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MATHEMATICS, Part -, , STANDARD NINE, , 9, , 2, , 61.00

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The Constitution of India, Chapter IV A, , Fundamental Duties, ARTICLE 51A, Fundamental Duties- It shall be the duty of every citizen of India(a), , to abide by the Constitution and respect its ideals and institutions,, the National Flag and the National Anthem;, , (b), , to cherish and follow the noble ideals which inspired our national, struggle for freedom;, , (c), , to uphold and protect the sovereignty, unity and integrity of India;, , (d), , to defend the country and render national service when called upon, to do so;, , (e), , to promote harmony and the spirit of common brotherhood amongst, all the people of India transcending religious, linguistic and regional, or sectional diversities, to renounce practices derogatory to the, dignity of women;, , (f), , to value and preserve the rich heritage of our composite culture;, , (g), , to protect and improve the natural environment including forests,, lakes, rivers and wild life and to have compassion for living, creatures;, , (h), , to develop the scientific temper, humanism and the spirit of inquiry, and reform;, , (i), , to safeguard public property and to abjure violence;, , (j), , to strive towards excellence in all spheres of individual and, collective activity so that the nation constantly rises to higher levels, of endeavour and achievement;, , (k), , who is a parent or guardian to provide opportunities for education, to his child or, as the case may be, ward between the age of six, and fourteen years.

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1, , Basic Concepts in Geometry, , Let’s study., , • Point, line and plane, • Betweenness, • Co-ordinates of a points and distance • Conditional statements, • Proof, Did you recognise the adjacent, picture ? It is a picture of pyramids, in Egypt, built 3000 years before, Christian Era. How the people were, able to build such huge structures in, so old time ? It is not possible to, build such huge structures without, developed knowledge of Geometry, and Engineering., The word Geometry itself suggests the origin of the subject. It is, generated from the Greek words Geo, (Earth) and Metria (measuring). So, it can be guessed that the subject must have evolved from the need of measuring the Earth,, that is land ., Geometry was developed in many nations in different periods and for different constructions. The first Greek mathematician, Thales, had gone to Egypt. It is said that he determined height of a pyramid by measuring its shadow and using properties of similar triangles., Ancient Indians also had deep knowledge of Geometry. In vedic period, people used, geometrical properties to build altars. The book shulba-sutra describes how to build different, shapes by taking measurements with the help of a string. In course of time, the mathematicians Aaryabhat, Varahamihir, Bramhagupta, Bhaskaracharya and many others have given, valuable contribution to the subject of Geometry., , Let’s learn., Basic concepts in geometry (Point, Line and Plane), We do not define numbers. Similarly we do not define a point, line and plane also. These, are some basic concepts in Geometry. Lines and planes are sets of points. Keep in mind that, the word ‘line’ is used in the sense ‘straight line’., , 1

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Co-ordinates of points and distance, Observe the following number line., , B C, A, -5 -4 -3 -2 -1, , O, 0, , D, 1, , 2, , E, 3, , 4, , 5, , Fig. 1.1, Here, the point D on the number line denotes the number 1. So, it is said that 1 is the, co-ordinate of point D. The point B denotes the number - 3 on the line. Hence the co-ordinate, of point B is - 3. Similarly the co-ordinates of point A and E are - 5 and 3 respectively., The point E is 2 unit away from point D. It means the distance between points D and E, is 2. Thus, we can find the distance between two points on a number line by counting number, of units. The distance between points A and B on the above number line is also 2., Now let us see how to find distance with the help of co-ordinates of points., To find the distance between two points, consider their co-ordinates and subtract the, smaller co-ordinate from the larger., The co-ordinates of points D and E are 1 and 3 respectively. We know that 3 > 1., Therefore, distance between points E and D = 3 - 1 = 2, The distance between points E and D is denoted as d (E,D). This is the same as l(ED),, that is, the length of the segment ED., d (C, D) = 1 - (- 2) , d (E, D) = 3 - 1 = 2, =1+2=3, \ l(ED) = 2, \ d (C, D) = l(CD) = 3, d (E, D) = l(ED) = 2, Similarly d (D, C) = 3, Similarly d (D, E) = 2, Now, let us find d(A,B). The co-ordinate of A is - 5 and that of B is - 3; - 3 > - 5, \ d (A, B) = - 3 - (- 5) = - 3+5 = 2., From the above examples it is clear that the distance between two distinct points is always, a positive number., Note that, if the two points are not distinct then the distance between them is zero., , , Remember this !, , • The distance between two points is obtained by subtracting the smaller co-ordiante from, •, , the larger co-ordinate., The distance between any two points is a non-negative real number., , 2

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Let’s learn., Betweenness, If P, Q, R are three distinct collinear points, there are three possibilities., , P, , Q, , R Q, Fig. 1.2, , R, , P, , R, , (i) Point Q is between, P and R , , P, , Q, , (ii) Point R is between, (iii) Point P is between, P and Q , R and Q, , If d (P, Q) + d (Q, R) = d (P, R) then it is said that point Q is between P and R. The, betweeness is shown as P - Q - R., Solved examples, Ex (1) On a number line, points A, B and C are such that, , d (A, B) = 5, d (B,C) = 11, , and d (A, C) = 6., , Which of the points is between the other two ?, Solution : Which of the points A, B and C is between the other two, can be decided as, follows., , B, , , d(B,C) = 11 . . . . (I), , d(A,B) + d(A,C)= 5+6 = 11 . . . . (II), , \ d (B, C) = d (A, B) + d (A, C) . . . . [from (I) and (II)], , 5, , A, , 6, Fig. 1.3, , C, , Point A is between point B and point C., Ex (2) U, V and A are three cities on a straight road. The distance between U and A is, 215 km, between V and A is 140 km and between U and V is 75 km. Which of, them is between the other two ?, Solution : d (U,A) = 215; d (V,A) = 140;, , d (U,V) = 75, , , d (U,V) + d (V,A) = 75 + 140 = 215;, , d (U,A) = 215, , \ d (U,A) = d (U,V) + d (V,A), , \ The city V is between the cities U and A., , 3

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Ex (3) The co-ordinate of point A on a number line is 5. Find the co-ordinates of points on, the same number line which are 13 units away from A., Solution : As shown in the figure, let us take points T and D to the left and right of A, respectively, at a distance of 13 units., , 5-13, , , , T, -8, , 5+13, D, 18, , A, 5, Fig. 1.4, , The co-ordinate of point T, which is to the left of A ,will be 5 - 13 = - 8, The co-ordinate of point D, which is to the right of A, will be 5 + 13 = 18, , \ the co-ordinates of points 13 units away from A will be - 8 and 18., Verify your answer : d (A,D) = d(A,T) = 13, , Activity, (1) Points A, B, C are given aside. Check,, with a stretched thread, whether the, , A, , three points are collinear or not. If they, are collinear, write which one of them, is between the other two., (2) Given aside are four points P, Q, R,, and S. Check which three of them are, collinear and which three are non, , the other two., , C, , Q, , S, R, , collinear. In the case of three collinear, points, state which of them is between, , B, , P, , (3) Students are asked to stand in a line for mass drill. How will you check whether the, students standing are in a line or not ?, (4) How had you verified that light rays travel in a straight line ?, Recall an experiment in science which you have done in a previous standard., , 4

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Practice set 1.1, 1., , Find the distances with the help of the number line given below., , Q P K J H, -5 -4 -3 -2 -1, , O A, 0 1, Fig. 1.5, , 2., , 3., , B, 2, , C, 3, , D, 4, , E, 5, , 6, , (i) d(B,E), , (ii) d(J, A) , , (iii) d(P, C) , , (iv) d(J, H), , (v) d(K, O) , , (vi) d(O, E) , , (vii) d(P, J) , , (viii) d(Q, B), , If the co-ordinate of A is x and that of B is y, find d(A, B) ., (i) x = 1, y = 7 , , (ii) x = 6, y = - 2 , , (iii) x = - 3, y = 7, , (iv) x = - 4, y = - 5, , (v) x = - 3, y = - 6 , , (vi) x = 4, y = - 8, , From the information given below, find which of the point is between the other two., If the points are not collinear, state so., , 4., , (i) d(P, R) = 7, , , d(P, Q) = 10, , , d(Q, R) = 3, , (ii) d(R, S) = 8, , , d(S, T) = 6, , , d(R, T) = 4, , (iii) d(A, B) = 16, , , d(C, A) = 9, , , d(B, C) = 7, , (iv) d(L, M) = 11, , , d(M, N) = 12,, , d(N, L) = 8, , (v) d(X, Y) = 15, , , d(Y, Z) = 7, , , d(X, Z) = 8, , (vi) d(D, E) = 5, , , d(E, F) = 8, , , d(D, F) = 6, , On a number line, points A, B and C are such that d(A,C) = 10, d(C,B) = 8, Find d(A, B) considering all possibilities., , 5., , Points X, Y, Z are collinear such that d(X,Y) = 17, d(Y,Z) = 8, find d(X,Z) ., , 6., , Sketch proper figure and write the answers of the following questions., , , , (i) If A - B - C, , and l(AC) = 11,, , l(BC) = 6.5, then l(AB) =?, , (ii) If R - S - T, , and l(ST) = 3.7,, , l(RS) = 2.5,, , then l(RT) =?, , (iii) If X - Y - Z and l(XZ) = 3 7 , l(XY) = 7 , then l(YZ) =?, 7., , Which figure is formed by three non-collinear points ?, , 5

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Let’s learn., In the book, Mathematics - Part I for std IX, we have learnt union and intersection of sets, in the topic on sets. Now, let us describe a segment, a ray and a line as sets of points., (1) Line segment :, The union set of point A, point B and points between, A and B is called segment AB. Segment AB is, written as seg AB in brief. Seg AB means seg BA., Point A and point B are called the end points of seg AB., The distance between the end points of a segment, is called the length of the segment. That is l(AB) = d (A,B), l(AB) = 5 is also written as AB = 5., (2) Ray AB :, Suppose, A and B are two distinct points. The, union set of all points on seg AB and the, points P such that A - B - P, is called ray AB., Here point A is called the starting point of ray AB., , A, , Fig. 1.6, , B, , B P, Fig. 1.7, , A, , (3) Line AB :, The union set of points on ray AB and opposite ray of ray AB is called line AB., The set of points of seg AB is a subset of points of line AB., (4) Congruent segments :, If the length of two segments is equal, then the two segments are congruent., If l(AB) = l(CD) then seg AB @ seg CD, , A, , B, , C, , Fig. 1.8, , (5) Properties of congruent segements :, , D, , (i) Reflexivity : seg AB @ seg AB, (ii) Symmetry : If seg AB @ seg CD then seg CD @ seg AB, (iii) Transitivity : If seg AB @ seg CD and seg CD @ seg EF then seg AB @ seg EF, (6) Midpoint of a segment :, If A-M-B and seg AM @ seg MB, then M is, called the midpoint of seg AB., Every segment has one and only one midpoint., , 6, , A, , M, Fig. 1.9, , B

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(7) Comparison of segments :, , If length of segment AB is less than the length of, segment CD, it is written as seg AB < seg CD or, seg CD > seg AB., , B, , A, C, , The comparison of segments depends upon their, lengths., , D, , Fig. 1.10, , (8) Perpendicularity of segments or rays :, , If the lines containing two segments, two rays or, a ray and a segment are perpendicular to each, other then the two segments, two rays or the, segment and the ray are said to be perpendicular, to each other., , C, B, , A, D, , In the figure 1.11, seg AB ^ line CD,, , seg AB ^ ray CD., , Fig. 1.11, , (9) Distance of a point from a line :, If seg CD ^ line AB and the point D lies on, , C, , line AB then the length of seg CD is called the, distance of point C from line AB., , A, , The point D is called the foot of the perpendicular., If l(CD) = a, then the point C is at a distance of ‘a’, from the line AB., , D B, Fig. 1.12, , Practice set 1.2, 1. The following table shows points on a number line and their co-ordinates. Decide, whether the pair of segments given below the table are congruent or not., Point, Co-ordinate, , A, , -3, , B, , C, , 5, , 2, , D, , -7, , E, , 9, , (i) seg DE and seg AB (ii) seg BC and seg AD (iii) seg BE and seg AD, 2. Point M is the midpoint of seg AB. If AB = 8 then find the length of AM., 3. Point P is the midpoint of seg CD. If CP = 2.5, find l(CD)., 4. If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments., , 7

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5. Write the answers to the following questions with reference to figure 1.13., (i) Write the name of the opposite ray of ray RP, (ii) Write the intersection set of ray PQ and ray RP., , T, , (iii) Write the union set of seg PQ and seg QR., , S R, , P, Fig. 1.13, , Q, , (iv) State the rays of which seg QR is a subset., (v) Write the pair of opposite rays with common end point R., (vi) Write any two rays with common end point S., (vii) Write the intersection set of ray SP and ray ST., 6. Answer the questions with the help of figure 1.14., , R, -6, , U, , Q, -4, , L, , P, -2, , A, 0, Fig. 1.14, , B, 2, , C, 4, , V, , D, 6, , (i) State the points which are equidistant from point B. , (ii) Write a pair of points equidistant from point Q., (iii) Find d (U,V), d (P,C), d (V,B), d (U, L)., , Let’s learn., Conditional statements and converse, The statements which can be written in the ‘If-then’ form are called conditional, statements. The part of the statement following ‘If’ is called the antecedent, and the part, following ‘then’ is called the consequent., For example, consider the statement : The diagonals of a rhombus are perpendicular, bisectors of each other., The statement can be written in the conditional form as, ‘If the given quadrilateral is, a rhombus then its diagonals are perpendicular bisectors of each other.’, If the antecedent and consequent in a given conditional statement are interchanged,, the resulting statement is called the converse of the given statement., If a conditional statement is true, its converse is not necessarily true. Study the following examples., Conditional statement : If a quadrilateral is a rhombus then its diagonals are perpendicular bisectors of each other., , 8

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Converse : If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it is a rhombus., In the above example, the statement and its converse are true., Now consider the following example,, Conditional statement : If a number is a prime number then it is even or odd., Converse : If a number is even or odd then it is a prime number., In this example, the statement is true, but its converse is false., , Let’s learn., Proofs, We have studied many properties of angles, triangles and quadrilaterals through activities., In this standard we are going to look at the subject of Geometry with a different point, of view, which was originated by the Greek mathematician Euclid, who lived in the third, century before Christian Era. He gathered the knowledge of Geometry prevailing at his time, and streamlined it. He took for granted some self evident geometrical statements which were, accepted by all and called them Postulates. He showed that on the basis of the postulates, some more properties can be proved logically., Properties proved logically are called Theorems., Some of Euclid’s postulates are given below., (1) There are infinite lines passing through a point., (2) There is one and only one line passing through two, points., (3) A circle of given radius can be drawn taking any point, as its centre., (4) All right angles are congruent with each other., (5) If two interior angles formed on one side of a, transversal of two lines add up to less than two, right angles then the lines produced in that direction, Euclid, intersect each other., We have verified some of these postulates through activities., A property is supposed to be true if it can be proved logically. It is then called a Theorem., The logical argument made to prove a theorem is called its proof., When we are going to prove that a conditional statement is true, its antecedent is called, ‘Given part’ and the consequent is called ‘the part to be proved’., There are two types of proofs, Direct and Indirect., Let us give a direct proof of the property of angles made by two intersecting lines., , 9

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Theorem : The opposite angles formed by two intersecting lines are of equal measures., Given : Line AB and line CD intersect at point O such that A - O - B, C - O - D., , A, , To prove : (i) ÐAOC = ÐBOD, (ii) ÐBOC =, , Proof :, , , , ÐAOD, , D, , C, O, , Fig. 1.15, , B, , ÐAOC + ÐBOC = 180° . . . . . . . (I) (angles in linear pair), ÐBOC + ÐBOD = 180° . . . . . . . (II) (angles in linear pair), ÐAOC + ÐBOC = ÐBOC + ÐBOD . . . . . . . [from (I) and (II)], \ ÐAOC = ÐBOD. . . . . . . eliminating ÐBOC., Similarly, it can be proved that ÐBOC = ÐAOD., , Indirect proof :, This type of proof starts with an assumption that the consequence is false. Using it and, the properties accepted earlier, we start arguing step by step and reach a conclusion. The, conclusion is contradictory with the antecedent or a property which is already accepted., Hence, the assumption that the consequent is false goes wrong. So it is accepted that the, consequent is true., Study the following example., Statement : A prime number greater than 2 is odd., Conditional statement : If p is a prime number greater than 2 then it is odd., Given :, , p is a prime number greater than 2. That is, 1 and p are the only divisors of p., , To prove : p is an odd number., Proof :, , Let us suppose that p is not an odd number., So p is an even number., , \ a divisor of p is 2 ..... (I), But it is given that p is a prime number greater than 2., , \ 1 and p are the only divisors of p ..... (II), Statements (I) and (II) are contradictory., \ the assumption , that p is not odd is false., This proves that a prime number greater than 2 is odd., , 10, , ....(given)

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Practice set 1.3, 1. Write the following statements in ‘if-then’ form., (i) The opposite angles of a parallelogram are congruent., (ii) The diagonals of a rectangle are congruent., (iii) In an isosceles triangle, the segment joining the vertex and the mid point of the, base is perpendicular to the base., 2. Write converses of the following statements., (i) The alternate angles formed by two parallel lines and their transversal are, congruent., (ii) If a pair of the interior angles made by a transversal of two lines are supplementary then the lines are parallel., (iii) The diagonals of a rectangle are congruent., , Problem set 1, 1., , Select the correct alternative from the answers of the questions given below., , (i) How many mid points does a segment have ?, , (A) only one, (B) two, (C) three, , (D) many, , (ii) How many points are there in the intersection of two distinct lines ?, (A) infinite, (B) two, (C) one, (D) not a single, (iii) How many lines are determined by three distinct points?, , (A) two, (B) three, (C) one or three, (D) six, (iv) Find d(A, B), if co-ordinates of A and B are - 2 and 5 respectively., (A) - 2, (B) 5, (C) 7 , (D) 3, (v) If P - Q - R and d(P,Q) = 2, d(P,R) = 10, then find d(Q,R)., (D) 20, , (A) 12, (B) 8, (C) 96, 2., , On a number line, co-ordinates of P, Q, R are 3, - 5 and 6 respectively. State with reason, whether the following statements are true or false., (i) d(P,Q) + d(Q,R) = d(P,R) (ii) d(P,R) + d(R,Q) = d(P,Q), (iii) d(R,P) + d(P,Q) = d(R,Q), (iv) d(P,Q) - d(P,R) = d(Q,R), , 3., , Co-ordinates of some pairs of points are given below. Hence find the distance between, each pair., (i) 3, 6 , (v) x + 3, x- 3 , , (ii) - 9, - 1 , (vi) - 25, - 47, , (iii) - 4, 5 , (vii) 80, - 85, , 11, , (iv) 0, - 2

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4., , Co-ordinate of point P on a number line is - 7. Find the co-ordinates of points on the, number line which are at a distance of 8 units from point P., , 5., , Answer the following questions., (i) If A - B - C and d(A,C) = 17, d(B,C) = 6.5 then d (A,B) = ?, (ii) If P - Q - R and d(P,Q) = 3.4, d(Q,R)= 5.7 then d(P,R) = ?, , 6., , Co-ordinate of point A on a number line is 1. What are the co-ordinates of points on the, number line which are at a distance of 7 units from A ?, , 7., , Write the following statements in conditional form., (i) Every rhombus is a square., (ii) Angles in a linear pair are supplementary., (iii) A triangle is a figure formed by three segments., (iv) A number having only two divisors is called a prime number., , 8., , Write the converse of each of the following statements., (i) If the sum of measures of angles in a figure is 1800, then the figure is a triangle., (ii) If the sum of measures of two angles is 900 then they are complement of each other., (iii) If the corresponding angles formed by a transversal of two lines are congruent then, the two lines are parallel., (iv) If the sum of the digits of a number is divisible by 3 then the number is divisible, by 3., , 9., , Write the antecedent (given part) and the consequent (part to be proved) in the following, statements., (i) If all sides of a triangle are congruent then its all angles are congruent., (ii) The diagonals of a parallelogram bisect each other., , 10*. Draw a labelled figure showing information in each of the following statements and, write the antecedent and the consequent., (i) Two equilateral triangles are similar., (ii) If angles in a linear pair are congruent then each of them is a right angle., (iii) If the altitudes drawn on two sides of a triangle are congruent then those two sides, are congruent., , qqq, 12

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2, , Parallel Lines, , Let’s study., , • Properties of angles formed by • Tests of parallelness of two lines, parallel lines and its transversal • Use of properties of parallel lines, Let’s recall., Parallel lines : The lines which are coplanar and do, , l, , not intersect each other are called parallel lines., m, Hold a stick across the horizontal parallel, bars of a window as shown in the figure., How many angles are formed ?, n, , · Do you recall the pairs of angles formed, , by two lines and their transversal ?, In figure 2.1, line n is a transversal of, line l and line m., Here, in all 8 angles are formed. Pairs of, angles formed out of these angles are as, , angles, , Ðd, Ðh, (ii) Ða,, (iii) Ðc,, (iv) Ðb,, (i), , h e, g f, , l, m, , Fig. 2.1, , follows :, Pairs of corresponding, , d a, c b, , Pairs of alternate interior, angles, (i) Ðc, Ðe (ii) Ðb, Ðh, Pairs of alternate exerior, angles, (i) Ðd, Ðf, (ii) Ða, Ðg, , Pairs of interior angles, on the same side of the, transversal, , Ðc, Ðh, (ii) Ðb, Ðe, , (i), , Some important properties :, (1) When two lines intersect, the pairs of opposite angles formed are congruent., (2) The angles in a linear pair are supplementary., , 13

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(3), , When one pair of corresponding angles is congruent, then all the remaining pairs of, corresponding angles are congruent., , (4), , When one pair of alternate angles is congruent, then all the remaining pairs of alternate, angles are congruent., , (5), , When one pair of interior angles on one side of the transversal is supplementary, then, the other pair of interior angles is also supplementary., , Let’s learn., Properties of parallel lines,   Activity, To verify the properties of angles formed by a transversal of two parallel lines., Take a piece of thick coloured paper. Draw a pair of parallel lines and a transversal, on it. Paste straight sticks on the lines. Eight angles will be formed. Cut pieces of, coloured paper, as shown in the figure, which will just fit at the corners of Ð1 and, Ð2. Place the pieces near different pairs of corresponding angles, alternate angles, and interior angles and verify their properties., 1, , 2, , 1, , 2, , 14

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Let’s learn., We have verified the properties of angles formed by a transversal of two parallel lines. Let, us now prove the properties using Euclid’s famous fifth postulate given below., If sum of two interior angles formed on one side of a transversal of two lines is less than, two right angles then the lines produced in that direction intersect each other., Interior angle theorem, Theorem : If two parallel lines are intersected by a transversal, the interior angles on either, side of the transversal are supplementary., Given : line l || line m and line n is their transversal. Hence as shown in the figure, Ða, Ðb are interior angles formed on, one side and Ðc, Ðd are interior angles, formed on other side of the transversal., To prove : Ða + Ðb = 180°, Ðd + Ðc = 180°, , n, d a, c b, , m, l, , Fig. 2.2, , Proof : Three possibilities arise regarding the sum of measures of Ða and, , Ðb., , (i) Ða + Ðb < 180° (ii) Ða + Ðb > 180° (iii) Ða + Ðb = 180°, Let us assume that the possibility (i) Ða + Ðb < 180° is true., Then according to Euclid’s postulate, if the line l and line m are produced will, intersect each other on the side of the transversal where Ða and Ðb are, formed., But line l and line m are parallel lines. ..........given, \ Ða + Ðb < 180° impossible ..........(I), Now let us suppose that, \, , Ða + Ðb > 180° is true., , Ða + Ðb >180°, But Ða + Ðd = 180°, , and Ðc + Ðb = 180° . . . . . angles in linear pairs, \ Ða + Ðd + Ðb + Ðc = 180° +180° = 360°, \Ðc + Ðd = 360° - (Ða + Ðb), If Ða + Ðb >180° then [360° - (Ða + Ðb)] < 180°, \Ðc + Ðd < 180°, 15

