , The second statement states that the length is 2 metres more than three times the width, which is a straight line written as l = 3b + 2, ... (2), Now we shall form table for the above equation (2)., b, 2, 4, 5, 8, 3b, 6, 12 15 24, 2, 2, 2, 2, 2, l= 3b + 2, 8, 14 17 26, Points: (2,8), (4,14), (5,17), (8,26), The solution is the point that is common to both the lines. Here we find it to be (4,14)., We can give the solution to be b = 4, l = 14., Verification :, 2(l+b) = 36, ...(1), 2(14+4) = 36, 2 × 18 = 36, 36 = 36 true, , l, 14, 14, 14, , =, =, =, =, , 3b + 2, 3(4) +2, 12 + 2, 14 true, , ...(2), , Exercise 3.10, 1., , Draw the graph for the following, (i) y = 2 x, , 2., , 3., , 3, (iii) y = x + 3 (iv) 3x + 2 y = 14, 2, , (ii) y = 4 x − 1, , Solve graphically, (i) x + y = 7; x − y = 3 , x y, x y, (iii) + = 1; + = 2, 2 4, 2 4, (v) y = 2 x + 1; y + 3x − 6 = 0, , (ii) 3x + 2 y = 4; 9 x + 6 y − 12 = 0, (iv) x − y = 0; y + 3 = 0, (vi) x = –3; y = 3, , Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour., If they drive in the same direction they will meet in 2 hours. Find their speed by using, graphical method., , Some special terminology, , We found that the graphs of equations within a system tell us how many solutions are, there for that system. Here is a visual summary., Intersecting lines, , l, , l, m, , One single solution, 124, , 3 Algebra_Term1.indd 124, , Parallel lines, , m, , No solution, , Coinciding lines, , l, , m, , Infinite number of solutions, , 9th Standard Mathematics, , 26-12-2019 13:40:51
Page 131 :
www.tntextbooks.in, , z, z, z, , When a system of linear equation has one solution (the graphs of the, equations intersect once), the system is said to be a consistent system., When a system of linear equation has no solution (the graphs of the equations, don’t intersect at all), the system is said to be an inconsistent system., When a system of linear equation has infinitely many solutions, the lines are the, same (the graph of lines are identical at all points), the system is consistent., , Solving by Substitution Method, In this method we substitute the value of one variable, by expressing it in terms of the, other variable to reduce the given equation of two variables into equation of one variable, (in order to solve the pair of linear equations). Since we are substituting the value of one, variable in terms of the other variable, this method is called substitution method., The procedure may be put shortly as follows:, Step 1: From any of the given two equations, find the value of one variable in terms of, the other., Step 2: Substitute the value of the variable, obtained in step 1 in the other equation, and solve it., Step 3: Substitute the value of the variable obtained in step 2 in the result of step 1 and, get the value of the remaining unknown variable., Example 3.48, substitution method., , Solve the system of linear equations x + 3 y = 16 and 2 x − y = 4 by, , Solution, Given, , x + 3 y = 16, 2x – y = 4, , Step 1, From equation (2), 2x − y = 4, –y = 4–2x, y = 2x − 4, , ...(3), , ... (1), ... (2), Step 2, Substitute (3) in (1), x + 3 y = 16, x + 3(2 x − 4) = 16, x + 6 x − 12 = 16, 7 x = 28, x=4, , Step 3, Substitute x = 4 in (3), y = 2x − 4, y = 2(4) − 4, y=4, , Solution, x=4, and, y=4, , Example 3.49, , The sum of the digits of a given two digit number is 5. If the digits are, reversed, the new number is reduced by 27. Find the given number., Solution, Let x be the digit at ten’s place and y be the digit at unit place., Given that x + y = 5 …… (1), Algebra 125, , 3 Algebra_Term1.indd 125, , 26-12-2019 13:40:54
Page 132 :
www.tntextbooks.in, , Tens, , Ones, , Value, , Given Number, , x, , y, , 10x + y, , New Number, (after reversal), , y, , x, , 10y+x, , Given, Original number − reversing number = 27, , , , (10x + y ) − (10 y + x ) = 27, 10 x − x + y − 10 y = 27, 9 x − 9 y = 27, , , ⇒ , Also from (1), y = 5 – x, , x− y=3, , ... (2), , ... (3), , Substitute (3) in (2) to get x − (5 − x ) = 3, , , Verification :, sum of the digits = 5, , x −5+ x = 3, , , , , x+y =5, , 2x = 8, , 4+1 =5, , x=4, , 5 = 5 true, , Substituting x = 4 in (3), we get y = 5 − x = 5 − 4, y = 1, , Original number –, reversed number = 27, 41 - 14 = 27, , Thus, 10 x + y = 10 × 4 + 1 = 40 + 1 = 41., , 27 = 27 true, , Therefore, the given two-digit number is 41., , Exercise 3.11, 1., , Solve, using the method of substitution, (i), , 2 x − 3 y = 7; 5x + y = 9 , , (ii) 1.5x + 0.1 y = 6.2; 3x − 0.4 y = 11.2, , (iii) 10% of x + 20% of y = 24; 3x − y = 20 (iv), , 2 x − 3 y = 1; 3 x − 8 y = 0, , 2., , Raman’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 Raman., , 3., , The middle digit of a number between 100 and 1000 is zero and the sum of the other, digit is 13. If the digits are reversed, the number so formed exceeds the original, number by 495. Find the number., , 126, , 3 Algebra_Term1.indd 126, , 9th Standard Mathematics, , 26-12-2019 13:40:59
Page 133 :
www.tntextbooks.in, , Solving by Elimination Method, This is another algebraic method for solving a pair of linear equations. This method is, more convenient than the substitution method. Here we eliminate (i.e. remove) one of the, two variables in a pair of linear equations, so as to get a linear equation in one variable which, can be solved easily., The various steps involved in the technique are given below:, Step 1: Multiply one or both of the equations by a suitable number(s) so that either the, coefficients of first variable or the coefficients of second variable in both the, equations become numerically equal., Step 2: Add both the equations or subtract one equation from the other, as obtained, in step 1, so that the terms with equal numerical coefficients cancel mutually., Step 3: Solve the resulting equation to find the value of one of the unknowns., Step 4: Substitute this value in any of the two given equations and find the value of the, other unknown., Example 3.50, , Given 4a + 3b = 65 and a + 2b = 35 solve by elimination method., , Solution, 4a + 3b = 65 .....(1), , Given,, , Verification :, 4a+3b = 65 ...(1), , a + 2b = 35 .....(2), (2) × 4 gives, Already (1) is, , 4(5)+3(15) = 65, , 4a + 8b = 140, (–) (–), (–), 4a + 3b = 65, , 20 + 45 = 65, 65 = 65 True, , 5b = 75 which gives b = 15, Put b = 15 in (2):, , a + 2b = 35, 5 + 2(15) = 35, , a + 2(15) = 35 which simplifies to a = 5, , 5+30 = 35, , Thus the solution is a = 5, b = 15., Example 3.51, , ...(2), , 35 = 35 True, , Solve for x and y: 8 x − 3 y = 5xy , 6 x − 5 y = −2 xy by the method of, , elimination., Solution, The given system of equations are 8 x − 3 y = 5xy, 6 x − 5 y = −2 xy, , ...(1), ...(2), , Observe that the given system is not linear because of the occurrence of xy term. Also, note that if x =0, then y =0 and vice versa. So, (0,0) is a solution for the system and any other, solution would have both x ≠ 0 and y ≠ 0., Algebra 127, , 3 Algebra_Term1.indd 127, , 26-12-2019 13:41:02
Page 134 :
www.tntextbooks.in, , Let us take up the case where x ≠ 0, y ≠ 0., Dividing both sides of each equation by xy,, 8 x 3 y 5xy, 8 3, − =5, , we get,, ...(3), −, =, y x, xy xy xy, 6 x 5 y −2 xy, 6 5, −, − = −2 ...(4), =, y x, xy xy, xy, 1, 1, Let a = , b = ., x, y, (3)&(4) respectively become, 8b − 3a = 5 ...(5), , , 6b − 5a = −2 ...(6), which are linear equations in a and b., To eliminate a, we have,, , (5) × 5 ⇒, , (6) × 3 ⇒, , 40b − 15a = 25 .....(7), 18b − 15a = −6 .....(8), , 11 22 , Now proceed as in the previous example to get the solution , ., 23 31 , 11 22 , Thus, the system have two solutions , and (0, 0) ., 23 31 , , Exercise 3.12, 1., , Solve by the method of elimination, (i) 2x–y = 3;, , 3x + y = 7 , , (ii) x–y = 5;, , 3x + 2y = 25, , x y, x y, + = 14;, + = 15 , (iv) 3(2 x + y ) = 7 xy ; 3(x + 3 y ) = 11xy, 10 5, 8 6, 4, 3, (v) + 5 y = 7; + 4 y = 5 (vi) 13x + 11 y = 70; 11x + 13 y = 74, x, x, The monthly income of A and B are in the ratio 3:4 and their monthly expenditures, are in the ratio 5:7. If each saves ₹ 5,000 per month, find the monthly income of each., , (iii), , 2., 3., , Five years ago, a man was seven times as old as his son, while five year hence, the man, will be four times as old as his son. Find their present age., , Solving by Cross Multiplication Method, The substitution and elimination methods involves many arithmetic operations,, whereas the cross multiplication method utilize the coefficients effectively, which simplifies, the procedure to get the solution. This method of cross multiplication is so called because, we draw cross ways between the numbers in the denominators and cross multiply the, coefficients along the arrows ahead. Now let us discuss this method as follows:, 128, , 3 Algebra_Term1.indd 128, , 9th Standard Mathematics, , 26-12-2019 13:41:09
Page 136 :
www.tntextbooks.in, , x, y, 1, =, =, (−4)(−5) − (3)(−10) (−10)(4) − (−5)(3) (3)(3) − (4)(−4), x, y, 1, =, =, Verification :, (20) − ( −30) (−40) − (−15) (9) − (−16), 3x–4y =, x, y, 1, =, =, 3(2)–4(–1) =, 20 + 30 −40 + 15 9 + 16, x, y, 1, 6+4 =, =, =, 50 −25 25, 10 =, 50, −25, 4x + 3y =, Therefore, we get, x= ;, y=, 25, 25, 4(2) + 3(–1) =, x = 2;, y = −1, 8–3 =, Thus the solution is x = 2, y = –1., 5 =, Example 3.53, , 10 ...(1), 10, 10, 10 True, 5, , ...(2), , 5, 5, 5 True, , Solve by cross multiplication method : 3x + 5 y = 21; −7 x − 6 y = −49, , Solution, The given system of equations are 3x + 5 y − 21 = 0; −7 x − 6 y + 49 = 0, Now using the coefficients for cross multiplication, we get,, x, 5, –6, ⇒, , y, –21, 49, , 1, 3, –7, , 5, –6, , x, y, 1, =, =, (5)(49) − ( −6)(−21) ( −21)( −7) − (49)(3) (3)( −6) − ( −7)(5), x, y 1, = =, 119 0 17, x, 1, y 1, ⇒, =, ,, =, 119 17, 0 17, 119, 0, ⇒, x=, ,, y=, 17, 17, ⇒, x=7,, y=0, , Verification :, 3x+5y = 21 ...(1), 3(7)+5(0) = 21, 21 + 0 = 21, 21 = 21 True, –7x – 6y = –49 ...(2), –7(7) – 6(0) = –49, –49 = –49, –49 = –49 True, , Note, y 1, 0, y, =, is to mean y = . Thus,, is only a notation and it is not division by, 0 17, 17, 0, zero. It is always true that division by zero is not defined., , Here, , 130, , 3 Algebra_Term1.indd 130, , 9th Standard Mathematics, , 26-12-2019 13:41:21
Page 137 :
www.tntextbooks.in, , Exercise 3.13, 1., , Solve by cross-multiplication method, (ii) 6 x + 7 y − 11 = 0 ; 5x + 2 y = 13, (i) 8 x − 3 y = 12 ; 5x = 2 y + 7, 2 3, 3 1, + = 5; − +9 = 0, x y, x y, Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins, totalling ₹ 220, how many coins of each kind does she have., It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter, is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half, of the pool is filled. How long would each pipe take to fill the swimming pool., , (iii), 2., 3., , 3.8.3 Consistency and Inconsistency of Linear Equations in Two Variables, Consider linear equations in two variables say, a1x + b1 y + c1 = 0 ...(1), a2 x + b2 y + c2 = 0 ...(2) where a1 , a2 , b1 , b2 , c1 and c2 are real numbers., Then the system has :, a1 b1, ≠, (Consistent), a2 b2, a, b c, (ii) an Infinite number of solutions if 1 = 1 = 1 (Consistent), a2 b2 c2, a1 b1 c1, = ≠, (Inconsistent), (iii) no solution if, a2 b2 c2, (i), , a unique solution if, , Example 3.54, , Check whether the following system of equation is consistent or, inconsistent and say how many solutions we can have if it is consistent., (i) 2x – 4y = 7, (ii) 4x + y = 3, (iii) 4x +7 = 2 y, x – 3y = –2, 8x + 2y = 6, 2x + 9 = y, Solution, Sl., No, , Pair of lines, , (i) 2x–4y = 7, x – 3y = –2, (ii) 4x + y = 3, 8x + 2y = 6, (iii) 4x + 7= 2y, 2x + 9 = y, , Graphical, representation, , Algebraic, interpretation, , a1 b1, ≠, a2 b2, , Intersecting, lines, , Unique, solution, , 3 1, =, 6 2, , a1 b1 c1, = =, a2 b2 c2, , Coinciding, lines, , 7, 9, , a1 b1 c1, = ≠, a2 b2 c2, , Parallel lines, , Infinite, many, solutions, No, solution, , a1, a2, , b1, b2, , c1, c2, , 2, =2, 1, , −4 4, =, −3 3, , 7 −7, =, −2 2, , 4 1, =, 8 2, , 1, 2, , 4, =2, 2, , 2, =2, 1, , Compare the, ratios, , Algebra 131, , 3 Algebra_Term1.indd 131, , 26-12-2019 13:41:28
Page 139 :
www.tntextbooks.in, , Activity - 3, 1., , Find the value of k for the given system of linear equations satisfying the condition below:, (i) 2 x + ky = 1; 3x − 5 y = 7 has a unique solution, (ii) kx + 3 y = 3; 12 x + ky = 6 has no solution, (iii) (k − 3)x + 3 y = k ; kx + ky = 12 has infinite number of solution, , 2., , Find the value of a and b for which the given system of linear equation has infinite, number of solutions 3x − (a + 1) y = 2b − 1, 5x + (1 − 2a) y = 3b, Activity - 4, , For the given linear equations, find another linear equation satisfying each of the given condition, Given linear, equation, 2x+3y = 7, 3x–4y = 5, y–4x = 2, 5y–2x = 8, , Unique Solution, 3x+4y = 8, , Another linear equation, Infinite many, solutions, 4x+6y = 14, , No solution, 6x+9y = 15, , Exercise 3.14, Solve by any one of the methods, 1., , The sum of a two digit number and the number formed by interchanging the digits is, 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times, the sums of the digits of the first number. Find the first number., , 2., , The sum of the numerator and denominator of a fraction is 12. If the denominator is, 1, increased by 3, the fraction becomes . Find the fraction., 2, ABCD is a cyclic quadrilateral such that ∠ A = (4y + 20) ° , ∠ B = (3y –5) ° ,, ∠ C =(4x) ° and ∠ D = (7x + 5) ° . Find the four angles., On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains ₹2000., But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains Rs.1500 on the, transaction. Find the actual price of the T.V. and the fridge., Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the, ratio becomes 4 : 5. Find the numbers., , 3., 4., , 5., 6., , 4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indians and 5, Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long, would it take for 1 Chinese to do it?, Algebra 133, , 3 Algebra_Term1.indd 133, , 26-12-2019 13:41:39
Page 140 :
www.tntextbooks.in, , Exercise 3.15, Multiple choice questions, 1., , If x3 + 6x2 + kx + 6 is exactly divisible by (x + 2), then k= ?, (1) –6, (2) –7, (3) –8, (4) 11, , 2., , The root of the polynomial equation 2x + 3 = 0 is, (1) 13, , (2) – 13, , (3) – 23, , (4) – 23, , 3., , The type of the polynomial 4–3x3 is, (1) constant polynomial, (2) linear polynomial, (3) quadratic polynomial, (4) cubic polynomial., , 4., , If x51 + 51 is divided by x + 1, then the remainder is, (1) 0, (2) 1, (3) 49, (4) 50, , 5., , The zero of the polynomial 2x+5 is, 5, (1) 2, , 5, (2) – 2, , (3) 2, 5, , (4) – 2, 5, , 6., , The sum of the polynomials p(x) = x3 – x2 – 2, q(x) = x2–3x+ 1, (1) x3 – 3x – 1 (2) x3 + 2x2 – 1 (3) x3 – 2x2 – 3x (4) x3 – 2x2 + 3x –1, , 7., , Degree of the polynomial (y3–2)(y3 + 1) is, (1) 9, (2) 2, (3) 3, , 8., , 9., 10., , Let the polynomials be, (A) –13q5 + 4q2 + 12q, (C) 4q8 – q6 + 2, Then ascending order of their, (1) A,B,D,C, (2) A,B,C,D, , (4) 6, , (B) (x2 +4 )(x2 + 9), (D) - 57 y12 + y3 + y5, degree is, (3) B,C,D,A, (4) B,A,C,D, , If p(a ) = 0 then (x - a ) is a ___________ of p(x), (1) divisor, (2) quotient, (3) remainder, Zeros of (2 - 3x ) is ___________, , 2, 3, 11. Which of the following has x - 1 as a factor?, (2) 3x - 3, (3) 4x - 3, (1) 2x - 1, (1) 3, , 12., 13., , (2) 2, , 2, , (4), , 3, 2, , (4) 3x - 4, (4) p(–3), , 2, , (x + y )(x − xy + y ) is equal to, 3, , 3 Algebra_Term1.indd 134, , (3), , If x - 3 is a factor of p(x ) , then the remainder is, (1) 3, (2) –3, (3) p(3), (1) (x + y ), , 134, , (4) factor, , 3, , (2) (x - y ), , (3) x 3 + y 3, , (4) x 3 - y 3, , 9th Standard Mathematics, , 26-12-2019 13:41:41
Page 141 :
www.tntextbooks.in, , 14., , 2, , (a + b − c) is equal to __________, 2, , 2, , (1) (a − b + c), , (2) (−a − b + c), , 2, , (3) (a + b + c), , 2, , (4) (a - b - c), , 15., , If (x + 5) and (x - 3) are the factors of ax 2 + bx + c , then values of a, b and c are, (1) 1,2,3, (2) 1,2,15, (3) 1,2, −15, (4) 1, −2,15, , 16., , Cubic polynomial may have maximum of ___________ linear factors, (1) 1, (2) 2, (3) 3, (4) 4, , 17., , Degree of the constant polynomial is __________, (1) 3, (2) 2, (3) 1, , 19., , (4) 0, , Find the value of m from the equation 2 x + 3 y = m . If its one solution is x = 2 and, y = −2 ., (1) 2 , (2) −2, (3) 10, (4) 0, , 20., , Which of the following is a linear equation, 2, 1, (1) x + = 2, (2) x(x − 1) = 2 (3) 3x + 5 =, (4) x 3 − x = 5, 3, x, 21. Which of the following is a solution of the equation 2 x − y = 6, (1) (2,4), (2) (4,2), (3) (3, −1), (4) (0,6), 22., , If (2,3) is a solution of linear equation 2 x + 3 y = k then, the value of k is, (1) 12, (2) 6, (3) 0 , (4) 13, , 23., , Which condition does not satisfy the linear equation ax + by + c = 0, (1) a ≠ 0 , b = 0 (2) a = 0 , b ≠ 0, (3) a = 0 , b = 0 , c ≠ 0, , (4) a ≠ 0 , b ≠ 0, , 24., , Which of the following is not a linear equation in two variable, (1) ax + by + c = 0 (2) 0 x + 0 y + c = 0, (3) 0 x + by + c = 0 (4) ax + 0 y + c = 0, , 25., , The value of k for which the pair of linear equations 4 x + 6 y − 1 = 0 and, 2 x + ky − 7 = 0 represents parallel lines is, (1) k = 3, (2) k = 2, (3) k = 4, (4) k = −3, , 26. A pair of linear equations has no solution then the graphical representation is, (1), (2) , (3) , (4), l, l, , m, , m, , l, , m, , m, l, , Algebra 135, , 3 Algebra_Term1.indd 135, , 26-12-2019 13:41:48
Page 142 :
www.tntextbooks.in, , 27., , a1 b1, ≠, where a1x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 then the given pair of linear, a2 b2, equation has __________ solution(s), If, , (1) no solution, , (2) two solutions (3) unique, , (4) infinite, , a1 b1 c1, = ≠, where a1x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 then the given pair of, a2 b2 c2, linear equation has __________ solution(s), (1) no solution, (2) two solutions (3) infinite, (4) unique, 28., , 29., 30., , If, , GCD of any two prime numbers is __________, (1) −1, (2) 0, (3) 1, 2, , (4) 2, , 2, , 4, 4, The GCD of x - y and x - y is, 4, , (1) x - y, , 4, , 2, , (2) x - y, , 2, , 2, , (3) (x + y ), , 4, , (4) (x + y ), , Points to Remember, , , An algebraic expression of the form p^ xh = an x n + an - 1 x n - 1 + ... + a2 x2 + a1 x + a0, is called Polynomial in one variable x of degree ‘n’ where a0, a1, a2, ...an are, constants (an ! 0) and n is a whole number., , , , Let p(x) be a polynominal. If p(a) = 0 then we say that ‘a’ is a zero of the, polynomial p(x), , , , If x = a statisfies the polynominal p(x) = 0 then x = a is called a root of the, polynominal equation p(x) = 0., , , , Remainder Theorem:, , If a polynomial p(x) of degree greater than or equal, , to one is divided by a linear polynomial (x–a), then the remainder is p(a),, where a is any real number., , 136, , 3 Algebra_Term1.indd 136, , , , Factor Theorem, If p(x ) is divided by (x - a ) and the remainder p(a ) = 0, then (x - a ) is a, factor of the polynomial p(x ), , , , Solution of an equation is the set of all values that when substituted for unknowns, make an equation true., , , , An equation with two variable each with exponent as 1 and not multiplied with, each other is called a linear equation with two variables., , , , Linear equation in two variables has infinite number of solutions., , , , The graph of a linear equation in two variables is a straight line., , , , Simultaneous linear equations consists of two or more linear equations with the, same variables., , 9th Standard Mathematics, , 26-12-2019 13:41:51
Page 143 :
www.tntextbooks.in, , ICT Corner-1, Expected Result is shown in this picture, Step – 1 Open the Browser and copy and paste the Link given, below (or) by typing the URL given (or) Scan the QR Code., Step - 2 GeoGebra Work Book called “Polynomials and Quadratic, Equations” will appear. There are several work sheets in this work, Book. Open the worksheet named “Zeroes: Quadratic Polynomial”., Step-3 Drag the sliders a, b and c to change the quadratic co-efficient. Follow the, changes in points A and B where the curve cuts the x-axis. These points are called, Zeroes of a Polynomial., Browse in the link, , Zeroes of Polynomials : https://ggbm.at/tgu3PpWm, , ICT Corner-2, Expected Result is shown in this picture, Step – 1 Open the Browser, type the URL Link given below (or) Scan the QR, Code. GeoGebra work sheet named “Algebraic Identities” will open. In the work, sheet, there are many activities on Algebraic Identities., In the first activity diagrammatic approach for (a+b)2 is given. Move the, sliders a and b and compare the areas with the Identity given., Step - 2 Similarly move the sliders a and b and compare the areas with the, remaining Identities., Browse in the link, , Algebraic Identities: https://ggbm.at/PyUj657Y or Scan the QR Code., , ICT Corner-3, Expected Result is shown, in this picture, Step – 1, , Open the Browser by typing the URL Link given below, (or) Scan the QR Code. GeoGebra work sheet named “Algebra” will, open. There are three worksheets under the title Solving by rule of, cross multiplication, Graphical method and Chick-Goat puzzle., , Step - 2, , Move the sliders or type the respective values in the respective boxes to, change the equations. Work out the solution and check the solutions. In third title click, on new problem and solve. Move the slider to see the steps., Browse in the link, , Algebra: https://ggbm.at/qampr4ta or Scan the QR Code., Algebra 137, , 3 Algebra_Term1.indd 137, , 26-12-2019 13:41:52
Page 144 :
www.tntextbooks.in, , 4, , 65 o, , Fig. 4.54, , GEOMETRY, Inspiration is needed in geometry just as much as in poetry., - Alexander Pushkin, Thales (Pronounced THAYLEES) was, born in the Greek city of Miletus. He, was known for theoretical and practical, understanding of geometry, especially, triangles. He used geometry to solve many problems such, as calculating the height of pyramids and the distance of, , Thales, (BC (BCE) 624 – 546), , ships from the sea shore. He was one of the so-called Seven, Sagas or Seven Wise Men of Greece and many regarded, him as the first philosopher in the western tradition., , Learning Outcomes, Â To understand theorems on linear pairs and vertically opposite angles., Â To understand the angle sum property of triangle., Â To understand the properties of quadrilaterals and use them in problem solving., Â To understand, interpret and apply theorems on the chords and the angles subtended, by arcs of a circle, Â To understand, interpret and apply theorems on the cyclic quadrilaterals., Â To construct and locate centroid, orthocentre, circumcentre and incentre of a, triangle., , 4.1 Introduction, In geometry, we study shapes. But what is there to study in shapes, you may ask., Think first, what are all the things we do with shapes? We draw shapes, we compare, shapes, we measure shapes. What do we measure in shapes?, 138, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 138, , 26-12-2019 11:41:40
Page 145 :
www.tntextbooks.in, , Take some shapes like this:, In both of them, there is a curve forming the, shape: one is a closed curve, enclosing a, region, and the other is an open curve. We, can use a rope (or a thick string) to measure, the length of the open curve and the length, of the boundary of the region in the case of, the closed curve., , Fig. 4.1, , Curves are tricky, aren’t they? It is so much, easier to measure length of straight lines using the scale, isn’t it? Consider the two shapes, below., Now we are going to focus our attention only on shapes made up of, straight lines and on closed figures. As you will see, there is plenty, of interesting things to do. Fig.4.2 shows an open figure., Fig. 4.2, , We not only want to draw such shapes, we, want to compare them, measure them and do much more. For, doing so, we want to describe them. How would you describe these, closed shapes? (See Fig 4.3) They are all made up of straight lines, and are closed., , Fig. 4.3, , 4.2 Types of Angles-Recall, Plumbers measure the angle between connecting pipes to, make a good fitting. Wood workers adjust their saw blades, to cut wood at the correct angle. Air Traffic Controllers, , acute, an angle that is, less than 90o, , Fig. 4.4, , (ATC) use angles to direct planes. Carom and billiards, players must know their angles to plan their shots. An, angle is formed by two rays that share a common end, point provided that the two rays are non-collinear., Acute Angle, , Right Angle, , P, , B, , Obtuse Angle, , an angle that is, exactly 90o, Straight Angle, , Fig. 4.5, Reflex Angle, , Y, , q, q, , q, , q, O, A Q, O, q = 90o, 0 < q < 90o, , right, , O, , 90o < q < 180o X P, , q, 0, q = 180o, , O, T, , L, , M 180o < q < 360o, Geometry 139, , 4-Geometry_Term1.indd 139, , 26-12-2019 11:41:41
Page 146 :
www.tntextbooks.in, , Complementary Angles, Two angles are Complementary if their sum is 90°., For example, if ∠ABC=64° and ∠DEF=26°, then, angles ∠ABC and ∠DEF are complementary to each, other because ∠ABC + ∠DEF = 90°, B, , C, F, 64, , o, , A, , Supplementary Angles, , Z, 110, , 70o, Y, B, Fig. 4.7, , A, , D, , E, , Two angles are Supplementary if their sum is 180°., For example if ∠ABC=110° and ∠XYZ=70°, , C, , o, , Fig. 4.6, , 26o, , X, , Here ∠ABC + ∠XYZ = 180°, , ∴∠ABC and ∠XYZ are supplementary to each other, , Adjacent Angles, Two angles are called adjacent angles if, (i), , They have a common vertex., , C, , B, , (ii) They have a common arm., (iii) The common arm lies between the two non-common arms., , O, , common, arm, , Fig. 4.8 A, , Linear Pair of Angles, If a ray stands on a straight line then the sum of two adjacent angle is 180°. We then say that, the angles so formed is a linear pair., ∠AOC + ∠BOC=180°, , X, C, , O, , A, , O, , Z, B, Fig. 4.9, , Vertically Opposite Angles, , ∴∠AOC and ∠BOC form a linear pair, Y, , ∠XOZ + ∠YOZ = 180°, , ∠XOZ and ∠YOZ form a linear pair, , If two lines intersect each other, then vertically, , S, , opposite angles are equal., In this figure ∠POQ = ∠SOR, , ∠POS = ∠QOR, , 140, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 140, , P, , O, , R, , Q, Fig. 4.10, , 26-12-2019 11:41:42
Page 147 :
www.tntextbooks.in, , 4.2.1 Transversal, A line which intersects two or more lines at distinct points is called a transversal of those, lines., Case (i), , When a transversal intersect two lines, we get eight angles., , In the figure the line l is the transversal for the lines, , l, , m and n, 2, , Corresponding Angles: ∠1 and ∠5, ∠2 and ∠6,, , (i), , ∠3 and ∠7, ∠4 and ∠8, , (ii) Alternate Interior Angles:, and ∠5, , 1, 3, , m, , 4, 5, , 6, , ∠4 and ∠6, ∠3, , n, , 8, , 7, Fig. 4.11, , (iii) Alternate Exterior Angles: ∠1 and ∠7, ∠2 and ∠8, , (iv) ∠4 and ∠5,∠3 and ∠6 are interior angles on the same side of the transversal., , (v) ∠1 and ∠8, ∠2 and ∠7 are exterior angles on the same side of the transversal., , Case (ii) If a transversal intersects two parallel lines. The transversal forms different pairs, of angles., Interior, ∠2, ∠7, ∠2 = ∠7, Corresponding, angles, ∠3, ∠4, ∠3 = ∠4, , Interior, ∠2, ∠5, , ∠2 + ∠5 =180, , Exterior, ∠3, ∠6, , Alternate, angles, Identify the, other pair, , Find the, other pair, , Angles, , Consecutive, angles, , Can you find the other pair?, o, , Fig. 4.12, , ∠3 = ∠6, Vertically, opposite angles, ∠1, ∠7, , l, 1, , 3, 7, , m, , 5, 2, , 4, 8, , n, , 6, , ∠1 = ∠7, Exterior, ∠1, ∠6, , ∠1 + ∠6 =180o, Geometry 141, , 4-Geometry_Term1.indd 141, , 26-12-2019 11:41:42
Page 148 :
www.tntextbooks.in, , 4.2.2 Triangles, Activity 1, 1. Take three different colour sheets; place one, over the other and draw a triangle on the top, sheet. Cut the sheets to get triangles of different, colour which are identical. Mark the vertices, and the angles as shown. Place the interior, angles ∠1, ∠2 and ∠3 on a straight line,, adjacent to each other, without leaving any, gap. What can you say about the total measure, of the three angles ∠1, ∠2 and ∠3?, 2 opposite, interior, angles, , Can you use, Fig. 4.13, the same figure, exterior angle, to explain the “Exterior angle property” of a triangle?, , a, b, , c, , d, , If a side of a triangle is stretched, the exterior angle so, formed is equal to the sum of the two interior opposite, angles. That is d=a+b (see Fig 4.14), , Fig. 4.14, , 4.2.3 Congruent Triangles, Two triangles are congruent if the sides and angles of one triangle are equal to the, corresponding sides and angles of another triangle., Rule, , Diagrams, A, , Reason, , Q, , R, , BC = QR, , SSS, B, , AC = PR, , P, , C, , A, , ΔABC ≅ ΔPQR, AB = XY, , X, , SAS, B, , 142, , AB = PQ, , C, , Y, , Z, , ∠BAC = ∠YXZ, AC = XZ, , ΔABC ≅ ΔXYZ, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 142, , 26-12-2019 11:41:43
Page 149 :
www.tntextbooks.in, , ASA, , A, , AB = PQ, , R, , C, , ∠B = ∠Q, , Q, , B, , AAS, , ∠A = ∠P, , P, , M, , N, , A, , B, , ΔABC ≅ ΔPQR, ∠A = ∠M, ∠B = ∠N, , BC = NO, O, , C, , RHS, , A, , ΔABC ≅ ΔMNO, ∠ACB = ∠PRQ = 90°(R), , P, , AB = PQ hypotenuse (H), C, , B, , AC = PR, , Q, , R, , (S), , ΔABC ≅ ΔPQR, , Exercise 4.1, 1., , In the figure, AB is parallel to CD, find x, (i), , A, , (ii), , B, , A, , B, , T, , xo, , 150o, C, , A, , 48o, , 140o, xo, , (iii), , D, , C, , E, , B, , 53o, , xo, 24o, , D, , C, , 38o, , D, , 2., , The angles of a triangle are in the ratio 1: 2 : 3, find the measure of each angle of the, triangle., , 3., , Consider the given pairs of triangles and say whether each pair is that of congruent, triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’, and also say if a small modification would make them congruent:, (i), , R, , (ii), , Q, , D, , C, , A, , Y, , (iii), P, , X, , P, B, , C, , A, , B, , Z, Geometry 143, , 4-Geometry_Term1.indd 143, , 26-12-2019 11:41:44
Page 150 :
www.tntextbooks.in, , (iv) A, , D, , (vi), , A, , A, , O, , B, , O, C, C, , 4., , D, , (v), , B, , M, , C, , B, , ΔABC and ΔDEF are two triangles in which, AB=DF, ∠ACB=70°, ∠ABC=60°; ∠DEF=70° and ∠EDF=60°., Prove that the triangles are congruent., , 5., , Find all the three angles of the ΔABC, A, (x+35)°, (4x–15)°, , (2x–5)°, B, , C, , D, , 4.3 Quadrilaterals, Activity 2, Four Tamil Nadu State Transport buses take, the following routes. The first is a one-way, journey, and the rest are round trips. Find the, places on the map, put points on them and, connect them by lines to draw the routes. The, places connecting four different routes are, given as follows., (i), , Nagercoil, Tirunelveli, Virudhunagar,, Madurai, , (ii) Sivagangai, Puthukottai,, Dindigul, Sivagangai, , Thanjavur,, , (iii) Erode, Coimbatore, Dharmapuri, Karur,, Erode, (iv) Chennai,, Cuddalore,, Vellore, Chennai, , Krishnagiri,, , Fig. 4.15, 144, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 144, , 26-12-2019 11:41:45
Page 151 :
www.tntextbooks.in, , You will get the following shapes., (i), , Fig. 4.16, , (ii), , , , Fig. 4.17 , , (iii), , (iv), , Fig. 4.18 , , Fig. 4.19, , Label the vertices with city names, draw the shapes exactly as they are shown on the map, without rotations., We observe that the first is a single line, the four points are collinear. The other three are, closed shapes made of straight lines, of the kind we have seen before. We need names to call, such closed shapes, we will call them polygons from now on., , Fig. 4.20, , How do polygons look? They have sides, with points at either end. We call these points, as vertices of the polygon. The sides are line segments joining the vertices. The word poly, stands for many, and a polygon is a many-sided figure., Note, Concave polygon: Polygon having any, one of the interior angle greater than 180o, Convex Polygon: Polygon having each, interior angle less than 180o, (Diagonals should be inside the polygon), , How many sides can a polygon have? One? But, that is just a line segment. Two ? But how can you, get a closed shape with two sides? Three? Yes, and, this is what we know as a triangle. Four sides?, Squares and rectangles are examples of polygons, with 4 sides but they are not the only ones., Here (Fig. 4.20) are some examples of 4-sided, polygons. We call them quadrilaterals., , 4.3.1 Special Names for Some Quadrilaterals, 1., , A parallelogram is a quadrilateral in which opposite sides are parallel and equal., , 2., , A rhombus is a quadrilateral in which opposite sides are parallel and all sides are, equal., , 3., , A trapezium is a quadrilateral in which one pair of opposite sides are parallel., , Draw a few parallellograms, a few rhombuses (correctly called rhombii, like cactus and, cactii) and a few trapeziums (correctly written trapezia)., Geometry 145, , 4-Geometry_Term1.indd 145, , 26-12-2019 11:41:45
Page 152 :
www.tntextbooks.in, , Fig. 4.21, , The great advantage of knowing properties of quadrilateral is that we can see the relationships, among them immediately., ÂÂ Every parallelogram is a trapezium, but not necessarily the other way., ÂÂ Every rhombus is a parallelogram, but not necessarily the other way., ÂÂ Every rectangle is a parallelogram, but not necessarily the other way., ÂÂ Every square is a rhombus and hence every square is a parallelogram as well., For “not necessarily the other way” mathematicians usually say “the converse is not true”., A smart question then is: just when is the other way also true? For instance, when is a, parallelogram also a rectangle? Any parallelogram in which all angles are also equal is a, rectangle. (Do you see why?) Now we can observe many more interesting properties. For, instance, we see that a rhombus is a parallelogram in which all sides are also equal., Note, You know bi-cycles and tri-cycles? When we attach bi or tri to the front of any word, they, stand for 2 (bi) or 3 (tri) of them. Similarly quadri stands for 4 of them. We should really, speak of quadri-cycles also, but we don’t. Lateral stands for sideways, thus quadrilateral, means a 4-sided figure. You know trilaterals; they are also called triangles !, After 4 ? We have: 5 – penta, 6 – hexa, 7 – hepta, 8 – octa, 9 – nano, 10 – deca. Conventions, are made by history. Trigons are called triangles, quadrigons are called quadrilaterals., Continuing in th same way we get pentagons, hexagons, heptagons, octagons, nanogons and, decagons. Beyond these, we have 11-gons, 12-gons etc. Perhaps you can draw a 23-gon !, , 4.3.2 More Special Names, When all sides of a quadrilateral are equal, we call it equilateral. When all angles of a, quadrilateral are equal, we call it equiangular. In triangles, we talked of equilateral triangles, as those with all sides equal. Now we can call them equiangular triangles as well!, We thus have:, A rhombus is an equilateral parallelogram., A rectangle is an equiangular parallelogram., A square is an equilateral and equiangular parallelogram., 146, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 146, , 26-12-2019 11:41:45
Page 153 :
www.tntextbooks.in, , Here are two more special quadrilaterals, called kite and isosceles trapezium., , Fig. 4.22, , 4.3.3 Types of Quadrilaterals, Progress Check, , Quadrilateral, , Answer the following question., (i), , (ii) A quadrilateral is a ______________ if a, pair of opposite sides are equal and parallel., , trapezium, , kite, , Are the opposite angles of a rhombus equal?, , (iii) Are the opposite sides of a kite equal?, (iv) Which is an equiangular but not an, equilateral parallelogram?, (v), , parallelogram, , bus, m, o, rh, , (vi) Which is an equilateral and equiangular, parallelogram?, , rectangle, , square, , Which is an equilateral but not an, equiangular parallelogram?, , (vii) __________ is a rectangle, a rhombus and a, parallelogram., , Fig. 4.23, , Activity 3, Step – 1: Cut out four different quadrilaterals from coloured glazed papers., D, , C, , A, , B, , D, , A, , C, , B, , D, , C, , A, , B, , D, , C, , A, , B, , Fig. 4.24, Geometry 147, , 4-Geometry_Term1.indd 147, , 26-12-2019 11:41:46
Page 154 :
www.tntextbooks.in, , Step – 2: Fold the quadrilaterals along their respective diagonals. Press to make, creases. Here, dotted line represent the creases., D, , C, , A, , B, , D, , C, , A, , B, , D, , C, , A, , B, , D, , C, , A, , B, , Fig. 4.25, , Step – 3: Fold the quadrilaterals along both of their diagonals. Press to make creases., D, , C, , D, , C, , O, A, , D, , O, B, , C, , D, , C, , O, , A, , B, , A, , O, B, , A, , B, , Fig. 4.26, , We observe that two imposed triangles are congruent to each other. Measure the, lengths of portions of diagonals and angles between the diagonals., Also do the same for the quadrilaterals such as Trapezium, Isosceles Trapezium and Kite., From the above activity, measure the lengths of diagonals and angles between the diagonals, and record them in the table below:, S., No., , Name of the, quadrilateral, , 1, , Trapezium, , 2, , Isosceles, Trapezium, , 3, , Parallelogram, , 4, , Rectangle, , 5, , Rhombus, , 6, , Square, , 7, , Kite, , 148, , Length along diagonals, AC, , BD, , OA, , OB, , OC, , Measure of angles, OD + AOB + BOC + COD + DOA, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 148, , 26-12-2019 11:41:46
Page 155 :
www.tntextbooks.in, , Activity 4, Angle sum for a polygon, , D, , , , Draw any quadrilateral ABCD., Mark a point P in its interior., , P, , Join the segments PA, PB, PC and PD., , C, , A, , You have 4 triangles now., , How much is the sum of all the angles of the 4, triangles?, How much is the sum of the angles at P?, , B, , Fig. 4.27, , Can you now find the ‘angle sum’ of the quadrilateral ABCD?, Can you extend this idea to any polygon?, , Thinking Corner, 1., , If there is a polygon of n sides (n ≥ 3), then the sum of all interior angles is, (n–2) # 180°, , 2., , For the regular polygon (All the sides of a polygon are equal in size), ÂÂ Each interior angle is, , ^ n - 2h, , n, , # 180°, , ÂÂ Each exterior angle is 360c, n, ÂÂ The sum of all the exterior angles formed by producing the sides of a convex, polygon in the order is 360° ., ÂÂ If a polygon has ‘n’ sides , then the number of diagonals of the polygon is, n^ n - 3h, 2, , Geometry 149, , 4-Geometry_Term1.indd 149, , 26-12-2019 11:41:47
Page 156 :
www.tntextbooks.in, , 4.3.4 Properties of Quadrilaterals, Name, , Diagram, , Sides, , E, , F, , Parallelogram, L, , I, , H, , Rhombus, , A, , Diagonals, , Opposite sides Opposite angles Diagonals bisect, are parallel, are equal and, each other., and equal, sum of any two, adjacent angles, is 180°, All sides are, equal and, opposite sides, are parallel, , T, , Angles, , Opposite angles Diagonals bisect, are equal and, each other at right, sum of any two angle., adjacent angles, is 180°, , M, D, , C, , Trapezium, A, , T, , One pair of, The angles at, Diagonals need not, opposite sides the ends of each be equal, are parallel, non-parallel, sides are, supplementary, B, , S, , Isosceles, Trapezium, B, , C, , E, , F, , H, , Kite, , G, , 150, , One pair of, opposite sides, are parallel and, non-parallel, sides are equal, in length., , Diagonals are of, The angles at, the ends of each equal length., parallel sides, are equal., , Two pairs of, adjacent sides, are equal, , One pair of, 1. Diagonals, intersect at right, opposite angles, angle., are equal, 2. Shorter diagonal, bisected by, longer diagonal, 3. Longer diagonal, divides the, kite into two, congruent, triangles, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 150, , 26-12-2019 11:41:47
Page 157 :
