1 Similarity. QR is the base BC AD A( D ABC) A( D PQR) = QR PS. Let s study.

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2 1 Similarity Let s study. Ratio of areas of two triangles asic proportionality theorem onverse of basic proportionality theorem Tests of similarity of triangles roperty of an angle bisector of a triangle roperty of areas of similar triangles The ratio of the intercepts made on the transversals by three parallel lines Let s recall. We have studied Ratio and roportion. The statement, the numbers a and b are in the ratio m is also written as, the numbers a and b are in proportion m:n. n For this concept we consider positive real numbers. We know that the lengths of line segments and area of any figure are positive real numbers. We know the formula of area of a triangle. rea of a triangle = 1 ase Height Let s learn. Ratio of areas of two triangles Let s find the ratio of areas of any two triangles. Ex. In, is the height and is the base. In R, S is the height and R is the base ( ) ( R) = 1 1 R S 1 Fig. 1.1 Fig. 1. S R

3 ( ) \ = ( R) R S Hence the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corrosponding heights. ase of a triangle is b 1 and height is h 1. ase of another triangle is b and height is h. Then the ratio of their areas = b 1 h 1 b h Suppose some conditions are imposed on these two triangles, ondition 1 If the heights of both triangles are equal then- h h Fig. 1.3 ( ) ( R) = h R h = R S Fig. 1.4 ( ) \ ( R) = b 1 b roperty The ratio of the areas of two triangles with equal heights is equal to the ratio of their corresponding bases. R ondition If the bases of both triangles are equal then - ( ) ( ) = h 1 h h 1 ( ) h ( ) = h 1 h Fig. 1.5 roperty The ratio of the areas of two triangles with equal bases is equal to the ratio of their corresponding heights.

4 ctivity : Fill in the blanks properly. (i) (ii) L R Fig. 1.6 M Fig.1.7 N ( ) ( ) = = ( LMN) ( MN) = = (iii) Solved Examples Ex. (1) M is the midpoint of seg and seg M is a median of \ ( M) ( M) = State the reason. = = M Fig. 1.8 In adjoining figure E ^ seg, seg F ^ line, ( ) E = 4, F = 6, then find ( ). E F Fig.1.9 Solution ( ) ( ) = E... bases are equal, hence areas proportional to F heights. = 4 6 = 3 3

5 Ex. () In point on side is such that = 6, = 15. Find ( ) : ( ) and ( ) : ( ). Solution oint is common vertex of, and and their bases are collinear. Hence, heights of these three triangles are equal = 15, = 6 \ = - = 15-6 = 9 ( ) ( ) = = ( ) ( ) = =... heights equal, hence areas proportional to bases = heights equal, hence areas proportional to bases. 9 6 = 3 Fig Ex. (3) c is a parallelogram. is any point on side. Find two pairs of triangles with equal areas. Fig Solution c is a parallelogram. \ and onsider and. oth the triangles are drawn in two parallel lines. Hence the distance between the two parallel lines is the height of both triangles. In and, common base is and heights are equal. Hence, ( ) = ( ) In and, is common base and and heights are equal. \ ( ) = ( ) 4

6 Ex.(4) In adjoining figure in, point is on side. If = 16, = 9 and ^, then find the following ratios. (i) ( ) ( ) (ii) ( ) ( ) Fig. 1.1 (iii) ( ) ( ) Solution In point and are on side, hence is common vertex of,, and and their sides,, and are collinear. Heights of all the triangles are equal. Hence, areas of these triangles are proportinal to their bases. = 16, = 9 \ = 16-9 = 7 ( ) \ ( ) = = triangles having equal heights 16 ( ) ( ) = = triangles having equal heights 16 ( ) ( ) = = triangles having equal heights Remember this! Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights. reas of triangles with equal heights are proportional to their corresponding bases. reas of triangles with equal bases are proportional to their corresponding heights. ractice set ase of a triangle is 9 and height is 5. ase of another triangle is 10 and height is 6. Find the ratio of areas of these triangles. 5

7 . In figure 1.13 ^, ^, ( ) = 4, = 8, then find ( ). Fig T R S Fig In adjoining figure, ^,, then find ( ) ( ). 3. In adjoining figure 1.14 seg S ^ seg R seg T ^ seg R. If R = 6, S = 6 and R = 1, then find T. Fig In adjoining figure ^, ^ then find following ratios. (i) ( ) ( ) (ii) ( ) ( ) Fig (iii) ( ) ( ) (iv) ( ) ( ) 6

8 Let s learn. asic proportionality theorem Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion. Given In line l line and line l intersects and in point and respectively l To prove = onstruction: raw seg and seg Fig roof and have equal heights. ( ) \ ( ) =... (I) (areas proportionate to bases) ( ) and ( ) =... (II) (areas proportionate to bases) seg is common base of and. seg seg, hence and have equal heights. ( ) = ( )... (III) ( ) ( ) \ = = ( ) ( ) onverse of basic proportionality theorem... [from (I), (II) and (III)]... [from (I) and (II)] Theorem : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. In figure 1.18, line l interesects the side and side of in the points and respectively and =, hence line l seg. 7

9 This theorem can be proved by indirect method. l ctivity : raw a. isect Ð and name the point of intersection of and the angle bisector as. Measure the sides. = cm = cm = cm = cm Find ratios and. Fig You will find that both the ratios are almost equal. isect remaining angles of the triangle and find the ratios as above. You can verify that the ratios are equal. Let s learn. Fig roperty of an angle bisector of a triangle Theorem : The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides. Given : In, bisector of Ð interesects seg in the point E. To prove : onstruction : E = E raw a line parallel to ray E, passing through the point. Extend so as to intersect it at point. E Fig

10 roof : ray E ray and is transversal, \ Ð E = Ð... (corresponding angles)...(i) Now taking as transversal Ð E = Ð... (alternate angle)...(ii) ut Ð Ð E... (given)...(iii) \ Ð... [from (I), (II) and (III)] In, side...(sides opposite to congruent angles) \ =...(IV) Now in, seg E seg... (construction) \ E E =...(asic proportionality theorem)..(v) \ E E =... [from (IV) and (V)] For more information : Write another proof of the theorem yourself. raw M ^ and N ^. Use the following properties and write the proof. (1) The areas of two triangles of equal heights are M proportional to their bases. N Fig. 1.1 () Every point on the bisector of an angle is equidistant M from the sides of the angle. N Fig. 1. 9

11 onverse of angle bisector theorem If in, point on side such that =, then ray bisects Ð. roperty of three parallel lines and their transversals ctivity: raw three parallel lines. Label them as l, m, n. raw transversals t 1 and t. and are intercepts on transversal t 1. and R are intercepts on transversal t. Find ratios roof : raw seg, which intersects line m at point. In, \ =..... (I) (asic proportionality theorem) In R, R \ =..... (II) (asic proportionality theorem) R \ = = from (I) and (II). R \ 10 t 1 t and. You will find that they are almost equal. R R l m n = R Fig. 1.3 Theorem The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corrosponding intercepts made on any other transversal by the same parallel lines. Given : line l line m line n t 1 t t 1 and t are transversals. Transversal t 1 intersects the l lines in points,, and t intersects the lines in points, m, R. To prove : = R n R Fig. 1.4

12 Remember this! (1) asic proportionality theorem. In, if seg seg Fig. 1.5 then = () onverse of basic proportionality theorem. In R, if S S = T TR then seg ST seg R. S T Fig. 1.6 R Fig. 1.7 (3) Theorem of bisector of an angle of a triangle. If in, is bisector of Ð, then = (4) roperty of three parallel lines and their transversals. If line X line Y line Z l and line l and line m are their transversals then = XY YZ X Y Z Fig. 1.8 m 11

13 Solved Examples Ex. (1) In, E If = 5.4 cm, = 1.8 cm E = 7. cm then find E. Solution : In, E \ = E... asic proportionality theorem E \ 18. = E \ E 5.4 = \ E = E =.4 cm =.4 Ex. () In R, seg RS bisects Ð R. If R = 15, R = 0 S = 1 then find S. Solution : In R, seg RS bisects Ð R. R R = S property of angle bisector S 15 = 1 0 S 1 0 S = 15 \ S = 16 ctivity : S E Fig. 1.9 Fig O O In the figure 1.31, EF If = 5.4, E = 9, = 7.5 then find F Solution : EF R E F = F..... ( ) Fig = F \ F = 1

14 ctivity : E Fig. 1.3 roof : In, ray bisects Ð. \ =.. (I)(ngle bisector theorem) In, E E E =..... (II) ( ) = from (I) and (II) E In, ray bisects Ð. --, side E side, -E- then prove that, = E E ractice set Given below are some triangles and lengths of line segments. Identify in which figures, ray M is the bisector of Ð R. (1) () (3) R 3 5 M 1 5 R 6 M 8 M R 10 Fig Fig Fig In R, M = 15, = 5 R = 0, NR = 8. State whether line NM is parallel to side R. Give reason. R N Fig M 13

15 M 3. In MN, N is a bisector of Ð N. If MN = 5, N = 7 M =.5 then find N Fig Fig In trapezium, side side side, = 15, = 1, = 14, find. N 5 M 40 Fig In figure 1.41, if FE 14 then find x and E Measures of some angles in the figure are given. rove that = 6. Find using given information in the figure. 8 Fig Fig x F E

16 L 6 T 8 M N 10 Fig In, seg bisects Ð. If = x, = x + 5, = x, = x +, then find the value of x. X R E F Fig In LMN, ray MT bisects Ð LMN If LM = 6, MN = 10, TN = 8, then find LT. x x + 5 Fig In the figure 1.44, X is any point in the interior of triangle. oint X is joined to vertices of triangle. Seg seg E, seg R seg EF. Fill in the blanks to prove that, seg R seg F. x - x + roof : In XE, E... \ X =... (I) (asic proportionality theorem) E In XEF, R EF... \ =...(II) \ =... from (I) and (II) \ seg R seg E... (converse of basic proportionality theorem) 11 «. In, ray bisects Ð and ray E bisects Ð. If seg then prove that E. 15

17 Let s recall. Similar triangles In and EF, if Ð, Ð E, Ð F and E = EF = F E F then and EF are similar Fig triangles. and EF are similar is expressed as ~ EF Tests of similarity of triangles Let s learn. For similarity of two triangles, the necessary conditions are that their corresponding sides are in same proportion and their corresponding angles are congruent. Out of these conditions; when three specifc conditions are fulfilled, the remaining conditions are automatically fulfilled. This means for similarity of two triangles, only three specific conditions are sufficient. Similarity of two triangles can be confirmed by testing these three conditions. The groups of such sufficient conditions are called tests of similarity, which we shall use. test for similarity of triangles For a given correspondence of vertices, when corresponding angles of two triangles are congruent, then the two triangles are similar. In and R, in the correspondence «R if Ð, Ð and Ð R then ~ R. Fig R 16

18 For more information roof of test M N Fig R Given : In and R, Ð, Ð, Ð R. To prove ~ R Let us assume that is bigger than R. Mark point M on, and point N on such that M = and N = R. Show that R. Hence show that MN. Now using basic proportionality theorem, M = N M N That is M M = N... (by invertendo) N M+M M = N +N N \ M = N \ = R Similarly it can be shown that \ = R = R... (by componendo) = R \ ~ R test for similarity of triangles: We know that for a given correspondence of vertices, when two angles of a triangle are congruent to two corresponding angles of another triangle, then remaining angle of first triangle is congruent to the remaining angle of the second triangle. This means, when two angles of one triangle are congruent to two corresponding angles of another triangle then this condition is sufficient for similarity of two triangles. This condition is called test of similarity. 17

19 SS test of similarity of triangles For a given correspondence of vertices of two triangles, if two pairs of corresponding sides are in the same proportion and the angles between them are congruent, then the two triangles are similar. For example, if in KLM and R RST, K Ð Ð RST 1 L 1 5 M S Fig T KL RS = LM ST = 3 Therefore, KLM ~ RST SSS test for similarity of triangles For a given correspondence of vertices of two triangles, when three sides of a triangle are in proportion to corresponding three sides of another triangle, then the two triangles are similar. X For example, if in R and XYZ, If YZ = R XY = R R XZ Y Z then R ~ ZYX Fig roperties of similar triangles (1) ~ - Reflexivity () If ~ EF then EF ~ - Symmetry (3) If ~ EF and EF ~ GHI, then ~ GHI - Transitivity Solved Examples Ex. (1) In XYZ, Ð Y = 100, Ð Z = 30, In LMN, Ð M = 100, Ð N = 30, re XYZ and LMN similar? If yes, by which test? X 18 Y L Z Fig M N

20 Solution : In XYZ and LMN, Ð Y = 100, Ð M = 100, \ Ð Ð M ÐZ = 30, Ð N = 30, \ Ð Ð N \ XYZ ~ LMN... by test. Ex. () re two triangles in figure 1.51 similar, according to the information given? If yes, by which test? Solution : In MN and UVW M UV = 6 =, MN 3 1 VW = 10 = 5 1 \ M UV = MN VW M 6 10 N Fig V 3 U 5 W and Ð Ð V MN ~ UVW... Given... SS test of similarity Ex. (3) an we say that the two triangles in figure 1.5 similar, according to information given? If yes, by which test? Solution : XYZ and MN, XY MN = 14 1 = 3, YZ N = 0 30 = 3 It is given that Ð Ð. ut Ð Z and Ð are not included angles by sides which are in proportion. \ XYZ and MN can not be said to be similar. Y 14 X 1 0 Z N 30 Fig. 1.5 M 19

21 Ex. (4) In the adjoining figure ^, ^, -, - -, then prove that and are similar. Solution : In and Ð = (I) Ð = (II) \ Ð...from(I) and (II) Ð... ( ) Fig \ ~... test Ex. (5) iagonals of a quadrilateral intersect in point. If =, =, then prove that =. Given : = = To prove : = Fig roof : = \ = 1... (I) = \ = 1... (II) \ =...from (I) and (II) In and, =... proved Ð... opposite angles \ ~... (SS test of similarity) \ = =... correponding sides are ut = 1 \ = \ = 1 0 proportional

22 ractice set In figure1.55, Ð =75, Ð E=75 state which two triangles are similar and by which test? lso write the similarity of these two 75 triangles by a proper one to one correspondence. 75 E Fig L 3 5. re the triangles in figure 1.56 similar? If yes, by which test? 8 R Fig M 4 N 3. s shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time? Fig R x Fig In, ^, ^ - -, - - then prove that, ~. If = 7, =8, =1 then find. 1

23 5. Given : In trapezium RS, side side SR, R = 5, S = 5 then prove that, SR = 5 O Fig c is a parallelogram point E is on side. Line E intersects ray in point T. rove that E E = E TE. S Fig R 6. In trapezium, (Figure 1.60) side side, diagonals and intersect in point O. If = 0, = 6, O = 15 then find O. E T Fig In the figure, seg and seg intersect each other in point and =. rove that, ~ Fig In the figure, in, point on side is such that, Fig Ð = Ð. rove that, =

24 Let s learn. Theorem of areas of similar triangles Theorem : When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides. S R Fig Given : ~ R, ^, S ^ R ( ) To prove: ( R) = = R = R ( ) roof : = ( R) R S = R... (I) S In and S, Ð = Ð... given Ð = Ð S = 90 \ ccording to test ~ S \ S =... (II) ut ~ R \ =... (III) R From (I), (II) and (III) ( ) ( R) = R S = R R = R = = R 3

25 Solved Examples Ex. (1) ~ R, () = 16, () = 5, then find the value of ratio. Solution ~ R \ ( ) ( R) = \ 16 5 = \ = theorem of areas of similar triangles... taking square roots Ex. () Ratio of corresponging sides of two similar triangles is 5, If the area of the small triangle is 64 sq.cm. then what is the area of the bigger triangle? Solution ssume that ~ R. is smaller and R is bigger triangle. ( ) \ ( R) = ( ) = 4... ratio of areas of similar triangles () \ ( R) = ( R) = ( R) = 4 = 400 \ area of bigger triangle = 400 sq.cm. Ex. (3) In trapezium, side side, diagonal and intersect each other at point. ( ) Then prove that ( ) =. Solution In trapezium side side In and Ð... alternate angles Ð... opposite angles \ ~... test of similarity Fig ( ) ( ) = theorem of areas of similar triangles 4

26 ractice set The ratio of corresponding sides of similar triangles is 3 5; then find the ratio of their areas.. If ~ R and : = :3, then fill in the blanks. ( ) ( R) = = 3 = 3. If ~ R, ( ) = 80, ( R) = 15, then fill in the blanks. ( ) (.... ) = \ = 4. LMN ~ R, 9 (R ) = 16 (LMN). If R = 0 then find MN. 5. reas of two similar triangles are 5 sq.cm. 81 sq.cm. If a side of the smaller triangle is 1 cm, then find corresponding side of the bigger triangle. 6. and EF are equilateral triangles. If () ( EF) = 1 and = 4, find E. 7. In figure 1.66, seg seg E, ( F) = 0 units, F =, then find ( c E) by completing the following activity. ( F) = 0 units, F =, Let us assume = x. \ F = x F = + = + = 3x In FE and F, Ð Ð... corresponding angles Ð Ð... corresponding angles \ FE ~ F... test \ ( FE) ( F ) = = ( 3x ) ( ) x = 9 4 ( FE) = 9 4 ( F ) = 9 4 = (c E) = ( FE) - ( F) = - = 5 E Fig F

27 1. Select the appropriate alternative. (1) In and R, in a one to one correspondence R = = R then () R ~ () R ~ () ~ R () ~ R () If in EF and R, Ð, Ð Ð E then which of the following statements is false? () () E EF R = F R = F () E = EF R () EF R = E roblem set 1 R (3) In and EF Ð = Ð E, Ð F = Ð and = 3E then which of the statements regarding the two triangles is true? ()The triangles are not congruent and not similar () The triangles are similar but not congruent. () The triangles are congruent and similar. () None of the statements above is true. (4) and EF are equilateral triangles, () (EF) = 1 If = 4 then what is length of E? () () 4 () 8 () 4 6 Fig E F R Fig Fig E Fig R E F F