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\ In that case line l and line m produced will intersect each other on the same, side of the transversal where Ðc and Ðd are formed., , \Ðc + Ðd < 180 is impossible., , , That is Ða + Ðb >180° is impossible...... (II), , \ the remaining possibility,, , , Ða + Ðb = 180° is true......from (I) and (II), , , \ Ða + Ðb = 180° Similarly, Ðc + Ðd = 180°, , , Note that, in this proof, because of the contradictions we have denied the, , possibilities Ða + Ðb >180° and Ða + Ðb <180°., Therefore, this proof is an example of indirect proof., Corresponding angles and alternate angles theorems, Theorem : The corresponding angles formed by a transversal of two parallel lines are of, n, equal measure., : line l || line m, Given, line n is a transversal., , a, c, b, , Ða = Ðb, : Ða + Ðc = 180° ........(I) angles in linear pair, Fig. 2.3, Ðb + Ðc = 180° ..............(II) property of interior angles of parallel lines, Ða + Ðc = Ðb + Ðc .......from (I) and (II), \ Ða = Ðb, , l, m, , To prove :, Proof, , Theorem : The alternate angles formed by a transversal of two parallel lines are of equal, measures., n, Given, : line l || line m, line n is a transversal., , d c, b, , To prove :, , Ðd = Ðb, , Proof, , Ðd + Ðc = 180°................(I) angles in linear pair, , :, , Fig. 2.4, , Ðc + Ðb = 180° ..............(II) property of interior angles of parallel line, Ðd + Ðc = Ðc + Ðb................from (I) and (II), \ Ðd = Ðb, , 16, , l, m

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Practice set 2.1, 1., , In figure 2.5, line RP || line MS and line DK, is their transversal. ÐDHP = 85°, Find the measures of following angles., (i) ÐRHD (ii) ÐPHG, (iii) ÐHGS (iv) ÐMGK, , R, , H, , D, 85° P, , G, , M, , S, , K, , Fig. 2.5, , p, , 2., , In figure 2.6, line p || line q and, line l and line m are transversals., Measures of some angles are shown., Hence find the measures of, Ða, Ðb, Ðc, Ðd., , n, , a, b, , 110°, , b, l, , 115°, , c, , d, , m, , Fig. 2.6, , p, 45°, , c, , a, , q, , 3., , l, m, , In figure 2.7, line l || line m and, line n || line p. Find Ða, Ðb, Ðc from, the given measure of an angle., , Fig. 2.7, 4*. In figure 2.8, sides of ÐPQR and ÐXYZ, are parallel to each other. Prove that,, ÐPQR @ ÐXYZ, , P, , X, , Y, Q, , R, , Fig. 2.8, , 17, , Z

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5. , In figure 2.9, line AB || line CD and, line PQ is transversal. Measure of one, of the angles is given., Hence find the measures of the, , P, , following angles. , , (i) ÐART, , (iii) ÐDTQ, , R, 105°, , A, T, , C, , (ii) ÐCTQ, , B, D, , Q, , (iv) ÐPRB, , Fig. 2.9, , Let’s learn., Use of properties of parallel lines, Let us prove a property of a triangle using the properties of angles made by a transversal, of parallel lines., Theorem : The sum of measures of all angles of a triangle is 180°., Given :, , D ABC is any triangle., , To prove : ÐABC + ÐACB + ÐBAC = 180°., , A, , Construction : Draw a line parallel to seg BC and passing, , , through A. On the line take points P and Q, , , , such that, P - A - Q., , B, , Fig. 2.10, , Proof : Line PQ || line BC and seg AB is a transversal., , \ ÐABC = ÐPAB.......alternate angles.....(I), line PQ || line BC and seg AC is a transversal., \ ÐACB = ÐQAC.......alternate angles.....(II), \ From I and II ,, ÐABC + ÐACB = ÐPAB + ÐQAC . . . (III), , P, , B, , Adding ÐBAC to both sides of (III)., , A, , C, , Q, , Fig. 2.11, , C, , ÐABC + ÐACB + ÐBAC = ÐPAB + ÐQAC + ÐBAC, , , , , , = ÐPAB +, , ÐBAC + ÐQAC, , , , = ÐPAC +, , ÐQAC ...(Q ÐPAB + ÐBAC = ÐPAC) , , = 180° , , ...Angles in linear pair, , That is, sum of measures of all three angles of a triangle is 180°., , 18

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Let’s discuss., In fig. 2.12, How will you decide, whether line l and line m are parallel or not ?, , l, , Fig. 2.12, , m, , Let’s learn., Tests for parallel lines, Whether given two lines are parallel or not can be decided by examining the angles, formed by a transversal of the lines., (1) If the interior angles on the same side of a transversal are supplementary then the lines, are parallel., (2) If one of the pairs of alternate angles is congruent then the lines are parallel., (3) If one of the pairs of corresponding angles is congruent then the lines are parallel., Interior angles test, Theorem : If the interior angles formed by a transversal of two distinct lines are, supplementary, then the two lines are parallel. , Given, , : Line XY is a transversal of line AB and, line CD., , A, , ÐBPQ + ÐPQD = 180°, , To prove : line AB || line CD, Proof, , means line AB and CD intersect at point T., So D PQT is formed., , Fig. 2.13, , X, , A, , line AB and line CD are not parallel,, , B, D, , Y, , Let us suppose that the statement to be, proved is wrong. That is, we assume,, , X, , Q, , C, , : We are going to give an indirect proof., , , , P, , P, Q, , C, Y, , B T, D, , Fig. 2.14, , , \ ÐTPQ + ÐPQT + ÐPTQ = 180° ..........sum of angles of a triangle, but ÐTPQ + ÐPQT = 180° ...........given, , That is the sum of two angles of the triangle is 180°., , But sum of three angles of a triangle is 180°., \, , ÐPTQ = 0°., , 19

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\ line PT and line QT means line AB and line CD are not distinct lines., But, we are given that line AB and line CD are distinct lines., , \ we arrive at a contradiction., \ our assumption is wrong. Hence line AB and line CD are parallel., Thus it is proved that if the interior angles formed by a transversal are supplementary,, then the lines are parallel., This property is called interior angles test of parallel lines., Alternate angles test, Theorem : If a pair of alternate angles formed by a transversal of two lines is congruent, then the two lines are parallel., Given, : Line n is a transversal of line l and line m., n, , Ða and Ðb is a congruent pair of alternate angles., l, That is, Ða = Ðb, a c, To prove : line l || line m, b, m, Proof, : Ða + Ðc = 180° . ....angles in linear pair, Ða = Ðb .......... given, , , \ Ðb + Ðc = 180°, , , Fig. 2.15, But Ðb and Ðc are interior angles on the same side of the transversal., , \ line l || line m ....... interior angles test, , This property is called the alternate angles test of parallel lines., Corresponding angles Test, Theorem : If a pair of corresponding angles formed by a transversal of two lines is congruent, then the two lines are parallel., Given, , : Line n is a transversal of line l and line m., , , Ða and Ðb is a congruent pair of corresponding angles., , That is, Ða = Ðb, To prove :, , n, , line l || line m, , Ða + Ðc = 180° ..........angles in linear pair, , Ða = Ðb .......... given, \ Ðb + Ðc = 180°, , Proof, , a, c, , :, , That is a pair of interior angles on the same, side of the transversal is congruent., , , \ line l || line m ..........interior angles test , , b, , Fig. 2.16, , This property is called the corresponding angles test of parallel lines., , 20, , l, , m

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Corollary I : If a line is perpendicular to two lines in a plane, then the two lines are parallel, to each other., n, Given, : Line n ^ line l and line n ^ line m, To prove : line l || line m, Proof, : line n ^ line l and line n ^ line m ...given, \ Ða = Ðc = 90°, , Ða and Ðc are corresponding, , angles formed by transversal n of, line l and line m., , \ line l || line m, , , a, , l, , c, , m, , Fig. 2.17, , ....corresponding angles test, , Corollary II : If two lines in a plane are parallel to a third line in the plane then those two, lines are parallel to each other. Write the proof of the corollary., , Practice set 2.2, l, , 1. In figure 2.18, y = 108° and x = 71°, Are the lines m and n parallel ? Justify ?, , m, , x, y, , Fig. 2.18, , n, , n, a, , 2. In figure 2.19, if Ða @ Ðb then, prove that line l || line m., , l, m, , b, l, , Fig. 2.19, 3. In figure 2.20, if Ða @ Ðb and, Ðx @ Ðy then prove that line l || line n., , b, , m, , K, , a, x, , D, A, , 50°, C, , Fig. 2.21, , y, , E, 100°, , B, , n, , Fig. 2.20, , 4. In figure 2.21, if ray BA || ray DE,, ÐC = 50° and ÐD = 100°. Find the measure, of ÐABC., (Hint : Draw a line passing through point C and, parallel to line AB.), , 21

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In figure 2.22, ray AE || ray BD,, ray AF is the bisector of ÐEAB and, ray BC is the bisector of ÐABD., Prove that line AF || line BC., , 5. , F, E, , x x, , B, y y, , A, C, , D, , Fig. 2.22, 6. A transversal EF of line AB and, line CD intersects the lines at point P, , A, , and Q respectively. Ray PR and ray QS, are parallel and bisectors of ÐBPQ and, , C, , ÐPQC respectively., Prove that line AB || line CD., , P, , E, R, , S, F, , Q, , B, D, , Fig. 2.23, , Problem set 2, Select the correct alternative and fill in the blanks in the following statements., (i) If a transversal intersects two parallel lines then the sum of interior angles on the, , same side of the transversal is ............, (B) 90°, (C) 180°, (D) 360°, , (A) 0°, (ii) The number of angles formed by a transversal of two lines is ............, , (A) 2, (B) 4, (C) 8, (D) 16, (iii) A transversal intersects two parallel lines. If the measure of one of the angles is, 40° then the measure of its corresponding angle is ............., (B) 140°, (C) 50°, (D) 180°, , (A) 40°, 1., , (iv), , (v), , , , In D ABC, ÐA = 76°, ÐB = 48°, \ ÐC = .............., (B) 56°, (C) 124°, (D) 28°, (A) 66°, Two parallel lines are intersected by a transversal. If measure of one of the, alternate interior angles is 75° then the measure of the other angle is ............., (B) 15°, (C) 75°, (D) 45°, (A) 105°, , 2*. Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior, and exterior of ÐQPR respectively. Ray PB and ray PA are perpendicular to each other., , Draw a figure showing all these rays and write (i) A pair of complementary angles (ii) A pair of supplementary angles., (iii) A pair of congruent angles., , 22

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3. Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular, to the other line also., 4. In figure 2.24, measures of some angles, are shown. Using the measures find the, measures of Ðx and Ðy and hence, show that line l || line m., Q, , A, , B, , x, , E, , 50°, , y, , m, , Fig. 2.24, , is their transversal. If y : z = 3 : 7 then find, , D, , z, , l, , x, , 5. Line AB || line CD || line EF and line QP, , y, C, , 130°, , the measure of Ðx. (See figure 2.25.), , F, P, , Fig. 2.25, , p, , 6. In figure 2.26, if line q || line r,, line p is their, transversal and if, a = 80° find the values of f and g., , b a, c d, , q, , f e, gh, , r, , Fig. 2.26, A, , P, , B, , 7. In figure 2.27, if line AB || line CF, and line BC || line ED then prove that, ÐABC = ÐFDE., , C, , E, , D, F, , Fig. 2.27, , P, Q, , A, , 8. In figure 2.28, line PS is a transversal, , Y, , X, , of parallel line AB and line CD. If Ray, QX, ray QY, ray RX, ray RY are angle, , C, , R, , bisectors, then prove that ¨ QXRY is a, , B, D, , S, , rectangle., Fig. 2.28, , 23, , qqq

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3, , Triangles, , Let’s study., , • Median of a triangle, • Property of median on hypotenuse, , • Theorem of remote interior angles, •, •, •, , of a triangle, Congruence of triangles, Theorem of an isosceles triangle, Property of 30°- 60°- 90° angled, triangle, , •, •, •, , of a right angled triangle, Perpendicular bisector theorem, Angle bisector theorem, Similar triangles, , Activity :, Draw a triangle of any measure on a thick paper. Take a point T on ray QR as shown, in fig. 3.1. Cut two pieces of thick paper which will exactly fit the corners of ÐP and ÐQ., See that the same two pieces fit exactly at the corner of ÐPRT as shown in the figure., , P, , Q, , Fig. 3.1, , R, , T, , Let’s learn., Theorem of remote interior angles of a triangle, Theorem : The measure of an exterior angle of a triangle is equal to the sum of its remote, interior angles., P, Given : ÐPRS is an exterior angle of D PQR., Fig. 3.2, To prove : ÐPRS = ÐPQR + ÐQPR, Proof, : The sum of all angles of a triangle is 180°., \ ÐPQR + ÐQPR + ÐPRQ = 180°.......(I), , Q, R S, ÐPRQ + ÐPRS = 180°.....angles in linear pair......(II), \ from (I) and (II), ÐPQR + ÐQPR + ÐPRQ = ÐPRQ + ÐPRS, \ ÐPQR + ÐQPR = ÐPRS .......eliminating ÐPRQ from both sides, \ the measure of an exterior angle of a triangle is equal to the sum of its remote, interior angles., , 24

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Use your, brain power!, , Can we give an alternative proof of the theorem drawing a, line through point R and parallel to seg PQ in figure 3.2 ?, , Let’s learn., Property of an exterior angle of triangle, The sum of two positive numbers a and b, that is (a + b) is greater than a and greater, than b also. That is, a + b > a, a + b > b, P, Using this inequality we get one property, relaed to exterior angle of a triangle., If ÐPRS is an exterior angle of D PQR then, ÐPRS > ÐP , ÐPRS > ÐQ, S, R, Q, Fig., 3.3, \ an exterior angle of a triangle is greater than, its remote interior angle., Solved examples, Ex (1) The measures of angles of a triangle are in the ratio 5 : 6 : 7. Find the measures., Solution : Let the measures of the angles of a triangle be 5x, 6x, 7x., \ 5x + 6x + 7x = 180°, 18x = 180°, , , , x = 10°, , , , 5x = 5 ´ 10 = 50°, 6x = 6 ´ 10 = 60°, 7x = 7 ´ 10 = 70°, , \ the measures of angles of the triangle are 50°, 60° and 70°., Ex (2) Observe figure 3.4 and find the measures of ÐPRS and ÐRTS., Solution : ÐPRS is an exterior angle of D PQR., So from the theorem of remote interior angles,, , P, 30°, , , ÐPRS = ÐPQR + ÐQPR, = 40°, , , , T, , + 30°, , = 70°, In D RTS, , Q, , \, , + ÐRTS +, , R, , 20°, , Fig. 3.4, , , ÐTRS + ÐRTS + ÐTSR =, , , 40°, , ........ sum of all angles of a triangle, , = 180°, , \ ÐRTS + 90° = 180°, \ ÐRTS =, , , , 25, , S

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Ex (4) In figure 3.7, bisectors of, , ÐB and ÐC of D ABC intersect at point P., A, ÐBAC., , 1, Prove that ÐBPC = 90 +, 2, , P, , Complete the proof filling in the blanks., Proof : In D ABC,, , Fig. 3.7, ...... sum of measures of angles of a triangle, , ÐBAC + ÐABC + ÐACB =, , , , \, , 1, 2, , ÐBAC +, , C, , B, , 1, 2, , ÐABC +, , 1, 2, , ÐACB =, , 1, 2, , ´, ....multiplying each term by 1, , 2, , \, , 1, 2, , ÐBAC + ÐPBC + ÐPCB = 90°, , \ ÐPBC + ÐPCB = 90° In D BPC, , 1, 2, , ÐBAC ......(I), , ÐBPC + ÐPBC + ÐPCB = 180° ......sum of measures of angles of a triangle, \ÐBPC +, , = 180° ......from (I), , 1, ÐBAC), 2, , , = 180° - 90° + 1 ÐBAC, 2, = 90° + 1 ÐBAC, 2, , , , \ ÐBPC = 180° - (90° -, , Practice set 3.1, 1., , In figure 3.8, ÐACD is an exterior angle of D ABC., , A, , ÐB = 40°, ÐA = 70°. Find the measure of ÐACD., 2., , In D PQR, ÐP = 70°,, , 3., , The measures of angles of a triangle are x°, (x-20)°, (x-40)°., Find the measure of each angle., , 4., , ÐQ = 65° then find ÐR., , B, , Fig. 3.8, , C, , D, , The measure of one of the angles of a triangle is twice the measure of its smallest, angle and the measure of the other is thrice the measure of the smallest angle. Find the, measures of the three angles., , 27

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T, , 5., , 100°, , In figure 3.9, measures of some angles, are given. Using the measures find the, , E, y, z, , x, N, , values of x, y, z., , M, , 140°, R, , Fig. 3.9, D, , 6., , B, , In figure 3.10, line AB || line DE. Find, the measures of ÐDRE and ÐARE, using given measures of some angles., , R, 40°, , 70°, , A, , 7., , 8., , In D ABC, bisectors of, of ÐAOB., , E, , Fig. 3.10, , ÐA and ÐB intersect at point O. If ÐC = 70°. Find measure, , In Figure 3.11, line AB || line CD and, line PQ is the transversal. Ray PT, and ray QT are bisectors of ÐBPQ and, ÐPQD respectively., Prove that mÐPTQ = 90°., , A, , B, , P, T, , C Q, , D, , Fig. 3.11, , 9., , a, , Using the information in figure 3.12,, find the measures of Ða, Ðb and Ðc., , b, 70°, , 10. In figure 3.13, line DE || line GF , ray EG and ray FG are bisectors of, ÐDEF and ÐDFM respectively., Prove that,, (i) ÐDEG =, , 1, 2, , 100°, , c, , Fig. 3.12, , D, , ÐEDF (ii) EF = FG., , E, , 28, , G, , F, , Fig. 3.13, , M

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Let’s learn., Congruence of triangles, We know that, if a segment placed upon another fits with it exactly then the two, segmetns are congruent. When an angle placed upon another fits with it exactly then the two, angles are congruent. Similarly, if a triangle placed upon another triangle fits exactly with it, then the two triangles are said to be congruent. If D ABC and D PQR are congruent is, written as D ABC @ D PQR., A1, , A3, , A2, , C1, , B1, , B2, , C2, , B3, , Activity : Draw D, , A, , C, , B, A4, , A, , B, , C3, , C, , C4, , B4, , Fig. 3.14, ABC of any measure on a card-sheet and cut it out., , Place it on a card-sheet. Make a copy of it by drawing its border. Name it as D A1B1C1, , Now slide the D ABC which is the cut out of a triangle to some distance and make one, more copy of it. Name it D A2B2C2., Then rotate the cut out of triangle ABC a little, as shown in the figure, and make another, copy of it. Name the copy as D A3B3C3 . Then flip the triangle ABC, place it on another, card-sheet and make a new copy of it. Name this copy as D A4B4C4 ., Have you noticed that each of D A1B1C1, D A2B2C2, D A3B3C3 and D A4B4C4 is, congruent with D ABC ? Because each of them fits exactly with D ABC., Let us verify for D A3B3C3 . If we place ÐA upon ÐA3 , ÐB upon ÐB3 and ÐC, upon ÐC3, then only they will fit each other and we can say that D ABC @ D A3B3C3 ., We also have AB = A3B3 , BC = B3C3 , CA = C3A3 ., Note that, while examining the congruence of two triangles, we have to write their, angles and sides in a specific order, that is with a specific one-to-one correspondence., If D ABC @ D PQR, then we get the following six equations :, ÐA = ÐP, ÐB = ÐQ, ÐC = ÐR . .. . (I) and AB = PQ, BC = QR, CA = RP . . . . (II), This means, with a one-to-one correspondence between the angles and the sides of two, triangles, we get three pairs of congruent angles and three pairs of congruent sides., , 29

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Given six equations above are true for congruent triangles. For this let us see three, specific equations are true then all six equations become true and hence two triangles, congruent., (1) In a correspondence, if two angles of DABC are equal to two angles of DPQR, and the sides included by the respective pairs of angles are also equal, then the, two triangles are congruent., A, , P, R, , CQ, , B, , This property is called as, angle-side-angle test, which, in short we write A-S-A test., , Fig. 3.15, , In a correspondence, if two sides of D ABC are equal to two sides of D PQR and, the angles included by the respective pairs of sides are also equal, then the two, triangles are congruent., , (2), , A, , P, , C, , This property is called as, side-angle-side test, which in, short we write S-A-S test., R, , Q, , B, , Fig. 3.16, , (3) In a correspondence, if three sides of D ABC are equal to three sides of, D PQR , then the two triangles are congruent., A, , Q, , B, , R, , C, , This property is called as, side-side-side test, which in, short we write S-S-S test., , P, , Fig. 3.17, , (4) If in D ABC and D PQR, ÐB and ÐQ are right angles, hypotenuses are equal, and AB = PQ, then the two triangles are congruent., P, , A, , This property is called the, hypotenuse side test., B, , C, , Q, , R, , Fig. 3.18, , 30

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Remember this !, We have constructed triangles using the given information about parts of triangles. (For, example, two angles and the included side, three sides, two sides and an included angle)., We have experienced that the triangle constructed with any of these information is unique., So if by some one-to-one correspondence between two triangles, these three parts of one, triangle are congruent with corresponding three parts of the other triangle then the two, triangles are congruent. Then we come to know that in that correspondence their three, angles and three sides are congruent. If two triangles are congruent then their respective, angles and respective sides are congruent. This property is useful to solve many problems, in Geometry., , Practice set 3.2, 1., , In each of the examples given below, a pair of triangles is shown. Equal parts of triangles, in each pair are marked with the same signs. Observe the figures and state the test by, which the triangles in each pair are congruent., , (i), P, , A, , B, , (ii), , C, , Q, , R, Y, , Z, , N, , M, , By . . . . . . . . . . test, , By . . . . . . . . . . test, , D ABC @ D PQR, , D XYZ @ D LMN, , (iii), , (iv), , L, , M, , T, , S, , P, , Q, , L, , X, , R, , T, , U, N, , R, , By . . . . . . . . . . test, , By . . . . . . . . . . test, , D PRQ @ D STU, , D LMN @ D PTR, Fig. 3.19, , 31, , P

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2., , Observe the information shown in pairs of triangles given below. State the test by which, the two triangles are congruent. Write the remaining congruent parts of the triangles., , (i) , , (ii), , A, , R, , P, , P, , T, C Q, , B, , Fig. 3.20, , R, , Fig. 3.21, From the information shown in the figure,,, In D PTQ and D STR, seg PT @ seg ST, ÐPTQ @ ÐSTR....vertically opposite angles, seg TQ @ seg TR, \ D PTQ @ D STR ......., test, \ÐTPQ @, ..... corresponding, angles of congruent, @ ÐTRS, and, triangles., , From the information shown in the figure,, in D ABC and D PQR, ÐABC @ ÐPQR, seg BC @ seg QR, ÐACB @ ÐPRQ, \ D ABC @ D PQR ......., test, \ÐBAC @, .......corresponding, , angles of congruent triangles., seg AB @, , @, , and, 3., , C, , 5., , }, , corresponding, .... sides of congruent, seg PR, triangles, , }, , seg PQ @, , , From the information shown in the, figure, state the test assuring the, congruence of D ABC and D PQR., Write the remaining congruent parts of, the triangles., A, , B P, , Fig. 3.22, , S, , Q, , 4., , corresponding sides of, congruent triangles., , As shown in the following figure, in, D LMN and D PNM, LM = PN,, LN = PM. Write the test which assures, the congruence of the two triangles., Write their remaining congruent parts., , L, , Q, R, , P, M, B, , In figure 3.24, seg AB @ seg CB, and seg AD @ seg CD., Prove that, D ABD @ D CBD, , Fig. 3.23, , N, , C, , A, , D, , Fig. 3.24, , Please note : corresponding sides of congruent triangles in short we write c.s.c.t. and, corresponding angles of congruent triangles in short we write c.a.c.t., , 32