www.tntextbooks.in, , Note, (i), , A rectangle is an equiangular parallelogram., , (ii) A rhombus is an equilateral parallelogram., (iii) A square is an equilateral and equiangular parallelogram., (iv) A square is a rectangle , a rhombus and a parallelogram., Progress Check, 1. State the reasons for the following., (i) A square is a special kind of a rectangle., (ii) A rhombus is a special kind of a parallelogram., (iii) A rhombus and a kite have one common property., (iv) A square and a rhombus have one common property., 2. What type of quadrilateral is formed when the following pairs of congruent triangles, are joined together?, (i) Equilateral triangle., (ii) Right angled triangle., (iii) Isosceles triangle., 3., , Identify which ones are parallelograms and which are not., (i), (ii), (iii), (iv), , 4., , 5., , (v), , (vi), , Which ones are not quadrilaterals?, (i), , (ii), , (iii), , (iv), , (v), , (vi), , (vii), , (viii), , Identify which ones are trapeziums and which are not., (i), , (ii), , (iii), , (iv), , (v), , (vi), , Geometry 151, , 4-Geometry_Term1.indd 151, , 26-12-2019 11:41:48
Page 158 :
www.tntextbooks.in, , 4.3.5 Properties of Parallelogram, We can now embark on an interesting journey. We can tour among lots of quadrilaterals,, noting down interesting properties. What properties do we look for, and how do we know, they are true?, For instance, opposite sides of a parallelogram are parallel, but are they also equal? We, could draw any number of parallelograms and verify whether this is true or not. In fact,, we see that opposite sides are equal in all of them. Can we then conclude that opposite, sides are equal in all parallelograms? No, because we might later find a parallelogram,, one which we had not thought of until then, in which opposite sides are unequal. So, we, need an argument, a proof., C, , D, , Consider the parallelogram ABCD in the given Fig. 4.28., We believe that AB = CD and AD = BC, but how can we be, sure? We know triangles and their properties. So we can try, and see if we can use that knowledge. But we don’t have any, triangles in the parallelogram ABCD., C, , D, , A, , Fig. 4.28, , B, , This is easily taken care of by joining AC. (We could equally well, have joined BD, but let it be AC for now.) We now have 2 triangles, ADC and ABC with a common side AC. If we could somehow, prove that these two triangles are congruent, we would get, , A, , Fig. 4.29, , B, , AB = CD and AD = BC, which is what we want!, , Is there any hope of proving that DADC and DABC are, congruent? There are many criteria for congruence, it is not clear which one is relevant here., So far we have not used the fact that ABCD is a parallelogram at all. So we need to use the, facts that AB < DC and AD < BC to show that DADC and DABC are congruent. From sides, being parallel we have to get into some angles being equal. Do we know any such properties?, we do, and that is all about transversals!, Now we can see it clearly. AD < BC and AC is a transversal, hence + DAC = + BCA. Similarly,, AB < DC, AC is a transversal, hence + BAC = + DCA. With AC as common side, the ASA, criterion tells us that DADC and DABC are congruent, just what we needed. From this we, can conclude that AB = CD and AD = BC., Thus opposite sides are indeed equal in a parallelogram., The argument we now constructed is written down as a formal proof in the following, manner., 152, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 152, , 26-12-2019 11:41:48
Page 159 :
www.tntextbooks.in, , Theorem 1, In a parallelogram, opposite sides are equal, Given, , ABCD is a parallelogram, , To Prove, , AB=CD and DA=BC, , C, , D, , Construction Join AC, Proof, A, , Since ABCD is a parallelogram, , B, , Fig. 4.30, , AD < BC and AC is the transversal, , , ∠DAC = ∠BCA, , →(1) (alternate angles are equal), , ∠BAC = ∠DCA , , →(2) (alternate angles are equal), , ∠DAC = ∠BCA , , from (1), , ∠DCA = ∠BAC , , from (2), , AB < DC and AC is the transversal, , , In ΔADC and ΔCBA, , , AC is common, , , ΔADC b ΔCBA , , Hence, , AD = CB and DC = BA, , (By ASA), (Corresponding sides are equal), , Along the way in the proof above, we have proved another property that is worth, recording as a theorem., , Theorem 2, A diagonal of a parallelogram divides it into two congruent triangles., Notice that the proof above established that ∠DAC = ∠BCA and ∠BAC = ∠DCA., Hence we also have, in the figure above,, ∠BCA + ∠BAC = ∠DCA + ∠DAC, , C, , D, , But we know that:, , ∠B + ∠BCA + ∠BAC = 180, , and ∠D + ∠DCA + ∠DAC = 180, , Therefore we must have that ∠B = ∠D., , A, , Fig. 4.31, , B, , With a little bit of work, proceeding similarly, we could have shown that ∠A = ∠C as well., Thus we have managed to prove the following theorem:, , Geometry 153, , 4-Geometry_Term1.indd 153, , 26-12-2019 11:41:48
Page 160 :
www.tntextbooks.in, , Theorem 3, The opposite angles of a parallelogram are equal., Now that we see congruence of triangles as a good “strategy”, we can look for more triangles., Consider both diagonals AC and DB. We already know that DADC and DCBA are congruent., By a similar argument we can show that DDAB and DBCD are congruent as well. Are there, more congruent triangles to be found in this figure ?, C, D, Yes. The two diagonals intersect at point O. We now see 4 new, DAOB, DBOC, DCOD and DDOA. Can you see any congruent, , O, , pairs among them?, Since AB and CD are parallel and equal, one good guess is, B, that ΔAOB and ΔCOD are congruent. We could again try A, Fig. 4.32, the ASA crierion, in which case we want ∠OAB = ∠OCD and, ∠ABO = ∠CDO. But the first of these follows from the fact that ∠CAB = ∠ACD, (which we already established) and observing that ∠CAB and ∠OAB are the same (and, so also ∠OCD and ∠ACD). We now use the fact that BD is a transversal to get that, ∠ABD = ∠CDB, but then ∠ABD is the same as ∠ABO, ∠CDB is the same as ∠CDO, and we, are done., Again, we need to write down the formal proof, and we have another theorem., , Theorem 4, The diagonals of a parallelogram bisect each other., It is time now to reinforce our concepts on, ÂÂ Each pair of its opposite sides are parallel., parallelograms. Consider each of the given, ÂÂ Each pair of opposite sides is equal., statements, in the adjacent box, one by one., ÂÂ All of its angles are right angles., Identifiy the type of parallelogram which, ÂÂ Its diagonals bisect each other., satisfies each of the statements. Support, ÂÂ The diagonals are equal., your answer with reason., ÂÂ The diagonals are perpendicular and, Now we begin with lots of interesting properties, equal., of parallelograms. Can we try and prove some ÂÂ The diagonals are perpendicular bisectors, property relating to two or more parallelograms, ? A simple case to try is when two parallelograms, share the same base, as in Fig.4.33, F, , D, , E, , C, , of each other., ÂÂ Each pair of its consecutive angles is, supplementary., , We see parallelograms ABCD and ABEF are on the, common base AB. At once we can see a pair of triangles, for being congruent DADF and DBCE. We already have, , A, 154, , Fig. 4.33, , B, , that AD = BC and AF = BE. But then since AD< BC and, AF< BE, the angle formed by AD and AF must be the same, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 154, , 26-12-2019 11:41:49
Page 161 :
www.tntextbooks.in, , as the angle formed by BC tand BE. Therefore ∠DAF = ∠CBE. Thus DADF and DBCE are, , congruent., , That is an interesting observation; can we infer anything more from this ? Yes, we know, that congruent triangles have the same area. This makes us think about the areas of the, parallelograms ABCD and ABEF., Area of ABCD = area of quadrilateral ABED + area of DBCE, = area of quadrilateral ABED + area of DADF, = area of ABEF, Thus we have proved another interesting theorem:, , Theorem 5:, Parallelograms on the same base and between the same parallels are equal in area., In this process, we have also proved other interesting statements. These are called Corollaries,, which do not need separate detailed proofs., Corollary 1: Triangles on the same base and between the same parallels are equal in area., Corollary 2: A rectangle and a parallelogram on the same base and between the same, parallels are equal in area., These statements that we called Theorems and Corollaries, hold for all parallelograms,, however large or small, with whatever be the lengths of sides and angles at vertices., Example 4.1, D, , In a parallelogram ABCD, the bisectors of the consecutive, angles ∠A and ∠B insersect at P. Show that ∠APB = 90°, , Solution, ABCD is a parallelogram AP and BP are bisectors of, consecutive angles ∠A and ∠B., Since the consecutive angles of a parallelogram are, supplementary, ∠A + ∠B = 180°, , C, , P, A, , Fig. 4.34, , B, , 1 ∠A + 1 ∠ B = 180 O, 2, 2, 2, , ( ∠ PAB + ∠ PBA = 90°, , In ΔAPB,, ∠ PAB + ∠ APB + ∠ PBA = 180° (angle sum property of triangle), ∠ APB = 180° – [∠ PAB + ∠ PBA], = 180° – 90° = 90°, Hence Proved., Geometry 155, , 4-Geometry_Term1.indd 155, , 26-12-2019 11:41:49
Page 162 :
www.tntextbooks.in, , Example 4.2, , Q, , D, , C, , In the Fig.4.35 ABCD is a parallelogram, P and Q are the midpoints of sides AB and DC respectively. Show that APCQ is a, parallelogram., A, , Solution, Since P and Q are the mid points of, , B, , P, , Fig. 4.35, , AB and DC respectively, AP = 12 AB and, QC = 12 DC, , Therefore, , But, , AB = DC, , (, , 1, 1, AB = DC, 2, 2, AP = QC, , (, , Also ,, , AB || DC, , (, , AP || QC, , (1), , (Opposite sides of a parallelogram are equal), , (2), , (3) [a ABCD is a parallelogram], , Thus , in quadrilateral APCQ we have AP= QC and AP || QC [from (2) and ( 3)], Hence , quadrilateral APCQ is a parallelogram., Example 4.3, ABCD is a parallelogram Fig.4.36 such that ∠BAD = 120o and AC bisects ∠BAD show that, ABCD is a rhombus., Solution, Given, , ∠BAD = 120° and AC bisects ∠BAD, ∠BAC = 12 # 120° = 60°, , ∠1 = ∠2 = 60°, , Δ ABC is isosceles triangle, (, , AB = BC, , 3, , 2, A, , AD || BC and AC is the traversal, ∠2 = ∠4 = 60°, , D, , 1, , Fig. 4.36, , C, 4, , B, , [ a ∠1 = ∠4 = 60 °], , Parallelogram ABCD is a rhombus., 156, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 156, , 26-12-2019 11:41:50
Page 163 :
www.tntextbooks.in, , Example 4.4, , In a parallelogram ABCD , P and Q are the points on line DB such, that PD = BQ show that APCQ is a parallelogram, Solution, ABCD is a parallelogram., OA = OC and, A, , OB = OD ( a Diagonals bisect each other), now, , B, , Q, , OB + BQ = OD + DP, O, , OQ = OP and OA = OC, , D, , APCQ is a parallelogram., , C, , P, , Fig. 4.37, , Exercise 4.2, 1., , The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find all the angles., , 2., , In a quadrilateral ABCD, ∠A = 72° and ∠C is the supplementary of ∠A. The other two, angles are 2x–10 and x + 4. Find the value of x and the measure of all the angles., , 3., , ABCD is a rectangle whose diagonals AC and BD intersect at O. If ∠OAB =46°,, find ∠OBC, , 4., , The lengths of the diagonals of a Rhombus are 12 cm and 16 cm . Find the side of the, rhombus., , 5., , Show that the bisectors of angles of a parallelogram form a rectangle ., , 6., , If a triangle and a parallelogram lie on the same base and between the same parallels,, then prove that the area of the triangle is equal to half of the area of parallelogram., , 7., , Iron rods, a, b, c, d,, e, and f are making a, design in a bridge as, shown in the figure., If a || b , c || d , e || f , find, the marked angles between, (i), , e, d, , b, , b and c, , (ii) d and e, , c, , 75o, , 30o, , a, , f, , (iii d and f, (iv) c and f, , Fig. 4.38, Geometry 157, , 4-Geometry_Term1.indd 157, , 26-12-2019 11:41:50
Page 164 :
www.tntextbooks.in, , 8., , A, , In the given Fig. 4.39, ∠ A = 64° , ∠ ABC = 58°. If BO and CO, are the bisectors of ∠ ABC and ∠ ACB respectively of ΔABC,, find x° and y°, , 64o, , O, xo, 58, , B, , 9., , C, , Fig. 4.39, , A, , In the given Fig. 4.40, if AB = 2, BC = 6, AE = 6,, BF = 8, CE = 7, and CF = 7, compute the ratio of the, area of quadrilateral ABDE to the area of ΔCDF. (Use, congruent property of triangles)., , yo, , o, , B, , C, , D, E, , Fig. 4.40, A, , E, , 4, , F, B, , 6, , 3, , 10. In the Fig. 4.41 , ABCD is a rectangle and EFGH is, a parallelogram. Using the measurements given in, the figure, what is the length d of the segment that is, perpendicular to HE and FG ?, , 5, , H, , d, F, , 5, 3, D, , 6, , 4, , G, , Fig. 4.41, , C, , 11. In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and, meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and, meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO, 9, is of the area of the parallelogram ABCD., 8, D, C, , M, , Q, , 158, , A, , O, , Fig. 4.42, , N, , B, , P, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 158, , 26-12-2019 11:41:51
Page 165 :
www.tntextbooks.in, , 4.4 Parts of a Circle, Circles are geometric shapes you can see all around you. The significance of the concept, of a circle can be well understood from the fact that the wheel is one of the ground-breaking, inventions in the history of mankind., , Fig. 4.43, , A circle, you can describe, is the set of all, Note, points in a plane at a constant distance from a fixed, point. The fixed point is the centre of the circle;, the constant distance corresponds to a radius of A circle notably differs from, a polygon. A polygon (for, the circle., example, a quadrilateral) has, A line that cuts the circle in two points is edges and corners while, a, called a secant of the circle., circle is a ‘smooth’ curve., A line segment whose end points lie on the, circle is called a chord of the circle., A chord of a circle that has the centre is called a diameter of the circle. The, circumference of a circle is its boundary. (We use the term perimeter in the case of polygons)., S, R, O, , A, , r, , In Fig.4.44, we see that all the line segments meet at two points on the circle., These line segments are called the chords of the circle. So, a line segment, joining any two points on the circle is called a chord of the circle. In this figure, B, AB, PQ and RS are the chords of the circle., , P, Q, , Fig. 4.44, , S, , Now place four points P, R, Q and S on the same circle, (Fig.4.45), then PRQ and QSP are the continuous, parts (sections) of the circle. These parts (sections) are, , and QSP, or simply by PQ, and QP, . This continuous P, to be denoted by PRQ, , part of a circle is called an arc of the circle. Usually the arcs are denoted in, anti-clockwise direction., , Q, R, , Fig. 4.45, , Now consider the points P and Q in the circle (Fig.4.45). It divides the whole circle into, , and, two parts. One is longer and another is shorter. The longer one is called major arc QP, , ., shorter one is called minor arc PQ, , Geometry 159, , 4-Geometry_Term1.indd 159, , 26-12-2019 11:41:52
Page 166 :
www.tntextbooks.in, , Now in (Fig.4.46), consider the, region which is surrounded by, , Major, Segment, O, , Note, , , the chord PQ and major arc QP, . This is called the major segment A diameter of a circle is:, Q, P, Minor, Segment, of the circle. In the same way, the zz the line segment which bisects the, Fig. 4.46, segment containing the minor, circle., arc and the same chord is called the minor segment., zz the largest chord of a circle., C, , D, O, , A, , B, , Fig. 4.47, , and, In (Fig.4.47), if two arcs AB, zz a line of symmetry for the circle., , CD of a circle subtend the same zz twice in length of a radius in a, angle at the centre, they are said, circle., to be congruent arcs and we, write,, , º CD, implies m AB, = mCD, , AB, , implies ∠AOB = ∠COD, Now, let us observe (Fig.4.48). Is there any special name for the region, surrounded by two radii and arc? Yes, its name is sector. Like segment, we, find that the minor arc corresponds to the minor sector and the major arc, corresponds to the major sector., , Major Sector, O, A, , Minor, Sector, , B, , Fig. 4.48, , Concentric Circles, Circles with the same centre but different radii are said to be concentric., Here are some real-life examples:, , An Archery target, , Congruent Circles, , A carom board coin, , Water ripples, , Fig. 4.49, , Two circles are congruent if they are copies of one another or identical. That is, they, have the same size. Here are some real life examples:, , 160, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 160, , 26-12-2019 11:41:53
Page 167 :
www.tntextbooks.in, , Thinking Corner, Draw four congruent circles, as shown. What, do you infer?, The two wheels of a bullock cart, , The Olympic rings, , Fig. 4.50, , Position of a Point with respect to a Circle, Consider a circle in a plane (Fig.4.51). Consider any point P on the circle. If the distance, Outside the, from the centre O to the point P is OP, then, (i), , OP = radius (If the point P lies on the circle), , (ii), , OP < radius (If the point P Point lies inside the circle), , (iii), , OP > radius(If the point P lies outside the circle), , P, , So, a circle divides the plane on which it lies into three parts., , Circle, P, , Inside the, Circle, O, , P On the, Circle, , Fig. 4.51, , Progress Check, , Say True or False, 1. Every chord of a circle contains exactly two points of the circle., 2. All radii of a circle are of same length., 3. Every radius of a circle is a chord., 4. Every chord of a circle is a diameter., 5. Every diameter of a circle is a chord., 6. There can be any number of diameters for a circle., 7. Two diameters cannot have the same end-point., 8. A circle divides the plane into three disjoint parts., 9. A circle can be partitioned into a major arc and a, minor arc., , Thinking Corner, 1. How many sides does, a circle have ?, 2. Is circle, a polygon?, , 10. The distance from the centre of a circle to the circumference is that of a diameter, , Geometry 161, , 4-Geometry_Term1.indd 161, , 26-12-2019 11:41:54
Page 168 :
www.tntextbooks.in, , 4.4.1 Circle Through Three Points, We have already learnt that there is one and only one line passing through two points., In the same way, we are going to see how many circles can be drawn through a given point,, and through two given points. We see that in both cases there can be infinite number of, circles passing through a given point P (Fig.4.52) , and through two given points A and B, (Fig.4.53)., A, P, , A, , B, , C, , B, Fig. 4.52, , Fig. 4.53, , Fig. 4.54, , Now consider three collinear points A, B and C (Fig.4.14). Can we draw a circle passing, through these three points? Think over it. If the points are collinear, we can’t?, If the three points are non collinear, they form a triangle, , C, , B, , (Fig.4.55). Recall the construction of the circumcentre. The intersecting, point of the perpendicular bisector of the sides is the circumcentre and, the circle is circumcircle., , O, A, , Fig. 4.55, , Therefore from this we know that, there is a unique circle which passes through A, B, and C. Now, the above statement leads to a result as follows., Theorem 6 There is one and only one circle passing through three non-collinear points., , 4.5 Properties of Chords of a Circle, In this chapter, already we come across lines, angles, triangles and quadrilaterals., Recently we have seen a new member circle. Using all the properties of these, we get some, standard results one by one. Now, we are going to discuss some properties based on chords, of the circle., Considering a chord and a perpendicular line from the centre to a chord, we are going, to see an interesting property., , 4.5.1 Perpendicular from the Centre to a Chord, Consider a chord AB of the circle with centre O. Draw OC ^ AB and join the points, OA,OB . Here, easily we get two triangles DAOC and DBOC (Fig.4.56)., 162, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 162, , 26-12-2019 11:41:55
Page 169 :
www.tntextbooks.in, , Can we prove these triangles are congruent? Now we try to prove, this using the congruence of triangle rule which we have already learnt., ∠OCA = ∠OCB = 90° (OC ⊥ AB ) and OA = OB is the radius of the, circle. The side OC is common. RHS criterion tells us that DAOC and, DBOC are congruent. From this we can conclude that AC = BC . This, argument leads to the result as follows., , O, A, , B, , C, , Fig. 4.56, , Theorem 7 The perpendicular from the centre of a circle to a chord bisects the chord., Converse of Theorem 7 The line joining the centre of the circle and the midpoint of a chord, is perpendicular to the chord., Example 4.5, , Find the length of a chord which is at a distance of 2 11 cm from the, centre of a circle of radius 12cm., Solution, Let AB be the chord and C be the mid point of AB, , 12, A, , Therefore, OC ^ AB, , 2 11cm, , B, , C, , Fig. 4.57, , Join OA and OC., , Note, , OA is the radius, Given OC = 2 11cm and OA = 12cm, In a right DOAC ,, using Pythagoras Theorem, we get,, AC 2 = OA2 − OC 2, = 122 − (2 11)2, = 144 − 44, = 100cm, , AC, , O, , cm, , 2, , = 100cm, , AC = 10cm, Therefore, length of the chord AB = 2AC, , Pythagoras theorem, One of the most important and, well known results in, C, geometry is Pythagoras, Theorem. “In a right, angled triangle, the, A, B square of the hypotenuse, is equal to the sum of the squares of, the other two sides”., In right ∆ ABC , BC 2 = AB 2 + AC 2 ., Application of this theorem is most, useful in this unit., , = 2 × 10cm = 20cm, Example 4.6, , In the concentric circles, chord AB of the, outer circle cuts the inner circle at C and D as shown in the diagram., Prove that, AB − CD = 2AC, , O, A, , C, , M, , D, , B, , Fig. 4.58, Geometry 163, , 4-Geometry_Term1.indd 163, , 26-12-2019 11:41:56
Page 170 :
www.tntextbooks.in, , Solution, Given : Chord AB of the outer circle cuts the inner circle at C and D., To prove, , : AB − CD = 2AC, , Construction : Draw OM ^ AB, Proof, , : Since, OM ^ AB (By construction), , Also, OM ^ CD, Therefore, AM = MB ... (1) (Perpendicular drawn from centre to chord bisect it), CM = MD ... (2), Now, AB – CD = 2AM–2CM, = 2(AM–CM), , from (1) and (2), , AB – CD = 2AC, Progress Check, 1. The radius of the circle is 25 cm and the length of one of its chord is 40cm. Find, the distance of the chord from the centre., 2. Draw three circles passing through the points P and Q, where PQ = 4cm., , 4.5.2 Angle Subtended by Chord at the Centre, Instead of a single chord we consider two equal chords., Now we are going to discuss another property., , C, D, , Let us consider two equal chords in the circle with, centre O. Join the end points of the chords with the centre, to get the triangles DAOB and DOCD , chord AB = chord, , O, A, , CD (because the given chords are equal). The other sides, are radii, therefore OA=OC and OB=OD. By SSS rule, the, triangles are congruent, that is ∆OAB ≡ ∆OCD . This, , B, Fig. 4.59, , gives mÐAOB = m∠COD. Now this leads to the following, result., Theorem 8 Equal chords of a circle subtend equal angles at the centre., 164, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 164, , 26-12-2019 11:41:57
Page 171 :
www.tntextbooks.in, , Activity - 5, Procedure, , O, , 1. Draw a circle with centre O and with suitable radius., 2. Make it a semi-circle through folding. Consider the point A, B on it., 3., 4., , A, B, D, , Make crease along AB in the semi circles and open it., We get one more crease line on the another part of semi circle,, name it as CD (observe AB = CD), , 5. Join the radius to get the DOAB and DOCD ., 6. Using trace paper, take the replicas of triangle DOAB and DOCD., , C, O, A, B, D, , 7. Place these triangles DOAB and DOCD one on the other., Observation, , C, , 1. What do you observe? Is ∆OAB ≡ ∆OCD ?, 2. Construct perpendicular line to the chords AB and CD passing, through the centre O. Measure the distance from O to the chords., , O, A, B, , Fig. 4.60, , Now we are going to find out the length of the chords AB and CD, given the angles, C, subtended by two chords at the centre of the circle are equal. That is,, ∠AOB = ∠COD and the two sides which include these angles of the D, DAOB and DCOD are radii and are equal., By SAS rule, ∆AOB ≡ ∆COD . This gives chord AB = chord CD., Now let us write the converse result as follows:, , Converse of theorem 8 , , O, , A, B, , Fig. 4.61, , If the angles subtended by two chords at the centre of a circle are, equal, then the chords are equal., In the same way we are going to discuss about the distance from the centre, when the, C, equal chords are given. Draw the perpendicular OL ^ AB and OM ^ CD., From theorem 7, these perpendicular divides the chords equally. So D M, AL = CM . By comparing the, DOAL and DOCM , the angles, O, ∠OLA = ∠OMC = 90° and OA =OC are radii. By RHS rule, the, A, ∆OAL ≡ ∆OCM . It gives the distance from the centre OL = OM and, L, B, write the conclusion as follows., Fig. 4.62, Theorem 9 Equal chords of a circle are equidistant from the centre., Let us know the converse of theorem 9, which is very useful in solving problems., Geometry 165, , 4-Geometry_Term1.indd 165, , 26-12-2019 11:41:59
Page 172 :
www.tntextbooks.in, , Converse of theorem 9, The chords of a circle which are equidistant from the centre are equal., , 4.5.3 Angle Subtended by an Arc of a Circle, Activity - 6, Procedure :, 1. Draw three circles of any radius with centre O on a chart paper., 2., , From these circles, cut a semi-circle, a minor segment and a major segment., , 3. Consider three points on these segment and name them as A, B and C., A, , B, , A, , C, , C, , O, , Semi circle, O, , O, , O, C, , B, , B, , C, , A, , A, , B, C, , Fig. 4.63, , 4., , (iv) Cut the triangles and paste it on the graph sheet, so that the point A coincides with the origin as, shown in the figure., , Major, segment, O, , B, , A, , C, , Observation :, , Minor, segment, , 0, , (i), , Angle in a Semi-Circle is ______ angle., , (ii), , Angle in a major segment is ____ angle., , (iii), , Angle in a minor segment is ____ angle., , A, , B, , Fig. 4.64, , Now we are going to verify the relationship between the angle subtended by an arc at, the centre and the angle subtended on the circumference., , 4.5.4 Angle at the Centre and the Circumference, Let us consider any circle with centre O. Now place the points A, B and C on the, circumference., C, , C, , C, , B, , A, O, A, , D, X, , Fig. 4.65, 166, , O, , A, , O, , B, , B, D, X, , Fig. 4.66, , D, X, , Fig. 4.67, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 166, , 26-12-2019 11:42:01
Page 173 :
www.tntextbooks.in, , is a minor arc in Fig.4.65, a semi circle in Fig.4.66 and a major arc in Fig.4.67., Here AB, , The point C makes different types of angles in different positions (Fig. 4.65 to 4.67). In all, subtends ÐAOB at the centre and ÐACB at a point on the circumference, these circles, AXB, , of the circle., We want to prove ∠AOB = 2∠ACB . For this purpose extend CO to D and join CD., since OA = OC, , ∠OCA =∠OAC, , (radii), , Exterior angle = sum of two interior opposite angles., ∠AOD = ∠OAC +∠OCA, = 2∠OCA, .... (1), , Progress Check, , Similarly,, , i. Draw the outline of different size, of bangles and try to find out the, centre of each using set square., , ∠BOD = ∠OBC +∠OCB, = 2∠OCB, .... (2), , ii. Trace the given cresent and, complete as full moon using ruler, and compass., , From (1) and (2),, ∠AOD +∠BOD = 2(∠OCA +∠OCB ), , Finally we reach our result ∠AOB = 2∠ACB ., From this we get the result as follows :, Fig. 4.68, , Theorem 10, , The angle subtended by an arc of the circle at the centre is double the angle subtended, by it at any point on the remaining part of the circle., Note, z, , Angle inscribed in a semicircle is a right angle., , z, , Equal arcs of a circle subtend equal angles at the centre., , Example 4.7, , Find the value of x° in the following figures:, , Q, , (i), , (ii), , 260c, , 50c, xc, , M, R, , Fig. 4.69, , xc, , L, , Z, X, , 63c, , xo, , C, , o, , Y, , Fig. 4.71, , 20 o, , O, , O, , N, , Fig. 4.70, , (iv) A, , xc, , O, , O, P, , (iii), , 35, B, , Fig. 4.72, Geometry 167, , 4-Geometry_Term1.indd 167, , 26-12-2019 11:42:03
Page 174 :
www.tntextbooks.in, , Solution, Using the theorem 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 a circle., Q, , (i), , ∠POR = 2 PQR, x ° = 2 × 50°, x ° = 100°, , (ii), , ∠MNL =, , 50c, O, P, , =, , xc, R, , 1, Reflex ÐMOL, 2, 1, × 260°, 2, , x° = 130°, , 260c, O, M, , L, , xc, N, , (iii), , XY is the diameter of the circle., , Therefore ∠XZY = 90°, (Angle on a semi – circle), In DXYZ, , (iv) OA = OB = OC (Radii), In ∆OAC ,, ∠OAC = ∠OCA = 20°, In ∆OBC ,, , Z, X, , O, , 20 o, , O, , xo, , C, , o, , 35, , ∠OBC = ∠OCB = 35° B, , xc, , x ° + 63° + 90° = 180°, x ° = 27°, , 63c, , A, , Y, , (angles opposite to equal sides are equal), ∠ACB = ∠OCA + ∠OCB, x ° = 20° + 35°, x ° = 55°, , Example 4.8, , If O is the centre of the circle and ∠ABC = 30° then find ÐAOC ., , (see Fig. 4.73), , Solution, Given, , ∠AOC, , = 2∠ABC, , O, , A, , ∠ABC = 30°, , (The angle subtended by an, arc at the centre is double the, angle at any point on the, circle), , o, , 30, , C, , B, , Fig. 4.73, , = 2 × 30°, = 60°, , Now we shall see, another interesting theorem. We have learnt that minor arc subtends, obtuse angle, major arc subtends acute angle and semi circle subtends right angle on the, circumference. If a chord AB is given and C and D are two different points on the circumference, of the circle, then find ÐACB and ÐADB . Is there any difference in these angles?, 168, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 168, , 26-12-2019 11:42:05
Page 175 :
www.tntextbooks.in, , 4.5.5 Angles in the same segment of a circle, Consider the circle with centre O and chord AB. C and D are the, points on the circumference of the circle in the same segment. Join the, radius OA and OB., , D, , A, , 1, ∠AOB = ∠ACB (by theorem 10), 2, 1, and ∠AOB = ∠ADB (by theorem 10), 2, ∠ACB = ∠ADB, , C, , O, , B, , Fig. 4.74, , This conclusion leads to the new result., Theorem 11 Angles in the same segment of a circle are equal., Example 4.9, , In the given figure, O is the center of the circle. If the measure of, P, , ∠OQR = 48° , what is the measure of ÐP ?, , Solution, Given ∠OQR = 48° ., , O, , Therefore, ÐORQ also is 48° . (Why?__________), Q, , ÐQOR = 180° − (2 × 48°) = 84° ., , R, , Fig. 4.75, , The central angle made by chord QR is twice the inscribed angle at P., 1, 2, , Thus, measure of ÐQPR = × 84° = 42° ., , Exercise 4.3, 1., , The diameter of the circle is 52cm and the length of one of its chord is 20cm. Find the, distance of the chord from the centre., , 2., , The chord of length 30 cm is drawn at the distance of 8cm from the centre of the, circle. Find the radius of the circle, , 3., , Find the length of the chord AC where AB and CD are the two diameters, perpendicular to each other of a circle with radius 4 2 cm and also find ÐOAC, and ÐOCA ., , 4., , A chord is 12cm away from the centre of the circle of radius 15cm. Find the length of, the chord., Geometry 169, , 4-Geometry_Term1.indd 169, , 26-12-2019 11:42:06
Page 176 :
www.tntextbooks.in, , 5., , In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such, that AB = 16 cm and CD = 12 cm determine the distance between the two chords?, , 6. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between, their centres is 4 cm. Find the length of the common chord., 7. Find the value of x° in the following figures:, A, , (i), , , , P, , (ii), , x°, O, , (iv), , 60, , o, , 80o, Q, , C, , 120o, , Z, , x°, , M, , O, , 70o, , N, , R, , X, Y, , x°, , o, , B, , O, , x°, , P, , (iii), , 30, , o, , 30, , , , (v), , A, o, , 140 x °, o, O 100, , B, , O, , C, , C, , 8., , In the given figure, ∠CAB = 25° ,, , A, , B, , P, O, , find ÐBDC , ÐDBA and ÐCOB, , D, , 4.6 Cyclic Quadrilaterals, , Now, let us see a special quadrilateral with its properties called “Cyclic Quadrilateral”., A quadrilateral is called cyclic quadrilateral if all its four vertices lie on the circumference of, the circle. Now we are going to learn the special property of cyclic quadrilateral., Consider the quadrilateral ABCD whose vertices lie on a circle., We want to show that its opposite angles are supplementary. Connect, the centre O of the circle with each vertex. You now see four radii OA,, OB, OC and OD giving rise to four isosceles triangles OAB, OBC,, OCD and ODA. The sum of the angles around the centre of the circle, is 360° . The angle sum of each isosceles triangle is 180°, Thus, we get from the figure,, , D, C, , Fig. 4.76, D, , 2×( Ð 1+ Ð 2+ Ð 3+ Ð 4) + Angle at centre O = 4× 180°, , , 4 3, , You now interpret this as, , ( Ð 1+ Ð 2+ Ð 3+ Ð 4) = 180° ., , 3, , O, , 2×( Ð 1+ Ð 2+ Ð 3+ Ð 4) + 360° = 720°, , Simplifying this,, , 170, , B, , A, , 4, , A, , 1, , 1, , C, , 2, 2, , B, , Fig. 4.77, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 170, , 26-12-2019 11:42:09
Page 177 :
www.tntextbooks.in, , (i) ( Ð 1+ Ð 2) + ( Ð 3+ Ð 4) = 180° (Sum of opposite angles B and D), (ii) ( Ð 1+ Ð 4) + ( Ð 2+ Ð 3) = 180° (Sum of opposite angles A and C), Now the result is given as follows., Theorem 12 Opposite angles of a cyclic quadrilateral are supplementary., Let us see the converse of theorem 12, which is very useful in solving problems, Converse of Theorem 12 If a pair of opposite angles of a quadrilateral is supplementary,, then the quadrilateral is cyclic., Activity - 7, Procedure, , D, , A, , 1. Draw a circle of any radius with centre O., 2. Mark any four points A, B, C and D on the boundary. Make a cyclic, quadrilateral ABCD and name the angles as in Fig. 4.78, , B, , (iii) ∠C + ∠A = _____ (iv) ∠D + ∠B = _____, , Example 4.10, , 2, A, , B, 4, , 3, C, , A, 3, , 2, 1, , D, 4, , O, , 3, , 2, B, , 4, , 1, C, , Fig. 4.80, , If PQRS is a cyclic quadrilateral in which ∠PSR = 70° and, , ∠QPR = 40° , then find ÐPRQ (see Fig. 4.81)., , Solution, , R, , S, , 70o, O, , PQRS is a cyclic quadrilateral, Given ∠PSR = 70°, , D, , Fig. 4.79, , 2. Sum of opposite angles of a cyclic quadrilateral is _________., 3. The opposite angles of a cyclic quadrilateral is _______., , Fig. 4.78, , 1, , Observe and complete the following:, 1. (i) ∠A + ∠C = _____ (ii) ∠B + ∠D = _____, , C, , 2, , 4. Make the cutout of the angles A, B, C and D as in Fig. 4.79, 5. Paste the angle cutout Ð1, Ð2, Ð3 and Ð4 adjacent to the angles, opposite to A, B, C and D as in Fig. 4.80, , 3, , O, , 3. Make a replica of the cyclic quadrilateral ABCD with the help of, tracing paper., , 6. Measure the angles ∠1 + ∠3, and ∠2 + ∠4 ., , 4, , 1, , o, , 40, , Q, , P, , Fig. 4.81, Geometry 171, , 4-Geometry_Term1.indd 171, , 26-12-2019 11:42:11
Page 178 :
www.tntextbooks.in, , ∠PSR + ∠PQR = 180° (state reason________), 70° + ∠PQR = 180°, ÐPQR = 180° − 70°, ÐPQR = 110°, , In DPQR we have,, ∠PQR + ∠PRQ + ∠QPR = 180°, , (state reason_________), , 110° + ∠PRQ + 40° = 180°, ÐPRQ = 180° − 150°, ÐPRQ = 30°, , Exterior Angle of a Cyclic Quadrilateral, An exterior angle of a quadrilateral is an angle in its exterior, formed by one of its sides and the extension of an adjacent side., Let the side AB of the cyclic quadrilateral ABCD be extended, to E. Here ÐABC and ÐCBE are linear pair, their sum is 180° and, , C, , D, , E, , O, A, , B, , Fig. 4.82, , the angles ÐABC and ÐADC are the opposite angles of a cyclic, quadrilateral, and their sum is also 180° . From this, ∠ABC + ∠CBE = ∠ABC + ∠ADC, and finally we get ∠CBE = ∠ADC . Similarly it can be proved for other angles., Theorem 13 If one side of a cyclic quadrilateral is produced then the exterior angle is equal, to the interior opposite angle., Progress Check, 1. If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is, _____________., 2. As the length of the chord decreases, the distance from the centre _____________., 3. If one side of a cyclic quadrilateral is produced then the exterior angle is _____________, to the interior opposite angle., 4. Opposite angles of a cyclic quadrilateral are _____________., Example 4.11, , In the figure given, find the value of x° and y° ., , Solution, By the exterior angle property of a cyclic quadrilateral,, we get, y° = 100° and, x° + 30° = 60° and so x° = 30°, 172, , 100o, , xo, 30o, , 60o, yo, , Fig. 4.83, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 172, , 26-12-2019 11:42:14
Page 179 :
www.tntextbooks.in, , Exercise 4.4, 1., , Find the value of x in the, given figure., , D, , xo, , A, , E, , 2., , In the given figure, AC is the diameter of the circle with centre, O. If ∠ADE = 30° ; ∠DAC = 35° and ∠CAB = 40° ., Find (i) ÐACD, , (ii) ÐACB, , C, , 120o, , B, , O, , D, , o, 30, o, , 35, 40o, , A, , C, , O, , (iii) ÐDAE, , B, , 4y–4 o, , Find all the angles of the given cyclic quadrilateral, ABCD in the figure., , C, , o, 6x–4, , 3., , 7x+2 o, , D, , 2y+4o, , A, , B, C, , 4., , In the given figure, ABCD is a cyclic quadrilateral where diagonals, intersect at P such that ∠DBC = 40° and ∠BAC = 60° find, (i) ÐCAD , , 5., , 6., , 7., , D, , (ii) ÐBCD, , In the given figure, AB and CD are the parallel chords of a circle, with centre O. Such that AB = 8cm and CD = 6cm. If, OM ^ AB and OL ^ CD distance between LM is 7cm. Find the, radius of the circle?, The arch of a bridge has dimensions as shown, where the arch, measure 2m at its highest point and its width is 6m. What is the, radius of the circle that contains the arch?, , P, A, , B, , 8cm, M, , A, , B, , O 7cm, C, , L, 6cm, , D, , C, 2m, D, 6m, , A, , In figure, ∠ABC = 1200 , where A,B and C are points on the circle, with centre O. Find ÐOAC ?, A school wants to conduct tree plantation programme. For, this a teacher allotted a circle of radius 6m ground to nineth, , B, , O, A, , 8., , 40 o, , 60o, , 120o, , C, , B, , Geometry 173, , 4-Geometry_Term1.indd 173, , 26-12-2019 11:42:16
Page 180 :
www.tntextbooks.in, , C, A, , B, , P, O, , D, , 9., , standard students for planting sapplings. Four students plant, trees at the points A,B,C and D as shown in figure. Here, AB = 8m, CD = 10m and AB ^ CD. If another student places a flower, pot at the point P, the intersection of AB and CD, then find the distance, from the centre to P., , In the given figure, ∠POQ = 1000 and ∠PQR = 300 , then find, ÐRPO ., , R, , O, 100o, P, , 4.7 Practical Geometry, , 30o, , Q, , Practical geometry is the method of applying the rules of geometry dealt with the properties of, points, lines and other figures to construct geometrical figures. “Construction” in Geometry, means to draw shapes, angles or lines accurately. The geometric constructions have been, discussed in detail in Euclid’s book ‘Elements’. Hence these constructions are also known, as Euclidean constructions. These constructions use only compass and straightedge (i.e., ruler). The compass establishes equidistance and the straightedge establishes collinearity., All geometric constructions are based on those two concepts., It is possible to construct rational and irrational numbers using straightedge and a compass as, seen in chapter II. In 1913 the Indian mathematical Genius, Ramanujam gave a geometrical, construction for 355/113 =π. Today with all our accumulated skill in exact measurements., it is a noteworthy feature that lines driven through a mountain meet and make a tunnel.In, the earlier classes, we have learnt the construction of angles and triangles with the given, measurements., In this chapter we are going to learn to construct Centroid, Orthocentre, Circumcentre and, Incentre of a triangle by using concurrent lines., , 4.7.1 Construction of the Centroid of a Triangle, Centroid, , A, , The point of concurrency of the medians of a triangle is called the, centroid of the triangle and is usually denoted by G., , B, , G, , Fig. 4.84, , C, , Activity 8, Objective, , To find the mid-point of a line segment using paper folding, , Procedure, , Make a line segment on a paper by folding it and name it PQ. Fold the line, segment PQ in such a way that P falls on Q and mark the point of intersection, of the line segment and the crease formed by folding the paper as M. M is the, midpoint of PQ., , 174, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 174, , 26-12-2019 11:42:17