28 (5) In figure 1.71, seg XY seg, then which of the following statements is true? () = X Y () X Y = Y X () X X = Y () Y = X. In, - and = 7, = 0 then find following ratios. (1) ( ) ( ) () ( ) ( ) (3) ( ) ( ) X Y Fig Fig Ratio of areas of two triangles with equal heights is 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle? 4. In figure 1.73, Ð = Ð = 90 = 6, = ( ) then ( ) =? Fig In figure 1.74, M = 10 cm (S) = 100 sq.cm M N S (RS) = 110 sq.cm then find NR. R Fig MNT ~ RS. Length of altitude drawn from point T is 5 and length of altitude ( MNT) drawn from point S is 9. Find the ratio ( RS). 7

29 7. In figure 1.75, and E seg E side If = 5, = 3, = 6.4 then find E. 5 x E Fig x S R 8. In the figure 1.76, seg, seg, seg R and seg S are perpendicular to line. = 60, = 70, = 80, S = 80 then find, R and RS. Fig In R seg M is a median. ngle bisectors of ÐM and ÐMR intersect side and side R in points X and Y respectively. rove that XY R. omplete the proof by filling in the boxes. In M, ray MX is bisector of ÐM. X O O x x M Fig Y R \ =... (I) theorem of angle bisector. In MR, ray MY is bisector of ÐMR. \ =... (II) theorem of angle bisector. M ut M = M MR... M is the midpoint R, hence M = MR. \ X X = Y YR \ XY R... converse of basic proportionality theorem. 8

30 10. In fig 1.78, bisectors of Ð and Ð of intersect each other in point X. Line X intersects side in point Y. = 5, = 4, = 6 then find X XY. Fig In fig 1.80, XY seg. 9 Y Fig In c, seg seg. iagonal and diagonal intersect each other in point. Then show that = If X = 3X and XY = 9. omplete the activity to find the value of. X ctivity : X = 3X \ X X = Y Fig X +X + =... by componendo. X X =... (I) ~ YX... test of similarity. \ X =... corresponding sides of similar triangles. XY \ = \ =...from (I) 9 G F 13 «. In figure1.81, the vertices of square EFG are on the sides of. Ð = 90. Then prove that E = E (Hint Show that G is similar to FE. Use G = FE = E.) Fig X rrr E

31 ythagoras Theorem Let s study. ythagorean triplet Similarity and right angled triangles Theorem of geometric mean ythagoras theorem pplication of ythagoras theorem pollonius theorem Let s recall. ythagoras theorem : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaning two sides. R In R Ð R = 90 l(r) = l() + l(r) We will write this as, Fig..1 R = + R The lengths, R and R of R can also be shown by letters r, p and q. With this convention, refering to figure.1, ythagoras theorem can also be stated as q = p + r. ythagorean Triplet : In a triplet of natural numbers, if the square of the largest number is equal to the sum of the squares of the remaining two numbers then the triplet is called ythagorean triplet. For Example: In the triplet ( 11, 60, 61 ), 11 = 11, 60 = 3600, 61 = 371 and = 371 The square of the largest number is equal to the sum of the squares of the other two numbers. \ 11, 60, 61 is a ythagorean triplet. Verify that (3, 4, 5), (5, 1, 13), (8, 15, 17), (4, 5, 7) are ythagorean triplets. Numbers in ythagorean triplet can be written in any order. 30

32 For more information Formula for ythagorean triplet: If a, b, c are natural numbers and a > b, then [(a + b ),(a - b ),(ab)] is ythagorean triplet. (a + b ) = a 4 + a b + b 4... (I) (a - b ) = a 4 - a b + b 4... (II) (ab) = 4a b... (III) \by (I), (II) and (III), (a + b ) = (a - b ) + (ab) \[(a + b ), (a - b ), (ab)] is ythagorean Triplet. This formula can be used to get various ythagorean triplets. For example, if we take a = 5 and b = 3, a + b = 34, a - b = 16, ab = 30. heck that (34, 16, 30) is a ythagorean triplet. ssign different values to a and b and obtain 5 ythagorean triplet. \ Last year we have studied the properties of right angled triangle with the angles and (I)roperty of triangle. If acute angles of a right angled triangle are 30 and 60, then the side opposite 3 30 angle is half of the hypotenuse and the side opposite to 60 angle is times the hypotenuse. See figure.. In LMN, Ð L = 30, Ð N = 60, Ð M = 90 L \side opposite 30 angle = MN = 1 LN 30 3 side opposite 60 angle = LM = LN If LN = 6 cm, we will find MN and LM. MN = 1 LN LM = 3 LN M Fig.. N = 1 6 = 3 6 = 3 cm = 3 3 cm 31

33 (II ) roperty of If the acute angles of a right angled triangle are 45 and 45, then each of the perpendicular sides is Z 45 1 times the hypotenuse. See Figure.3. In XYZ, XY = XZ = \ XY = XZ = 1 ZY 1 ZY 1 ZY If ZY = 3 cm then we will find XY and 45 ZX X Y 1 Fig..3 XY = XZ = 3 XY = XZ = 3cm In 7 th standard we have studied theorem of ythagoras using areas of four right angled triangles and a square. We can prove the theorem by an alternative method. ctivity: Take two congruent right angled triangles. Take another isosceles right angled triangle whose congruent sides are equal to the hypotenuse of the two congruent right angled triangles. Join these triangles to form a trapezium rea of the trapezium = 1 (sum of the lengths of parallel sides) height Using this formula, equating the area of trapezium with the sum of areas of the three right angled triangles we can prove the theorem of ythagoras. x z z y y Fig..4 3 x

34 Theorem : In a right angled triangle, if the altitude is drawn to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other. roof : In and Ð...(common angle) Ð... (each 90 ) ~... ( test)... (I) Let s learn. Now we will give the proof of ythagoras theorem based on properties of similar triangles. For this, we will study right angled similar triangles. Similarity and right angled triangle Given : In, Ð = 90, seg ^ seg, -- To prove : ~ ~ ~ 33 Fig..5 In and Ð...(common angle) Ð... (each 90 ) ~... ( test)... (II) \ ~ from (I) and (II)...(III) \ from (I), (II) and (III), ~ ~...(transitivity) Theorem of geometric mean In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided. R roof In right angled triangle R, seg S ^ hypotenuse R SR ~ S... ( similarity of right triangles ) S S S S SR S SR S S = S SR \ seg S is the geometric mean of seg S and SR. S Fig..6

35 ythagoras Theorem In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides. Given : In, Ð = 90 Fig..7 To prove : = + onstruction :raw perpendicular seg on side. --. roof : In right angled, seg ^ hypotenuse... (construction) \ ~ ~... (similarity of right angled triangles) ~ Similarly, ~ = = - corresponding = = -corresponding sides sides = =... (I) dding (I) and (II) + = + = ( + ) =... (--) \ + = \ = + onverse of ythagoras theorem = =... (II) In a triangle if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle. Given In, = + To prove Ð = 90 Fig Fig..9 R

36 onstruction : raw R such that, =, = R, Ð R = 90. roof : In R, Ð = 90 R = + R... (ythagoras theorem) = +... (construction)...(i) =... (given)...(ii) \ R = \ R =... (III) R... (SSS test) \ Ð = Ð R = 90 Remember this! (1) (a) Similarity and right angled triangle In R Ð = 90, seg S ^ seg R, R ~ S ~ SR. Thus all the S right angled triangles in the figure are similar to one another. R Fig..10 (b) Theorem of geometric mean In the above figure, S ~ SR \ S = S SR \ seg S is the geometric mean of seg S and seg SR () ythagoras Theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides. (3) onverse of ythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle (4) Let us remember one more very useful property. In a right angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is 30. This property is the converse of theorem. 35

37 Solved Examples Ex. (1) See fig.11. In, Ð = 90, Ð = 30, =14, then find and Solution : In, Fig..11 Ð = 90, Ð = 30, \ Ð = 60 y theorem, = 1 = 3 = 1 14 = 3 14 = 7 = 7 3 Ex. () See fig.1, In, seg ^ seg, Ð = 45, = 5 and = 8 then find and. Solution : In 5 Fig Ð = 90, Ð = 45, \ Ð = 45 = = 8... by theorem = 8 \ = 8 = + = = 13 Ex. (3) In fig.13, Ð R = 90,seg N ^ seg R, N = 9, NR = 16. Find N. Solution : In R, seg N ^ seg R N = N NR... theorem of geometric 9 mean N \N = N NR 16 = 9 16 = 3 4 R = 1 Fig

38 Ex. (4) See figure.14.in R, Ð R = 90, seg S ^ seg R then find x, y, z. Solution : In R, Ð R = 90, seg S ^ seg R S = S SR... (theorem of geometric mean) = 10 8 = 5 8 z 10 = 5 16 = 4 5 \ x = 4 5 In SR, by ythagoras theorem R = S + SR ( ) + 8 = 4 5 = = = 144 \ R = 1 x y Fig..14 S 8 In S, by ythagoras theorem = S + S = 4 5 ( ) + 10 = = = 180 = 36 5 \ = 6 5 R Hence x = 4 5, y = 1, z = 6 5 Ex. (5) In the right angled triangle, sides making right angle are 9 cm and 1 cm. Find the length of the hypotenuse Solution: In R,Ð = 90 R = + R (ythagoras theorem) = = R = 5 R R = 15 1 Hypotenuse= 15 cm Fig

39 Ex. (6) In LMN,l = 5, m = 13, n = 1. State whether LMN is a right angled triangle or not. Solution : l = 5, m = 13, n = 1 l = 5, m = 169, n = 144 \ m = l + n \ by converse of ythagoras theorem LMN is a right angled triangle. Ex. (7) See fig.16. In, seg ^ seg. rove that: + = + Solution : ccording to ythagoras theorem, in = + \ = -... (I) In = + \ = -... (II) \ - = -...from I and II \ + = + Fig..16 ractice set.1 1. Identify, with reason, which of the following are ythagorean triplets. (i)(3, 5, 4) (ii)(4, 9, 1) (iii)(5, 1, 13) (iv) (4, 70, 74) (v)(10, 4, 7) (vi)(11, 60, 61) M. In figure.17, Ð MN = 90, seg N ^seg M, M = 9, = 4, find N M Fig..18 R 38 N Fig In figure.18, Ð R = 90, seg M ^ seg R and -M-R, M = 10, M = 8, find R.

40 4. See figure.19. Find R and S using the information given in SR. S 6 R 30 Fig In figure.1, Ð FE = 90, FG ^ E, If G = 8, FG = 1, find (1) EG () F and (3) EF 8 Fig For finding and with the help of information given in figure.0, complete following activity. =... \ Ð = \ = = 6. Find the side and perimeter of a square whose diagonal is 10 cm. R E = 8 = = Fig..1 M Fig.. G Find the diagonal of a rectangle whose length is 35 cm and breadth is 1 cm. 9 «. In the figure., M is the midpoint of R. Ð R = 90. rove that, = 4M - 3R 10 «. Walls of two buildings on either side of a street are parellel to each other. ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4. m. Find the width of the street. F 39

41 Let s learn. pplication of ythagoras theorem In ythagoras theorem, the relation between hypotenuse and sides making right angle i.e. the relation between side opposite to right angle and the remaining two sides is given. In a triangle, relation between the side opposite to acute angle and remaining two sides and relation of the side opposite to obtuse angle with remaining two sides can be determined with the help of ythagoras theorem. Study these relations from the following examples. Ex. (1) In, Ð is an acute angle, seg ^ seg. rove that: = + - In the given figure let = c, = b, = p, = a, = x, \ = a - x In, by ythagoras theorem c b c = (a-x) + p c = a - ax + x +... (I) In, by ythagoras theorem a - x x b = p + Fig..3 p = b -... (II) Substituting value of p from (II) in (I), c = a - ax + x + b - x \ c = a + b - ax \ = + - Ex. () In, Ð is obtuse angle, seg ^ seg. rove that: = + + In the figure seg ^ seg Let = p, = b, = c, c p = a and = x. b = a + x In, by ythagoras theorem, x a c = (a + x) + p Fig..4 c = a + ax + x + p... (I) 40

42 Similarly, in b = x + p \ p = b - x... (II) \ substituting the value of p from (II) in (I) \ c = a + ax + b \ = + + pollonius theorem In, if M is the midpoint of side, then + = M + M M Fig Given In, M is the midpoint of side. To prove + = M + M onstruction raw seg ^ seg roof If seg M is not perpendicular to seg then out of Ð M and Ð M one is obtuse angle and the other is acute angle In the figure, Ð M is obtuse angle and Ð M is acute angle. From examples (1) and () above, = M + M + M M... (I) and = M + M - M M \ = M + M - M M ( M = M)...(II) \ adding (I) and (II) + = M + M Write the proof yourself if seg M ^ seg. From this example we can see the relation among the sides and medians of a triangle. This is known as pollonius theorem. Solved Examples Ex. (1) In the figure.6, seg M is a median of R. M = 9 and + R = 90, then find R. Solution In R, seg M is a median. M is the midpoint of seg R. \

43 9 M Fig..6 R M = MR = 1 R + R = M + M (by pollonius theorem) 90 = 9 + M 90 = 81 + M 90 = 16 + M M = M = 18 M = 64 M = 8 \ R = M = 8 = 16 Ex. () rove that, the sum of the squares of the diagonals of a rhombus is equal to the sum of the squares of the sides. S Given : c RS is a rhombus. iagonals R and S intersect each other at T point T To prove : S + SR + R + = R + S R Fig..7 roof : iagonals of a rhombus bisect each other. \ by pollonius theorem, + S = T + T... (I) R + SR = RT + T... (II) \ adding (I) and (II), + S + R + SR = (T + RT ) + 4T = (T + T ) + 4T... (RT = T) = 4T + 4T = (T) + (T) = R + S (The above proof can be written using ythagoras theorem also.) 4

44 1. In R, point S is the midpoint of side R.If = 11,R = 17, S =13, find R.. In, = 10, = 7, = 9 then find the length of the median drawn from point to side 3. In the figure.8 seg S is the median of R and T ^ R. rove that, (1) R = S + R ST + ii) = S - R ST + R R 43 T S R Fig In, point M is the midpoint of side. If, + = 90 cm, M = 8 cm, find. M Fig..9 5 «. In figure.30, point T is in the interior of rectangle RS, rove that, TS + T = T + TR T (s shown in the figure, draw seg side SR and -T-) S Fig..30 R roblem set 1. Some questions and their alternative answers are given. Select the correct alternative. ractice set. (1) Out of the following which is the ythagorean triplet? () (1, 5, 10) () (3, 4, 5) () (,, ) () (5, 5, ) () In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse? () 15 () 13 () 5 () 1

45 (3) Out of the dates given below which date constitutes a ythagorean triplet? () 15/08/17 () 16/08/16 () 3/5/17 () 4/9/15 (4) If a, b, c are sides of a triangle and a + b = c, name the type of triangle. () Obtuse angled triangle () cute angled triangle () Right angled triangle () Equilateral triangle (5) Find perimeter of a square if its diagonal is 10 cm. ()10 cm () 40 cm () 0 cm () 40 cm (6) ltitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude. () 9 cm () 4 cm () 6 cm () 6 cm (7) Height and base of a right angled triangle are 4 cm and 18 cm find the length of its hypotenus () 4 cm () 30 cm () 15 cm () 18 cm (8) In, = 6 3 cm, = 1 cm, = 6 cm. Find measure of Ð. () 30 () 60 () 90 () 45. Solve the following examples. (1) Find the height of an equilateral triangle having side a. () o sides 7 cm, 4 cm, 5 cm form a right angled triangle? Give reason. (3) Find the length a diagonal of a rectangle having sides 11 cm and 60cm. (4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 1 cm. (5) side of an isosceles right angled triangle is x.find its hypotenuse. (6) In R; = 8, R = 5, R = 3. Is R a right angled triangle? If yes, which angle is of 90? 3. In RST, Ð S = 90, Ð T = 30, RT = 1 cm then find RS and ST. 4. Find the diagonal of a rectangle whose length is 16 cm and area is 19 sq.cm. 5 «. Find the length of the side and perimeter of an equilateral triangle whose height is 3 cm. 6. In seg is a median. If = 18, + = 60 Find. 44

46 7 «. is an equilateral triangle. oint is on base such that = 1 3, if = 6 cm find. 8. From the information given in the figure.31, prove that a a M = N = 3 a M a S a R a N Fig rove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. 10. ranali and rasad started walking to the East and to the North respectively, from the same point and at the same speed. fter hours distance between them was 15 km. Find their speed per hour. 11 «. In, Ð = 90, seg L and seg M are medians of. Then prove that M 4(L + M ) = 5 L Fig Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is14 cm. Find the length of the other diagonal. 13. In, seg ^ seg = 3. rove that = + Fig «. In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid. 45

47 15. In a trapezium, seg seg seg ^ seg, seg ^ seg, If = 15, = 15 and = 5. Find (c ) Fig «. In the figure.35, R is an equilatral triangle. oint S is on seg R such that S = 1 R. 3 rove that : 9 S = S T Fig «. Seg M is a median of R. If = 40, R = 4 and M = 9, find R. 18. Seg M is a median of. If =, = 34, = 4, find M R IT Tools or Links Obtain information on the life of ythagoras from the internet. repare a slide show. rrr 46

48 3 ircle Let s study. ircles passing through one, two,three points Secant and tangent ircles touching each other rc of a circle Inscribed angle and intercepted arc yclic quadrilateral Secant tangent angle theorem Theorem of intersecting chords Let s recall. You are familiar with the concepts regarding circle, like - centre, radius, diameter, chord, interior and exterior of a circle. lso recall the meanings of - congruent circles, concentric circles and intersecting circles. congruent circles concentric circles intersecting circles Recall the properties of chord studied in previous standard and perform the activity below. ctivity I : In the adjoining figure, seg E is a chord of a circle with centre. seg F ^ seg E. If diameter of the F E circle is 0 cm, E =16 cm find F. Fig. 3.1 Recall and write theorems and properties which are useful to find the solution of the above problem. (1) The perpendicular drawn from centre to a chord () (3) Using these properties, solve the above problem. 47