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6., , In figure 3.25, ÐP @ ÐR, seg PQ @ seg RQ, Prove that,, D PQT @ D RQS, , Q, S, , T, , Fig. 3.25, , P, , R, , Let’s learn., Isosceles triangle theorem, Theorem : If two sides of a triangle are congruent then the angles opposite to them are, congruent., : In D ABC, side AB @ side AC, Given, A, To prove : ÐABC @ ÐACB, Construction : Draw the bisector of ÐBAC, , which intersects side BC at point D., Proof, : In D ABD and D ACD, B, C, D, seg AB @ seg AC ....... given, ÐBAD @ ÐCAD........construction, , Fig. 3.26, seg AD @ seg AD ....... common side, \ D ABD @ D ACD ......, \ÐABD @, ....... (c.a.c.t.), , \ÐABC @ ÐACB , , Q B-D-C, , Corollary : If all sides of a triangle are congruent then its all angles are congruent., (write the proof of this corollary.), Converse of isosceles triangle theorem, Theorem : If two angles of a triangle are congruent then the sides opposite to them are, congruent., Given : In D PQR, ÐPQR @ ÐPRQ, P, To prove : Side PQ @ side PR, Construction : Draw the bisector of ÐP, , , intersecting side QR at point M, , Proof : In D PQM and D PRM, , ÐPQM, , @, ........ given, , ÐQPM @ ÐRPM........, seg PM @, ....... common side, \ D PQM @ D PRM ......, test, , \ seg PQ @ seg PR....... c.s.c.t., 33, , Q, , M, , Fig. 3.27, , R

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Corollary :, , , , , If three angles of a triangle are congruent then its three sides also are congruent., (Write the proof of this corollary yourself.), Both the above theorems are converses of each other also., Similarly the corollaries of the theorems are converses of each other., , Use your brain power!, (1), (2), , Can the theorem of isosceles triangle be proved doing a different construction ?, Can the theorem of isosceles triangle be proved without doing any construction ?, , Let’s learn., Property of 30° - 60° - 90° triangle, Activity I, , A, , Every student in the group should draw a, , 60°, , right angled triangle, one of the angles, measuring 30°. The choice of lengths of sides, should be their own. Each one should measure, , 30°, , B, , the length of the hypotenuse and the length of, , C, , Fig. 3.28, , the side opposite to 30° angle., , One of the students in the group should fill in the following table., Triangle Number, , 1, , 2, , 3, , 4, , Length of the side, opposite to 30° angle, Length of the hypotenuse, Did you notice any property of sides of right angled triangle with one of the angles, measuring 30°?, Activity II, The measures of angles of a set square in your compass box are 30°,60° and 90°., Verify the property of the sides of the set square., Let us prove an important property revealed from these activities., , 34

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Theorem : If the acute angles of a right angled triangle have measures 30° and 60°, then, the length of the side opposite to 30° angle is half the length of the hypotenuse., (Fill in the blanks and complete the proof .), A, Given, , : In D ABC, , 60°, , , ÐB = 90°, ÐC = 30°, ÐA = 60°, To prove : AB =, , 1, AC, 2, , 30°, , B, , Construction : Take a point D on the extended, seg AB such that AB = BD. Draw seg DC., , Fig. 3.29, , C, , A, , 60°, D ABC and D DBC, seg AB @ seg DB ..........., 30°, B, C, ÐABC @ ÐDBC ........, seg BC @ seg BC ............, D, \ D ABC @ D DBC ....., Fig. 3.30, \ ÐBAC @ ÐBDC ........ (c.a.c.t.), In D ABC, ÐBAC = 60° \ ÐBDC = 60°, ÐDAC = ÐADC = ÐACD = 60° ...Q sum of angles of D ADC is 180°, , \ D ADC is an equilateral triangle., , \ AC = AD = DC ........ corollary of converse of isosceles triangle theorem, , Proof, , :, , 1, AD........ construction, 2, \AB = 1 AC ........ Q AD = AC, , 2, , , , But AB =, , Activity, With the help of the Figure 3.29 above fill in the blanks and complete the proof of the, following theorem., Theorem : If the acute angles of a right angled triangle have measures 30°and 60° then, the length of the side opposite to 60° angle is, Proof : In the above theorem we have proved AB =, AB2 + BC2 =, , .............. Pythagoras theorem, , 1, 2, 2, 4 AC + BC =, , \ BC2 = AC2 -, , 1, AC, 2, , 3, 2 ´ hypotenuse, , 1, 2, 4 AC, , \ BC2 =, \ BC = 23 AC, 35

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Activity : Complete the proof of the theorem., Theorem : If measures of angles of a triangle are 45°, 45°, 90° then the length of each, side containing the right angle is 1 ´ hypotenuse., A, Proof : In D ABC, ÐB = 90° and, , 2, , ÐA = ÐC = 45°, , 45°, , \ BC = AB, By Pythagoras theorem, , \ 2AB2 =, \ AB2 =, \ AB =, , 45°, , B, , AB2 + BC2 =, AB2 +, = AC2 ... Q (BC = AB), , C, , Fig. 3.31, , 1, AC, 2, , This property is called 45°- 45°- 90° theorem., , Remember this !, (1) If the acute angles of a right angled triangle are 30°, 60°then the length of side, , opposite to 30° angle is half of hypotenuse and the length of side opposite to 60° angle, is, , 3, hypotenuse . This property is called 30°-60°-90°theorem., 2, , (2) If acute angles of a right angled triangle are 45°, 45° then the length of each side, hypotenuse, containing the right angle is, ., 2, , This property is called 45°-45°-90° theorem., , Let’s recall., Median of a triangle, The segment joining a vertex and the mid-point of the side opposite to it is called a, Median of the triangle., , A, , In Figure 3.32, point D is the mid point of side BC., , \ seg AD is a median of D ABC., , B, , D, , Fig. 3.32, , 36, , C

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Activity I, , : Draw a triangle ABC. Draw medians, , AD, BE and CF of the triangle. Let their point of, , B, , concurrence be G, which is called the centroid of, , D, , the triangle. Compare the lengths of AG and GD, , F, , with a divider. Verify that the length of AG is twice, the length of GD. Similarly, verify that the length of, BG is twice the length of GE and the length of CG, is twice the length of GF. Hence note the following, , C, , G, E, , A, , Fig. 3.33, , property of medians of a triangle., The point of concurrence of medians of a triangle divides each median in the ratio 2 : 1., , Activity II : Draw a triangle ABC on a card, board. Draw its medians and denote their point of, concurrence as G. Cut out the triangle., Now take a pencil. Try to balance the triangle, on the flat tip of the pencil. The triangle is balanced, only when the point G is on the flat tip of the pencil., This activity shows an important property of, the centroid (point of concurrence of the medians), of the triangle., , Fig. 3.34, , Let’s learn., Property of median drawn on the hypotenuse of right triangle, Activity : In the figure 3.35, D ABC is a right angled triangle. seg BD is the median on, B, hypotenuse., Measure the lengths of the following segments., AD =........., , DC =............, , BD = ............, , From the measurements verify that BD =, , 1, AC ., 2, , A, , D, , Fig. 3.35, Now let us prove the property, the length of the median is half the length of the, hypotenuse., , 37, , C

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Theorem : In a right angled triangle, the length of the median of the hypotenuse is half the, length of the hypotenuse., Given, , : In D ABC,, , E, , A, , ÐB = 90°, seg BD is the median., , To prove : BD = 1 AC, , D, , B, , 2, , Construction : Take point E on the ray BD such that B - D - E, , , C, , Fig. 3.36, , and l(BD) = l(DE). Draw seg EC., , : (Main steps are given. Write the steps in between with reasons and complete the, proof.), , D ADB @ D CDE .......... by S-A-S test, , Proof, , , , || line EC ..........by test of alternate angles, D ABC @ D ECB .......... by S-A-S test, , , , BD =, , line AB, , 1, AC, 2, , Remember this !, In a right angled triangle, the length of, the median on its hypotenuse is half the length of the hypotenuse., , Practice set 3.3, A, , 1., , Find the values of x and y using the, , x, , information shown in figure 3.37., , Find the measure of ÐABD and mÐACD., , 50°, , B 60°, , y C, , Fig. 3.37, D, , 2., , The length of hypotenuse of a right angled triangle is 15. Find the length of median of its, hypotenuse., , 3., , In D PQR, ÐQ = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS)., , 4., , In figure 3.38,, , P, , point G is the point of, , concurrence of the medians of, , D PQR ., , G, , If GT = 2.5, find the lengths of PG and PT., Q, , T, , Fig. 3.38, , 38, , R

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Let’s recall., Activity : Draw a segment AB of convenient length. Lebel its, mid-point as M. Draw a line l passing through the point M, and perpendicular to seg AB., Did you notice that the line l is the perpendicular bisector, of seg AB ?, Now take a point P anywhere on line l. Compare the, distance PA and PB with a divider. What did you find ? You A, should have noticed that PA = PB. This observation shows, that any point on the perpendicular bisector of a segment is, equidistant from its end points., Now with the help of a compass take any two points like C, and D, which are equidistant from A and B. Did all such, points lie on the line l ? What did you notice from the, observation ? Any point equidistant from the end points of, a segment lies on the perpendicular bisector of the segment., These two properties are two parts of the perpendicular, bisector theorem. Let us now prove them., , D, P, M, , B, , C, , l, Fig. 3.39, , Let’s learn., Perpendicular bisector theorem, Part I, , Given, , : Every point on the perpendicular bisector of a, segment is equidistant from the end points of, the segment., , l, , : line l is the perpendicular bisector of seg AB at point M., , Point P is any point on l,, To prove : PA = PB, , A, , Construction : Draw seg AP and seg BP., Proof, , P, , M, , : In D PMA and D PMB, , seg PM @ seg PM ....... common side, , ÐPMA @ ÐPMB .........each is a right angle, seg AM @ seg BM .......given, Fig. 3.40, , 39, , B

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\ D PMA @ D PMB ...... S-A-S test, \ seg PA @ seg PB .......c.s.c.t., \ l (PA) = l (PB), , , , Hence every point on the perpendicular bisector of a segment is equidistant from, the end points of the segment., , Part II : Any point equidistant from the end points of a segment lies on the perpendicular, bisector of the segment., Given : Point P is any point equidistant from, the end points of seg AB. That is, PA = PB., To prove : Point P is on the perpendicular bisector of seg AB., Construction : Take mid-point M of seg AB and draw line PM., Proof, , P, , : In D PAM and D PBM, , seg PA @ seg PB ........., seg AM @ seg BM ......., seg PM @, , M, , A, , B, , ....... common side, , \D PAM @ D PBM ......, test., \ ÐPMA @ ÐPMB.......c.a.c.t., Fig. 3.41, But ÐPMA +, = 180°, ÐPMA + ÐPMA = 180° ........ (Q ÐPMB = ÐPMA), 2 ÐPMA =, , \ ÐPMA = 90°, \ seg PM ^ seg AB , ......(1), But Point M is the midpoint of seg AB., , ......construction .... (2), , \ line PM is the perpendicular bisector of seg AB. So point P is on the, perpendicular bisector of seg AB., Angle bisector theorem, Part I, , : Every point on the bisector of an angle is equidistant, from the sides of the angle., , :, Given, , , To prove :, Proof, , Ray QS is the bisector of ÐPQR ., Point A is any point on ray QS, seg AC ^ ray QR, seg AB ^ ray QP, seg AB @ seg AC, , B, , S, , A, Q, , : Write the proof using test of congruence of triangles., , 40, , P, , C, , Fig. 3.42, , R

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Part II, , : Any point equidistant from sides of an angle is on the bisector of the angle., , Given, , : A is a point in the interior of ÐPQR., seg AC ^ ray QR, and AB = AC, , P, , seg AB ^ ray QP, , B, , To prove : Ray QA is the bisector of ÐPQR., That is ÐBQA = ÐCQA, , Proof, , D, A, , Q, , R, , C, , Fig. 3.43, : Write the proof using proper test of congruence of triangles., , Let’s recall., , X, , Activity, As shown in the figure, draw D XYZ, such that XZ > side XY, Find which of ÐZ and ÐY is greater., , Y, , Z, , Fig. 3.44, , Let’s learn., Properties of inequalities of sides and angles of a triangle, Theorem : If two sides of a triangle are unequal, then the angle opposite to the greater side, is greater than angle opposite to the smaller side. , Given, , : In D XYZ, side XZ > side XY, , To prove : ÐXYZ > ÐXZY, Construction : Take point P on side XZ such that, , XY = XP, Draw seg YP., : In D XYP, Proof, XY = XP .........construction, , X, P, Y, , Fig. 3.45, , Z, , \ ÐXYP = ÐXPY.....isosceles triangle theorem .....(I), ÐXPY is an exterior angle of D YPZ., \ ÐXPY > ÐPZY .........exterior angle theorem, \ ÐXYP > ÐPZY ..........from (I), \ ÐXYP + ÐPYZ > ÐPZY ........If a > b and c > 0 then a + c > b, \ ÐXYZ > ÐPZY, that is ÐXYZ > ÐXZY, 41

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Theorem : If two angles of a triangle are unequal then the side opposite to the greater angle, is greater than the side opposite to smaller angle. , The theorem can be proved by indirect proof. Complete the following proof by, filling in the blanks., Given, , : In DABC,, , ÐB > ÐC, , A, , To prove : AC > AB, Proof, , : There, , are, , only, , three, , possibilities, , B, , regarding the lengths of side AB and, side AC of, , D ABC., , (i) AC < AB , , C, , (iii), , (ii), , (i) Let us assume that AC < AB., If, , Fig. 3.46, , (ii) If AC = AB, then ÐB = ÐC, , two sides of a triangle are, , unequal then the angle opposite to, , But, , greater side is, , This also creates a contradiction., , ., , \ ÐC >, But ÐC < ÐB ......... (given), <, , ...... (given), , \, =, is wrong, \ AC > AB is the only remaining, , This creates a contradiction., , \, , >, , possibility., , \ AC > AB, , is wrong., , Let’s recall., As shown in the adjacent picture, there, is a shop at A. Sameer was standing at C. To, reach the shop, he choose the way C, , ®A, , ® B ® A, because he knew, that the way C ® A was shorter than the way, C ® B ® A. So which property of a triangle, instead of C, , A, , B, , had he realised ?, The sum of two sides of a triangle is, greater than its third side., Let us now prove the property., , C, 42

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Theorem : The sum of any two sides of a triangle is greater than the third side., Given, , D ABC is any triangle., , :, , D, , To prove : AB + AC > BC, AB + BC > AC, AC + BC > AB, , A, , Construction : Take a point D on ray BA such that AD = AC., Proof, , : In D ACD, AC = AD ..... construction, B, C, \ ÐACD = ÐADC ...... c.a.c.t., Fig. 3.47, \ ÐACD + ÐACB > ÐADC, \ ÐBCD > ÐADC, \ side BD > side BC .........the side opposite to greater angle is greater, \ BA + AD > BC ..............Q BD = BA + AD, , , , , , BA + AC > BC ........ Q AD = AC, Similarly we can prove that AB + BC > AC, and BC + AC > AB., , Practice set 3.4, 1., , In figure 3.48, point A is on the bisector of ÐXYZ., If AX = 2 cm then find AZ., , X, A, Y, , Z, , Fig. 3.48, , T, , 2., S, , P, R, , In figure 3.49, ÐRST = 56°, seg PT ^ ray ST,, seg PR ^ ray SR and seg PR @ seg PT, Find the measure of ÐRSP., State the reason for your answer., , Fig. 3.49, 3., , In D PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest, angle of the triangle., , 4., , In D FAN, ÐF = 80°, ÐA = 40° . Find out the greatest and the smallest side of the, triangle. State the reason., , 5., , Prove that an equilateral triangle is equiangular., , 43

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6., , 7., , Prove that, if the bisector of ÐBAC of, is an isosceles triangle., , D ABC is perpendicular to side BC, then D ABC, P, , In figure 3.50, if seg PR @ seg PQ,, show that seg PS > seg PQ., , Q, , R, , Fig. 3.50, , S, , A, , 8., , E, , In figure 3.51, in D ABC, seg AD, and seg BE are altitudes, and AE = BD., Prove that seg AD @ seg BE, , B, , D, , Fig. 3.51, , C, , Let’s learn., Similar triangles, Observe the following figures., , The pairs of figures shown in each part have the same shape but their sizes are different., It means that they are not congruent., Such like looking figures are called similar figures., , 44

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We find similarity in a photo and its enlargement, also we find similarity between a roadmap and the roads., The proportionality of all sides is an important property of similarity of two figures., But the angles in the figures have to be of the same measure. If the angle between this roads, is not the same in its map, then the map will be misleading., , ICT Tools or Links, Take a photograph on a mobile or a computer. Recall what you do to reduce it or to, enlarge it. Also recall what you do to see a part of the photograph in detail., Now we shall learn properties of similar triangles through an activity., Activity : On a card-sheet, draw a triangle of sides 4 cm, 3 cm and 2 cm. Cut it out. Make, 13 more copies of the triangle and cut them out from the card sheet., Note that all these triangular pieces are congruent. Arrange them as shown in the, following figure and make three triangles out of them., , P, , D, 4, A, 4, B, , 3, , 2, , 4, , 2, C, , E, , Fig. 3.52, 1 triangle, , 3, , 4, 4, , 2, 3, , Fig. 3.53, , 2, , ´, , 4, Q, , F, , 2, , 4 triangles, , 3, , 2, 3, , 3, , Fig. 3.54, , R, , 9 triangles, , D ABC and D DEF are similar in the correspondence ABC « DEF., ÐA @ ÐD, ÐB @ ÐE, ÐC @ ÐF, and, , AB, =, DE, , 4 1, = ;, 8 2, , BC, =, EF, , 3 1, = ;, 6 2, , AC =, DF, , 2 1, = ,, 4 2, , ........the corresponding sides are in proportion., , D DEF and D PQR. Are their angles congruent and sides, proportional in the correspondence DEF « PQR ?, Similarly, consider, , 45

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Remember this !, y•y If corresponding angles of two triangles are equal then the two triangles are similar., y•y If two triangles are similar then their corresponding sides are in proportion and, corresponding angles are congruent., , Ex., , P, , Some information is shown, , in D ABC and, , O, , D PQR in, , A, , figure 3.57. Observe it., Hence find the lengths of, side AC and PQ., , 7.5, , O, , 3, 4, , C, , Solution : The sum of all angles of a triangle is 180°., , B, , R, , 6, , Q, , Fig. 3.57, , It is given that,, , ÐA = ÐP and ÐB = ÐQ, , \ ÐC = ÐR, , \ D ABC and D PQR are equiangular triangles., \ there sides are propotional., AB, BC AC, \ , =, =, PQ, QR PR, 3, 4, AC, \ , =, =, PQ, 6, 7.5, , , , \ 4 ´ PQ = 18, 18, \ PQ =, = 4.5, 4, Similarly 6 ´ AC = 7.5 ´ 4, , \ AC =, , 7.5 ´ 4, 30, =, =5, 6, 6, , Practice set 3.5, 1., 2., 3., , If D XYZ ~ D LMN, write the corresponding angles of the two triangles and also write, the ratios of corresponding sides., In D XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If D XYZ, PQ = 8 cm then find the lengths of remaining sides of D PQR., , ~ D PQR and, , Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles, by the same signs. Show the lengths of corresponding sides by numbers in proportion., , 47

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Let’s recall., While preparing a map of a locality, you have to show the distances between different, spots on roads with a proper scale. For example, 1 cm = 100 m, 1 cm = 50 m etc. Did you, think of the properties of triangle ? Keep in mind that side opposite to greater angle is greater., Project :, Prepare a map of road surrounding your school or home, upto a distance of about 500, metre., How will you measure the distance between two spots on a road ?, While walking, count how many steps cover a distance of about two metre. Suppose,, your three steps cover a distance of 2 metre. Considering this proportion 90 steps means 60, metre. In this way you can judge the distances between different spots on roads and also the, lengths of roads. You have to judge the measures of angles also where two roads meet each, other. Choosing a proper scale for lengths of roads, prepare a map. Try to show shops,, buildings, bus stops, rickshaw stand etc. in the map., A sample map with legend is given below., , 6, 2, , 1, 7, , 2, 1, 4, , 6, 3, , 5, , 4, , 6, , 3, Legend: 1. Book store, 5. Medical store, , 2. Bus stop, 6. Restaurant, , 3. Stationery shop, 7. Cycle shop, , 48, , 4. Bank

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Problem set 3, 1., , Choose the correct alternative answer for the following questions., (i) If two sides of a triangle are 5 cm and 1.5 cm, the lenght of its third side cannot be, ........, , , , (A) 3.7 cm, , (B) 4.1 cm, , (C) 3.8 cm (D) 3.4 cm, , (ii) In D PQR, If ÐR > ÐQ then . . . . . . . . ., , , (A) QR > PR, , (B) PQ > PR, , (C) PQ < PR, , (D) QR < PR, , (iii) In D TPQ, ÐT = 65°, ÐP = 95° which of the following is a true statement ?, , 2., , (A) PQ < TP, , (B) PQ < TQ, , (C) TQ < TP < PQ, , (D) PQ < TP < TQ, , D ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that, BD = CE., P, , 3., , In D PQR, If PQ > PR and bisectors of, intersect at S. Show that SQ > SR., , ÐQ and ÐR, , S, Q, , 4., , R, , Fig. 3.58, A, , In figure 3.59, point D and E are on, side BC of D ABC,, such that BD = CE and AD = AE., Show that D ABD @ D ACE., B, , D, , Fig. 3.59, , 5., , E, , C, , P, , In figure 3.60, point S is any point, on side QR of D PQR., Prove that : PQ + QR + RP > 2PS, , Q, , 49, , S, , Fig. 3.60, , R

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In figure 3.61, bisector of, , 6., , ÐBAC, , A, , intersects side BC at point D., Prove that AB > BD, B, , C, , D, , Fig. 3.61, S, , In figure 3.62, seg PT is the bisector of ÐQPR., , 7. , , A line parallel to seg PT and passing through R, , P, , intersects ray QP at point S. Prove that PS = PR., , Q, , T, , R, , Fig. 3.62, C, , 8. In figure 3.63, seg AD ^ seg BC., , seg AE is the bisector of ÐCAB and, C - E - D., , D, , Prove that, , ÐDAE =, , E, 1, (Ð C 2, , ÐB ), A, , B, , Fig. 3.63, , Use your brain power!, We have learnt that if two triangles are equiangular then their sides are in, proportion. What do you think if two quadrilaterals are equiangular ? Are their, sides in proportion? Draw different figures and verify., Verify the same for other polygons., , qqq, 50