Page 181 :
www.tntextbooks.in, , Example 4.12, , Construct the centroid of, , Rough Diagram, , DPQR whose sides are PQ = 8cm; QR = 6cm; RP = 7cm., , Solution, , Step 1 : Draw DPQR using the given, measurements PQ = 8cm, QR = 6cm and RP = 7cm and, construct the perpendicular, bisector of any two sides (PQ, and QR) to find the mid-points, M and N of PQ and QR, respectively., Fig. 4.85, , Step 2 : Draw the medians PN and, RM and let them meet at G., The point G is the centroid, of the given DPQR ., Note, z Three medians can be drawn in a triangle, z The centroid divides each median in the ratio 2:1, , Fig. 4.86, , from the vertex., z The centroid of any triangle always lie inside the, triangle., z Centroid is often described as the triangle’s centre, of gravity (where the triangle balances evenly) and, also as the barycentre., , A, , 4.7.2 Construction of Orthocentre of a Triangle, Orthocentre, The orthocentre is the point of concurrency of the altitudes of a, triangle. Usually it is denoted by H., , H, B, , Fig. 4.87, , C, , Geometry 175, , 4-Geometry_Term1.indd 175, , 26-12-2019 11:42:17
Page 182 :
www.tntextbooks.in, , Activity 9, Objective, , To construct a perpendicular to a line segment from an external point using, paper folding., , Procedure, , Draw a line segment AB and mark an external point P. Move B along BA till, the fold passes through P and crease it along that line. The crease thus formed, is the perpendicular to AB through the external point P., , Activity 10, Objective, , To locate the Orthocentre of a triangle using paper folding., , Procedure, , Using the above Activity with any two vertices of the triangle as external points,, construct the perpendiculars to opposite sides. The point of intersection of, the perpendiculars is the Orthocentre of the given triangle., , Example 4.13, , Construct ΔPQR whose sides are PQ = 6 cm + Q = 600 and QR = 7 cm, R, and locate its Orthocentre., Solution, Step 1 Draw the ΔPQR with the given measurements., 7cm, H, , R, , R, , 60 O, , P, , 6cm, , Q, , Rough Diagram, , 7cm, , 7cm, , H, 60 O, , 60 O, , P, , 6cm, , Fig. 4.88, , Q, , P, , Q, , 6cm, , Fig. 4.89, , Step 2:, , Construct altitudes from any two vertices (say) R and P, to their opposite sides PQ and QR, respectively., The point of intersection of the altitude H is the Orthocentre of the given ΔPQR., 176, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 176, , 26-12-2019 11:42:18
Page 183 :
www.tntextbooks.in, , Note, Where do the Orthocentre lie in the given triangles., , Orthocentre, , Acute Triangle, , Obtuse Triangle, , Right Triangle, , Inside of Triangle, , Outside of Triangle, , Vertex at Right Angle, , H, H, H, , Exercise 4.5, 1., 2., 3., 4., 5., 6., 7., 8., , Construct the DLMN such that LM=7.5cm, MN=5cm and LN=8cm. Locate its, centroid., Draw and locate the centroid of the triangle ABC where right angle at A, AB = 4cm, and AC = 3cm., Draw the DABC , where AB = 6cm, ∠B = 110° and AC = 9cm and construct the, centroid., Construct the DPQR such that PQ = 5cm, PR= 6cm and ∠QPR = 60° and locate its, centroid., Draw ΔPQR with sides PQ = 7 cm, QR = 8 cm and PR = 5 cm and construct its, Orthocentre., Draw an equilateral triangle of sides 6.5 cm and locate its Orthocentre., Draw ΔABC, where AB = 6 cm, + B = 1100 and BC = 5 cm and construct its, Orthocentre., Draw and locate the Orthocentre of a right triangle PQR where PQ = 4.5 cm,, QR = 6 cm and PR = 7.5 cm., , 4.7.3 Construction of the Circumcentre of a Triangle, Circumcentre, The Circumcentre is the point of concurrency of the Perpendicular, bisectors of the sides of a triangle., It is usually denoted by S., , Circumcircle, The circle passing through all the three vertices of the triangle with, circumcentre (S) as centre is called circumcircle., , S, , Fig. 4.90, , Geometry 177, , 4-Geometry_Term1.indd 177, , 26-12-2019 11:42:19
Page 184 :
www.tntextbooks.in, , Circumcircle, , Circumradius, The line segment from any vertex of a triangle, to the Circumcentre of a given triangle is called, circumradius of the circumcircle., , Circumcentre, S, , r, , Circumradius, , Activity 11, Objective, , To construct a perpendicular bisector, of a line segment using paper folding., , Procedure, , Make a line segment on a paper by folding it and name it as PQ. Fold PQ in, such a way that P falls on Q and thereby creating a crease RS. This line RS is, the perpendicular bisector of PQ., , Fig. 4.91, , Activity 12, Objective, , To locate the circumcentre of a triangle using paper folding., , Procedure, , Using Activity 12, find the perpendicular bisectors for any two sides of the, given triangle. The meeting point of these is the circumcentre of the given, triangle., C, , Example 4.14, Construct the circumcentre of the ΔABC with AB = 5 cm, + A = 600 and, + B = 800. Also draw the circumcircle and find the circumradius of the, ΔABC., , 60o, A, , 80o, , 5cm, , B, , Rough Diagram, , Solution, Step 1 Draw the ΔABC with the given measurements, C, , Step 2, Construct the perpendicular bisector of any two sides (AC, and BC) and let them meet at S which is the circumcentre., Step 3, S as centre and SA = SB = SC as radius,, draw the Circumcircle to passes through A,B and C., Circumradius = 3.9 cm., 178, , 80o, , 60o, 5cm, , A, , Fig. 4.92, , B, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 178, , 26-12-2019 11:42:19
Page 185 :
www.tntextbooks.in, , C, , 3.9, , cm, , S, , 60o, A, , 80o, 5cm, , B, , Fig. 4.93, , Note, Where do the Circumcentre lie in the given triangles., Acute Triangle, Inside of Triangle, Circumcentre, , Obtuse Triangle, Outside of Triangle, , Right Triangle, Midpoint of Hypotenuse, , S, S, , S, , 4.7.4 Construction of the Incircle of a Triangle, Incentre, The incentre is (one of the triangle’s points of concurrency formed, by) the intersection of the triangle’s three angle bisectors., The incentre is the centre of the incircle ; It is usually denoted by I; it, is the one point in the triangle whose distances to the sides are equal., Example 4.15, , Construct the incentre of DABC with AB = 6 cm,, ∠B = 65° and AC = 7 cm Also draw the incircle and measure its radius., , Fig. 4.94, Rough Diagram, , Solution, 65o, , Geometry 179, , 4-Geometry_Term1.indd 179, , 26-12-2019 11:42:20
Page 186 :
www.tntextbooks.in, , Step 1 : Draw the DABC with AB = 6cm, ∠B = 65° and, AC = 7cm, , 65o, , Fig. 4.95, , Step 2 : Construct the angle, bisectors of any two angles (A, and B) and let them meet at I., Then I is the incentre of DABC., Draw perpendicular from I to, any one of the side (AB) to meet, AB at D., , 65 o, , Fig. 4.96, , Step 3: With I as centre and ID, as radius draw the circle. This circle touches all the, sides of the triangle internally., , Step 4: Measure inradius, In radius = 1.9 cm., 65 o, , Fig. 4.97, , Exercise 4.6, 1, 2, 3., 4., 5., , 180, , Draw a triangle ABC, where AB = 8 cm, BC = 6 cm and + B = 700 and locate its, circumcentre and draw the circumcircle., Construct the right triangle PQR whose perpendicular sides are 4.5 cm and 6 cm. Also, locate its circumcentre and draw the circumcircle., Construct ΔABC with AB = 5 cm + B = 1000 and BC = 6 cm. Also locate its circumcentre, draw circumcircle., Construct an isosceles triangle PQR where PQ= PR and + Q = 500, QR = 7cm. Also, draw its circumcircle., Draw an equilateral triangle of side 6.5 cm and locate its incentre. Also draw the, incircle., , 9th Standard Mathematics, , 4-Geometry_Term1.indd 180, , 26-12-2019 11:42:21
Page 187 :
www.tntextbooks.in, , 6., , Draw a right triangle whose hypotenuse is 10 cm and one of the legs is 8 cm. Locate, its incentre and also draw the incircle., , 7., , Draw DABC given AB = 9 cm, ∠CAB = 115° and ∆ABC = 40° . Locatge its incentre, and also draw the incircle. (Note: You can check from the above examples that the, incentre of any triangle is always in its interior)., , 8., , Construct DABC in which AB = BC = 6cm and ∠B = 80° . Locate its incentre and, draw the incircle., , Exercise 4.7, Multiple Choice Questions, 1., , 2., , 3., , 4., , The exterior angle of a triangle is equal to the sum of two, (1) Exterior angles, , (2) Interior opposite angles, , (3) Alternate angles, , (4) Interior angles, , 6., , B, , 108o, , In the given figure CE || DB then the value of x is, o, , The correct statement out of the following is, (1), , ΔABC , ΔDEF, , (2) ΔABC , ΔDEF, , (3), , ΔABC , ΔFDE, , (4) ΔABC , ΔFED, , D, , C, , A, , B, , B, , 30o, , 110o, , E, , 75o, , xo, , 60o, , D, , C, , C, 60, , 70o, B, , D, , O, , A, , (2) 30°, (4) 85°, , 42o, , C, , ABCD is a square, diagonals AC and BD meet at O., The number of pairs of congruent triangles with vertex O are, (1) 6, (2) 8, (3) 4, (4) 12, , (1) 45°, (3) 75°, 5., , A, , In the quadrilateral ABCD, AB = BC and, AD = DC Measure of ∠BCD is, (1) 150°, (2) 30°, (3) 105°, (4) 72°, , If the diagonal of a rhombus are equal, then the rhombus is a, (1) Parallelogram but not a rectangle, (2) Rectangle but not a square, (3) Square, (4) Parallelogram but not a square, , D, 60o, , o, , 70o, , 50o, A, , E, , 50o, F, , Geometry 181, , 4-Geometry_Term1.indd 181, , 26-12-2019 11:42:22
Page 188 :
www.tntextbooks.in, , 7., , If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is, (1) ∠C + ∠D , , (3) 12 +C + 13 +D , 8., , (2) 12 (∠C + ∠D), (4) 13 +C + 12 +D, , The interior angle made by the side in a parallelogram is 90° then the parallelogram is a, (1) rhombus (2) rectangle (3) trapezium (4) kite, , 9., , Which of the following statement is correct?, (1) Opposite angles of a parallelogram are not equal., (2) Adjacent angles of a parallelogram are complementary., (3) Diagonals of a parallelogram are always equal., (4) Both pairs of opposite sides of a parallelogram are always equal., , 10. The angles of the triangle are 3x–40, x+20 and 2x–10 then the value of x is, (1) 40°, (2) 35° , (3) 50° (4) 45°, 11., , PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70° ,, then ∠ORS =, (1) 60°, (2) 70°, (3) 55°, (4) 80°, , 12., , A chord is at a distance of 15cm from the centre of the circle of radius 25cm. The, length of the chord is, (1) 25cm, , 13., , (3) 40cm, , C, , (4) 18cm, , 40c, , In the figure, O is the centre of the circle and, ∠ACB= 40° then ∠AOB =, (1) 80°, , 14., , (2) 85°, , (3) 70°, , O, , (4) 65°, , A, B, , In a cyclic quadrilaterals ABCD, ∠A = 4x , ∠C = 2x the value of x is, (1) 30°, , 15., , (2) 20cm, , (2) 20°, , (3) 15°, , In the figure, O is the centre of a circle and diameter AB bisects A, the chord CD at a point E such that CE=ED=8 cm and EB=4cm., The radius of the circle is, (1) 8cm (2) 4cm (3) 6cm (4)10cm, S, , 16., , D, , (4) 25°, , In the figure, PQRS and PTVS are two cyclic, quadrilaterals, If ∠QRS =100°, then ∠TVS =, , 100o, , O, , C, , R, V, , (1) 80° (2) 100° (3) 70° (4) 90°, 17., , 182, , If one angle of a cyclic quadrilateral is 75° , then the, opposite angle is, (1) 100° (2) 105° (3) 85°, (4) 90°, , B, , E, , T, P, , Q, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 182, , 26-12-2019 11:42:24
Page 189 :
www.tntextbooks.in, , 18., , 19., , In the figure, ABCD is a cyclic quadrilateral in which D, DC produced to E and CF is drawn parallel to AB, such that ∠ADC = 80° and ∠ECF = 20°, then, ∠BAD = ?, (1) 100° (2) 20°, (3) 120°, (4) 110°, AD is a diameter of a circle and AB is a chord If, AD = 30 cm and AB = 24 cm then the distance of AB, from the centre of the circle is, (1) 10cm, (2) 9cm, (3) 8cm, , 20., , C, , 80o, , O, , F, , 100o, , A, , B, , (4) 6cm., , In the given figure, If OP = 17cm, PQ = 30 cm and OS is, perpendicular to PQ , then RS is, (1) 10cm, (3) 7cm, , E, , 20o, , (2) 6cm, (4) 9cm., , m, 7c, , O, , 1, P, , R, , Q, , S, , Points to Remember, In a parallelogram the opposite sides and opposite angles are equal, The diagonals of a parallelogram bisect each other., The diagonals of a parallelogram divides it into two congruent triangles, A quadrilateral is a parallelogram if its opposite sides are equal ., Parallelogram on the same base and between same parallel are equal in area., Triangles on the same base and between same parallel are equal in area., Parallelogram is a rhombus if its diagonals are perpendicular., There is one and only one circle passing through three non-collinear points., Equal chords of a circle subtend equal angles at the centre., Perpendicular from the centre of a circle to a chord bisects the chord., Equal chords of a circle are equidistant from the centre., The angle substended 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., , The angle in a semi circle is a right angle., Angles in the same segment of a circle are equal., The sum of either pair of opposite angle of a cyclic quadrilateral is 180o., If one side of a cyclic quadrilateral is produced then the exterior angle is equal to the, , interior opposite angle., , Geometry 183, , 4-Geometry_Term1.indd 183, , 26-12-2019 11:42:26
Page 190 :
www.tntextbooks.in, , ICT Corner-1, Expected Result is shown in this picture, , Step – 1, , Open the Browser and copy and paste the Link given below, (or) by typing the URL given (or) Scan the QR Code., Step - 2, GeoGebra worksheet “Properties: Parallelogram” will appear. There are two sliders, named “Rotate” and “Page”, Step-3, Drag the slider named “Rotate “and see that the triangle is doubled as parallelogram., Step-4, Drag the slider named “Page” and you will get three pages in which, the Properties are explained., Browse in the link, , Properties: Parallelogram: https://www.geogebra.org/m/m9Q2QpWD, , ICT Corner-2, , Expected Result is shown, in this picture, Step – 1, Open the Browser type the URL Link given below (or) Scan the QR Code. GeoGebra work sheet, named “Angles in a circle” will open. In the work sheet there are two activities on Circles., The first activity is the relation between Angle at the circumference and the angle at the centre., You can change the angle by moving the slider. Also, you can drag on the point A, C and D to, change the position and the radius. Compare the angles at A and O., Step - 2, The second activity is “Angles in the segment of a circle”. Drag the points, B and D and check the angles. Also drag “Move” to change the radius and, chord length of the circle., Browse in the link, , Angle in a circle: https://ggbm.at/yaNUhv9S or Scan the QR Code., 184, , 9th Standard Mathematics, , 4-Geometry_Term1.indd 184, , 26-12-2019 11:42:26
Page 191 :
Y, www.tntextbooks.in, H, , (0, y ), 2, , G, , B(x, (x , y ), , S, , 2, , 5, , (0, y), , (0, y ), 1, , 1, , COORDINATEP (x(x, y), GEOMETRY, r, , Divide each difficulty into as many parts as is, P, feasible and necessary to resolve it., A((xx , y ), 1, , Rene Descartes, (AD(CE)1596-1650), , 2, , - Rene Descartes, , 1, , The French Mathematician Rene Descartes (pronounced, A′, X′ CART”), O, “DAYdeveloped, a new branch, P ′ of Mathematics, (x1 , 0), B′, (x, 0), Y′, known as Analytical, Geometry or Coordinate, Geometry, (x2 , 0), X, which combined all arithmetic, algebra and geometry, of the past ages in a single technique of visualising as, points on a graph and equations as geometrical shapes., The fixing of a point position in the plane by assigning, two numbers, coordinates, giving its distance from two, lines perpendicular to each other, was entirely Descartes’, invention., , Learning Outcomes, , , , , , , , To understand the Cartesian coordinate system., To identify the abscissa, ordinate and coordinates of any given point., To find the distance between any two points in the Cartesian plane using formula., To understand the mid-point formula and use it in problem solving., To derive the section formula and apply this in problem solving., To understand the centroid formula and to know its applications., , 5.1 Mapping the Plane, How do you write your address? Here is one., Sarakkalvilai Primary School, 135, Sarakkalvilai,, Sarakkalvilai Housing Board Road,, Keezha Sarakkalvilai,, Nagercoil 629002, Kanyakumari Dist., Tamil Nadu, India., Coordinate Geometry 185, , 5-Coordinat_1.indd 185, , 26-12-2019 11:47:54
Page 192 :
www.tntextbooks.in, , Somehow, this information is enough for, anyone in the world from anywhere to locate, the school one studied. Just consider there are, crores and crores of buildings on the Earth., But yet, we can use an address system to locate, a particular person’s place of study, however, interior it is., How is this possible? Let us work out the, procedure of locating a particular address. We, know the World is divided into countries. One, among them is India. Subsequently India is, divided into States. Among these States we can, locate our State Tamilnadu., Further going deeper, we find our State, is divided into Districts. Districts into taluks,, taluks into villages proceeding further in this, Fig. 5.1, way, one could easily locate “Sarakkalvilai”, among the villages in that Taluk. Further among the roads in that village, ‘Housing Board, Road’ is the specific road which we are interested to explore. Finally we end up the search, by locating Primary School building bearing the door number 135 to enable us precisely, among the buildings in that road., In New York city of USA, there is an area, called Manhattan. The map shows Avenues, run in the North – South direction and the, Streets run in the East – West direction. So,, if you know that the place you are looking, for is on 57th street between 9th and 10th, Avenues, you can find it immediately on the, map. Similarly it is easy to find a place on 2nd, Avenue between 34th and 35th streets. In fact,, New Yorkers make it even simpler. From the, , Fig. 5.2, , door number on a street, you can actually calculate which avenues it lies between, and, from the door number on an avenue, you can calculate which streets it stands., All maps do just this for us. They help us in finding our way and locate a place easily, by using information of any landmark which is nearer to our search to make us understand, whether we are near or far, how far are we, or what is in between etc. We use latitudes, 186, , 5-Coordinat_1.indd 186, , 9th Standard Mathematics, , 26-12-2019 11:47:55
Page 193 :
www.tntextbooks.in, , (east – west, like streets in Manhattan) and longitudes (north – south, like avenues in, Manhattan) to pin point places on Earth. It is interesting to see how using numbers in maps, helps us so much., This idea, of using numbers to map places, comes from geometry. Mathematicians, wanted to build maps of planes, solids and shapes of all kinds. Why would they want such, maps? When we work with a geometric figure, we want to observe wheather a point lies, inside the region or outside or on the boundary. Given two points on the boundary and a, point outside, we would like to examine which of the two points on the boundary is closer, to the one outside, and how much closer and so on. With solids like cubes, you can imagine, how interesting and complicated such questions can be., Mathematicians asked such questions and answered them only to develop their own, understanding of circles, polygons and spheres. But the mathematical tools and techniques, were used to find immense applications in day to day life. Mapping the world using latitudes, and longitudes would not have been developed at all in 18th century, if the co-ordinate, system had not been developed mathematically in the 17th century., You already know the map of the real number system is the number line. It extends, infinitely on both directions. In between any two points, on a number line, there lies infinite, number of points. We are now going to build a map of the plane so that we can discuss about, the points on the plane, of the distance between the points etc. We can then draw on the, plane all the geometrical shapes we have discussed so far, precisely., Arithmetic introduced us to the world of numbers and operations on them. Algebra, taught us how to work with unknown values and find them using equations. Geometry, taught us to describe shapes by their properties. Co-ordinate geometry will teach us how to, use numbers and algebraic equations for studying geometry and beautiful integration of, many techniques in one place. In a way, that is also great fun as an activity. Can’t wait? Let, us plunge in., , 5.2 Devising a Coordinate, System, You ask your friend to draw a, rectangle on a blank sheet of paper,, 5 cm by 3 cm. He says, “Sure, but, where on this sheet?” How would, you answer him?, Now look at the picture. How, will you describe it to another, person?, , Fig. 5.3, Coordinate Geometry 187, , 5-Coordinat_1.indd 187, , 26-12-2019 11:47:55
Page 194 :
www.tntextbooks.in, , Let us analyse the given picture (Fig. 5.3). Just like that a particular house is to be, pointed out, it is going to be a difficult task. Instead, if any place or an object is fixed for, identification then it is easy to identify any other place or object relative to it. For example,, you fix the flag and talk about the house to the left of it, the hotel below it, the antenna on the, house to right of it etc., As shown in Fig. 5.4 draw two perpendicular lines in such a way that the flag is pointed, out at near the intersection point. Now if you tell your friend the total length and width, of the picture frame,, keeping the flagpole as, landmark, you can also, say 2 cm to the right, 3 cm, above etc. Since you know, directions, you can also, say 2 cm east, 3 cm north., This is what we are, going to do. A number line, is usually represented as, horizontal line on which, the positive numbers, always lie on the right, , Fig. 5.4, , side of zero, negative on, , the left side of zero. Now consider another copy of the number line, but drawn vertically: the, positive integers are represented above zero and the negative integers are below zero (fig 5.5)., Where do the two number, , Y, , lines meet? Obviously at zero for, , 3, , both lines. That will be our location, where the “flag” is fixed. We can talk, , 2, , of other numbers relative to it, on, , 1, , both the lines. But now you see that, we talk not only of numbers on the, two number lines but lots more !, Suppose we go 2 to, the right and then 3 to the, top. We would call this place, ( " 2, - 3)., , Xʹ –4 –3, , –2, , O, , –1, , 1, , 2, , 3, , 4, , X, , –1, –2, –3, , Yl, , Fig. 5.5, 188, , 5-Coordinat_1.indd 188, , 9th Standard Mathematics, , 26-12-2019 11:47:55
Page 195 :
www.tntextbooks.in, , All this vertical, horizontal, up, down etc is all very cumbersome. We simply say (2,3) and, understand this as 2 to the right and then 3 up. Notice that we would reach the same place if we first, went 3 up and then 2 right, so for us, the instruction (2,3) is not the same as the instruction (3,2)., What about (–2,3)? It would mean 2 left and 3, up. From where? Always from (0,0). What, about the instruction (2,–3)? It would mean 2, right and 3 down. We need names for, horizontal and vertical number lines too. We, call the horizontal number line the x-axis and, 3cm, the vertical number line the y-axis. To the, right we mark it as X, to the left as X l, to the, top as Y, to the bottom as Y l., 2cm, , The x-co-ordinate is called the abscissa, and the y-co-ordinate is called the ordinate., We call the meeting point of the axes (0,0) the, origin., Fig. 5.6, , Now we can describe any point on a sheet, of paper by a pair (x,y). However, what do (1, 2), etc mean on our paper? We need to choose some, convenient unit and represent these numbers., For instance we can choose 1 unit to be 1 cm., Thus (2,3) is the instruction to move 2 cm to the, , Note, Whether we place (0,0) at the centre of the, sheet, or somewhere else does not matter,, (0,0) is always the origin for us, and all, “instructions” are relative to that point. We, usually denote the origin by the letter ‘O’ ., , right of (0,0) and then to move 3 cm up. Please remember that the choice of units is arbitrary: if we, fix 1 unit to be 2 cm, our figures will be larger, but the relative distances will remain the same., In fact, we now have a language to describe all the infinitely many points on the plane,, not just our sheet of paper !, Since the x-axis and the y-axis divide the plane into four regions, we call them quadrants., (Remember, quadrilateral has 4 sides, quadrants are 4 regions.) They are usually numbered, as I, II, III and IV, with I for upper east side, II for upper west side, III for lower west side, and IV for lower east side, thus making an anti-clockwise tour of them all., Note, Why this way (anti-clockwise) and not clockwise, or not starting from any of the other, quadrants? It does not matter at all, but it is good to follow some convention, and this is what, we have been doing for a few centuries now., Coordinate Geometry 189, , 5-Coordinat_1.indd 189, , 26-12-2019 11:47:56
Page 196 :
www.tntextbooks.in, , Region Quadrant, , X l OY, , I, II, , Nature, of x,y, x>0, y>0, x<0, y>0, , Signs of the coordinates, (+,+), (–,+), , X lO Y l, , III, , x<0, y<0, , (–,–), , XO Y l, , IV, , x>0, y<0, , (+,–), , XOY, , 3, , I Quadrant, (+, +), , 2, , zz (x, y) ≠ (y, x) unless x= y, zz A plane with the rectangular, , coordinate system is called the, Cartesian plane., , 1, –2, , O, , –1, , 1, , 2, , 3, , 4, , –1, III Quadrant, (–, –), , x axis,the value of y, coordinate (ordinate) is zero, that is, P (x,0)., the value of x coordinate, (abscissa) is zero., that is, Q (0, y), , 4, , X′ –4 –3, , zz For any point P on the, , zz For any point Q on the y axis,, , Y, , II Quadrant, (–, +), , Note, , –2, , IV Quadrant, (+, –), , –3, –4, , Y′, , X, , 5.2.1 Plotting Points in, Cartesian Coordinate Plane, To plot the points (4, 5) in the, Cartesian coordinate plane. We, follow the x – axis until we reach, 4 and draw a vertical line at x = 4., , Fig. 5.7, , Similarly, we follow the y – axis until we reach 5, and draw a horizontal line at y = 5., The intersection of these two lines is the position, of (4, 5) in the Cartesian plane., , Y, , p(4, 5), , 5, 4, 3, 2, , This point is at a distance of 4 units from, the y-axis and 5 units from the x-axis. Thus the, 1, position of (4, 5) is located in the Cartesian, plane., O, X′ –1, , 1, , 2, , 3, , 4, , X, , –1, , Y′, Fig. 5.8, , 190, , 5-Coordinat_1.indd 190, , 9th Standard Mathematics, , 26-12-2019 11:47:56
Page 197 :
www.tntextbooks.in, , Example 5.1, (a) (3,–8), , In which quadrant does the following points lie?, (b) (–1,–3), , (c) (2, 5), , (d) (–7, 3), , Solution, (a) The x- coordinate is positive and y – coordinate is negative. So, point(3,–8) lies in the, IV quadrant., (b) The x-coordinate is negative and y – coordinate is negative. So, point(–1,–3) lies in the, III quadrant., (c) The x-coordinate is positive and y – coordinate is positive. So point(2,5) lies in the, I quadrant., (d) The x-coordinate is negative and y – coordinate is positive. So, point(–7,3) lies in, the II quadrant, Example 5.2, , Plot the points A(2,4), B(–3,5), C(–4,–5), D(4,–2) in the Cartesian plane., , Solution, (i) To plot (2, 4), draw a vertical line at x = 2 and draw a horizontal line at y = 4. The, intersection of these two lines is the position of (2, 4) in the Cartesian plane. Thus,, the Point A (2, 4) is located in the I quadrant of Cartesian plane., (ii) To plot (–3, 5), draw a vertical, line at x = –3 and draw a, horizontal line at y = 5. The, intersection of these two lines, , Y, , B(–3, 5), , 5, , is the position of (–3, 5) in, , 3, , the Cartesian plane. Thus, the, , 2, , Point B (–3, 5) is located in the, II quadrant of Cartesian plane., (iii) To plot (–4, –5), draw a, vertical line at x =–4 and draw, , 1, , X′, –4, , –3, , –2, , O, , –1, , –2, , intersection of these two lines, is the position of (-4, 5) in, , –3, , the Cartesian plane. Thus, the, , –4, , Point C (–4, –5) is located in, , –5, , plane., , X, 1, , 2, , 3, , 4, , –1, , a horizontal line at y = –5. The, , the III quadrant of Cartesian, , A(2, 4), , 4, , C(–4, –5), , D(4, –2), , Y′, Fig. 5.9, Coordinate Geometry 191, , 5-Coordinat_1.indd 191, , 26-12-2019 11:47:56
Page 198 :
www.tntextbooks.in, , (iv) To plot (4, –2), draw a vertical line at x = 4 and draw a horizontal line at y = –2. The, Intersection of these two lines is the position of (4,–2) in the Cartesian plane. Thus, the, Point D (4,–2) is located in the IV quadrant of Cartesian plane., Example 5.3, , Plot the following points A(2,2), B(–2,2), C(–2,–1), D(2,–1) in the, Cartesian plane . Discuss the type of the diagram by joining all the points taken in order., Solution, Point, , A, , B, , C, , D, , Quadrant, , I, , II, , III, , IV, , ABCD is a rectangle., Can you find the length, breath and area of the rectangle?, Y, , B(–2, 2), , A(2, 2), , 2, 1, , X′, –5, , –4, , –3, , –2, , O, , –1, , C(–2, –1), , X, 1, , –1, , 2, , 3, , 4, , D(2, –1), , –2, , Note, , Y′, Fig. 5.10, , Exercise 5.1, 1., , 2., , Y, , P, , 4, , Plot the following points, in the coordinate system, and identify the quadrants, P(–7,6), Q(7,–2), R(–6,–7),, X′, S(3,5) and T(3,9), Write down the abscissa, and ordinate of the, following from fig 5.11., (i) P, (ii) Q, (iii) R, (iv) S, , Q, , 3, 2, 1, –5, , –4, , –3, , –2, , O, , –1, , X, 1, , 2, , 3, , 4, , –1, –2, , S, , R, , –3, , Y′, Fig. 5.11, , 192, , 5-Coordinat_1.indd 192, , 9th Standard Mathematics, , 26-12-2019 11:47:57
Page 199 :
www.tntextbooks.in, , 3. Plot the following points in the coordinate plane and join them. What is your conclusion, about the resulting figure?, (i) (–5,3) (–1,3) (0,3) (5,3) , (ii) (0,–4) (0,–2) (0,4) (0,5), 4., , Plot the following points in the coordinate plane. Join them in order. What type of, geometrical shape is formed?, (i) (0,0) (–4,0) (–4,–4) (0,–4) , (ii) (–3,3) (2,3) (–6,–1) (5,–1), Activity - 1, Plot the following points on a, graph sheet by taking the scale, as 1cm = 1 unit., , Y, 3, 2, , Find how far the points are, from each other?, A (1, 0) and D (4, 0). Find, AD and also DA., Is AD = DA?, , 1, , A(1,0), , X′ –4 –3, , You plot another set of, points and verify your Result., , –2, , O, , –1, , 1, , D(4,0), 2, , 3, , 4, , X, , –1, , Y′, Fig. 5.12, , 5.3 Distance between any Two Points, , Akila and Shanmugam are friends living on the same street in Sathyamangalam. Shanmugam’s, house is at the intersection of one street with another street on which there is a library. They, both study in the same school, and that is not far from Shanmugam’s house. Try to draw a, picture of their houses, library and school by yourself before looking at the map below., Consider the school as the origin. (We can do this ! That is the whole point about the coordinate language we are using.), Y, Now fix the scale as 1 unit = 50 metres. Here are, 4, some questions for you to answer by studying the, given figure (Fig 5.13)., 3, 1. How far is Akila’s house from Shanmugam’s, 2, house?, 2. How far is the library from Shanmugam’s, 1, house?, X′, X, 3. How far is the school from Shanmugam’s and, O 1, –1, 2, 3, 4, Akila’s house?, –1, 4. How far is the library from Akila’s house?, Y′, 5. How far is Shanmugam’s house from Akila’s, Fig. 5.13, house?, Coordinate Geometry 193, , 5-Coordinat_1.indd 193, , 26-12-2019 11:47:57
Page 200 :
www.tntextbooks.in, , Question 5 is not needed after answering, question 1. Obviously, the distance from, point A to B is the same as the distance, from point B to A, and we usually call it, the distance between points A and B. But, as mathematicians we are supposed to note, down properties as and when we see them,, so it is better to note this too: distance (A,B), = distance (B,A). This is true for all points A, and B on the plane, so of course question 5, is same as question 1., , Note, The equation distance(A,B)=distance(B,A), is not always obvious. Suppose that the, road from A to B is a one-way street on, which you cannot go the other way? Then, the distance from B to A might be longer !, But we will avoid all these complications, and assume that we can go both ways., , What about the other questions? They are not the same. Since we know that the two, houses are on the same street which is running north – south, the y-distance tells us the, answer to question 1. Similarly, we know that the library and Shanmugam’s house are on the, same street running east – west, we can take the x-distance to answer question 2., Questions 3 and 4 depend on what kind of routes are available. If we assume that the, only streets available are parallel to the x and y axes at the points marked 1, 2, 3 etc then we, answer these questions by adding the x and y distances. But consider the large field east of, Akila’s house., If she can walk across the field, of course she would prefer it. Now there are many ways, of going from one place to another, so when we talk of the distance between them, it is not, precise. We need some way to fix what we mean. When there are many routes between A and, B, we will use distance(A,B) to denote the distance on the shortest route between A and B., Once we think of distance(A,B) as the “straight line distance” between A and B, there is, an elegant way of understanding it for any points A and B on the plane. This is the important, reason for using the co-ordinate system at all ! Before that, 2 more questions from our, example., 1., , With the school as origin, define the coordinates of the two houses, the school, and the library., , 2., , Use the coordinates to give the distance between any one of these and another., , The “straight line distance” is usually called “as the crow flies”. This is to mean that we, don’t worry about any obstacles and routes on the ground, but how we would get from A to, B if we could fly. No bird ever flies on straight lines, though., We can give a systematic answer to this: given any two points A = (x,y) and, B = ( xl , yl ) on the plane, find distance(A,B). It is easy to derive a formula in terms of the four, numbers x, y, xl and yl . This is what we set out to do now., 194, , 5-Coordinat_1.indd 194, , 9th Standard Mathematics, , 26-12-2019 11:47:57
Page 201 :
www.tntextbooks.in, , 5.3.1 Distance Between Two Points on the Coordinate Axes, Points on x – axis: If two points lie on the, x- axis, then the distance between them is equal, to the difference between the x- coordinates., , Y, 3, 2, , Consider two points A (x1,0) and B (x2,0) on, the x-axis ., The distance of B from A is, , 1, , X′, , AB = OB – OA = x2–x1 if x2>x1 or, , , , (x1, 0), A, , = x1–x2 if x1>x2, AB = |x2–x1|, , O 1, , –1, , 2, , 5, , –2, , Y′, Fig. 5.14, , Y, , 5, , Q (0, y2), |y2–y1|, , 4, , Consider two points P(0,y1) and Q(0,y2), , 3, , The distance Q from P is, PQ = OQ – OP., , 2, , = y2 – y1 if y2> y1 or, , P (0, y1), , 1, , X′, , O 1, , –1, , PQ = |y2–y1|, , X, 2, , 3, , 4, , –1, , (Read as modulus or absolute value of y2–y1), , Y′, Fig. 5.15, , 5.3.2 Distance Between Two Points Lying, on a Line Parallel to Coordinate Axes, , Distance AB = Distance between PQ = |x2–x1|, [The difference between x coordinates], Similarly consider the line joining the two, points, , 4, , |x2–x1|, , Points on y – axis: If two points lie on y-axis then, the distance between them is equal to the difference, between the y-coordinates., , Consider the points A(x1, y1) and B(x2, y1). Since X′, the y - coordinates are equal the points lie on a, line parallel to x- axis. From A and B draw AP, and BQ perpendicular to x- axis respectively., Observe the given figure (Fig. 5.16), it is obvious, that the distance AB is same as the distance PQ, , 3, , –1, , (Read as modulus or absolute value of x2–x1), , = y1–y2 if y1> y2, , (x2, 0), B X, , Y, 1, , |x2–x1|, , O 1, , –1, , 2P, , 3, , 4, , Q, 5, , X, , –1, –2, , A(x1, y1), , B(x2, y1), , –3, –4, , Y′, Fig. 5.16, Coordinate Geometry 195, , 5-Coordinat_1.indd 195, , 26-12-2019 11:47:57
Page 203 :
www.tntextbooks.in, , = ON – OM, = x2 – x1 , , (Measuring the distance from O), ……..(1), , And RQ = NQ – NR, = NQ – MP, , (Opposite sides of the rectangle MNRP), , = y2 – y1 , , ………..(2), , Step 2, , Triangle PQR is right angled at R. (PR = NQ), PQ2 = PR2 + RQ2 , d2, , = (x2– x1)2 + (y2– y1 )2, , d, , = ^ x2 –x1h2 + ^ y2 –y1h2, , (By Pythagoras theorem), (Taking positive square root), , Note, Distance between two points, zz Given two points P (x1, y1) and Q (x2, y2), the distance between these points is given, , by the formula d = ^ x2 –x1h2 + ^ y2 –y1h2 ., zz The distance between PQ = The distance between QP, i.e. ^ x2 –x1h2 + ^ y2 –y1h2 = ^ x1 –x2h2 + ^ y1 –y2h2, zz The distance of a point P (x1, y1) from the origin O (0,0) is OP =, , x12 + y12, , 5.3.4 Properties of Distances, , We have already seen that distance (A,B) = distance (B,A) for any points A, B on the, plane. What other properties have you noticed? In case you have missed them, here are some:, distance (A,B) = 0 exactly when A and B denote the identical point: A = B., distance (A,B) >0 for any two distinct points A and B., Now consider three points A, B and C. If we are given their co-ordinates and we find, that their x-co-ordinates are the same then we know that they are collinear, and lie on a line, parallel to the y-axis. Similarly, if their y-co-ordinates are the same then we know that they, are collinear, and lie on a line parallel to the x-axis. But these are not the only conditions., Points (0,0), (1,1) and (2,2) are collinear as well. Can you think of what relationship should, exist between these coordinates for the points to be collinear?, The distance formula comes to our help here. We know that when A, B and C are the, vertices of a triangle, we get,, distance(A,B) + distance(B,C) >distance(A,C) (after renaming the vertices suitably)., When do three points on the plane not form a triangle? When they are collinear, of, course. In fact, we can show that when,, Coordinate Geometry 197, , 5-Coordinat_1.indd 197, , 26-12-2019 11:47:59
Page 204 :
www.tntextbooks.in, , distance(A,B) + distance(B,C) = distance(A,C), the points A, B and C must be collinear., Similarly, when A, B and C are the vertices of a right angled triangle, +ABC = 90c we, know that:, distance(AB)2+ distance(BC)2 = distance(AC)2, with appropriate naming of vertices. We can also show that the converse holds: whenever, the equality here holds for A, B and C, they must be the vertices of a right angled triangle., The following examples illustrate how these properties of distances are useful for answering, questions about specific geometric shapes., Example 5.4, , Y, 4, , Find the distance between the points, (–4, 3), (2,–3)., , P(–4, 3), , 3, 2, , Solution, The distance between the points, (-4, 3), (2,-3) is, d = (x2 - x1) + (y2 - y1), 2, , 1, , X′, –4, , –3, , –2, , 2, , O, , –1, , –2, , = ^62 + (- 6) 2h = ^36 + 36h, , –3, , = ^36 # 2h, , –4, , 3, , Q(2, –3), , Fig. 5.19, , Show that the following points A(3,1) , B(6,4) and C(8,6) lies on a, , straight line., , Collinear points, , Solution, Using the distance formula, we have, AB = ^6 - 3h2 + ^4 - 1h2 = 9 + 9, , =, , BC = ^8 - 6h2 + ^6 - 4h2 = ^4 + 4h, , 18 = 3 2, =, , AC = ^8 - 3h2 + ^6 - 1h2 = 25 + 25 =, AB + BC = 3 2 + 2 2 = 5 2 = AC, Therefore the points lie on a straight line., , 5-Coordinat_1.indd 198, , 2, , Y′, , = 6 2, , 198, , 1, , –1, , = (2 + 4) 2 + (- 3 - 3) 2, , Example 5.5, , X, , 8, , = 2 2, , 50 = 5 2, , To show the collinearity of, three points, we prove that, the sum of the distance, between two pairs of, points is equal to the third, pair of points., In otherwords, points A,, B, C are collinear if, AB +BC = AC, , 9th Standard Mathematics, , 26-12-2019 11:48:00
Page 205 :