49 (1) () ctivity II : In the adjoining figure, seg R is a chord of the circle with O R centre O. is the midpoint of the chord R. If R = 4, O = 10, find radius of the circle. Fig. 3. To find solution of the problem, write the theorems that are useful. Using these theorems solve the problems. ctivity III : In the adjoining figure, M is the centre of the circle and S seg is a diameter. M seg MS ^ chord seg MT ^ chord T Ð. rove that : chord. Fig. 3.3 To solve this problem which of the following theorems will you use? (1) The chords which are equidistant from the centre are equal in length. () ongruent chords of a circle are equidistant from the centre. Which of the following tests of congruence of triangles will be useful? (1) SS, () S, (3) SSS, (4) S, (5) hypotenuse-side test. Using appropriate test and theorem write the proof of the above example. Let s learn. ircles passing through one, two, three points In the adjoining figure, point lies in a plane. ll the three circles with centres,, R pass through point. How many more such circles may pass through point? R correct. If your answer is many or innumerable, it is Infinite number of circles pass through a point. Fig

50 In the adjoining figure, how many circles pass through points and? How many circles contain all the three points,,? erform the activity given below and try to Fig. 3.5 find the answer. ctivityi: raw segment. raw perpendicular bisector l of the l segment. Take point on the line l as centre, as radius and draw a circle. Observe that the circle passes through point also. Find the reason. (Recall the property of perpendicular Fig. 3.6 bisector of a segment.) Taking any other point on the line l, if a circle is drawn with centre and radius, will it pass through? Think. How many such circles can be drawn, passing through and? Where will their centres lie? ctivity II : Take any three non-collinear points. What should be done to draw a circle passing through all these points? raw a circle passing through these points. Is it possible to draw one more circle passing through these three Fig. 3.7 points? Think of it. ctivity III : Take 3 collinear points, E, F. Try to draw a circle passing through these points. If you are not able to draw a circle, think of the reason. Let s recall. (1) Infinite circles pass through one point. () Infinite circles pass through two distinct points. (3) There is a unique circle passing through three non-collinear points. (4) No circle can pass through 3 collinear points. 49

51 Let s learn. Secant and tangent l m n Fig. 3.8 In the figure above, not a single point is common in line l and circle with centre. Two points and R are common to both, the line n and the circle with centre. and R are intersecting points of line n and the circle. Line n is called a secant of the circle. Let us understand an important property of a tangent from the following activity. raw a sufficiently large circle with centre O. raw radius O. raw a line ^ seg O. It intersects the circle at O points,. Imagine the line slides towards point such that all the time it remains parallel to its original position. Obviously, while the line slides, points and approach each other along the Fig. 3.9 circle. t the end, they get merged in point, but the angle between the radius O and line will remain a right angle. t this stage the line becomes a tangent of the circle at. So it is clear that, the tangent at any point of a circle is perpendicular to the radius at that point. This property is known as tangent theorem. oint is common to both, line m and circle with centre. Here, line m is called a ctivity : tangent of the circle and point is called the point of contact. 50 R

52 Tangent theorem Theorem : tangent at any point of a circle is perpendicular to the radius at the point of contact. There is an indirect proof of this theorem. For more information Given : Line l is a tangent to the circle with centre O at the point of contact. To prove : line l ^ radius O. roof : ssume that, line l is not l perpendicular to seg O. Suppose, seg O is drawn perpendicular to line l. Of course is not same as. Now take a point on line l such that -- and =. O Fig Now in, O and O seg... (construction) Ð Ð O... (each right angle) seg seg O \ O... (SS test) \ O = O ut seg O is a radius. \seg O must also be radius. \ lies on the circle. That means line l intersects the circle in two distinct points and. ut line l is a tangent.... (given) \ it intersects the circle in only one point. Fig Our assumption that line l is not perpendicular to radius O is wrong. \ line l ^ radius O. O l 51

53 Let s recall. Which theorems do we use in proving that hypotenuse is the longest side of a right angled triangle? Fig. 3.1 Let s learn. onverse of tangent theorem Theorem: line perpendicular to a radius at its point on the circle is a tangent to the circle. Given : M is the centre of a circle M seg MN is a radius. Line l ^ seg MN at N. N l Fig To prove : Line l is a tangent to the circle. roof : Take any point, other that N, on the line l. raw seg M. Now in MN, Ð N is a right angle. \ seg M is the hypotenuse. \ seg M > seg MN. s seg MN is radius, point can t be on the circle. \ no other point, except point N, of line l is on the circle. \ line l intersects the circle in only one point N. \ line l is a tangent to the circle. Let s discuss. In figure 3.14, is a point on the circle with centre. The tangent of the circle passing through is to be drawn. There are infinite lines passing through the point. Which of them will be the tangent? an the number of tangents through be more than one? Fig

54 oint lies in the interior of the circle. an you draw tangents to the circle through? oint is in the exterior of the circle. an you draw a tangent to the circle through? If yes, how many such tangents are possible? From the discussion you must have understood that two tangents can be drawn to a circle from the point outside the circle as shown in the figure. In the adjoing figure line and line. touch the circle at points and. Seg Fig and seg are called tangent segments. Tangent segment theorem Theorem Tangent segments drawn from an external point to a circle are congruent. Observe the adjoining figure. Write given and to prove. raw radius and radius and complete the following proof of the theorem. roof : In and, radii of the same circle. seg Ð = Ð = tangent theorem \ seg Fig Solved Examples Ex. (1) In the adjoining figure circle with centre touches the sides of Ð at and. If Ð = 5, find measure of Ð. Solution : The sum of all angles of a quadrilateral is 360. \ Ð + Ð + Ð + Ð = 360 Fig \ Ð = Tangent theorem \ Ð + 3 = 360 \ Ð = = 18 53

55 Eg. () oint O is the centre of a circle. Line a and line b are parallel tangents to the circle at and. rove that segment is a diameter of the circle. Solution raw a line c through O which is parallel to line a. raw radii O and O. Now, Ð OT = Tangent theorem \ÐSO = Int. angle property... (I) line a line c... construction line a line b... given \ line b line c \ÐSO = Int. angle property...(ii) \ From (I) and (II), Ð SO + Ð SO = = 180 \ ray O and ray O are opposite rays. \, O, are collinear points. \ seg is a diameter of the circle. When a motor cycle runs on a wet road in rainy season, you may have seen water splashing from its wheels. Those splashes are like tangents of the circle of the wheel. Find out the reason from your science teacher. Observe the splinters escaping from a splintering wheel in iwali fire works and while sharpning a knife. o they also look like tangents? O Fig T S R irection of running wheel b a c Remember this! (1) Tangent theorem : The tangent at any point of a circle is perpendicular to the radius through the point of contact. () line perpendicular to a radius at its point on the circle, is a tangent to the circle. (3) Tangent segments drawn from an external point to a circle are congruent. 54

56 ractice set In the adjoining figure the radius of a circle with centre is 6 cm, line is a tangent at. nswer the following questions. (1) What is the measure of Ð? Why? () What is the distance of point from line? Why? (3) d(,)= 6 cm, find d(,). (4) What is the measure of Ð? Why? O M N Fig. 3.0 R Fig In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then (1) What is the length of each tangent segment? () What is the measure of ÐMRO? (3) What is the measure of Ð MRN? M 3. Seg RM and seg RN are tangent segments of a circle with centre O. rove that seg OR bisects ÐMRN O R as well as ÐMON. N Fig What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer. IT Tools or Links With the help of Geogebra software, draw a circle and its tangents from a point in its exterior. heck that the tangent segments are congrurent. 55

57 Let s learn. Touching circles ctivity I : Take three collinear points X-Y-Z as shown in figure 3.. raw a circle with centre X and radius XY. raw another circle with centre Z and radius YZ. Note that both the circles intersect each other at the single point Y. raw a line through point Y and perpendicular to seg XZ. Note that this line is a common tangent of the two circles. ctivity II : Take points Y-X-Z as shown in the figure 3.3. raw a circle with centre Z and radius ZY. lso draw a circle with centre X and radius XY. Note that both the circles intersect each other at the point Y. raw a line perpendicular to seg YZ through point Y, that is the common tangent for the circles. Y X Y Z Fig. 3. X Z Fig. 3.3 You must have understood, the circles in both the figures are coplaner and intersect at one point only. Such circles are said to be circles touching each other. Touching circles can be defined as follows. If two circles in the same plane intersect with a line in the plain in only one point, they are said to be touching circles and the line is their common tangent. The point common to the circles and the line is called their common point of contact. 56

58 l p R T S K N M Fig. 3.4 Fig. 3.5 In figure 3.4, the circles with centres R and S touch the line l in point T. So they are two touching circles with l as common tangent. They are touching externally. In figure 3.5 the circles with centres M, N touch each other internally and line p is their common tangent. Let s think. (1) The circles shown in figure 3.4 are called externally touching circles. why? () The circles shown in figure 3.5 are called internally touching circles. why? (3) In figure 3.6, the radii of the circles with centers and are 3 cm and 4 cm respectively. Find - (i) d(,) in figure 3.6 (a) (ii) d(,) in figure 3.6 (b) Theorem of touching circles Theorem : If two circles touch each other, their point of contact lies on the line joining their centres. l l (a) (b) Fig

59 Given : is the point of contact of the two circles with centers,. To prove : oint lies on the line. roof : Let line l be the common tangent passing through, of the two touching circles. line l ^ seg, line l ^ seg. \ seg ^ line l and seg ^ line l. Through, only one line perpendicular to line l can be drawn. \ points,, are collinear. E Remember this! (1) The point of contact of the touching circles lies on the line joining their centres. () If the circles touch each other externally, distance between their centres is equal to the sum of their radii. (3) The distance between the centres of the circles touching internally is equal to the difference of their radii. ractice set Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.. Two circles of radii 5.5 cm and 4. cm touch each other externally. Find the distance between their centres. 3. If radii of two circles are 4 cm and.8 cm. raw figure of these circles touching each other - (i) externally (ii) internally. 4. In fig 3.7, the circles with centres and touch each other at R. line passing through R meets the circles at and respectively. rove that - (1) seg seg, () R ~ R, and (3) Find Ð R if Ð R = 35 l Fig R Fig «. In fig 3.8 the circles with centres and touch each other at E. Line l is a common tangent which touches the circles at and respectively. Find the length of seg if the radii of the circles are 4 cm, 6 cm.

60 Let s recall. rc of a circle Y secant divides a circle in two parts. ny one of these two parts and the common points of the circle and the k secant constitute an arc of the circle. The points of intersection of circle X Fig. 3.9 and secant are called the end points of the arcs. In figure 3.9, due to secant k we get two arcs of the circle with centre - arc Y, arc X. If the centre of a circle is on one side of the secant then the arc on the side of the centre is called major arc and the arc which is on the other side of the centre is called minor arc. In the figure 3.9 arc Y is a major arc and arc X is a minor arc. If there is no confusion then the name of a minor arc is written using its end points only. For example, the arc X in figure 3.9, is written as arc. Here after, we are going to use the same convention for writing the names of arcs. entral angle Major arc O q Minor arc Fig When the vertex of an angle is the centre of a circle, it is called a central angle. In the figure 3.30, O is the centre of a circle and Ð O is a central angle. Like secant, a central angle also divides a circle into two arcs. Measure of an arc To compare two arcs, we need to know their measures. Measure of an arc is defined as follows. 59

61 (1) Measure of a minor arc is equal to the measure of its corresponding central angle. In figure 3.30 measure of central Ð O is q. \ measure of minor arc is also q. () Measure of major arc = measure of corresponding minor arc. In figure 3.30 measure of major arc = measure of minor arc = q (3) Measure of a semi circular arc, that is of a semi circle is 180. (4) Measure of a complete circle is 360. ongruence of arcs Let s learn. When two coplanar figures coincide with each other, they are called congruent figures. We know that two angles of equal measure are congruent. Similarily, are two arcs of the same measure congruent? Find the answer of the question by doing the following activity. ctivity : raw two circles with centre, as shown in the figure. raw Ð E, Ð FG of G the same measure and Ð IJ of different measure. I rms of Ð E intersect inner circle 70 F at and. 40 o you notice that the measures of J 70 arcs and E are the same? o they E coincide? No, definitely not. Now cut and seperate the sectors Fig E; -FG and -IJ. heck whether the arc E, arc FG and arc IJ coincide with each other. id you notice that equality of measures of two arcs is not enough to make the two arcs congruent? Which additional condition do you think is neccessary to make the two arcs congruent? From the above activity - Two arcs are congruent if their measures and radii are equal. rc E and arc GF are congruent is written in symbol as arc arc GF. 60

62 roperty of sum of measures of arcs E Fig. 3.3 ut arc and arc E have many points in common. [ll points on arc.] So m(arc E) ¹ m(arc ) + m(arc E). 61 In figure 3.3, the points,,,, E are concyclic. With these points many arcs are formed. There is one and only one common point to arc and arc E. So measure of arc E is the sum of measures of arc and arc E. m(arc ) + m(arc E) = m(arc E) Theorem : The chords corresponding to congruent arcs of a circle ( or congruent circles) are congruent. Given : In a circle with centre arc E To rove : chord E E roof : (Fill in the blanks and complete the proof.) In and E, side (...) side...(...) Ð E measures of congruent Fig arcs E (...) \ chord E (...) Theorem : orresponding arcs of congruent chords of a circle (or congruent circles) are congruent. Given : O is the centre of a circle chord RS. S To prove : rc arc RNS O N roof : onsider the following statements and write the proof. R Two arcs are congruent if their measures M and radii are equal. rc M and arc RNS are arcs of the same circle, hence Fig have equal radii.

63 Their measures are same as the measures of their central angles. To obtain central angles we have to draw radii O, O, OR, OS. an you show that O and ORS are congruent? rove the above two theorems for congruent circles. Let s think. While proving the first theorem of the two, we assume that the minor arc and minor arc E are congruent. an you prove the same theorem by assuming that corresponding major arcs congruent? In the second theorem, are the major arcs corresponding to congruent chords congruent? Is the theorem true, when the chord and chord RS are diameters of the circle? Solved Examples Ex. (1),, are any points on the circle with centre O. (i) Write the names of all arcs formed due to these points. (ii) If m arc () = 110 and m arc () = 15, find measures of all remaining arcs. Solution (i) Names of arcs - arc, arc, arc, arc, arc, arc (ii) m(arc ) = m(arc ) + m(arc ) = = 35 m (arc ) = m (arc ) = = 15 Similarly, m(arc ) = = 35 and m(arc ) = = 50 O Fig

64 Ex. () In the figure 3.36 a rectangle RS is inscribed in a circle with centre T. rove that, (i) arc SR (ii) arc arc R Solution : (i) c RS in a rectangle. \ chord SR... opposite sides T of a rectangle S \ arc SR... arcs corresponding R to congruent chords. (ii) chord chord R... Opposite sides of Fig a rectangle \ arc arc R... arcs corresponding to congruent chords. \ measures of arcs S and R are equal Now, m(arc S) + m(arc ) = m(arc ) + m(arc R) \ m(arc S) = m(arc R) \ arc arc R Remember this! (1) n angle whose vertex is the centre of a circle is called a central angle. () efinition of measure of an arc - (i) The measure of a minor arc is the measure of its central angle. (ii) Measure of a major arc = measure of its corresponding minor arc. (iii) measure of a semicircle is 180. (3) When two arcs are of the same radius and same measure, they are congruent. (4) When only one point is common to arc, and arc E of the same circle, m(arc ) + m(arc E) = m(arc E) (5) hords of the same or congruent circles are equal if the related arcs are congruent. (6) rcs of the same or congruent circles are equal if the related chords are congruent. ractice set In figure 3.37, points G,, E, F are concyclic points of a circle with centre. Ð EF = 70, m(arc GF) = 00 find m(arc E) and m(arc EF). 63 E F G Fig. 3.37

65 «. In fig 3.38 RS is an equilateral R S triangle. rove that, (1) arc arc arc R () m(arc RS) = 40. Fig In fig 3.39 chord, rove that, arc Fig Let s learn. We have learnt some properties relating to a circle and points as well as lines (tangents). Now let us learn some properties regarding circle and angles with the help of some activities. ctivity I : raw a sufficiently large circle of any radius as shown in the figure raw a chord and central Ð. Take any point F on the major arc and point E on the minor arc. (1) Measure Ð and Ð and compare H G the measures. () Measure Ð and Ð E. dd the measures. I E Fig

66 (3) Take points F, G, H on the arc. Measure Ð F, Ð G, Ð H. ompare these measures with each other as well as with measure of Ð. (4) Take any point I on the arc E. Measure Ð I and compare it with Ð E. From the activity you must have noticed- (1) The measure Ð is twice the measure of Ð. () The sum of the measures of Ð and Ð E is180. (3) The angles Ð H, Ð, Ð F and Ð G are of equal measure. (4) The measure of Ð E and Ð I are equal. ctivity II : R raw a sufficiently large circle with centre as S shown in the figure raw any diameter. Now take points R,S,T on both the semicircles. Measure Ð R, Ð S, Ð T. Note that each is a right angle. T Fig The properties you saw in the above activities are theorems that give relations between circle and angles. Let us learn some definitions required to prove the theorems. Inscribed angle In figure 3.4, is the centre of a circle. The vertex, of Ð lies on the circle. The arms of Ð intersect the circle at and. Such an angle is called an angle inscribed in the circle or in the arc. In figure 3.4, Ð is inscribed in the arc. Fig

67 Intercepted arc Observe all figures (i) to (vi) in the following figure (i) (ii) (iii) (iv) (v) (vi) Fig In each figure, the arc of a circle that lies in the interior of the Ð is an arc intercepted by the Ð. The points of intersection of the circle and the angle are end points of that intercepted arc. Each side of the angle has to contain an end point of the arc. In figures 3.43 (i), (ii) and (iii) only one arc is intercepted by that angle; and in (iv), (v) and (vi), two arcs are intercepted by the angle. lso note that, only one side of the angle touches the circle in (ii) and (v), but in (vi) both sides of the angle touch the circle. In figure 3.44, the arc is not intercepted arc, as arm does not contain any end point of the arc. 66 Fig Inscribed angle theorem The measure of an inscribed angle is half of the measure of the arc intercepted by it. Given In a circle with centre O, x Ð is inscribed in arc. rc is intercepted by the O x x angle. To prove Ð = 1 m(arc ) x onstruction raw ray O. It E intersects the circle at E. Fig raw radius O.