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4, , Constructions of Triangles, , Let’s study., To construct a triangle, if following information is given., • Base, an angle adjacent to the base and sum of lengths of two remaining sides., • Base, an angle adjacent to the base and difference of lengths of remaining two, sides., • Perimeter and angles adjacent to the base., , Let’s recall., In previous standard we have learnt the following triangle constructions., * To construct a triangle when its three sides are given., * To construct a triangle when its base and two adjacent angles are given., * To construct a triangle when two sides and the included angle are given., * To construct a right angled triangle when its hypotenuse and one side is given., Perpendicular bisector Theorem, • Every point on the perpendicular bisector of a, segment is equidistant from its end points., , Q, , • Every point equidistant from the end points of, , a segment is on the perpendicular bisector of, the segment., , A, , T, l, , B, , Fig. 4.1, , Let’s learn., Constructions of triangles, To construct a triangle, three conditions are required. Out of three sides and three angles of a triangle two parts and some additional information about them is given, then we can, construct a triangle using them., We frequently use the following property in constructions., If a point is on two different lines then it is the intersecrtion of the two lines., , 51

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Construction I, To construct a triangle when its base, an angle adjacent to the base and the sum, of the lengths of remaining sides is given., Ex. Construct D ABC in which BC = 6.3 cm, ÐB = 75° and AB + AC = 9 cm., Solution : Let us first draw a rough figure of expected triangle., Explanation : As shown in the rough figure,, first we draw seg BC = 6.3 cm of length. On, the ray making an angle of 75° with seg BC,, mark point D such that , BD = AB + AC = 9 cm, Now we have to locate point A on, ray BD., BA + AD = BA + AC = 9, , A, , 6.3 cm, , C, , Rough figure 4.2, , 9 cm, , B, , D, , 75°, , \ AD = AC, \ point A is on the perpendicular, , A, , bisector of seg CD., , B, , \ the point of intersection of ray, , 6.3 cm, , Rough figure 4.3, , BD and the perpendicular bisector of seg CD is point A., , P, , Steps of construction, (1) Draw seg BC of length 6.3 cm., (2) Draw ray BP such that, mÐ PBC = 75°., (3) Mark point D on ray BP such, that d(B,D) = 9 cm, (4) Draw seg DC., (5) Construct the perpendicular, bisector of seg DC ., (6) Name the point of intersection, of ray BP and the perpendicular, bisector of CD as A., (7) Draw seg AC., , D, , A, , D ABC is the required triangle., 75°, , B, , 6.3 cm, , Fair fig. 4.4, , 52, , C, , C

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Practice set 4.1, 1., 2., 3., 4., , Construct D, Construct D, Construct D, Construct D, is 10 cm, , PQR, in which QR = 4.2 cm, mÐQ = 40° and PQ + PR = 8.5 cm, XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. ÐXYZ = 50°, ABC, in which BC = 6.2 cm, ÐACB = 50°, AB + AC = 9.8 cm, ABC, in which BC = 3.2 cm, ÐACB = 45° and perimeter of D ABC, , Construction II, To construct a triangle when its base, angle adjacent to the base and difference, between the remaining sides is given., Ex (1) Construct D ABC, such that BC = 7.5 cm, ÐABC = 40°, AB - AC = 3 cm., Solution : Let us draw a rough figure., A, Explanation : AB - AC = 3 cm \ AB > AC, Draw seg BC. We can draw the ray BL such, 40°, that Ð LBC = 40°. We have to locate point, C, B, 7.5 cm, A on ray BL. Take point D on ray BL such, Rough Figure 4.5, A, that BD = 3 cm., Now , B-D-A and BD = AB - AD = 3., It is given that AB - AC = 3, \ AD = AC, D, , \ point A is on the perpendicular bisector, of seg DC., , B, , 40°, , C, , \ point A is the intersection of ray BL and, , Rough figure 4.6, , the perpendicular bisector of seg DC., , A, , Steps of construction, (1) Draw seg BC of length7.5 cm., (2) Draw ray BL such that, Ð LBC = 40°, (3) Take point D on ray BL such that, BD = 3 cm., (4) Construct, the, perpendicular, bisector of seg CD., (5) Name the point of intersection of, ray BL and the perpendicular, bisector of seg CD as A., (6) Draw seg AC., D ABC is required triangle., , L, , D, 40°, , B, , C, , 7.5 cm, , Fair figure 4.7, , 53

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Ex. 2 Construct D ABC, in which side BC = 7 cm, ÐB = 40° and AC - AB = 3 cm., Solution : Let us draw a rough figure., A, seg BC = 7 cm. AC > AB., We can draw ray BT such that, 40°, Ð TBC = 40° , C, B, 7 cm, Point A is on ray BT. Take point D on, Rough figure 4.8, opposite ray of ray BT such that, T, A, BD = 3 cm., Now AD = AB + BD = AB + 3 = AC, (Q AC - AB = 3 cm.), , \ AD = AC, \ point A is on the perpendicular, , 7 cm, , B, S, , bisector of seg CD., , C, , D, , Rough figure 4.9, , Steps of construction, (1) Draw BC of length 7 cm., , T, , (2) Draw ray BT such that, Ð TBC = 40°, , (3) Take point D on the opposite, ray BS of ray BT such that, , A, , BD = 3 cm., (4) Construct perpendicular, , B, , bisector of seg DC., , m, 3c, , (5) Name the point of intersection, , of DC as A., (6) Draw seg AC., , 7 cm, , D, , of ray BT and, the perpendicular bisector, , 40°, , S, , Fair fig. 4.10, , D ABC is the required triangle., , Practice set 4.2, 1. Construct D XYZ, such that YZ = 7.4 cm, ÐXYZ = 45° and XY - XZ = 2.7 cm., , 2. Construct D PQR, such that QR = 6.5 cm, ÐPQR = 40° and PQ - PR = 2.5 cm., , 3. Construct D ABC, such that BC = 6 cm, ÐABC = 100° and AC - AB = 2.5 cm., , 54, , C

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Construction III, To construct a triangle, if its perimeter, base and the angles which include the base are, given., Ex. Construct D ABC such that AB + BC + CA = 11.3 cm, ÐB = 70°,, Solution : Let us draw a rough figure., , P, , 35°, , ÐC = 60°., , A, , B, , 70°, , 60°, , 11.3 cm, , C, , 30°, , Q, , Rough Fig. 4.11, Explanation : As shown in the figure, points P and Q are taken on line BC such that,, PB = AB, CQ = AC, , \ PQ = PB + BC + CQ = AB + BC +AC = 11.3 cm., Now in DPBA, PB = BA, \ ÐAPB = ÐPAB and ÐAPB + ÐPAB = extieror angleABC = 70°, ......theorem of remote interior angles, , \ ÐAPB = ÐPAB = 35° Similarly, ÐCQA = ÐCAQ = 30°, Now we can draw D PAQ, as its two angles and the included side is known., Since BA = BP , point B lies on the perpendicular bisector of seg AP., , Similarly, CA = CQ, therefore point C lies on the perpendicular bisector of seg AQ, , \ by constructing the perpendicular bisectors of seg AP and AQ we can get points, B and C, where the perpendicular bisectors intersect line PQ., Steps of construction, (1) Draw seg PQ of 11.3 cm length., , (5) Draw the perpendicular bisector of, seg AP and seg AQ. Name the points as, B and C respectively where the, perpendicular bisectors intersect line, PQ., (6) Draw seg AB and seg AC., , (2) Draw a ray making angle of 35° at, point P., (3) Draw another ray making an angle of, 30° at point Q., (4) Name the point of intersection of the, two rays as A., , D ABC is the required triangle., 55

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A, , P, , 35°, , B, , 70°, , 60°, , C, , 30°, , 11.3 cm, Final Fig. 4.12, , Q, , Practice set 4.3, 1. Construct D PQR, in which ÐQ = 70°, ÐR = 80° and PQ + QR + PR = 9.5 cm., 2. Construct D XYZ, in which ÐY = 58°, ÐX = 46° and perimeter of triangle is 10.5 cm., 3. Construct D LMN, in which ÐM = 60°, ÐN = 80° and LM + MN + NL = 11 cm., , Problem set 4, 1., 2., 3., , Construct D XYZ, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, ÐXYZ = 45°., Construct D ABC, in which ÐB = 70°, ÐC = 60°, AB + BC + AC = 11.2 cm., The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4., Construct the triangle., , 4., , Construct D PQR, in which PQ - PR = 2.4 cm, QR = 6.4 cm and, , ÐPQR = 55°., , ICT Tools or Links, Do constructions of above types on the software Geogebra and enjoy the, constructions. The third type of construction given above is shown on Geogebra by a, different method. Study that method also., , qqq, , 56

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5, , Quadrilaterals, , Let’s study., , • Parallelogram, • Tests of parallelogram, • Rhombus, , • Rectangles, • Square, • Trapezium, , • Mid point theorem, , Let’s recall., B, , 1., , Write the following pairs considering, C, , A, , Pairs of adjacent sides:, (1) ... , ..., (2) ... , ..., (3) ... , ..., (4) ... , ..., , �ABCD, , Pairs of adjacent angles :, (1) ... , ..., (2) ... , ..., (3) ... , ..., (4) ... , ..., , Pairs of opposite sides (1) ..... , ..... (2) ..... , ....., Pairs of opposite angles (1) ..... , ..... (2) ..... , ....., , D, , Fig. 5.1, , Let’s recall types of quadrilaterals and their properties ., I am a quadrilateral, , My both pairs of opposite sides are parallel., , My all angles are, right angles., , My all sides are, equal in length., , My properties, , My properties, , My properties, , · Opposite sides congruent · Opposite sides ....., · Opposite angels ......, · Diagonals ....., · Diagonals ....., , · Opposite sides ....., · Diagonals ....., , My only one pair of opposite sides is parallel., , 57, , My all angles are equal, and all sides are equal., , My properties, , · Diagonals .....

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You know different types of quadrilaterals and their properties. You have learned then, through different activities like measuring sides and angles, by paper folding method etc., Now we will study these properties by giving their logical proofs., A property proved logically is called a proof., In this chapter you will learn that how a rectangle, a rhombus and a square are, parallelograms. Let us start our study from parallelogram., , , Let’s learn., , Parallelogram, A quadrilateral having both pairs of opposite sides parallel is called a parallelogram., For proving the theorems or for solving the problems we need to draw figure of a, parallelogram frequently. Let us see how to draw a parallelogram., Suppose we have to draw a parallelogram �ABCD., Method I :, � Let us draw seg AB and seg BC of any, length and making an angle of any measure, A, with each other., � Now we want seg AD and seg BC, parallel to each other. So draw a line parallel, B, to seg BC through the point A., Fig. 5.2, � Similarly we will draw line parallel to, AB through the point C. These lines will, intersect in point D., So constructed quadrilateral ABCD will be a parallelogram., Method II :, � Let us draw seg AB and seg BC of, any length and making angle of any measure, between them., • Draw an arc with compass with centre, A and radius BC., • Similarly draw an arc with centre C, and radius AB intersecting the arc previously, drawn., • Name the point of intersection of two, arcs as D., Draw seg AD and seg CD., Quadrilateral so formed is a, parallelogram ABCD, , D, , C, , A, , B, , 58, , D, , Fig. 5.3, , C

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In the second method we have actually drawn �ABCD in which opposite sides, are equal. We will prove that a quadrilateral whose opposite sides are equal, is a parallelogram., Activity I Draw five parallelograms by taking various measures of, , lengths and angles., For the proving theorems on parallelogram, we use congruent triangles. To understand, how they are used, let’s do the following activity., Activity II, , • Draw a parallellogram ABCD on a, , card sheet. Draw diagonal AC. Write the, names of vertices inside the triangle as, shown in the figure. Then cut is out., , B, B, , C C, , C, , I, , • Fold the quadrilateral on the diagonal, AC and see whether DADC and DCBA, , A, , match with each other or not., , A A, , II, , D, D, , Fig. 5.4, D, , • Cut �ABCD along diagonals AC, and separate DADC and DCBA. By, rotating and flipping DCBA, check, whether it matchs exactly with DADC., , B, B, , What did you find ? Which sides of, DCBA match with which sides of DADC ?, Which angles of DCBD match with, which angles of DADC ?, , A, , C, , Fig. 5.5, B, , C, A, , D, , Side DC matches with side AB and, side AD matches with side CB. Similarly ∠ B matches with ∠ D., , A, C, , So we can see that opposite sides and, angles of a parallelogram are congruent., , We will prove these properties of a parallelogram., , 59, , Fig. 5.6

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Theorem 1. Opposite sides and opposite angles of a parallelogram are congruent., , Given, : �ABCD is a parallelogram., D, C, , It means side AB || side DC,, ° ´, side AD || side BC., ´ °, To prove : seg AD ≅ seg BC ; seg DC ≅ seg AB, A, B, , ∠ADC ≅ ∠CBA, and ∠DAB ≅ ∠BCD., Fig. 5.7, , Construction : Draw diagonal AC., , || seg AB and diagonal AC is a transversal., ∴ ∠DCA ≅ ∠BAC ................(1), and ∠DAC ≅ ∠BCA ..............(2), ..... alternate angles, Now , in ∆ADC and ∆CBA,, , ∠DAC ≅ ∠BCA, .......... from (2), ∠DCA ≅ ∠BAC, .......... from (1), , .......... common side, seg AC ≅ seg CA, ∴ ∆ADC ≅ ∆CBA, .......... ASA test, ∴side AD ≅ side CB, .......... c.s.c.t., and side DC ≅ side AB .......... c.s.c.t.,, .......... c.a.c.t., Also, ∠ADC ≅ ∠CBA, Similarly we can prove ∠DAB ≅ ∠BCD., Proof : seg DC, , }, , Use your brain power!, In the above theorem, to prove ∠DAB ≅ ∠BCD, is any change in, the construction needed ? If so, how will you write the proof making the change ?, To know one more property of a parallelogram let us do the following activity., Activity : Draw a parallelogram PQRS. Draw, diagonals PR and QS. Denote the, intersection of diagonals by letter, O. Compare the two parts of each, diagonal with a divider. What do, you find ?, , 60, , P, , S, , X, , O, Q, , X, , Fig. 5.8, , R

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Theorem : Diagonals of a parallelogram bisect each other., P, , S, , X, , Q, , Given, , O, X, , Fig. 5.9, , R, , :, , �PQRS is a parallelogram. Diagonals PR, , and QS intersect in point O., To prove : seg PO ≅ seg RO,, seg SO ≅ seg QO., , Proof : In ∆POS and ∆ROQ, ∠OPS ≅ ∠ORQ .......... alternate angles, , side PS ≅ side RQ ......... opposite sides of parallelogram , ∠PSO ≅ ∠RQO .......... alternate angles, , ∴∆POS ≅ ∆ROQ ....... ASA test, , ∴seg PO ≅ seg RO .............., , ..... corresponding sides of, and seg SO ≅ seg QO .........., congruent triangles, , }, , Remember this !, , •, •, •, •, , Adjacent angles of a parallelogram are supplementary., Opposite sides of a parallelogram are congruent., Opposite angles of a parallelogram are congruent., Diagonals of a parallelogram bisect each other., , Solved Examples, Ex (1) �PQRS is a parallelogram. PQ = 3.5, PS = 5.3, , ∠Q = 50° then find the lengths of, , remaining sides and measures of remaining angles., , P, , 5.3, , S, , �PQRS is a parallelogram., \∠ Q + ∠ P = 180° ........ interior angles are, 50°, supplementary., \ 50° + ∠ P = 180°, Q, R, Fig. 5.10, \ ∠ P = 180° - 50° = 130° , Now , ∠ P = ∠ R and ∠ Q = ∠ S ........opposite angles of a parallelogram., \ ∠ R = 130° and ∠ S = 50°, 3.5, , Solution :, , Similarly, PS = QR and PQ = SR ........opposite sides of a parallelogram., , , \ QR = 5.3 and SR = 3.5, , 61

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Ex (2), , �ABCD is a parallelogram. If Ð A = (4x +13)° and Ð D = (5x -22)° then find, the measures of Ð B and Ð C., , Solution : Adjacent angles of a parallelogram are supplementary., Ð A and Ð D are adjacent angles., D, C, 5x - 22, \ (4x +13)°+ (5x - 22)° = 180, \ 9x - 9 = 180, 4x +13, A, \ 9x = 189, Fig. 5.11, \ x = 21, \ Ð A = 4x +13 = 4 ´ 21 + 13 = 84+13 = 97° \ Ð C = 97°, Ð D = 5x - 22 = 5 ´ 21 - 22 = 105 - 22 = 83° \ Ð B = 83°, , B, , Practice set 5.1, 1. Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠XYZ = 135°, then what is the measure of ∠XWZ and ∠YZW ?, If l(OY)= 5 cm then l(WY)= ?, 2. , In a parallelogram ABCD, If ∠A = (3x + 12)°, ∠B = (2x - 32)° then find the value, of x and then find the measures of ∠C and ∠D., 3. Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side, by 25 cm. Find the lengths of all sides., 4. If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the, measures of all angles of the parallelogram., 5*. Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and, AB = 13 then show that �ABCD is a rhombus., 6. In the figure 5.12, �PQRS and �ABCR, are two parallelograms., If ∠P = 110° then find the measures of, all angles of �ABCR., , Q, , P, , 110°, , S, , B, , A, , C, R, , Fig. 5.12, 7. In figure 5.13 �ABCD is a parallelogram. Point E is on the ray AB such, that BE = AB then prove that line ED, bisects seg BC at point F., , B, , A, F, D, , 62, , C, , Fig. 5.13, , E

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Let’s recall., Tests for parallel lines, 1., , If a transversal interesects two lines and a pair of corresponding angles is congruent, then those lines are parallel., , 2., , If a transversal intersects two lines and a pair of alternate angles is corgruent then, those two lines are parallel., , 3., , If a transversal intersects two lines and a pair of interior angles is supplementary then, those two lines are parallel., , Let’s learn., Tests for parallelogram, Suppose, in �PQRS, PS = QR and PQ = SR, S, P, and we have to prove that �PQRS is a, parallelogram. To prove it, which pairs of, sides of �PQRS should be shown parallel ?, R, Q, Which test can we use to show the sides, Fig. 5.14, parallel ? Which line will be convenient as a, transversal to obtain the angles necessary to, apply the test ?, Theorem : If pairs of opposite sides of a quadrilateral are congruent then that quadrilateral, is a parallelogram., P, S, : In �PQRS, Given, side PS @ side QR, side PQ @ side SR, To prove : �PQRS is a parallelogram., Construction : Draw diagonal PR., Proof, : In D SPR and D QRP, side PS @ side QR ........given, R, Q, Fig., 5.15, side SR @ side QP ........ given, side PR @ side RP ........ common side, \D SPR @ D QRP ...... sss test, , \∠ SPR @ ∠ QRP ....... c.a.c.t., Similarly, ∠PRS @ ∠RPQ ..... c.a.c.t., ∠SPR and ∠QRP are alternate angles formed by the transversal PR of, , seg PS and seg QR., , 63

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\ side PS || side QR ......(I) alternate angles test for parallel lines., Similarly, , ∠PRS and ∠RPQ are the alternate angles formed by transversal PR of, , seg PQ and seg SR., , \ side PQ || side SR ......(II) .....alternate angle test, \ from (I) and (II) �PQRS is a parallelogram., On page 56, two methods to draw a parallelogram are given. In the second method, actually we have drawn a quadrilateral of which opposite sides are equal. Did you now, understand why such a quadrilateral is a parallelogram ?, Theorem : If both the pairs of opposite angles of a quadrilateral are congruent then it is, a parallelogram., H, , G, , E, , Fig. 5.16, , Proof : Let, , F, , �EFGH ∠ E @ ∠ G, and ∠.......... @ ∠.........., To prove : �EFGH is a ............, , Given, , : In, , ∠ E = ∠ G = x and ∠ H = ∠ F = y, , Sum of all angles of a quadrilateral is ............., , \ ∠ E +, , ∠ G + ∠ H + ∠ F = ........., , , \ x + y + .......... + .......... = .........., , \ �x +, , �y = ....., , \ x + y = 180°, , \ ∠ G + ∠ H = .........., , ∠ G and ∠ H are interior angles formed by transversal HG of seg HE, and seg GF., , \ side HE, , || side GF .......... (I) interior angle test for parallel lines. , Similarly, ∠ G + ∠ F = .........., , \ side .......... || side .......... .......... (II) interior angle test for parallel lines., , \ From (I) and (II), �EFGH is a .................... ., , 64

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If the diagonals of a quadrilateral bisect each other then it is a parallelogram., Diagonals of �ABCD bisect each other in the point E., It means seg AE @ seg CE, A, B, and seg BE @ seg DE, �ABCD is a parallelogram., E, Find the answers for the following questions, C, D, and write the proof of your own., Fig. 5.17, 1. Which pair of alternate angles should be shown, congruent for proving seg AB || seg DC ?, Which transversal will form a pair of alternate, angles ?, 2. Which triangles will contain the alternate angles formed by the transversal?, 3. Which test will enable us to say that the two triangles congruent ?, 4. Similarly, can you prove that seg AD || seg BC?, The three theorems above are useful to prove that a given quadrilateral, is a parallelogram. Hence they are called as tests of a parallelogram., One more theorem which is useful as a test for parallelogram is given below., , Theorem :, Given, :, , , To prove :, Proof, :, , Theorem : A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and, C, D, congruent., Given, : In �ABCD, seg CB @ seg DA and seg CB || seg DA, To prove : �ABCD is a parallelogram., B, A, Construction : Draw diagonal BD., Fig. 5.18, Write the complete proof which is given in short., D CBD @ D ADB .......SAS test, , \ ∠CDB @ ∠ABD ..... c.a.c.t., , \ seg CD || seg BA ..... alternate angle test for parallel lines, , , Remember this !, •, •, •, •, , A quadrilateral is a parallelogram if its pairs of opposite angles are congruent., A quadrilateral is a parallelogram if its pairs of opposite sides are congruent., A quadrilateral is a parallelogram if its diagonals bisect each other., A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent., These theorems are called tests for parallelogram., , Let’s recall., Lines in a note book are parallel. Using these lines, how can we draw a parallelogram ?, , 65

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Solved examples �PQRS is parallelogram. M is the midpoint of side PQ and N is the mid point, Ex (1), of side RS. Prove that �PMNS and �MQRN are parallelograms., P, S, Given, : � PQRS is a parallelogram., M and N are the midpoints of, side PQ and side RS respectively., M, N, To prove : �PMNS is a parallelogram., �MQRN is a parallelogram., , R, Q, Fig. 5.19, Proof, : side PQ || side SR, \ side PM || side SN ...... ( P-M-Q; S-N-R) ......(I), , , side PQ @ side SR., 1, 2, , \ side PQ =, , , 1, side SR, 2, , \ side PM @ side SN ..... ( M and N are midpoints.)......(II), , \ From (I) and (II), �PMNQ is a parallelogram,, , Similarly, we can prove that �MQRN is parallelogram., Ex (2) Points D and E are the midpoints of side AB and side AC of D ABC respectively., Point F is on ray ED such that ED = DF. Prove that �AFBE is a parallelogram., For this example write ‘given’ and ‘to prove’ and complete the proof given below., Given : ------------------To prove : ---------------------A, , Proof : seg AB and seg EF are, seg AD, seg, , of, , �AFBE., , @ seg DB......., @ seg, , , \ Diagonals of, , F, , D, , E, , .......construction., , �AFBE, , each other, , , \ �AFBE is a parallelogram ...by, , test., , B, , C, , Fig. 5.20, , Ex (3) Prove that every rhombus is a parallelogram., Given, , :, , To prove :, Proof, , B, , �ABCD is a rhombus., , C, , �ABCD is parallelogram., , : seg AB @ seg BC @ seg CD @ seg DA (given), , \side AB @ side CD and side BC @ side AD, , , A, , D, , Fig. 5.21, \ �ABCD is a parallelogram..... opposite side test for parallelogram, , , 66

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Practice set 5.2, A, , 1. In figure 5.22, �ABCD is a parallelogram, P and Q are midpoints of side AB, and DC respectively, then prove �APCQ, is a parallelogram., , D, , P, , Q, , B, , C, , Fig. 5.22, , 2. Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram., D, , 3. In figure 5.23, G is the point of concurrence of medians of D DEF. Take point, H on ray DG such that D-G-H and, DG = GH, then prove that �GEHF is a, parallelogram., , G, F, , E, H, , A, , 4. Prove that quadrilateral formed by the, intersection of angle bisectors of all, angles of a parallelogram is a rectangle., (Figure 5.24), , Fig. 5.23, B, , P, , Q, , S, R, , C, , D, , Fig. 5.24, 5. In figure 5.25, if points P, Q, R, S are, on the sides of parallelogram such that, AP = BQ = CR = DS then prove that, �PQRS is a parallelogram., , A, , P, , B, Q, , S, D, , R, , C, , Fig. 5.25, , Let’s learn., Properties of rectangle, rhombus and square, Rectangle, rhombus and square are also parallelograms. So the properties that opposite, sides are equal, opposite angles are equal and diagonals bisect each other hold good in, these types of quadrilaterals also. But there are some more properties of these quadrilaterals., Proofs of these properties are given in brief. Considering the steps in the given proofs,, write the proofs in detail., , 67

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Theorem : Diagonals of a rectangle are congruent., , �ABCD is a rectangle., To prove : Diagonal AC @ diagonal BD, , Given, , A, , D, , :, , : Complete the proof by giving suitable reasons., B, D ADC @ D DAB ...... SAS test, \ diagonal AC @ diagonal BD..... c.s.c.t., , , Proof, , Fig. 5.26, , C, , Theorem : Diagonals of a square are congruent., Write ‘Given’, ‘To prove’ and ‘proof’ of the theorem., Theorem : Diagonals of a rhombus are perpendicular bisectors of each other., Given, : �EFGH is a rhombus, To prove : (i) Diagonal EG is the perpendicular, bisector of diagonal HF., (ii) Diagonal HF is the perpendicular, bisector of diagonal EG., Proof, , @ seg EH, seg GF @ seg GH, , : (i) seg EF, , } given, , E, , F, , H, , Fig. 5.27, G, , Every point which is equidistant from end points of a segment is on the, perpendicular bisector of the segment., \ point E and point G are on the perpendicular bisector of seg HF., , One and only one line passes through two distinct points., \ line EG is the perpendicular bisector of diagonal HF., , \ diagonal EG is the perpendicular bisector of diagonal HF., , (ii) Similarly, we can prove that diagonal HF is the perpendicular bisector, , of EG., Write the proofs of the following statements., � Diagonals of a square are perpendicular bisectors of each other., � Diagonals of a rhombus bisect its opposite angles., � Diagonals of a square bisect its opposite angles., , Remember this !, � Diagonals of a rectangle are congruent., � Diagonals of a square are congruent., � Diagonals of a rhombus are perpendicular, bisectors of each other., , 68, , � Diagonals of a rhombus bisect the pairs, of opposite angles., � Diagonals of a square are perpendicular, bisectors of each other., � Diagonals of a square bisect opposite, angles.