www.tntextbooks.in, , Example 5.6, , Show that the points A(7,10), B(-2,5), C(3,-4) are the vertices of a, , right angled triangle., Solution, Here A = (7, 10), B = (-2, 5), C = (3,-4), AB =, , (x2 - x1) 2 + (y2 - y1) 2, , = ^- 2 - 7h2 + ^5 - 10h2, = (- 9) 2 + (- 5) 2, = ^81 + 25h, = 106, AB2 = 106 , BC =, , ... (1), , (x2 - x1) 2 + (y2 - y1) 2, , = ^3 - (- 2)h2 + ^- 4 - 5h2 = (5) 2 + (- 9) 2, = 25 + 81 = 106, BC2 = 106 ... (2), AC =, , (x2 - x1) 2 + (y2 - y1) 2, , = ^3 - 7h2 + ^- 4 - 10h2 = (- 4) 2 + (- 14) 2, = 16 + 196 = 212, , Right angled triangle, We know that the sum, of the squares of, two sides is equal, to the square of the, third side,which is the, hypotenuse of a right, angled triangle., , AC2 = 212 ... (3), From (1), (2) & (3) we get,, , , AB2 + BC2 = 106 + 106 = 212 = AC2, , Since AB2 + BC2 = AC2, `, , DABC is a right angled triangle, right angled at B., Coordinate Geometry 199, , 5-Coordinat_1.indd 199, , 26-12-2019 11:48:01
Page 206 :
www.tntextbooks.in, , Example 5.7, , Show that the points A(–4,–3), B(3,1), C(3,6), D(–4,2) taken in that, order form the vertices of a parallelogram., Solution, Let A(–4, –3), B(3, 1), C(3, 6), D(–4, 2) be the four vertices of any quadrilateral ABCD., Using the distance formula,, Parallelogram, 2, 2, Let, d = (x2 - x1) + (y2 - y1), We know that opposite, 2, 2, AB = ^3 + 4h + ^1 + 3h = 49 + 16 = 65, sides are equal, BC =, , ^3 - 3h2 + ^6 - 1h2, , =, , 0 + 25 =, , , CD =, , ^- 4 - 3h2 + ^2 - 6h2, , =, , ^- 7h2 + ^- 4h2 =, , , AD =, , ^- 4 + 4h2 + ^2 + 3h2 =, , 25 = 5, , 49 + 16 =, , 65, , ^0h2 + ^5h2 =, , 25 = 5, , AB = CD = 65 and BC = AD = 5, Here, the opposite sides are equal. Hence ABCD is a parallelogram., Example 5.8, Calculate the distance between the points A (7, 3) and B which lies on the x-axis whose, abscissa is 11., Solution, Since B is on the x-axis, the y-coordinate of B is 0., So, the coordinates of the point B is (11, 0), By the distance formula the distance between the points A (7, 3), B (11, 0) is, d = ^ x2 - x1h2 + ^ y2 - y1h2, AB =, , ]11 - 7g2 + ]0 - 3g2 = ]4g2 + ]- 3g2 = 16 + 9 = 25, , =5, , 200, , 5-Coordinat_1.indd 200, , 9th Standard Mathematics, , 26-12-2019 11:48:02
Page 207 :
www.tntextbooks.in, , Example 5.9, , Find the value of ‘a’ such that PQ = QR where P, Q, and R are the, , points whose coordinates are (6, –1), (1, 3) and (a, 8) respectively., Solution, Given P (6, –1), Q (1, 3) and R (a, 8), , , PQ =, , , , QR =, , , , ]1 - 6g2 + ]3 + 1g2 = ]- 5g2 + ]4g2 = 41, ]a - 1g2 + ]8 - 3g2 = ]a - 1g2 + ]5 g2, , Given PQ = QR, , ]a - 1g2 + ]5 g2, , Therefore, , 41 =, , , , 41 = (a –1)2 + 25 [Squaring both sides], , , , , , , , , , (a–1)2 + 25, (a–1)2, (a–1)2, (a–1), a, a, a, , Example 5.10, , = 41, = 41 - 25, = 16, = + 4 [taking square root on both sides], = 1+4, = 1 + 4 or a = 1 – 4, = 5 or a = –3, , Let A(2, 2), B(8, –4) be two given points in a plane. If a point P lies on, , the X- axis (in positive side), and divides AB in the ratio 1: 2, then find the coordinates of P., Solution, Given points are A(2, 2) and B(8, –4) and let P = (x, 0) [P lies on x axis], By the distance formula, , , d =, , , , AP =, , ]x - 2g2 + ]0 - 2g2 =, , , , BP =, , ]x - 8g2 + ]0 + 4g2 = x 2 - 16x + 64 + 16 = x 2 - 16x + 80, , Given, , i.e., , ^ x2 - x1h2 + ^ y2 - y1h2, , AP : PB = 1 : 2, AP, 1, BP = 2, , x 2 - 4x + 4 + 4 =, , x 2 - 4x + 8, , ( a BP = PB), , 2AP = BP, squaring on both sides,, 2, , 4AP = BP, , 2, , Coordinate Geometry 201, , 5-Coordinat_1.indd 201, , 26-12-2019 11:48:03
Page 208 :
www.tntextbooks.in, , 4(x2 – 4x + 8) = (x2 – 16x + 80), 4x2 – 16x + 32 = x2 – 16x + 80, , , 3x2 – 48 = 0, , , , 3x2 = 48, , , , x2 = 16, , , x=+4, As the point P lies on x-axis (positive side), its x- coordinate cannot be –4., Hence the coordinates of P is(4, 0), , Example 5.11, , Show that (4, 3) is the centre of the circle passing through the points, (9, 3), (7,–1), (–1,3). Also find its radius., Solution, Let P(4, 3), A(9, 3), B(7, –1) and C(–1, 3), If P is the centre of the circle which passes through the points A, B, and C, then P is, equidistant from A, B and C (i.e.) PA = PB = PC, By distance formula,, , , d = ^ x2 - x1h2 + ^ y2 - y1h2, , , , AP = PA =, , ]4 - 9g2 + ]3 - 3g2 = ]- 5g2 + 0 = 25 = 5, , , , BP = PB =, , ]4 - 7g2 + ]3 + 1g2 = ]3 g2 + ]4g2 = 9 + 16 = 25 = 5, , , , CP = PC =, , ]4 + 1g2 + ]3 - 3g2 = ]5 g2 + 0 = 25 = 5, , PA = PB = PC = 5, Radius = 5, Therefore P is the centre of the circle, passing through A, B and C, , Exercise 5.2, 1., , Find the distance between the following pairs of points., (i) (1, 2) and (4, 3), (ii) (3,4) and (– 7, 2), (ii) (a, b) and (c, b), , (iv) (3,– 9) and (–2, 3), , 2., , Determine whether the given set of points in each case are collinear or not., (i) (7,–2),(5,1),(3,4), (ii) (a,–2), (a,3), (a,0), , 3., , Show that the following points taken in order form an isosceles triangle., (i), , 202, , 5-Coordinat_1.indd 202, , A (5,4), B(2,0), C (–2,3), , (ii) A (6,–4), B (–2, –4), C (2,10), , 9th Standard Mathematics, , 26-12-2019 11:48:04
Page 209 :
www.tntextbooks.in, , 4., , Show that the following points taken in order form an equilateral triangle in each case., (i), , 5., , Show that the following points taken in order form the vertices of a parallelogram., (i), , 6., , A(2, 2), B(–2, –2), C^- 2 3 , 2 3 h (ii) A ^ 3 , 2 h , B (0,1), C(0,3), , A(–3, 1), B(–6, –7), C (3, –9), D(6, –1) (ii), , A (–7, –3), B(5,10), C(15,8), D(3, –5), , Verify that the following points taken in order form the vertices of a rhombus., (i), , A(3,–2), B (7,6),C (–1,2), D (–5, –6), , A (1,1), B(2,1),C (2,2), D(1,2), , (ii), , 7., , A (–1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’., , 8., , The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3), is 10 units, What are the coordinates of A?, , 9., , The point (x, y) is equidistant from the points (3,4)and(–5,6). Find a relation between, x and y., , 3, 10. Let A(2, 3) and B(2,–4) be two points. If P lies on the x-axis, such that AP = 7 AB,, find the coordinates of P., 11. Show that the point (11,2) is the centre of the circle passing through the points (1,2),, (3,–4) and (5,–6), 12. The radius of a circle with centre at origin is 30 units. Write the coordinates of the, points where the circle intersects the axes. Find the distance between any two such, points., , 5.4 The Mid-point of a Line Segment, Imagine a person riding his two-wheeler on a straight road, towards East from his college to village A and then to village B. At, some point in between A and B, he suddenly realises that there is, not enough petrol for the journey. On the way there is no petrol, bunk in between these two places. Should he travel back to A or, just try his luck moving towards B? Which would be the shorter, distance? There is a dilemma. He has to know whether he crossed, the half way mid-point or not., , Fig. 5.20, O, , A, , M, , B, , x1, , x, , x2, , Fig. 5.21, , X, , The above Fig. 5.21 illustrates the situation. Imagine college as origin O from which the, distances of village A and village B are respectively x1 and x2 (x1 < x2 ) . Let M be the midpoint of AB then x can be obtained as follows., Coordinate Geometry 203, , 5-Coordinat_1.indd 203, , 26-12-2019 11:48:04
Page 210 :
www.tntextbooks.in, , AM = MB and so,, and this is simplified to, , x − x1 = x2 − x, x=, , x1 + x2, , 2, Now it is easy to discuss the general case. If A x1, y1 , B x2, y2 are any two points and, M ( x , y ) is the mid-point of the line segment AB, then M ′ is the mid-point of AC (in the, Fig. 5.22). In a right triangle the perpendicular bisectors of the sides intersect at the midpoint of the hypotenuse. (Also, this property is due to similarity among the two coloured, triangles shown; In such triangles, the corresponding sides will be proportional)., Y, , (, , y, 2 , , , (Using similarity property), Let us take the point M as M(x,y), , +, , X′ O, , Y′, , 2, , , x1 +, x, 2 2, , y1, , (0, y1), , M, , ), , Another way of solving, , B(x2, y2), , (0, y2), , ) (, , D x2 , y1 + y2 , , , 2, , AM ′ MM ′ AM, =, =, MD, BD, MB, x − x1 y − y1 1, =, = (AM = MB), x2 − x y2 − y 1, , M′, A(x1, y1), , x1 + x2 , , y1 , , , 2, , C(x2, y1), , (x1, 0), , x1 + x2 , , 0, , , 2, , (x2, 0) X, , Now, ∆AMM ′ and ∆MBD are, similar. Therefore,, , x − x1, =1, x2 − x, x + x1, 2 x = x2 + x1 ⇒ x = 2, 2, y + y1, Similarly, y = 2, 2, Consider,, , Fig. 5.22, , x-coordinate of M = the average of x-coordinates of A and C =, y-coordinate of M = the average of y-coordinates of B and C =, The mid-point M of the line segment, joining the points A (x1 , y1 ) and B (x2 , y2 ) is, , x1 + x2, 2, y1 + y2, , and similarly,, , 2, , Thinking Corner, , x + x2 y1 + y2 , M 1, ,, 2, 2 , , If D is the mid-point of AC, and C is the mid-point of, AB, then find the length of, AB if AD = 4cm., , For example, The mid-point of the line segment joining, , x1 + x2 y1 + y2 , where x1 = −8 , x2 = 4,, ,, 2, 2 , , the points ( −8, −10) and ( 4, −2) is given by , , 204, , 5-Coordinat_1.indd 204, , 9th Standard Mathematics, , 26-12-2019 11:48:14
Page 211 :
www.tntextbooks.in, , y1 = −10 and y2 = −2 ., −8 + 4 −10 − 2 , The required mid-point is , ,, or ( −2, −6) ., 2, 2 , Let us now see the application of mid-point formula in our real life situation, consider, the longitude and latitude of the following cities., Name of the city Longitude Latitude, , Kuthethoor, , 80.27° E, , 13.00° N, , Mangaluru, (Kuthethoor), , 74.85° E, , 13.00° N, , Bengaluru, (Rajaji Nagar), , 77.56° E, , 13.00° N, , Let us take the longitude and latitude, of Chennai (80.27° E, 13.00° N) and, Mangaluru (74.85° E , 13.00° N) as pairs. Since, Bengaluru is, , located in the middle of, , Chennai, , Mangaluru,, , and, , we, , have, , i an S, A r ab, , ea, , Chennai, (Besant Nagar), , Besant Nagar, , Bay of Bengal, , Rajaji Nagar, , N, (Not to Scale), , to, , Fig. 5.23, , find the average of the coordinates, that is, 80.27 + 74.85 13.00 + 13.00 . This gives (77.56°E, 13.00°N) which is the longitude and, ,, , , 2, 2, latitude of Bengaluru. In all the above examples, the point exactly in the middle is the, mid-point and that point divides the other two points in the same ratio., Example 5.12, , The point (3, −4 ) is the centre of a circle. If AB is a diameter of the, circle and B is (5, −6) , find the coordinates of A., , Solution Let the coordinates of A be (x1 , y1 ) and the given point is B (5, −6) . Since the, centre is the mid-point of the diameter AB, we have, x1 + x2, =3, 2, x1 + 5 = 6, , y1 + y2, = −4, 2, y1 − 6 = −8, , x1 = 6 − 5, x1 = 1, Therefore, the coordinates of A is (1, −2) ., , y1 = −8 + 6, , A (x1, y1), (3, –4), , y1 = −2, , B (5, –6), Fig. 5.24, , Coordinate Geometry 205, , 5-Coordinat_1.indd 205, , 26-12-2019 11:48:18
Page 212 :
www.tntextbooks.in, , Progress Check, (i) Let X be the mid-point of the line segment joining A(3, 0) and B(−5, 4), and Y be the mid-point of the line segment joining P(−11, −8) and, Q(8, −2) . Find the mid-point of the line segment XY., (ii) If (3, x ) is the mid-point of the line segment joining the points A(8, −5), and B(−2,11) , then find the value of ‘x ’., Example 5.13, , If (x,3), (6,y), (8,2) and (9,4) are the vertices of a parallelogram taken, in order, then find the value of x and y., Let A(x,3), B(6,y), C(8,2) and D(9,4) be the vertices of the parallelogram, Solution, D (9, 4), C (8, 2), ABCD. By definition, diagonals AC and BD bisect each other., Mid-point of AC = Mid-point of BD, 6 + 9 y + 4, x + 8 3 + 2, ,, ,, , = , , 2, 2, 2, 2 , equating the coordinates on both sides, we get, x + 8 15, =, 2, 2, x + 8 = 15, , and, , A (x, 3), , B (6, y), , Fig. 5.25, , 5 y+4, =, 2, 2, 5= y+4, y =1, , x=7, , Hence, x = 7 and y = 1., Thinking Corner, A(6,1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD taken in, order. Find the fourth vertex D. If (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) and (x 4 , y 4 ) are the four, vertices of the parallelogram, then using the given points, find the value of, (x1 + x3 − x2 , y1 + y3 − y2 ) and state the reason for your result., Example 5.14, , Find the points which divide the line segment joining A(−11,4) and, ), B(9,8) into four equal parts., B (9, 8, R, , Q, , 4), A (–11, 206, , 5-Coordinat_1.indd 206, , P, , Fig. 5.26, , 9th Standard Mathematics, , 26-12-2019 11:48:22
Page 213 :
www.tntextbooks.in, , Solution, Let P, Q, R be the points on the line segment joining A(−11,4) and B(9,8) such that, AP = PQ = QR = RB ., Here Q is the mid-point of AB, P is the mid-point of AQ and R is the mid-point of QB., −11 + 9 4 + 8 −2 12 , ,, Q is the mid-point of AB = , = , = ( −1, 6), 2, 2 2 2, −12 10 , −11 − 1 4 + 6 , P is the mid-point of AQ = , ,, =, ,, = ( −6, 5), , 2 2 , 2, 2 , −1 + 9 6 + 8 , ,, R is the mid-point of QB = , , 2, 2 , , 8 14 , = , , 2 2 , , = ( 4, 7 ), , Hence the points which divides AB into four equal parts are P(–6, 5), Q(–1, 6) and R(4, 7)., Example 5.15, , The mid-points of the sides of a triangle, are (5,1), (3, −5) and (−5, −1). Find the coordinates of the, vertices of the triangle., , A (x1, y1), (–5, –1), , Solution Let the vertices of the ∆ABC be A(x1 , y1 ), B(x2 , y2 ), , and C(x3 , y3 ) and the given mid-points of the sides AB, BC C (x3, y3) (3, –5), and CA are (5,1), (3, −5) and (−5, −1) respectively. Therefore,, Fig. 5.27, x1 + x2, = 5 ⇒ x1 + x2 = 10, 2, x2 + x3, = 3 ⇒ x2 + x3 = 6, 2, x3 + x1, = −5 ⇒ x3 + x1 = –10, 2, Adding (1), (2) and (3), , ...(1), ...(2), ...(3), , B (x2, y2), , ⇒ y1 + y2 = 2, , …(5), , ⇒ y2 + y3 = –10, , …(6), , ⇒, , …(7), , y3 + y1 = –2, , Adding (5), (6) and (7),, , 2 x1 + 2 x2 + 2 x3 = 6, x1 + x2 + x3 = 3, , y1 + y2, =1, 2, y2 + y3, = −5, 2, y3 + y1, = −1, 2, , (5, 1), , 2 y1 + 2 y2 + 2 y3 = −10, ...(4), , y1 + y2 + y3 = −5, , (4) − (2) ⇒ x1 = 3 − 6 = −3, , (8) − (6) ⇒ y1 = −5 + 10 = 5, , (4) − (3) ⇒ x2 = 3 + 10 = 13, , (8) − (7) ⇒ y2 = −5 + 2 = −3, , (4) − (1) ⇒ x3 = 3 − 10 = −7, , (8) − (5) ⇒ y3 = −5 − 2 = −7, , ...(8), , Therefore the vertices of the triangles are A (−3, 5), B (13, −3) and C (−7, −7)., Coordinate Geometry 207, , 5-Coordinat_1.indd 207, , 26-12-2019 11:48:30
Page 214 :
www.tntextbooks.in, , Thinking Corner, A, , If (a1 , b1 ), (a2 , b2 ) and (a3 , b3 ) are the mid-points of the sides of a, triangle, using the mid-points given in example 5.15 find the (a3, b3), (a1, b1), value of (a1 + a3 − a2 , b1 + b3 − b2 ), (a1 + a2 − a3 , b1 + b2 − b3 ) and, (a2 + a3 − a1 , b2 + b3 − b1 ) . Compare the results. What do you, C, B, (a2, b2), observe? Give reason for your result?, Fig. 5.28, , Exercise 5.3, 1., , Find the mid-points of the line segment joining the points, (i) (−2,3) and (−6,−5), , 3 −11, 1 3, , 2 7, 2 7 , The centre of a circle is (−4,2). If one end of the diameter of the circle is (−3,7), then, find the other end., (iii) (a,b) and (a+2b,2a−b), , 2., , (ii) (8,−2) and (−8,0), (iv) , − and ,, , , , , 3., , If the mid-point (x,y) of the line joining (3,4) and (p,7) lies on 2 x + 2 y + 1 = 0 , then, what will be the value of p?, , 4., , The mid-point of the sides of a triangle are (2,4), (−2,3) and (5,2). Find the coordinates, of the vertices of the triangle., , 5., , O(0,0) is the centre of a circle whose one chord is AB, where the points A and B are, (8,6) and (10,0) respectively. OD is the perpendicular from the centre to the chord, AB. Find the coordinates of the mid-point of OD., , 6., , The points A(−5, 4) , B(−1, −2) and C(5, 2) are the vertices of an isosceles right-angled, triangle where the right angle is at B. Find the coordinates of D so that ABCD is a, square., , 7., , The points A(−3, 6) , B(0, 7) and C(1, 9) are the mid-points of the sides DE, EF and FD, of a triangle DEF. Show that the quadrilateral ABCD is a parallellogram., , 8., , A(−3, 2) , B(3, 2) and C(−3, −2) are the vertices of the right triangle, right angled at A., Show that the mid-point of the hypotenuse is equidistant from the vertices., , 5.5 Points of Trisection of a Line Segment, The mid-point of a line segment is the point of bisection, which means dividing into, two parts of equal length. Suppose we want to divide a line segment into three parts of equal, length, we have to locate points suitably to effect a trisection of the segment., 208, , 5-Coordinat_1.indd 208, , 9th Standard Mathematics, , 26-12-2019 11:48:34
Page 215 :
www.tntextbooks.in, , Points of trisection, , Mid-point (bisection), A, , M, Fig. 5.29, , B, , P, , A, , AM = MB, , Q, , Fig. 5.30, , B, , AP = PQ = QB, , For a given line segment, there are two points of trisection. The method of obtaining, this is similar to that of what we did in the case of locating the point of bisection, (i.e., the mid-point). Observe the given Fig. 5.31. Here P and Q are the points of trisection, , of the line segment AB where A is ( x1 , y1 ) and B is ( x2 , y2 ) . Clearly we know that, P is the, mid-point of AQ and Q is the mid-point of PB. Now consider the DACQ and ∆PDB (Also,, can be verified using similarity property of triangles which will be dealt in detail in higher, classes)., A′P ′ = P ′Q ′ = Q ′B ′, , Y, , B(x2, y2), , B ′′, , y2 − y1, 3, (c, d)), , Q ′′, , y2 − y1, 3, (a, b), , P ′′, , Note that when we, divide the segment, into 3 equal parts, we, are also dividing the, horizontal and vertical, legs into three equal, parts., , Q, , D, , P, , y2 − y1, 3, A′′, , C, , A(x1, y1), , y1, , X′, , x1, , O, Y′, , A′, , x2 − x1 P ′, 3, , x2 − x1 Q ′, 3, , x2 − x1, 3, , B′, , X, , Fig. 5.31, , If P is (a, b), then, a = OP ′ = OA′ + A′P ′, = x1 +, , x2 − x1, x + 2 x1, = 2, ;, 3, 3, , b = PP ′ = OA′′ + A′′P ′′, = y1 +, , y2 − y1, y + 2 y1, = 2, 3, 3, , x + 2 x1 y2 + 2 y1 , ,, Thus we get the point P as 2, , , 3, 3 , , 5-Coordinat_1.indd 209, , Coordinate Geometry 209, , 26-12-2019 11:48:41
Page 216 :
www.tntextbooks.in, , If Q is (c, d), then, d = OQ ′′ = OB ′′ − Q ′′B ′′, , c = OQ ′ = OB ′ − Q ′B ′, x − x1 2 x2 + x1, =, = x2 − 2, ;, 3 , 3, , Example 5.16, , y − y1 2 y2 + y1, =, = y2 − 2, 3 , 3, , 2 x + x1 2 y2 + y1 , ,, Thus the required point Q is 2, , , 3, 3 , Find the points of trisection of the line segment joining (−2, −1) and (4, 8) ., B, , Solution Let A (−2, −1) and B (4, 8), are the given points., , (4, 8), , Q, P, , Let P (a, b) and Q(c, d ) be the A, points of trisection of AB, so that (–2, –1), , Fig. 5.32, , AP = PQ = QB ., By the formula proved above,, P is the point, , Progress Check, , x2 + 2 x1 y2 + 2 y1 , 4 + 2(−2) 8 + 2(−1) , ,, ,, = , , , 3, 3, 3, 3, , (i) Find the coordinates of the, points of trisection of the, line segment joining ( 4, −1), , 4−4 8−2 , ,, = , = (0, 2), 3, 3 , Q is the point, , and ( −2, −3) ., , (ii) Find the coordinates of, points of trisection of the, line segment joining the, , 2(4) − 2 2( 8) − 1 , 2 x2 + x1 2 y2 + y1 , ,, ,, = , , , 3, 3, 3, 3 , , , 5.6 Section Formula, , 8 − 2 16 − 1, ,, = , = (2, 5), 3, 3 , , point (6, −9) and the origin., , We studied bisection and trisection of a given line segment. These are only particular, cases of the general problem of dividing a line segment joining two points x1, y1 and, , (, , ( x2 , y2 ) in the ratio m : n., , ), , Given a segment AB and a positive real number r., O, , A, x1, , P, r units, , x, , 1 unit, , B, x2, , Fig.5.33, , We wish to find the coordinate of point P which divides AB in the ratio r :1., This means, 210, , 5-Coordinat_1.indd 210, , AP r, = or AP = r (PB) ., PB 1, , 9th Standard Mathematics, , 26-12-2019 11:48:47
Page 217 :
www.tntextbooks.in, , This means that x – x1 = r(x2 – x), x=, , Solving this,, , rx2 + x1, , ….. (1), r +1, We can use this result for points on a line to the general case as follows., (0, y2), , Y, , B(x2, y2), 1, , (0, y), , P (x, y), r, P, , (0, y1), , X′ O, , A(x1, y1), , Y′, , A′, , P′, , (x1, 0), , (x, 0), , B′, , (x2, 0) X, , Fig. 5.34, , Taking AP : PB = r :1 , we get A′P ′ : P ′B ′ = r :1 ., Therefore, , A′P ′ = r (P ′B ′), , (x − x1 ) = r (x2 − x ), rx + x1, which gives, x = 2, r +1, Thus,, , … [see (1)], , Precisely in the same way we can have y =, AP, , ry2 + y1, r +1, , If P is between A and B, and, = r , then we have the, PB, formula,, rx + x1 ry2 + y1, ,, P is 2, r +1, r +1, If r is taken as, , , ., , Thinking Corner, (i) What happens when, m = n = 1? Can you, identify it with a result, already proved?, (ii) AP : PB = 1 : 2 and, AQ : QB = 2:1., What is AP : AB? What, is AQ : AB?, , mx + nx1 my2 + ny1 , ,, , then the section formula is 2, , which is the, m + n , n, m +n, , m, , standard form., Coordinate Geometry 211, , 5-Coordinat_1.indd 211, , 26-12-2019 11:48:51
Page 218 :
www.tntextbooks.in, , Note, zz The line joining the points ( x1 , y1 ) and ( x2 , y2 ) is divided by x-axis in the ratio, , − y1, −x1, and by y-axis in the ratio, ., y2, x2, zz If three points are collinear, then one of the points divide the line segment joining, the other two points in the ratio r : 1., , zz Remember that the section formula can be used only when the given three points, , are collinear., zz This formula is helpful to find the centroid, incenter and excenters of a triangle., , It has applications in physics too; it helps to find the center of mass of systems,, equilibrium points and many more., Example 5.17, , Find the coordinates of the point which divides the line segment, joining the points (3,5) and (8,−10) internally in the ratio 3:2., Solution, 6, , Y, A (3, 5), , 4, 3, , 2, , X′ O, , 2, , 4, , 6, , 8, , 10, , 12, , X, , –2, –4, , P (x, y), , –6, , 2, , –8, B (8, –10), , –10, , Y′, Fig. 5.35, , Let A(3,5), B(8,−10) be the given points and let the point P(x,y) divides the line segment, AB internally in the ratio 3:2., mx + nx1 my2 + ny1 , By section formula, P (x , y ) = P 2, ,, m+n, m + n , 212, , 5-Coordinat_1.indd 212, , 9th Standard Mathematics, , 26-12-2019 11:48:52
Page 219 :
www.tntextbooks.in, , Here x1 = 3, y1 = 5, x2 = 8, y2 = −10 and m = 3, n = 2, 3(8) + 2(3) 3(−10) + 2(5) , 24 + 6 −30 + 10 , ,, = P, ,, Therefore P (x , y ) = P , , = P 6, − 4, , 3+2, 5, 3+2, 5, , (, , Example 5.18, , In what ratio does, the point P(–2, 4) divide the line segment, joining the points A(–3, 6) and B(1, –2), internally?, , A (–3, 6), m, , 6, , P (–2, 4), , (, , ), , 4, 3, 2, , n, , 1, , By section formula,, , = P −2, 4, , Y, , 5, , Solution, Given points are A(–3, 6) and, B(1, –2). P(–2, 4) divide AB internally in, the ratio m : n., , mx + nx1 my2 + ny1 , P (x , y ) = P 2, ,, m+n, m + n , , ), , X′, , –3, , –2, , –1 O, –1, , .....(1), , 1, , 2, , 3, , X, , B (1, –2), , –2, , Y′, , Here x1 = −3, y1 = 6, x2 = 1, y2 = −2, , Fig. 5.36, , m(1) + n(−3) m(−2) + n(6) , (1) ⇒ , ,, = P ( −2, 4), , m+n, m+n , Equating x-coordinates, we get, m − 3n, = −2 or m − 3n = −2m − 2n, m+n, , 3m = n, , , , m, 1, =, n, 3, m : n = 1: 3, , Hence P divides AB internally in the ratio 1:3., , Note, We may arrive at the same, result by also equating the, y-coordinates., Try it., , Example 5.19, , What are the coordinates of B if point P(−2,3) divides the line segment, joining A(−3,5) and B internally in the ratio 1:6?, Solution, Let A(−3,5) and B(x2 , y2 ) be the given two points., Coordinate Geometry 213, , 5-Coordinat_1.indd 213, , 26-12-2019 11:48:57
Page 220 :
www.tntextbooks.in, , Given P(−2,3) divides AB internally in the ratio 1:6., mx + nx1 my2 + ny1 , By section formula, P 2, = P(−2, 3), ,, m+n, m + n , 1(x ) + 6(−3) 1( y2 ) + 6(5) , P 2, ,, , 1+ 6, 1 + 6 , , = P(−2, 3), , Equating the coordinates, x2 − 18, = −2, 7, x2 − 18 = −14, , y2 + 30, =3, 7, y2 + 30 = 21, , x2 = 4, , , y2 = −9, , Therefore, the coordinate of B is (4, −9), , Exercise 5.4, 1., 2., 3., , 4., 5., 6., 7., , Find the coordinates of the point which divides the line segment joining the points, A(4, −3) and B(9, 7) in the ratio 3:2., In what ratio does the point P(2, −5) divide the line segment joining A(−3, 5) and, B(4, −9) ., Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in, 2, such a way that AP = AB., 5, Find the coordinates of the points of trisection of the line segment joining the points, A(−5, 6) and B(4, −3) ., The line segment joining A(6,3) and B(−1, −4) is doubled in length by adding half of, AB to each end. Find the coordinates of the new end points., Using section formula, show that the points A(7, −5), B(9, −3) and C(13,1) are, collinear., A line segment AB is increased along its length by 25% by producing it to C on the, side of B. If A and B have the coordinates (−2, −3) and (2,1) respectively, then find the, coordinates of C., , 5.7 The Coordinates of the Centroid, , A (x1, y1), , Consider a ∆ABC whose vertices are A(x1 , y1 ),, B(x2 , y2 ) and C(x3 , y3 )., , F, , Let AD, BE and CF be the medians of the ∆ABC., x + x3 y2 + y3 , ,, The mid-point of BC is D 2, 2, 2 , 214, , 5-Coordinat_1.indd 214, , 9th Standard Mathematics, , B, (x2, y2), , 2, G, 1, D, Fig. 5.37, , E, , C, (x3, y3), , 26-12-2019 11:49:03
Page 222 :
www.tntextbooks.in, , Note, zz The medians of a triangle are concurrent and the point of concurrence, the, , centroid G, is one-third of the distance from the opposite side to the vertex, along the median., A, , zz The centroid of the triangle obtained by joining the, , mid-points of the sides of a triangle is the same as the, F, (a1, b1), centroid of the original triangle., zz If (a1, b1), (a2, b2) and (a3, b3) are the mid-points of the, , sides of a triangle ABC then its centroid G is given by, , E, , C, , D, (a2, b2), , (a3, b3), B, , Fig. 5.39, , a +a +a b +b +b , G 1 2 3 , 1 2 3, , , 3, 3, , Example 5.20, , Find the centroid of the triangle whose veritices, are A(6, −1), B(8, 3) and C(10, −5)., Solution, The centroid G(x , y ) of a triangle whose vertices are (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ) is given by, x + x2 + x3 y1 + y2 + y3 , G(x , y ) = G 1, ,, B (8, 3), , , 3, 3, We have (x1 , y1 ) = (6, −1); (x2 , y2 ) = (8, 3);, (x3 , y3 ) = (10, −5), The centroid of the triangle, , , A (6, –1), , 6 + 8 + 10 −1 + 3 − 5 , G( x , y ) = G , ,, , , 3, 3, , , Note, , 24 −3 , = G , = G (8, −1), 3 3, , C (10, –5), Fig. 5.40, , zz The Euler line of a triangle is the line that passes through, , the orthocenter (H), centroid (G) and the circumcenter, (S). G divides the line segment HS in the ratio 2:1 from, the orthocenter. That is centroid divides orthocenter, and circumcenter internally in the ratio 2:1 from the, Orthocentre., zz In an equilateral triangle, orthocentre, incentre, centroid, and circumcentre are all the same., 216, , 5-Coordinat_1.indd 216, , S, G, H, , Fig. 5.41, , 9th Standard Mathematics, , 26-12-2019 11:49:12
Page 223 :
www.tntextbooks.in, , Example 5.21, , If the centroid of a triangle is at ( −2,1) and two of its vertices are, (1, −6) and ( −5, 2) , then find the third vertex of the triangle., Solution Let the vertices of a triangle be, A(1, −6), B(−5, 2) and C(x3 , y3 ), Given the centroid of a triangle as ( −2,1) we get,, x1 + x2 + x3, = −2, 3, 1 − 5 + x3, = −2, 3, −4 + x3 = −6, , y1 + y2 + y3, =1, 3, −6 + 2 + y3, =1, 3, −4 + y3 = 3, , x3 = −2, , y3 = 7, , Therefore, third vertex is (−2,7)., Thinking Corner, (i), , Master gave a trianglular plate with vertices A(5, 8), B(2,, 4), C(8, 3) and a stick to a student. He wants to balance, the plate on the stick. Can you help the boy to locate that, point which can balance the plate., , (ii) Which is the centre of gravity for this triangle? why?, , Fig. 5.42, , Exercise 5.5, 1., , Find the centroid of the triangle whose vertices are, (i) (2,−4), (−3,−7) and (7,2), , (ii) (−5,−5), (1,−4) and (−4,−2), , 2., , If the centroid of a triangle is at (4,−2) and two of its vertices are (3,−2) and (5,2) then, find the third vertex of the triangle., , 3., , Find the length of median through A of a triangle whose vertices are A(−1,3), B(1,−1), and C(5,1)., , 4., , The vertices of a triangle are (1,2), (h,−3) and (−4,k). If the centroid of the triangle is, at the point (5,−1) then find the value of, , 5., , (h + k )2 + (h + 3k )2 ., , Orthocentre and centroid of a triangle are A(−3,5) and B(3,3) respectively. If C is the, circumcentre and AC is the diameter of this circle, then find the radius of the circle., Coordinate Geometry 217, , 5-Coordinat_1.indd 217, , 26-12-2019 11:49:15
Page 224 :
www.tntextbooks.in, , 6., 7., , ABC is a triangle whose vertices are A(3, 4), B(−2, −1) and C(5, 3) . If G is the centroid, and BDCG is a parallelogram then find the coordinates of the vertex D., 3 , If , 5 ,, 2 , , −9 , 13 −13 , are mid-points of the sides of a triangle, then find, 7, and ,, 2, 2 2 , , the centroid of the triangle., , Exercise 5.6, Multiple Choice Questions, 1., , If the y-coordinate of a point is zero, then the point always lies ______, (1)in the I quadrant, , 2., , (2) in the II quadrant, , The points (–5, 2) and (2, –5) lie in the ________, (1) same quadrant, (3) II and IV quadrant respectively, , 3., , (4) Rhombus, , (3) only S, , (4) P and Q, , (2) (0, 4), , (3) (1, 4), , (4) (4, 2), , 56, , (3) 10, , (4), , 2, , If the points A (2,0), B (-6,0), C (3, a–3) lie on the x-axis then the value of a is _____, (2) 2, , (3) 3, , (4) –6, , If ( x+2, 4) = (5, y–2), then the coordinates (x,y) are _____, (1) (7, 12), , 9., , (2) Q and R, , (2), , (1) 0, 8., , (3) Trapezium, , The distance between the two points ( 2, 3 ) and ( 1, 4 ) is ______, (1) 2, , 7., , (2) Rectangle, , The point whose ordinate is 4 and which lies on the y-axis is _______________, (1)( 4, 0 ), , 6., , II and III quadrant respectively, IV and II quadrant respectively, , If P( –1,1), Q( 3,–4), R( 1, –1), S(–2, –3) and T( –4, 4) are plotted on a graph paper,, then the points in the fourth quadrant are __________, (1) P and T, , 5., , (2), (4), , On plotting the points O(0,0), A(3, – 4), B(3, 4) and C(0, 4) and joining OA, AB, BC, and CO, which of the following figure is obtained?, (1) Square, , 4., , (3)on x-axis (4) on y-axis, , (2) (6, 3), , (3) (3, 6), , (4) (2, 1), , If Q1, Q2, Q3, Q4 are the quadrants in a Cartesian plane then Q2 + Q3 is ___________, (1) Q1 , Q2, , (2) Q2 , Q3, , (3) Null set, , (4) Negative x-axis., , 10. The distance between the point ( 5, –1 ) and the origin is _________, (1), 218, , 5-Coordinat_1.indd 218, , 24, , (2), , 37, , (3), , 26, , (4) 17, , 9th Standard Mathematics, , 26-12-2019 11:49:17
Page 225 :
www.tntextbooks.in, , 11. The coordinates of the point C dividing the line segment joining the points P(2,4) and, Q(5,7) internally in the ratio 2:1 is, 7 11, (2) (3,5), (3) (4,4), (4) (4,6), , 2 2, a b, 12. If P , is the mid-point of the line segment joining A(−4,3) and B(−2,4) then (a,b) is, 3 2, 7, 7, , , (1) (−9,7), (2) −3, , (3) (9, −7), (4) 3, − , , , 2, 2, (1) ,, , , 13. In what ratio does the point Q(1,6) divide the line segment joining the points P(2,7), and R(−2,3), (1) 1:2, , (2) 2:1, , (3) 1:3, , (4) 3:1, , 14. If the coordinates of one end of a diameter of a circle is (3,4) and the coordinates of its, centre is (−3,2), then the coordinate of the other end of the diameter is, (1) (0,−3), , (2) (0,9), , (3) (3,0), , (4) (−9,0), , 15. The ratio in which the x-axis divides the line segment joining the points A(a1 , b1 ) and, B(a2 , b2 ) is, (1) b1 : b2, , (2) −b1 : b2, , (3) a1 : a2, , (4) −a1 : a2, , 16. The ratio in which the x-axis divides the line segment joining the points (6,4) and, (1, −7) is, (1) 2:3, , (2) 3:4, , (3) 4:7, , (4) 4:3, , 17. If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3,4),, (1,1) and (2,−3) respectively, then the vertices A and B of the triangle are, (1) (3,2), (2,4), , (2) (4,0), (2,8), , (3) (3,4), (2,0), , (4) (4,3), (2,4), , 18. The mid-point of the line joining (−a,2b) and (−3a,−4b) is, (1) (2a,3b), , (2) (−2a, −b), , (3) (2a,b), , (4) (−2a, −3b), , 19. In what ratio does the y-axis divides the line joining the points (−5,1) and (2,3), internally, (1) 1:3, , (2) 2:5, , (3) 3:1, , (4) 5:2, , 20. If (1,−2), (3,6), (x,10) and (3,2) are the vertices of the parallelogram taken in order,, then the value of x is, (1) 6 , , (2) 5, , (3) 4, , (4) 3, , Coordinate Geometry 219, , 5-Coordinat_1.indd 219, , 26-12-2019 11:49:19
Page 227 :
www.tntextbooks.in, , ICT Corner-2, Expected Result is shown in this picture, , Step – 1, Open the Browser by typing the URL Link given below (or) Scan the QR Code. GeoGebra, work sheet named “Co-ordinate Geometry” will open. There are two worksheets under the, title Distance Formula and Section Formula., Step - 2, Move the sliders of the respective values to change the points and ratio., Work out the solution and check and click on the respective check box and, check the answer., Browse in the link, , Co-Ordinate Geometry: https://ggbm.at/sfszfe24 or Scan the QR Code., , Activity, Plot the points A( 1, 0), B ( –7, 2), C (–3, 7) on a graph sheet and join them to form a, triangle., Plot the point G ( –3, 3)., Join AG and extend it to intersect BC at D., Join BG and extend it to intersect AC at E., What do you infer when you measure the distance between BD and DC and the distance, between CE and EA?, Using distance formula find the lengths of CG and GF, where F in on AB., Write your inference about AG: GD, BG: GE and CG: GF., , Note: G is the centroid of the triangle and AD, BE and CF are the three medians of the, triangle., Coordinate Geometry 221, , 5-Coordinat_1.indd 221, , 26-12-2019 11:49:23
Page 228 :
www.tntextbooks.in, B, , 6, , P (x , y ), 1, O, , q, , x, , y, Q, , A, , TRIGONOMETRY, There is perhaps nothing which so occupies the middle, position of mathematics as Trigonometry.- J. F. Herbart, Euler, like Newton, was the greatest, mathematician of his generation. He, studied all areas of mathematics and, continued to work hard after he had, gone blind. Euler made discoveries in, many areas of mathematics, especially, , Leonhard, Euler, Aryabhatta, (AD, (CE), 1707, (A.D (CE) 476- –1783), 550), , Calculus and Trigonometry. He was the first to prove, several theorems in Geometry., , Learning Outcomes, , ÂÂ To understand the relationship among various trigonometric ratios., ÂÂ To recognize the values of trigonometric ratios and their reciprocals., ÂÂ To use the concept of complementary angles., ÂÂ To understand the usage of trigonometric tables., , 6.1 Introduction, Trigonometry (which comes from Greek words trigonon means triangle and metron, means measure) is the branch of mathematics that studies the relationships involving, lengths of sides and measures of angles of triangles. It is a useful tool for engineers,, scientists, and surveyors and is applied even in seismology and navigation., Observe the three given right angled triangles; in particular scrutinize their, measures., , The corresponding angles shown in the three triangles are of the same, , size. Draw your attention to the lengths of “opposite” sides (meaning the side opposite, to the given angle) and the “adjacent” sides (which is the side adjacent to the given, angle) of the triangle., 222, , 6 Trigonometry.indd 222, , 9th Standard Mathematics, , 26-12-2019 14:14:05
Page 229 :
www.tntextbooks.in, , I, , 3.5 units, , 35°, , A 5 units, , 7 units, , C, , 14 units, , F, , 35°, , 35°, , B D, , EG, , 10 units, , H, , 20 units, , Fig. 6.1, opposite side , , What can you say about the ratio adjacent side in each case? Every right angled, , , triangle given here has the same ratio 0.7 ; based on this finding, now what could be the, C, , length of the side marked ‘x’ in the Fig 6.2? Is it 15?, remarkable, , ratios, , stunned, , early, , 10.5 units, , Such, , mathematicians and paved the way for the subject of, trigonometry., , 35°, , There are three basic ratios in trigonometry, each, , A, , B, , x units, , Fig. 6.2, , of which is one side of a right-angled triangle divided, by another., , sine, , cosine, , tangent, , Short form, , sin, , cos, , tan, , C, , Related, measurements, , po, hy, , te n, , us, , e, , q, A, , Relationship, , sinθ =, , B, , opposite side, hypotenuse, , opposite side of angle θ, , Name of the, angle, , C, , C, , hy, , te, po, , nu, , se, , q, A, , A, , adjacent side of angle θ, , cosθ =, , B, , adjacent side, hypotenuse, , q, adjacent side of angle θ, , tanθ=, , B, , opposite side of angle θ, , The three ratios are:, , opposite side, adjacent side, , Trigonometry 223, , 6 Trigonometry.indd 223, , 26-12-2019 14:14:06
Page 230 :
www.tntextbooks.in, , For the measures in the figure, compute sine, cosine and tangent, , ratios of the angle q ., , 35 units, , R, , Solution, In the given right angled, triangle, note that for the given angle, q , PR is the ‘opposite’ side and PQ is, the ‘adjacent’ side., sinθ =, , opposite side PR 35, = =, hypotenuse, QR 37, , cosθ =, , adjacent side PQ 12, = =, hypotenuse, QR 37, , tanθ =, , opposite side PR 35, = =, adjacent side PQ 12, , 37 u, , P, 12 units, , Example 6.1, , n it s, , q, , Fig. 6.3, , Q, , It is enough to leave the ratios as fractions. In case, if you want to simplify each ratio, neatly in a terminating decimal form, you may opt for it, but that is not obligatory., Note, • Since trigonometric ratios are defined in terms of ratios of sides, they are unitless numbers., • Ratios like sinθ, cosθ, tanθ are not to be treated like (sin)×(θ), (cos)×(θ),(tan)×(θ)., , Thinking Corner, The given triangles ABC,DEF, and GHI have measures, , I, , 3-4-5, 6-8-10 and 12-16-20., Are they all right triangles?, How do you know?, The angles at the vertices B, E and H, are of equal size (each angle is equal to θ)., With these available details, fill up the, following table and comment on the ratios, that you get., 224, , 6 Trigonometry.indd 224, , F, , 6 3, D, , 10, , 5, A, , G, , 20, , C, , 12, , 4, , q, , B, , 8, 16, , q, , Fig. 6.4, , E, , q, H, , 9th Standard Mathematics, , 26-12-2019 14:14:07
Page 231 :
www.tntextbooks.in, , In ∆ABC, 3, sin q =, 5, cos q = ?, tanq =, , 3, 4, , In ∆DEF, 6, sin =, q = ?, 10, cos q = ?, , In ∆GHI, 12, sin=, q = ?, 20, cos q = ?, , tan q = ?, , tan q = ?, , Reciprocal ratios, We defined three basic trigonometric ratios namely, sine, cosine and tangent. The, reciprocals of these ratios are also often useful during calculations. We define them as follows:, Basic Trigonometric, Ratios, opposite side, sin θ =, hypotenuse, cos θ =, tan θ =, , Its reciprocal, cosecant θ =, , adjacent side, opposite side, adjacent side, , hypotenuse, opposite side, , cotangent θ =, , cosec θ =, , hypotenuse, adjacent side, , secant θ =, , hypotenuse, , Short form, , sec θ =, , adjacent side, , cot θ =, , opposite side, , hypotenuse, opposite side, hypotenuse, adjacent side, adjacent side, opposite side, , From the above ratios we can observe easily the following relations:, 1, 1, 1, cosec θ =, sec θ =, cot θ =, tanq, sinq, cos q, 1, 1, 1, cos θ =, sinθ =, tan θ =, secq, cosecq, cot q, (sin q ) (cosec q ) 1 . We usually write this as sin θ cosec θ 1., (cos q ) (sec q ) 1 . We usually write this as cos θ sec θ = 1., (tan q ) (cot q ) 1. We usually write this as tan θ cot θ = 1., Example 6.2, , Find the six trigonometric ratios, of the angle q using the given diagram., 7, Solution, By Pythagoras theorem,, = (25)2 − 72, = 625 − 49 = 576 = 24, , 25, , q, , A, , Fig. 6.5, , C, opposite side of, angle θ, , AB = BC 2 − AC 2, , C, , 7, A, , B, , hypotenuse, , 25, 24, , adjacent side of angle θ, , Fig. 6.6, , q, , B, Trigonometry 225, , 6 Trigonometry.indd 225, , 26-12-2019 14:14:12
Page 232 :