68 roof In O, side side O... radii of the same circle. \ Ð O = Ð O... theorem of isosceles triangle. Let Ð O = Ð O = x... (I) Now, Ð EO = Ð O + Ð O... exterior angle theorem of a triangle. = x + x = x ut Ð EO is a central angle. \ m(arc E) = x... definition of measure of an arc... (II) \ from (I) and (II). Ð O = Ð E = 1 m(arc E)... (III) Similarly, drawing seg O, we can prove Ð E = 1 m(arc E)... (IV) \Ð E + Ð E = 1 m(arc E) + 1 m(arc E)... from (III) and (IV) \ Ð = 1 [m(arc E) + m(arc E)] = 1 [m(arc E)] = 1 [m(arc )]... (V) Note that we have to consider three cases regarding the position of the centre of the circle and the inscribed angle. The centre of the circle lies (i) on one of the arms of the angle (ii) in the interior of the angle (iii) in the exterior of the angle. Out of these, first two are proved in (III) and (V). We will prove now the third one. In figure 3.46, Ð = Ð E - Ð E O = 1 m(arc E) - 1 m(arc E)... from (III) = 1 [m(arc E) - m(arc E)] Fig [m(arc )]... (VI) = 1 The above theorem can also be stated as follows. The measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre. The corollaries of the above theorem can also be stated in similar language. 67 E

69 orollaries of inscribed angle theorem 1. ngles inscribed in the same arc are congruent. Write given and to prove with the help of the figure S Think of the answers of the following questions and write the proof. (1) Which arc is intercepted by Ð R? R () Which arc is intercepted by Ð SR? (3) What is the relation between an T Fig inscribed angle and the arc intersepted by it?. ngle inscribed in a semicircle is a right angle. With the help of figure 3.48 write given, to prove and the proof. M yclic quadrilateral Fig X Fig If all vertices of a quadrilateral lie on the same circle then it is called a cyclic quadrilateral. Theorem of cyclic quadrilateral Theorem: Opposite angles of a cyclic quadrilateral are supplementry. Fill in the blanks and complete the following proof. Given c is cyclic. To prove Ð + Ð = + Ð = 180 roof rc is intercepted by the inscribed angle Ð. \ Ð = 1... (I) Similarly, is an inscribed angle. It intercepts arc. 68

70 \ = 1 m(arc )... (II) \ mð + = m(arc )... from (I) & (II) = 1 [ + m(arc )] = arc and arc constitute a complete circle. = Similarly we can prove, Ð + Ð =. orollary of cyclic quadrilateral theorem n exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. Write the proof of the theorem yourself. Let s think. In the above theorem, after proving Ð + Ð = 180, can you use another way to prove Ð + Ð = 180? onverse of cyclic quadrilateral theorem Theorem : If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic. Try to prove this theorem by indirect method. From the above converse, we know that if opposite angles of a quadrilateral are supplementary then there is a circumcircle for the quadrilateral. For every triangle there exists a circumcircle but there may not be a circumcircle for every quadrilateral. The above converse gives us the condition to ensure the existence of circumcircle of a quadrilateral. With one more condition four non-collinear points are concyclic. It is stated in the following theorem. 69

71 Theorem : If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic. Given : oints and lie on the same side x of the line. Ð x To prove : oints,,, are concyclic. Fig Let s think. (That is, c is cyclic.) This theorem can be proved by indirect method. The above theorem is converse of a certain theorem. State it. Solved Examples Ex. (1) In figure 3.51, chord chord LN L Ð L = 35 find (i) m(arc MN) 35 (ii) m(arc LN) Solution (i) Ð L = 1 m(arc MN)...inscribed angle theorem. M N \ 35 = 1 m(arc MN) Fig \ 35 = m(arc MN) = 70 (ii) m(arc MLN) = m(arc MN)... definition of measure of arc = = 90 Now, chord chord LN \ arc arc LN but m(arc LM) + m(arc LN) = m(arc MLN) = arc addition property m(arc LM) = m(arc LN) = 90 = 145 or, (ii) chord chord LN \ Ð M = Ð N... isosceles trinangle theorem. \ Ð M = = 145 \ Ð M =

72 Now, m(arc LN) = Ð M = 145 = 145 Ex. () In figure 3.5, chords and RS intersect at T. (i) Find m(arc S) if mð ST = 58, mð SR = 4. (ii) Verify, S Ð ST = 1 4 [m(arc R) + m(arc S)] 58 (iii) rove that : R T 1 ÐST = [m(arc R) + m(arc S)] for any measure of Ð ST. Fig. 3.5 (iv) Write in words the property in (iii). Solution (i) ÐS = ÐST = 58-4 = exterior angle theorem. m(arc S) = ÐS = 34 = 68 (ii) m(arc R) = ÐSR = 4 = 48 Now, 1 [m(arc R) + m(arc S)] = 1 [ ] = = 58 = Ð ST (iii) Fill in the blanks and complete the proof of the above property. ÐST = ÐS +... exterior angle theorem of a triangle = 1 m(arc S) +... inscribed angle theorem = 1 [ + ] (iv) If two chords of a circle intersect each other in the interior of a circle then the measure of the angle between them is half the sum of measures of arcs intercepted by the angle and its opposite angle. 71

73 Ex. (3) rove that, if two lines containing chords of a circle intersect each other outside the circle, then the measure of angle between them is half the difference in measures of the arcs intercepted by the angle. Fig E Given : hords and intersect at E in the exterior of the circle. To prove : Ð E = 1 [m(arc )-m(arc )] onstruction: raw seg. onsider angles of E and its exterior angle and write the proof. Remember this! (1) The measure of an inscribed angle is half the measure of the arc intercepted by it. () ngles inscribed in the same arc are congruent. (3) ngle inscribed in a semicircle is a right angle. (4) If all vertices of a quadrilateral lie on the same circle then the quadrilateral is called a cyclic quadrilateral. (5) Opposite angles of a cyclic quadrilateral are supplementary. (6) n exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. (7) If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic. (8) If two points on a given line subtend equal angles at two different points which lie on the same side of the line, then those four points are concyclic. (9) In figure 3.54, E (i) Ð E = [m(arc ) + m(arc )] 1 (ii) Ð E= 1 [m(arc ) + m(arc )] Fig

74 (10) In figure 3.55, Ð E = 1 [m(arc ) - m(arc )] E Fig In figure 3.56, in a circle with centre O, length of chord is equal to the radius of the circle. Find measure of each of the following. (1) Ð O ()Ð (3) arc (4) arc. S R Fig ractice set In figure 3.57, c RS is cyclic. side R. Ð SR = 110, Find- (1) measure of Ð R () m(arc R) (3) m(arc R) (4) measure of Ð R 3. c MRN is cyclic, Ð R = (5x - 13), Ð N = (4x + 4). Find measures of Ð R and Ð N. 4. In figure 3.58, seg RS is a diameter of the circle with centre O. oint T lies in the exterior of the circle. rove that Ð RTS is an acute angle. 5. rove that, any rectangle is a cyclic quadrilateral. R O Fig O Fig T S

75 X Z W T Fig Y 6. In figure 3.59, altitudes YZ and XT of WXY intersect at. rove that, (1) c WZT is cyclic. () oints X, Z, T, Y are concyclic. N 7. In figure 3.60, m(arc NS) = 15, m(arc EF) = 37, find the measure Ð NMS. M E F S E Fig In figure 3.61, chords and E intersect at. If Ð E = 108, m(arc E) = 95, find m(arc ). Fig Let s learn. ctivity : raw a circle as shown in figure 3.6. raw a chord. Take any point on the circle. raw inscribed Ð, measure it and note the measure. Fig. 3.6 Now as shown in figure 3.63, draw a tangent of the same circle, measure angle Ð and note the measure. Fig

76 You will find that Ð = Ð. You know that Ð = 1 m(arc ) From this we get Ð = 1 m(arc ). Now we will prove this property. Theorem of angle between tangent and secant If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc. M M x M x y F E (i) (ii) (iii) Fig Given Let Ð be an angle, where vertex lies on a circle with centre M. Its side touches the circle at and side intersects the circle at. rc is intercepted by Ð. To prove Ð = 1 m(arc ) roof onsider three cases. (1) In figure 3.64 (i), the centre M lies on the arm of Ð, Ð = Ð M = tangent theorem (I) arc is a semicircle. M \ m(arc ) = definition of measure of arc (II) From (I) and (II) Ð = 1 m(arc ) Fig. 3.64(i) () In figure 3.64 (ii) centre M lies in the exterior of Ð, raw radii M and M. Now, Ð M = Ð M... isosceles triangle theorem Ð M = tangent theorem... (I) 75

77 let Ð M = Ð M = x and Ð = y. Ð M = (x + x) = x Ð M = Ð M + Ð = x + y \ x + y = 90 \ x + y = 180 In M, x + Ð M = 180 \ x + y = x + Ð M \ y = Ð M \ y = Ð = 1 Ð M = 1 m(arc ) (3) With the help of fig 3.64 (iii), Fill in the blanks and write proof. Ray is the opposite ray of ray. M x x y Fig. 3.64(ii) Now, ÐE = 1 m( \ = 1 )... proved in (ii). m(arc F)... linear pair = 1 [360 - m( )] \ Ð = m(arc ) \ - Ð = - 1 m( ) \ Ð = 1 m(arc ) E lternative statement of the above theorem. Fig. 3.64(iii) In the figure 3.65, line is a secant and line is a tangent. The arc is intercepted by Ð. hord divides the circle in two parts. These are opposite arcs of each other. Now take any point T on the arc opposite to arc. T From the above theorem, Ð = 1 m (arc ) = Ð T. Fig \ the angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle. 76 F M

78 onverse of theorem of the angle between tangent and secant line is drawn from one end point of a chord of a circle and if the angle between the chord and the line is half the measure of the arc intercepted by that angle then that line is a tangent to the circle. In figure 3.66, U O S If Ð R = 1 m(arc S), [or Ð T = 1 m(arc U)] T R Fig then line TR is a tangent to the circle. This property is used in constructing a tangent to the given circle. n indirect proof of this converse can be given. Theorem of internal division of chords Suppose two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord. Given hords and of a circle with centre intersect at point E. E To prove E E = E E onstruction raw seg and seg. Fig roof In E and E, Ð Ð E... opposite angles Ð Ð E... angles inscribed in the same arc \ E ~ E... test \ E E E E \ E E = E E... corresponding sides of similar triangles Let s think. We proved the theorem by drawing seg and seg in figure 3.67, an the theorem be proved by drawing seg and seg, instead of seg and seg? 77

79 For more information In figure 3.67 point E divides the chord into segments E and E. E E is the area of a rectangle having sides E and E. Similarly E divides into segments E and E. E E is the area of a rectangle of sides E and E. We have proved that E E = E E. So the above theorem can be stated as, If two chords of a circle intersect in the interior of a circle then the area of the rectangle formed by the segments of one chord is equal to the area of similar rectangle formed by the other chord. Theorem of external division of chords If secants containing chords and of a circle intersect outside the circle in point E, then E E = E E. Write given and to prove with the help of the statement of the theorem and figure E onstruction raw seg and seg. Fig Fill in the blanks and complete the proof. roof In E and E, Ð common angle Ð Ð E...( ) \ E ~...( ) \ (E) =... corresponding sides of similar triangles \ = E E 78

80 Tangent secant segments theorem oint E is in the exterior of a circle. secant through E intersects the circle at points and, and a tangent through E touches the circle at point T, then E E = ET. Write given and to prove with reference to E T Fig the statement of the theorem. onstruction raw seg T and seg T. roof In ET and ET, Ð Ð TE... common angle Ð Ð ET... tangent secant theorem \ ET ~ ET... similarity \ ET E E ET \ E E = ET Remember this!... corresponding sides (1) In figure 3.70, E E = E E This property is known as theorem of chords intersecting inside the circle. Fig (3) In figure 3.7, E E = ET This property is known as tangent secant segments theorem. E 79 E Fig () In figure 3.71, E E = E E This property is known as theorem of chords intersecting outside the circle. E T Fig. 3.7

81 Solved Examples Ex. (1) In figure 3.73, seg S is a tangent segment. Line R is a secant. If = 3.6, R = 6.4, find S. Solution S = R... tangent secant segments S theorem = ( + R) = 3.6 [ ] = = 36 \S = 6 R Fig Ex. () In figure 3.74, chord MN and chord M RS intersect each other at point. S If R = 6, S = 4, MN = 11 N find N. Solution y theorem on interesecting chords, N M = R S... (I) R Fig let N = x. \M = 11 - x substituting the values in (I), x (11 - x) = 6 4 \ 11x - x - 4 = 0 \ x - 11x + 4 = 0 \ (x - 3) (x - 8) = 0 \ x - 3 = 0 or x - 8 = 0 \ x = 3 or x = 8 \ N = 3 or N = 8 80

82 Ex. (3) In figure 3.75, two circles intersect each other in points X and Y.Tangents drawn from a point M on line XY touch the circles at and. rove that, seg seg M. Solution Fill in the blanks and write proof. Fig Line MX is a common... of the two circles. \M = MY MX... (I) Similarly... =......, tangent secant segment theorem...(ii) \From (I) and (II)... = M \ M = M seg seg M Ex. (4) R T S O Fig Y M In figure 3.76, seg is a diameter of a circle with centre O. R is any point on the circle. seg RS ^ seg. rove that, SR is the geometric mean of S and S. [That is, SR = S S] Solution Write the proof with the help of the following steps. (1) raw ray RS. It intersects the circle at T. () Show that RS = TS. (3) Write a result using theorem of intersection of chords inside the circle. (4) Using RS = TS complete the proof. X Let s think. (1) In figure 3.76, if seg R and seg R are drawn, what is the nature of R? () Have you previously proved the property proved in example (4)? 81

83 ractice set In figure 3.77, ray touches the circle at point. = 1, R = 8, find S and RS. S R Fig R M S N Fig In figure 3.78, chord MN and chord RS intersect at point. (1) If R = 15, S = 4, M = 8 find N () If RS = 18, M = 9, N = 8 find S 3. In figure 3.79, O is the centre of the circle and is a point of contact. seg OE ^ seg, = 1, = 8, find (1) () (3) E. O E Fig T S Fig In figure 3.80, if = 6, R = 10, S = 8 find TS. R 5. In figure 3.81, seg EF is a diameter and seg F is a tangent segment. The radius of the circle is r. rove that, E GE = 4r E G H F Fig

84 roblem set 3 1. Four alternative answers for each of the following questions are given. hoose the correct alternative. (1) Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centers? () 4.4 cm () 8.8 cm (). cm () 8.8 or. cm () Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 1, what is the radius of each circle? () 6 cm () 1 cm () 4 cm () can t say (3) circle touches all sides of a parallelogram. So the parallelogram must be a,.... () rectangle () rhombus () square () trapezium (4) Length of a tangent segment drawn from a point which is at a distance 1.5 cm from the centre of a circle is 1 cm, find the diameter of the circle. () 5 cm () 4 cm () 7 cm () 14 cm (5) If two circles are touching externally, how many common tangents of them can be drawn? () One () Two () Three () Four (6) Ð is inscribed in arc of a circle with centre O. If Ð = 65, find m(arc ). () 65 () 130 () 95 () 30 (7) hords and of a circle intersect inside the circle at point E. If E = 5.6, E = 10, E = 8, find E. () 7 () 8 () 11. () 9 (8) In a cyclic c, twice the measure of Ð is thrice the measure of Ð. Find the measure of Ð? () 36 () 7 () 90 () 108 9) «oints,, are on a circle, such that m(arc ) = m(arc ) = 10. No point, except point, is common to the arcs. Which is the type of? () Equilateral triangle () Scalene triangle () Right angled triangle 83 () Isosceles triangle

85 (10) Seg XZ is a diameter of a circle. oint Y lies in its interior. How many of the following statements are true? (i) It is not possible that Ð XYZ is an acute angle. (ii) Ð XYZ can t be a right angle. (iii) Ð XYZ is an obtuse angle. (iv) an t make a definite statement for measure of Ð XYZ. () Only one () Only two () Only three () ll. Line l touches a circle with centre O at point. If radius of the circle is 9 cm, answer the following. (1) What is d(o, ) =? Why? () If d(o, ) = 8 cm, where does the point lie? (3) If d() = 15 cm, How many locations of point R are line on line l? t what distance will each of them be from point? M L Fig K 3. In figure 3.83, M is the centre of the circle and seg KL is a tangent segment. o Fig. 3.8 If MK = 1, KL = 6 3 then find - (1) Radius of the circle. () Measures of Ð K and Ð M. l 4. In figure 3.84, O is the centre of the circle. Seg, seg are tangent segments. Radius of the circle is r and l() = r, rove that, c O is a square. O Fig

86 H G E T Fig In figure 3.86, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and.5 cm. Find the answers to the following questions hence find the ratio MS:SR. (1) Find the length of segment MT () Find the length of seg MN (3) Find the measure of Ð NSM. F 5. In figure 3.85, c is a 85 parallelogram. It circumscribes the circle with cnetre T. oint E, F, G, H are touching points. If E = 4.5, E = 5.5, find. Fig In the adjoining figure circles with centres X and Y touch each other at point Z. secant passing through X Z intersects the circles at points and Y Z respectively. rove that, radius X radius Y. Fill in the blanks and complete the Fig proof. onstruction raw segments XZ and.... roof y theorem of touching circles, points X, Z, Y are.... \ Ð opposite angles Let Ð XZ = Ð ZY = a... (I) Now, seg seg XZ... (...) \ Ð XZ =... = a... (isosceles triangle theorem) (II) similarly, seg (...) \ Ð ZY =... = a... (...) (III) T R N S M

87 \ from (I), (II), (III), Ð XZ =... \ radius X radius Y... (...) 8. In figure 3.88, circles with centres X and Y touch internally at point Z. Seg Z is a chord of bigger circle and it itersects smaller circle at point. rove that, seg X seg Y. Y X Z Fig In figure 3.89, line l touches the l R O Fig S circle with centre O at point. is the mid point of radius O. RS is a chord through such that chords RS line l. If RS = 1 find the radius of the circle. 10 «. In figure 3.90, seg is a diameter of a circle with centre. Line is a tangent, which touches the circle at point T. seg ^ line and seg ^ line. rove that, seg. T Fig «. raw circles with centres, and each of radius 3 cm, such that each circle touches the other two circles. 1 «. rove that any three points on a circle cannot be collinear. 86