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Practice set 5.3, 1. Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find the length, of BO and if ∠CAD = 35° then find the measure of ∠ACB., 2. In a rhombus PQRS if PQ = 7.5 then find the length of QR., If ∠QPS = 75° then find the measure of ∠PQR and ∠SRQ., 3. Diagonals of a square IJKL intersects at point M, Find the measures of ∠IMJ, ∠JIK, and ∠LJK ., 4. Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter., 5. State with reasons whether the following statements are ‘true’ or ‘false’., (i) Every parallelogram is a rhombus., (ii) Every rhombus is a rectangle., (iii) Every rectangle is a parallelogram., (iv) Every squre is a rectangle., (v) Every square is a rhombus., (vi) Every parallelogram is a rectangle., , Let’s learn., Trapezium, When only one pair of opposite sides of a quadrilateral is parallel then the quadrilateral, is called a trapezium., B, , A, , In the adjacent figure only side AB and, side DC of �ABCD are parallel to each, other. So this is a trapezium. ÐA and ÐD, is a pair of adjacent angles and so is the, pair of ÐB and ÐC. Therefore by property, of parallel lines both the pairs are, supplementary., , D, , C, , Fig. 5.28, , P, , If non-parallel sides of a trapezium are, congruent then that quadrilateral is called, as an Isoceles trapezium., S, , Q, , Fig. 5.29, , R, , The segment joining the midpoints of non parallel sides of a trapezium, is called the median of the trapezium., , 69

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PQ =, , 1, PR ...... (construction), 2, , \ PQ =, , , 1, BC, 2, , a PR = BC, , Converse of midpoint theroem, Theorem : If a line drawn through the midpoint of one side of a triangle and parallel to, the other side then it bisects the third side. , For this theorem ‘Given’, To prove’, ‘construction’ is given below. Try to, write the proof., Given, , : Point D is the midpoint of side AB of D ABC. Line l passing through the, point D and parallel to side BC intersects side AC in point E., A, , To prove : AE = EC, passing through the point C. Name the, , B, , Fig. 5.35, , point of intersection where this line and, , F, , E, , D, , Construction : Draw a line parallel to seg AB, , l, , C, , line l will intersect as F., Proof, , : Use the construction and line l || seg BC which is given. Prove, , , D ADE, , @ D CFE and complete the proof., , Ex (1) Points E and F are mid points of seg AB and seg AC of, , D ABC respectively. If, , EF = 5.6 then find the length of BC., Solution : In, , D ABC, point E and F are midpoints of, , A, , side AB and side AC respectively., EF =, , 1, BC .......midpoint theorem, 2, , 5.6 =, , 1, BC, 2, , E, B, , \ BC = 5.6 ´ 2 = 11.2, , Ex (2) Prove that the quadrilateral formed by joining the, , F, C, , Fig. 5.36, , midpoints of sides of a quadrilateral in order is a parallelogram., Given, : �ABCD is a quadrilateral., P, Q, R, S are midpoints of the, sides, AB, BC, CD and AD, respectively., To prove : �PQRS is a parallelogram., Construction : Draw diagonal BD, , A, S, D, , 72, , P, , B, , Q, R, , Fig. 5.37, C

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Proof, , : In, , D ABD, the midpoint of side AD is S and the midpoint of side AB is P., , \ by midpoint theorem, PS, , , || DB and PS =, , 1, BD ........... (1), 2, , In D DBC point Q and R are midpoints of side BC and side DC respectively., , \QR || BD and QR =, , \PS, , \, , , 1, BD ...........by midpoint theorem (2), 2, , || QR and PS = QR ................ from (1) and (2), , �PQRS is a parallelogram., Practice set 5.5, A, , 1. In figure 5.38, points X, Y, Z are the midpoints, of side AB, side BC and side AC of D ABC, respectively. AB = 5 cm, AC = 9 cm and, BC = 11 cm. Find the length of XY, YZ, XZ., , Z, , X, Y, , B, , Fig. 5.38, L, , S, , 2. In figure 5.39, �PQRS and �MNRL are, rectangles. If point M is the midpoint of side PR, 1, then prove that, (i) SL = LR, (ii) LN = 2 SQ., , C, R, N, , M, P, , Q, , Fig. 5.39, A, , 3. In figure 5.40, DABC is an equilateral traingle., Points F,D and E are midpoints of side AB, side, BC, side AC respectively. Show that D FED is, an equilateral traingle., , F, B, P, , 4. In figure 5.41, seg PD is a median of D PQR., Point T is the mid point of seg PD. Produced, , PM, 1, QT intersects PR at M. Show that PR = 3 ., , [Hint : draw DN, , || QM.], , E, , Fig. 5.40, M, T, , Q, , C, , D, , D, , N, R, , Fig. 5.41, , Problem set 5, 1., , Choose the correct alternative answer and fill in the blanks., , (i) If all pairs of adjacent sides of a quadrilateral are congruent then it is called ...., (A) rectangle (B) parallelogram (C) trapezium, (D) rhombus, , 73

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(ii) If the diagonal of a square is 12 2 cm then the perimeter of square is ......, (A) 24 cm (B) 24 2 cm (C) 48 cm (D) 48 2 cm, (iii) If opposite angles of a rhombus are (2x)° and (3x - 40)° then value of x is ..., (A) 100, , ° (B) 80 ° (C) 160 ° (D) 40 °, , 2. Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal., 3. If diagonal of a square is 13 cm then find its side., 4. Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm., Find the length of its each side., 5. Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find, the length of side PQ., , ÐQMR = 50° then, , 6. Diagonals of a rectangle PQRS are intersecting in point M. If, find the measure of ÐMPS., 7., , In the adjacent Figure 5.42, if, , B, , then prove that,, seg BC, , Q, , || seg QR and seg BC @ seg QR., , C, , 8*. In the Figure 5.43,�ABCD is a trapezium., AB, , of seg AD and seg BC respectively., 1, PQ = (AB + DC)., 2, , 9., , In the adjacent figure 5.44,, trapezium. AB, , �ABCD is a, , respectively then prove that MN, , || AB., 74, , Q, , Fig. 5.43, , M, A, , C, , C, , D, , || DC. Points M and N are, , midpoints of diagonal AC and DB, , B, , P, D, , || AB and, , R, , Fig. 5.42, , A, , || DC. Points P and Q are midpoints, , Then prove that, PQ, , P, , A, , || seg PQ , seg AB @ seg PQ,, seg AC || seg PR, seg AC @ seg PR, seg AB, , N, , Fig. 5.44, , B

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Activity, To verify the different properties of quadrilaterals, Material : A piece of plywood measuring about 15 cm´ 10 cm, 15 thin screws,, twine, scissors., , Note : On the plywood sheet, fix five screws in a, •, •, •, •, •, horizontal row keeping a distance of 2cm between, any two adjacent screws. Similarly make two more, •, •, •, •, •, rows of screws exactly below the first one. Take care screw thread, that the vertical distance between any two adjacent, screws is also 2cm., Fig. 5.45, , With the help of the screws, make different types of, quadrilaterals of twine. Verify the properties of sides and angles of the quadrilaterals., Additional information, You know the property that the point of concurrence of medians of a triangle, divides the medians in the ratio 2 : 1. Proof of this property is given below., A, , Given : seg AD and seg BE are the medians of D ABC, which intersect at point G., , E, , To prove : AG : GD = 2 : 1, Construction :, , , Take point F on ray AD such that, G-D-F and GD = DF, , G, B, , D, , Proof : Diagonals of �BGCF bisect each other, .... given and construction, F, \ �BGCF is a parallelogram., Fig. 5.46, \ seg BE || set FC, Now point E is the midpoint of side AC of D AFC . .......... given, seg EB || seg FC, , , C, , Line passing through midpoint of one side and parallel to the other side, bisects the third side., , \ point G is the midpoint of side AF., \ AG = GF, But GF = 2GD ....... construction, \ AG = 2 GD, \, , AG 2, GD = 1 i.e. AG : GD = 2 : 1, 75, , qqq

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6, , Circle, Let’s study., , • Incircle, • Circumcircle, , • Circle, • Property of chord of the circle, , Let’s recall., , C, , B, P, , In adjoining figure, observe the circle with center P. With, reference to this figure, complete the following table., , A, , --- seg PA, ------chord, --diameter radius centre, , D, Fig. 6.1, , --central, angle, , ÐCPA, ---, , Let’s learn., , Circle, Let us describe this circle in terms of a set of points., l The set of points in a plane which are equidistant from a fixed point in the plane is called, a circle., Some terms related with a circle., l The fixed point is called the centre of the circle., l The segment joining the centre of the circle and a point on the circle is called a radius of, the circle., l The distance of a point on the circle from the centre of the circle is also called the radius, of the circle., l The segment joining any two points of the circle is called a chord of the circle., l A chord passing through the centre of a circle is called a diameter of the circle., A diameter is a largest chord of the circle., Circles in a plane, Congruent, Concentric, Circles intersecting, Circles intersecting, circles, circles, in a point, in two points, , · the same, , radii, , · the same centre, · different centres,, , different radii, Fig. 6.2, , different radii, only, one common point, , 76, , · different centres,, , different radii, two, common points

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Let’s learn., , Properties of chord, Activity I : Every student in the group will do this, activity. Draw a circle in your notebook. Draw any, chord of that circle. Draw perpendicular to the chord, through the centre of the circle. Measure the lengths, of the two parts of the chord. Group leader will prepare a table and other students will write their observations in it., Length, , Student, , l (AP), l (PB), , 1, , 2, , 3, , O, A, , P, , B, , Fig. 6.3, 4, , 5, , 6, , ...... cm, ...... cm, , Write theproperty which you have observed., Let us write the proof of this property., Theorem : A perpendicular drawn from the centre of a circle on its chord bisects the chord., Given, : seg AB is a chord of a circle with centre O., seg OP ^ chord AB, To prove : seg AP @ seg BP, O, : Draw seg OA and seg OB, Proof, In D OPA and D OPB, B, A, P, ÐOPA @ ÐOPB . . . . . . . . . . . seg OP ^ chord AB, seg OP @ seg OP . . . . . . . . . . . . common side, Fig. 6.4, hypotenuse OA @ hypotenuse OB . . . . . . . . . . . radii of the same circle, \ D OPA @ D OPB . . . . . . . . . hypotenuse side theorem, seg PA @ seg PB . . . . . . . . . . . . c.s.c.t., Activity II : Every student from the group will do, this activity. Draw a circle in your notebook., Draw a chord of the circle. Join the midpoint, of the chord and centre of the circle. Measure, the angles made by the segment with the chord., Discuss about the measures of the angles with, your friends., Which property do the observations suggest ?, , 77, , Fig. 6.5

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Theorem : The segment joining the centre of a circle and the midpoint of its chord is, perpendicular to the chord., Given, : seg AB is a chord of a circle with centre O and P is the midpoint of chord AB of, the circle. That means seg AP @ seg PB., To prove : seg OP ^ chord AB, Proof, : Draw seg OA and seg OB., O, In D AOP and D BOP, seg OA @ seg OB . . . . . . . . . . . . radii of the same circle, B, A, P, seg OP @ seg OP. . . . . . . . . . . . . common sides, Fig. 6.6, seg AP @ seg BP . . . . . . . . . . . . . given, \ D AOP @ D BOP . . . . . . . . . SSS test, , \ ÐOPA @ ÐOPB . . . . . . . . . . c.a.c.t. . . .(I), , , ÐOPA + ÐOPB = 180° . . . angles in linear pair, \ ÐOPB + ÐOPB = 180° . . . . . . from (I), , \ 2 ÐOPB = 180°, , , \ ÐOPB = 90°, , \ seg OP ^ chord AB, Now, , Solved examples, Ex (1) Radius of a circle is 5 cm. The length of a chord of the circle is 8 cm. Find the distance, of the chord from the centre., Let us draw a figure from the given information., Solution : , , O is the centre of the circle., O, , Length of the chord is 8 cm., seg, OM ^ chord PQ., P, Q, M, , Fig. 6.7, We know that a perpendicular drawn from the centre of a circle on its chord bisects, the chord., \ PM = MQ = 4 cm, Radius of the circle is 5 cm, means OQ = 5 cm .... given, In the right angled D OMQ using Pythagoras’ theorem,, OM2 + MQ2 = OQ2, \ OM2 + 42 = 52, \ OM2 = 52 - 42 = 25 - 16 = 9 = 32, \ OM = 3, Hence distance of the chord from the centre of the circle is 3 cm., , 78

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Ex (2) Radius of a circle is 20 cm. Distance of a chord from the centre of the circle is 12 cm., Find the length of the chord., Solution : Let the centre of the circle be O. Radius = OD = 20 cm., Distance of the chord CD from O is12 cm. seg OP ^ seg CD, , \ OP = 12 cm, Now CP = PD ...... perpendicular drawn from the, , centre bisects the chord, In the right angled D OPD, using Pythagoras’ theorem, OP2 + PD2 = OD2, (12)2 + PD2 = 202, , PD2 = 202 - 122, , PD2 = (20+12) (20-12), , O, C, , P, , D, , Fig. 6.8, , = 32 ´ 8 = 256, , , , \ PD = 16 , , \ CP = 16, , CD = CP + PD = 16 + 16 = 32, \ the length of the chord is 32 cm., , Practice set 6.1, 1., 2., 3., 4., 5., , 6., , Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is, 12 cm. Find the diameter of the circle., Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the, distance of the chord from the centre., Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find, the length of the chord., Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find, the distance of the chord from the centre of the circle., In figure 6.9, centre of two circles is O. Chord, O, AB of bigger circle intersects the smaller circle, in points P and Q. Show that AP = BQ, A P, Q B, Prove that, if a diameter of a circle bisects two, chords of the circle then those two chords are, Fig. 6.9, parallel to each other., , Activity I, (1), (2), (3), (4), , Draw circles of convenient radii., Draw two equal chords in each circle., Draw perpendicular to each chord from the centre., Measure the distance of each chord from the centre., , 79

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Let’s learn., Relation between congruent chords of a circle and their distances from the centre,    Activity II : Measure the lengths of the perpendiculars on chords in the following figures., N, , L, , A, , P, , O, , M, B, , M, , Figure (i) , , T, , Figure (ii) , , Figure (iii), , Did you find OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii) ?, Write the property which you have noticed from this activity., , Let’s learn., Properties of congruent chords, Theorem : Congruent chords of a circle are equidistant from the centre of the circle., : In a circle with centre O, Given, A, , chord AB @ chord CD, OP ^ AB, OQ ^ CD, To prove : OP = OQ, Construction : Join seg OA and seg OD., , P, , B, , O, , C, Q, , Fig. 6.10, D, 1, 1, : AP = AB, DQ = CD . . . perpendicular drawn from the centre, 2, 2, , Proof, , , of a circle to its chord bisects the chord., , AB = CD . . . . . . . . . . . . . . . . given, , \ AP = DQ, \ seg AP @ seg DQ . . . . . . . . . (I) . . . segments of equal lengths, , In right angled D APO and right angled D DQO, seg AP @ seg DQ . . . . . . . . . . . from (I), hypotenuse OA @ hypotenuse OD . . . . . . . . . . radii of the same circle, \ D APO @ D DQO . . . . . . . hypotenuse side theorem, , seg OP @ seg OQ . . . . . . . . . . . c.s.c.t., \ OP = OQ . . . . . . . . . . . . . Length of congruent segments., Congruent chords in a circle are equidistant from the centre of the circle., , 80

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Theorem : The chords of a circle equidistant from the centre of a circle are congruent., Given, : In a circle with centre O, P, B, A, seg OP ^ chord AB, seg OQ ^ chord CD, C, O, and OP = OQ, To prove : chord AB @ chord CD, Q, Construction : Draw seg OA and seg OD., Fig. 6.11, D, Proof, : (Complete the proof by filling in the gaps.), In right angled D OPA and right D OQD, hypotenuse OA @ hypotenuse OD . . . . . . ., seg OP @ seg OQ . . . . . . . . given, , \ D OPA @ D OQD . . . . . . ., , , \ seg AP @ seg QD . . . . . . . . c.s.c.t., , \ AP = QD . . . . . . . . . . . . . (I), , But AP =, , 1, 1, AB, and DQ = CD . . . . ., 2, 2, , and AP = QD . . . . . . . . . . from (I), \ AB = CD, , \ seg AB @ seg CD, , Note that both the theorems are converses of each other., , Remember this !, Congruent chords of a circle are equidistant from the centre of the circle., The chords equidistant from the centre of a circle are congruent., Activity : The above two theorems can be proved for two congruent circles also., 1. Congruent chords in congruent circles are equidistant from their respective centres., 2. Chords of congruent circles which are equidistant from their respective centres are, congruent., Write ‘Given’, ‘To prove’ and the proofs of these theorems., Solved example, Ex., In the figure 6.12, O is the centre of the circle and , AB = CD. If OP = 4 cm, find the length of OQ., Solution : O is the centre of the circle,, chord AB @ chord CD .....given, OP ^ AB, OQ ^ CD, , A, , B, , P, O, , C, , Q, , D, , Fig. 6.12, , 81

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OP = 4 cm, means distance of AB from the centre O is 4 cm., The congruent chords of a circle are equidistant from the centre of the circle., , \ OQ = 4 cm, , Practice set 6.2, 1., , Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the, distance of these chords from the centre of the circle ?, , 2., , In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre., Find the lengths of the chords., , 3., , Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC, is the bisector of ÐNPM., , Let’s recall., In previous standard we have verified the property that the angle bisectors of a triangle, are concurrent. We denote the point of concurrence by letter I., , Let’s learn., Incircle of a triangle, , A, , In fig. 6.13, bisectors of all angles of a DABC, meet in the point I. Perpendiculars on three sides, are drawn from the point of concurrence., , R, , P, , IP ^ AB,, , I, , B, , Q, , C, , IQ ^ BC,, , IR ^ AC, , We know that, every point on the angle, bisector is equidistant from the sides of the angle., , Fig. 6.13, Point I is on the bisector of ÐB., , \ IP = IQ., Point I is on the bisector of ÐC \ IQ = IR, \ IP = IQ= IR, , That is point I is equidistant from all the sides of DABC., \ if we draw a circle with centre I and radius IP, it will touch the sides AB, AC,, , BC of DABC internally., This circle is called the Incircle of the triangle ABC., , 82

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Let’s learn., To construct the incircle of a triangle, Ex. Construct D PQR such that PQ = 6 cm, ÐQ = 35°,, , R, , QR = 5.5 cm. Draw incircle of D PQR., , 5.5, , I, , Draw a rough figure and show all measures in it., (1) Construct D PQR of given measures., , 65°, , P, , (2) Draw bisectors of any two angles of the triangle., , cm, Q, , M 6 cm, , Rough fig. 6.14, , (3) Denote the point of intersection of angle, , R, , bisectors as I., , 5.5, , (4) Draw perpendicular IM from the point I, , cm, , I, , to the side PQ., (5) Draw a circle with centre I and radius IM., , P, , M, , 65°, , 6 cm, , Q, , Fig. 6.15, , Remember this !, The circle which touches all the sides of a triangle is called incircle of the triangle, and the centre of the circle is called the incentre of the triangle., , Let’s recall., In previous standards we have verified the property that perpendicular bisectors of, sides of a triangle are concurrent. That point of concurrence is denoted by the letter C., , Let’s learn., , P, , In fig. 6.16, the perpendicular bisectors of, sides of D PQR are intersecting at point, C. So C is the point of concurrence of, perpendicular bisectors., , C, Q, , R, Fig. 6.16, , 83

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Circumcircle, Point C is on the perpendicular bisectors of the sides of triangle PQR. Join PC, QC and, RC. We know that, every point on the perpendicular bisector is equidistant from the end points, of the segment., Point C is on the perpendicular bisector of seg PQ. \ PC = QC . . . . . . I, Point C is on the perpendicular bisector of seg QR., , \ QC = RC . . . . . . II, , \ PC = QC = RC . . . . . From I and II, \ the circle with centre C and radius PC will pass through all the vertices of D PQR., This circle is called as the circumcircle., , Remember this !, Circle passing through all the vertices of a triangle is called circumcircle of the triangle, and the centre of the circle is called the circumcentre of the triangle., , Let’s learn., To draw the circumcircle of a triangle, Ex. Construct D DEF such that DE = 4.2 cm, ÐD = 60°, ÐE = 70° and draw circumcircle, of it. Draw rough figure. Write the given measures., Rough figure, , F, F, D, , 60°, , C, , 70°, , E, , Fig. 6.17, C, 70°, , 60°, D, , Steps of construction :, (1) Draw D DEF of given measures., (2) Draw perpendicular bisectors of any two, sides of the triangle., (3) Name the point of intersection of, perpendicular bisectors as C., (4) Join seg CF., (5) Draw circle with centre C and radius CF., , 4.2 cm, , E, , Fig. 6.18, , 84