www.tntextbooks.in, , The six trignometric ratios are, , tanq =, secq =, , opposite side, , =, , 7, 25, , cos q =, , =, , 7, 24, , cosecq =, , hypotenuse 25, =, adjacent side 24, , cotq =, , hypotenuse, opposite side, adjacent side, , Example 6.3, , adjacent side, hypotenuse, , hypotenuse 25, =, opposite side 7, adjacent side, opposite side, , an g, , C, , nt, ace, a dj, , B, Fig. 6.7, , AC = 13, opposite side, =, hypotenuse, , 2, 13, , =, cos A, , hypotenuse, 13, =, opposite side, 2, , sec A =, cot A =, , If sec q =, , Let BC = 13 and AB = 5, hypotenuse, BC 13, =, =, adjacent side AB 5, , By the Pythagoras theorem,, , 226, , 6 Trigonometry.indd 226, , 9th, , 2, , adjacent side, =, hypotenuse, , opp, , 2 si d e, , te, o si, , a, of, , ng l, , eΑ, , 3, 13, , hypotenuse, 13, =, adjacent side, 3, adjacent side, opposite side, , =, , 3, 2, C, , 13, 2 sin q 3 cos q, , then show that, 3, 5, 4 sin q 9 cos q, , Solution:, , AC BC AB, , se, , 3, , 32 22 9 4 13, , 2, , t e nu, , of, si d e, , AC = AB 2 + BC 2, , secq =, , 24, 7, , le A, , 2, tan A =, =, adjacent side 3, By Pythagoras theorem,, , Example 6.4, , =, , hy p, o, , A, , opposite side, , cosecA =, , 24, 25, , 2, , then find all the other trigonometric ratios., 3, A, , If tan A =, , Solution, , =, sin A, , =, , hypotenuse, , opposite side of angle θ, , sinq =, , 13, , q, A, , 5, , B, , adjacent side of angle θ, , Fig. 6.8, , Standard Mathematics, , 26-12-2019 14:14:17
Page 233 :
www.tntextbooks.in, , , , 132 52, , , , 169 25 =, , 144 = 12, , AC 12, AB 5, =, ; cos=, q =, BC 13, BC 13, , Therefore, sinq, =, , 12, 5, 24 15, 2 3 , 2 sin q 3 cos q, 13, 13 13 = 9= 3 = RHS, LHS , , 48 45 3, 4 sin q 9 cos q 4 12 9 5, 13, 13, 13, Note: We can also take the angle ‘ q ’ at the vertex ‘C’ and proceed in the same way., Exercise 6.1, 1., , From the given figure, find all the trigonometric ratios of angle B., 40, , A, 9, , B, , 41, C, , 2., , A, , From the given figure, find the values of, (i) sin B (ii) sec B, , (iii) cot B, , (iv) cosC (v) tanC, , (vi) cosecC, , 13, , B, , 5, , D, , 3., , If 2 cos q = 3 , then find all the trigonometric ratios of angle q ., , 4., , If cos A =, , 5., 6., 7., , C, , 16, , 3, sin A − cos A, , then find the value of, ., 5, 2 tan A, 2x, If cos A , , then find the values of sinA and tanA in terms of x., 1 x2, a, If sinq , , then show that b sin q = a cos q ., 2, 2, a b, 4 sin A 3 cos A, If 3 cot A = 2 , then find the value of, ., 2 sin A 3 cos A, C, 8 cos q 2 sin q, . , 4 cos q 2 sin q, q f, , 8., , If cos q : sin q = 1 : 2, then find the value of, , 9., , From the given figure, prove that θ φ 90 ., Also prove that there are two other right angled, triangles. Find sin a, cos b and tanf ., , 15, , 20, , 12, a, A, , b, 9, , D, , 16, , B, , Trigonometry 227, , 6 Trigonometry.indd 227, , 26-12-2019 14:14:23
Page 234 :
www.tntextbooks.in, , A boy standing at a point O finds his kite flying at a point P with distance OP=25m., , 10., , It is at a height of 5m from the ground. When the thread is extended by 10m from P,, it reaches a point Q. What will be the height QN of the kite from the ground? (use, trigonometric ratios), , 10m, , P, , 25m, , Q, , h, , 5m, , O, , M, , N, , 6.2 Trigonometric Ratios of Some Special Angles, The values of trigonometric ratios of certain angles can be obtained geometrically., Two special triangles come to help here., C, , 6.2.1 Trigonometric ratios of 45°, 45°, , Consider a triangle ABC with angles 45° , 45° and 90° as, shown in the figure 6.9., It is the shape of half a square, cut along the square’s, diagonal. Note that it is also an isosceles triangle (both legs have, the same length, a units)., , a 2, , a, , 45°, A, , B, , a, Fig. 6.9, , Use Pythagoras theorem to check if the diagonal is of, length a 2 ., Now, from the right-angled triangle ABC,, =, sin 45o, , opposite side BC, a, = = =, hypotenuse AC a 2, , cos, =, 45o, , adjacent side AB, a, = = =, hypotenuse, AC a 2, , =, tan, 45o, , opposite side BC a, =, = = 1, adjacent side AB a, , 1, 2, 1, , The reciprocals of these ratio, can be easily foundout to be, cosec 45 2 ;, , 2, , sec 45 2 and, cot 45 1, P, , 6.2.2 Trigonometric Ratios of 30° and 60°, Consider an equilateral triangle PQR of side length 2 units., , 30, , Draw a bisector of ∠P. Let it meet QR at M., , , 2, , Knowing PQ and QM, we can find PM, using Pythagoras, theorem,, , 6 Trigonometry.indd 228, , 2, 3, , PQ = QR = RP = 2 units., , QM = MR = 1 unit (Why?), , 228, , °, , 60°, Q, , 1, , M, , 1, , R, , Fig. 6.10, , 9th Standard Mathematics, , 26-12-2019 14:14:25
Page 235 :
www.tntextbooks.in, , we find that PM =, , 3 units., , Now, from the right-angled triangle PQM,, opposite side QM 1, =, =, hypotenuse, PQ 2, adjacent side PM, 3, cos 30o =, =, =, hypotenuse, PQ, 2, opposite side QM, 1, tan30o =, =, =, adjacent side PM, 3, sin30o, =, , The reciprocals of these ratio, can be easily foundout to be, 2, cosec 30° = 2, sec 30° =, 3, and cot 30° = 3, , We will use the same triangle but the other angle of measure 60° now., sin 60o =, , cos 60o =, tan 60o =, , opposite side, hypotenuse, adjacent side, hypotenuse, opposite side, adjacent side, , =, =, =, , PM, PQ, QM, PQ, PM, QM, , =, =, =, , 3, , The reciprocals of these ratio, can be easily foundout to be, , 2, 1, , 2, , cosec60 , , 2, 3, 1, , 3, , and cot 60 , , = 3, , 1, 3, , Y, , 6.2.3 Trigonometric ratios of 0° and 90°, To find the trigonometric ratios of, 0° and 90° , we take the help of what is, known as a unit circle., A unit circle is a circle of unit radius, (that is of radius 1 unit), centred at the, origin., , ; sec 60 2, , P(cos q , sin q ), sinq, , q, , O, , Q, , cos q, , X, , Why make a circle where the radius, is 1unit?, This means that every reference, triangle that we create here has a, hypotenuse of 1unit, which makes it so, much easier to compare angles and ratios., We will be interested only in the, positive values since we consider ‘lengths’, and it is hence enough to concentrate on, the first quadrant., , Fig. 6.11, B, , P (x , y ), 1, y, , q, , O, , Q, x, Fig. 6.12, , A, , Trigonometry 229, , 6 Trigonometry.indd 229, , 26-12-2019 14:14:28
Page 236 :
www.tntextbooks.in, , We can see that if P(x,y) be any point on the unit circle in the first quadrant and, POQ q, PQ y, OQ x, PQ y, =, = y ; cos, =, q=, =, ; tan q, =, x=, OP 1, OP 1, OQ x, , sin, =, q, , When q 0 , OP coincides with OA, where A is (1,0) giving x = 1, y = 0 ., We get thereby,, , , sin 0° = 0 ;, , , , cos 0° = 1 ;, , , , tan 0° =, , 0, 1, , =0 ;, , cosec 0° = not defined (why?), sec 0° = 1, cot 0° = not defined (why?), , When q 90 , OP coincides with OB, where B is (0,1) giving x = 0, y = 1 ., Hence,, , , sin90 1 ; cosec90° = 1, , , , cos 90 0 ;, , sec90° = not defined, , 1, tan90° = = not defined ;, cot 90 0, 0, Let us summarise all the results in the table given below:, , 230, , 6 Trigonometry.indd 230, , q, , 0°, , 30°, , 45°, , 60°, , sinq, , 0, , 1, 2, , 1, , cos q, , 1, , 2, , 3, 2, 1, 2, , tanq, , 0, , 1, , 3, , cosecq, , not defined, , secq, , 1, , cot q, , not defined, , Trigonometric, ratio, , 3, 2, 1, 3, 2, 2, , 2, 1, , 2, , 3, , 2, , 3, , 1, , 2, 3, 2, 1, 3, , 90°, 1, 0, not, defined, 1, not, defined, 0, , 9th Standard Mathematics, , 26-12-2019 14:14:33
Page 237 :
www.tntextbooks.in, , Example 6.5, (iii), , Evaluate:, , (i) sin 30 cos 30 , Note, , tan 45 , tan 30 tan 60 , 2, , (ii) tan 60˚ cot 60˚, , (i) (sin q )2 is written as sin2 q sin q sin q , , 2, , (iv) sin 45 cos 45 , , (ii) (sin q )2 is not written as sinq 2 , because, it may mean as sin (q × q ) ., , Solution, , 3, 1 3, 1, (i) sin 30 cos 30 , , 2, 2 2, 1, 1, (ii) tan 60˚ cot 60˚ 3 , 3, 1, 1, tan 45 , 1, 3, (iii), , , , =, 2, 1 3, tan 30 tan 60 , 4, 1, 3, 1 3, , 3, 3 1, 3, 2, 2, 12, 12, 1 1, 1 1 , 2, 2, (iv) sin 45 cos 45 , , 1, , , 2, 2, 2 2, 2 2, 2, 2, , , , , , Thinking Corner, The set of three numbers are called as Pythagorean triplets as they form the sides, of a right angled triangle. For example,, (i) 3, 4, 5, , (ii) 5, 12, 13, , (iii) 7, 24, 25, , Multiply each number in any of the above Pythagorean triplet by a non-zero, constant. Verify whether each of the resultant set so obtained is also a Pythagorean, triplet or not., Example 6.6, (i), , Find the values of the following:, , (cos 0 sin 45 sin 30 )(sin 90 cos 45 cos 60 ), , 3, (ii) tan2 60 2 tan2 45 cot 2 30 2 sin2 30 cosec2 45 , 4, Solution, (i) (cos 0 sin 45 sin 30 )(sin 90 cos 45 cos 60 ), 1 1, 1 1 , , 1 , 1 , , 2 2 , 2 2, , , 6 Trigonometry.indd 231, , Trigonometry 231, , 26-12-2019 14:14:37
Page 238 :
www.tntextbooks.in, , 2 2 2 2 2 2 2 2 3 2 2 3 2 2 , , , , , , 2 2, 2 2, 2 2 2 2 , , , 18 4, 14, 7, , , , 2, 42 4, 4 2, , , , 3, (ii) tan2 60 2 tan2 45 cot 2 30 2 sin2 30 cosec2 45 , 4, 2, 2, 2, 2, 1 3, 2, 3 2(1)2 3 2 , 2 4, 1 3, 3 2 3 , 2 2, 4, 2 2 2 0, 2, , , , , , , , Note, (i) In a right angled triangle, if the angles are in the ratio 45° : 45° : 90° , then the, sides are in the ratio 1 : 1 : 2 ., (ii) Similarly, if the angles are in the ratio 30° : 60° : 90° , then the sides are in the, ratio 1 : 3 : 2 ., (The two set squares in your geometry box is one of the best example for the above two, types of triangles)., Exercise 6.2, 1., , 2., , Verify the following equalities:, (i) sin2 60 cos2 60 1 , , (iii), , cos 90 1 2 sin2 45 2 cos2 45 1, , (ii) 1 tan2 30 sec2 30 , , (iv), , sin 30 cos 60 cos 30 sin 60 sin 90 , , Find the value of the following:, (i), , tan 45 sec 60 5 sin 90 , , , cosec30 cot 45 2 cos 0 , , (ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° − cos 45°), (iii) sin2 30 2 cos3 60 3 tan 4 45 , 3., , Verify cos 3 A 4 cos3 A 3 cos A , when A 30 , , 4., , Find the value of 8 sin 2 x cos 4 x sin 6 x, when x 15 , , 232, , 6 Trigonometry.indd 232, , 9th Standard Mathematics, , 26-12-2019 14:14:40
Page 239 :
www.tntextbooks.in, , 6.3 Trigonometric Ratios for Complementary Angles, , C, q, , Recall that two acute angles are said to be complementary if, the sum of their measures is equal to 90° ., What can we say about the acute angles of a right-angled, triangle?, In a right angled triangle the sum of the two acute angles is A, equal to 90°. So, the two acute angles of a right angled triangle are, always complementary to each other., , 90 q, Fig. 6.13, , B, , In the above figure 6.13, the triangle is right-angled at B. Therefore, if ∠C is q , then, ∠A = 90° − q ., We find that, AB, AC, BC, cos q , AC, AB, tan q , BC, , sin q , , Similarly for the angle ( 90° – θ ), We have, AC , AB , , AC , sec q , ....(1), BC , BC , cot q , AB , , cosec q , , BC, AC, AB, cos(90 q ) , AC, BC, tan(90 q ) , AB, , sin(90 q ) , , AC , BC , , AC , sec(90 q ) , .....(2), AB , AB , cot(90 q ) , BC , , cosec(90 q ) , , Comparing (1) and (2), we get, sinq, , =, , cos(90 q ), , cosecq, , =, , sec(90 q ), , cos q, , =, , sin(90 q ), , secq, , =, , cosec(90 q ), , tanq, , =, , cot(90 q ), , cot q, , =, , tan(90 q ), , Example 6.7, , Express (i) sin74° in terms of cosine (ii) tan12° in terms of, cotangent (iii) cosec39° in terms of secant, Solution, (i), , sin74° sin(90 16 ), , (since, 90 16 74 ), , RHS is of the form sin(90 q ) cos q, Therefore sin74° cos16 , (ii) tan12° tan(90 78 ), , (since, 12 90 78 ), , RHS is of the form tan(90 q ) cot q, Therefore tan12° cot 78 , Trigonometry 233, , 6 Trigonometry.indd 233, , 26-12-2019 14:14:46
Page 240 :
www.tntextbooks.in, , (iii) cosec39° cosec(90 51 ), , (since 39 90 51 ), , RHS is of the form cosec(90 q ) sec q, Therefore cosec39° sec51 , Example 6.8, , Evaluate:, , (i), , Solution, (i), , sin 49°, cos 41°, , (ii), , sec63°, cosec 27°, , sin 49°, cos 41°, , sin 49 sin(90 41 ) cos 41 , since 49 41 90 (complementary),, cos 41 , Hence on substituting sin 49 cos 41 we get,, 1, cos 41 , sec63°, (ii), cosec 27°, sec 63 sec(90 27 ) cosec 27 , here 63° and 27° are complementary angles., we have, Example 6.9, , cosec 27 , sec63 , , 1, cosec 27 cosec 27 , Find the values of (i) tan 7° tan 23° tan 60° tan 67° tan 83°, (ii), , Solution, (i), , cos 35 sin 12 cos 18 , , , sin 55 cos 78 sin 72 , , tan 7° tan 23° tan 60° tan 67° tan 83°, tan 7 tan 83 tan 23 tan 67 tan 60 , , (Grouping complementary angles), , tan 7 tan(90 7 )tan 23 tan(90 23 )tan 60 , (tan 7 .cot 7 )(tan 23 .cot 23 )tan 60 , (1) (1) tan 60 , tan60 , (ii), , 234, , 6 Trigonometry.indd 234, , = 3, , cos 35 sin 12 cos 18 , , , sin 55 cos 78 sin 72 , , , since, , , cos(90 55 ) sin(90 78 ) cos(90 72 ) cos 35° = cos(90° − 55°), , , , sin 12° = sin(90° − 78°) , sin 55 , cos 78 , sin 72 , , , cos 18° = cos(90° − 72°) , , 9th Standard Mathematics, , 26-12-2019 14:14:50
Page 241 :
www.tntextbooks.in, , , , sin 55 cos 78 sin 72 , , , sin 55 cos 78 sin 72 , , 1 1 1 1, Example 6.10, , (i) If cosec A sec 34 , then find A (ii) If tan B cot 47 , then find B., , Solution, , We know that cosec A sec(90 A), , (i), , (ii) We know that tan B cot(90 B), cot(90 B) cot 47 , , sec(90 A) sec(34 ), , 90 B 47 , , 90 A 34 , , B = 90° − 47°, , We get, , A= 90° − 34°, , We get, , B 43 , , A 56 , Exercise 6.3, Find the value of the following:, 2, , 2, , cos 70 cos 59 , cos q, cos 47 sin 72 , , , , (i) , 2 cos2 45 (ii), 8 cos2 60 , , , sin 20 sin 31 sin(90 q ), sin 43 cos 18 , (iii) tan 15° tan 30° tan 45° tan 60° tan 75°, (iv), , cot q, cos(90° − q )tan q sec(90° − q ), +, tan(90° − q ) sin(90° − q )cot(90° − q )cosec(90° − q ), , Thinking Corner, (i) What are the minimum and maximum values of sinq ?, (ii) What are the minimum and maximum values of cos q ?, , 6.4 Method of using Trigonometric Table, We have learnt to calculate the trigonometric ratios for angles 0°, 30°, 45°, 60° and, 90°. But during certain situations we need to calculate the trigonometric ratios of all the, other acute angles. Hence we need to know the method of using trigonometric tables., One degree (1°) is divided into 60 minutes ( 60′ ) and one minute ( 1′ ) is divided into, 60 seconds ( 60′′ ). Thus, 1° = 60′ and 1 60 ., Trigonometry 235, , 6 Trigonometry.indd 235, , 26-12-2019 14:14:54
Page 242 :
www.tntextbooks.in, , The trigonometric tables give the values, correct to four places of decimals for the, angles from 0° to 90° spaced at intervals of 60′ . A trigonometric table consists of three parts., A column on the extreme left which contains degrees from 0° to 90°, followed by ten, columns headed by 0′ , 6′ , 12′ , 18′ , 24′ , 30′ , 36′ , 42′ , 48′ and 54′ ., Five columns under the head mean difference has values from 1,2,3,4 and 5., For angles containing other measures of minutes (that is other than 0′ , 6′ , 12′ , 18′ ,, 24′ , 30′ , 36′ , 42′ , 48′ and 54′ ), the appropriate adjustment is obtained from the mean, difference columns., The mean difference is to be added in the case of sine and tangent while it is to be, subtracted in the case of cosine., Now let us understand the calculation of values of trigonometric angle from the, following examples., Example 6.11, , Find the value of sin 64 34 ., , Solution, 0′, 6′ 12′ 18′ 24′, 0.0° 0.1° 0.2° 0.3° 0.4°, 64°, , 30′, 0.5°, , 36′ 42′ 48′ 54′, 0.6° 0.7° 0.8° 0.9°, , 0.9026, , Mean Difference, 1, , 2, , 3, , 4, , 5, , 5, , Write 64 34 64 30 4 , From the table we have, sin64 30 = 0.9026, Mean difference for, Example 6.12, , 4′ =, 5(Mean difference to be added for sine), sin64 34 = 0.9031, , Find the value of cos19 59 , , Solution, 0′, 6′ 12′ 18′ 24′ 30′ 36′ 42′ 48′, 0.0° 0.1° 0.2° 0.3° 0.4° 0.5° 0.6° 0.7° 0.8°, 19°, , 54′, 0.9°, 0.9403, , Mean Difference, 1 2 3 4 5, 5, , Write 19 59 19 54 5 , From the table we have,, cos19 54 = 0.9403, Mean difference for, , 236, , 6 Trigonometry.indd 236, , 5′ =, 5 (Mean difference to be subtracted for cosine), cos19 59 = 0.9398, , 9th Standard Mathematics, , 26-12-2019 14:14:59
Page 243 :
www.tntextbooks.in, , Example 6.13, , Find the value of tan70 13 , , Solution, 0′, , 6′, , 0.0° 0.1°, 70°, , 24′, , 30′, , 36′, , 42′, , 48′, , 54′, , Mean Difference, , 12′, , 18′, , 0.2°, , 0.3° 0.4° 0.5° 0.6° 0.7° 0.8° 0.9° 1, , 2.7776, , 2, , 3, , 4, , 5, , 26, , Write 70 13 70 12 1 , From the table we have, tan70 12 = 2.7776, 1′ =, Mean difference for, 26 (Mean difference to be added for tan), tan70 13 = 2.7802, Example 6.14, , Find the value of, (i) sin 38 36 tan 12 12 , , (ii) tan 60 25 cos 49 20 , , Solution, (i), , sin 38 36 tan 12 12 , , (ii), , tan 60 25 cos 49 20 , , sin38 36 = 0.6239, , tan60 25 1.7603 0.0012 = 1.7615, , tan12 12 = 0.2162, , cos 49 20 0.6521 0.0004 = 0.6517, tan 60 25 cos 49 20 = 1.1098, , sin38 36 + tan12 12 = 0.8401, Example 6.15, , Find the value of q if, (ii) cos q = 0.7656, (i) sin q = 0.9858, , Solution, (i) sin q = 0.9858 0.9857 0.0001, From the sine table 0.9857 80 18 , Mean difference 1 =, , 2′, , 0.9858 80 20 , sin q = 0.9858 sin80 20 , q 80 20 , , (ii) cos q = 0.7656 0.7660 0.0004, From the cosine table, 0.7660 = 40°0′, Mean difference 4 =, , 2′, , 0.7656 = 40°2′, cos q = 0.7656 = cos 40°2′, q = 40°2′, , Trigonometry 237, , 6 Trigonometry.indd 237, , 26-12-2019 14:15:05
Page 244 :
www.tntextbooks.in, , Example 6.16, , Find the area of the right angled triangle with hypotenuse 5cm and, one of the acute angle is 48 30 , Solution, From the figure,, AB, AC, AB, sin 48 30 =, 5, AB, 0.7490 =, 5, 5 × 0.7490 = AB, , BC, AC, BC, cos 48 30 =, 5, BC, 0.6626 =, 5, 0.6626 × 5 = BC, , sinq =, , A, , cos q =, , 5cm, , 48 30 , B, , BC = 3.313 cm, , AB = 3.7450 cm, , Fig. 6.14, , C, , 1, Area of right triangle = bh, 2, 1, BC AB, 2, 1, 3.3130 3.7450, 2, 1.6565 3.7450 = 6.2035925 cm2, Activity, Observe the steps in your home. Measure the breadth and the height of one step., Enter it in the following picture and measure the angle (of elevation) of that step., , A, A, , h, , h, B, , (i), (ii), , 238, , 6 Trigonometry.indd 238, , b, , C, , B, , b, , q, , C, , Compare the angles (of elevation) of different steps of same height and same, breadth and discuss your observation., Sometimes few steps may not be of same height. Compare the angles (of, elevation) of different steps of those different heights and same breadth and, dicuss your observation., , 9th Standard Mathematics, , 26-12-2019 14:15:09
Page 245 :
www.tntextbooks.in, , Exercise 6.4, 1., , Find the value of the following:, (i) sin 49° (ii) cos 74 39 (iii) tan54 26 (iv) sin21 21 (v) cos 33 53 (vi) tan70 17 , , 2., , 3., , Find the value of q if, (i) sin q = 0.9975, , (ii) cos q = 0.6763, , (iv) cos q = 0.0410, , (v) tan q = 7.5958, , (iii) tan q = 0.0720, , Find the value of the following:, (i) sin 65 39 cos 24 57 tan 10 10 , , (ii) tan 70 58 cos 15 26 sin 84 59 , , 4., , Find the area of a right triangle whose hypotenuse is 10cm and one of the acute, angle is 24 24 , , 5., , Find the angle made by a ladder of length 5m, with the ground, if one of its end is 4m away, from the wall and the other end is on the wall., , 6., , H, , In the given figure, HT shows the height of a, tree standing vertically. From a point P, the, angle of elevation of the top of the tree (that is, ∠P ) measures 42° and the distance to the tree, is 60 metres. Find the height of the tree., , 42°, T, , 60 m, , P, , Exercise 6.5, Multiple choice questions, 1., , 2., 3., , 4., , If sin30 x and cos 60 y , then x 2 + y 2 is, 1, (1), (2) 0, (3) sin90°, 2, , (4) cos 90°, , If tan q cot 37 , then the value of q is, (1) 37°, (2) 53°, (3) 90°, , (4) 1°, , The value of tan 72° tan18° is, (1) 0, (2) 1, , (4) 72°, , The value of, (1) cos 60°, , (3) 18°, , 2 tan 30 , is equal to, 1 tan2 30 , (2) sin60°, (3) tan60°, , (4) sin30°, Trigonometry 239, , 6 Trigonometry.indd 239, , 26-12-2019 14:15:14
Page 246 :
www.tntextbooks.in, , If 2 sin 2q = 3 , then the value of q is, , 5., , (1) 90°, , (2) 30°, , (3) 45°, , (4) 60°, , The value of 3 sin 70 sec 20 2 sin 49 sec 51 is, , 6., , (1) 2, 7., , (2) 3, , (3) 5, , (4) 6, , 1 tan2 45 , The value of, is, 1 tan2 45 , (1) 2, (2) 1, , (3) 0, , (4), , (1) 0, , (3) 2, , (4) 3, , 1, 2, The value of cosec(70 q ) sec(20 q ) tan(65 q ) cot(25 q ) is, , 8., 9., , (2) 1, , The value of tan1° tan 2° tan 3° ... tan 89° is, (1) 0, , (2) 1, , (3) 2, , (4), , 3, 2, , 1, 1, and cos b = , then the value of α + β is, 2, 2, (2) 90°, (3) 30°, (4) 60°, , Given that sina =, , 10., , (1) 0°, , Points to Remember, zz Trigonometric ratios are, , opposite side, , sinθ =, , hypotenuse, adjacent side, , cos θ =, tan θ =, , hypotenuse, opposite side, adjacent side, , cosec θ =, , hypotenuse, opposite side, , sec θ, , =, , hypotenuse, adjacent side, , cot θ, , =, , adjacent side, opposite side, , zz Reciprocal trigonometric ratios, , 1, , cos ec q, 1, cosec θ =, , sinq, zz Complementary angles, sinθ =, , 240, , 6 Trigonometry.indd 240, , 1, , sec q, 1, sec θ =, , cos q, , 1, cot q, 1, cot θ =, tanq, , cos θ =, , tan θ =, , sinq, , =, , cos(90 q ), , cosecq, , =, , sec(90 q ), , cos q, , =, , sin(90 q ), , secq, , =, , cosec(90 q ), , tanq, , =, , cot(90 q ), , cot q, , =, , tan(90 q ), , 9th Standard Mathematics, , 26-12-2019 14:15:18
Page 247 :
www.tntextbooks.in, , ICT Corner, Expected Result is shown, in this picture, , Step – 1, Open the Browser by typing the URL Link given below (or) Scan the QR Code. GeoGebra work, sheet named “Trigonometry” will open. There are three worksheets under the title Trigonometric, ratios and Complementary angles and kite problem., Step - 2, Move the sliders of the respective values to change the points and ratio. Work out the solution and, check., For the kite problem click on “NEW PROBLEM” to change the question and work it out. Click the, check box for solution to check your answer., Step 1, , Step 2, , Browse in the link, , Trigonometry: https://ggbm.at/hkwnccr6 or Scan the QR Code., , Trigonometry 241, , 6 Trigonometry.indd 241, , 26-12-2019 14:15:18
Page 248 :
www.tntextbooks.in, , Degree, , NATURAL SINES, 0´, , 6´, , 12´, , 18´, , 24´, , 30´, , 36´, , 42´, , 48´, , 54´, , 0.0˚, , 0.1˚, , 0.2˚, , 0.3˚, , 0.4˚, , 0.5˚, , 0.6˚, , 0.7˚, , 0.8˚, , 0.9˚, , 1, , 2, , 3, , 4, , 5, , 0, 1, 2, 3, 4, , 0.0000, 0.0175, 0.0349, 0.0523, 0.0698, , 0.0017, 0.0192, 0.0366, 0.0541, 0.0715, , 0.0035, 0.0209, 0.0384, 0.0558, 0.0732, , 0.0052, 0.0227, 0.0401, 0.0576, 0.0750, , 0.0070, 0.0244, 0.0419, 0.0593, 0.0767, , 0.0087, 0.0262, 0.0436, 0.0610, 0.0785, , 0.0105, 0.0279, 0.0454, 0.0628, 0.0802, , 0.0122, 0.0297, 0.0471, 0.0645, 0.0819, , 0.0140, 0.0314, 0.0488, 0.0663, 0.0837, , 0.0157, 0.0332, 0.0506, 0.0680, 0.0854, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 15, , 5, 6, 7, 8, 9, , 0.0872, 0.1045, 0.1219, 0.1392, 0.1564, , 0.0889, 0.1063, 0.1236, 0.1409, 0.1582, , 0.0906, 0.1080, 0.1253, 0.1426, 0.1599, , 0.0924, 0.1097, 0.1271, 0.1444, 0.1616, , 0.0941, 0.1115, 0.1288, 0.1461, 0.1633, , 0.0958, 0.1132, 0.1305, 0.1478, 0.1650, , 0.0976, 0.1149, 0.1323, 0.1495, 0.1668, , 0.0993, 0.1167, 0.1340, 0.1513, 0.1685, , 0.1011, 0.1184, 0.1357, 0.1530, 0.1702, , 0.1028, 0.1201, 0.1374, 0.1547, 0.1719, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 14, 14, 14, 14, 14, , 10, 11, 12, 13, 14, , 0.1736, 0.1908, 0.2079, 0.2250, 0.2419, , 0.1754, 0.1925, 0.2096, 0.2267, 0.2436, , 0.1771, 0.1942, 0.2113, 0.2284, 0.2453, , 0.1788, 0.1959, 0.2130, 0.2300, 0.2470, , 0.1805, 0.1977, 0.2147, 0.2317, 0.2487, , 0.1822, 0.1994, 0.2164, 0.2334, 0.2504, , 0.1840, 0.2011, 0.2181, 0.2351, 0.2521, , 0.1857, 0.2028, 0.2198, 0.2368, 0.2538, , 0.1874, 0.2045, 0.2215, 0.2385, 0.2554, , 0.1891, 0.2062, 0.2233, 0.2402, 0.2571, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 8, 8, , 12, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 15, 16, 17, 18, 19, , 0.2588, 0.2756, 0.2924, 0.3090, 0.3256, , 0.2605, 0.2773, 0.2940, 0.3107, 0.3272, , 0.2622, 0.2790, 0.2957, 0.3123, 0.3289, , 0.2639, 0.2807, 0.2974, 0.3140, 0.3305, , 0.2656, 0.2823, 0.2990, 0.3156, 0.3322, , 0.2672, 0.2840, 0.3007, 0.3173, 0.3338, , 0.2689, 0.2857, 0.3024, 0.3190, 0.3355, , 0.2706, 0.2874, 0.3040, 0.3206, 0.3371, , 0.2723, 0.2890, 0.3057, 0.3223, 0.3387, , 0.2740, 0.2907, 0.3074, 0.3239, 0.3404, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 5, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 20, 21, 22, 23, 24, , 0.3420, 0.3584, 0.3746, 0.3907, 0.4067, , 0.3437, 0.3600, 0.3762, 0.3923, 0.4083, , 0.3453, 0.3616, 0.3778, 0.3939, 0.4099, , 0.3469, 0.3633, 0.3795, 0.3955, 0.4115, , 0.3486, 0.3649, 0.3811, 0.3971, 0.4131, , 0.3502, 0.3665, 0.3827, 0.3987, 0.4147, , 0.3518, 0.3681, 0.3843, 0.4003, 0.4163, , 0.3535, 0.3697, 0.3859, 0.4019, 0.4179, , 0.3551, 0.3714, 0.3875, 0.4035, 0.4195, , 0.3567, 0.3730, 0.3891, 0.4051, 0.4210, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 13, , 25, 26, 27, 28, 29, , 0.4226, 0.4384, 0.4540, 0.4695, 0.4848, , 0.4242, 0.4399, 0.4555, 0.4710, 0.4863, , 0.4258, 0.4415, 0.4571, 0.4726, 0.4879, , 0.4274, 0.4431, 0.4586, 0.4741, 0.4894, , 0.4289, 0.4446, 0.4602, 0.4756, 0.4909, , 0.4305, 0.4462, 0.4617, 0.4772, 0.4924, , 0.4321, 0.4478, 0.4633, 0.4787, 0.4939, , 0.4337, 0.4493, 0.4648, 0.4802, 0.4955, , 0.4352, 0.4509, 0.4664, 0.4818, 0.4970, , 0.4368, 0.4524, 0.4679, 0.4833, 0.4985, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 11, 10, 10, 10, 10, , 13, 13, 13, 13, 13, , 30, 31, 32, 33, 34, , 0.5000, 0.5150, 0.5299, 0.5446, 0.5592, , 0.5015, 0.5165, 0.5314, 0.5461, 0.5606, , 0.5030, 0.5180, 0.5329, 0.5476, 0.5621, , 0.5045, 0.5195, 0.5344, 0.5490, 0.5635, , 0.5060, 0.5210, 0.5358, 0.5505, 0.5650, , 0.5075, 0.5225, 0.5373, 0.5519, 0.5664, , 0.5090, 0.5240, 0.5388, 0.5534, 0.5678, , 0.5105, 0.5255, 0.5402, 0.5548, 0.5693, , 0.5120, 0.5270, 0.5417, 0.5563, 0.5707, , 0.5135, 0.5284, 0.5432, 0.5577, 0.5721, , 3, 2, 2, 2, 2, , 5, 5, 5, 5, 5, , 8, 7, 7, 7, 7, , 10, 10, 10, 10, 10, , 13, 12, 12, 12, 12, , 35, 36, 37, 38, 39, , 0.5736, 0.5878, 0.6018, 0.6157, 0.6293, , 0.5750, 0.5892, 0.6032, 0.6170, 0.6307, , 0.5764, 0.5906, 0.6046, 0.6184, 0.6320, , 0.5779, 0.5920, 0.6060, 0.6198, 0.6334, , 0.5793, 0.5934, 0.6074, 0.6211, 0.6347, , 0.5807, 0.5948, 0.6088, 0.6225, 0.6361, , 0.5821, 0.5962, 0.6101, 0.6239, 0.6374, , 0.5835, 0.5976, 0.6115, 0.6252, 0.6388, , 0.5850, 0.5990, 0.6129, 0.6266, 0.6401, , 0.5864, 0.6004, 0.6143, 0.6280, 0.6414, , 2, 2, 2, 2, 2, , 5, 5, 5, 5, 4, , 7, 7, 7, 7, 7, , 10, 9, 9, 9, 9, , 12, 12, 12, 11, 11, , 40, 41, 42, 43, 44, , 0.6428, 0.6561, 0.6691, 0.6820, 0.6947, , 0.6441, 0.6574, 0.6704, 0.6833, 0.6959, , 0.6455, 0.6587, 0.6717, 0.6845, 0.6972, , 0.6468, 0.6600, 0.6730, 0.6858, 0.6984, , 0.6481, 0.6613, 0.6743, 0.6871, 0.6997, , 0.6494, 0.6626, 0.6756, 0.6884, 0.7009, , 0.6508, 0.6639, 0.6769, 0.6896, 0.7022, , 0.6521, 0.6652, 0.6782, 0.6909, 0.7034, , 0.6534, 0.6665, 0.6794, 0.6921, 0.7046, , 0.6547, 0.6678, 0.6807, 0.6934, 0.7059, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 7, 7, 6, 6, 6, , 9, 9, 9, 8, 8, , 11, 11, 11, 11, 10, , 242, , 6 Trigonometry.indd 242, , Mean Difference, , 9th Standard Mathematics, , 26-12-2019 14:15:19
Page 249 :
www.tntextbooks.in, , NATURAL SINES, 6´, , 12´, , 18´, , 24´, , 30´, , 36´, , 42´, , 48´, , 54´, , Mean Difference, , Degree, , 0´, 0.0˚, , 0.1˚, , 0.2˚, , 0.3˚, , 0.4˚, , 0.5˚, , 0.6˚, , 0.7˚, , 0.8˚, , 0.9˚, , 1, , 2, , 3, , 4, , 5, , 45, 46, 47, 48, 49, , 0.7071, 0.7193, 0.7314, 0.7431, 0.7547, , 0.7083, 0.7206, 0.7325, 0.7443, 0.7559, , 0.7096, 0.7218, 0.7337, 0.7455, 0.7570, , 0.7108, 0.7230, 0.7349, 0.7466, 0.7581, , 0.7120, 0.7242, 0.7361, 0.7478, 0.7593, , 0.7133, 0.7254, 0.7373, 0.7490, 0.7604, , 0.7145, 0.7266, 0.7385, 0.7501, 0.7615, , 0.7157, 0.7278, 0.7396, 0.7513, 0.7627, , 0.7169, 0.7290, 0.7408, 0.7524, 0.7638, , 0.7181, 0.7302, 0.7420, 0.7536, 0.7649, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 10, 10, 10, 10, 9, , 50, 51, 52, 53, 54, , 0.7660, 0.7771, 0.7880, 0.7986, 0.8090, , 0.7672, 0.7782, 0.7891, 0.7997, 0.8100, , 0.7683, 0.7793, 0.7902, 0.8007, 0.8111, , 0.7694, 0.7804, 0.7912, 0.8018, 0.8121, , 0.7705, 0.7815, 0.7923, 0.8028, 0.8131, , 0.7716, 0.7826, 0.7934, 0.8039, 0.8141, , 0.7727, 0.7837, 0.7944, 0.8049, 0.8151, , 0.7738, 0.7848, 0.7955, 0.8059, 0.8161, , 0.7749, 0.7859, 0.7965, 0.8070, 0.8171, , 0.7760, 0.7869, 0.7976, 0.8080, 0.8181, , 2, 2, 2, 2, 2, , 4, 4, 4, 3, 3, , 6, 5, 5, 5, 5, , 7, 7, 7, 7, 7, , 9, 9, 9, 9, 8, , 55, 56, 57, 58, 59, , 0.8192, 0.8290, 0.8387, 0.8480, 0.8572, , 0.8202, 0.8300, 0.8396, 0.8490, 0.8581, , 0.8211, 0.8310, 0.8406, 0.8499, 0.8590, , 0.8221, 0.8320, 0.8415, 0.8508, 0.8599, , 0.8231, 0.8329, 0.8425, 0.8517, 0.8607, , 0.8241, 0.8339, 0.8434, 0.8526, 0.8616, , 0.8251, 0.8348, 0.8443, 0.8536, 0.8625, , 0.8261, 0.8358, 0.8453, 0.8545, 0.8634, , 0.8271, 0.8368, 0.8462, 0.8554, 0.8643, , 0.8281, 0.8377, 0.8471, 0.8563, 0.8652, , 2, 2, 2, 2, 1, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 4, , 7, 6, 6, 6, 6, , 8, 8, 8, 8, 7, , 60, 61, 62, 63, 64, , 0.8660, 0.8746, 0.8829, 0.8910, 0.8988, , 0.8669, 0.8755, 0.8838, 0.8918, 0.8996, , 0.8678, 0.8763, 0.8846, 0.8926, 0.9003, , 0.8686, 0.8771, 0.8854, 0.8934, 0.9011, , 0.8695, 0.8780, 0.8862, 0.8942, 0.9018, , 0.8704, 0.8788, 0.8870, 0.8949, 0.9026, , 0.8712, 0.8796, 0.8878, 0.8957, 0.9033, , 0.8721, 0.8805, 0.8886, 0.8965, 0.9041, , 0.8729, 0.8813, 0.8894, 0.8973, 0.9048, , 0.8738, 0.8821, 0.8902, 0.8980, 0.9056, , 1, 1, 1, 1, 1, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 6, 6, 5, 5, 5, , 7, 7, 7, 6, 6, , 65, 66, 67, 68, 69, , 0.9063, 0.9135, 0.9205, 0.9272, 0.9336, , 0.9070, 0.9143, 0.9212, 0.9278, 0.9342, , 0.9078, 0.9150, 0.9219, 0.9285, 0.9348, , 0.9085, 0.9157, 0.9225, 0.9291, 0.9354, , 0.9092, 0.9164, 0.9232, 0.9298, 0.9361, , 0.9100, 0.9171, 0.9239, 0.9304, 0.9367, , 0.9107, 0.9178, 0.9245, 0.9311, 0.9373, , 0.9114, 0.9184, 0.9252, 0.9317, 0.9379, , 0.9121, 0.9191, 0.9259, 0.9323, 0.9385, , 0.9128, 0.9198, 0.9265, 0.9330, 0.9391, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 4, 3, 3, 3, 3, , 5, 5, 4, 4, 4, , 6, 6, 6, 5, 5, , 70, 71, 72, 73, 74, , 0.9397, 0.9455, 0.9511, 0.9563, 0.9613, , 0.9403, 0.9461, 0.9516, 0.9568, 0.9617, , 0.9409, 0.9466, 0.9521, 0.9573, 0.9622, , 0.9415, 0.9472, 0.9527, 0.9578, 0.9627, , 0.9421, 0.9478, 0.9532, 0.9583, 0.9632, , 0.9426, 0.9483, 0.9537, 0.9588, 0.9636, , 0.9432, 0.9489, 0.9542, 0.9593, 0.9641, , 0.9438, 0.9494, 0.9548, 0.9598, 0.9646, , 0.9444, 0.9500, 0.9553, 0.9603, 0.9650, , 0.9449, 0.9505, 0.9558, 0.9608, 0.9655, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 2, 2, , 4, 4, 3, 3, 3, , 5, 5, 4, 4, 4, , 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, , 0.9659, 0.9703, 0.9744, 0.9781, 0.9816, 0.9848, 0.9877, 0.9903, 0.9925, 0.9945, , 0.9664, 0.9707, 0.9748, 0.9785, 0.9820, 0.9851, 0.9880, 0.9905, 0.9928, 0.9947, , 0.9668, 0.9711, 0.9751, 0.9789, 0.9823, 0.9854, 0.9882, 0.9907, 0.9930, 0.9949, , 0.9673, 0.9715, 0.9755, 0.9792, 0.9826, 0.9857, 0.9885, 0.9910, 0.9932, 0.9951, , 0.9677, 0.9720, 0.9759, 0.9796, 0.9829, 0.9860, 0.9888, 0.9912, 0.9934, 0.9952, , 0.9681, 0.9724, 0.9763, 0.9799, 0.9833, 0.9863, 0.9890, 0.9914, 0.9936, 0.9954, , 0.9686, 0.9728, 0.9767, 0.9803, 0.9836, 0.9866, 0.9893, 0.9917, 0.9938, 0.9956, , 0.9690, 0.9732, 0.9770, 0.9806, 0.9839, 0.9869, 0.9895, 0.9919, 0.9940, 0.9957, , 0.9694, 0.9736, 0.9774, 0.9810, 0.9842, 0.9871, 0.9898, 0.9921, 0.9942, 0.9959, , 0.9699, 0.9740, 0.9778, 0.9813, 0.9845, 0.9874, 0.9900, 0.9923, 0.9943, 0.9960, , 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, , 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, , 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, , 85, 86, 87, 88, 89, , 0.9962, 0.9976, 0.9986, 0.9994, 0.9998, , 0.9963, 0.9977, 0.9987, 0.9995, 0.9999, , 0.9965, 0.9978, 0.9988, 0.9995, 0.9999, , 0.9966, 0.9979, 0.9989, 0.9996, 0.9999, , 0.9968, 0.9980, 0.9990, 0.9996, 0.9999, , 0.9969, 0.9981, 0.9990, 0.9997, 1.0000, , 0.9971, 0.9982, 0.9991, 0.9997, 1.0000, , 0.9972, 0.9983, 0.9992, 0.9997, 1.0000, , 0.9973, 0.9984, 0.9993, 0.9998, 1.0000, , 0.9974, 0.9985, 0.9993, 0.9998, 1.0000, , 0, 0, 0, 0, 0, , 0, 0, 0, 0, 0, , 1, 1, 0, 0, 0, , 1, 1, 1, 0, 0, , 1, 1, 1, 0, 0, , Trigonometry 243, , 6 Trigonometry.indd 243, , 26-12-2019 14:15:20
Page 250 :
www.tntextbooks.in, , NATURAL COSINES, (Numbers in mean difference columns to be subtracted, not added), 6´, , 12´, , 18´, , 24´, , 30´, , 36´, , 42´, , 48´, , 54´, , Mean Difference, , Degree, , 0´, 0.0˚, , 0.1˚, , 0.2˚, , 0.3˚, , 0.4˚, , 0.5˚, , 0.6˚, , 0.7˚, , 0.8˚, , 0.9˚, , 1, , 2, , 3, , 4, , 5, , 0, 1, 2, 3, 4, , 1.0000, 0.9998, 0.9994, 0.9986, 0.9976, , 1.0000, 0.9998, 0.9993, 0.9985, 0.9974, , 1.0000, 0.9998, 0.9993, 0.9984, 0.9973, , 1.0000, 0.9997, 0.9992, 0.9983, 0.9972, , 1.0000, 0.9997, 0.9991, 0.9982, 0.9971, , 1.0000, 0.9997, 0.9990, 0.9981, 0.9969, , 0.9999, 0.9996, 0.9990, 0.9980, 0.9968, , 0.9999, 0.9996, 0.9989, 0.9979, 0.9966, , 0.9999, 0.9995, 0.9988, 0.9978, 0.9965, , 0.9999, 0.9995, 0.9987, 0.9977, 0.9963, , 0, 0, 0, 0, 0, , 0, 0, 0, 0, 0, , 0, 0, 0, 1, 1, , 0, 0, 1, 1, 1, , 0, 0, 1, 1, 1, , 5, 6, 7, 8, 9, , 0.9962, 0.9945, 0.9925, 0.9903, 0.9877, , 0.9960, 0.9943, 0.9923, 0.9900, 0.9874, , 0.9959, 0.9942, 0.9921, 0.9898, 0.9871, , 0.9957, 0.9940, 0.9919, 0.9895, 0.9869, , 0.9956, 0.9938, 0.9917, 0.9893, 0.9866, , 0.9954, 0.9936, 0.9914, 0.9890, 0.9863, , 0.9952, 0.9934, 0.9912, 0.9888, 0.9860, , 0.9951, 0.9932, 0.9910, 0.9885, 0.9857, , 0.9949, 0.9930, 0.9907, 0.9882, 0.9854, , 0.9947, 0.9928, 0.9905, 0.9880, 0.9851, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 2, 2, 2, , 2, 2, 2, 2, 2, , 10, 11, 12, 13, 14, , 0.9848, 0.9816, 0.9781, 0.9744, 0.9703, , 0.9845, 0.9813, 0.9778, 0.9740, 0.9699, , 0.9842, 0.9810, 0.9774, 0.9736, 0.9694, , 0.9839, 0.9806, 0.9770, 0.9732, 0.9690, , 0.9836, 0.9803, 0.9767, 0.9728, 0.9686, , 0.9833, 0.9799, 0.9763, 0.9724, 0.9681, , 0.9829, 0.9796, 0.9759, 0.9720, 0.9677, , 0.9826, 0.9792, 0.9755, 0.9715, 0.9673, , 0.9823, 0.9789, 0.9751, 0.9711, 0.9668, , 0.9820, 0.9785, 0.9748, 0.9707, 0.9664, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 3, 3, 3, , 3, 3, 3, 3, 4, , 15, 16, 17, 18, 19, , 0.9659, 0.9613, 0.9563, 0.9511, 0.9455, , 0.9655, 0.9608, 0.9558, 0.9505, 0.9449, , 0.9650, 0.9603, 0.9553, 0.9500, 0.9444, , 0.9646, 0.9598, 0.9548, 0.9494, 0.9438, , 0.9641, 0.9593, 0.9542, 0.9489, 0.9432, , 0.9636, 0.9588, 0.9537, 0.9483, 0.9426, , 0.9632, 0.9583, 0.9532, 0.9478, 0.9421, , 0.9627, 0.9578, 0.9527, 0.9472, 0.9415, , 0.9622, 0.9573, 0.9521, 0.9466, 0.9409, , 0.9617, 0.9568, 0.9516, 0.9461, 0.9403, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 3, 3, 3, , 3, 3, 3, 4, 4, , 4, 4, 4, 5, 5, , 20, 21, 22, 23, 24, , 0.9397, 0.9336, 0.9272, 0.9205, 0.9135, , 0.9391, 0.9330, 0.9265, 0.9198, 0.9128, , 0.9385, 0.9323, 0.9259, 0.9191, 0.9121, , 0.9379, 0.9317, 0.9252, 0.9184, 0.9114, , 0.9373, 0.9311, 0.9245, 0.9178, 0.9107, , 0.9367, 0.9304, 0.9239, 0.9171, 0.9100, , 0.9361, 0.9298, 0.9232, 0.9164, 0.9092, , 0.9354, 0.9291, 0.9225, 0.9157, 0.9085, , 0.9348, 0.9285, 0.9219, 0.9150, 0.9078, , 0.9342, 0.9278, 0.9212, 0.9143, 0.9070, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 4, , 4, 4, 4, 5, 5, , 5, 5, 6, 6, 6, , 25, 26, 27, 28, 29, , 0.9063, 0.8988, 0.8910, 0.8829, 0.8746, , 0.9056, 0.8980, 0.8902, 0.8821, 0.8738, , 0.9048, 0.8973, 0.8894, 0.8813, 0.8729, , 0.9041, 0.8965, 0.8886, 0.8805, 0.8721, , 0.9033, 0.8957, 0.8878, 0.8796, 0.8712, , 0.9026, 0.8949, 0.8870, 0.8788, 0.8704, , 0.9018, 0.8942, 0.8862, 0.8780, 0.8695, , 0.9011, 0.8934, 0.8854, 0.8771, 0.8686, , 0.9003, 0.8926, 0.8846, 0.8763, 0.8678, , 0.8996, 0.8918, 0.8838, 0.8755, 0.8669, , 1, 1, 1, 1, 1, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 6, 6, , 6, 6, 7, 7, 7, , 30, 31, 32, 33, 34, , 0.8660, 0.8572, 0.8480, 0.8387, 0.8290, , 0.8652, 0.8563, 0.8471, 0.8377, 0.8281, , 0.8643, 0.8554, 0.8462, 0.8368, 0.8271, , 0.8634, 0.8545, 0.8453, 0.8358, 0.8261, , 0.8625, 0.8536, 0.8443, 0.8348, 0.8251, , 0.8616, 0.8526, 0.8434, 0.8339, 0.8241, , 0.8607, 0.8517, 0.8425, 0.8329, 0.8231, , 0.8599, 0.8508, 0.8415, 0.8320, 0.8221, , 0.8590, 0.8499, 0.8406, 0.8310, 0.8211, , 0.8581, 0.8490, 0.8396, 0.8300, 0.8202, , 1, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 5, 5, 5, 5, , 6, 6, 6, 6, 7, , 7, 8, 8, 8, 8, , 35, 36, 37, 38, 39, , 0.8192, 0.8090, 0.7986, 0.7880, 0.7771, , 0.8181, 0.8080, 0.7976, 0.7869, 0.7760, , 0.8171, 0.8070, 0.7965, 0.7859, 0.7749, , 0.8161, 0.8059, 0.7955, 0.7848, 0.7738, , 0.8151, 0.8049, 0.7944, 0.7837, 0.7727, , 0.8141, 0.8039, 0.7934, 0.7826, 0.7716, , 0.8131, 0.8028, 0.7923, 0.7815, 0.7705, , 0.8121, 0.8018, 0.7912, 0.7804, 0.7694, , 0.8111, 0.8007, 0.7902, 0.7793, 0.7683, , 0.8100, 0.7997, 0.7891, 0.7782, 0.7672, , 2, 2, 2, 2, 2, , 3, 3, 4, 4, 4, , 5, 5, 5, 5, 6, , 7, 7, 7, 7, 7, , 8, 9, 9, 9, 9, , 40, 41, 42, 43, 44, , 0.7660, 0.7547, 0.7431, 0.7314, 0.7193, , 0.7649, 0.7536, 0.7420, 0.7302, 0.7181, , 0.7638, 0.7524, 0.7408, 0.7290, 0.7169, , 0.7627, 0.7513, 0.7396, 0.7278, 0.7157, , 0.7615, 0.7501, 0.7385, 0.7266, 0.7145, , 0.7604, 0.7490, 0.7373, 0.7254, 0.7133, , 0.7593, 0.7478, 0.7361, 0.7242, 0.7120, , 0.7581, 0.7466, 0.7349, 0.7230, 0.7108, , 0.7570, 0.7455, 0.7337, 0.7218, 0.7096, , 0.7559, 0.7443, 0.7325, 0.7206, 0.7083, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 9, 10, 10, 10, 10, , 244, , 6 Trigonometry.indd 244, , 9th Standard Mathematics, , 26-12-2019 14:15:21
Page 251 :
www.tntextbooks.in, , NATURAL COSINES, (Numbers in mean difference columns to be subtracted, not added), 6´, , 12´, , 18´, , 24´, , 30´, , 36´, , 42´, , 48´, , 54´, , Mean Difference, , Degree, , 0´, 0.0˚, , 0.1˚, , 0.2˚, , 0.3˚, , 0.4˚, , 0.5˚, , 0.6˚, , 0.7˚, , 0.8˚, , 0.9˚, , 45, 46, 47, 48, 49, , 0.7071, 0.6947, 0.6820, 0.6691, 0.6561, , 0.7059, 0.6934, 0.6807, 0.6678, 0.6547, , 0.7046, 0.6921, 0.6794, 0.6665, 0.6534, , 0.7034, 0.6909, 0.6782, 0.6652, 0.6521, , 0.7022, 0.6896, 0.6769, 0.6639, 0.6508, , 0.7009, 0.6884, 0.6756, 0.6626, 0.6494, , 0.6997, 0.6871, 0.6743, 0.6613, 0.6481, , 0.6984, 0.6858, 0.6730, 0.6600, 0.6468, , 0.6972, 0.6845, 0.6717, 0.6587, 0.6455, , 0.6959, 0.6833, 0.6704, 0.6574, 0.6441, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 7, 7, , 8, 8, 9, 9, 9, , 10, 11, 11, 11, 11, , 50, 51, 52, 53, 54, , 0.6428, 0.6293, 0.6157, 0.6018, 0.5878, , 0.6414, 0.6280, 0.6143, 0.6004, 0.5864, , 0.6401, 0.6266, 0.6129, 0.5990, 0.5850, , 0.6388, 0.6252, 0.6115, 0.5976, 0.5835, , 0.6374, 0.6239, 0.6101, 0.5962, 0.5821, , 0.6361, 0.6225, 0.6088, 0.5948, 0.5807, , 0.6347, 0.6211, 0.6074, 0.5934, 0.5793, , 0.6334, 0.6198, 0.6060, 0.5920, 0.5779, , 0.6320, 0.6184, 0.6046, 0.5906, 0.5764, , 0.6307, 0.6170, 0.6032, 0.5892, 0.5750, , 2, 2, 2, 2, 2, , 4, 5, 5, 5, 5, , 7, 7, 7, 7, 7, , 9, 9, 9, 9, 9, , 11, 11, 12, 12, 12, , 55, 56, 57, 58, 59, , 0.5736, 0.5592, 0.5446, 0.5299, 0.5150, , 0.5721, 0.5577, 0.5432, 0.5284, 0.5135, , 0.5707, 0.5563, 0.5417, 0.5270, 0.5120, , 0.5693, 0.5548, 0.5402, 0.5255, 0.5105, , 0.5678, 0.5534, 0.5388, 0.5240, 0.5090, , 0.5664, 0.5519, 0.5373, 0.5225, 0.5075, , 0.5650, 0.5505, 0.5358, 0.5210, 0.5060, , 0.5635, 0.5490, 0.5344, 0.5195, 0.5045, , 0.5621, 0.5476, 0.5329, 0.5180, 0.5030, , 0.5606, 0.5461, 0.5314, 0.5165, 0.5015, , 2, 2, 2, 2, 3, , 5, 5, 5, 5, 5, , 7, 7, 7, 7, 8, , 10, 10, 10, 10, 10, , 12, 12, 12, 12, 13, , 60, 61, 62, 63, 64, , 0.5000, 0.4848, 0.4695, 0.4540, 0.4384, , 0.4985, 0.4833, 0.4679, 0.4524, 0.4368, , 0.4970, 0.4818, 0.4664, 0.4509, 0.4352, , 0.4955, 0.4802, 0.4648, 0.4493, 0.4337, , 0.4939, 0.4787, 0.4633, 0.4478, 0.4321, , 0.4924, 0.4772, 0.4617, 0.4462, 0.4305, , 0.4909, 0.4756, 0.4602, 0.4446, 0.4289, , 0.4894, 0.4741, 0.4586, 0.4431, 0.4274, , 0.4879, 0.4726, 0.4571, 0.4415, 0.4258, , 0.4863, 0.4710, 0.4555, 0.4399, 0.4242, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 10, 10, 10, 10, 11, , 13, 13, 13, 13, 13, , 65, 66, 67, 68, 69, , 0.4226, 0.4067, 0.3907, 0.3746, 0.3584, , 0.4210, 0.4051, 0.3891, 0.3730, 0.3567, , 0.4195, 0.4035, 0.3875, 0.3714, 0.3551, , 0.4179, 0.4019, 0.3859, 0.3697, 0.3535, , 0.4163, 0.4003, 0.3843, 0.3681, 0.3518, , 0.4147, 0.3987, 0.3827, 0.3665, 0.3502, , 0.4131, 0.3971, 0.3811, 0.3649, 0.3486, , 0.4115, 0.3955, 0.3795, 0.3633, 0.3469, , 0.4099, 0.3939, 0.3778, 0.3616, 0.3453, , 0.4083, 0.3923, 0.3762, 0.3600, 0.3437, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 13, 14, 14, 14, 14, , 70, 71, 72, 73, 74, , 0.3420, 0.3256, 0.3090, 0.2924, 0.2756, , 0.3404, 0.3239, 0.3074, 0.2907, 0.2740, , 0.3387, 0.3223, 0.3057, 0.2890, 0.2723, , 0.3371, 0.3206, 0.3040, 0.2874, 0.2706, , 0.3355, 0.3190, 0.3024, 0.2857, 0.2689, , 0.3338, 0.3173, 0.3007, 0.2840, 0.2672, , 0.3322, 0.3156, 0.2990, 0.2823, 0.2656, , 0.3305, 0.3140, 0.2974, 0.2807, 0.2639, , 0.3289, 0.3123, 0.2957, 0.2790, 0.2622, , 0.3272, 0.3107, 0.2940, 0.2773, 0.2605, , 3, 3, 3, 3, 3, , 5, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 75, 76, 77, 78, 79, , 0.2588, 0.2419, 0.2250, 0.2079, 0.1908, , 0.2571, 0.2402, 0.2233, 0.2062, 0.1891, , 0.2554, 0.2385, 0.2215, 0.2045, 0.1874, , 0.2538, 0.2368, 0.2198, 0.2028, 0.1857, , 0.2521, 0.2351, 0.2181, 0.2011, 0.1840, , 0.2504, 0.2334, 0.2164, 0.1994, 0.1822, , 0.2487, 0.2317, 0.2147, 0.1977, 0.1805, , 0.2470, 0.2300, 0.2130, 0.1959, 0.1788, , 0.2453, 0.2284, 0.2113, 0.1942, 0.1771, , 0.2436, 0.2267, 0.2096, 0.1925, 0.1754, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 8, 8, 9, 9, 9, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 80, 81, 82, 83, 84, , 0.1736, 0.1564, 0.1392, 0.1219, 0.1045, , 0.1719, 0.1547, 0.1374, 0.1201, 0.1028, , 0.1702, 0.1530, 0.1357, 0.1184, 0.1011, , 0.1685, 0.1513, 0.1340, 0.1167, 0.0993, , 0.1668, 0.1495, 0.1323, 0.1149, 0.0976, , 0.1650, 0.1478, 0.1305, 0.1132, 0.0958, , 0.1633, 0.1461, 0.1288, 0.1115, 0.0941, , 0.1616, 0.1444, 0.1271, 0.1097, 0.0924, , 0.1599, 0.1426, 0.1253, 0.1080, 0.0906, , 0.1582, 0.1409, 0.1236, 0.1063, 0.0889, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 14, 14, 14, 14, 14, , 85, 86, 87, 88, 89, , 0.0872, 0.0698, 0.0523, 0.0349, 0.0175, , 0.0854, 0.0680, 0.0506, 0.0332, 0.0157, , 0.0837, 0.0663, 0.0488, 0.0314, 0.0140, , 0.0819, 0.0645, 0.0471, 0.0297, 0.0122, , 0.0802, 0.0628, 0.0454, 0.0279, 0.0105, , 0.0785, 0.0610, 0.0436, 0.0262, 0.0087, , 0.0767, 0.0593, 0.0419, 0.0244, 0.0070, , 0.0750, 0.0576, 0.0401, 0.0227, 0.0052, , 0.0732, 0.0558, 0.0384, 0.0209, 0.0035, , 0.0715, 0.0541, 0.0366, 0.0192, 0.0017, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 15, , 1, , 2, , 3, , 4, , 5, , Trigonometry 245, , 6 Trigonometry.indd 245, , 26-12-2019 14:15:23