88 T 13. In figure 3.91, line R touches the circle at point. nswer the following questions with the help of S the figure. (1) What is the sum of Ð T and Ð TS? () Find the angles which are R congruent to Ð. Fig (3) Which angles are congruent to Ð TS? (4) ÐTS = 65, find the measure of ÐTS and arc TS. (5) If Ð = 4 and ÐSR = 58 find measure of ÐTS. 14. In figure 3.9, O is the centre of a circle, chord RS If Ð OR = 70 and (arc RS) = 80, find - (1) m(arc R) () m(arc S) (3) m(arc SR) O Fig. 3.9 R S Y W T Fig Z X 15. In figure 3.93, m(arc WY) = 44, m(arc ZX) = 68, then (1) Find the measure of Ð ZTX. () If WT = 4.8, TX = 8.0, YT = 6.4, find TZ. (3) If WX = 5, YT = 8, YZ = 6, find WT. 87

89 16. In figure 3.94, (1) m(arc E) = 54, m(arc ) = 3, find measure of ÐE. () If = 4., = 5.4, E = 1.0, find (3) If = 3.6, = 9.0, = 5.4, find E E Fig In figure 3.95, chord EF chord GH. rove that, chord chord FH. Fill in the blanks and write the proof. roof : raw seg GF. Ð EFG = Ð FGH... (I) Ð EFG =... inscribed angle theorem (II) Ð FGH =... inscribed angle theorem (III) \ m(arc EG) = from (I), (II), (III). chord chord FH... G E Fig F H R Fig O 18. In figure 3.96 is the point of contact. (1) If m(arc R) = 140, Ð OR = 36, find m(arc ) () If O = 7., O = 3., find OR and R (3) If O = 7., OR = 16., find R. 19. In figure 3.97, circles with centres and touch internally at point E. lies on the inner circle. hord E of the outer circle intersects inner circle at point. rove that, seg seg. 88 E Fig. 3.97

90 0. In figure 3.98, seg is a diameter of a circle with centre O. The bisector of Ð intersects the circle at point. rove that, seg. omplete the following proof by filling in the blanks. roof raw seg O. Fig Ð =... angle inscribed in semicircle Ð =... is the bisector of Ð m(arc ) =... inscribed angle theorem Ð O =... definition of measure of an arc (I) seg seg O... (II) \ line O is of seg... From (I) and (II) \ seg O 1. In figure 3.99, seg MN is a chord of a circle with centre O. MN = 5, L is a point on chord MN such that ML = 9 and d(o,l) = 5. M L O Find the radius of the circle. N Fig S R Fig «. In figure 3.100, two circles intersect each other at points S and R. Their common tangent touches the circle at points,. rove that, Ð R + Ð S =

91 4 «. In figure 3.10, two circles intersect each other at points and E. Their common secant through E intersects the circles at points and. The tangents of the circles at points and intersect each other at point. rove that c is cyclic. R F Fig M N O E Fig S 3 «. In figure 3.101, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and, respectively. rove that : seg S seg R. E Fig «. In figure 3.103, seg ^ side, seg E ^ side, seg F ^ side. onit O is the orthocentre. rove that, point O is the incentre of EF. IT Tools or Links Use the geogebra to verify the properties of chords and tangents of a circle. rrr 90

92 4 Geometric onstructions Let s study. onstruction of a triangle similar to the given triangle * To construct a triangle, similar to the given triangle, bearing the given ratio with the sides of the given triangle. (i) When vertices are distinct (ii) When one vertex is common onstruction of a tangent to a circle. * To construct a tangent at a point on the circle. (i) Using centre of the circle. (ii) Without using the centre of the circle. * To construct tangents to the given circle from a point outside the circle. Let s recall. In the previous standard you have learnt the following constructions. Let us recall those constructions. To construct a line parallel to a given line and passing through a given point outside the line. To construct the perpendicular bisector of a given line segment. To construct a triangle whose sides are given. To divide a given line segment into given number of equal parts To divide a line segment in the given ratio. To construct an angle congruent to the given angle. In the ninth standard you have carried out the activity of preparing a map of surroundings of your school. efore constructing a building we make its plan. The surroundings of a school and its map, the building and its plan are similar to each other. We need to draw similar figures in Geography, architecture, machine drawing etc. triangle is the simplest closed figure. We shall learn how to construct a triangle similar to the given triangle. 91

93 Let s learn. onstruction of Similar Triangle To construct a triangle similar to the given triangle, satisfying the condition of given ratio of corresponding sides. The corresponding sides of similar triangles are in the same proportion and the corresponding angles of these triangles are equal. Using this property, a triangle which is similar to the given triangle can be constructed. Ex. (1) ~ R, in, = 5.4 cm, = 4. cm, = 6.0 cm. = 3. onstruct and R. R Fig. 4.1 Rough Figure onstruct of given measure. and R are similar. \ their corresponding sides are proportional. R R 3... (I) s the sides,, are known, we can find the lengths of sides, R, R. Using equation [I] R 6.0 R 3 \ = 3.6 cm, R =.8 cm and R = 4.0 cm 9

94 4 cm 6 0 cm 8 cm R 4 0 cm 5 4 cm Fig cm For More Information While drawing the triangle similar to the given triangle, sometimes the lengths of the sides we obtain by calculation are not easily measureable by a scale. In such a situation we can use the construction To divide the given segment in the given number of eqaul parts. For example, if length of side is cm, then by dividing the line segment 3 of length 11.6 cm in three equal parts, we can draw segment. If we know the lengths of all sides of R, we can construct R. In the above example (1) there was no common vertex in the given triangle and the triangle to be constructed. If there is a common vertex, it is convenient to follow the method in the following example. Ex.() onstruct any. onstruct such that : = 5:3 and ~ nalysis : s shown in fig 4.3, let the points,, and,, be collinear. ~ \ Ð = Ð Fig. 4.3 = = = 5 3 Rough Figure \ sides of are longer than corresponding sides of. \ the length of side will be equal to 3 parts out of 5 equal parts of side. So if we construct, point will be on the side, at a distance equal to 3 parts from. Now is the point of intersection of and a line through, parallel to. 93

95 = 3 = 5 i.e, = = Taking inverse Steps of construction : (1) onstruct any. () ivide segment in 5 equal parts. (3) Name the end point of third part of seg as \ = 3 5 (4) Now draw a line parallel to through. Name the point where the parallel line intersects as. (5) is the required trinangle similar to Fig. 4.4 Note : To divide segment, in five equal parts, it is convenient to draw a ray from, on the side of line in which point does not lie. Take points T 1, T, T 3, T 4, T on 5 the ray such that T 1 = T 1 T = T T 3 = T 3 T 4 = T 4 T 5 Join T 5 and draw lines parallel to T 5 through T 1, T, T 3, T 4. T 1 T T3 T 4 T 5 Fig. 4.5 Let s think can also be constructed as shown in the adjoining figure. What changes do we have to make in steps of construction in that case? 94 Fig. 4.6

96 Ex. (3) onstruct similar to such that : = 5:7 nalysis : Let points,, as well as points,, be collinear. ~ and : = 5:7 \ sides of are smaller than sides of and Ð Let us draw a rough figure with these considerations. Now = 5 7 \ If seg is divided into 5 equal parts, then seg will be 7 times each part of seg. \ let us divide side of in 5 equal parts and locate point at a distance equal to 7 such parts from on ray. line through point parallel to seg is drawn it will intersect ray at point. ccording to the basic proportionality theorem we will get as described. Fig. 4.7 Rough Figure Steps of construction : (1) onstruct any. () ivide segment into 5 five equal parts. Fix point on ray such that length of is seven times of each equal part of seg (3) raw a line parallel to side, through. Name the point of intersection of the line and ray as. We get the required similar to Fig

97 ractice set ~ LMN. In, = 5.5 cm, = 6 cm, = 4.5 cm. onstruct and LMN such that = 5. MN 4. R ~ LTR. In R, = 4. cm, R = 5.4 cm, R = 4.8 cm. onstruct R and LTR, such that = 3. LT 4 3. RST ~ XYZ. In RST, RS = 4.5 cm, ÐRST = 40, ST = 5.7 cm onstruct RST and XYZ, such that RS = 3 XY MT ~ HE. In MT, M = 6.3 cm, ÐTM = 50, T = 5.6 cm. M H = 7 5. onstruct HE. onstruction of a tangent to a circle at a point on the circle (i) Using the centre of the circle. nalysis : Suppose we want to construct a tangent l passing through a point on the circle with centre. We shall use the property that a line perpendicular to the l radius at its outer end is a tangent to the circle. If is a radius with point on the circle, line l through and perpendicular to is the tangent at. For this we will use the construction of drawing a Fig. 4.9 perpendicular to a line through a point on it. For convenience we shall draw ray Steps of construction (1) raw a circle with centre. Take any point on the circle. l () raw ray. (3) raw line l perpendicular to ray X through point. X Line l is the required tangent to the circle at point. Fig

98 ii) Without using the centre of the circle. Example : onstruct a circle of any radius. Take any point on it. onstruct a tangent to the circle without using centre of the circle. nalysis : s shown in the figure, let line l be the tangent to the circle at point. Line is l Fig a chord and Ð is an inscribed angle. Now by tangent- secant angle theorem, Ð. y converse of tangent- secant theorem, if we draw the line such that, Ð, then it will be the required tangent. Steps of onstruction : (1) raw a circle of a suitable radius. Take any point on it. () raw chord and an inscribed Ð. (3) With the centre and any M convenient radius draw an arc intersecting the sides of Ð N R in points M and N. (4) Using the same radius and centre, draw an arc intersecting the chord at point R. Fig. 4.1 (5) Taking the radius equal to d(mn) and centre R, draw an arc intersecting the arc drawn in the previous step. Let be the point of intersection of these arcs. raw line. Line is the required tangent to the circle. Note that Ð MN and Ð in the above figure are congruent. If we draw seg MN and seg R, then MN and R are congruent by SSS test. \ Ð Ð 97

99 To construct tangents to a circle from a point outside the circle. nalysis : O Fig s shown in the figure let be a point in the exterior of the circle. Let and be the tangents to the circle with the centre O, touching the circle in points and respectively. So if we find points and on the circle, we can construct the tangents and. If O and O are the radii of the circle, then O ^ line and seg O ^ line. O and O are right angled triangles and seg O is their common hypotenuse. If we draw a circle with diameter O, then the points where it intersects the circle with centre O, will be the positions of points and respectively, because angle inscribed in a semicircle is a right angle. Steps of onstruction (1) onstruct a circle of any radius with centre O. () Take any point in the exterior of the O circle. M (3) raw segment O. raw perpendicular bisector of seg O to get its midpoint M. (4) raw a circle with radius OM and centre M Fig (5) Name the points of intersection of the two circles as and. (6) raw line and line. ractice set onstruct a tangent to a circle with centre and radius 3. cm at any point M on it.. raw a circle of radius.7 cm. raw a tangent to the circle at any point on it. 3. raw a circle of radius 3.6 cm. raw a tangent to the circle at any point on it without using the centre. 4. raw a circle of radius 3.3 cm raw a chord of length 6.6 cm. raw tangents to the circle at points and. Write your obeservation about the tangents.

100 5. raw a circle with radius 3.4 cm. raw a chord MN of length 5.7 cm in it. construct tangents at point M and N to the circle. 6. raw a circle with centre and radius 3.4 cm. Take point at a distance 5.5 cm from the centre. onstruct tangents to the circle from point. 7. raw a circle with radius 4.1 cm. onstruct tangents to the circle from a point at a distance 7.3 cm from the centre. roblem set 4 1. Select the correct alternative for each of the following questions. (1) The number of tangents that can be drawn to a circle at a point on the circle is.... () 3 () () 1 () 0 () The maximun number of tangents that can be drawn to a circle from a point out side it is.... () () 1 () one and only one () 0 (3) If ~ R and = 7, then... 5 () is bigger. () R is bigger. () oth triangles will be equal. () an not be decided.. raw a circle with centre O and radius 3.5 cm. Take point at a distance 5.7 cm from the centre. raw tangents to the circle from point. 3. raw any circle. Take any point on it and construct tangent at without using the centre of the circle. 4. raw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. raw tangents from point R. 5. raw a circle with centre. raw an arc of 100 measure. raw tangents to the circle at point and point. 6. raw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point such that E-F- and F = 4.1 cm. raw tangents to the circle from point. 7. ~ LN. In, = 5.1cm, Ð = 40, =4.8 cm, = 4. LN 7 onstruct and LN. 8. onstruct Y such that, Y = 6.3 cm, Y = 7. cm, = 5.8 cm. If YZ Y = 6 5, then construct XYZ similar to Y. 99 rrr

101 5 o-ordinate Geometry Let s study. istance formula Section formula Slope of a line Let s recall. We know how to find the distance between any two points on a number line. If co-ordinates of points, and R are -1,-5 and 4 respectively then find the length of seg, seg R. O R Fig. 5.1 If x 1 and x are the co-ordinates of points and and x > x 1 then length of seg = d(,) = x - x 1 s shown in the figure, co-ordinates of points, and R are -1,-5 and 4 respectively. \ d(, ) = (-1)-(-5) = = 4 and d(, R) = 4 - (-5) = = 9 Using the same concept we can find the distance between two points on the same axis in XY-plane. Let s learn. (1) To find distance between any two points on an axis. Two points on an axis are like two points on the number line. Note that points on the X-axis have co-ordinates such as (, 0), ( -5, 0), (8, 0). Similarly points on the Y-axis have co-ordinates such as (0, 1), (0, 17 ), (0, -3). art of the X-axis which shows negative co-ordinates is OX and part of the Y-axis which shows negative co-ordinates is OY. 100

102 i) To find distance between two points on X-axis. Y ii) To find distance between two points on Y-axis. Y X` O Y` (x 1, 0) (x, 0) Fig. 5. In the above figure, points (x 1, 0 ) and (x, 0) are on X- axis such that, x > x 1 X (0, y ) (0, y 1 ) X` O Y` Fig. 5.3 In the above figure, points (0, y 1 ) and (0, y ) are on Y- axis such that, y > y 1 X \ d(, ) = x - x 1 \ d(,) = y - y 1 () To find the distance between two points if the segment joining these points is parallel to any axis in the XY plane. X` Y O Y` (x 1, y 1 ) L(x 1, 0) (x, y 1 ) X M(x, 0) Fig. 5.4 Fig. 5.5 i) In the figure, seg is parallel to X- axis. \ y co-ordinates of points and are equal raw seg L and seg M perpendicular to X-axis \ c ML is a rectangle. \ = LM ut, LM = x - x 1 \ d(,) = x - x (0, y ) R (0, y 1 ) S X` Y Y` O (x 1, y ) (x 1, y 1 ) ii) In the figure seg is parallel to Y- axis. \ x co-ordinates of points and are equal raw seg R and seg S perpendicular to Y-axis. \ c SR is a rectangle \ = RS ut, RS = y - y 1 \ d(,) = y - y 1 X

103 ctivity: In the figure, seg Y-axis and seg X-axis. o-ordinates of points and are given. To find, fill in the boxes given below. is a right angled triangle. ccording to ythagoras theorem, () + () = We will find co-ordinates of point to find the lengths and, X- axis \ y co-ordinate of = Y- axis \ x co-ordinate of = = 3 - = = - = 4 \ = + = \ = 17 X` (-, ) Y Fig Y` (, 3) X istance formula X` Y O (x 1, y 1 ) (x, y ) (x, y 1 ) X Let s learn. In the figure 5.7, (x 1, y 1 ) and (x, y ) are any two points in the XY plane. From point draw perpendicular on X-axis. Similarly from point draw perpendicular on seg. seg is parallel to Y-axis. \ the x co-ordinate of point is x. seg is parallel to X-axis. \ the y co-ordinate of point is y 1. Y` Fig. 5.7 \ =d(, ) = x - x 1 ; = d(, ) = y - y 1 In right angled triangle, + = x x y y ( ) +( ) \ = ( x x ) +( y y ) 1 1 This is known as distance formula.

104 Note that, ( x x ) +( y y ) ( x x ) +( y y ) In the previous activity, we found the lengths of seg and seg and then used ythagoras theorem to find the length of seg. Now we will use distance formula to find. (, 3) and (-, ) is given Y Let (x 1, y 1 ) and (x, y ). x 1 =, y 1 = 3, x = -, y = = ( x x1) +( y y1) = ( ) +( 3) = ( 4) +( 1) X` (-, ) O (, 3) X = = 17 seg Y-axis and seg X-axis. \ co-ordinates of point are (, ). Y` Fig. 5.8 \ = ( x x ) +( y y ) = ( ) +( 3) = 0+ 1 = = ( x x1) +( y y1) = ( ) + ( ) = ( 4) + 0 = 4 In the Figure 5.1, distance between points and is found as (-1) - (-5) = 4. In XY- plane co-ordinates of these points are (-1, 0) and (-5, 0). Verify that, using the distance formula we get the same answer. o-ordinates of origin are (0, 0). Hence if co-ordinates of point are (x, y) then d(o, ) = x + y. If points (x 1, y 1 ), (x, y ) lie on the XY plane then d(, ) = ( x x1) +( y y1) Remember this! that is, = ( x x1) +( y y1) = ( x x ) +( y y )

105 Solved Examples Ex. (1) Find the distance between the points (-1, 1) and (5,-7). Solution Suppose co-ordinates of point are (x 1, y 1 ) and of point are (x, y ). x 1 = -1, y 1 = 1, x = 5, y = -7 Ex. () ccording to distance formula, d(, ) = ( x x1) +( y y1) 104 ( ) + ( ) = = ( 6) +( 8) = d(, ) = 100 = 10 \ distance between points and is 10. Show that points (-3, ), (1, -) and (9, -10) are collinear. Solution If the sum of any two distances out of d(, ), d(, ) and d(, ) is equal to the third, then the three points, and are collinear. \ we will find d(, ), d(, ) and d(, ). o-ordinates of o-ordinates of istance formula (-3, ) (1, -) d(,) = ( x x1) +( y y1) (x 1, y 1 ) (x, y ) \ d(, ) = 1 ( 3) + ( )... from distance formula = ( 1+ 3) +( 4) = = 3 = 4...(I) d(, ) = ( 9 1) + ( 10 + ) = = 8...(II) and d(, ) = ( 9+ 3) +( 10 ) = = 1...(III) \ from(i), (II) and (III) = 1 \ d(, ) + d(, ) = d(, ) \ oints,, are collinear.