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Activity :, Draw different triangles of different measures and draw incircles and circumcircles of, them. Complete the table of observations and discuss., Type of triangle, , Equilateral triangle, , Isosceles triangles, , Scalene triangle, , Position of incenter, , Inside the triangle, , Inside the triangle, , Inside the triangle, , Position of, circumcentre, , Inside the triangle, , Inside, outside, on the triangle, , Type of triangle, , Acute angled, triangle, , Right angled, triangle, , Obtuse angled, triangle, , Position of incentre, Position of, circumcircle, , Midpoint of, hypotenuse, , Remember this !, , ·, , Incircle of a triangle touches all sides of, the triangle from inside., , ·, , Circumcentre of an acute angled, triangle lies inside the triangle., , ·, , For construction of incircle of a triangle, we have to draw bisectors of any two, angles of the triangle., , ·, , Circumcentre of a right angled triangle, is the midpoint of its hypotenuse., , ·, , Circumcentre of an obtuse angled, triangle lies in the exterior of the, triangle., , ·, , Circumcircle of a triangle passes, through all the vertices of a triangle., , ·, , For construciton of a circumcircle of a, triangle we have to draw perpendicular, bisectors of any two sides of the triangle., , • Incentre of any triangle lies in the, interior of the triangle., , Activity : Draw any equilateral triangle. Draw incircle and circumcircle of it. What did you, observe while doing this activity ?, (1) While drawing incircle and circumcircle, do the angle bisectors and perpendicular, bisectors coincide with each other ?, (2) Do the incentre and circumcenter coincide with each other ? If so, what can be the, reason of it ?, (3) Measure the radii of incircle and circumcircle and write their ratio., , 85

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Remember this !, , · The perpendicular bisectors and angle bisectors of an equilateral triangle are coincedent., · The incentre and the circumcentre of an equilateral triangle are coincedent., · Ratio of radius of circumcircle to the radius of incircle of an equilateral triangle is 2 : 1, Practice set 6.3, 1. Construct D ABC such that ÐB =100°, BC = 6.4 cm, ÐC = 50° and construct its, incircle., 2. Construct D PQR such that ÐP = 70°, ÐR = 50°, QR = 7.3 cm. and construct its, circumcircle., 3. Construct D XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its, incircle., 4. In D LMN, LM = 7.2 cm, ÐM = 105°, MN = 6.4 cm, then draw D LMN and construct, its circumcircle., 5. Construct D DEF such that DE = EF = 6 cm, ÐF = 45° and construct its circumcircle., , Problem set 6, 1., , Choose correct alternative answer and fill in the blanks., (i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence, the length of the chord is ........., (A) 16 cm, (B) 8 cm, (C) 12 cm (D) 32 cm, (ii) The point of concurrence of all angle bisectors of a triangle is called the ......, (A) centroid, (B) circumcentre, (C) incentre (D) orthocentre, (iii) The circle which passes through all the vertices of a triangle is called ....., , (A) circumcircle (B) incircle, (C) congruent circle (D) concentric circle, (iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5, cm, then the radius of that circle is ...., , (A) 12 cm, (B) 13 cm (C) 14 cm (D) 15 cm, (v), , The length of the longest chord of the circle with radius 2.9 cm is ....., (A) 3.5 cm, (B) 7 cm, (C) 10 cm (D) 5.8 cm, , (vi) Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P, will lie., , (A) on the centre (B) Inside the circle (C) outside the circle(D) on the circle, (vii) The lengths of parallel chords which are on opposite sides of the centre of a circle, are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these, chords is ....., (A) 2 cm , (B) 1 cm, (C) 8 cm, (D) 7 cm, , 86

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2. Construct incircle and circumcircle of an equilateral D DSP with side 7.5 cm. Measure, the radii of both the circles and find the ratio of radius of circumcircle to the radius of, incircle., 3., , Construct D NTS where NT = 5.7 cm, TS = 7.5 cm and, and circumcircle of it., , ÐNTS = 110° and draw incircle, Q, , 4., , R, , In the figure 6.19, C is the centre of the circle., seg QT is a diameter, CT = 13, CP = 5, find the length of chord RS., , S, , P, C, , Fig. 6.19, , T, 5., , ‍ , , In the figure 6.20, P is the centre of the circle., Chord AB and chord CD intersect, on the diameter at the point E., If ÐAEP @ ÐDEP, then prove that AB = CD., , B, , E, , C, , ° °, , D, , P, , A, Fig. 6.20, , C, 6., , In the figure 6.21, CD is a diameter, of the circle with centre O. Diameter, CD is perpendicular to chord AB at, point E. Show that D ABC is an, isosceles triangle., , O, A, , E, D, , B, Fig. 6.21, , ICT Tools or Links, Draw different circles with Geogebra software. Verify and experience the, properties of chords. Draw circumcircle and incircle of different triangles., Using ‘Move Option’ experience how the incentre and circumcentre changes if, the size of a triangle is changed., , qqq, 87

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7, , Co-ordinate Geometry, Let’s, study., F, , • Axis, Origin, Quadrant, • Co-ordinates of a point in a plane., • To plot a point., , • Line parallel to X-axis., • Line parallel to Y-axis., • Equation of a line., , Chintu and his friends were playing, cricket on the ground in front of a big, building, when a visitor arrived., Visitor : Hey Chintu, Dattabhau lives, here, doesn’t he ?, Chintu : Yes, on the second floor. See, that window ? Thats his flat., Visitor : But there are five windows on, the second floor. It could be, any of them !, Chintu ः His window is the third one, from the left, on the second, floor., Chintu’s description of the location of Dattabhau’s flat is in fact, based on the most basic, concept in Co-ordinate Geometry., It did not suffice to give only the floor number to locate the house. Its serial number from, the left or from the right also needed to be given. That is two numbers had to be given in a, specific sequence. Two ordinal numbers namely, second from the ground and third from, the left had to be used., Let’s learn., Axis, origin, quadrants, We could give the location of Dattabhau’s house using two ordinal numbers. Similarly, the, location of a point can be fully described using its distances from two mutually perpendicular, lines., To locate a point in a plane, a horizontal number line is drawn in the plane. This number, line is called the X-axis., , 88

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Rene Descartes (1596-1650), Rene Descartes, a French mathematician of the 17th Century,, proposed the co-ordinate system to describe the position of a, point in a plane accurately. It is called the Cartesian co-ordinate, system. Obviously the word Cartesian is derived from his name., He brought about a revolution in the field of mathematics by, establishing the relationship between Algebra and Geometry., The Cartesian co-ordinate system is the foundation, of Analytical Geometry. La Geometric was Descartes’, first book on mathematics. In it, he used algebra for the, study of geometry and proposed that a point in a plane, can be represented by an ordered pair of real numbers. This ordered pair is the ‘Cartesian, Co-ordinates’ of a point., Co-ordinate geometry has used in a variety of fields such as Physics, Engineering,, Nautical Science, Siesmology and Art. It plays an important role in the development of, technology in Geogebra. We see the inter-relationship between Algebra and Geometry, quite clearly in the software Geogebra;the very name being a combination of the words, ‘Geometry’ and ‘Algebra’., Another number line intersecting the, X-axis at point marked O and perpendicular, to the X-axis, is the Y-axis. Generally,, the number O is represented by the same, point on both the number lines. This point, is called the origin and is shown by the, letter O., On the X-axis, positive numbers are, shown on the right of O and negative numbers, on the left., On the Y-axis, positive numbers are shown, above O and negative numbers below it., The X and Y axes divide the plane into, four parts, each of which is called a Quadrant., As shown in the figure, the quadrants are, numbered in the anti-clocksise direction., The points on the axes are not included in, the quadrants., , 89, , Y -axis, , 3, Second quadrant, II, , First quadrant, I, , 2, 1, , -3, , -2, , -1, , Third quadrant, III, , 0, -1, -2, -3, , O, , 1, , 2, , 3 X-axis, , Fourth quadrant, IV, , Fig.7.1

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The Co-ordinates of a point in a plane, Y, 4, Q (-3, 2), , X -4, ¢, , -3 -2 -1, , P (2, 3), , 2 R, 1, , S, , N, , 3, , The point P is shown in the plane, determined by the X-axis and the Y-axis., Its position can be determined by its, distance from the two axes. To find these, distances, we draw seg PM ^ X-axis and, seg PN ^ Y-axis., Co-ordinate of point M on X-axis is 2, and co-ordinate of point N on Y-axis is 3., Therefore x co-ordinate of point P is 2, and y co-ordinate of point P is 3., The, convention, for, describing, the position of a point is to mention, , -1, , O, M, 0 1 2, , 3, , 4 X, , -2, -3, -4, Y¢, , Fig.7.2, x co-ordinate first. According to this convention the order of co-ordinates of point P is, decided as 2, 3. The position of the point P in brief, is described by the pair (2, 3), The order of the numbers in the pair (2, 3) is important. Such a pair of numbers is, called an ordered pair., To describe the position of point Q, we draw seg QS ^ X-axis and seg QR ^ Y-axis., The co-ordinate of point Q on the X-axis is -3 and the co-ordinate on the Y-axis is 2., Hence the co-ordinates of point Q are (-3, 2)., Y, , Ex. Write the co-ordinates of points E, F, G, T in, the figure alongside., Solution :, , · The co-ordinates of point E are (2, 1), · The co-ordinates of point F are (-3, 3), · The co-ordinates of point G are (-4, -2)., · The co-ordinates of point T are (3, -1), , 4, F·, , 3, 2, E, , 1, -4, G·, , -3, , -2, , -1, , -1, , X, 0 1, , -2, -3, -4, , Fig.7.3, , 90, , 2, , 3, T, , 4

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Let’s learn., Co-ordinates of points on the axes, Y, , (0,-3), , The x co-ordinate of point M is its, distance from the Y-axis. The distance of, point M from the X-axis is zero. Hence,, the y co-ordinate of M is 0., Thus, the co-ordinates of point M on, the X-axis are (3,0)., The y co-ordinate of point N on the, Y-axis is 4 units from the X-axis because, N is at a distance of 4. Its x co-ordinate, is 0 because its distance from the Y-axis, is zero., Hence, the co-ordinates of point, , (0,-4), , N on the Y-axis are (0, 4)., , (0,4), , ·P, , N, , (0,3), , Quadrant II, (-,+), , Quadrant I, (+,+), , (0,2), (0,1), , O, , (-4,0) (-3,0) (-2,0) (-1,0) (0,0) (1,0), , M, , (2,0) (3,0) (4,0), , X, , (0,-1), , Quadrant III, (-,-), , (0,-2), , Quadrant IV, (+,-), , Fig.7.4, Now the origin ‘O’ is on X-axis as well as on Y-axis. Hence, its distance from X-axis and, Y-axis is zero. Therefore, the co-ordinates of O are (0, 0)., One and only one pair of co-ordinates (ordered pair) is associated with every point in, a plane., Let’s Remember, , · The y co-ordinate of every point on the X-axis is zero., · The x co-ordinate of every point on the Y-axis is zero., · The coordinates of the origin are (0, 0)., In which quadrant or on which axis are the points given below ?, A(5,7), B(-6,4), C(4,-7), D(-8,-9), P(-3,0), Q(0,8), Solution : The x co-ordinate of A (5, 7) is positive and its y co-ordinate is positive.., \ point A is in the first quadrant., The x co-ordinate of B (-6, 4) is negative and y co-ordinate is positive., \ point B is in the second quadrant., ., The x co-ordinate of C (4, -7) is positive and y co-ordinate is negative., \ point C is in the fourth quadrant., The x co-ordinate of D (-8, -7) is negative and y co-ordinate is negative., \ point D is in the third quadrant., Ex., , 91

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The y co-ordinate of P(-3,0) is zero \ point P is on the X-axis., The x co-ordinate of Q (0,8) is zero \ point Q is on the Y-axis., Activity, , As shown in fig. 7.5, ask girls to sit in lines so as to form the X-axis and, Y-axis., , • Ask some boys to sit at the positions, •, , •, , marked by the coloured dots in the four, quadrants., Now, call the students turn by turn using, the initial letter of each student’s name., As his or her initial is called, the student, stands and gives his or her own, co-ordinates. For example Rajendra, (2, 2) and Kirti (-1, 0), Even as they have fun during this field, activity, the students will learn how to, , R, K, , Fig. 7.5, , state the position of a point in a plane., , Let’s learn., To plot the points of given co-ordinates, Suppose we have to plot the points, , Y, , P (4,3) and Q (-2,2), 4, , Steps for plotting the points, (i), , Draw X-axis and Y-axis on the plane., Show the origin., , Q·(-2,2), , (ii) To find the point P (4,3), draw a line, parallel to the Y-axis through the point, on X axis which represents the number 4., Through the point on Y-axis which, represents the number 3 draw a line parallel to, the X-axis ., , P (4,3), , 3, , ·, , 2, 1, , -2, , -1, , -1, , 0, , -2, -3, , Fig. 7.6, , 92, , 1, , 2, , 3, , 4, , X

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(iii), , The point of intersection of these two lines parallel to the Y and X-axis respectively, is, the point P (4,3). In which quadrant does this point lie ? , , (iv), , In the same way, plot the point Q (-2, 2) . Is this point in the second quadrant ?, Using the same method, plot the points, , Ex., , R(-3, -4), S(3, -1), , In which quadrants or on which axis are the points given below ?, (i) (5, 3), , (ii) (-2, 4), , (iii) (2, -5), , (iv) (0, 4), , (v) (-3, 0), , (vi) (-2, 2.5), , (vii) (5, 3.5), , (viii) (-3.5, 1.5), , (x) (2, -4 ), , (ix) (0, -4), Solution :, , Quadrant / axis, , (i), , coordinates, (5,3), , Quadrant I, , (ii), , (-2,4), , (iii), , co-ordinates, , Quadrant / axis, , (vi), , (-2, -2.5), , Quadrant III, , Quadrant II, , (vii), , (5,3.5), , Quadrant I, , (2,-5), , Quadrant IV, , (viii), , (-3.5,1.5), , Quadrant II, , (iv), , (0,4), , Y-axis, , (ix), , (0, -4), , Y-axis, , (v), , (-3,0), , X-axis, , (x), , (2,-4 ), , Quadrant IV, , Practice set 7.1, 1., , State in which quadrant or on which axis do the following points lie., , · A(-3, 2),, · E(37, 35),, · M(12, 0),, 2., , · B(-5, -2), · K(3.5, 1.5), · D(2, 10),, · F(15, -18), · G(3, -7),, · H(0, -5),, · N(0, 9),, · P(0, 2.5),, · Q(-7, -3), , In which quadrant are the following points ?, (i), , whose both co-ordinates are positive., , (ii) whose both co-ordinates are negative., (iii) whose x co-ordinate is positive, and the y co-ordinate is negative., (iv) whose x co-ordinate is negative and y co-ordinate is positive., 3., , Draw the co-ordinate system on a plane and plot the following points., L(-2, 4),, , M(5, 6),, , N(-3, -4),, , P(2, -3),, , 93, , Q(6, -5),, , S(7, 0), T(0, -5)

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Let’s learn., Lines parallel to the X-axis, , · On a graph paper, plot the following points, A (5, 4), B (2, 4), C (-2, 4), D (-4, 4), E (0, 4), F (3, 4), , · Observe the co-ordinates of the, , Y, , given points., , · Did, , you, notice, that, the, y co-ordinates of all the points are, equal ?, , D, , 5, , C, , 4, , A, , ·, , 3, , · All the points are collinear., · To which axis is this line parallel ?, · The, y co-ordinate of every, , point on the line DA is 4. It is, constant. Therefore the line DA is, described by the equation y = 4. If, the y co-ordinate of any point is 4,, will be on the line DA., , F, , B, , E, , 2, 1, -4, , -3, , -2, , 0, , -1, -1, , 1, , 2, , 3, , 4, , 5, , X, , -2, -3, -4, , The equation of the line parallel to, the X axis at a distance of 4 units, , Fig. 7.7, , from the X-axis is y = 4., , Let’s discuss., , · Can we draw a line parallel to the X-axis at a distance of 6 units from it and below the, X-axis ?, , 1, , · Will all of the points (-3, -6), (10, -6), ( 2 , -6) be on that line ?, · What would be the equation of this line ?, , Remember this !, If b > 0, and we draw the line y = b through the point (0, b), it will be above the X-axis, and parallel, to it. If b < 0, then the line y = b will be below the X-axis and parallel to it., The equation of a line parallel to the X-axis is in the form y = b., , 94

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Let’s learn., Lines parallel to the Y-axis, , · On a graph paper, plot the following points., P(-4, 3), Q(-4, 0),, , R(-4, 1),, , S(-4, -2), T(-4, 2),, , U(-4, -3), , · Observe the co-ordinates of the points., · Did you notice that the x co-ordinate of, ·, ·, ·, , all the points are the same ?, Are all the points collinear ?, To which axis is this line parallel ?, The x co-ordinate of every point on, the line PS is - 4. It is constant. Therefore, the line PS can be described by the, equation x = - 4. Every point whose, x co-ordinate is - 4 lies on the line PS., The equation of the line parallel to the, Y-axis at a distance of 4 units and to the, , Y, P, , 3, , T, , 2, , R, Q, , 1, -4, , -3, , -2, , -1, -1, , S, , -2, , U, , -3, -4, , left of Y-axis is x = - 4., , 0, , 1, , 2, , 3, , X, , Fig. 7.8, , Let’s discuss., , · Can we draw a line parallel to the Y-axis at a distance of 2 units from it and to its right ?, 1, · Will all of the points (2,10), (2,8), (2, - 2 ) be on that line ?, · What would be the equation of this line ?, Remember this !, If we draw the line x = a parallel to the Y-axis passing through the point (a, 0) and, if a > 0 then the line will be to the right of the Y-axis. If a < 0, then the line will be to the, left of the Y-axis., The equation of a line parallel to the Y-axis is in the form x = a., , 95

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Remember this !, (1) The y co-ordinate of every point on the X-axis is zero. Conversely, every point whose, y co-ordinate is zero is on the X-axis. Therefore, the equation of the X axis is y = 0., (2) The x co-ordinate of every point on the Y-axis is zero. Conversely, every point whose, x co-ordinate is zero is on the Y-axis. Therefore, the equation of the Y-axis is x = 0., , Let’s learn., Graph of a linear equations, , Y, , Ex., , 3, 2, , x=2, , Draw the graphs of the equations, x = 2 and y = -3., Solution : (i) On a graph paper draw the X-axis and, the Y-axis., (ii) Since it is given that x = 2, draw a line on the, right of the Y-axis at a distance of 2 units, from it and parallel to it., (iii) Since it is given that y = -3, draw a line, below the X-axis at a distance of 3 units, from it and parallel to it., (iv) These lines, parallel to the two axes, are the, graphs of the given equations., (v) Write the co-ordinates of the point P, the, point of intersection of these two lines., (vi) Verify that the co-ordinates of the point P, are (2, -3)., , 1, -3, , -2, , 0, , -1, , 2, , 1, , 3, , -1, -2, , y = -3, , P, , -3, , Fig. 7.9, , The graph of a linear equation in the general form., Activity : On a graph paper, plot the, points (0,1) (1,3) (2,5). Are they, collinear ? If so, draw the line that, passes through them., · Through which quadrants does this, line pass ?, · Write the co-ordinates of the point at, which it intersects the Y-axis., · Show any point in the third quadrant, which lies on this line. Write the, co-ordinates of the point., , 96, , Y, 5, , (2, 5), , 4, 3, , (1, 3), , 2, 1 (0, 1), -3 -2 -1, , -1, , 0 1, , 2, , 3, , X, , -2, -3, -4, -5, , Fig. 7.10, , X

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Ex. 2x - y + 1 = 0 is a linear equation in two variables in general form. Let us draw the graph, of this equation., Solution : 2x - y + 1 = 0 means y = 2x + 1, Let us assume some values of x and find the corresponding values of y., For example, If x = 0, then substituting this value of x in the equation we get y = 1., Similarly, let us find the values of y when 0, 1, 2, 1 , -2 are some values of x and, 2, , write these values in the table below in the form of ordered pairs., x, , 0, , 1, , 2, , 1, 2, , -2, , y, , 1, , 3, , 5, , 2, , -3, , (x, y), , (0,1), , (1,3), , (2,5), , ( 1 , 2), , ( -2, -3), , 2, , Now, let us plot these points. Let us verify that these points are collinear. Let us draw that, line. The line is the graph of the equaiton 2x - y + 1 = 0., , ICT Tools or Links, , , Use the Software Geogebra to draw the X and Y-axis. Plot several points., , Find and study the co-ordinates of the points in ‘Algebraic view’. Read the equations of, lines that are parallel to the axes. Use the ‘move’ option to vary the positions of the lines., What are the equations of the X-axis and the Y-axis ?, , Practice set 7.2, 1. , On a graph paper plot the points A (3,0), B(3,3), C(0,3). Join A, B and B, C. What is the, figure formed ?, 2. Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its, left., 3. Write the equation of the line parallel to the X-axis at a distance of 5 units from it and, below the X-axis., 4. The point Q( -3, -2) lies on a line parallel to the Y-axis. Write the equation of the line, and draw its graph., , 97

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5. X-axis and line x = -4 are parallel lines. What is the distance between them?, 6. Which of the equations given below have graphs parallel to the X-axis, and which, ones have graphs parallel to the Y-axis ?, (i) x = 3 , , (ii) y - 2 = 0, , (iii) x + 6 = 0, , (iv) y = -5, , 7. On a graph paper, plot the points A(2, 3), B(6, -1) and C(0, 5). If those points are, collinear then draw the line which includes them. Write the co-ordinates of the points at, which the line intersects the X-axis and the Y-axis., 8. Draw the graphs of the following equations on the same system of co-ordinates. Write, the co-ordinates of their points of intersection., x + 4 = 0,, , y - 1 =0,, , 2x + 3 = 0,, , 3y - 15 =0, , 9. Draw the graphs of the equations given below., (i) x + y = 2, , (ii) 3x - y = 0, , (iii) 2x + y = 1, Problem set 7, , 1., , Choose the correct alternative answer for the following quesitons., , (i) What is the form of co-ordinates of a point on the X-axis ?, , (A) (b, b) (B) (o, b) (C) (a, o) (D) (a, a), (ii) Any point on the line y = x is of the form ....., , (A) (a, a) (B) (o, a) (C) (a, o) (D) (a, - a), (iii) What is the equation of the X-axis ?, (B) y = 0, (C) x + y = 0 (D) x = y, , (A) x = 0, (iv) In which quadrant does the point (-4, -3) lie ?, , (A) First (B) Second (C) Third (D) Fourth, (v) What is the nature of the line which includes the points (-5,5), (6,5), (-3,5), (0,5) ?, , , (A) Passes through the origin, , (B) Parallel to Y-axis, , , , (C) Parallel to X-axis , , (D) None of these, , (vi) Which of the points P (-1,1), Q (3,-4), R(1,-1), S (-2,-3), T (-4,4) lie in the fourth, quadrant ?, , , (A) P and T, , (B) Q and R, , (C) only S, , 98, , (D) P and R