Page 252 :
www.tntextbooks.in, , NATURAL TANGENTS, 6´, , 12´, , 18´, , 24´, , 30´, , 36´, , 42´, , 48´, , 54´, , Mean Difference, , Degree, , 0´, 0.0˚, , 0.1˚, , 0.2˚, , 0.3˚, , 0.4˚, , 0.5˚, , 0.6˚, , 0.7˚, , 0.8˚, , 0.9˚, , 1, , 2, , 3, , 4, , 5, , 0, 1, 2, 3, 4, , 0.0000, 0.0175, 0.0349, 0.0524, 0.0699, , 0.0017, 0.0192, 0.0367, 0.0542, 0.0717, , 0.0035, 0.0209, 0.0384, 0.0559, 0.0734, , 0.0052, 0.0227, 0.0402, 0.0577, 0.0752, , 0.0070, 0.0244, 0.0419, 0.0594, 0.0769, , 0.0087, 0.0262, 0.0437, 0.0612, 0.0787, , 0.0105, 0.0279, 0.0454, 0.0629, 0.0805, , 0.0122, 0.0297, 0.0472, 0.0647, 0.0822, , 0.0140, 0.0314, 0.0489, 0.0664, 0.0840, , 0.0157, 0.0332, 0.0507, 0.0682, 0.0857, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 15, , 5, 6, 7, 8, 9, , 0.0875, 0.1051, 0.1228, 0.1405, 0.1584, , 0.0892, 0.1069, 0.1246, 0.1423, 0.1602, , 0.0910, 0.1086, 0.1263, 0.1441, 0.1620, , 0.0928, 0.1104, 0.1281, 0.1459, 0.1638, , 0.0945, 0.1122, 0.1299, 0.1477, 0.1655, , 0.0963, 0.1139, 0.1317, 0.1495, 0.1673, , 0.0981, 0.1157, 0.1334, 0.1512, 0.1691, , 0.0998, 0.1175, 0.1352, 0.1530, 0.1709, , 0.1016, 0.1192, 0.1370, 0.1548, 0.1727, , 0.1033, 0.1210, 0.1388, 0.1566, 0.1745, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 15, , 10, 11, 12, 13, 14, , 0.1763, 0.1944, 0.2126, 0.2309, 0.2493, , 0.1781, 0.1962, 0.2144, 0.2327, 0.2512, , 0.1799, 0.1980, 0.2162, 0.2345, 0.2530, , 0.1817, 0.1998, 0.2180, 0.2364, 0.2549, , 0.1835, 0.2016, 0.2199, 0.2382, 0.2568, , 0.1853, 0.2035, 0.2217, 0.2401, 0.2586, , 0.1871, 0.2053, 0.2235, 0.2419, 0.2605, , 0.1890, 0.2071, 0.2254, 0.2438, 0.2623, , 0.1908, 0.2089, 0.2272, 0.2456, 0.2642, , 0.1926, 0.2107, 0.2290, 0.2475, 0.2661, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 16, , 15, 16, 17, 18, 19, , 0.2679, 0.2867, 0.3057, 0.3249, 0.3443, , 0.2698, 0.2886, 0.3076, 0.3269, 0.3463, , 0.2717, 0.2905, 0.3096, 0.3288, 0.3482, , 0.2736, 0.2924, 0.3115, 0.3307, 0.3502, , 0.2754, 0.2943, 0.3134, 0.3327, 0.3522, , 0.2773, 0.2962, 0.3153, 0.3346, 0.3541, , 0.2792, 0.2981, 0.3172, 0.3365, 0.3561, , 0.2811, 0.3000, 0.3191, 0.3385, 0.3581, , 0.2830, 0.3019, 0.3211, 0.3404, 0.3600, , 0.2849, 0.3038, 0.3230, 0.3424, 0.3620, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 7, , 9, 9, 10, 10, 10, , 13, 13, 13, 13, 13, , 16, 16, 16, 16, 16, , 20, 21, 22, 23, 24, , 0.3640, 0.3839, 0.4040, 0.4245, 0.4452, , 0.3659, 0.3859, 0.4061, 0.4265, 0.4473, , 0.3679, 0.3879, 0.4081, 0.4286, 0.4494, , 0.3699, 0.3899, 0.4101, 0.4307, 0.4515, , 0.3719, 0.3919, 0.4122, 0.4327, 0.4536, , 0.3739, 0.3939, 0.4142, 0.4348, 0.4557, , 0.3759, 0.3959, 0.4163, 0.4369, 0.4578, , 0.3779, 0.3979, 0.4183, 0.4390, 0.4599, , 0.3799, 0.4000, 0.4204, 0.4411, 0.4621, , 0.3819, 0.4020, 0.4224, 0.4431, 0.4642, , 3, 3, 3, 3, 4, , 7, 7, 7, 7, 7, , 10, 10, 10, 10, 11, , 13, 13, 14, 14, 14, , 17, 17, 17, 17, 18, , 25, 26, 27, 28, 29, , 0.4663, 0.4877, 0.5095, 0.5317, 0.5543, , 0.4684, 0.4899, 0.5117, 0.5340, 0.5566, , 0.4706, 0.4921, 0.5139, 0.5362, 0.5589, , 0.4727, 0.4942, 0.5161, 0.5384, 0.5612, , 0.4748, 0.4964, 0.5184, 0.5407, 0.5635, , 0.4770, 0.4986, 0.5206, 0.5430, 0.5658, , 0.4791, 0.5008, 0.5228, 0.5452, 0.5681, , 0.4813, 0.5029, 0.5250, 0.5475, 0.5704, , 0.4834, 0.5051, 0.5272, 0.5498, 0.5727, , 0.4856, 0.5073, 0.5295, 0.5520, 0.5750, , 4, 4, 4, 4, 4, , 7, 7, 7, 8, 8, , 11, 11, 11, 11, 12, , 14, 15, 15, 15, 15, , 18, 18, 18, 19, 19, , 30, 31, 32, 33, 34, , 0.5774, 0.6009, 0.6249, 0.6494, 0.6745, , 0.5797, 0.6032, 0.6273, 0.6519, 0.6771, , 0.5820, 0.6056, 0.6297, 0.6544, 0.6796, , 0.5844, 0.6080, 0.6322, 0.6569, 0.6822, , 0.5867, 0.6104, 0.6346, 0.6594, 0.6847, , 0.5890, 0.6128, 0.6371, 0.6619, 0.6873, , 0.5914, 0.6152, 0.6395, 0.6644, 0.6899, , 0.5938, 0.6176, 0.6420, 0.6669, 0.6924, , 0.5961, 0.6200, 0.6445, 0.6694, 0.6950, , 0.5985, 0.6224, 0.6469, 0.6720, 0.6976, , 4, 4, 4, 4, 4, , 8, 8, 8, 8, 9, , 12, 12, 12, 13, 13, , 16, 16, 16, 17, 17, , 20, 20, 20, 21, 21, , 35, 36, 37, 38, 39, , 0.7002, 0.7265, 0.7536, 0.7813, 0.8098, , 0.7028, 0.7292, 0.7563, 0.7841, 0.8127, , 0.7054, 0.7319, 0.7590, 0.7869, 0.8156, , 0.7080, 0.7346, 0.7618, 0.7898, 0.8185, , 0.7107, 0.7373, 0.7646, 0.7926, 0.8214, , 0.7133, 0.7400, 0.7673, 0.7954, 0.8243, , 0.7159, 0.7427, 0.7701, 0.7983, 0.8273, , 0.7186, 0.7454, 0.7729, 0.8012, 0.8302, , 0.7212, 0.7481, 0.7757, 0.8040, 0.8332, , 0.7239, 0.7508, 0.7785, 0.8069, 0.8361, , 4, 5, 5, 5, 5, , 9, 9, 9, 9, 10, , 13, 14, 14, 14, 15, , 18, 18, 18, 19, 20, , 22, 23, 23, 24, 24, , 40, 41, 42, 43, 44, , 0.8391, 0.8693, 0.9004, 0.9325, 0.9657, , 0.8421, 0.8724, 0.9036, 0.9358, 0.9691, , 0.8451, 0.8754, 0.9067, 0.9391, 0.9725, , 0.8481, 0.8785, 0.9099, 0.9424, 0.9759, , 0.8511, 0.8816, 0.9131, 0.9457, 0.9793, , 0.8541, 0.8847, 0.9163, 0.9490, 0.9827, , 0.8571, 0.8878, 0.9195, 0.9523, 0.9861, , 0.8601, 0.8910, 0.9228, 0.9556, 0.9896, , 0.8632, 0.8941, 0.9260, 0.9590, 0.9930, , 0.8662, 0.8972, 0.9293, 0.9623, 0.9965, , 5, 5, 5, 6, 6, , 10, 10, 11, 11, 11, , 15, 16, 16, 17, 17, , 20, 21, 21, 22, 23, , 25, 26, 27, 28, 29, , 246, , 6 Trigonometry.indd 246, , 9th Standard Mathematics, , 26-12-2019 14:15:24
Page 253 :
www.tntextbooks.in, , Degree, , NATURAL TANGENTS, 0´, , 6´, , 12´, , 18´, , 24´, , 30´, , 36´, , 42´, , 48´, , 54´, , Mean Difference, , 0.0˚, , 0.1˚, , 0.2˚, , 0.3˚, , 0.4˚, , 0.5˚, , 0.6˚, , 0.7˚, , 0.8˚, , 0.9˚, , 1, , 2, , 3, , 4, , 5, , 45, 46, 47, 48, 49, , 1.0000, 1.0355, 1.0724, 1.1106, 1.1504, , 1.0035, 1.0392, 1.0761, 1.1145, 1.1544, , 1.0070, 1.0428, 1.0799, 1.1184, 1.1585, , 1.0105, 1.0464, 1.0837, 1.1224, 1.1626, , 1.0141, 1.0501, 1.0875, 1.1263, 1.1667, , 1.0176, 1.0538, 1.0913, 1.1303, 1.1708, , 1.0212, 1.0575, 1.0951, 1.1343, 1.1750, , 1.0247, 1.0612, 1.0990, 1.1383, 1.1792, , 1.0283, 1.0649, 1.1028, 1.1423, 1.1833, , 1.0319, 1.0686, 1.1067, 1.1463, 1.1875, , 6, 6, 6, 7, 7, , 12, 12, 13, 13, 14, , 18, 18, 19, 20, 21, , 24, 25, 25, 27, 28, , 30, 31, 32, 33, 34, , 50, 51, 52, 53, 54, , 1.1918, 1.2349, 1.2799, 1.3270, 1.3764, , 1.1960, 1.2393, 1.2846, 1.3319, 1.3814, , 1.2002, 1.2437, 1.2892, 1.3367, 1.3865, , 1.2045, 1.2482, 1.2938, 1.3416, 1.3916, , 1.2088, 1.2527, 1.2985, 1.3465, 1.3968, , 1.2131, 1.2572, 1.3032, 1.3514, 1.4019, , 1.2174, 1.2617, 1.3079, 1.3564, 1.4071, , 1.2218, 1.2662, 1.3127, 1.3613, 1.4124, , 1.2261, 1.2708, 1.3175, 1.3663, 1.4176, , 1.2305, 1.2753, 1.3222, 1.3713, 1.4229, , 7, 8, 8, 8, 9, , 14, 15, 16, 16, 17, , 22, 23, 24, 25, 26, , 29, 30, 31, 33, 34, , 36, 38, 39, 41, 43, , 55, 56, 57, 58, 59, , 1.4281, 1.4826, 1.5399, 1.6003, 1.6643, , 1.4335, 1.4882, 1.5458, 1.6066, 1.6709, , 1.4388, 1.4938, 1.5517, 1.6128, 1.6775, , 1.4442, 1.4994, 1.5577, 1.6191, 1.6842, , 1.4496, 1.5051, 1.5637, 1.6255, 1.6909, , 1.4550, 1.5108, 1.5697, 1.6319, 1.6977, , 1.4605, 1.5166, 1.5757, 1.6383, 1.7045, , 1.4659, 1.5224, 1.5818, 1.6447, 1.7113, , 1.4715, 1.5282, 1.5880, 1.6512, 1.7182, , 1.4770, 1.5340, 1.5941, 1.6577, 1.7251, , 9, 10, 10, 11, 11, , 18, 19, 20, 21, 23, , 27, 29, 30, 32, 34, , 36, 38, 40, 43, 45, , 45, 48, 50, 53, 56, , 60, 61, 62, 63, 64, , 1.7321, 1.8040, 1.8807, 1.9626, 2.0503, , 1.7391, 1.8115, 1.8887, 1.9711, 2.0594, , 1.7461, 1.8190, 1.8967, 1.9797, 2.0686, , 1.7532, 1.8265, 1.9047, 1.9883, 2.0778, , 1.7603, 1.8341, 1.9128, 1.9970, 2.0872, , 1.7675, 1.8418, 1.9210, 2.0057, 2.0965, , 1.7747, 1.8495, 1.9292, 2.0145, 2.1060, , 1.7820, 1.8572, 1.9375, 2.0233, 2.1155, , 1.7893, 1.8650, 1.9458, 2.0323, 2.1251, , 1.7966, 1.8728, 1.9542, 2.0413, 2.1348, , 12, 13, 14, 15, 16, , 24, 26, 27, 29, 31, , 36, 38, 41, 44, 47, , 48, 51, 55, 58, 63, , 60, 64, 68, 73, 78, , 65, 66, 67, 68, 69, , 2.1445, 2.2460, 2.3559, 2.4751, 2.6051, , 2.1543, 2.2566, 2.3673, 2.4876, 2.6187, , 2.1642, 2.2673, 2.3789, 2.5002, 2.6325, , 2.1742, 2.2781, 2.3906, 2.5129, 2.6464, , 2.1842, 2.2889, 2.4023, 2.5257, 2.6605, , 2.1943, 2.2998, 2.4142, 2.5386, 2.6746, , 2.2045, 2.3109, 2.4262, 2.5517, 2.6889, , 2.2148, 2.3220, 2.4383, 2.5649, 2.7034, , 2.2251, 2.3332, 2.4504, 2.5782, 2.7179, , 2.2355, 2.3445, 2.4627, 2.5916, 2.7326, , 17, 18, 20, 22, 24, , 34, 37, 40, 43, 47, , 51, 55, 60, 65, 71, , 68, 73, 79, 87, 95, , 85, 92, 99, 108, 119, , 70, 71, 72, 73, 74, , 2.7475, 2.9042, 3.0777, 3.2709, 3.4874, , 2.7625, 2.9208, 3.0961, 3.2914, 3.5105, , 2.7776, 2.9375, 3.1146, 3.3122, 3.5339, , 2.7929, 2.9544, 3.1334, 3.3332, 3.5576, , 2.8083, 2.9714, 3.1524, 3.3544, 3.5816, , 2.8239, 2.9887, 3.1716, 3.3759, 3.6059, , 2.8397, 3.0061, 3.1910, 3.3977, 3.6305, , 2.8556, 3.0237, 3.2106, 3.4197, 3.6554, , 2.8716, 3.0415, 3.2305, 3.4420, 3.6806, , 2.8878, 3.0595, 3.2506, 3.4646, 3.7062, , 26, 29, 32, 36, 41, , 52, 58, 64, 72, 81, , 78, 87, 96, 108, 122, , 104, 116, 129, 144, 163, , 131, 145, 161, 180, 204, , 75, 76, 77, 78, 79, , 3.7321, 4.0108, 4.3315, 4.7046, 5.1446, , 3.7583, 4.0408, 4.3662, 4.7453, 5.1929, , 3.7848, 4.0713, 4.4015, 4.7867, 5.2422, , 3.8118, 4.1022, 4.4373, 4.8288, 5.2924, , 3.8391, 4.1335, 4.4737, 4.8716, 5.3435, , 3.8667, 4.1653, 4.5107, 4.9152, 5.3955, , 3.8947, 4.1976, 4.5483, 4.9594, 5.4486, , 3.9232, 4.2303, 4.5864, 5.0045, 5.5026, , 3.9520, 4.2635, 4.6252, 5.0504, 5.5578, , 3.9812, 4.2972, 4.6646, 5.0970, 5.6140, , 46, 53, , 93, 107, , 139, 160, , 186, 213, , 232, 267, , 80, 81, 82, 83, 84, , 5.6713, 6.3138, 7.1154, 8.1443, 9.5144, , 5.7297, 6.3859, 7.2066, 8.2636, 9.6768, , 5.7894, 6.4596, 7.3002, 8.3863, 9.8448, , 5.8502, 6.5350, 7.3962, 8.5126, 10.0187, , 5.9124, 6.6122, 7.4947, 8.6427, 10.1988, , 5.9758, 6.6912, 7.5958, 8.7769, 10.3854, , 6.0405, 6.7720, 7.6996, 8.9152, 10.5789, , 6.1066, 6.8548, 7.8062, 9.0579, 10.7797, , 6.1742, 6.9395, 7.9158, 9.2052, 10.9882, , 6.2432, 7.0264, 8.0285, 9.3572, 11.2048, , 85, 86, 87, 88, 89, , 11.4301, 14.3007, 19.0811, 28.6363, 57.2900, , 11.6645, 14.6685, 19.7403, 30.1446, 63.6567, , 11.9087, 15.0557, 20.4465, 31.8205, 71.6151, , 12.1632, 15.4638, 21.2049, 33.6935, 81.8470, , 12.4288 12.7062, 15.8945 16.3499, 22.0217 22.9038, 35.8006 38.1885, 95.4895 114.5887, , 12.9962, 16.8319, 23.8593, 40.9174, 143.2371, , 13.2996, 17.3432, 24.8978, 44.0661, 190.9842, , 13.6174, 17.8863, 26.0307, 47.7395, 286.4777, , 13.9507, 18.4645, 27.2715, 52.0807, 572.9572, , Trigonometry 247, , 6 Trigonometry.indd 247, , 26-12-2019 14:15:25
Page 254 :
www.tntextbooks.in, , 7, , MENSURATION, The most beautiful plane figure is the circle and the, most beautiful solid figure is the sphere. - Pythagoras., , Heron of Alexandria was a Greek mathematician. He, wrote books on mathematics, mechanics and physics., His famous book ‘Metrica’ consists of three volumes., This book shows the way to calculate area and volume, of plane and solid figures. Heron has derived the, formula for the area of triangle when three sides are, Heron, A.D (C.E) 10-75, , given., , Learning Outcomes, ÂÂ To use Heron’s formula for calculating area of triangles and, quadrilaterals., , ÂÂ To find Total Surface Area (TSA), Lateral Surface Area (LSA) and, Volume of cuboids and cubes., , 7.1 Introduction, Mensuration is the branch of mathematics which deals with the study of areas and, volumes of different kinds of geometrical shapes. In the broadest sense, it is all about the, process of measurement., Mensuration is used in the field of architecture, medicine, construction, etc. It is, necessary for everyone to learn formulae used to find the perimeter and area of two, dimensional figures as well as the surface area and volume of three dimensional solids in, day to day life. In this chapter we deal with finding the area of triangles (using Heron’s, formula), surface area and volume of cuboids and cubes., For a closed plane figure (a quadrilateral or a triangle), what do we call the distance, around its boundary? What is the measure of the region covered inside the boundary?, In general, the area of a triangle is calculated by the formula, 248, , 7 Mensuration.indd 248, , 9th Standard Mathematics, , 26-12-2019 14:16:17
Page 255 :
www.tntextbooks.in, , Area =, , 1, × Base × Height sq. units, 2, , 1, × b × h sq. units, 2, where, b is base and h is height B, of the triangle., That is, A =, , A, , A, , h, , h, b, Fig. 7.1, , C, , B, , b, Fig. 7.2, , C, , From the above, we know how, to find the area of a triangle when its ‘base’ and ‘height’ (that is altitude) are given., , 7.2 Heron’s Formula, How will you find the area of a triangle, if the height is not known, but the lengths of the three sides are known?, For this, Heron has given a formula to find the area of a triangle., If a, b and c are the sides of a triangle, then, the area of a triangle = s(s − a)(s − b)(s − c) sq.units., a+b+c, where s =, , ‘s’ is the semi-perimeter (that is half, 2, of the perimeter) of the triangle., , A, , c, , B, , b, , a, , C, , Fig. 7.3, , Note, If we assume that the sides are of equal length that is a = b = c, then Heron’s formula, 3 2, will be, a sq.units, which is the area of an equilateral triangle., 4, , Example 7.1, , The lengths of sides of a triangular field are 28 m, 15 m and 41 m., Calculate the area of the field. Find the cost of levelling the field at the rate of ₹ 20 per m2., Solution, Let a = 28 m, b = 15 m and c = 41m, Then, s =, , a+b+c, 28 + 15 + 41 84, =, =, = 42 m, 2, 2, 2, , Area of triangular field, , = s(s − a)(s − b)(s − c), = 42(42 − 28)(42 − 15)(42 − 41), = 42 × 14 × 27 × 1, Mensuration 249, , 7 Mensuration.indd 249, , 26-12-2019 14:16:20
Page 256 :
www.tntextbooks.in, , = 2 × 3 × 7 × 7 × 2 × 3 × 3 × 3 ×1, = 2×3×7×3, = 126 m2, Given the cost of levelling is ₹ 20 per m2., The total cost of levelling the field = 20 × 126 = ₹ 2520., Example 7.2, , Three different triangular plots are available for sale in a locality., Each plot has a perimeter of 120 m. The side lengths are also given:, Shape of plot, , Perimeter, , Length of sides, , Right angled triangle, , 120 m, , 30 m, 40 m, 50 m, , Acute angled triangle, , 120 m, , 35 m, 40 m, 45 m, , Equilateral triangle, , 120 m, , 40 m, 40 m, 40 m, , Help the buyer to decide which among these will be more spacious., Solution, For clarity, let us draw a rough figure indicating the measurements:, , B, , Fig. 7.4, , 40 m, , m, 40, , 30 m, , 35 m, C, , C, , B, , Fig. 7.5, , m, , 40 m, , 40, , m, , m, , 50, , 45, , B, , A, , A, , A, , 40 m, , C, , Fig. 7.6, , 30 + 40 + 50, = 60 m, 2, 35 + 40 + 45, = 60 m, Fig.7.5, s =, 2, 40 + 40 + 40, Fig.7.6, s =, = 60 m, 2, Note that all the semi-perimeters are equal., , (i) The semi-perimeter of Fig.7.4, s =, , (ii) Area of triangle using Heron’s formula:, In Fig.7.4, Area of triangle = 60(60 − 30)(60 − 40)(60 − 50), = 60 × 30 × 20 × 10, 250, , 7 Mensuration.indd 250, , 9th Standard Mathematics, , 26-12-2019 14:16:21
Page 257 :
www.tntextbooks.in, , = 30 × 2 × 30 × 2 × 10 × 10, = 600 m2, In Fig.7.5, Area of triangle = 60(60 − 35)(60 − 40)(60 − 45), = 60 × 25 × 20 × 15, = 20 × 3 × 5 × 5 × 20 × 3 × 5, = 300 5, , ( Since, , 5 = 2.236 ), , = 670.8 m2, In Fig.7.6, Area of triangle = 60(60 − 40)(60 − 40)(60 − 40), = 60 × 20 × 20 × 20, = 3 × 20 × 20 × 20 × 20, = 400 3, , ( Since, , 3 = 1.732 ), , = 692.8 m2, We find that though the perimeters are same, the areas of the three triangular plots, are different. The area of the triangle in Fig 7.6 is the greatest among these; the buyer can, be suggested to choose this since it is more spacious., Note, If the perimeter of different types of triangles have the same value, among all the, types of triangles, the equilateral triangle possess the greatest area. We will learn more, about maximum areas in higher classes., , 7.3 Application of Heron’s Formula in Finding Areas of Quadrilaterals, A plane figure bounded by four line segments is, called a quadrilateral., Let ABCD be a quadrilateral. To find the area of, a quadrilateral, we divide the quadrilateral into two triangular, parts and use Heron’s formula to calculate the area of the, triangular parts., In Fig 7.7,, , C, , D, , A, Fig. 7.7, , B, , Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD, , Mensuration 251, , 7 Mensuration.indd 251, , 26-12-2019 14:16:24
Page 258 :
www.tntextbooks.in, , Example 7.3, , A farmer has a field in the shape of a rhombus. The perimeter of, the field is 400m and one of its diagonal is 120m. He wants to divide the field into two, equal parts to grow two different types of vegetables. Find the area of the field., C, , D, , Solution, Let ABCD be the rhombus., , 12, , 0m, , Its perimeter = 4 × side = 400 m, Therefore, each side of the rhombus = 100 m, , A, , 100 m, Fig. 7.8, , Given the length of the diagonal AC = 120 m, , B, , In ∆ABC, let a =100 m, b =100 m, c =120 m, s=, , a + b + c 100 + 100 + 120, =, = 160 m, 2, 2, , Area of ∆ABC = 160(160 − 100)(160 − 100)(160 − 120), = 160 × 60 × 60 × 40, = 40 × 2 × 2 × 60 × 60 × 40, = 40 × 2 × 60 = 4800 m2, Therefore, Area of the field ABCD = 2 × Area of ∆ABC = 2 × 4800 = 9600 m2, , Exercise 7.1, 1., , Using Heron’s formula, find the area of a triangle whose sides are, (i) 10 cm, 24 cm, 26 cm, , (ii) 1.8m, 8 m, 8.2 m, , The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area and, cost of levelling the ground at the rate of ₹ 20 per m2., , 3., , The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5:12:13, then, find the area of the plot., , 4., , Find the area of an equilateral triangle whose perimeter is 180 cm., , 5., , An advertisement board is in the form of an isosceles triangle with perimeter 36m, and each of the equal sides are 13 m. Find the cost of painting it at ₹ 17.50 per, B, square metre., Find the area of the unshaded region., , 4, , 16, , cm, , 12, , cm, , 6., , m, 2c, , D, , 2., , C, 252, , 7 Mensuration.indd 252, , 34 cm, , A, , 9th Standard Mathematics, , 26-12-2019 14:16:25
Page 259 :
www.tntextbooks.in, , 7., , Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm,, CD = 9 cm, AD = 14 cm and diagonal BD = 15 cm., , 8., , A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and, 17 m and the angle between the first two sides is a right angle. Find the area of the park., , 9., , A land is in the shape of rhombus. The perimeter of the land is 160 m and one of the, diagonal is 48 m. Find the area of the land., , 10., , The adjacent sides of a parallelogram measures 34 m, 20 m and the measure of one, of the diagonal is 42 m. Find the area of parallelogram., , 7.4 Surface Area of Cuboid and Cube, We have learnt in the earlier classes about 3-Dimension structures., The 3D shapes are those which do not lie completely in a plane. Any 3D, shape has dimensions namely length, breadth and height., 7.4.1 Cuboid and its Surface Area, Cuboid: A cuboid is a closed solid figure bounded by, six rectangular plane regions. For example, match box,, Brick, Book., A cuboid has 6 faces, 12 edges and 8 vertices., Ultimately, a cuboid has the shape of a rectangular box., , Face, Vertex, , Total Surface Area (TSA) of a cuboid is the sum of, the areas of all the faces that enclose the cuboid. If we, , Edge, , leave out the areas of the top and bottom of the cuboid, , Fig. 7.9, , we get what is known as its Lateral Surface Area (LSA)., In the Fig 7.10, l, b and h represents length, breadth and height respectively., (i) Total Surface Area (TSA) of a cuboid, Top and bottom, , 2 × lb, , Front and back, , 2 × bh, , Left and Right sides, , 2 × lh, , b, , l, , h, , = 2 (lb + bh + lh ) sq. units., (ii) Lateral Surface Area (LSA) of a cuboid, Front and back, , 2 × bh, , Left and Right sides, , 2 × lh, , Fig. 7.10, , = 2 (l+b)h sq. units., Mensuration 253, , 7 Mensuration.indd 253, , 26-12-2019 14:16:27
Page 260 :
www.tntextbooks.in, , We are using the concept of Lateral Surface Area (LSA) and Total Surface Area (TSA), in real life situations. For instance a room can be cuboidal in shape that has different, length, breadth and height. If we require to find areas of only the walls of a room, avoiding, floor and ceiling then we can use LSA. However if we want to find the surface area of the, whole room then we have to calculate the TSA., If the length, breadth and height of a cuboid are l, b and h respectively. Then, (i) Total Surface Area = 2 (lb + bh + lh ) sq.units., (ii) Lateral Surface Area = 2 (l+b)h sq.units., Note, The top and bottom area in a cuboid is independent of height. The total area of, , top and bottom is 2lb. Hence LSA is obtained by removing 2lb from 2(lb+bh+lh)., The units of length, breadth and height should be same while calculating surface, , area of the cuboid., Example 7.4, , Find the TSA and LSA of a cuboid whose length, breadth and height, are 7.5 m, 3 m and 5 m respectively., Solution, Given the dimensions of the cuboid;, that is length (l) = 7.5 m, breadth (b) = 3 m and height (h) = 5 m., TSA = 2(lb + bh + lh), , = 2(22.5 + 15 + 37.5), = 2 × 75, = 150 m2, , 5m, , = 2 [(7.5 × 3) + (3 × 5) + (7.5 × 5)], , 3m, , 7.5m, Fig. 7.11, , LSA = 2(l + b) × h, = 2(7.5 + 3) × 5, = 2 × 10.5 × 5, = 105 m2, 254, , 7 Mensuration.indd 254, , 9th Standard Mathematics, , 26-12-2019 14:16:29
Page 261 :
www.tntextbooks.in, , Example 7.5, , The length, breadth and height of a hall are 25 m, 15 m and 5 m, respectively. Find the cost of renovating its floor and four walls at the rate of ₹80 per m2., Solution, Here, length (l ) = 25 m, breadth (b) =15 m, height (h) = 5 m., 15 m, , Area of four walls = LSA of cuboid, = 2(l + b) × h, , 25, , = 2(25 + 15) × 5, , m, , 5m, , = 80 × 5 = 400 m2, Area of the floor = l × b, = 25 × 15, = 375 m2, , Fig. 7.12, , Total renovating area of the hall, = (Area of four walls + Area of the floor), = (400 + 375) m2 = 775 m2, Therefore, cost of renovating at the rate of ₹80 per m2 = 80 × 775, = ₹ 62,000, 7.4.2 Cube and its Surface Area, Cube: A cuboid whose length, breadth and height are all equal is called as a cube., That is a cube is a solid having six square faces. Here are some real-life examples., , Dice, , Ice cubes, , Sugar cubes, , Fig. 7.13, , C, , D, , A cube being a cuboid has 6 faces, 12 edges and, 8 vertices., Consider a cube whose sides are ‘a’ units as shown in, the Fig 7.14. Now,, , B, , A, a, , H, , G, , a, E, , a, Fig. 7.14, , F, , Mensuration 255, , 7 Mensuration.indd 255, , 26-12-2019 14:16:31
Page 262 :
www.tntextbooks.in, , (i) Total Surface Area of the cube, = sum of area of the faces (ABCD+EFGH+AEHD+BFGC+ABFE+CDHG), = (a2 + a2 + a2 + a2 + a2 + a2 ), = 6a2 sq. units, (ii) Lateral Surface Area of the cube, = sum of area of the faces (AEHD+BFGC+ABFE+CDHG), = (a2 + a2 + a2 + a2 ), = 4a2 sq. units, Thinking Corner, , If the side of a cube is a units, then,, (i) The Total Surface Area = 6a2 sq.units, (ii) The Lateral Surface Area = 4a2 sq.units, Example 7.6, , Can you get these formulae, from the corresponding, formula of Cuboid?, , Find the Total Surface Area and Lateral Surface Area of the cube,, , whose side is 5 cm., , The side of the cube (a), , = 5 cm, , Total Surface Area, , = 6a2 = 6(52 ) = 150 sq. cm, , Lateral Surface Area, , = 4a2 = 4(52 ) = 100 sq. cm, , Example 7.7, , 5 cm, , Solution, , 5 cm, , 5 cm, Fig. 7.15, , A cube has the Total Surface Area of 486 cm2. Find its lateral surface, , area., Solution, Here, Total Surface Area of the cube = 486 cm2, 486, and so, a2 = 81 ., 6, The side of the cube = 9 cm, 6a2 = 486 ⇒, , a2 =, , This gives a = 9., , Lateral Surface Area = 4a2 = 4 × 92 = 4 × 81 = 324 cm2, 256, , 7 Mensuration.indd 256, , 9th Standard Mathematics, , 26-12-2019 14:16:33
Page 263 :
www.tntextbooks.in, , Example 7.8, , Two identical cubes of side 7 cm are joined end to end. Find the, Total and Lateral surface area of the new resulting cuboid., Solution, Side of a cube = 7 cm, Now length of the resulting cuboid (l) = 7+7 =14 cm, Breadth (b) = 7 cm, Height (h) = 7 cm, So, Total Surface Area = 2(lb + bh + lh), = 2 (14 × 7 ) + (7 × 7 ) + (14 × 7 ), = 2(98 + 49 + 98), 7, , 7, , = 2 × 245, 7, , = 490 cm2, , 7, , 7, , 7, 7, , Fig. 7.16, , 7, 14, , Lateral Surface Area = 2(l + b) × h, = 2(14 + 7) × 7 = 2 × 21 × 7, = 294 cm2, , Exercise 7.2, 1., , Find the Total Surface Area and the Lateral Surface Area of a cuboid whose, dimensions are: length = 20 cm, breadth = 15 cm and height = 8 cm, , 2., , The dimensions of a cuboidal box are 6 m × 400 cm × 1.5 m., painting its entire outer surface at the rate of ₹22 per m2., , 3., , The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the, walls and ceiling at the rate of ₹8.50 per m2., , 4., , Find the TSA and LSA of the cube whose side is (i) 8 m (ii) 21 cm (iii) 7.5 cm, , 5., , If the total surface area of a cube is 2400 cm2 then, find its lateral surface area., , 6., , A cubical container of side 6.5 m is to be painted on the entire outer surface. Find, the area to be painted and the total cost of painting it at the rate of ₹24 per m2., , 7., , Three identical cubes of side 4 cm are joined end to end. Find the total surface area, and lateral surface area of the new resulting cuboid., , Find the cost of, , Mensuration 257, , 7 Mensuration.indd 257, , 26-12-2019 14:16:35
Page 264 :
www.tntextbooks.in, , 7.5 Volume of Cuboid and Cube, All of us have tasted 50 ml and 100 ml of ice cream. Take one such, 100 ml ice cream cup. This cup can contain 100 ml of water, which means, that the capacity or volume of that cup is 100 ml. Take a 100 ml cup and, find out how many such cups of water can fill a jug. If 10 such 100 ml cups, , (, , ), , can fill a jug then the capacity or volume of the jug is 1 litre 10 × 100 ml = 1000 ml = 1l ., Further check how many such jug of water can fill a bucket. That is the capacity or volume, of the bucket. Likewise we can calculate the, volume or capacity of any such things., , Note, Unit Cube :, , Volume is the measure of the amount of, space occupied by a three dimensional solid., Cubic centimetres (cm3) , cubic metres (m3) are, some cubic units to measure volume., , A cube with side, 1 unit., , 1, 1, 1, , Fig. 7.17, , Volume of the solid is the product of ‘base area’ and ‘height’. This can easily be, understood from a practical situation. You might have seen the bundles of A4 size paper., Each paper is rectangular in shape and has an area (=lb). When you pile them up, it, becomes a bundle in the form of a cuboid; h times lb make the cuboid., , Fig. 7.18, , Let the length, breadth and height of a cuboid be, l, b and h respectively., Then, volume of the cuboid, V = (cuboid’s base area) × height, = (l × b) × h = lbh cubic units, , height, , 7.5.1 Volume of a Cuboid, , length, , br, , d, ea, , th, , Fig. 7.19, , Note, The units of length, breadth and height should be same while calculating the volume, of a cuboid., 258, , 7 Mensuration.indd 258, , 9th Standard Mathematics, , 26-12-2019 14:16:36
Page 265 :
www.tntextbooks.in, , Example 7.9, , The length, breadth and height of a cuboid is 120 mm, 10 cm and, 8 cm respectively. Find the volume of 10 such cuboids., Solution, Since both breadth and height are given in cm, it is necessary to convert the length, also in cm., 120, = 12 cm and take b = 10 cm, h = 8 cm as such., 10, Volume of a cuboid = l × b × h, So we get, l = 120 mm =, , cm, , Volume of 10 such cuboids = 10 × 960, , 10, , = 960 cm3, , 8 cm, , = 12 × 10 × 8, , = 9600 cm3, Example 7.10, volume is 35840, , cm3., , 12 cm, Fig. 7.20, , The length, breadth and height of a cuboid are in the ratio 7:5:2. Its, Find its dimensions., , Solution, Let the dimensions of the cuboid be, l = 7 x , b = 5x and h = 2 x., Given that volume of cuboid = 35840 cm3, l × b × h = 35840, , T HINKING C ORNER, Each cuboid given below has the, same volume 120 cm3. Can you find, the missing dimensions?, , (7 x )(5x )(2 x ) = 35840, 70 x 3 = 35840, , (i), 5 cm, , 35840, 70, 3, x = 512, x3 =, , x =, , 3, , 8×8×8, , ?, Fig. 7.21, , (ii), , 8c, , m, , Length of cuboid = 7 x = 7 × 8 = 56 cm, , 4 cm, , x = 8 cm, , Breadth of cuboid = 5x = 5 × 8 = 40 cm, Height of cuboid = 2 x = 2 × 8 = 16 cm, , ?, Fig. 7.22, , 6c, , m, , Mensuration 259, , 7 Mensuration.indd 259, , 26-12-2019 14:16:39
Page 266 :
www.tntextbooks.in, , Example 7.11, , 1.6 m, , The dimensions of a fish tank, are 3.8 m × 2.5 m × 1.6 m. How many litres of water, it can hold?, 2.5, , Solution, , 3.8 m, Fig. 7.23, , m, , Length of the fish tank l =3.8 m, Breadth of the fish tank b =2.5 m ,, Volume of the fish tank, , Height of the fish tank h =1.6 m, , = l ×b×h, = 3. 8 × 2. 5 × 1. 6, = 15.2 m3, = 15.2 × 1000 litres, = 15200 litres, , Note, A few important conversions, 1 cm3 =1 ml, 1000 cm3 =1 litre,, 1m3 =1000 litres, , Example 7.12, , The dimensions of a sweet box are 22 cm × 18 cm × 10 cm. How, many such boxes can be packed in a carton of dimensions 1 m × 88 cm × 63 cm?, Solution, Here, the dimensions of a sweet box are Length (l) = 22cm, breadth (b) = 18cm,, height (h) = 10 cm., Volume of a sweet box = l × b × h, = 22 × 18 × 10 cm3, The dimensions of a carton are, 63 cm, , Length (l) = 1m= 100 cm, breadth (b) = 88 cm,, height (h) = 63 cm., Volume of the carton = l × b × h, , 88 cm, , 1m, Fig. 7.24, , = 100 × 88 × 63 cm3, volume of the carton, The number of sweet boxes packed =, volume of a sweet box, 100 × 88 × 63, =, 22 × 18 × 10, = 140 boxes, , 260, , 7 Mensuration.indd 260, , a, , It is easy to get the volume of a cube whose side is a units. Simply, put l = b = h = a in the formula for the volume of a cuboid. We get, volume of cube to be a3 cubic units., , a, , 7.5.2 Volume of a Cube, a, , Fig. 7.25, , 9th Standard Mathematics, , 26-12-2019 14:16:40
Page 267 :
www.tntextbooks.in, , If the side of a cube is ‘a’ units then the Volume of the cube (V) = a3 cubic units., , Note, For any two cubes, the following results are true., Ratio of surface areas, = (Ratio of sides)2, Ratio of volumes, = (Ratio of sides)3, (Ratio of surface areas)3 = (Ratio of volumes)2, , Example 7.13, , Find the volume of cube whose side is 10 cm., 10 cm, , Solution, Given that side (a) = 10 cm, = 10 × 10 × 10, , 10 cm, Fig. 7.26, , = 1000 cm3, Example 7.14, , 10, , cm, , volume of the cube = a3, , A cubical tank can hold 64,000 litres of water. Find the length of its, , side in metres., Solution, Let ‘a’ be the side of cubical tank., Here, volume of the tank = 64, 000 litres, i.e., a3 = 64, 000 =, , 64000, [since,1000 litres=1m3 ], 1000, , a3 = 64 m3, a = 3 64, , a=4 m, , Therefore, length of the side of the tank is 4 metres., Example 7.15, , The side of a metallic cube is 12 cm. It is melted and formed into a, cuboid whose length and breadth are 18 cm and 16 cm respectively. Find the height of the, cuboid., Mensuration 261, , 7 Mensuration.indd 261, , 26-12-2019 14:16:42
Page 268 :
www.tntextbooks.in, , Solution, Side (a) = 12 cm, , Cuboid, length, , (l) = 18cm, , 12 cm, , Cube, , breadth (b) = 16cm, height (h) = ?, 16, , ?, , Here, Volume of the Cuboid = Volume of the Cube, , cm, , l × b × h = a3, , 18 cm, , 18 × 16 × h = 12 × 12 × 12, , Fig. 7.27, , 12 × 12 × 12, h=, 18 × 16, h = 6 cm, Therefore, the height of the cuboid is 6 cm., , Activity, Take some square sheets of paper / chart paper of given dimension 18 cm × 18 cm., Remove the squares of same sizes from each corner of the given square paper and fold, up the flaps to make a open cuboidal box. Then tabulate the dimensions of each of the, cuboidal boxes made. Also find the volume each time and complete the table. The side, measures of corner squares that are to be removed is given in the table below., Dimensions of, boxes, l, b, h, , Volume, V, , 2 cm, 3 cm, , 18 cm, , Side of, the corner, square, , 4 cm, 5 cm, Observe the above table and answer the following:, (i) What is the greatest possible volume?, , 18 cm, Fig. 7.28, , (ii) What is the side of the square that when removed produces the greatest volume?, , 262, , 7 Mensuration.indd 262, , 9th Standard Mathematics, , 26-12-2019 14:16:43
Page 269 :
www.tntextbooks.in, , Exercise 7.3, 1., , Find the volume of a cuboid whose dimensions are, (i) length = 12 cm, breadth = 8 cm, height = 6 cm, (ii) length = 60 m, breadth = 25 m, height = 1.5 m, , 2., , The dimensions of a match box are 6 cm × 3.5 cm × 2.5 cm. Find the volume of a, packet containing 12 such match boxes., , 3., , The length, breadth and height of a chocolate box are in the ratio 5:4:3. If its volume, is 7500 cm3, then find its dimensions., , 4., , The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find, the capacity of the pond in litres., , 5., , The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be, required to build a wall of 20 m length, 48 cm breadth and 6 m height?, , 6., , The volume of a container is 1440 m3. The length and breadth of the container are 15 m, and 8 m respectively. Find its height., , 7., , Find the volume of a cube each of whose side is (i) 5 cm (ii) 3.5 m (iii) 21 cm, , 8., , A cubical milk tank can hold 125000 litres of milk. Find the length of its side in, metres., , 9., , A metallic cube with side 15 cm is melted and formed into a cuboid. If the length, and height of the cuboid is 25 cm and 9 cm respectively then find the breadth of the, cuboid., , Exercise 7.4, Multiple choice questions, 1., , The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is, (1) 60 cm, , 2., , (4) 15 cm, , (2) 6 cm2, , (3) 9 cm2, , (4) 12 cm2, , The perimeter of an equilateral triangle is 30 cm. The area is, (1) 10 3 cm2, , 4., , (3) 30 cm, , If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is, (1) 3 cm2, , 3., , (2) 45 cm, , (2) 12 3 cm2, , (3) 15 3 cm2 (4) 25 3 cm2, , The lateral surface area of a cube of side 12 cm is, (1) 144 cm2, , (2) 196 cm2, , (3) 576 cm2, , (4) 664 cm2, Mensuration 263, , 7 Mensuration.indd 263, , 26-12-2019 14:16:43
Page 270 :
www.tntextbooks.in, , 5., , If the lateral surface area of a cube is 600 cm2, then the total surface area is, (1) 150 cm2, , 6., , (3) 360 cm2, , (4) 600 cm2, , (3) 6:9, , (4) 16:36, , The volume of a cuboid is 660 cm3 and the area of the base is 33 cm2. Its height is, (2) 12 cm, , (3) 20 cm, , (4) 22 cm, , The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is, (1) 75 litres, , 10., , (2) 300 cm2, , (2) 4:9, , (1) 10 cm, 9., , (4) 1350 cm2, , If the ratio of the sides of two cubes are 2:3, then ratio of their surface areas will be, (1) 4:6, , 8., , (3) 900 cm2, , The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is, (1) 280 cm2, , 7., , (2) 400 cm2, , (2) 750 litres, , (3) 7500 litres, , (4) 75000 litres, , The number of bricks each measuring 50 cm × 30 cm × 20 cm that will be required, to build a wall whose dimensions are 5 m × 3 m × 2 m is, (1) 1000, , (2) 2000, , (3) 3000, , (4) 5000, , Points to Remember, , , If a, b and c, , , , If the length, breadth and height of the cuboid are l, b and h respectively, then, , are the sides of a triangle, then the area of a triangle, a+b+c, = s(s − a)(s − b)(s − c) sq.units, where s =, ., 2, , (i) Total Surface Area(TSA) = 2 (lb + bh + lh ) sq.units, (ii) Lateral Surface Area(LSA) = 2 (l + b ) h sq.units, , , If the side of a cube is ‘a’ units, then, (i) Total Surface Area(TSA) = 6a2 sq.units, (ii) Lateral Surface Area(LSA) = 4a2 sq.units, , 264, , 7 Mensuration.indd 264, , , , If the length, breadth and height of the cuboid are l, b and h respectively, then, the Volume of the cuboid (V ) = lbh cu.units, , , , If the side of a cube is ‘a’ units then, the Volume of the cube (V ) = a3 cu.units., , 9th Standard Mathematics, , 26-12-2019 14:16:45
Page 271 :
www.tntextbooks.in, , ICT Corner, Expected Result is shown, in this picture, , Step – 1, Open the Browser by typing the URL Link given below (or) Scan the QR Code. GeoGebra work, sheet named “Mensuration” will open. There are two worksheets under the title CUBE and, CUBOID., Step - 2, Click on “New Problem”. Volume, Lateral surface and Total surface area are asked. Work out the, solution, and click on the respective check box and check the answer., Step 1, , Step 2, , Browse in the link, , Mensuration: https://ggbm.at/czsby7ym or Scan the QR Code., Mensuration 265, , 7 Mensuration.indd 265, , 26-12-2019 14:16:46
Page 272 :
www.tntextbooks.in, , 8, , N, , m, 2, c, l , f, , X , , fx, f, , STATISTICS, , “Lack of statistics is to hide inconvenient facts.”, - Albert Bertilsson, Sir Ronald Aylmer Fisher was a British, Statistician and Biologist., , Also he, , was known as the Father of Modern, Statistics and Experimental Design., Fisher did experimental agricultural, Sir Ronald Aylmer Fisher, (AD (CE) 1890 - 1962), , research, which saved millions from starvation. He was, awarded the Linnean Society of London’s prestigious, Darwin-Wallace Medal in 1958., , Learning Outcomes, To recall different types of averages known already., To recall the methods of computing the Mean, Median and Mode for ungrouped, data., To compute the Mean, Median and Mode for the grouped data., , 8.1 Introduction, Statistics is the science of collecting, organising, analysing and interpreting data, in order to make decisions. In everyday life, we come across a wide range of quantitative, and qualitative information. These have profound impact on our lives., Data means the facts, mostly numerical, that are gathered; statistics implies collection, of data. We analyse the data to make decisions. The methods of statistics are tools to help, us in this., , 266, , 8-Stats.indd 266, , 9th Standard Mathematics, , 26-12-2019 14:18:22
Page 273 :