106 Ex. (3) Verify, whether points (6, -6), (3, -7) and R(3, 3) are collinear. Solution = ( 6 3) + ( 6+ 7)...by distance formula = ( 3) + () 1 = (I) R = ( 3 3) +( 7 3) = ( 0) +( 10) = (II) R = ( 3 6) + ( 3+ 6) = 3 9 ( ) + ( ) = (III) From I, II and III out of 10, 100 and 90, 100 is the largest number. Now we will verify whether 100 For this compare 100 You will see that ( ) and ( ) are equal. ( ) and ( ). ( ) > ( 100 ) \ + R ¹ R \ points (6, -6), (3, -7) and R(3, 3) are not collinear. Ex. (4) Show that points (1, 7), (4, ), (-1, -1) and (-4, 4) are vertices of a square. Solution In a quadrilateral, if all sides are of equal length and both diagonals are of equal length, then it is a square. \ we will find lengths of sides and diagonals by using the distance formula. Suppose, (1, 7), (4, ), (-1, -1) and (-4,4) are the given points = ( ) +( ) = 9+ 5 = = ( ) + ( + ) = = 34 = = = ( 1+ 4) +( 1 4) ( 1+ 4) +( 7 4) ( ) + ( + ) = 9+ 5 = 34 = = 34 = = = ( ) +( ) = = 68 \ = = = and = 105 Fig. 5.9

107 So, the lengths of four sides are equal and two diagonals are equal. \ (1,7), (4,), (-1,-1) and (-4,4) are the vertices of a square. Ex. (5) Find the co-ordinates of a point on Y- axis which is equidistant from M (-5,-) and N(3,) Solution Let point (0, y) on Y- axis be equidistant from M(-5,-) and N(3,). \ M = N \ M = N \ [0 -(-5)] + [y -(-)] = (0-3) + (y - ) \ 5 + (y + ) = 9 + y - 4y + 4 \ 5 + y + 4y + 4 = 13 + y - 4y ` \ 8y = -16 \ y = - \ the co-ordinates of the point on the Y-axis which is equidistant from M and N are M (0, -). Ex. (6) (-3, -4), (-5, 0), (3, 0) are the vertices of. Find the co-ordinates of the circumcentre of. Solution Let, (a, b) be the circumcentre of. \ point is equidistant from, and. \ = =... (I) \ = (a + 3) + (b + 4) = (a + 5) + (b - 0) \a + 6a b + 8b + 16 = a + 10a + 5 +b \ -4a + 8b = 0 \ a - b = 0... (II) Similarly =... (I) From \ (a + 3) + (b + 4) = (a - 3) + (b - 0) \a + 6a b + 8b + 16 = a - 6a b \ 1a + 8b = -16 \ 3a + b = (III) Solving (II) and (III) we get a = -1, b = - 1 \ co-ordinates of circumcentre are (-1, - 1 ). (-5,0) (a,b) (-3,-4) Fig (3,0) 106

108 Ex. (7) If point (x, y) is equidistant from points (7, 1) and (3, 5), show that y = x-. Solution Let point (x, y) be equidistant from points (7, 1) and (3, 5) \ = \ = \ (x - 7) + (y - 1) = (x - 3) + (y - 5) \ x - 14x y - y + 1 = x - 6x y - 10y +5 \- 8x + 8y = -16 \ x - y = \ y = x - Ex. (8) Find the value of y if distance between points (,-) and (-1, y)is 5. Solution = [(-1) - ] + [y - (-)]... by distance formula \ 5 = (-3) + (y + ) \ 5 = 9 + (y + ) \ 16 = (y + ) \ y + = ± 16 \ y + = ± 4 \ y = 4 - or y = -4 - \ y = or y = -6 \value of y is or -6. ractice set Find the distance between each of the following pairs of points. (1) (, 3), (4, 1) () (-5, 7), (-1, 3) (3) R(0, -3), S(0, 5 ) (4) L(5, -8), M(-7, -3) (5) T(-3, 6), R(9, -10) (6) W( -7, 4), X(11, 4). etermine whether the points are collinear. (1) (1, -3), (, -5), (-4, 7) () L(-, 3), M(1, -3), N(5, 4) (3) R(0, 3), (, 1), S(3, -1) (4) (-, 3), (1, ), R(4, 1) 3. Find the point on the X-axis which is equidistant from (-3, 4) and (1, -4). 4. Verify that points (-, ), (, ) and R(, 7) are vertices of a right angled triangle. 107

109 5. Show that points (, -), (7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram. 6. Show that points (-4, -7), (-1, ), (8, 5) and (5, -4) are vertices of a rhombus. 7. Find x if distance between points L(x, 7) and M(1, 15) is Show that the points (1, ), (1, 6), (1 + 3, 4) are vertices of an equilateral triangle. Let s recall. roperty of intercepts made by three parallel lines : In the figure line l line m line n, line p and line q are transversals, p q E F l m n then E EF Fig Let s learn. ivision of a line segment 6 10 In the figure, = 6 and = 10. Fig. 5.1 \ = 6 10 = 3 5 In other words it is said that point divides the line segment in the ratio 3 5. Let us see how to find the co-ordinates of a point on a segment which divides the segment in the given ratio. 108

110 Let s learn. Section formula In the figure 5.13, point on the seg in XY plane, divides seg in the ratio m n. ssume (x Y 1, y 1 ) (x, y ) and (x, y) (x, y raw seg, seg and seg ) perpendicular to X- axis. \ (x 1, 0); (x, 0) and (x, 0). \ = x - x 1 }... (I) X` X and = x O (x 1, 0) - x (x, 0) (x, 0) Y` seg seg seg. Fig \ y the property of intercepts of three parallel lines, = = m n Now = x - x 1 and = x - x... from (I) \ x x 1 m = x x n \ n(x - x 1 ) = m (x - x) \ nx - nx 1 = mx - mx \ mx + nx = mx + nx 1 \ x(m + n) = mx + nx 1 mx + nx1 \ x = m+ n Similarly drawing perpendiculars from points, and to Y- axis, my + ny we get, y = 1. m+ n \ co-ordinates of the point, which divides the line segment joining the points (x 1, y 1 ) and (x, y ) in the ratio m n are given by mx + nx1 my + ny1,. m+ n m+ n 109 (x 1, y 1 ) (x, y)

111 o-ordinates of the midpoint of a segment If (x 1, y 1 )and (x, y ) are two points and (x, y) is the midpoint of (x 1, y 1 ) (x, y) seg then m = n. (x, y ) \values of x and y can be written as Fig x = mx + nx m+ n 1 y = my + ny m+ n 1 = mx + mx m+ m 1 ( ) = m x + x m = x + x 1 1 \ m = n \ co-ordinates of midpoint are x x y y +, = my + my m+ m 1 ( ) = m y + y m = y + y 1 This is called as midpoint formula. In the previous standard we have shown that a + b 1 \ m = n is the midpoint of the segment joining two points indicating rational numbers a and b on a number line. Note that it is a special case of the above midpoint formula. Solved Examples Ex. (1) If (3,5), (7,9) and point divides seg in the ratio 3 then find co-ordinates of point. Solution : In the given example let (x 1, y 1 )= (3, 5) and (x, y )= (7, 9). m n = 3 ccording to section formula, x = mx + nx m+ n 1 = \ o-ordinates of are 3 5, my y = m + + ny n 1 =

112 Ex. () Find the co-ordinates of point if is the midpoint of a line segment with (-4,) and (6,). Solution : In the given example, suppose (-4,) (x, y) (6,) Fig (-4, ) = (x 1, y 1 ) ; (6, ) = (x, y ) and co-ordinates of are (x, y) \ according to midpoint theorem, x = x y = y + x 1 + y 1 = 4+ 6 = + = = 1 = 4 = \ co-ordinates of midpoint are (1,). Let s recall. We know that, medians of a triangle are concurrent. The point of concurrence (centroid) divides the median in the ratio 1. Let s learn. entroid formula Suppose the co-ordinates of vertices of a triangle are given. Then we will find the co-ordinates of the centroid of the triangle. (x 1, y 1 ) In, (x 1, y 1 ), (x, y ), (x 3, y 3 ) are the vertices. Seg is a median and G(x, y) is the centroid. G(x, y) 1 is the mid point of line segment. (x, y ) (x 3, y 3 ) Fig

113 \ co-ordinates of point are x = x x + 3 +, y = y y 3 oint G(x, y) is centroid of triangle. \ G G = 1 \ according to section formula, x = x + x 1 x = x + x + x = x + x + x midpoint theorem y = y + y 1 y = y + y + y = y + y + y Thus if (x 1, y 1 ), (x, y ) and (x 3, y 3 )are the vertices of a triangle then the x1+ x + x3 y1+ y + y3 co-ordinates of the centroid are,. 3 3 This is called the centroid formula. Remember this! Section formula The co-ordinates of a point which divides the line segment joined by two distinct points (x 1, y 1 ) and (x, y ) in the ratio m n are mx + nx m+ n my + ny,. m+ n 1 1 Midpoint formula The co-ordinates of midpoint of a line segment joining two distinct points x x y y (x 1, y 1 ) and (x, y ) are 1+ 1+,. entroid formula If (x 1, y 1 ), (x, y ) and (x 3, y 3 ) are the vertices of a triangle then co-ordinates of the centroid are x1+ x + x3 y1+ y + y3,

114 Solved Examples Ex. (1) If point T divides the segment with (-7,4) and (-6,-5) in the ratio 7, find the co-ordinates of T. Solution Let the co-ordinates of T be (x, y). \ by the section formula, x = mx nx + 1 = 7 ( 6 )+ ( 7 ) m+ n 7+ (-6, -5) T y = my + ny m+ n 1 = = ( )+ ( ) = (-7, 4) Fig (x, y) 7 = = -7 9 = -3 56,. 9 \ co-ordinates of point T are 3 Ex. () If point (-4, 6) divides the line segment with (-6, 10) and (r, s) in the ratio 1, find the co-ordinates of. Solution y section formula -4 = r + 1 (-6) + 1 \ -4 = r \ -1 = r - 6 \ r = -6 \ r = -3 s = + 1 s + 10 \ 6 = 3 \ 18 = s + 10 \ s = 8 \ s = 4 \ co-ordinates of point are (-3, 4). Ex. (3) (15,5), (9,0) and --. Find the ratio in which point (11,15) divides segment. Solution Suppose, point (11,15) divides segment in the ratio m n \ by section formula, 113

115 x = mx nx + 1 m+ n m + n m+ n \ 11 = 9 15 \ 11m + 11n = 9m + 15n \ m = 4n Similarly, find the ratio using y co-ordinates. Write the conclusion. \ m n = 4 = 1 \ The required ratio is 1. Ex. (4) Find the co-ordinates of the points of trisection of the segment joining the points (,-) and (-7,4). (The two points that divide the line segment in three equal parts are called as points of trisection of the segment.) Solution Let points and be the points of trisection of the line segment joining the points and. oint and divide line segment into three parts. = =... (I) = + = = = 1 + Fig oint divides seg in the ratio 1. ( )+ 1+ x co-ordinate of point = From (I) 7+ 4 = 3 = = -3 3 = -1 y co-ordinate of point = ( ) = 0 = oint divides seg in the ratio 1. \ = 1 x co-ordinate of point = ( 7 )+ 1 = y co-ordinate of point = 4 1 = = -1 3 = 6 3 = = -4 \ co-ordinates of points of trisection are (-1, 0) and (-4, ).

116 For more information : See how the external division of the line segment joining points and takes place. Let us see how the co-ordinates of point can be found out if divides the line segment joining points (-4, 6) and (5, 10) in the ratio 3 1 externally. = 3 that is is larger than and = 3 that is = 3k, = k, then = k 1 \ = (5, 10) 1 (-4, 6) Now point divides seg in the ratio : 1. Fig We have learnt to find the coordinates of point if co-ordinates of points and are known. ractice set Find the coordinates of point if divides the line segment joining the points (-1,7) and (4,-3) in the ratio 3.. In each of the following examples find the co-ordinates of point which divides segment in the ratio a b. (1) (-3, 7), (1, -4), a b = 1 () (-, -5), (4, 3), a b = 3 4 (3) (, 6), (-4, 1), a b = 1 3. Find the ratio in which point T(-1, 6)divides the line segment joining the points (-3, 10) and (6, -8). 4. oint is the centre of the circle and is a diameter. Find the coordinates of point if coordinates of point and are (, -3) and (-, 0) respectively. 5. Find the ratio in which point (k, 7) divides the segment joining (8, 9) and (1, ). lso find k. 6. Find the coordinates of midpoint of the segment joining the points (, 0) and (0, 16). 7. Find the centroids of the triangles whose vertices are given below. (1)(-7, 6), (, -), (8, 5) () (3, -5), (4, 3), (11, -4) (3) (4, 7), (8, 4), (7, 11) 115

117 8. In, G (-4, -7) is the centroid. If (-14, -19) and (3, 5) then find the co-ordinates of. 9. (h, -6), (, 3) and (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1, 5). Find h and k. 10. Find the co-ordinates of the points of trisection of the line segment with (, 7) and (-4, -8). 11. If (-14, -10), (6, -) is given, find the coordinates of the points which divide segment into four equal parts. 1. If (0, 10), (0, 0) are given, find the coordinates of the points which divide segment into five congruent parts. Let s learn. Slope of a line When we walk on a plane road we need not exert much effort but while climbing up a slope we need more effort. In science, we have studied that while climbing up a slope we have to work against gravitational force. In co-ordinate geometry, slope of a line is an important concept. We will learn it through the following activity. ctivity I : In the figure points (-, -5), (0,-), (,1), (4,4), E(6,7) lie on line l. Observe the table which is made with the help of coordinates of these points on line l. 116 X (-, -5) Y Y l E 7 (6, 7) (4, 4) 3 1 (, 1) (0, -) X Fig. 5.0

118 Sr. No. First point Second point o-ordinates of first point (x 1, y 1 ) o-ordinates of second point (x, y ) y x - y - x E (, 1) (6, 7) = 6 4 = 3 (-, -5) (4, 4) 3 (4, 4) (-, -5) ( ) ( ) = 9 = = - 9 = Fill in the blank spaces in the above table. Similarly take some other pairs of points on line l and find the ratio y y - 1 for each pair. x - x 1 From this activity, we understand that for any two points (x 1, y 1 ) and (x, y ) on line l, the ratio y y - 1 is constant. x - x 1 If (x 1, y 1 ) and (x, y ) are any two points on line l, the ratio y slope of the line l. Generally slope is shown by letter m. \ m = y - y x - x 1 1 x - y - x 1 1 is called the 117

119 ctivity II In the figure, some points on line l, t and n are given. Find the slopes of those lines. Now you will know, (1) Slopes of line l and line t are positive. () Slope of line n is negative. (3) Slope of line t is more than slope of line l. (4) Slopes of lines l and t which make acute angles with X- axis, are positive. Y (3,4) l (-1,0) (6,1) 0 (4,0) X n (5) Slope of line n making obtuse angle with X- axis is negative. Fig. 5.1 Y Slopes of X-axis, Y-axis and lines parallel to axes. l In the figure 5., (x 1, 0) and (x, 0) m are two points on the X- axis. Slope of X- axis = 0-0 = 0 (0, y ) x - x 1 (0, y 1 ) In the same way (0, y 1 ) and (0, y ) are two points on the Y- axis. 0 X (x 1, 0) (x, 0) Slope of Y- axis = y - y = y - y 1 0, ut division by 0 is not possible. Fig. 5. \slope of Y- axis can not be determined. Now try to find the slope of any line like line m which is parallel to X- axis. It will come out to be 0. Similarly we cannot determine the slope of a line like line l which is parallel to Y- axis t Slope of line using ratio in trigonometry In the figure 5.3, points (x 1, y 1 ) and (x, y ) are on line l. Line l intersects X axis in point T. seg S ^ X- axis, seg R ^ seg S \ seg R seg TS... corresponding angle test \ R = y - y 1 and R = x - x 1 118

120 \ = tanq... (II) \ From (I) and (II), y - y \ R R R = y R x - y - x (I) Line T makes an angle q with the X- axis. 1 = tanq x - x (y 1 - y 1 ) \ m = tanq q R Now seg R seg TS, line l is transversal (x - x 1 ) \ ÐR = ÐTS...corresponding angles T From this, we can define slope in q O S X another way. The tan ratio of an angle made by the line with the positive direction l of X-axis is called the slope of that line. Fig. 5.3 When any two lines have same slope, these lines make equal angles with the positive direction of X- axis. \ These two lines are parallel. Slope of arallel Lines ctivity : In the figure 5. both line l and line t make angle q with the positive direction of X- axis. \ line l line t... corresponding angle test onsider, point (-3, 0) and point (0, 3) on line l Find the slope of line l. Slope of line l = y y - 1 x - x 1 - = = - = In the similar way, consider suitable points on the line t and find the slope of line t. From this, you can verify that parallel lines have equal slopes. 119 X` Y q (-3,0) 0 (x 1, y 1 ) Y (0,3) Y` Fig. 5.4 q (x, y ) l t X

121 Here q = 45. Use slope, m = tanq and verify that slopes of parallel lines are equal. Similarly taking q = 30, q = 60 verify that slopes of parallel lines are equal. Remember this! The slope of X- axis and of any line parallel to X- axis is zero. The slope of Y- axis and of any line parallel to Y- axis cannot be determined. Solved Examples EX. (1) Find the slope of the line passing through the points (-3, 5), and (4, -1) Solution Let, x 1 = -3, x = 4, y 1 = 5, y = -1 EX. () \ Slope of line = y - y x 1 - x 1 = ( 3 ) = -6 7 Show that points (-, 3), (1, ), R(4, 1) are collinear. Solution (-, 3), (1, ) and R(4, 1) are given points slope of line = y x - y - x 1 1 = 3 1 ( ) = Slope of line R = y x - y - x 1 1 = 1 - = Slope of line and line R is equal. ut point lies on both the lines. \ oint,, R are collinear. EX. (3) If slope of the line joining points (k, 0) and (-3, -) is 7 Solution (k, 0) and (-3, -) Slope of line = = k -3 -k ut slope of line is given to be 7. \ - = -3 -k 7 \ k = 4 then find k. 10