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2. Some points are shown in the figure 7.11., , Y, 3, , With the help of it answer the following questions :, (i), , 2, , Q, , Write the co-ordinates of the points, , P, , 1, , Q and R., -4, , (ii) Write the co-ordinates of the points, , -3 -2 -1, , T and M., , (iv) Which are the points whose x and, , 2, , M, 3, , 4, , X, R, , -2, , S, , (iii) Which point lies in the third quadrant ?, , 0 1, -1 T, -3, , Fig.7.11, , y co-ordinates are equal ?, , 3. Without plotting the points on a graph, state in which quadrant or on which axis do the, following point lie., (i) (5, -3), , (ii) (-7, -12) , , (iii) (-23, 4), , (iv) (-9, 5), , (v) (0, -3) , , (vi) (-6, 0), , Y, , 4. Plot the following points on the one and the, same co-ordinate system., A(1, 3), B(-3, -1), C(1, -4),, D(-2, 3), E(0, -8), F(1, 0), , 3, , L, , 2, , P, , 1, , 5. In the graph alongside, line LM is parallel, , to the Y-axis. (Fig. 7.12), (i) What is the distance of line LM from, the Y-axis ?, (ii) Write the co-ordinates of the points, P, Q and R., (iii) What is the difference between the, , -4, , -3 -2 -1, , -1, , 0 1, , 2 3, , R, , X, , Q, , -2, -3, , M, , -4, , Fig.7.12, , x co-ordinates of the points L and M?, , 6. How many lines are there which are parallel to X-axis and having a distance 5 units?, 7*. If ‘a’ is a real number, what is the distance between the Y-axis and the line x = a ?, qqq, 99

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8, , Trigonometry, , Let’s study., , • Introduction of Trigonometry • Relations among Trigonometric Ratios, • Trigonometic Ratios, • Trigonometric Ratios of Particular Angles, Introduction to Trigonometry, How far is this ship, from the seashore ?, , What will be, the height of, this tree?, How to, measure it?, , We can measure distances by using a rope or by walking on ground, but how to measure, the distance between a ship and a light house? How to measure the height of a tall tree?, Observe the above pictures. Questions in the pictures are related to mathematics., Trigonometry, a branch of mathematics, is useful to find answers to such questions., Trigonometry is used in different branches of Engineering, Astronomy, Navigation etc., The word Trigonometry is derived from three Greek words ‘Tri’ means three, ‘gona’, means sides and ‘metron’ means measurements., , Let’s recall., We have studied triangle. The subject trigonometry starts with right angled triangle,, theorem of Pythagoras and similar triangles, so we will recall these topics., , · In D ABC, ÐB is a right angle and side, , A, , AC opposite to ÐB, is hypotenuse., Side opposite to ÐA is BC and side, opposite to ÐC is AB., Using Pythagoras’ theorem, we can, write the following statement for this, triangle., (AB)2 + (BC)2 = (AC)2, , B, , 100, , Fig. 8.1, , C

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A, , · If DABC ~ DPQR then their corresponding, sides are in the same proportions., AB, PQ, , So =, , P, , BC AC, =, QR PR, , C Q, , B, , R, , Fig. 8.2, Let us see how to find the height of a tall tree using properties of similar triangles., Activity : This experiment can be conducted on a clear sunny day., Look at the figure given alongside., Height of the tree is QR, height of the stick is BC., Thrust a stick in the ground as shown, in the figure. Measure its height and length, of its shadow. Also measure the length of, the shadow of the tree. Rays of sunlight, are parallel. So D PQR and D ABC are, equiangular, means similar triangles. Sides of, similar triangles are proportional., QR BC, ., =, PR AC, , Therefore, we get an equation,, height of the tree = QR =, , B, stick, , So we get, , Q, , R, , P, , BC, × PR, AC, , A, , C, , Fig.8.3, , We know the values of PR, BC and AC. Substituting these values in this equation,, we get length of QR, means height of the tree., , Use your brain power !, It is convenient to do this experiment between 11:30 am and 1:30 pm, instead of doing it in the morning at 8’O clock. Can you tell why ?, Activity : You can conduct this activity and find the height of a tall tree in your, surrounding. If there is no tree in the premises then find the height of a pole., Lamp, post, , stick, , Fig. 8.4, , 101

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Let’s learn., Terms related to right angled triangle, In right angled D ABC, ÐB = 90°, Ð A and Ð C are acute angles., C, , A, , Opposite side, , Hypotenuse, , of ÐA, , Adajcent side, of ÐC, , B, ¯, Adjacent side of Ð A, , A, , Fig. 8.5, Ex., , ¯, , ¯, , Hypotenuse, , C, , ¯, B, Opposite side of ÐC, , Fig. 8.6, , In the figure 8.7, D PQR is a right angled triangle. WriteP, , side opposite to Ð P =....., side opposite to Ð R =....., side adjacent to Ð P = ....., side adjacent to Ð R = ....., , Q, , Fig. 8.7, , R, , Trigonometic ratios, In the adjacent Fig.8.8 some right angled, triangles are shown. ÐB is their common, angle. So all right angled triangles are, similar., D PQB ~ D ACB, , \, , PB PQ BQ, = =, AB AC BC, , \, , PQ PB, =, AC AB, QB PB, =, BC AB, , \, \, , E, A, P, , PQ AC, =, ... alternando, PB AB, QB BC, =, ... alternando, PB AB, , 102, , B, , Q, , C, , Fig. 8.8, , F

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The figures of triangles in 8.9 and.8.10 are of the triangles separated from the figure 8.8., A, , Hypotenuse, , Hypotenuse, , ¯, , Opposite side, of ÐB, , B, , B, Q, ¯, Adjacent side of Ð B, , C, ¯, Adjacent side of Ð B, , In D ACB,, , In D PQB,, , Opposite side of ÐB, AC, =, Hypotenuse, AB, , PQ, Opposite side of ÐB, =, Hypotenuse, PB, , The ratios, , \, , of ÐB, , Fig.8.10, , Fig.8.9, (i), , Opposite side, , ¯, , P, , PQ, AC, and, are equal., PB, AB, , Opposite side of ÐB, PQ, AC, =, =, , Hypotenuse, PB, AB, , This ratio is called the ‘sine’ ratio of Ð B, and is written in brief as ‘sin B’., (ii) In D PQB and D ACB,, Adjacent side of ÐB, BC, BQ Adjacent side of ÐB, =, and, =, Hypotenuse, Hypotenuse, AB, PB, , \, , BQ, BC, Adjacent side of ÐB, =, =, Hypotenuse, PB, AB, , This ratio is called as the ‘cosine’ ratio of Ð B , and written in brief as ‘cos B’, PQ, BQ, , (iii) =, , AC, =, BC, , Opposite side of ÐB, Adjacent side of ÐB, , This ratio is called as the tangent ratio of Ð B, and written in brief as tan B., Ex. :, A, , q, B, , Fig.8.11, , C, , Sometimes we write measures of acute angles, of a right angled triangle by using Greek letters, q (Theta), a (Alpha), b (Beta) etc., In the adjacent figure of D ABC, measure of acute, angle C is denoted by the letter q. So we can write, the ratios sin C, cos C, tan C as sin q, cos q, tan q, respectively., , 103

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AB, ,, AC, , sin C = sin q =, , cos C = cos q =, , BC, ,, AC, , tan C = tan q =, , AB, BC, , Remember this !, , sin ratio =, , opposite side, hypotenuse, , ·, , sin q =, , opposite side of Ðq, hypotenuse, , · , , cos ratio =, , adjacent side, hypotenuse, , · , , cos q =, , adjacent side of Ðq, hypotenuse, , · , , tan ratio =, , opposite side, adjcent side, , · , , tan q =, , opposite side of Ðq, opposite side of Ðq, , ·, , Practice set 8.1, 1., , Fig. 8.12, 2., , In the Fig.8.12, ÐR is the right angle of, D PQR. Write the following ratios., (i) sin P (ii) cos Q (iii) tan P (iv) tan Q, , R, , P, , b, , Y, , Q, , In the right angled D XYZ, ÐXYZ = 90° and, a,b,c are the lengths of the sides as shown in, the figure. Write the following ratios,, (i) sin X (ii) tan Z (iii) cos X (iv) tan X., , X, c, , a, Z, , Fig. 8.13, , 3., , In right angled D LMN, ÐLMN = 90°, ÐL = 50° and ÐN = 40°,, write the following ratios., (i) sin 50° , (ii) cos 50°, (iii) tan 40° , (iv) cos 40°, , L, 50°, , M, , 40°, , Fig. 8.14, , N, , P, , 4., , q, a, , R, , Q, , Fig. 8.15, , S, , In the figure 8.15,.ÐPQR = 90°,, ÐPQS = 90°, ÐPRQ = a and ÐQPS = q, Write the following trigonometric ratios., (i) sin a, cos a, tan a, (ii) sin q, cos q, tan q, , 104

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Let’s learn., Relation among trigonometric ratios, P, , In the Fig.8.16, , D PMN is a right angled triangle., Ð M = 90°, ÐP and Ð N are complimentary, , (90- q)°, , angles., , \ If Ð N = q then Ð P = 90 - q, , q, , N, , M, , Fig.8.16, , sin q =, , PM, ........(1), PN, , sin (90 - q) =, , NM, ........(4), PN, , cos q =, , NM, .......(2), PN, , cos (90 - q) =, , PM, .......(5), PN, , tan q =, , PM, ........(3), NM, , tan (90 - q) =, , NM, ........(6), PM, , \ sin q = cos (90 - q) ........ from (1) and (5), cos q = sin (90 - q) ........ from (2) and (4), Also note that tan q, , , Similarly,, , ´ tan (90 - q) =, , PM, NM, , ´, , NM, ........ from (3) and (6), PM, , \ tan q ´ tan (90 - q) = 1, sin q, =, cos q, , PM, PN, NM, PN, , =, , PM, PN, , ´, , PN, NM, , =, , PM, = tan q, NM, , Remember this !, cos (90 - q) = sin q,, sin q, = tan q,, cos q, , sin (90 - q) = cos q, tan q, , 105, , ´ tan (90 - q) = 1

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* For more information, 1, 1, sin q = cosec q, cos q = sec q,, , 1, = cot q, tan q, , It means cosec q, sec q and cot q are inverse ratios of sin q,, cos q and tan q respectively., · sec q = cosec (90 - q) · cosec q = sec (90 - q), , · tan q = cot (90 - q), , · cot q = tan (90 - q), , Let’s recall., Theorem of 30°- 60°-90° triangle :, , We know that if the measures of angles of a triangle are 30°,60°, 90° then side opposite to 30°, angle is half of the hypotenuse and side opposite to 60° angle is, , 3, of hypotenuse., 2, , A, In the Fig. 8.17, D ABC is a right angled, , 60°, , triangle. ÐC = 30°, ÐA = 60°, ÐB = 90° ., C, , \ AB =, , B, , 30°, , 1, AC and BC = 3 AC, 2, 2, , Fig. 8.17, , Let’s learn., Trignometric ratios of 30° and 60° angles, P, , In right angled, 60°, , a, Q, , D PQR if ÐR = 30°,, , ÐP = 60°, ÐQ = 90° and PQ = a, , 2a, , 1, PR, 2, 1, a = PR, 2, , then PQ =, 30°, , 3a, , R, , Fig. 8.18, , \ PR = 2a, , \ If PQ = a, then PR = 2a and QR =, 106, , QR =, , 3, PR, 2, , QR =, , 3, 2, , ´ 2a, , QR = 3 a, 3a

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(II) Trigonometric ratios of 60° angle, , (I) Trigonometric ratios of the 30° angle, PQ, PR, , a 1, =, 2a 2, , sin 30° = =, , sin 60° =, , QR, =, PR, , QR, PR, , 3a, =, 2a, , 3, 2, , cos 60° =, , PQ, =, PR, , PQ, QR, , a, =, 3a, , 1, 3, , tan 60° =, , QR, =, PQ, , cos 30° = =, tan 30° = =, , 3a, 2a, , =, , 3, 2, , a, 1, =, 2a, 2, 3a, , a, , = 3, , In right angled D PQR, ÐQ = 90°. Therefore ÐP and ÐR are complimentary angles of each, other. Verify the relation between sine and cosine ratios of complimentary angles here also., sin q = cos (90 - q), sin 30° = cos (90°- 30°) = cos 60°, sin 30° = cos 60°, , cos q = sin (90 - q), cos 30° = sin (90°- 30°) = sin 60°, cos 30°= sin 60°, , Remember this !, , sin 30° =, sin 60° =, , 1, 2, , cos 30° =, , 3, 2, , cos 60° =, , 1, 3, , 3, 2, , tan 30° =, , 1, 2, , tan 60° = 3, , (III) Trigonometric ratios of the 45°angle, A, 45°, , 2a, , a, , B, , 45°, , a, Fig.8.19, , C, , In right angled D ABC, ÐB= 90°, ÐA =45°,, ÐC = 45° \ This is an isosceles triangle., Suppose AB = a then BC = a., Using Pythagoras’ theorem,, let us find the length of AC., AC2 = AB2 + BC2, = a2 + a2, AC2 = 2a2, \ AC = 2 a, , 107

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In the Fig. 8.19, ÐC = 45°, sin 45° =, cos 45° =, , AB, =, AC, BC, =, AC, , a, 2a, , a, 2a, , =, , =, , 1, , 2, , tan 45° =, , AB, =, BC, , a, a, , =1, , 1, 2, , Remember this !, sin 45° =, , 1, , , 2, , cos 45° =, , 1, , , 2, , tan 45° = 1, , (IV) Trigonometric ratios of the angle 0° and 90°, A, A, A, B, , B, , C, , C, , B, , C, , Fig.8.20, AC, . Keeping, AB, , In the right angled D ACB, ÐC = 90° and ÐB = 30°. We know that sin 30°=, , the length of side AB constant, if the measure of ÐB goes on decreasing the length of AC,, , which is opposite to ÐB also goes on decreasing. So as the measure of ÐB decreases, then, value of sin q also decreases., , \ when measure of ÐB becomes 0°, then length of AC becomes 0., , \ sin 0° =, , AC, 0, AB = AB = 0 \ sin 0° = 0, A, , A, , A, , B, , 30°, , C, , B, , 70°, , Fig.8.21, , 108, , C, , B, , 85°, C

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Now look at the Fig. 8.21. In this right angled triangle, as the measure of ÐB increases the, length of AC also increases. When measure of ÐB becomes 90°, the length of AC become, equal to AB., , \ sin 90° = AC , , \ sin 90° = 1, AB, We know the relations between trigonometric ratios of complimentary angles., sin q = cos (90 - q), , , , and cos q = sin (90 - q), , , \ cos 0° = sin (90 - 0)° = sin 90° = 1, and cos 90° = sin (90 - 90)° = sin 0°= 0, , Remember this !, sin 0° = 0, , , sin 90° = 1, , , cos 0° = 1, , , cos 90° = 0, , We know that,, sin q, sin 0, 0, tan q =, , \, tan 0 = cos 0 = = 0, cos q, 1, sin 90°, 1, But tan 90° = cos 90° =, 0, But we can not do the division of 1 by 0. Note that q is an acute angle. As it increases, and reaches the value of 90°, tan q also increases indefinitely. Hence we can not find the, definite value of tan 90., , Remember this !, Trigonometric ratios of particular ratios., Ratios, , Measures, of angles, , 0°, , 30°, , 45°, , 60°, , 90°, , sin, , 0, , 1, 2, , 1, 2, , 3, 2, , 1, , cos, , 1, , 3, 2, , 1, 2, , 1, 2, , 0, , tan, , 0, , 1, 3, , 1, , 109, , 3, , Undefined

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Solved Examples :, Ex. (1) Find the value of 2tan 45° + cos 30° - sin 60°, 2tan 45° + cos 30° - sin 60°, , Solution :, , = 2 ´ 1, , +, , 3, 2, , -, , 3, 2, , =2+0, = 2, Ex. (2) Find the value of, , cos 56°, sin 34°, , 56° + 34° = 90° means 56 and 34 are the measures of complimentary angles., , Solution :, , sin q = cos (90- q), , \ sin 34° = cos (90- 34)° = cos 56°, \ cos 56° = cos 56° = 1, sin 34°, , cos 56°, , Ex. 3 In right angled D ACB, If ÐC = 90°, AC = 3, BC = 4., , A, , Find the ratios sin A, sin B, cos A, tan B, Solution : In right angled D ACB, using Pythagoras’ theorem,, AB = AC +BC, = 32 + 42 = 52, \ AB = 5, 2, , 2, , 2, , 3, , BC, 4, AC, 3, = cos A =, =, sin A =, AB, 5, AB, 5, , and sin B =, , 3, AC, = , 5, AB, , tan B =, , then find cos q and tan q., P, , 5, 13, , In right angled D PQR, ÐR= q, sin q =, , Q, , Fig. 8.22, , AC, 3, =, BC, 4, , Ex. 4 In right angled triangle D PQR, ÐQ = 90°, ÐR= q and if sin q =, Solution :, , 4, , C, , q, Fig. 8.23, , \, , R, , 110, , 5, 13, , 5, PQ, =, PR, 13, , B

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Remember this !, ‘Square of’ sin q means (sin q)2. It is written as sin2 q., We have proved the equation sin2 q + cos2 q = 1 using Pythagoras’ theorem, where q is, an acute angle of a right angled triangle., Verify that the equation is true even when q = 0° or q = 90°., Since the equation sin2 q + cos2 q = 1 is true for any value of q. So it is a basic, trigonometrical identity., (i) 0 £ sin q £ 1,, , 0 £ sin2 q £ 1, , (ii) 0 £ cos q £ 1,, , 0 £ cos2 q £ 1, , Practice set 8.2, 1., , In the following table, a ratio is given in each column. Find the remaining two ratios in, the column and complete the table., 11, 61, , sin q, cos q, , 1, 3, , 35, 37, , tan q, 2., , 3, 5, , 1, 2, , 21, 20, , 1, , 8, 15, , Find the values of 4, tan 2 60° + 3 sin 2 60°, 5, tan 60, (iv), sin 60 + cos 60, , (i) 5 sin 30° + 3 tan 45° , , (ii), , (iii) 2 sin 30° + cos 0° + 3 sin 90° , (v) cos 2 45° + sin 2 30° , , 3., , If sin q =, , 4, then find cos q, 5, , 4., , If cos q =, , 15, then find sin q, 17, , (vi) cos 60°× cos 30° + sin 60°× sin 30°, , 112, , 1, 2 2

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Problem set 8, 1., , Choose the correct alternative answer for following multiple choice questions., (i) Which of the following statements is true ?, , (A) sin q = cos (90- q) , , (B) cos q = tan (90- q), , (C) sin q = tan (90- q) , , (D) tan q = tan (90- q), , (ii) Which of the following is the value of sin 90° ?, (A), , 3, 2, , (B) 0 , , (C), , 1, 2, , (D) 1, , (iii) 2 tan 45° + cos 45° - sin 45° = ?, (A) 0 , (B) 1 , (C) 2 , cos 28°, (iv), = ?, sin 62°, (A) 2 , (B) -1 , (C) 0 , 2., , (D) 3, , (D) 1, , T, , In right angled D TSU, TS = 5, ÐS = 90°,, SU = 12 then find sin T, cos T, tan T., Similarly find sin U, cos U, tan U., , S, 3., , In right angled D YXZ, ÐX = 90°, XZ = 8 cm,, YZ = 17 cm, find sin Y, cos Y, tan Y,, sin Z, cos Z, tan Z., , X, 8, Z, , 4., , Fig. 8.26, , In right angled D LMN, if ÐN = q, ÐM = 90°,, cos q = 24, 25 , find sin q and tan q, , L, , Fill in the blanks., , M, , U, Y, , 17, Fig. 8.27, , Similarly, find (sin2 q) and (cos2 q)., , 5., , (i) sin20° = cos, , ° , , (ii) tan30° ´ tan, , ° =1, , (iii) cos40° = sin, , Fig. 8.28, , N, , ° , qqq, 113

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9, , Surface Area and Volume, , Let’s study., , • Surface area of a cone, • Volume of a cone, , • Surface area of a sphere, • Volume of a sphere, , Let’s recall., We have learnt how to find the surface area and volume of a cuboid, a cube and a cylinder,, in earlier standard., • Length, breadth and height of a cuboid are l , b, h respectively., , Cuboid, , (i) Area of vertical surfaces of a cuboid = 2(l + b) ´ h, , h, b, , l, , Fig.9.1, , Cube, , Here we have considered only 4 surfaces into consideration., (ii) Total surface area of a cuboid = 2(lb + bh + lh), Here we have taken all 6 surfaces into consideration., (iii) Volume of a cuboid = l ´ b ´ h, • If l is the edge of a cube,, , l, , (i) Total surface area of a cube = 6l 2, (ii) Area of vertical surfaces of a cube = 4l 2, (iii) Volume of a cube = l 3, , Fig.9.2, Cylinder, , • Radius of cylinder is r and height is h., , h, r, , (i) Curved surface area of a cylinder = 2prh, (ii) Total surface area of a cylinder = 2pr(r + h), (iii) Volume of a cylinder = pr2h, , Fig.9.3, , 114

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Practice set 9.1, 1. Length, breadth and height of a cuboid shape box of medicine is 20cm, 12 cm and 10 cm, respectively. Find the surface area of vertical faces and total surface area of this box., 2. Total surface area of a box of cuboid shape is 500 sq. unit. Its breadth and height is 6 unit, and 5 unit respectively. What is the length of that box ?, 3. Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area, of the cube., 4. Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of, the cube., 5. Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5m and, 1.15m respectively. Find its length., 6. What will be the volume of a cube having length of edge 7.5 cm ?, 7. Radius of base of a cylinder is 20cm and its height is 13cm, find its curved surface area, and total surface area. (p = 3.14), 8. Curved surface area of a cylinder is 1980 cm2 and radius of its base is 15cm. Find the, height of the cylinder. (p =, , 22, )., 7, , Let’s learn., , Terms related to a cone and their relation, A, , h, O, , l, r B, , Fig.9.4, , A cone is shown in the adjacent Fig.9.4. Centre of the, circle, which is the base of the cone, is O and A is the, vertex (apex) of the cone. Seg OB is a a radius and, seg OA is perpendicular to the radius at O, means AO, is perpendicular height of the cone. Slant height of the, cone is the length of AB, which is shown by (l)., D AOB is a right angled triangle., \ by the Pythagoras’ theorem, AB 2 = AO 2 + OB 2, , \l 2 = h 2 + r 2, , That is, (slant height)2 = (Perpendicular height)2 + (Base radius)2, Surface area of a cone, A cone has two surfaces : (i) circular base and (ii) curved surface., Out of these two we can find the area of base of a cone because we know the formula for, the area of a circle., How to find the curved surface area of a cone ? How to derive a formula for it ?, , 115

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To find a formula for the curved surface area of, a cone, let us see the net of the curved surface,, which is a sector of a circle., If a cone is cut along edge AB,we get its net, as shown in fig.9.5., , A, B, , D, , Compare the figures 9.4 and 9.5, C, Have you noticed the following things ?, Fig.9.5, (i) Radius AB of the sector is the same as the slant height of the cones., (ii) Arc BCD of the sector is the same as circumference of the base of the cone., (iii) Curved surface area of cone = Area of sector A-BCD., It means to find the curved surface area of a cone we have to find the area of its net that, is the area of the sector., Try to understand, how it is done from the following activity., Activity : Look at the following figures., , l, , l, , l, , Cone, Fig. 9.6, , l, , Net of curved surface, Fig. 9.7, , Circumference of base of the circle = 2pr, , Pieces of the net, Fig. 9.8, , B, , A, , As shown in the Fig.9.8, make pieces of the net as, small as possible. Join them as shown in the Fig.9.9., , l, , By Joining the small pieces of net of the cone, we, get a rectangle ABCD approximately., Total length of AB and CD is 2pr., , D, , \ length of side AB of rectangle ABCD is pr, , pr, Fig. 9.9, , and length of side CD is also pr., Length of side BC of rectangle = slant height of cone = l., , Curved surface area of cone is equal to the area of the rectangle., \ curved surface area of cone = Area of rectangle = AB ´ BC = pr ´ l = prl, 116, , C