www.tntextbooks.in, , Cricket News, Team U19, , Matches Won, , India, Australia, Pakistan, Bangladesh, West Indies, South Africa, England, Sri Lanka, New Zealand, Zimbabwe, Ireland, Afghanistan, Namibia, Kenya, Canada, PNG, , 71, 67, 69, 64, 71, 61, 69, 68, 66, 62, 49, 24, 47, 17, 29, 41, , 52, 50, 50, 45, 44, 43, 40, 36, 30, 28, 16, 11, 9, 5, 4, 3, , Lost NR/Tied, 18, 15, 19, 17, 27, 17, 28, 31, 35, 34, 32, 13, 37, 12, 23, 38, , India’s, 2018 GDP Forecast, , Customer, Satisfaction Survey, Hotel Tamilnadu, , 0/1, 0/2, 0/0, 1/1, 0/0, 0/1, 0/1, 0/1, 0/1, 0/0, 1/0, 0/0, 1/0, 0/0, 1/1, 0/0, , Tell us how you were, satisfied with our service, , 7.3%, 7.5%, , Satisfied, , HSBC, , 7%, , Neutral, , Bank of America, , 7.2%, , Unsatisfied, , Merill Lynch, , 7.5%, , Annoyed, , Goldman Sachs, , 8%, , Progress Check, , For example, study the marks obtained by 50, students in mathematics in an examination, given, below:, 31, 60, 25, , World Bank, , 7.6%, , When data are initially collected and before it is, edited and not processed for use, they are known as Raw, data. It will not be of much use because it would be too, much for the human eye to analyse., , 49, 40, 58, , 7.4%, , Moody’s, , Identify the primary data, , (i), (ii), (iii), (iv), Secondary data are the data taken from figures (v), collected by someone else. For example, government- (vi), published statistics, available research reports etc.,, (vii), (viii), 8.2.1 Getting the Facts Sorted Out, , 44, 46, 65, , IMF, Morgan Stanley, , Primary data are first-hand original data that we, collect ourselves. Primary data collection can be done, in a variety of ways such as by conducting personal, interviews (by phone, mail or face-to-face), by conducting, experiments, etc.,, , 60, 43, 50, 44, , 7.2%, , Very Satisfied, , 8.2 Collection of Data, , 61, 52, 62, 42, , UN, , 60, 63, 35, , 79, 72, 54, , 62, 46, 59, , 39, 34, 43, , 51, 55, 46, , Customer surveys, Medical researches, Economic predictions, School results, Political polls, Marketing details, Sales forecasts, Price index details, Activity - 1, , Prepare an album of, pictures, tables, numeric, details etc that exhibit, data. Discuss how they, are related to daily life, situations., 67, 76, 58, , 65, 55, 58, , 43, 30, 56, , 54, 67, 59, , 51, 44, 59, , 42, 57, 45, , In this data, if you want to locate the five highest marks, is it going to be easy?, You have to search for them; in case you want the third rank among them, it is further, Statistics 267, , 8-Stats.indd 267, , 26-12-2019 14:18:23
Page 274 :
www.tntextbooks.in, , complicated. If you need how many scored less than, say 56, the task will be quite time, consuming., Hence arrangement of an array of marks will make the job simpler., With some difficulty you may note in the list that 79 is the highest mark and 25 is, the least. Using these you can subdivide the data into convenient classes and place each, mark into the appropriate class. Observe how one can do it., Class Interval, 25-30, , Marks, 30, 25, , 31-35, , 31, 34, 35, , 36-40, , 39, 40, , 41-45, , 44, 43, 42, 43, 44, 43, 45, 42, 44, , 46-50, , 49, 46, 46, 50, 46, , 51-55, , 51, 54, 51, 52, 55, 55, 54, , 56-60, , 60, 60, 60, 57, 58, 59, 58, 58, 56, 59, 59, , 61-65, , 61, 62, 65, 63, 62, 65, , 66-70, , 67, 67, , 71-75, , 72, , 76-80, , 79, 76, , From this table can you answer the questions raised above? To answer the question,, “how many scored below 56”, you do not need the actual marks. Class, Number, You just want “how many” were there. To answer such cases, which Interval of items, 2, often occur in a study, we can modify the table slightly and just note 25-30, 3, down how many items are there in each class. We then may have a 31-35, 36-40, 2, slightly simpler and more useful arrangement, as given in the table., This table gives us the number of items in each class; each such, number tells you how many times the required item occurs in the, class and is called the frequency in that class., The table itself is called a frequency table., , 41-45, 46-50, 51-55, 56-60, 61-65, 66-70, 71-75, 76-80, , 9, 5, 7, 11, 6, 2, 1, 2, , We use what are known as tally marks to compute the, frequencies. (Under the column ‘number of items’, we do not write, the actual marks but just tally marks). For example, against the, class 31-35, instead of writing the actual marks 31, 34, 35 we simply, put AAA. You may wonder if for the class 56-60 in the example one has to write A A A A A A A A A A A ,, making it difficult to count. To avoid confusion, every fifth tally mark is put across the, four preceding it, like this A A A A . For example, 11 can be written as A A A A A A A A A . The frequency, table for the above illustration will be seen as follows:, 268, , 8-Stats.indd 268, , 9th Standard Mathematics, , 26-12-2019 14:18:23
Page 275 :
www.tntextbooks.in, , Class, Tally Marks, Interval, , Frequency, , 25-30, , AA, , 2, , 31-35, , AAA, , 3, , 36-40, , AA, , 2, , 41-45, , AAAA AAaa, , 9, , 46-50, , AAAA, , 5, , 51-55, , AAAA AA, , 7, , 56-60, , AAAA AAAA A, , 11, , 61-65, , AAAA A, , 6, , 66-70, , AA, , 2, , 71-75, , A, , 1, , 76-80, , AA, , 2, , TOTAL, , 50, , Note, Consider any class, say 56- 60; then 56 is called, the lower limit and 60 is called the upper limit, of the class., Progress Check, Form a frequency table for the following data:, 23, 52, 62, 21, 17, 80, 11, , 44, 37, 58, 39, 28, 46, 57, , 12, 77, 69, 80, 65, 30, , 11, 97, 24, 65, 35, 49, , 45, 82, 12, 54, 55, 50, , 55, 56, 99, 44, 68, 61, , 79, 28, 55, 59, 84, 59, , 20, 71, 78, 65, 97, 33, , 8.3 Measures of Central Tendency, It often becomes necessary in everyday life to express a quantity that is typical for a, given data. Suppose a researcher says that on an average, people watch TV serials for 3 hours, per day, it does not mean that everybody does so; some may watch more and some less. The, average is an acceptable indicator of the data regarding programmes watched on TV., Averages summarise a large amount of data into a single value and indicate that, there is some variability around this single value within the original data., A mathematician’s view of an average is slightly different from that of the commoner., There are three different definitions of average known as the Mean, Median and Mode., Each of them is found using different methods and when they are applied to the same, set of original data they often result in different average values. It is important to figure, out what each of these measures of average tells you about the original data and consider, which one is the most appropriate to calculate., , 8.4 Arithmetic Mean, 8.4.1 Arithmetic Mean-Raw Data, The Arithmetic Mean of a data is the most commonly used of all averages and is, found by adding together all the values and dividing by the number of items., For example, a cricketer, played eight (T20) matches and scored the following scores, 25, 32, 36, 38, 45, 41, 35, 36., Statistics 269, , 8-Stats.indd 269, , 26-12-2019 14:18:23
Page 276 :
www.tntextbooks.in, , Then, the mean of his scores (that is the arithmetic average of the scores) is, obtained by, x, + + + + + + +, X = ∑ = 25 32 36 38 45 41 35 36 = 288 = 36 ., 8, 8, n, , In general, if we have n number of observations x1, x2, x3, …, xn then their arithmetic, mean denoted by X (read as X bar) is given by, , X=, , 1 n, sum of all the observations, X = xi, n i 1, number of observations, , We express this as a formula:, , x, X=∑, , Note, , n, , It does not matter which, Assumed Mean method: Sometimes we can make, number is chosen as the, calculations easy by working from an entry that we guess, assumed mean; we need a, to be the right answer. This guessed number is called, number that would make, the assumed mean., our calculations simpler., In the example above on cricket scores, let us Perhaps a choice of number, assume that 38 is the assumed mean. We now list the that is closer to most of the, differences between the assumed mean and each score entries would help; it need, not even be in the list given., entered:, 25–38 = –13, 32–38 = –6, 36–38 = –2, 38–38 = 0,, 45–38 = 7, 41–38 = 3, 35–38 = –3, 36–38 = –2, The average of these differences is, , 13 6 2 0 7 3 3 2 16, , 2, 8, 8, , We add this ‘mean difference’ to the assumed mean to get the correct mean., Thus the correct mean = Assumed Mean +Mean difference = 38 – 2 = 36 ., This method will be very helpful when large numbers are involved., 8.4.2 Arithmetic Mean-Ungrouped Frequency Distribution, Consider the following list of heights (in cm) of 12 students who are going to take, part in an event in the school sports., 140, 142, 150, 150, 140, 148, 140, 147, 145, 140, 147, 145., How will you find the Mean height?, There are several options., (i), 270, , 8-Stats.indd 270, , You can add all the items and divide by the number of items., , 9th Standard Mathematics, , 26-12-2019 14:18:25
Page 277 :
www.tntextbooks.in, , , (ii), , , 140 + 142+150 + 150 + 140 + 148+140 + 147+145+140 + 147+145 1734, =, = 144.5, 12, 12, , You can use Assumed mean method. Assume, 141 as the assumed mean., Then the mean will be given by, , (−1) + (1) + (9) + (9) + (−1) + (7) + (−1) + (6) + (4) + (−1) + (6) + (4), 12, , −4 + 46, 42, = 141 +, = 141 +, = 141 + 3.5 = 144.5, 12, 12, = 141 +, , (iii), , A third method is to deal with an ungrouped frequency distribution. You find, that 140 has occurred 4 times, (implying 4 is the frequency of 140), 142 has, occurred only once (indicating that 1 is the frequency of 142) and so on. This, enables us to get the following frequency distribution., Height(cm), , 140, , 142, , 150, , 148, , 145, , 147, , No. of students, , 4, , 1, , 2, , 1, , 2, , 2, , , , You find that there are four 140s; their total will be 140 × 4 = 560, , , , There is only one 142; so the total in this case is, , 142 × 1 = 142, , , , There are two 150s; their total will be, , 150 × 2 = 300 etc., , , , These details can be neatly tabulated as follows:, Height (x) Frequency (f), , fx, , 140, , 4, , 560, , 142, , 1, , 142, , 150, , 2, , 300, , 148, , 1, , 148, , 145, , 2, , 290, , 147, , 2, , 294, , 12, , 1734, , Sum of all fx, No.of items, , Mean =, =, , 1734, = 144.5 cm, 12, , Looking at the procedure in general terms, you can obtain a formula, for ready use. If x1, x2, x3, … xn are n observations whose corresponding, frequencies are f1, f2, f3, … fn then the mean is given by, n, , X=, , f1 x 1 + f2 x 2 +… + fn x n , = i 1n, f1 + f2 +… fn, , fi x i, , f, i 1, , i, , =∑, , fx, , Note, Study each step, and understand, the meaning of, each symbol., , ∑f, , Can you adopt the above method combining with the assumed mean method? Here, is an attempt in that direction:, Statistics 271, , 8-Stats.indd 271, , 26-12-2019 14:18:27
Page 278 :
www.tntextbooks.in, , (iv) Let the assumed mean be 145. Then we can prepare the following table:, d = deviation from, the assumed mean, , Frequency (f), , 140, , 140 – 145 = – 5, , 4, , –20, , 142, , 142– 145 = – 3, , 1, , – 3, , 150, , 150 – 145 = + 5, , 2, , +10, , 148, , 148 – 145 = + 3, , 1, , + 3, , 145 – 145 =, , 0, , 2, , 0, , 147 – 145 = + 2, , 2, , + 4, , Height(x), , 145 (Assumed), 147, , Total, , ∑f, , = 12, , fd, , ∑ fd = –23 +17 = –6, , Arithmetic mean = Assumed mean + Average of the sum of deviations, = A , , fd, f, , 6 , , = 145 + = 145.0 – 0.5 = 144.5, 12 , , When large numbers are involved, this method could be useful., 8.4.3 Arithmetic Mean-Grouped Frequency Distribution, When data are grouped in class intervals and presented in the form of a frequency, table, we get a frequency distribution like this one:, Age (in years), , 10-20, , 20-30, , 30 – 40, , 40 – 50, , 50 - 60, , Number of customers, , 80, , 120, , 50, , 22, , 8, , The above table shows the number of customers in the various age groups. For, example, there are 120 customers in the age group 20 – 30, but does not say anything, about the age of any individual. (When we form a grouped frequency table the identity of, the individual observations is lost). Hence we need a value that represents the particular, class interval. Such a value is called mid value (mid-point or class mark) The mid-point, or class mark can be found using the formula given below., +, Mid Value = UCL 2 LCL ,, UCL – Upper Class Limit, LCL – Lower Class Limit, In grouped frequency distribution, arithmetic mean may be computed by applying, any one of the following methods., , (i) Direct Method, 272, , 8-Stats.indd 272, , (ii) Assumed Mean Method, , (iii) Step Deviation Method, , 9th Standard Mathematics, , 26-12-2019 14:18:28
Page 279 :
www.tntextbooks.in, , Direct Method, When direct method is used, the formula for finding the arithmetic mean is, X, , fx, = ∑, , ∑f, , Where x is the mid-point of the class interval and f is the corresponding frequency, Steps, (i), , Obtain the mid-point of each class and denote it by x, , (ii) Multiply those mid-points by the respective frequency of each class and obtain, the sum of fx, (iii) Divide ∑fx by ∑f to obtain mean, Example 8.1, , The following data gives the number of residents in an area based, on their age. Find the average age of the residents., Age, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 2, , 6, , 9, , 7, , 4, , 2, , Number of Residents, Solution, Age, , Number of, Residents(f), , Midvalue(x), , fx, , 0-10, , 2, , 5, , 10, , 10-20, , 6, , 15, , 90, , 20-30, , 9, , 25, , 225, , 30-40, , 7, , 35, , 245, , 40-50, , 4, , 45, , 180, , 50-60, , 2, , 55, , 110, , ∑ f = 30, Mean= X, , fx, = ∑ =, , ∑f, , fx 860, , 860, = 28.67, 30, , Hence the average age = 28.67., Assumed Mean Method, We have seen how to find the arithmetic mean of a grouped data quickly using the, Statistics 273, , 8-Stats.indd 273, , 26-12-2019 14:18:28
Page 280 :
www.tntextbooks.in, , direct method formula. However, if the observations are large, finding the products of, the observations and their corresponding frequencies, and then adding them is not only, difficult and time consuming but also has chances of errors. In such cases, we can use the, Assumed Mean Method to find the arithmetic mean of grouped data., Steps, 1., , Assume any value of the observations as the Mean (A). Preferably, choose the, middle value., , 2., , Calculate the deviation d = x − A for each class, , 3., , Multiply each of the corresponding frequency ‘f ’ with ‘d’ and obtain Sfd, , 4., , Apply the formula X, , Example 8.2, , = A+ ∑, , fd, , ∑f, , Find the mean for the following frequency table:, , Class, Interval, , 100-120, , 120-140, , 140-160, , 160-180, , 180-200, , 200-220, , 220-240, , Frequency, , 10, , 8, , 4, , 4, , 3, , 1, , 2, , Solution, Let Assumed mean A = 170, Class Frequency, Interval, f, , Mid, value, x, , d = x–A, d = x–170, , fd, , 100-120, , 10, , 110, , –60, , –600, , 120-140, , 8, , 130, , –40, , –320, , 140-160, , 4, , 150, , –20, , –80, , 160-180, , 4, , 170, , 0, , 0, , 180-200, , 3, , 190, , 20, , 60, , 200-220, , 1, , 210, , 40, , 40, , 220-240, , 2, , 230, , 60, , 120, , ∑ f = 32, , Mean X, , = A+ ∑, , fd, , ∑f, , 780 , , 32 , , = 170 , , Therefore , X = 170–24.375, = 145.625, , ∑ fd = –780, , Step Deviation Method, In order to simplify the calculation, we divide the deviation by the width of class, x− A, intervals (i.e. calculate, ) and then multiply by c in the formula for getting the mean, c, of the data. The formula to calculate the Arithmetic Mean is, 274, , 8-Stats.indd 274, , 9th Standard Mathematics, , 26-12-2019 14:18:29
Page 281 :
www.tntextbooks.in, , Example 8.3, , fd, , x A, X =A , c , where d , c, f, , , Find the mean of the following distribution using Step Deviation Method., , Class Interval, , 0-8, , 8-16, , 16-24, , 24-32, , 32-40, , 40-48, , Frequency ( f ), , 10, , 20, , 14, , 16, , 18, , 22, , Solution, Let Assumed mean A = 28, class width c =8, Class, Interval, , Mid Value, x, , Frequency, f, , 0-8, , 4, , 10, , –3, , –30, , 8-16, , 12, , 20, , –2, , –40, , 16-24, , 20, , 14, , –1, , –14, , 24-32, , 28, , 16, , 0, , 0, , 32-40, , 36, , 18, , 1, , 18, , 40-48, , 44, , 22, ∑ f = 100, , 2, , 44, ∑ fd = –22, , Mean, fd, X = A+ ∑ ×c, ∑f, , −22 , = 28 + , ×8, 100 , , = 28 − 1.76= 26.24, , d=, , x −A, c, , fd, , Note, When xi and fi are small, then Direct Method is the, , appropriate choice., When xi and fi are numerically large numbers, then, Assumed Mean Method or Step Deviation Method, can be used., When class sizes are unequal and d numerically, large,we can still use Step Deviation Method., , 8.4.4 A special property of the Arithmetic Mean, 1., , The sum of the deviations of the entries from the arithmetic mean is always zero., , If x 1, x 2 , x 3 ,..., x n are n observations taken from the arithmetic mean X, then x 1 X x 2 X x 3 X ... x n X =0 .Hence, 2., , n, , (x, i 1, , i, , X) 0, , If each observation is increased or decreased by k (constant) then the arithmetic, mean is also increased or decreased by k respectively., , 3., , If each observation is multiplied or divided by k, k ¹ 0, then the arithmetic mean, is also multiplied or divided by the same quantity k respectively., Statistics 275, , 8-Stats.indd 275, , 26-12-2019 14:18:30
Page 282 :
www.tntextbooks.in, , Example 8.4, , Find the sum of the deviations from the arithmetic mean for the, following observations:, 21, 30, 22, 16, 24, 28, 18, 17, Solution, , 8, , X=, , ∑x, i =1, , i, , n, , =, , 21 + 30 + 22 + 16 + 24 + 28 + 18 + 17 176, =, = 22, 8, 8, , Deviation of an entry xi from the arithmetic mean X is xi − X , i = 1, 2, ...8., Sum of the deviations, = (21-22)+(30-22)+(22-22)+(16-22)+(24-22)+(28-22)+(18-22)+(17-22), 8, , (xi - X ) = 0, = 16–16 = 0. or equivalently, /, i=1, , Hence, we conclude that sum of the deviations from the Arithmetic Mean is zero., Example 8.5, , The arithmetic mean of 6 values is 45 and if each value is increased, by 4, then find the arithmetic mean of new set of values., 6, Solution, xi, ∑, i =1, = 45., Let x1, x2, x3, x4, x5, x6 be the given set of values then, 6, , If each value is increased by 4, then the mean of new set of values is, 6, , New A.M., , X, , / (xi + 4), , i=1, , =, , 6, ( x + 4) + ( x2 + 4) + ( x3 + 4) + ( x4 + 4) + ( x5 + 4) + ( x6 + 4), = 1, 6, 6, , =, , ∑ x + 24, i =1, , i, , 6, , 6, , =, , X = 45 + 4 = 49., Example 8.6, , ∑x, i =1, , 6, , i, , +4, , Progress Check, Mean of 10 observations is, 48 and 7 is subtracted to each, observation, then mean of new, observation is __________, , If the arithmetic mean of 7 values, is 30 and if each value is divided by 3, then find the, arithmetic mean of new set of values, Progress Check, Solution, Let X represent the set of seven values 1. The Mean of 12 numbers is 20., x1 , x2 , x3 , x4 , x5 , x6 , x7., If each number is multiplied by, 7, xi, ∑, 6, then the new mean is __, 7, i =1, = 30 or ∑ xi = 210, Then, X=, 2. The Mean of 30 numbers is 16., 7, i =1, If each number is divided by 4,, If each value is divided by 3, then the mean of, then the new mean is___, new set of values is, 276, , 8-Stats.indd 276, , 9th Standard Mathematics, , 26-12-2019 14:18:31
Page 283 :
www.tntextbooks.in, , x1 x2 x3 x4 x5 x6 x7 , + + + + + + , 3 3 3 3 3 3 3, =, 7, , xi, , 7, , ∑3, i =1, , 7, 7, , =, , ∑x, i =1, , 21, , i, , =, , 210, = 10, 21, , Aliter, If Y is the set of values obtained by dividing each value of X by 3., Then, Y =, , X 30, = = 10., 3, 3, , Example 8.7, , The average mark of 25 students was found to be 78.4. Later on, it, , was found that score of 96 was misread as 69. Find the correct mean of the marks., Solution, =, n 25, =, , X 78.4, Given that the total number of students, , Progress Check, , There are four numbers., If we leave out any one, Correct Sx incorrect x wrong entry correct entry number, the average of the, remaining three numbers, 1960 69 96 1987, will be 45, 60, 65 or 70., correct x 1987, What is the average of all, Correct X , =, = 79.48, 25, n, four numbers?, , So, Incorrect, , x X n, , 78.4 25 1960, , Exercise 8.1, 1., , In a week, temperature of a certain place is measured during winter are as follows, 26oC, 24oC, 28oC, 31oC, 30oC, 26oC, 24oC. Find the mean temperature of the week., , 2., , The mean weight of 4 members of a family is 60kg.Three of them have the weight, 56kg, 68kg and 72kg respectively. Find the weight of the fourth member., , 3., , In a class test in mathematics, 10 students scored 75 marks, 12 students scored 60, marks, 8 students scored 40 marks and 3 students scored 30 marks. Find the mean, of their score., , 4., , In a research laboratory scientists treated 6 mice with lung cancer using natural, medicine. Ten days later, they measured the volume of the tumor in each mouse and, given the results in the table., Statistics 277, , 8-Stats.indd 277, , 26-12-2019 14:18:38
Page 284 :
www.tntextbooks.in, , Mouse marking, Tumor Volume(mm3), , 1, , 2, , 3, , 4, , 5, , 6, , 145, , 148, , 142, , 141, , 139, , 140, , Find the mean., 5., , 6., , If the mean of the following data is 20.2, then find the value of p, Marks, , 10, , 15, , 20, , 25, , 30, , No. of students, , 6, , 8, , p, , 10, , 6, , In the class, weight of students is measured for the class records. Calculate mean, weight of the class students using Direct method., Weight in kg, No. of students, , 7., , 25-35, , 35-45, , 45-55, , 55-65, , 65-75, , 4, , 11, , 19, , 14, , 0, , 2, , Calculate the mean of the following distribution using Assumed Mean Method:, Class Interval, Frequency, , 8., , 15-25, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 5, , 7, , 15, , 28, , 8, , Find the Arithmetic Mean of the following data using Step Deviation Method:, Age, No. of persons, , 15-19, , 20-24, , 25-29, , 30-34, , 35-39, , 40-44, , 4, , 20, , 38, , 24, , 10, , 9, , 8.5 Median, The arithmetic mean is typical of the data because it ‘balances’ the numbers; it, is the number in the ‘middle’, pulled up by large values and pulled down by smaller, values., Suppose four people of an office have incomes of `5000, `6000, `7000 and `8000., Their mean income can be calculated as, , 5000 + 6000 + 7000 + 8000, which gives `6500. If a, 4, , fifth person with an income of ` 29000 is added to this group, then the arithmetic mean, of all the five would be, , 55000, 5000 + 6000 + 7000 + 8000 + 29000, =, = `11000. Can one say that, 5, 5, , the average income of `11000 truly represents the income status of the individuals in the, office? Is it not, misleading? The problem here is that an extreme score affects the Mean, and can move the mean away from what would generally be considered the central area., In such situations, we need a different type of average to provide reasonable answers., , 278, , 8-Stats.indd 278, , 9th Standard Mathematics, , 26-12-2019 14:18:40
Page 285 :
www.tntextbooks.in, , Median is the value which occupies the middle position when all the observations are, arranged in an ascending or descending order. It is a positional average., For example, the height of nine students in a class are 122 cm, 124 cm, 125 cm, 135, cm, 138 cm, 140cm, 141cm, 147 cm, and 161 cm., (i) Usual calculation gives Arithmetic Mean to be 137 cm., (ii), , If the heights are neatly arranged in, say, ascending order, as follows, 122 cm, 124 cm, 125 cm, 135 cm, 138 cm, 140cm, 141cm, 147 cm, 161 cm, one, can observe the value 138 cm is such that equal number of items lie on either, side of it. Such a value is called the Median of given readings., , (iii), , Suppose a data set has 11 items arranged in order. Then the median is the 6th, item because it will be the middlemost one. If it has 101 items, then 51st item, will be the Median., If we have an odd number of items, one can find the middle one easily. In general,, th, , n + 1, item., if a data set has n items and n is odd, then the median will be the , 2 , , (iv), , If there are 6 observations in the data, how will you find the Median? It will be, the average of the middle two terms. (Shall we denote it as 3.5th term?) If there, are 100 terms in the data, the Median will be 50.5th term!, In general, if a data set has n items and n is even, then the Median will be the, th, , th, , n , n, , average of and + 1 items., 2, , 2 , , Example 8.8, , The following are scores obtained by 11 players in a cricket match, , 7, 21, 45, 12, 56, 35, 25, 0, 58, 66, 29. Find the median score., Solution, Let us arrange the values in ascending order., 0,7,12,21,25,29,35,45,56,58,66, The number of values = 11 which is odd, th, , Median, , 11 + 1, , = , 2 , , value, , th, , 12 , = value = 6th value = 29, 2, , 8-Stats.indd 279, , Statistics 279, , 26-12-2019 14:18:40
Page 286 :
www.tntextbooks.in, , Example 8.9, , For the following ungrouped data 10, 17, 16, 21, 13, 18, 12, 10, 19, 22., , Find the median., Solution, Arrange the values in ascending order., 10, 10, 12, 13, 16, 17, 18, 19, 21, 22., The number of values = 10, th, , 10 , th, + 1j values, Median = Average of and ` 10, 2, 2, , = Average of 5th and 6th values, +, = 16.5, = 16 2 17 = 33, 2, , Example 8.10, , The following table represents the marks obtained by a group of 12, students in a class test in Mathematics and Science., Marks, 52, (Mathematics), , 55, , 32, , 30, , 60, , 44, , 28, , 25, , 50, , 75, , 33, , 62, , Marks, (Science), , 42, , 48, , 49, , 27, , 25, , 24, , 19, , 28, , 58, , 42, , 69, , 54, , Indicate in which subject, the level of achievement is higher?, Solution, Let us arrange the marks in the two subjects in ascending order., Marks, (Mathematics), , 25 28, , 30, , 32, , 33, , 44, , 50, , 52, , 55, , 60, , 62, , 75, , Marks (Science), , 19 24, , 25, , 27, , 28, , 42, , 42, , 48, , 49, , 54, , 58, , 69, , Since the number of students is 12, the marks of the middle-most student would be, the mean mark of 6th and 7th students., , 44 + 50, = 47, 2, 42 + 42, = 42, Median mark in Science, =, 2, Here the median mark in Mathematics is greater than the median mark in Science., Therefore, the level of achievement of the students is higher in Mathematics than Science., Therefore , Median mark in Mathematics, , 280, , 8-Stats.indd 280, , =, , 9th Standard Mathematics, , 26-12-2019 14:18:41
Page 287 :
www.tntextbooks.in, , 8.5.1 Median-Ungrouped Frequency Distribution, (i), , Arrange the data in ascending ( or) decending order of magnitude., , (ii), , Construct the cumulative frequency distribution. Let N be the total frequency., th, , (iii), , (iv), , N + 1, , If N is odd, median = , 2 , , observation., , th, th, , N, , N , , , observation + + 1 observation , , 2, 2, If N is even, median = , , , 2, , Example 8.11, , Calculate the median for the following data:, , Height (cm), 160, No. of Students 12, , 150, 8, , 152, 4, , 161, 4, , 156, 3, , 154, 3, , 155, 7, , Solution, Let us arrange the marks in ascending order and prepare the following data:, Height (cm), 150, , Number of, students (f), 8, , Cumulative, frequency (cf), 8, , 152, , 4, , 12, , 154, , 3, , 15, , 155, , 7, , 22, , 156, , 3, , 25, , 160, , 12, , 37, , 161, , 4, , 41, , Here N = 41, th, , th, , N + 1, 41 + 1, value = size of , value = size of 21st value., Median = size of , , , 2 , 2 , , If the 41 students were arranged in order (of height), the 21st student would be the, middle most one, since there are 20 students on either side of him/her. We therefore need, to find the height against the 21st student. 15 students (see cumulative frequency) have, height less than or equal to 154 cm. 22 students have height less than or equal to 155 cm., This means that the 21st student has a height 155 cm., Therefore, Median = 155 cm, 8.5.2 Median - Grouped Frequency Distribution, In a grouped frequency distribution, computation of median involves the following, Steps, (i) Construct the cumulative frequency distribution., Statistics 281, , 8-Stats.indd 281, , 26-12-2019 14:18:41
Page 288 :
www.tntextbooks.in, , th, , N , (ii) Find term., 2, , (iii) The class that contains the cumulative frequency, , (iv) Find the median by using the formula:, N, , − m , , , Median = l + 2, ×c, f, , Where l = Lower limit of the median class,, , N, is called the median class., 2, , f = Frequency of the median class, , c = Width of the median class,, , N = The total frequency, , f , , m = cumulative frequency of the class preceeding the median class, Example 8.12, , The following table gives the weekly expenditure of 200 families., Find the median of the weekly expenditure., Weekly, expenditure (`), Number of, families, Solution, , 0-1000, , 1000-2000, , 2000-3000, , 3000-4000, , 4000-5000, , 28, , 46, , 54, , 42, , 30, , Weekly, Expenditure, 0-1000, , Number of, families (f), 28, , Cumulative, frequency (cf), 28, , 1000-2000, , 46, , 74, , 2000-3000, , 54, , 128, , 3000-4000, , 42, , 170, , 4000-5000, , 30, , 200, , N=200, th, , th, , N , 200 , Median class = value = value, 2, 2 , , = 100 th value, Median class = 2000 – 3000, N, = 100 l = 2000, 2, , m = 74, c = 1000, f = 54, , Median, 282, , 8-Stats.indd 282, , N, , m, 2, c, =l , f, , Progress Check, 1. The median of the first four, whole numbers _____., 2. If 4 is also included to, the collection of first, four whole numbers then, median value is ________., 3. The difference between, two median is _____., , 9th Standard Mathematics, , 26-12-2019 14:18:42
Page 289 :
www.tntextbooks.in, , = 2000 + ` 10054 74 j # 1000, -, , 26 #, = 2000 + ` 54, j 1000 = 2000 + 481.5, , = 2481.5, Example 8.13, , The Median of the following data is 24. Find the value of x., , Class Interval (CI), Frequency (f), Solution, , 0 - 10, 6, , Class, Interval (CI), 0-10, 10-20, 20-30, , 10 - 20, 24, , 20 - 30, x, , 30 - 40, 16, , Frequency, (f), 6, 24, , Cumulative, frequency (cf), 6, 30, , 30-40, , x, 16, , 30 + x, , 40-50, , 9, , 55 + x, , 40 - 50, 9, , 46 + x, , N = 55 + x, Since the median is 24 and median class is 20 – 30, l = 20, , N = 55 + x, m = 30, c = 10, f = x, N, , m, 2, c, Median = l , f, 55 x, , 30 , , 2, 10, 24 = 20 , x, 5 x − 25, (after simplification), 4 =, x, , 4x = 5x – 25, 5x – 4x = 25, x = 25, , Note, The median is a good, measure of the average, value when the data, include extremely high, or low values, because, these have little influence, on the outcome., , Exercise 8.2, 1., , Find the median of the given values : 47, 53, 62, 71, 83, 21, 43, 47, 41., , 2., , Find the Median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51, Statistics 283, , 8-Stats.indd 283, , 26-12-2019 14:18:45
Page 290 :
www.tntextbooks.in, , 3., 4., , 5., , The median of observation 11, 12, 14, 18, x+2, x+4, 30, 32, 35, 41 arranged in, ascending order is 24. Find the values of x., A researcher studying the behavior of mice has recorded the time (in seconds) taken, by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33,, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching, its food., The following are the marks scored by the students in the Summative Assessment exam, Class, , 0-10, , No. of Students, 2, Calculate the median., 6., , 8.6, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 7, , 15, , 10, , 11, , 5, , The mean of five positive integers is twice their median. If four of the integers are 3,, 4, 6, 9 and median is 6, then find the fifth integer., , Mode, (i), , The votes obtained by three candidates in an election are as follows:, , Name of the, Candidate, Mr. X, Mr. Y, Mr. Z, Total, , Votes Polled, 4, 12, 006, 9, 87, 991, 7, 11, 973, 21, 11, 970, , Who will be declared as the winner? Mr. Y will be the winner, because the number, of votes secured by him is the highest among all the three candidates. Of course, the votes, of Mr. Y do not represent the majority population (because there are more, votes against him). However, he is declared winner because the mode of, selection here depends on the highest among the candidates., (ii), , An Organisation wants to donate sports shoes of same size to, maximum number of students of class IX in a School. The, distribution of students with different shoe sizes is given below., , Shoe Size, , 5, , 6, , 7, , 8, , 9, , 10, , No. of Students, , 10, , 12, , 27, , 31, , 19, , 1, , If it places order, shoes of only one size with the manufacturer, which size of the, shoes will the organization prefer?, In the above two cases, we observe that mean or median does not fit into the situation., We need another type of average, namely the Mode., The mode is the number that occurs most frequently in the data., 284, , 8-Stats.indd 284, , 9th Standard Mathematics, , 26-12-2019 14:18:45
Page 291 :
www.tntextbooks.in, , When you search for some good video about Averages on You Tube, you look to, watch the one with maximum views. Here you use the idea of a mode., 8.6.1 Mode - Raw Data, For an individual data mode is the value of the variable which occurs most frequently., Example 8.14, , In a rice mill, seven labours are receiving the daily wages of `500,, `600, `600, `800, `800, `800 and `1000, find the modal wage., Solution, In the given data `800 occurs thrice.Hence the mode is ` 800., Example 8.15, , Find the mode for the set of values 17, 18, 20, 20, 21, 21, 22, 22., , Solution, In this example, three values 20, 21, 22 occur two times each. There are three modes, for the given data!, Note, A distribution having only one mode is called unimodal., A distribution having two modes is called bimodal., A distribution having Three modes is called trimodal., A distribution having more than three modes is called multimodal., , 8.6.2 Mode for Ungrouped Frequency Distribution, In a ungrouped frequency distribution, the value of the item having maximum, frequency is taken as the mode., Example 8.16, , A set of numbers consists of five 4’s, four 5’s, nine 6’s,and six 9’s., , What is the mode., Solution, Size of item, , 4, , 5, , 6, , 9, , Frequency, , 5, , 4, , 9, , 6, , 6 has the maximum frequency 9. Therefore 6 is the mode., Statistics 285, , 8-Stats.indd 285, , 26-12-2019 14:18:45
Page 292 :
www.tntextbooks.in, , 8.6.3 Mode – Grouped Frequency Distribution:, In case of a grouped frequency distribution, the exact values of the variables are not, known and as such it is very difficult to locate mode accurately. In such cases, if the class, intervals are of equal width, an appropriate value of the mode may be determined by, f f1 , c, 2 f f1 f 2 , , , Mode = l , , The class interval with maximum frequency is called the modal class., Where l - lower limit of the modal class;, , f - frequency of the modal class, , f1 - frequency of the class just preceding the modal class, f2 - frequency of the class succeeding the modal class, c - width of the class interval, Example 8.17, , Find the mode for the following data., , Marks, , 1-5, , 6-10, , 11-15, , 16-20, , 21-25, , 7, , 10, , 16, , 32, , 24, , No. of students, Solution, Marks, , f, , 0.5-5.5, , 7, , 5.5-10.5, , 10, , 10.5-15.5, , 16, , 15.5-20.5, , 32, , 20.5-25.5, , 24, , Note, To convert discontinuous, class interval into continuous, class interval, 0.5 is to be, substracted at the lower limit, and 0.5 is to be added at the, upper limit for each class, interval., , Modal class is 16 -20 since it has the maximum frequency., l = 15.5, f = 32, f1 = 16, f2=24, c = 20.5–15.5 = 5, , , , × c, 2 f − f1 − f2 , , Mode = l + , , f − f1, , 32 - 16, j# 5, 64 - 16 - 24, # 5 = 15.5 + 3.33 =18.83., = 15.5 + ` 16, 24 j, , = 15.5 + `, , 8.6.4 An Empirical Relationship between Mean, Medan and Mode, We have seen that there is an approximate relation that holds among the three, averages discussed earlier, when the frequencies are nearly symmetrically distributed., Mode 3 Median – 2 Mean, 286, , 8-Stats.indd 286, , 9th Standard Mathematics, , 26-12-2019 14:18:47
Page 293 :
www.tntextbooks.in, , Example 8.18, , In a distribution, the mean and mode are 66 and 60 respectively., , Calculate the median., Solution, Given,, Using,, , Mean = 66 and Mode = 60., 3Median – 2Mean, , Mode, , 3Median – 2(66), , 60, , 60 +132, , 3 Median, , 192, 3, , Therefore, Median, , 64, , Exercise 8.3, 1., , The monthly salary of 10 employees in a factory are given below :, `5000, `7000, `5000, `7000, `8000, `7000, `7000, `8000, `7000, `5000, Find the mean, median and mode., , 2., , Find the mode of the given data : 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1, , 3., , For the data 11, 15, 17, x+1, 19, x–2, 3 if the mean is 14 , find the value of x. Also find, the mode of the data., , 4., , The demand of track suit of different sizes as obtained by a survey is given below:, Size, , 38, , 39, , 40, , 41, , 42, , 43, , 44, , 45, , No. of Persons, , 36, , 15, , 37, , 13, , 26, , 8, , 6, , 2, , Which size is in greater demanded?, 5., , Find the mode of the following data:, Marks, Number of students, , 6., , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 22, , 38, , 46, , 34, , 20, , Find the mode of the following distribution:, Weight(in kgs), Number of students, , 25-34, , 35-44, , 45-54, , 55-64, , 65-74, , 75-84, , 4, , 8, , 10, , 14, , 8, , 6, Statistics 287, , 8-Stats.indd 287, , 26-12-2019 14:18:47
Page 294 :
www.tntextbooks.in, , Exercise 8.4, Multiple choice questions, 1., , Let m be the mid point and b be the upper limit of a class in a, continuous frequency distribution. The lower limit of the class is, (1) 2m - b, , 2., , 8-Stats.indd 288, , (2) 1,3,3,3,5, , (3) 1,1,2,5,6, , (4) 1,1,2,1,5., , (2) n-1, , (3) n, , (4) n+1., , (2) 36, , (3) 26, , (4) 34, , (2) 11, , (3) 13, , (4) 15., , (2) 13, , (3) 17, , (4) 21, , (2) 46, , (3) 48, , (4) 52., , The mean of a set of numbers is X . If each number is multiplied by z, the mean is, (1) X +z, , 288, , (4) Median., , The mean of the square of first 11 natural numbers is, (1) 26, , 10., , (3) mode, , The mean of 5, 9, x, 17, and 21 is 13, then find the value of x, (1) 9, , 9., , (2) range, , If the mean of five observations x, x+2 , x+4 , x+6 , x+8 , is 11 , then the mean of first, three observations is, (1) 9, , 8., , (4) 98., , The mean of a,b,c,d and e is 28 . If the mean of a, c and e is 24 , then mean of b and, d is_, (1) 24, , 7., , (3) 99, , The algebraic sum of the deviations of a set of n values from their mean is, (1) 0, , 6., , (2) 100, , For which set of numbers do the mean, median and mode all have the same values?, (1) 2,2,2,4, , 5., , (4) m-2b., , A particular observation which occurs maximum number of times in a given data, is called its, (1) Frequency, , 4., , (3) m-b, , The mean of a set of seven numbers is 81. If one of the numbers is discarded , the, mean of the remaining numbers is 78 . The value of discarded number is, (1) 101, , 3., , (2) 2m+b, , (2) X - z, , (3) z X, , (4) X, , 9th Standard Mathematics, , 26-12-2019 14:18:49
Page 295 :
www.tntextbooks.in, , Project, 1., , Prepare a frequency table of the top speeds of 20 different land animals., Find mean, median and mode. Justify your answer., , 2., , From the record of students particulars of the class,, (i) Find the mean age of the class ( using class interval), (ii) Calculate the mean height of the class( using class intervals), , Points to Remember, The information collected for a definite purpose is called data., The data collected by the investigator are known as primary data. When the, , information is gathered from an external source,the data are called secondary data., Initial data obtained through unorganized form are called Raw data., Mid Value =, , UCL + LCL, (where UCL–Upper Class Limit, LCL–Lower Class Limit)., 2, , Size of the class interval = UCL – LCL., The mean for grouped data:, , Direct Method, X , , fx, f, , Assumed Mean Method, X A , , fd, , f, , Step-Deviation Method, fd , X A c, f, , , The cumulative frequency of a class is the frequency obtained by adding the, , frequency of all up to the classes preceeding the given class., , N, , m, 2, c ., Formula to find the median for grouped data: Median = l , f, f f1 , c ., 2 f f1 f 2 , , , Formula to find the mode for grouped data: Mode = l , , Statistics 289, , 8-Stats.indd 289, , 26-12-2019 14:18:52
Page 296 :