122 EX. (4) If (6, 1), (8, ), (9, 4) and (7, 3) are the vertices of c, show that c is a parallelogram. Solution : You know that Slope of line = y y - 1 x - x Slope of line = - 1 Slope of line = 4 Slope of line = 3 4 Slope of line = 3 1 = = (I) =... (II)... (III) =... (IV) Slope of line = Slope of line... From (I) and (III) \ line line Slope of line = Slope of line... From (II) and (IV) \ line line oth the pairs of opposite sides of the quadrilateral are parallel \ c is a parallelogram. ractice set ngles made by the line with the positive direction of X-axis are given. Find the slope of these lines. (1) 45 () 60 (3) 90. Find the slopes of the lines passing through the given points. (1) (, 3), (4, 7) () (-3, 1), (5, -) (3) (5, -), (7, 3) (4) L (-, -3), M (-6, -8) (5) E(-4, -), F (6, 3) (6) T (0, -3), S (0, 4) 3. etermine whether the following points are collinear. (1) (-1, -1), (0, 1), (1, 3) () (-, -3), E(1, 0), F(, 1) (3) L(, 5), M(3, 3), N(5, 1) (4) (, -5), (1, -3), R(-, 3) (5) R(1, -4), S(-, ), T(-3, 4) (6) (-4, 4), K(-, 5 ), N(4, -) 4. If (1, -1), (0, 4), (-5, 3) are vertices of a triangle then find the slope of each side. 5. Show that (-4, -7), (-1, ), (8, 5) and (5, -4) are the vertices of a parallelogram. 11

123 6. Find k, if R(1, -1), S (-, k) and slope of line RS is Find k, if (k, -5), (1, ) and slope of the line is Find k, if RS and (, 4), (3, 6), R(3, 1), S(5, k). roblem set 5 1. Fill in the blanks using correct alternatives. (1) Seg is parallel to Y-axis and coordinates of point are (1,3) then co-ordinates of point can be.... () (3,1) () (5,3) () (3,0) () (1,-3) () Out of the following, point... lies to the right of the origin on X- axis. () (-,0) () (0,) () (,3) () (,0) (3) istance of point (-3,4) from the origin is.... () 7 () 1 () 5 () -5 (4) line makes an angle of 30 with the positive direction of X- axis. So the slope of the line is.... () 1 () () 1 3. etermine whether the given points are collinear. 3 (1) (0,), (1,-0.5), (,-3) () (1, ), (, 8 5 ), R(3, 6 5 ) (3) L(1,), M(5,3), N(8,6) 1 () 3 3. Find the coordinates of the midpoint of the line segment joining (0,6) and (1,0). 4. Find the ratio in which the line segment joining the points (3,8) and (-9, 3) is divided by the Y- axis. 5. Find the point on X-axis which is equidistant from (,-5) and (-,9). 6. Find the distances between the following points. (i) (a, 0), (0, a) (ii) (-6, -3), (-1, 9) (iii) R(-3a, a), S(a, -a) 7. Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1), (0,-) and (1,3)

124 8. In the following examples, can the segment joining the given points form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle. (1) L(6,4), M(-5,-3), N(-6,8) () (-,-6), (-4,-), R(-5,0) (3) (, ), ( -, - ), ( - 6, 6 ) 9. Find k if the line passing through points (-1,-3) and (4, k)has slope Show that the line joining the points (4, 8) and (5, 5) is parallel to the line joining the points (,4) and (1,7). 11. Show that points (1,-), (5,), R(3,-1), S(-1,-5) are the vertices of a parallelogram 1. Show that the c RS formed by (,1), (-1,3), R(-5,-3) and S(-,-5) is a rectangle 13. Find the lengths of the medians of a triangle whose vertices are (-1, 1), (5, -3) and (3, 5). 14 «. Find the coordinates of centroid of the triangles if points (-7, 6), E(8, 5) and F(, -) are the mid points of the sides of that triangle. 15. Show that (4, -1), (6, 0), (7, -) and (5, -3)are vertices of a square. 16. Find the coordinates of circumcentre and radius of circumcircle of if (7, 1), (3, 5) and (, 0) are given. 17. Given (4,-3), (8,5). Find the coordinates of the point that divides segment in the ratio «. Find the type of the quadrilateral if points (-4, -), (-3, -7) (3, -) and (, 3) are joined serially. 19 «. The line segment is divided into five congruent parts at,, R and S such that ---R-S-. If point (1, 14) and S(4, 18) are given find the coordinates of,, R,. 0. Find the coordinates of the centre of the circle passing through the points (6,-6), (3,-7)and R(3,3). 1 «. Find the possible pairs of coordinates of the fourth vertex of the parallelogram, if three of its vertices are (5,6), (1,-)and (3,-).. Find the slope of the diagonals of a quadrilateral with vertices (1,7), (6,3), (0,-3) and (-3,3). 13 rrr

125 6 Trigonometry Let s study. Trigonometric ratios ngle of elevation and angle of depression Trigonometric identities roblems based on heights and distances Let s recall. 1. Fill in the blanks with reference to figure 6.1. sin q =, cos q =, Fig. 6.1 q tan q =. omplete the relations in ratios given below. (i) sin q = cosq (ii) sin q = cos (90 - ) (iii) cos q = sin (90 - ) (iv) tan q x tan (90 - q) = 3. omplete the equation. sin q + cos q = 4. Write the values of the following trigonometric ratios. 1 (i) sin30 = (ii) cos30 = (iii) tan30 = (iv) sin60 = (v) cos45 = (vi) tan45 = In std IX, we have studied some trigonometric ratios of some acute angles. Now we are going to study some more trigonometric ratios of acute angles. 14

126 Let s learn. cosec, sec and cot ratios Multiplicative inverse or the reciprocal of sine ratio is called cosecant ratio. It is 1 written in brief as cosec. \ cosecq = sinq Similarly, multiplicative inverses or reciprocals of cosine and tangent ratios are called secant and cotangent ratios respectively. They are written in brief as sec and cot. 1 \ secq = cosq and cotq = 1 tanq In figure 6., sinq = \ cosecq = It means, cosecq = 1 sinq 1 = = hypotenuse opposite side tanq = \ cot q = cot q = = 1 tanq 1 adjacent side = opposite side cosq = 1 secq = cosq 1 = = It means, hypotenuse secq = adjacent side You know that, tanq = sinq cosq 1 \ cot q = tanq 1 = sinq cosq = cos q sin q \ cot q = cos q sin q 15 q Fig. 6.

127 Remember this! The relation between the trigonometric ratios, according to the definitions of cosec, sec and cot ratios 1 sinq 1 cosq 1 tanq = cosec q \ sin q cosec q = 1 = sec q \ cos q sec q = 1 = cot q \ tan q cot q = 1 For more information : The great Indian mathematician ryabhata was born in in Kusumpur which was near atna in ihar. He has done important work in rithmetic, lgebra and Geometry. In the book ryabhatiya he has written many mathematical formulae. For example, (1) In an rithmetic rogression, formulae for n th term and the sum of first n terms. () The formula to approximate (3) The correct value of p upto four decimals, p = In the study of stronomy he used trigonometry and the sine ratio of an angle for the first time. omparing with the mathematics in the rest of the world at that time, his work was great and was studied all over India and was carried to Europe through Middle East. Most observers at that time believed that the earth is immovable and the Sun, the Moon and stars move arround the earth. ut ryabhata noted that when we travel in a boat on the river, objects like trees, houses on the bank appear to move in the opposite direction. Similarly, he said the Sun, Moon and the stars are observed by people on the earth to be moving in the opposite direction while in reality the Earth moves! On 19 pril 1975, India sent the first satellite in the space and it was named ryabhata to commemorate the great Mathematician of India. 16

128 * The table of the values of trigonometric ratios of angles 0,30,45,60 and 90. Trigonometric ngle (q) ratio sin q 0 cos q 1 tan q 0 cosec q 1 = sin q sec q 1 = cos q cot q 1 = tan q Not defined 1 3 Not defined Not defined 3 Not defined Let s learn. Trigonometric identities In the figure 6.3, is a right angled triangle, Ð= 90 (i) sinq = (ii) cosq = (iii) tanq = (v) secq = y ythagoras therom, + =.....(I) (iv) cosecq = (vi) cotq = ividing both the sides of (1) by + = Fig. 6.3 q 17

129 \ + = 1 \ + 1 \(sinq) + (cosq) = 1... [(sinq) is written as sin q and (cosq) is written as cos q.] sin q + cos q = 1... (II) Now dividing both the sides of equation (II) by sin q sin sin θ cos θ 1 + = θ sin θ sin θ 1 + cot q = cosec q... (III) ividing both the sides of equation (II) by cos q sin θ cos θ + cos θ cos θ 1 cos θ tan q + 1 = sec q 1 + tan q = sec q... (IV) Relations (II),(III), and (IV) are fundamental trigonometric identities. Solved Examples Ex. (1) If sinq = 0 9 then find cosq Solution Method I We have sin q + cos q = cos q = cos q = 1 cos q = = Taking square root of both sides. cosq = 1 9 Method II sinq = 0 9 from figure, sinq = \ = 0k and = 9k 0k Let = x. ccording to ythagoras therom, + = (0k) + x = (9k) 400k + x = 841k x = 841k - 400k = 441k \ x = 1k \ cos q = = 1k 9k = 1 9 9k q x Fig

130 Ex. () If secq = 5, find the value of tanq. 7 Solution Method I Method II we have, 1+ tan q = sec q \ 1+ tan q = 5 7 \ tan q = = = \ tanq = 4 7 from the figure, sec q = R \ = 7k, R = 5k ccording to ythagoras therom, + R = R \ (7k) + R = (5k) \ R = 65k - 49k = 576k \ R = 4k Now, tan q = R = 4k 7k = 4 7 Ex. (3) If 5sinq - 1cosq = 0, find the values of secq and cosecq. Solution 5sinq - 1cosq = 0 R 5k q 7k Fig. 6.5 \ 5sinq = 1cosq \ sin q = 1 cosq 5 \ tanq = 1 5 we have, 1+ tan q = sec q \ = sec q \ = sec q \ = sec q \ sec q = \ secq = 13 5 \cosq = 5 13 Now, sin q + cos q = 1 \ sin q = 1 - cos q \sin q = = = \ sinq = 1 13 \ cosecq =

131 3 Ex. (4) cosq = then find the value of 1 secθ. 1+ cosecθ Solution : Method I cosq = 3 sin q + cos q = 1 \ sin q + 3 \ secq = = 1 \ sin q = = 1 4 \ sinq = 1 \ cosecq = \ 1 1 secθ = 3 1+ cosecθ 1+ = = Method II cosq = 3 we know that, cos 30 = \ q = 30 \ sec q = sec 30 = 3 cosec q = cosec 30 = \ 1 1 secθ = 3 1+ cosecθ 1+ = = Ex. (5) Show that sec x + tan x = Solution sec x + tan x = 1 sin x + cos x cos x 1+ sin x 1 sin x = 1+ sin x cos x = ( 1+ sin x ) cos x = 1 + sinx 1 sin x 1 sin x ( )( + ) = ( 1+ sin x)( 1+ sin x) ( 1 sin x)( 1+ sin x) = 1 + sin x 1 sin x 130

132 Ex. (6) Eliminate q from given equations. x = a cot q - b cosec q y = a cot q + b cosec q Solution : x = a cot q - b cosec q... (I) y = a cot q + b cosec q... (II) dding equations (I) and (II). x + y = a cot q \ cot q = x + y... (III) a Subtracting equation (II) from (I), y - x = b cosec q \ cosec q = y x... (IV) or b Now, cosec q - cot q = 1 y x y+ x \ = 1 \ b a ( y x) y+ x 4b 4a y x b ( ) y+ x a = 4 = 1 ractice set If sinq = 7, find the values of cosq and tanq. 5. If tanq = 3, find the values of secq and cosq If cotq = 40 9, find the values of cosecq and sinq. 4. If 5secq- 1cosecq = 0, find the values of secq, cosq and sinq. 5. If tanq = 1 then, find the values of 6. rove that: (1) sin q + cosq = secq cosq () cos q(1 + tan q) = 1 sinθ + cosθ secθ + cosecθ 131.

133 (3) 1 sinθ 1+ sinθ = secq - tanq (4) ( secq - cosq)( cotq + tanq) = tanq secq (5) cotq + tanq = cosecq secq (6) 1 secθ tanθ = secq + tanq (7) sec 4 q - cos 4 q = 1 - cos q (8) secq + tanq = (9) If tanq + (10) tan 1+ tan 1 cosθ 1 sinθ 1 tanq =, then show that 1 tan q + tan q = cot cot ( ) + ( + ) (11) sec 4 (1 - sin 4 ) - tan = 1 (1) tan θ tanθ + secθ + 1 = secθ 1 tanθ + secθ 1 = sin cos pplication of trigonometry Let s learn. Many times we need to know the height of a tower, building, tree or distance of a ship from the lighthouse or width of a river etc. We cannot measure them actually but we can find them with the help of trigonometric ratios. For the purpose of computation, we draw a rough sketch to show the given Fig. 6.6 information. Trees, hills or towers are vertical objects, so we shall represent them in the figure by segments which are perpendicular to the ground. We will not consider height of the observer and we shall normally regard observer s line of vision to be parallel to the horizontal level. 13

134 Let us study a few related terms. (i) Line of vision If the observer is standing at the location,looking at an object then the line is called line of the vision. (ii) ngle of elevation : If an observer at, observes the point which is at a level higher than and M is the horizontal line, then Ð M is called the angle of elevation. (iii) ngle of depression : If an observer at, observes the point which is at a level lower than and M is the horizontal line, the Ð M is called the angle of depression. When we see above the horizontal line, the angle formed is the angle of elevation. When we see below the horizontal line, the angle formed is the angle of depression. line of vision angle of elevation Fig. 6.7 angle of depression line of vision M M horizontal line horizontal line Fig. 6.8 Solved Examples Ex. (1) n observer at a distance of 10 m from a tree looks at the top of the tree, the angle of elevation is 60. What is the height of the tree? ( 3 = 1.73) Solution : In figure 6.9, = h = height of the tree. = 10 m, distance of the observer from the tree. ngle of elevation (q) = Ð = 60 from figure, tanq =... (I) tan 60 = 3... (II) \ = 3... from equation (I)and (II) m Fig. 6.9 \ = 3 = 10 3 \ = = 17.3 m \ height of the tree is 17.3m. 133

135 Ex. () From the top of a building, an observer is looking at a scooter parked at some distance away, makes an angle of depression of 30. If the height of the building is 40m, find how far the scooter is from the building. ( 3 = 1.73) M Solution: In the figure 6.10, is the 30 building. scooter is at which is x m away from the 40 m building. In figure, is the position of 30 the observer. x Fig M is the horizontal line and Ð M is the angle of depression. Ð M and Ð are alternate angles. from fig, tan30 = \ 1 3 = 40 x \ x = 40 3 = = m. \ the scooter is 69.0 m. away from the building. Ex. (3) To find the width of the river, a man observes the top of a tower on the h opposite bank making an angle of elevation of 61. When he moves 50m backword from bank and observes the same top of the tower, his line of vision makes an angle of elevation of 35. Find the height of the tower and width of the river. (tan61 = 1.8, tan35 = 0.7) x 50m Fig Solution : seg shows the tower on the opposite bank. is the top of the tower and seg shows the 134 width of the river. Let h be the height of the tower and x be the width of the river. from figure, tan 61 = h x

136 \ 1.8 = h x h = 1.8 x 10h = 18x... (I)... multipling by 10 In right angled, h tan 35 = 0.7 = x + 50 h x + 50 \ h = 0.7 (x + 50) \ 10h = 7 (x + 50)... (II) \ from equations (I) and (II), 18x = 7(x + 50) \ 18x = 7x \ 11x = 350 \ x = = 31.8 Now, h = 1.8x = = 57.8 m. \ width of the river = 31.8 m and height of tower = 57.8 m Ex. (4) Roshani saw an eagle on the top of a tree at an angle of elevation of 61, while she was standing at the door of her house. She went on the terrace of the house so that she could see it clearly. The terrace was at a height of 4m. While observing the eagle from there the angle of elevation was 5. t what height from the ground was the eagle? (Find the answer correct upto nearest integer) y T R S x Fig (tan 61 = 1.80, tan 5 = 1.8, tan 9 = 0.55, tan 38 = 0.78) 135

137 Solution : In figure 6.1, is the house and SR is the tree. The eagle is at R. raw seg T ^ seg RS. \ c TS is a rectangle. Let S = x and TR = y Ex. (5) Now in RS, Ð RS = = 9 and in RT, Ð RT = 90-5 = 38 \ tan Ð RS = tan9 = S RS \ 0.55 = x y + 4 \ x = 0.55(y + 4)... (I) Similarly, tan Ð RT = T RT \ tan 38 = x y... [ \ 0.78 = x y \ x = 0.78y... (II) \ S = T = x] \ 0.78y = 0.55(y + 4)... from (I) and (II) \ 78y = 55(y + 4) \ 78y = 55y + 0 \ 3y = 0 \ y = = 10 (upto nearest integer) \ RS = y + 4 = = 14 \ the eagle was at a height of 14 metre from the ground. tree was broken due to storm. Its broken upper part was so inclined that its top touched the ground making an angle of 30 with the ground.the distance from the foot of the tree and the point where the top touched the ground was 10 metre. What was the height of the tree. Solution: s shown in figure 6.13, suppose is the tree. It was broken at and its top touched at. 136

138 Ð = 30, = 10 m, = x m = = y m In right angled, tan30 = 13 = x 10 x = 10 3 y = 0 3 x + y = = 30 3 Fig x + y = 10 3 \ height of the tree was 10 3 m. ractice set person is standing at a distance of 80m from a church looking at its top. The angle of elevation is of 45. Find the height of the church.. From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. ( 3 =1.73) 3. Two buildings are facing each other on a road of width 1 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60. What is the height of the second building? 4. Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in metre. Find the angle made by the wire with the horizontal. 5. storm broke a tree and the treetop rested 0 m from the base of the tree, making an angle of 60 with the horizontal. Find the height of the tree. 6. kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60 with the ground. ssuming that the string is straight, find the length of the string. ( 3 =1.73) 137

139 roblem set 6 1. hoose the correct alternative answer for the following questions. (1) sinq cosecq =? () 1 () 0 () 1 () cosec45 =? () 1 (3) 1 + tan q =? () () 3 () () 3 () cot q () cosec q () sec q () tan q (4) When we see at a higher level, from the horizontal line,angle formed is.... () angle of elevation. () 0 () angle of depression. () straight angle.. If sinq = 11, find the values of cosq using trigonometric identity If tanq =, find the values of other trigonometric ratios. 4. If secq = 13 1, find the values of other trigonometric ratios. 5. rove the following. (1) secq (1 - sinq) (secq + tanq) = 1 () (secq + tanq) (1 - sinq) = cosq (3) sec q + cosec q = sec q cosec q (4) cot q - tan q = cosec q - sec q (5) tan 4 q + tan q = sec 4 q - sec q sinθ 1 + sin (6) 1 θ = sec q (7) sec 6 x - tan 6 x = 1 + 3sec x tan x (8) tanθ secθ +1 secθ 1 tanθ (9) tan 3 θ tan θ 1 1 = sec q + tanq 138