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Now, we can derive the formula for total surface area of a cone., Total surface area of cone = Curved surface area + Area of base, = prl + pr2, = pr(l + r), Did you notice a thing ? If a cone is not closed (Just like a cap of jocker or a cap in a birthday, party) it will have only one surface, which is the curved surface. Then we get the surface area, of the cone by the formula prl., Activity : Prepare a cylinder of a card sheet, keeping one of its faces open. Prepare an open, cone of card sheet which will have the same base-radius and the same height as that of the, cylinder., Pour fine sand in the cone till it just fills up the cone. Empty the cone in the cylinder., Repeat the procedure till the cylinder is just filled up with sand. Note how many coneful of, sand is required to fill up the cylinder., , r, h, , r, l, , h, , Fig. 9.10, To fill up the cylinder, three coneful of sand is required., , Let’s learn., Volume of a cone, If the base-radii and heights of a cone and a cylinder are equal then, 3 ´ volume of cone = volume of cylinder, \ 3 ´ volume of cone = pr2h, , \ volume of cone =, , 1, 3, , ´ pr2h, , Remember this !, (i) Area of base of a cone = pr2, , (ii) Curved surface area of a cone = prl, , (iii) Total surface area of a cone = pr (l + r) (iv) Volume of a cone =, , 117, , 1, ´ pr2h, 3

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Solved Examples :, Ex. (1) Radius of base (r) and perpendicular height (h) of cone is given., Find its slant height (l), (i) r = 6 cm, h = 8 cm,, (ii) r = 9 cm, h = 12 cm, Solution :, (i) r = 6 cm, h = 8 cm, , l2 = r2 + h2, \l2 = (6)2 + (8)2, \l2 = 36 + 64, \l2 = 100, \l = 10 cm, , (ii) r = 9 cm, h = 12 cm, l2 = r2 + h2, \l2 = (9)2 + (12)2, \l2 = 81 + 144, \l2 = 225, \l = 15 cm, , Ex. (2) Find (i) the slant height, (ii) the curved surface area and (iii) total surface, area of a cone, if its base radius is 12 cm and height is 16 cm. (p = 3.14), Solution :, (ii) Curved surface area = prl, (i) r = 12 cm, h = 16 cm, l2 = r2 + h2, = 3.14 ´ 12 ´ 20, 2, 2, 2, = 753.6 cm2, \l = (12) + (16), , \l2 = 144 + 256, \l2 = 400, \l = 20 cm, , (iii) Total surface area of cone, , = pr (l + r), , = 3.14 ´ 12(20+12), , = 3.14 ´ 12 ´ 32, , = 1205.76 cm2, , Ex. (3) The total surface area of a cone is 704 sq.cm and radius of its base is 7 cm, find the, slant height of the cone. (p =, Solution :, , 22, ), 7, , Total surface area of cone = pr (l + r), 22, 7, , \, , 704 =, , ´ 7 (l + 7), , \, , 704, =l+7, 22, , \, , 32 = l + 7, , \ 32 - 7 = l, \, , l = 25 cm, , 118

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Ex. (4) Area of the base of a cone is 1386 sq.cm and its height is 28 cm., 22, , ), Find its surface area. (p =, 7, Solution :, Area of base of cone = pr 2, 22, 7, , \, , 1386 =, , \, , 1386 × 7, = r2, 22, , \l 2 = (21)2 + (28)2, \l 2 = 441 + 784, \l 2 = 1225, \l = 35 cm, , ´ r2, , Surface area of cone = prl, , \ 63 ´ 7 = r 2, \, , 441 = r 2, , \, , r = 21 cm, , 22, ´ 21 ´ 35, 7, , , , =, , , , = 22 ´ 21 ´ 5 , , , , = 2310 sq. cm., , Practice set 9.2, 1., , Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of, the base of the cone., , 2., , Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is, 28 cm. (p =, , 22, ), 7, , 3., , Curved surface area of a cone is 251.2 cm2 and radius of its base is 8cm. Find its slant, height and perpendicular height. (p = 3.14 ), , 4., , What will be the cost of making a closed cone of tin sheet having radius of base 6 m, and slant height 8 m if the rate of making is Rs.10 per sq.m ?, , 5., , Volume of a cone is 6280 cubic cm and base radius of the cone is 30 cm. Find its, perpendicular height. (p = 3.14), , 6., , Surface area of a cone is 188.4 sq.cm and its slant height is 10cm. Find its perpendicular height (p = 3.14)., , 7., , Volume of a cone is 1212 cm3 and its height is 24cm. Find the surface area of the cone., (p = 22 ), 7, , 8., , The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the, total surface area of cone. (p =, , 9., , 22, ), 7, , There are 25 persons in a tent which is conical in shape. Every person needs an area of, 4 sq.m. of the ground inside the tent. If height of the tent is 18m, find the volume of the, tent., , 119

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10. In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is, 2.1m. and diameter of base is 7.2 m. Find the volume of the fodder. if it is to be covered, by polythin in rainy season then how much minimum polythin sheet is needed ?, (p =, , 22, and, 7, , 17.37 = 4.17.), , Let’s learn., Surface area of a sphere, Surface area of a sphere = 4pr2, , \ Surface area of a hollow hemisphere = 2pr2, , Fig. 9.11, , Total surface area of a solid hemisphere, = Surface area of hemisphere + Area of circle, = 2pr2 + pr2 = 3pr2, , Take a sweet lime (Mosambe), Cut it into, two equal parts., , Take one of the parts. Place its circular face on a, paper. Draw its circular border. Copy three more, such circles. Again, cut each half of the sweet lime, into two equal parts., , Now you get 4 quarters of sweet lime. Separate the, peel of a quarter part. Cut it into pieces as small as, possible. Try to cover one of the circles drawn, by the, small pieces., Observe that the circle gets nearly covered., The activity suggests that,, curved surface area of a sphere = 4 pr2., , 120

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Solved Examples :, (1), Find the surface area of a sphere having, radius 7 cm. (p =, Solution :, , (2) Find the radius of a sphere having, surface area 1256sq.cm.(p = 3.14), Solution : Surface Area of Sphere = 4pr2, \ 1256 = 4 ´ 3.14 ´ r2, , 22, ), 7, , Surface Area of sphere = 4pr2, 22, 7, 22, 7, , , , =4´, , , , =4´, , , , = 88 ´ 7, , ´ (7)2, , \ r2 =, , ´7´7, , , , =, , \ 100 = r2, \, 10 = r, \radius of the sphere is 10 cm., , = 616, , Surface Area of sphere = 616 sq.cm., Activity :, , x, , Make a cone and a hemisphere of cardsheet such that radii of cone and, , , , hemisphere are equal and height of cone is equal to radius of the hemisphere., , , , Fill the cone with fine sand. Pour the sand in the hemisphere. How, , , , many cones are required to fill the hemisphere completely ?, , r, r, , h, , Fig. 9.12, , \volume of sphere, , Two conefull of sand is required to fill the, hemisphere., \ 2 ´ volume of cone = volume of hemisphere., , = 2 ´ volume of hemisphere., , \ volume of hemisphere = 2 ´ volume of cone, , =, , 1, ´ pr2h, 3, 1, , = 2 ´ ´ pr2 ´ r, 3, 2, = pr3, , 3, , , , =2´, , 121, , 4 3, pr, 3, 4, 3, , \ volume of sphere = pr3

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Remember this !, 2, , • Volume of hemisphere = 3 pr3, • Total surface area of hemisphere = 2pr2 + pr2= 3pr2, Solved Examples :, Ex. (1) Find the volume of a sphere having, , 22, radius 21 cm. (p =, ), 7, 4, Solution : Volume of sphere = pr3, 3, 22, 4, , = ´, ´ (21)3, 7, 3, 22, 4, = ´, ´ 21 ´ 21 ´ 21, 7, 3, , = 88 ´ 441, , Ex. (2) Find the radius of a sphere, whose volume is, 113040 cubic cm. (p = 3.14), Solution : Volume of sphere =, 113040 =, , 4, 3, , 4 3, pr, 3, , ´ 3.14 ´ r3, , 113040 ´ 3, = r3, 4 ´ 3.14, 28260 ´ 3, = r3, 3.14, , \ 9000 ´ 3 = r3, \ r3 = 27000, \ r = 30 cm, \ radius of sphere is 30 cm., , \ volume of sphere = 38808 cubic cm., , Ex. (3) Find the volume of a sphere whose surface area is 314.sq.cm. (Take p = 3.14), Solution : Surface area of sphere = 4pr2, 314 = 4 ´ 3.14, , \, \, \, , Volume of sphere =, , ´ r2, , 4 3, pr, 3, , 314, 4 ´ 3.14 = r2, , =, , 4, 3, , ´ 3.14 ´ 53, , 31400, = r2, 4 ´ 314, 100, = r2, 4, , =, , 4, 3, , ´ 3.14 ´ 125, , 25, , r, , = 523.33 cubic cm., , = r2, = 5 cm, , 122

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Practice set 9.3, , 3., , Find the surface areas and volumes of spheres of the following radii., (i) 4 cm (ii) 9 cm (iii) 3.5 cm., (p = 3.14), If the radius of a solid hemisphere is 5cm, then find its curved surface area and total, surface area. (p = 3.14), If the surface area of a sphere is 2826 cm2 then find its volume. (p = 3.14), , 4., , Find the surface area of a sphere, if its volume is 38808 cubic cm. (p =, , 5., , Volume of a hemisphere is 18000 p cubic cm. Find its diameter., , 1., 2., , 22, ), 7, , Problem set 9, 1., , If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will, be pressed in its 500 rotations ?, , 2., , To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length,, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How, much maximum volume of water will be contained in it ?, , 3., , If the ratio of radius of base and height of a cone is 5:12 and its volume is 314 cubic, metre. Find its perpendicular height and slant height (p = 3.14)., , 4., , Find the radius of a sphere if its volume is 904.32 cubic cm. (p = 3.14), , 5., , Total surface area of a cube is 864 sq.cm. Find its volume., , 6., , Find the volume of a sphere, if its surface area is 154 sq.cm., , 7., , Total surface area of a cone is 616 sq.cm. If the slant height of the cone is three times the, radius of its base, find its slant height., , 8., , The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the, inner surface area of the well. Find the cost of plastering it from inside at the rate, Rs.52 per sq.m., , 9., , The length of a road roller is 2.1m and its diameter is 1.4m. For levelling a ground, 500 rotations of the road roller were required. How much area of ground was levelled, by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq. m., , qqq, 123

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Answers, 1., , Basic Concepts in Geometry, Practice set 1.1, , 1., , (i) 3, , (ii) 3, , (iii) 7, , (iv) 1, , (v) 3, , (vi) 5, , (vii) 2, , (viii) 7, , 2., , (i) 6, , 3., , (i) P-R-Q, , (ii) Non collinear, , (v) X-Y-Z, , (vi) Non collinear, , 4., , (ii) 8, , 18 and 2, , (iii) 10, , 5. 25 and 9, , (iv) 1, , (v) 3, , (vi) 12, , (iii) A-C-B, , 6. (i) 4.5, , (ii) 6.2, , (iv) Non collinear, (iii) 2 7, , 7. Triangle, , Practice set 1.2, 1., , (i) No, , (ii) No, , (iii) Yes, , 2. 4, , 3. 5, , 4. BP < AP < AB, , 5., , (i) Ray RS or Ray RT (ii) Ray PQ (iii) Seg QR (iv) Ray QR and Ray RQ etc., (v) Ray RQ and Ray RT etc.. (vi) Ray SR , Ray ST etc.., , 6., , (vii) Point S, , (i) Point A & Point C, Point D & Point P (ii) Point L & Point U, Point P & Point R, (iii), , d (U,V ) = 10, d ( P,C) = 6 , d (V,B) = 3 , d (U,L) = 2, Practice set 1.3, , 1., , (i), , , , If a quadrilateral is a parallelogram then opposite angles of that quadrilateral are, congruent., , (ii) If quadrilateral is a rectangle then diagonals are congruent., (iii) If a triangle is an isosceles then segment joining vertex of a triangle and mid point, , 2., , (i), , , , of the base is perpendicular to the base., If alternate angles made by two lines and its transversal are congruent then the, lines are parallel., , (ii) If two parallel lines are intersected by a transversal the interior angles so formal, , , are supplementary., , (iii) If the diagonals of a quaddrilateral are congruent then that quadrilateral is rectangle., Problem set 1, 1., 2., 3., 4., , (i) A (ii) C (iii) C (iv) C (v) B, (i) False, (ii) False, (iii) True (iv) False, (i) 3 (ii) 8 (iii) 9 (iv) 2 (v) 6 (vi) 22 (vii) 165, -15 and 1, 5. (i) 10.5 (ii) 9.1, 6. -6 and 8, , 124

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2. Parallel Lines, Practice set 2.1, 1., 2., 3., 5., , (i) 95° (ii) 95° (iii) 85° (iv) 85°, Ða = 70°, Ðb = 70°, Ðc = 115°, Ðd = 65°, Ða = 135°, Ðb = 135°, Ðc = 135°, (i) 75° (ii) 75° (iii) 105° (iv) 75°, Practice set 2.2, , 1., , 4. ÐABC = 130°, , No., , Problem set 2, 1., 5., , (i) C (ii) C (iii) A (iv) B (v) C, 6. f = 100° g = 80°, x = 1260°, , 4., , x = 130°, , y = 50°, , 3. Triangles, Practice set 3.1, 1., 5., 7., , 110°, 2. 45°, 3. 80°, 60°, 40°, 6. ÐDRE = 70°, ÐARE = 110°, 60°, 80°, 40°, ÐAOB = 125°, 9. 30°, 70°, 80°, , 4. 30°, 60°, 90°, , Practice set 3.2, 1., 2., 3., 4., , (i) SSC Test, (ii) SAS Test, (iii) ASA Test (iv) Hypotenuse Side Test., (i) ASA Test, ÐBAC @ ÐQPR ,side AB @ side PQ, side AC @ side PR, (ii) SAS Test, ÐTPQ @ ÐTSR, ÐTQP @ ÐTRS, side PQ @ side SR, Hypotenuse Side Test, ÐACB @ ÐQRP, ÐABC @ ÐQPR, side AC @ side QR, SSS Test, ÐMLN @ ÐMPN, ÐLMN @ ÐMNP,ÐLNM @ ÐPMN, Practice set 3.3, , 1., 2., , x = 50° , y = 60°, mÐABD = 110° , mÐACD = 110° ., 7.5 Units, , 3. 6.5 Units, , 4. l(PG) = 5 cm , l(PT) = 7.5 cm, , Practice set 3.4, 1., , 2 cm, , 2. 28°, , 3. ÐQPR, ÐPQR, , 4. greatest side NA, smallest side FN, , Practice set 3.5, 1., 2., , XY, YZ, XZ , ÐX @ ÐL, ÐY @ ÐM,, = =, LM MN LN, l(QR) = 12 cm, l(PR) = 10 cm, , 125, , ÐZ @ ÐN

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Problem set 3, 1., , (i) D, , (ii) B, , (iii) B, , 5. Quadrilaterals, Practice set 5.1, 1., 2., 3., 4., 6., , mÐXWZ = 135°, mÐYZW = 45° , l (WY) = 10 cm, x = 40° , ÐC = 132° , ÐD = 48°, 25 cm, 50 cm, 25 cm, 50 cm, 60°, 120°, 60°, 120°, , ÐA = 70° , ÐB = 110° , ÐC = 70° , ÐR = 110°, Practice set 5.3, , 1., 2., 3., 4., 5., , BO = 4 cm, ÐACB = 35°, QR = 7.5 cm, ÐPQR = 105°, ÐSRQ = 75°, ÐIMJ = 90°, ÐJIK = 45°, ÐLJK = 45°, side = 14.5 cm, Perimetere = 58 cm, (i) False (ii) False (iii) True (iv) True (v) True (vi) False, Practice set 5.4, , 1., , ÐJ = 127°, ÐL = 72°, , 2. ÐB = 108°, ÐD = 72°, Practice set 5.5, , 1., , XY = 4.5 cm,, YZ = 2.5 cm,, XZ = 5.5 cm, Problem set 5, , 1., (i) D, (ii) C, 4. 24 cm, 32 cm, 24 cm, 32 cm, , (iii) D, 2. 25 cm,, 3. 6.5 2 cm, 5. PQ = 26 cm, 6. ÐMPS = 65°, , 6. Circle, Practice set 6.1, 1., , 20 cm, , 2. 5 cm, , 3. 32 unit, , 4. 9 unit, , Practice set 6.2, 1., , 12 cm, , 2. 24 cm, Problem set 6, , 1., , (i) A (ii) C, , (iii) A (iv) B, , (v) D, , (vi) C, , 126, , (vii) D or B, , 2. 2:1, , 4. 24 units

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7. Co - ordinate Geometry, Practice set 7.1, 1., , 2., , point A : Quadrant II, point B : Quadrant III, point K : Quadrant I, point D : Quadrant I, point E : Quadrant I, point F : Quadrant IV, point G : Quadrant IV, point H : Y-Axis., point M : X-Axis, point N : Y-Axis, point P : Y-Axis, point Q : Quadrant III, (i) Quadrant I, (ii) Quadrant III, (iii) Quadrant IV (iv) Quadrant II, Practice set 7.2, , 1., 6., 7., 8., , Square, 2. x = -7, (i) Y-Axis,, (ii) X-axis,, To X-axis (5,0) , To Y-axis (0,5), (-4,1), (-1.5, 1), (-1.5,5), (-4,5), , 3. y = -5, (iii) Y-axis,, , 4. x = -3, (iv) X-axis,, , 5. 4, , Problem set 7, 1., 2., 3., 5., , (i) C, (ii) A (iii) B, (iv) C, (v) C, (vi) B, (i) Q (- 2,2), R(4,-1) (ii) T(0,-1), M(3,0) (iii) point S (iv) point O, (i) Quadrant IV (ii) Quadrant III, (iii) Quadrant II (iv) Quadrant II, (v) Y-axis (vi) X-axis, 7. |a|, (i) 3 (ii) P(3,2), Q(3,-1), R (3,0) (iii) 0 6. . y = 5, y = -5, , 8. Trigonometry, Practice set 8.1, 1., , (i), , QR, PQ, , (ii), , QR, PQ, , (iii), , QR, PR, , (iv), , 2., , (i), , a, , c, , (ii), , b, , a, , (iii), , b, , c, , (iv) a, b, , 3., , (i), , MN, LN, , (ii), , LM, LN, , (iii), , LM, MN, , (iv), , 4., , (i), , PQ RQ PQ, ,, ,, PR PR RQ, , (ii), , PR, QR, , MN, LN, , QS PQ QS, ,, ,, PS PS PQ, Practice set 8.2, , 1., , 2 21 8 1, 60 1, 3 20 15 4 2 2, ,, ,, ,, ; cos q :, ,, ,, ,, ,, , ,, 3, 61, 3 29 17 3, 2 2 29 17 5, , sin q :, , 12 1, ,, ,, 37, 2, , tan q :, , 12 11 1, ,, ,, ,, 3, 35 60, , 2,, , 3, 4, , 127

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2., , (i), , 93, 11, (ii), 20, 2, , (iii) 5, , (iv), , 2 3, 3 +1, , (v), , 3, 3, (vi), 2, 4, , 3., , 3, 5, , 4., , 8, 17, , Problem set 8, 1., 2., 3., 4., 5., , (i) A (ii) D (iii) C (iv) D, 12, 5, 12, 5, 5, 12, sin T =, , cos T =, , tan T =, , sin U =, , cos U =, , tan U =, 5, 12, 13, 13, 13, 13, 8, 8, 15, 15, 8, 15, sin Y =, , cos Y =, , tan Y =, , sin Z =, , cos Z =, , tan Z =, 15, 17, 17, 17, 17, 8, 7, 49, 576, 7, sin q =, , tan q =, , sin2 q =, , cos2 q =, 25, 625, 625, 24, (i) 70 , , (ii) 60 , , (iii) 50, , 9. Surface Area and Volume, Practice set 9.1, 1., , 640 sq.cm, 1120 sq.cm., , 2. 20 Unit , , 3. 81 sq.cm, 121.50 sq.cm., , 4., , 3600 sq.cm. , , 5. 20 m , , 6. 421.88 cubic cm, , 7., , 1632.80 sq.cm, 4144.80 sq.cm. , , 8. 21 cm, , Practice set 9.2, 1. 5 cm, , 2. 36960 cubic cm., , 5. 15 cm, , 6., , 8 cm, , 3. 10 cm, 6 cm, , 4. ` 2640, , 7. 550 sq.cm 8. 2816 sq.cm, 9856 cubic cm, , 9. 600 cubic metre 10. 28.51 cubic metre, 47.18 sq.m., Practice Set 9.3, 1. (i) 200.96 sq.cm, 267.95 cubic cm., , (ii) 1017.36 sq.cm, 3052.08 cubic cm., , (iii) 153.86 sq.m, 179.50 cubic cm., 2. 157 sq.cm, 235.5 sq.cm., , 3. 14130 cubic cm. 4. 5544 sq.cm., , 5. 60 cm, , Problem set 9, 1. 1980 sq.m. , 4. 6 cm , , 2. 96801.6 cubic cm. , 5. 1728 cubic cm. , , 3. 12 m, 13 m, 6. 179.67 cubic cm., , 7. 21 cm , , 8. 132 sq.m., ` 6864 9. 4620 sq.m, ` 32340, , qqq, 128

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Practical Notebook for Standard IX, , Practical Notebook Cum Journal - Mathematics, •, , English, Medium, , •, , •, , Price, ` 55.00, , •, •, •, , Based, on, Government, approved syllabus and textbook, Inclusion of practicals based, on all chapters as per, Evaluation scheme., With full of various activities,, pictures, figures/ diagrams, etc., Inclusion of objective/multiple, choice questions, Inclusion of useful questions, for oral examination, More questions for practice, and separate space for writing, answers, , Practical notebooks are available for sale in the regional depots of, the Textbook Bureau., (1) Maharashtra State Textbook Stores and Distribution Centre, Senapati Bapat Marg, Pune 411004  25659465, (2) Maharashtra State Textbook Stores and Distribution Centre, P-41, Industrial Estate, Mumbai - Bengaluru, Highway, Opposite Sakal Office, Kolhapur 416122  2468576 (3) Maharashtra State Textbook Stores and, Distribution Centre, 10, Udyognagar, S. V. Road, Goregaon (West), Mumbai 400062  28771842, (4) Maharashtra State Textbook Stores and Distribution Centre, CIDCO, Plot no. 14, W-Sector 12, Wavanja, Road, New Panvel, Dist. Rajgad, Panvel 410206  274626465 (5) Maharashtra State Textbook Stores and, Distribution Centre, Near Lekhanagar, Plot no. 24, 'MAGH' Sector, CIDCO, New Mumbai-Agra Road, Nashik, 422009  2391511 (6) Maharashtra State Textbook Stores and Distribution Centre, M.I.D.C. Shed no. 2 and 3,, Near Railway Station, Aurangabad 431001  2332171 (7) Maharashtra State Textbook Stores and Distribution, Centre, Opposite Rabindranath Tagore Science College, Maharaj Baug Road, Nagpur 440001,  2547716/2523078 (8) Maharashtra State Textbook Stores and Distribution Centre, Plot no. F-91, M.I.D.C.,, Latur 413531  220930 (9) Maharashtra State Textbook Stores and Distribution Centre, Shakuntal Colony,, Behind V.M.V. College, Amravati 444604  2530965, , E-learning material (Audio-Visual) for Standards One to, Twelve is available through Textbook Bureau, Balbharati..., • Register your demand by scanning the Q.R. Code, given alongside., • Register your demand for E-learning material by using, Google play store and downloading ebalbharati app., www.ebalbharati.in, www.balbharati.in, ebalbharati

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MATHEMATICS, Part -, , STANDARD NINE, , 9, , 2, , 61.00