www.tntextbooks.in, , ICT Corner, , Expected Result is, shown in this picture, , Step, Open the browser, type the URL Link given below (or) Scan the QR Code. GeoGebra, work sheet named “Mean by step deviation method” will open., In the work sheet Example 5.5 is given. Observe the steps. You can change the question, by typing new data “From”, “To” and “Frequency f ” in the spread sheet on Left hand, side. After that change, the Assumed mean on the right-hand side and check the, calculation., , Step 1, , Browse in the link, , Mean by step deviation method:, https://ggbm.at/NWcKTRtA or Scan the QR Code, 290, , 8-Stats.indd 290, , 9th Standard Mathematics, , 26-12-2019 14:18:52
Page 297 :
www.tntextbooks.in, , 9, , PROBABILITY, Probability theory is nothing more than common sense, - Pierre Simon Laplace., reduced to calculation., The, , statistical, , or, , empirical,, , attitude, , towards, , probability has been developed mainly by R.F.Fisher, and R.Von Mises. The notion of sample space comes, from R.Von Mises. This notion made it possible to, build up a strictly mathematical theory of probability, based on measure theory. Such an approach emerged, Richard Von Mises, (AD (CE) 1883-1953), , gradually in the last century under the influence of, many authors., , Learning Outcomes, , ÂÂ To understand the basic concepts of probability., ÂÂ To understand the classical and empirical approach of probability., ÂÂ To familiarise the types of events in probability., , 9.1 Introduction, To understand the notion of probability, we look into some real life situations that, involve some traits of uncertainty., A life-saving drug is administered to a patient admitted, in a hospital. The patient’s relatives may like to know the, probability with which the drug will work; they will be happy, if the doctor tells that out of 100 patients treated with the drug,, it worked well with more than 80 patients. This percentage of, success is illustrative of the concept of probability; it is based, on the frequency of occurrence. It helps one to arrive at a, conclusion under uncertain conditions. Probability is thus a, way of quantifying or measuring uncertainty., Probability 291, , 9 Probability.indd 291, , 26-12-2019 14:21:19
Page 298 :
www.tntextbooks.in, , You should be familiar with the usual complete pack of 52 playing cards. It has 4, suits(Hearts ♥, Clubs ♣, Diamonds ♦, Spades♠), each with 13 cards. Choose one of the, suits or cards, say spades. Keep these 13 cards facing downwards on the table. Shuffle, them well and pick up any one card. What is the, chance that it will be a King? Will the chances vary if, you do not want a King but an Ace? You will be quick, to see that in either case, the chances are 1 in 13, (Why?). It will be the same whatever single card you, choose to pick up. The word ‘Probability’ means, precisely the same thing as ‘chances’ and has the same, value, but instead of saying 1 in 13 we write it as a, 1, . (It would be easy to manipulate with, fraction, 13, fractions when we combine probabilities). It is ‘the, ratio of the favourable cases to the total number of possible cases’., Have you seen a, ‘dice’ ? (Some people Note, use the word ‘die’, In a fair die the sum of the, for a single ‘dice’;, numbers turning on the, we use ‘dice’ here,, opposite sides will always, both for the singular, be equal to 7., and plural cases). A, standard dice is a, cube, with each side having a different number of spots on it, ranging from one to six,, rolled and used in gambling and other games involving chance., If you throw a dice, what is the probability of getting a five? a two? a seven?, In all the answers you got for the questions raised above, did you notice anything, special about the concept of probability? Could there be a maximum value for probability?, or the least value? If you are sure of a certain occurrence what could be its probability?, For a better clarity, we will try to formalize the notions in the following paragraphs., , 9.2 Basic Ideas, , When we carry out experiments in science repeatedly under identical conditions, we, get almost the same result. Such experiments are known as deterministic. For example, the, experiments to verify Archimedes principle or to verify Ohm’s law are deterministic. The, outcomes of the experiments can be predicted well in advance., But, there are experiments in which the outcomes may be different even when, performed under identical conditions. For example, when a fair dice is rolled, a fair coin, is flipped or while selecting the balls from an urn, we cannot predict the exact outcome, 292, , 9 Probability.indd 292, , 9th Standard Mathematics, , 26-12-2019 14:21:19
Page 299 :
www.tntextbooks.in, , of these experiments; these are random experiments. Each performance of a random, experiment is called a trial and the result of each trial is called an outcome. (Note: Many, statisticians use the words ‘experiment’ and ‘trial’ synonymously.), Now let us see some of the important terms related to probability., Trial : Rolling a dice and flipping a coin are trials. A trial is an action which results in, one or several outcomes., Outcome : While flipping a coin we get Head or Tail . Head and Tail are called outcomes., The result of the trial is called an outcome., Sample point : While flipping a coin, each outcome H or T are the sample points. Each, outcome of a random experiment is called a sample point., Sample space : In a single flip of a coin, the collection of sample points is given by, , S = {H , T } ., , If two coins are tossed the collection of sample points S={HH,HT,TH,TT}., The set of all possible outcomes (or Sample points) of a random experiment is, called the Sample space. It is denoted by S. The number of elements in it are denoted by, n(S)., Event : If a dice is rolled, it shows 4 which is called an outcome (since, it is a result of a, single trial). In the same experiment the event of getting an even number is {2,4,6}. So, any subset of a sample space is called an event. Hence an event can be one or more than, one outcome., For example, (i), , Random experiment : Flipping a coin, Possible outcomes, , : Head(H ) or Tail(T ), , Sample space , , : S = {H,T}, , Subset of S , , : A={H} or A={T}, , Thus, in this example A is an event., When we roll a single dice, the collection of all sample points is, (ii) , S = {1,2,3,4,5,6}. (iii) When we select a day in a week the collection of sample points is, S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}., Probability 293, , 9 Probability.indd 293, , 26-12-2019 14:21:19
Page 300 :
www.tntextbooks.in, , Activity - 1, Perform the experiment of tossing two coins Perform the experiment of throwing two, at a time. List out the following in the above dice at a time. List out the following in, experiment., this experiment also., Random experiment, :, Random experiment :, Possible outcomes, , :, , Possible outcomes, , :, , Sample space, , :, , Sample space, , :, , Any three subsets of S, (or any 3 events), , :, , Any three subsets of S :, (or any 3 events), , Activity - 2, Each student is asked to flip a coin 10 times and tabulate the number of heads and tails, obtained in the following table., Number of tosses, , Number of times head, comes up, , Number of times tail, comes up, , , , , , , , (i) Fraction 1 :, (ii) Fraction 2 :, , Number of times head comes up, Total number of times the coin is tossed, Number of times tail comes up, , Total number of times the coin is tossed, Repeat it by tossing the coin 20, 30, 40, 50 times and find the fractions., , Number of times, head comes up, , Number of times, tail comes up, , Group, , Activity - 3, Divide the class students into groups of pairs. In each pair, the first one tosses a coin 50 times,, and the second one records the outcomes of tosses. Then prepare a table given below., , , , , , Number of times head comes up, , Number of times tail comes up, , Total number of times the coin is tossed Total number of times the coin is tossed, , 1, 2, 3, , 294, , 9 Probability.indd 294, , , , , , 9th Standard Mathematics, , 26-12-2019 14:21:20
Page 301 :
www.tntextbooks.in, , 9.3 Classical Approach, The chance of an event happening when expressed quantitatively is, probability., For example, An urn contains 4 Red balls and 6 Blue balls. You, choose a ball at random from the urn. What is the probability of, choosing a Red ball?, The phrase ‘at random’ assures you that each one of the 10 balls has the same chance, (that is, probability) of getting chosen. You may be blindfolded and the balls may be mixed, up for a “fair” experiment. This makes the outcomes “equally likely”., 4, 2, (You may also give it as or 0.4)., 10, 5, 6, 3, What would be the probability for choosing a Blue ball? It is, (or or 0.6)., 10, 5, Note that the sum of the two probabilities is 1. This means that no other outcome is, possible., The probability that the Red Ball is chosen is, , The approach we adopted in the above example is, classical. It is calculating a priori probability. (The Latin Thinking Corner, phrase a priori means ‘without investigation or sensory, experience’). Note that the above treatment is possible If the probability of success, of an experiment is 0.4, what, only when the outcomes are equally likely., is the probability of failure?, Classical probability is so named, because it, was the first type of probability studied formally by, mathematicians during the 17th and 18th centuries., Let S be the set of all equally likely outcomes of a random experiment. (S is called the, sample space for the experiment.), Let E be some particular outcome or combination of outcomes of an experiment., (E is called an event.), The probability of an event E is denoted as P(E)., P(E) =, , Number of favourable outcomes, Total number of outcomes, , =, , n(E ), n(S), , The empirical approach (relative frequency theory) of probability holds that if an, experiment is repeated for an extremely large number of times and a particular outcome, occurs at a percentage of the time, then that particular percentage is close to the probability, of that outcome., Probability 295, , 9 Probability.indd 295, , 26-12-2019 14:21:21
Page 302 :
www.tntextbooks.in, , 9.4 Empirical Approach, For example, A manufacturer produces 10,000 electric switches every month and1,000 of, them are found to be defective. What is the probability of the manufacturer producing a, defective switch every month?, Note, The required probability, according to relative, frequency concept, is nearly 1000 out of 10000,, which is 0.1, Let us formalize the definition: “If, the total, number of trials, say n, we find r of the outcomes in, an event E, then the probability of event E, denoted, by P(E), is given by, , The number of trials has to be, large to decide this probability., The larger the number of trials,, the better will be the estimate, of probability., , r, ., n, Is there a guarantee that this value will settle down to a constant value when the, number of trials gets larger and larger? One cannot say; the concept being experimental,, it is quite possible to get distinct relative frequency each time the experiment is, repeated., P (E) =, , , , However, there is a security range: the value of probability can at the least take the, value 0 and at the most take the value 1. We can state this mathematically as, 0 ≤ P ( E ) ≤ 1., Thinking Corner, , Let us look at this in a little detail., First, we know that r cannot be larger than n., This means, , r, n, , <1. That is P(E) < 1., , … (1), , For a question on probability the, student’s answer was, , 3, ., 2, , The teacher told that, the answer was wrong. Why?, , Next, if r = 0, it means either the event cannot, happen or has not occurred in a large number of trials. (Can you get a 7, when you roll a, dice?)., Thus , in this case, , 0, = = 0. , n n, r, , … (2), , Lastly, if r = n, the event must occur (in every trial or in a large number of trials)., In such a situation,, , n, = = 1., n n, r, , , , … (3), , (getting any number from 1 to 6 when you roll a dice), From (1), (2) and (3) we find 0 ≤ P (E ) ≤ 1., 296, , 9 Probability.indd 296, , 9th Standard Mathematics, , 26-12-2019 14:21:23
Page 303 :
www.tntextbooks.in, , Progress Check, A random experiment was conducted. Which of these cannot be considered as a, probability of an outcome?, (i) 1/5, , (ii) - 1/7, , (iii) 0.40, , (iv) - 0.52, , (vi) 1.3, , (vii) 1, , (viii) 72%, , (ix) 107%, , Example 9.1, , (v) 0, , When a dice is rolled, find the probability to get the number which, , is greater than 4?, Solution, Sample space S ={1,2,3,4,5,6}, Let E be the event of getting a number greater than 4, E ={5,6}, P(E) =, , Number of favourable outcomes, Total number of outcomes, , P(E) =, , n(E ) 2, = = 0.333..., n(S) 6, , Example 9.2, , In an office, where 42 staff members work, 7 staff members use cars,, 20 staff members use two-wheelers and the remaining 15 staff members use cycles. Find, the relative frequencies., Solution, Total number of staff members = 42., The relative frequencies:, Car users =, Two-wheeler users =, Cycle users =, , 7, 42, 20, 42, 15, 42, , 1, =, 6, 10, =, 21, 5, =, 14, , In this example note that the, total probability does not, exceed 1 that is,, 1 10 5, 7 20 15, =, +, + =1, + +, 6 21 14 42 42 42, , Example 9.3, , Team I and Team II play 10 cricket matches each of 20 overs. Their, total scores in each match are tabulated in the table as follows:, Probability 297, , 9 Probability.indd 297, , 26-12-2019 14:21:24
Page 304 :
www.tntextbooks.in, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Team I, , 200, , 122, , 111, , 88, , 156, , 184, , 99, , 199, , 121, , 156, , Team II, , 143, , 123, , 156, , 92, , 164, , 72, , 100, , 201, , 98, , 157, , Match numbers, , What is the relative frequency of Team I winning?, Solution, In this experiment, each trial is a match where Team I faces Team II., We are concerned about the winning status of Team I., There are 10 trials in total; out of which Team I wins in the 1st, 6th and 9th matches., 3, or 0.3., 10, (Note : The relative frequency depends on the sequence of outcomes that we observe, during the course of the experiment)., The relative frequency of Team I winning the matches =, , Exercise 9.1, 1., , You are walking along a street. If you just choose a stranger crossing you, what is the, probability that his next birthday will fall on a sunday?, , 2., , What is the probability of drawing a King or a Queen or a Jack from a deck of cards?, , 3., , What is the probability of throwing an even number with a single standard dice of, six faces?, , 4., , There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining, are Green then, what is the probability of picking out (i) a Blue ball, (ii) a Red ball, and (iii) a Green ball?, , 5., , When two coins are tossed, what is the probability that two heads are obtained?, , 6., , Two dice are rolled, find the probability that the sum is, i) equal to 1 ii) equal to 4, , 7., , 298, , 9 Probability.indd 298, , iii) less than 13, , A manufacturer tested 7000 LED lights at random and found that 25 of them were, defective. If a LED light is selected at random, what is the probability that the, selected LED light is a defective one., 9th Standard Mathematics, , 26-12-2019 14:21:24
Page 305 :
www.tntextbooks.in, , 8., , In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40, attempts tried by a team. Find the probability that the opponent team can convert, the attempt into a goal., , 9., , What is the probability that the spinner will not land on a, multiple of 3?, , 10., , Frame two problems in calculating probability, based on the, spinner shown here., , 9.5 Types of Events, We have seen some important cases of events already., When the likelihood of happening of two events are same they are known as equally, likely events., zz, , If we toss a coin, getting a head or a tail are equally likely events., , zz, , If a dice is rolled, then getting an odd number and getting an even number are, equally likely events, whereas getting an even number and getting 1 are not, equally likely events., , When probability is 1, the event is sure to happen. Such an event is called a sure or, certain event. The other extreme case is when the probability is 0, which is known as an, impossible event., If P(E) = 1 then E is called Certain event or Sure event., If P(E) = 0 then E is known is an Impossible event., Consider a “coin flip”. When you flip a coin, you cannot, get both heads and tails simultaneously. (Of course, the coin, must be fair; it should not have heads or tails on both sides!)., If two events cannot occur simultaneously (at the same time),, in a single trial they are said to be mutually exclusive events., Are rain and sunshine mutually exclusive? What about choosing, Kings and Hearts from a pack of 52 cards?, A dice is thrown. Let E be the event of getting an “even, face”. That is getting 2, 4 or 6. Then the event of getting an “odd, face” is complementary to E and is denoted by E ′ or E c . In the, above sense E and E ′ are complementary events., Probability 299, , 9 Probability.indd 299, , 26-12-2019 14:21:25
Page 306 :
www.tntextbooks.in, , Note, z The events E and E ′ are mutually exclusive. (how?), z The probability of E + the probability of E ′ =1. Also E and E ′ are mutually, , exclusive and exhaustive., , ( ), , ( ), , z Since P E + P E ′ = 1, if you know any one of them, you can find the other., , Progress Check, Which among the following are mutually exclusive?, Sl.No. Trial, 1, Roll a dice, , Event 1, getting a 5, , Event 2, getting an odd number, , 2, , Roll a dice, , getting a 5, , getting an even number, , 3, , Draw a card from, a standard pack, , getting a Spade Card, , getting a black, , 4, , Draw a card from, a standard pack, , getting a Picture Card, , getting a 5, , 5, , Draw a card from, a standard pack, , getting a Heart Card, , getting a 7, , Example 9.4, , 91, The probability that it will rain tomorrow is, . What is the, 100, probability that it will not rain tomorrow?, Solution, Let E be the event that it will rain tomorrow. Then E ′ is the event that it will not rain, tomorrow., Since P(E) = 0.91, we have P (E ′) = 1−0.91 (how?), = 0.09, Therefore, the probability that it will not rain tomorrow, = 0.09, , 300, , 9 Probability.indd 300, , 9th Standard Mathematics, , 26-12-2019 14:21:26
Page 307 :
www.tntextbooks.in, , Example 9.5, , In a recent year, of the 1184 centum scorers in various subjects in, tenth standard public exams, 233 were in mathematics. 125 in social science and 106 in, science. If one of the student is selected at random, find the probability of that selected, student,, (i) is a centum scorer in Mathematics, , (ii) is not a centum scorer in Science, , Solution, Total number of centum scorers =1184, Therefore n = 1184, (i) Let E1 be the event of getting a centum scorer in Mathematics., Therefore n(E1 ) = 233, That is, r1 = 233, r1 233, =, n 1184, (ii) Let E2 be the event of getting a centum scorer in Science., P(E1) =, , Therefore, , n(E2 ) = 106, That is, r2 = 106, r2 106, =, n 1184, P ( E 2 ′ ) = 1 − P ( E2 ), P(E2) =, , 106, 1184, 1078, =, 1184, =1−, , Exercise 9.2, 1., , A company manufactures 10000 Laptops in 6 months. Out of which 25 of them are, found to be defective. When you choose one Laptop from the manufactured, what, is the probability that selected Laptop is a good one., , 2., , In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their, voter ID card. If a youngster is selected at random, find the probability that the, youngster does not have their voter ID card., , 3., , x, The probability of guessing the correct answer to a certain question is . If the, 3, x, probability of not guessing the correct answer is , then find the value of x ., 5, Probability 301, , 9 Probability.indd 301, , 26-12-2019 14:21:28
Page 308 :
www.tntextbooks.in, , 4., , If a probability of a player winning a particular tennis match is 0.72. What is the, probability of the player loosing the match?, , 5., , 1500 families were surveyed and following data was recorded about their maids at, homes, Type of maids, , Only part time, , Only full time, , Both, , 860, , 370, , 250, , Number of families, , A family is selected at random. Find the probability that the family selected has, (i)Both types of maids, , (ii) Part time maids, , (iii)No maids, , Exercise 9.3, Multiple choice questions, 1., , A number between 0 and 1 that is used to measure uncertainty is, called, (1) Random variable, , 2., , 4., , (3) Simple event (4) Probability, , (3) 0 and n, , (4) 0 and ∞, , Probability lies between, (1) −1 and +1 (2) 0 and 1, , 3., , (2) Trial, , The probability based on the concept of relative frequency theory is called, (1) Empirical probability, , (2) Classical probability, , (3) Both (1) and (2), , (4) Neither (1) nor (2), , The probability of an event cannot be, (1) Equal to zero (2) Greater than zero (3) Equal to one (4) Less than zero, , 5., , The probability of all possible outcomes of a random experiment is always equal to, (1) One, , 6., , 9 Probability.indd 302, , (4) Less than one, , (2) 0, , (3) 1−A, , (4) 1−P(A), , Which of the following cannot be taken as probability of an event?, (1) 0, , 302, , (3) Infinity, , If A is any event in S and its complement is A’ then, P ( A′) is equal to, (1) 1, , 7., , (2) Zero, , (2) 0.5, , (3) 1, , (4) −1, , 9th Standard Mathematics, , 26-12-2019 14:21:29
Page 309 :
www.tntextbooks.in, , 8., , A particular result of an experiment is called, (1) Trial, , 9., , (3) Compound event, , (4) Outcome, , A collection of one or more outcomes of an experiment is called, (1) Event, , 10., , (2) Simple event, , (2) Outcome, , (3) Sample point (4) None of the above, , The six faces of the dice are called equally likely if the dice is, (1) Small, , (2) Fair, , (3) Six-faced, , (4) Round, , Points to Remember, , , If we are able to predict the exact outcome of an experiment then it is called, deterministic experiment., , , , If we cannot predict the exact outcome of an experiment then it is called, random experiment., , , , Sample space S for a random experiment is the set of all possible outcomes of, a random experiment., , , , An event is a particular outcome or combination of outcomes of an experiment., , , , Empirical probability states that probability of an outcome is close to the, percentage of occurrence of the outcome., , , , If the likelyhood of happening of two events are same then they are known as, equally likely events., , , , If two events cannot occur simultaneously in single trial then they are said to, be mutually exclusive events., , , , Two events E and E ′ are said to be complementary events if P (E ) + P (E ′) = 1., , , , An event which is sure to happen is called certain or sure event. The probability, of a sure event is always one., , , , An event which never happen is called impossible event. The probability of an, impossible event is always zero., , Probability 303, , 9 Probability.indd 303, , 26-12-2019 14:21:29
Page 310 :
www.tntextbooks.in, , ICT Corner, Expected Result is shown, in this picture, , Step – 1, Open the Browser by typing the URL Link given below (or) Scan the QR Code. GeoGebra work, sheet named “Probability” will open. There are two worksheets under the title Venn diagram and, Basic probability., Step - 2, Click on “New Problem”. Work out the solution, and click on the respective check box and check, the answer., Step 1, , Step 2, , Browse in the link, , Probability: https://ggbm.at/mj887yua or Scan the QR Code., , 304, , 9 Probability.indd 304, , 9th Standard Mathematics, , 26-12-2019 14:21:30
Page 311 :
www.tntextbooks.in, , ANSWERS, 1, , Set Language, Exercise 1.1, , 1. (i) set, , (ii) not a set, , 2. (i) {I, N, D, A}, , (iii) Set , , (iv) not a set, , (ii) {P, A, R, L, E, O, G, M}, , (iii) {M, I, S, P}, , (iv) {C, Z, E, H, O, S, L, V, A, K, I}, 3. (a) (i) True, (b) (i) A, , (ii) True, , (iii) False, , (iv) True, , (ii) C , , (iii) g , , (iv) ! , , 4. (i) A = {2, 4, 6, 8, 10, 12, 14, 16, 18} , (iii) C = {64, 125}, , , , (v) False, , (vi) False, , 1 /, (ii) B = % 12 , 14 , 16 , 18 , 10, , (iv) D = #- 4, - 3, - 2, - 1, 0,1, 2 -, , 5. (i) B = {x : x is an Indian player who scored double centuries in One Day International}, (ii) C = $ x : x =, , n, , n ! N . , n+ 1, , (iii) D = {x : x is a tamil month in a year}, , (iv) E = {x : x is an odd whole number less than 9}, 6. (i) P = The set of English months starting with letter ‘J’, (ii) Q = The set of Prime numbers between 5 and 31, (iii) R = The set of natural numbers less than 5, (iv) S = The set of English consonants, Exercise 1.2, , 1. (i) n(M) = 6, , (ii) n(P) =5, , (iii) n(Q) = 3, , (iv) n(R) = 10, , 2. (i) finite, , (ii) infinite, , (iii) infinite, , (iv) finite, , 3. (i) Equivalent sets, , (ii) Unequal sets (iii) Equal sets, , (v) n(S) =5, , (iv) Equivalent sets, Answers 305, , Answers_Combine.indd 305, , 26-12-2019 11:53:54
Page 312 :
www.tntextbooks.in, , 4. (i) null set, , (ii) null set, , (iii) singleton set, , 5. (i) overlapping, , (ii) disjoint, , (iii) overlapping, , 6. (i) {square, rhombus}, , (ii) {circle}, , (iii) {triangle}, , (iv) null set, , (iv) { }, , 7. { }, {a}, {a, b}, {a, {a, b}}, 8. (i) {{ }, {a}, {b}, {a, b}}, (ii) {{ }, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3 }, {1, 2, 3}}, (iii) {{ }, {p}, {q} {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s},, {q, r, s}, {p, q, r, s}}, , (iv) P(E)= {{ }}, , 9. (i) 8, 7, , (ii) 1024, 1023 , , 10. (i) 16, , (ii) 1, , (iii) 8, Exercise 1.3, , 1. (i) {2, 4, 7, 8, 10}, (iv) {4, 7} , , (ii) {3, 4, 6, 7, 9, 11}, , (iii) {2, 3, 4, 6, 7, 8, 9, 10, 11}, , (v) {2, 8, 10} , , (vi) {3, 6, 9, 11}, , (vii) {1, 3, 6, 9, 11, 12} , , (viii) {1, 2, 8, 10, 12}, , (ix) {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}, 2. (i) {2, 5, 6, 10, 14, 16}, {2, 14}, {6, 10}, {5, 16}, (ii) {a, b, c, e, i, o, u}, {a, e, u}, {b, c}, {i, o}, (iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 2, 3, 4, 5}, {6, 7, 8, 9, 10,}, {0}, (iv) {m, a, t, h, e, i, c, s, g, o, r, y}, {e, m, t,}, {a, h, i, c, s}, {g, o, r, y}, 3. (i) {a, c, e, g} (ii) {b, c, f, g}, (vi) {a, b, c, e, f, g}, , 306, , (iii) {a, b, c, e, f, g}, , (iv) {c, g}, , (v) {c, g} , , (vii) {b, d, f, h} (viii) {a, d, e, h}, , 9th Standard Mathematics, , Answers_Combine.indd 306, , 26-12-2019 11:53:54
Page 313 :
www.tntextbooks.in, , 4. (i) {0, 2, 4, 6} (ii) {1, 4, 6}, , (iii) {0, 1, 2, 4, 6}, , (iv) {4, 6}, , (vi) {0, 1, 2, 4, 6} , , (vii) {1, 3, 5, 7} , , (viii) {0, 2, 3, 5, 7}, , 5. (i) {1, 2, 7} (ii) {m, o, p, q, j}, , (iii) {6, 9, 10}, , 6. (i) Y–X, , (iii) ^ X - Yh , ^ Y - X h, , (ii) ^ X , Y hl , , 7. (i), , (ii), , , , A∪B, , (iii), , A∩B (A∩B)′, , , (v), , (iv), , , , (v) {4, 6} , , (vi), , A′∪B′ , , (B–A)′ , , A′∩B′, , (vii) (A∩B)′ = A′∪B′, Exercise 1.4, , 1. (i) {1,2,3,4,5,7,9,11}, , (ii) {2,5}, , (iii) {3,5 }, Exercise 1.5, , 1. (i), (iv), , {3, 4, 6}, , (ii), , {−1, 5, 7}, , (iii) {−3, 0, 1, 2, 3, 4, 5, 6, 7, 8}, , {−3, 0, 1, 2}, , (v), , {1, 2, 4, 6}, , (vi), , 2. (i) { a,b,c,d,e,f }, , (ii) {a,b,d }, , {4, 6}, , (vii), , (iii) { a,b,c,d,e,f }, , {−1, 3, 4, 6}, , (iv) {a, b, d }, , Exercise 1.6, , 1. (i) 15, 65, , (ii) 250, 600, , 4. (i) 17, , (ii) 22, , (iii) 47, , 5. (i) 10, , (ii) 10, , (iii) 25, , 6. 1000, , 7. 8, , 9. (i) 185, , (ii) 141, , (iii) 326, , 10. 70, , 12. (i) 5, , (ii) 7, , 11. x = 20 , y = 40 , z = 30, , 8. Not correct, , (iii)8, , 13. 5, Answers 307, , Answers_Combine.indd 307, , 26-12-2019 11:53:55
Page 314 :
www.tntextbooks.in, , Exercise 1.7, , 1. (2) 2. (1) 3. (3) 4. (2) 5. (4) 6. (1) 7. (2) 8. (4) 9. (3) 10. (4), 11. (2) 12. (1) 13. (1) 14. (3) 15. (4) 16. (1) 17. (4) 18. (3) 19. (3) 20. (1), , 2, , Real Numbers, Exercise 2.1, , 1., , 6 - 5 - 4, 1, D 2. - 11, ,, ,, , ..., 11, 11, 11, , 3., , 9 19 39 79 159, (i) 40, ;, , ,, ,, ,, 80 160 320 640, , The given answer is one of the answers. There can be many more answers, , (ii) 0.101, 0.102, ... 0.109, The given answer is one of the answers. There can be many more answers, 17 - 33, (iii) - 23 , - 54 , - 89 , - 16, ,, 32, The given answer is one of the answers. There can be many more answers, Exercise 2.2, , 1. (i) 0.2857142..., Non terminating and recurring, , (ii) –5. 27 , Non terminating and, recurring, , (iii) 7. 3 , Non terminating and recurring , , (iv) 1.635, Terminating, , 2. 0.076293, 6 3. 0.0303, 2.15, 4. (i) 24, 99, , (ii) 2325, 999, , 5. (i) Terminating, , (iii) - 1283, 250, , (ii) Terminating, , 5681, (iv) 143, (v), 45, 330, , (iii) Non terminating, , (vi) - 190924, 9000, , (iv) Non terminating, , Exercise 2.3, , 2. (i) 0.301202200222..., 0.301303300333..., , (ii) 0.8616611666111 ..., 0.8717711777111 ..., , (iii) 1.515511555..., 1.616611666..., 3. 2.2362, 2.2363, Exercise 2.5, , 1.(i) 54, , (ii) 5-1, , 2.(i) 42, , (ii) 4 2, , 308, , 3, , 1, , (iii) 5 2, , 3, , (iv) 5 2, , 5, , (iii) 4 2, , 9th Standard Mathematics, , Answers_Combine.indd 308, , 26-12-2019 11:53:56
Page 315 :
www.tntextbooks.in, , 3.(i) 7, , (ii) 9, , 1, , (iii), 1, , 1, 27, , (iv), , 10, , 25, 16, , 4.(i) 5 2, , (ii) 7 2, , (iii) 7 3, , (iv) 10, , 5.(i) 2, , (ii) 3, , (iii) 10, , (iv), , -14, 3, , 4, 5, , Exercise 2.6, , 1.(i) 21 3, , (ii) 3 3 5, , (iii) 26 3, , (iv) 8 3 5, , 2. (i), , (ii), , (iii) 30, , (iv) 49a - 25b, , 30, 5, 16, , (v), 4. (i), , 3, , 5, , 3.(i) 1.852, , (ii) 23.978, , 5 > 6 3 > 9 4 , , (ii), , 3>, , 2 3, , 5>, , 3 4, , 7, , 5. (i) yes, , (ii) yes, , (iii) yes, , (iv) yes, , 6. (i) yes, , (ii) yes, , (iii) yes, , (iv) yes, , Exercise 2.7, , 1.(i), , 2, 10, , 5, 3, , (iii), , 5 6, 6, , (iv), , (ii) 13 - 4 6, , (iii), , 9 + 4 30, 21, , (iv) -2 5, , (ii), , 4, (5 + 2 6 ), 3, 11, −4, 3. a =, ,b =, 3, 3, , 2. (i), , 4. x 2 +, , 1, = 18, x2, , 30, 2, , 5. 5.414, , Exercise 2.8, , 1. (i) 5.6943 ´ 1011, , (ii) 2.00057 ´ 103, , (iii) 6.0 × 10−7, , (iv) 9.000002 × 10−4, , 2. (i) 3459000, , (ii) 56780, , (iii) 0.0000100005, , 3. (i) 1.44 ´ 1028, , (ii) 8.0 × 10−60, , (iii) 2.5 × 10−36, , 4.(i) 7.0 ´ 109, , (ii) 9.4605284 ´ 1015 km, , 5. (i) 1.505 ´ 108, , (ii) 1.5522 ´ 1017, , (iv) 0.0000002530009, , (iii) 9.1093822 × 10−31 kg, , (iii) 1.224 ´ 107, , (iv) 1.9558 × 10−1, , Exercise 2.9, , 5. (4), , 6. (2), , 1. (4), , 2. (3) 3. (2) 4. (1), , 7. (2), , 8. (2), , 9. (4), , 10. (1), , 11. (4), , 12. (4) 13. (4) 14. (2) 15. (2) 16. (3) 17. (2) 18. (4) 19. (2) 20. (3), Answers 309, , Answers_Combine.indd 309, , 26-12-2019 11:53:58
Page 316 :
www.tntextbooks.in, , 3, , Algebra, Exercise 3.1, , 1. (i) not a polynomial, , (ii) polynomial , , (iii) not a polynomial, , (iv) polynomial, , (v) polynomial , , (vi) not a polynomial, , 2. Cooefficient of x2 , , Cooefficient of x, , 2, 5, , (i), , –3, , (ii) –2, , - 7, , (iii) p, , –1, , (iv), (v), , 3, , 1, , 3. (i) 7 (ii) 4, , (iii) 5, , (iv) 6, , (v) 4, , 4. Descending order , 7 x + 6x + x - 9, 7 4, 3, 2, (ii) - 2 x - 5x + 2 x + x, 6 2, 3, (iii) 7x - 5 x + 4x - 1, 7, 4, 3, 2, (iv) 9y + 5 y + y - 3 y - 11, 3, , (i), , 2, , 2, -7, 2, , Ascending order, - 9 + x + 6x2 +, , 7 x3, 7, x + 2 x2 - 5x3 - x4, 2, 6, 2, - 1 + 4x - x + 7x3, 5, 7, - 11 - y + y2 + 5 y3 + 9y4, 3, , 5. (i) 6x3+6x2–14x+17, 3, , (ii) 7x3+7x2+11x–8, 3, , (iii) 16x4–6x3–5x2+7x–6, 4, , 6. (i) 7x2+8, 2 , , (ii) –y3+6y2–14y+2, 3, , (iii) z5–6z4–6z2–9z+7, 5, , 7. x3–8x2+11x+7 , , 8. 2x4–3x3+5x2–5x+6, , 9. (i) 6x4+ 7x3–56x2–63x+18, 4, , (ii) 105x2–33x–18, 2, , (iii) 30x3–77x2+54x–7, 3, , 10. x2+y2+2xy, ` 225 , , 11. 9x2–4, 3596 sq. units, , 12. cubic polynomial or polynomial of degree 3, Exercise 3.2, 9, 4. (i) 56 (ii) –3 (iii) - 10, , 3. (i) 3 (ii) - 52 (iii) 32 (iv) 0, (iv) 94, , 6. (i) 2 (ii) 3, , (v) 1, , 1. (i) 6 (ii) –6 (iii) 3, (iii) 0, , 2. 1, (iv) 1, , (v) 0, , (vi) - ba, , Exercise 3.3, , 1. p(x) is not a multiple of g(x), 2. (i) Remainder : 0, , (ii) Remainder : 32, , (iii) Remainder : 62, , 3. Remainder : –143, , 4. Remainder : 2019, , 5. K = 8, , 310, , 9th Standard Mathematics, , Answers_Combine.indd 310, , 26-12-2019 11:54:01
Page 318 :
www.tntextbooks.in, , Exercise 3.6, , 1.(i) (x + 6)(x + 4) , (ii) (z + 6)(z − 2) , (iii) (p − 8)(p + 2) , (iv) (t - 9)(t - 8), (v) (y − 20)(y + 4), (vi) (a + 30)(a − 20), 2. (i) (2a + 5)(a + 2), (ii) (x − 7y )(5x + 6y ) (iii) (2x - 3)(4x - 3), (v) 3x 2 (3y + 2)2, , (iv) 2(3x + 2y )(x + 2y ), , (vi) (a + b + 6)(a + b + 3), , 3. (i) (p − q − 8)(p − q + 2) , , (, , (ii) (m + 6n )(m − 4n ) (iii) a + 5, , )(, , ), , 5a − 3 (iv) (a + 1)(a − 1)(a 2 − 2), , (v) m(4m + 5n )(2m − 3n ) , 2, , 1 1, (vi) + , x y , , Exercise 3.7, , 1. (i) Quotient : 4x2–6x–5, Remainder : 33, (iii) Quotient : 4x2+2x+1, Remainder : 0, , (ii) Quotient : 4y2–6y+5, Remainder : –10, (iv) Quotient : 8z2–6z+2, Remainder : 10, , 2. Length : x+4 3. Height : 5x–4 4. Mean : x2–5x+25, 5. (i) x 2 + 4x + 5, 12 , (ii) (x 2 - 1), - 2, (iii) 3x 2 − 11x + 40, − 125 , , (iv) 2x 3 −, , 51 109, x 2 3x, −, + ,, 2, 8, 32 32, , 6. 4x 3 − 2x 2 + 3, p = −2, q = 0, remainder=−10, 7. a = 20, b = 94 & remainder=388, Exercise 3.8, , 1.(i) (x − 2)(x + 3)(x − 4) , , (ii) (x + 1)(x − 2)(2x − 1), , (iii) (x − 1)(2x − 1)(2x + 3) , , (iv) (x + 2)(x + 3)(x − 4) , , (v) (x − 1)(x − 2)(x + 3), 312, , , , (vi) (x − 1)(x − 10)(x + 1), , 9th Standard Mathematics, , Answers_Combine.indd 312, , 26-12-2019 11:54:06
Page 319 :
www.tntextbooks.in, , Exercise 3.9, , 1. (i) p 5, (v) abc, 2. (i) 1, (v), , (x + 1)(x − 1), , (ii) 1, , (iii) 3a 2b 2c 3, , (vi) 7xyz 2, , (vii) 25ab, , (ii) a m+1, , (iii) (2a + 1), , (iv) 16x 6, (viii) 1, (iv) 1, , (vi) (a − 3x ), Exercise 3.10, , 2. (i) (5,2), (iv) (−3, −3), , (ii) Infinite number of solutions, , (iii) no solution, , (v) (1,3) , , (vi) (−3, 3), , 3. 75km/hr, 25km/hr, Exercise 3.11, , 1.(i)(2, −1), , (ii) (4,2), , (2) 45 , , (3) 409, , (iii) (40,100), , (iv), , (, , 8, 3, , ), , Exercise 3.12, , 1.(i) (2,1), 1 , (v) , −1, 3 , , (ii) (7,2), , (iii) (80,30) (iv), , (vi) (2,4), , 3, 1, 2 , , (2) `30000, `40000 (3) 75, 15, Exercise 3.13, , 1 1, (iii) − , , 2 3, (2) Number of 2 rupee coins 60; Number of 5 rupee coins 20, , 1.(i) (3,4), , (ii) (3, −1), , (3) Larger pipe 40 hours; Smaller pipe 60 hours, Exercise 3.14, , 1. 64, 5, 7, 3. ∠A = 120° , ∠B = 70° , ∠C = 60° , ∠D = 110°, 2., , 4. Price of TV = `20000; Price of fridge = `10000, 5. 40, 48, 6. 1 Indian – 18 days; 1 Chinese – 36 days, , Answers 313, , Answers_Combine.indd 313, , 26-12-2019 11:54:07
Page 320 :
www.tntextbooks.in, , Exercise 3.15, , 5. (2), , 6. (1), , 1. (4), , 2. (3) 3. (4) 4. (4), , 11. (2), , 12. (3) 13. (3) 14. (2) 15. (3) 16. (3) 17. (4) 18. (2) 19. (3) 20. (2), , 21. (4), , 22. (3) 23. (2) 24. (1) 25. (2) 26. (3) 27. (1) 28. (3) 29. (2), , 4, , 7. (4), , 8. (4), , 9. (4) 10. (3), , Geometry, Exercise 4.1, , 1. (i) 70°, , (ii) 288°, , (iii) 89° 2. 30°, 60°, 90° 5. 80°, 85 °, 15 °, Exercise 4.2, , 1., , (i) 40°, 80°, 100°, 140°, , 2. 62°, 114° , 66° 3. 44° 4. 10cm, , 7. (i) 30° (ii) 105° (iii) 75° (iv) 105° 8. 122° , 29o, 9. Ratios are equal 10. d = 7.6, Exercise 4.3, , 1. 24cm, , 2. 17cm, , 3. 8cm, 45°, 45°, , 4. 18cm, , 5. 14 cm, , 6. 6 cm, , 7. (i) 45°, , (ii) 10°, , (iii) 55°, , (iv) 120°, , (v) 60°, , 8. ∠BDC = 25°, ∠DBA = 65°, ∠COB = 50°, Exercise 4.4, , 1. 30°, , 2.(i) ∠ACD = 55°, , (ii) ∠ACB = 50°, , (iii) ∠DAE = 25°, , 3. ∠A = 64°; ∠B = 80°; ∠C = 116°; ∠D = 100°, 4.(i) ∠CAD = 40°, 7. ∠OAC = 30°, , (ii) ∠BCD = 80°, 8. 5.6m, , 5. Radius=5cm, , 6. 3.25m, , 9. ∠RPO = 60°, , Exercise 4.7, 2. (3) 3. (1) 4. (4), , 11. (3), , 12. (3) 13. (1) 14. (1) 15. (4) 16. (2) 17. (2) 18. (3) 19. (2) 20. (4), , 314, , 5. (4), , 6. (3), , 1. (2), , 7. (2), , 8. (2), , 9. (4), , 10. (2), , 9th Standard Mathematics, , Answers_Combine.indd 314, , 26-12-2019 11:54:09
Page 321 :
www.tntextbooks.in, , 5, , Coordinate Geometry, Exercise 5.1, , 1. P, (–7,6) = II Quadrant; Q(7,–2) = IV Quadrant; R(–6, –7) = III Quadrant;, S(3,5) = I Quadrant; and T (3,9) = I Quadrant, , 2. (i) P = (– 4,4), , (ii) Q = (3,3) , , 3. (i) Straight line parallel to x -axis, 4. (i) Square , , (iii) R=(4,–2) , , (iv) S = (–5,–3), , (ii) Straight line which lie on y -axis., , (ii) Trapezium, Exercise 5.2, , 1. (i) 10 units, , (ii) 2 26 units (iii) c–a (iv)13 units, , 2. (i) Collinear, , (ii) Collinear, , 7. 5 or 1, , 8. Coordinates of A (9, 9) or (–5,–5) 9. y = 4x+9 10. Coordinates of P(2,0), 12. 30 2, Exercise 5.3, , 1.(i) (−4, −1), , (ii) (0, −1), , 2. (−5, −3), , 3. P = −15, , 9 3, 5. , , 2 2, , 6. (1, 8), , (iii) (a + b, a), , (iv) (1, −1), , 4. (9, 3)(−5, 5) and (1,1), , Exercise 5.4, , 1. (7, 3), , 2. 5:2 , , 3. (3, 4), , 4. (−2, 3) , (1, 0), , 19 13 −9 −15 , 5. , , ,, 2 2 2 2 , , 7. (3, 2), , Exercise 5.5, , 1.(i) (2, −3), 4. 20, , −8 −11, (ii) ,, 3 3 , 5. 3, , 5, units, 2, , 2. (4, −6), , 3. 5 units, , 6. (1, 0), , 7. (5, −2), , Exercise 5.6, , 5. (2), , 6. (4), , 1. (3), , 2. (3) 3. (3) 4. (2), , 7. (3), , 8. (3), , 9. (3), , 10. (3), , 11. (4), , 12. (1) 13. (3) 14. (4) 15. (2) 16. (3) 17. (2) 18. (2) 19. (4) 20. (2), Answers 315, , Answers_Combine.indd 315, , 26-12-2019 11:54:12
Page 322 :
www.tntextbooks.in, , 6, , Trigonometry, Exercise 6.1, , 9, 40, 9, 41, 41, 40, ; cos B = ; tan B =, ; cosecB = ;sec B = ; cot B =, 41, 41, 40, 9, 40, 9, 12, 13, 5, 4, 2. (i) sin B =, (ii) sec B =, (iii) cot B =, (iv) cosC =, 13, 5, 12, 5, 3, 5, (v) tanC =, (vi) cosecC =, 4, 3, 1, 3, 1, 2, 2, ; tan q =, ; cot q = 3, 3. sin q = ; cos q =, ; cosecq = ;sec q =, 1, 2, 2, 3, 3, 1. sin B =, , 3, 4., 40, 1, 2, , 8., , 1 − x2, 1 − x2, ; tan A =, 5. sin A =, 2x, 1 + x2, , 4, 4, 4, 9. sin α = ; cos β = ; tan φ =, 5, 5, 3, , 7., , 1, 2, , 10. 7m, , Exercise 6.2, , 2.(i) 0, , (ii), , 7, 4, , (iii) 3 , , 4. 2, , Exercise 6.3, , 1.(i) 1, , (ii) 1, , (iii) 1 , , (iv) 2, , Exercise 6.4, , 1.(i) 0.7547, , (ii) 0.2648, , (iii) 1.3985, , (v) 0.8302, , (vi) 2.7907, , 2.(i) 85°57′ (or) 85°58′ (or) 85°59′, , (ii) 47°27′, , (iii) 4°7′, , (iv) 87°39′, , 3.(i) 1.9970, , (ii) 2.8659, , 4. 18.81 cm2, , (iv) 0.3641, (v) 82°30′, 5. 36°52′, , 6. 54.02 m, Exercise 6.5, , 1. (1) 2. (2) 3. (2) 4. (3) 5. (2) 6. (3) 7. (3) 8. (1) 9. (2) 10. (2), , 7, , Mensuration, Exercise 7.1, , 1.(i) 120 cm2, , (ii) 7.2 m2, , 4. 1558.8 cm2, , 5. ₹ 1050, , 8. 354m2, , 9. 1536 m2, , 316, , 2. 1320 m2, ₹26400, , 3. 12000 m2, , 6. 240 cm2, , 7. 138 cm2, , 10. 672 m2, , 9th Standard Mathematics, , Answers_Combine.indd 316, , 26-12-2019 11:54:16
Page 323 :
www.tntextbooks.in, , Exercise 7.2, , 1. 1160cm2, 560cm2, 4.(i) 384 m2, 256 m2, 5. 1600 cm2, , 2. ₹1716, , 3. ₹3349, , (ii) 2646 cm2, 1764 cm2, , (iii) 337.5 cm2, 225 cm2, , 6. 253.50m2, ₹6084, , 7. 224cm2, 128cm2, , Exercise 7.3, , 1.(i) 576 cm3, , (ii) 2250 m3, , 3. 25 cm, 20 cm, 15 cm , , 2. 630 cm3, 4. 2624000 litres, , 6. 12 m, , 7.(i) 125 cm3, , 8. 5 m, , 9. 15 cm, , (ii) 42.875 m3, , 5. 25000, (iii) 9261 cm3, , Exercise 7.4, , 1. (3) 2. (2) 3. (4) 4. (3) 5. (3) 6. (1) 7. (2) 8. (3) 9. (4) 10. (1), , 8, , Statistics, Exercise 8.1, , 1. 27°C, , 2. 44kg, , 3. 56.96 (or) 57 (approximately), , 4. 142.5 mm3, , 5. p = 20, , 6. 40.2, , 7. 29.29, , 8. 29.05, , Exercise 8.2, , 1. 47, , 2. 44, , 5. 31, , 6. 38, , 3. 21, , 4. 32, , Exercise 8.3, , 1. 6600, 7000, 7000, , 2. 3.1 and 3.3 (bimodal) , , 3. 15, , 4. 40, , 5. 24 , , 6. 58.5, , Exercise 8.4, , 1. (1) 2. (3) 3. (3) 4. (2) 5. (1) 6. (4) 7. (1) 8. (2) 9. (2) 10. (3), , Answers 317, , Answers_Combine.indd 317, , 26-12-2019 11:54:16
Page 324 :
www.tntextbooks.in, , 9, , Probability, Exercise 9.1, , 1, 7, 1, (ii), 8, 1, (ii), 12, 3, 9., 4, 1., , 1. 0.9975, 5.(i), , 1, 6, , 3, 13, 2, (iii), 3, 2., , (iii) 1, , 1, 2, 1, 5., 4, 1, 7., 280, 3., , 4.(i), , 5, 24, , 6.(i) 0, 8., , 1, 5, , Exercise 9.2, , 209, 400, 43, (ii), 75, 2., , 15, 8, 1, (iii), 75, 3., , 4. 0.28, , Exercise 9.3, , 1. (4) 2. (2) 3. (1) 4. (4) 5. (1) 6. (4) 7. (4) 8. (4) 9. (1) 10. (2), , 318, , 9th Standard Mathematics, , Answers_Combine.indd 318, , 26-12-2019 11:54:19