140 (10) sinθ cosθ sinθ + cosθ 1 secθ tanθ 6. boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30.Find the height of the building. 7. From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30.If the height of the light-house is 100 meters, then find how far the ship is from the light-house. 8. Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 1 meter, the angle of elevation of the top of the second building is 30.What is the height of the second building? 9. ladder on the platform of a fire brigade van can be elevated at an angle of 70 to the maximum. The length of the ladder can be extended upto 0m. If the platform is m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70 = 0.94) 10 «. While landing at an airport, a pilot made an angle of depression of 0. verage speed of the plane was 00 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 0 = 0.34) rrr 139

141 7 Mensuration Let s study. Mixed examples on surface area and volume of different solid figures rc of circle - length of arc rea of a sector rea of segment of a circle Let s recall. Last year we have studied surface area and volume of some three dimentional figures. Let us recall the formulae to find the surface areas and volumes. No. Three dimensional Formulae figure 1. uboid Lateral surface area = h ( l + b ) h Total surface area = (lb + bh + hl ) l b Volume = lbh. ube Lateral surface area = 4l l Total surface area = 6l Volume = l 3 3. ylinder r urved surface area = prh h Total surface area = pr ( r + h ) Volume = pr h 4. one Slant height (l) = h + r h r l urved surface area = prl Total surface area = pr (r + l) Volume = 1 3 pr h 140

142 No. Three dimensional Formulae figure 5. Sphere Surface area = 4 pr r Volume = 4 3 pr3 6. Hemisphere urved surface area = pr r Total surface area of a solid hemisphere = 3pr Volume = 3 pr3 Solve the following examples Ex. (1) 30 cm 0 cm Fig 7.1 Ex. () 0 cm 1 cm 10 cm The length, breadth and height of an oil can are 0 cm, 0 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm 3 ) The adjoining figure shows the measures of a Joker s cap. How much cloth is needed to make such a cap? Fig 7. Let s think. s shown in the adjacent figure, a sphere is placed in a cylinder. It touches the top, bottom and the curved surface of the cylinder. If radius of the base of the cylinder is r, (1) What is the ratio of the radii of the sphere and the cylinder? () What is the ratio of the curved surface area of the cylinder and the surface area of the sphere? (3) What is the ratio of the volumes of the cylinder and the sphere? Fig

143 ctivity : r r r Fig. 7.4 r Fig. 7.5 Fig. 7.6 s shown in the above figures, take a ball and a beaker of the same radius as that of the ball. ut a strip of paper of length equal to the diameter of the beaker. raw two lines on the strip dividing it into three equal parts. Stick it on the beaker straight up from the bottom. Fill water in the beaker upto the first mark of the strip from the bottom. ush the ball in the beaker slowly so that it touches its bottom. Observe how much the water level rises. You will notice that the water level has risen exactly upto the total height of the strip. Try to understand how we get the formula for the volume of a the sphere. The shape of the beaker is cylindrical. Therefore, the volume of the part of the beaker upto height r can be obtained by the formula of volume of a cylinder. Let us assume that the volume is V. \ V = pr h = p r r = pr 3 ( h = r) ut V = volume of the ball + volume of the water which was already in the beaker. = volume of the ball pr3 \ volume of the ball = V pr3 = pr 3-3 pr3 = 6pr3 - pr = 4pr3 3 Hence we get the formula of the volume of a sphere as V = 4 3 pr3 (Now you can find the answer of the question number 3 relating to figure 7.3)

144 Solved Examples Ex. (1) The radius and height of a cylindrical water reservoir is.8 m and 3.5 m respectively. How much maximum water can the tank hold? person needs 70 litre of water per day. For how many persons is the water sufficient for a day? (p = 7 ) Solution : (r) =.8 m, ( h) = 3.5 m, p = 7 apacity of the water reservoir = Volume of the cylindrical reservoir = pr h Ex. () = 7 \ the reservoir can hold 8640 litre of water = 86.4 m 3 = ( 1m 3 = 1000 litre) = litre. The daily requirement of water of a person is 70 litre. \ water in the tank is sufficient for = 13 persons. How many solid cylinders of radius 10 cm and height 6 cm can be made by melting a solid sphere of radius 30 cm? Solution : Radius of a sphere, r = 30 cm Radius of the cylinder, R = 10 cm Height of the cylinder, H = 6 cm Let the number of cylinders be n. Volume of the sphere = n volume of a cylinder Volume of the sphere \ n = Volume of a cylinder = 4 π 3 π ( r) ( R) 3 H \ 60 cylinders can be made. 143 = ( 30) 3 = = 60

145 Ex. (3) tent of a circus is such that its lower part is cylindrical and upper part is conical. The diameter of the base of the tent is 48 m and the height of the cylindrical part is 15 m. Total height of the tent is 33 m. Find area of canvas required to make the tent. lso find volume of air in the tent. Solution : Total height of the tent = 33 m. Let height of the cylindrical part be H \ H = 15 m. 18 m Let the height of the conical part be h \h = (33-15) = 18 m. 4 m Slant height of cone, l = r + ( h) = = = 900 = 30 m. anvas required for tent = urved surface area of the cylindrical part + urved surface area of conical part = prh + prl = pr (H + l) = 7 = 7 4 ( ) 4 60 = m Volume of air in the tent = volume of cylinder + volume of cone = pr H pr h 1 = pr H + 3 h = 7 4 ( ) = = 38,016 m 3 \ canvas required for the tent = m \ volume of air in the tent = 38,016 m Fig m

146 ractice set Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.. Find the volume of a sphere of diameter 6 cm. 3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm. 4. Find the surface area of a sphere of radius 7 cm. 5. The dimensions of a cuboid are 44 cm, 1 cm, 1 cm. It is melted and a cone of height 4 cm is made. Find the radius of its base cm 3.5 cm Fig 7.8 conical water jug 10 cm 7 cm Fig 7.9 cylindrical water pot Observe the measures of pots in figure 7.8 and 7.9. How many jugs of water can the cylindrical pot hold? 7. cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm.the cone is placed upon the cylinder. Volume of the solid figure so formed is 500 cm 3. Find the total height of the figure. 8. In figure 7.11, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy. 9. In the figure 7.1, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper? 10 cm Fig mm Fig Figure 7.13 shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.(p= 3.14) cm Fig cm 4 cm 4 cm 3 cm Fig. 7.13

147 11. Find the surface area and the volume of a beach ball shown in the figure. 4 cm Fig s showm in the figure, a cylindrical glass contains water. metal sphere of diameter cm is immersed in it. Find the volume of the water. 30 cm 14 cm Fig Let s learn. Frustum of a cone The shape of glass used to drink water as well as the shape of water it contains, are examples of frustum of a cone. Fig onical part Frustum h r l r 1 Fig Fig Fig Fig. 7.0 cone being cut Two parts of the Frustum glass placed cone upside down When a cone is cut parallel to its base we get two figures; one is a cone and the other is a frustum. below. Volume and surface area of a frustum can be calculated by the formulae given 146

148 Remember this! h = height of a frustum, l = slant height height of a frustum, r 1 and r = radii of circular faces of a frustum ( r 1 > r ) Slant height of a frustum = l = h +( r1 r) urved surface area of a frustum = pl ( r 1 + r ) Total surface area of a frustum =pl (r 1 + r ) + pr 1 + pr Volume of a frustum = 1 ph (r + r r 1 r ) h r r 1 Fig. 7.1 l Solved Examples Ex. (1) bucket is frustum shaped. Its height is 8 cm. Radii of circular faces are 1 cm and 15 cm. Find the capacity of the bucket. ( p = 7 ) Solution : r 1 = 15 cm, r = 1 cm, h = 8 cm Ex. () apacity of the bucket = Volume of frustum = 1 ph ( 3 r + r 1 + r 1 r ) = ( ) = 4 3 = 4 3 ( ) 549 = = cm 3 = litre \ capacity of the bucket is litre. 147 Fig. 7. Radii of the top and the base of a frustum are 14 cm, 8 cm respectively. Its height is 8 cm. Find its i) curved surface area ii) total surface area iii) volume. Solution : r 1 = 14 cm, r = 8 cm, h = 8 cm Slant height of the frustum = l = = = +( ) h r r 1 +( ) + = 10 cm cm 8 cm

149 urved surface area of the frustum = p(r 1 + r ) l = 3.14 (14 + 8) 10 = cm Total surface area of frustum = p l (r 1 + r ) + pr 1 + pr = (14 + 8) = = = cm Volume of the frustum = 1 3 ph(r 1 + r + r 1 r ) = ( ) = cm 3 ractice set The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many liters of water it can hold? (1 litre = 1000 cm 3 ). The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its i) curved surface area ii) total surface area. iii ) volume (p = 3.14) 3. The circumferences of circular faces of a frustum are 13 cm and 88 cm and its height is 4 cm. To find the curved surface area of the frustum complete the following activity. (p = 7 ). circumference 1 = pr 1 = 13 r 1 = 13 = p circumference = pr = 88 r = 88 p = slant height of frustum, l = h +( r r ) 1 4 cm r 1 r Fig. 7.3 = + = cm 148

150 curved surface area of the frustum = p(r 1 + r )l = p = sq.cm. Let s recall. omplete the following table with the help of figure 7.4. Type of arc Name of the arc Measure of the arc Minor arc arc X arc Y... Let s learn. Y O 100 X Fig. 7.4 Sector of a circle In the adjacent figure, the central angle divides the circular region in two parts. Each of the O q M Fig. 7.5 parts is called a sector of the circle. Sector of a circle is the part enclosed by two radii of the circle and the arc joining their end points. In the figure 7.5, O M and O- are two sectors of the circle. Minor Sector : Sector of a circle enclosed by two radii and their corresponding minor arc is called a minor sector. In the above figure O-M is a minor sector. Major Sector : Sector of a circle that is enclosed by two radii and their corresponding major arc is called a major sector. In the above figure, O- is a major sector. 149

151 rea of a sector Observe the figures below. Radii of all circles are equal. Observe the areas of the shaded regions and complete the following table. q = 360 q = 180 q = 90 q = 60 q r q r q r q r 1 = pr = 1 pr 3 = 1 4 pr Fig. 7.6 entral angle of a circle is = 360 = complete angle 4 = 1 6 pr Sector of circle entral angle of a circle is = 360, rea of a circle = pr Measure of arc of the sector q 360 rea of the sector = 1 1 pr pr 1 4 pr q q 360 q 360 pr From the above table we see that, if measure of an arc of a circle is q, then the q area of its corresponding sector is obtained by multiplying area of the circle by. q 360 rea of sector () = 360 pr From the formula, θ rea of sector ; that is π r 360 rea of circle = q

152 Length of an arc In the following figures, radii of all circles are equal. Observe the length of arc in each figure and complete the table. q = 360 q = 180 q = 90 q = 60 q q q q l l l l 3 4 l = pr l = 1 pr l 3 = 1 4 pr l 4 = 1 6 pr Length of the arc Fig. 7.7 ircumference of a circle = pr Measure of the arc (q) l l 180 l q 360 Length of the arc (l) = 1 1 pr = = pr 1 4 pr l l q q 360 q 360 pr The pattern in the above table shows that, if measure of an arc of a circle is q, then q its length is obtained by multiplying the circumference of the circle by. 360 q Length of an arc (l) = 360 pr From the formula, that is, l θ π r 360 Length of an arc ircumference = q 360

153 relation between length of an arc and area of the sector q rea of a sector, () = 360 pr... I q Length of an arc, (l) = 360 pr θ l \... II 360 π r l \ = π r...from I and II π r = 1 lr = lr \ rea of a sector = Length of the arc Radius Similarly, l θ πr πr 360 Solved Examples Ex. (1) The measure of a central angle of a circle is 150 and radius of the circle is 1 cm. Find the length of the arc and area of the sector 1 O 150 associated with the central angle. Solution : r = 1 cm, q = 150, p = 7 q rea of the sector, = 360 pr = = 1155 = cm q Length of the arc, l = 360 pr Fig. 7.8 = = 55 cm 15

154 ex. () In figure 7.9, is the centre of the circle of radius 6 cm. Seg R is a tangent at. If R = 1, find the area of the shaded region. ( 3 = 1.73) 6 1 Fig. 7.9 R Solution Radius joining point of contact of the tangent is perpendicular to the tangent. \ in R, Ð R = 90, = 6 cm, R = 1 cm \ = R If one side of a right anglesd triangle is half the hypoteneus then angle opposite to, that side is of 30 measure \ Ð R = 30 and Ð = 60 \ by theorem, R = \ R = 6 3 cm \( R) = 1 rea of a sector = R = = 18 3 = = cm (-) = 60 q 360 pr = = cm. = rea of shaded region = (R) - (-) = = 1.30 cm rea of the shaded region = 1.30 cm 3 R = 3 1 =

155 ctivity In figure 7.30, side of square is 7 cm. With centre and radius, sector - X is drawn. Fill in the following boxes properly and find out the area of the shaded region. Solution : rea of a square = = = 49 cm rea of sector (- X) = (shaded region) = 154 (Formula) = = 38.5 cm - (Formula) = cm - cm = cm ractice set Radius of a circle is 10 cm. Measure of an arc of the crcleis 54. Find the area of the sector associated with the arc. (p = 3.14 ). Measure of an arc of a circle is 80 cm and its radius is 18 cm. Find the length of the arc. (p = 3.14 ) 3. Radius of a sector of a circle is 3.5 cm and length of its arc is. cm. Find the area of the sector. 4. Radius of a circle is 10 cm. rea of a sector of the sector is 100 cm. Find the area of its corresponding major sector. (p = 3.14 ) 5. rea of a sector of a circle of radius 15 cm is 30 cm. Find the length of the arc of the sector. 6. In the figure 7.31, radius of the circle is 7 cm and m(arc MN) = 60, find (1) rea of the circle. () (O - MN). (3) (O - MN). M O 68 N 7 cm X Fig Fig. 7.31

156 7. In figure 7.3, radius of circle is 3.4 cm and perimeter of sector - is 1.8 cm. Find (-). O R X M N Y Fig In figure 7.34, if (-) = 154 cm 8. In figure 7.33 O is the centre of the sector. Ð RO = Ð MON = 60. OR = 7 cm, and OM = 1 cm. Find the lengths of arc RX and arc MYN. (p = 7 ) 3.4 cm radius of the circle is 14 cm, find (1) Ð. () l(arc ). Fig Radius of a sector of a circle is 7 cm. If measure of arc of the sector is - (1) 30 () 10 (3 ) three right angles; find the area of the sector in each case. Fig The area of a minor sector of a circle is 3.85 cm and the measure of its central angle is 36. Find the radius of the circle. 1. In figure 7.35, c RS is a rectangle. If = 14 cm, R = 1 cm, find the areas of the parts x, y and z. 13. LMN is an equilateral triangle. LM = 14 cm. s shown in figure, three sectors are drawn with vertices as centres and radius7 cm. z x y Fig L S R Find, (1) ( LMN) () rea of any one of the sectors. (3) Total area of all the three sectors. (4) rea of the shaded region. M Fig N 155

157 Segment of a circle Let s learn. Y O Major segment Segment of a circle is the region bounded by a chord and Minor segment its corresponding arc of the circle. X Fig Minor segment : The area enclosed by a chord and its corresponding minor arc is called a minor segment. In the figure, segment X is a minor segment. Major segment : The area enclosed by a chord and its corresponding major arc is called a major segment. In the figure, seg Y is a major segment. Semicircular segment : segment formed by a diameter of a circle is called a semicircular segment. rea of a Segment r Y X O T X O Fig Fig (segment X) = (O - X) - ( O) In the figure, seg T ^ radius O. = Now, in OT sin q = T O \ T = O sinq q 360 pr - ( O)... (I) 156 In figure 7.38, X is a minor segment and Y is a major segment. How can we calculate the area of a minor segment? In figure 7.39, draw radii O and O. You know how to find the area of sector O-X and O. We can get area of segment X by substracting area of the triangle from the area of the sector.

158 T = r sinq ( O = r) ( O) = 1 = 1 base height O T = 1 r r sinq = 1 r sinq... (II) From (I) and (II), (segment X) = q 360 pr - 1 r sinq pq = r sinq (Note that, we have studied the sine ratios of acute angles only. So we can use the above formula when q 90.) Solved Examples Ex. (1) In the figure 7.40, Ð O = 30, O = 1 cm. Find the area of the segment. (p = 3.14) Method I r = 1, q = 30, p = 3.14 (O-X) = q 360 pr = = = cm O 30 X Fig ( O) = 1 r sinq = 1 1 sin 30 = ( sin 30 = 1 ) = 36 cm 157

159 (segment X) = (O - X) - ( O) = = 1.68 cm Method II (segment X) = r pq sinq sin = = = = = = = 1.68 cm. 30 Ex. () The radius of a circle with centre is 10 cm. If chord of the circle substends a right angle at, find areas of the minor segment and the major segment. (p = 3.14) Solution : r = 10 cm, q = 90, p = 3.14 (-X) = () q 360 pr = = = 78.5 cm = 1 = 1 = 50 cm base height (minor segment) = (-X) - () = = 8.5 cm 158 X Fig. 7.41

160 (major segment) = (circle) - (minor segment) = = = 85.5 cm Ex. (3) regular hexagon is inscribed in a circle of radius 14 cm. Find the area of the region between the circle and the hexagon. (p = 7, 3 = 1.73) Solution : side of the hexagon = 14 cm (hexagon) = 6 = (side) = cm (circle) = pr = 7 = 616 cm Fig. 7.4 The area of the region between the circle and the hexagon = (circle) - (hexagon) = = cm O R ractice set In figure 7.43, is the centre of the circle. Ð = 45 and = 7 cm. Find the area of segment X. Fig Fig X. In the figure 7.44, O is the centre of the circle. m(arc R) = 60 O = 10 cm. Find the area of the shaded region. (p = 3.14, 3 = 1.73)