BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No

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1 BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to + level to C.B.S.E., New Delhi) Affiliation No CBSE Board Level X- S.A.- II Maths Chapterwise Printable Worksheets with Solution Session : 04-5 Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-840 Website: brilliantpublic@yahoo.com Ph , Mobile: ,

2 MATHEMATICS (Class X) Index: S.A.-II CBSE Chapter-wise Solved Test Papers. Quadratic Equations 00. Arithmetic Progressions 06. Coordinate Geometry Some Applications of Trigonometry Circles Constructions 7. Areas Related to Circles Surface Areas and Volumes Probability 00

3 CBSE TEST PAPER-0 CLASS X Mathematics (Quadratic Equation). If αand βare the roots of the equation 5x 7x + = 0, then the value of + is α β (a) 7 (b) 9 (c) 6 (d) 8. Construct the quadratic equation whose roots are and is (a) (c) x x x + = 0 (b) x 4 x + 9 = 0 (d) x + 4 x + 9 = 0 + x + = 0. If the roots of the quadratic equation. ax + bx + c = 0are equal then (a) b = 4bc (b) a = 4bc (c) c = 4ab (d) b = 4ac 4. If the quadratic equation ax + bx + c = 0has a real root, then b 4acmust be (a) 0 (b) = 0 (c) 0 (d) >0 5. Determine whether the given values are solution of the equation or not x + x 4 = 0, x =, x = 6. If x = and and R. x = are solution of the equation 5x + px + R = 0find the value of P 5 7. Find K for which the given value is solution of the equation 8. If α, β are roots of the quadratic equation that α + β = 4 9. Given that one root of the quadratic equation other, show that b = 6ac Kx x ax k x a [] [] + + = 0, = [] + 4x + 4 = 0, find the value of Ksuch + + = 0is three times the ax bx c 0. One fourth of a herd of camels was in the forest. Twice the square root of the herd had gone to mountains and the remaining 5 camels were seen on the bank of a river. Find the total no. of camels.. A farmer wishes to grow a 00m rectangular vegetable garden. Since he has with him only 0 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.. Some student planned a picnic. The budget for food was Rs But 5 of these failed to go and then cost of food for each member is increased by Rs 5. How many students attended the picnic [] [] [4] [4] [4]

4 CBSE TEST PAPER-0 CLASS - X Mathematics (Quadratic Equation) Ans. Ans. Ans. Ans4. (a) (c) (d) (a) [ANSWERS] Ans5. Put x = L.H.S = ( ) + ( ) 4 = + 4 = 4 4 = 0 Yes x = L.H.S = ( ) + ( ) 4 = = 8 8 = 0 Yes Ans6. x = and Eq. 5x + Px + R = 0 ( ) P ( ) R = 0 ( ii) x = are solution of the 5 ( ) P + R = 0... i 5 + p + R = R P + = P + 5R = 0 P 5 R... On solving (i) and R R = P =

5 Ans7. Ans8. Ans9. x ax k + + = 0 x = a ( ) ( ) a + a a + k = 0 a a k + = 0 K = a Kx + 4x + 4 = 0 4 b α + β = [ k a 4 c αβ = [ k a α + β = 4(Given) ( ) α + β αβ = = 4 k k 6 8 = 4 k k 6 8k 4 = k 6 8k = 4k 4k + 8k 6 = 0 k + k = 0 0 k + k k = k ( k ) ( k ) ( k )( k + ) = = 0 k = Let one root = α Other root = α b ( α ) + α = a b 4α = a b α =...( i) 4a

6 c α. α = a c α = a c α = a b c = 4a a b c = a ac 6a b = 6 [from(i) Ans0. Let total no. of camels = x No. of camels see in the forest = 4 x No. of camels gone to mountains = x No. of camels on the bank of river = 5 x ATQ + x + 5 = x 4 x 8 x 60 = 0 Put x = y x = y y 8y 60 = 0 y y y = 0 y ( y ) ( y ) ( y 6)( y + 0) = = 0 0 y = 6, y = 6 x = y x = ( 6) x = 6 Ans. Let length = x Width = y x + y + y = 0 y = 0 x y = 0 x 4

7 xy = 00(Given) 0 x x = 00 0x x = 00 x 0x + 00 = 0 x x x = 0 x( x ) ( x ) ( x 0)( x 0) = 0 x = 0 x = = 0 x = 0 y = 0(Neglect) x = 0 y = 5 Ans. Let no. of student = x Cost per head= 500 x When 5 failed to go New cost per head = 500 x ATQ = 5 x 5 x 500x 500x = x 5x 500 x 5x = 5 x 5x = 500 x 5x 500 = 0 x x x = 0 x( x ) ( x ) ( x 5)( x + 0) = = 0 x = 5, x = 0 No of student attended the picnic = 5 5 = 0 5

8 CBSE TEST PAPER-0 CLASS X Mathematics (Quadratic Equation). Which of the following is quadratic equation? (a) x x x 5 = 0 (b) x 5x + 9 = x 7x + (c) x + = x x x. Factor of a x abx + b = 0is (a) b b,, (b) b a,, a a a b (d) x + x + = 0 b a (c),, a b a a (d),, b b. Value of xfor x 8x + 5 = 0is quadratic formula is (a), (b) 5, (c) 5 (d), 4. Discriminate of x x = is 0 (a) 0 (b) (c) (d) 5 5. Solve abx 9a x + 8b x 6ab = 0 [] 6. Solve for x by quadratic formula ( ) [] [] P x + p q x q = 0 [] 7. Find the value of K for which the quadratic equation distinct root. kx + x + = 0, has real and [] 8. If -4 is a root of the quadratic equation x + px 4, and the quadratic equation [] x px k + + = 0has equal root, find the value of K. 9. Solve for x x+ x : = [] 0. If I had walked km per hr faster, I would have taken 0 mints. Less to walk km Find the rate of my walking.. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.. A plane left 0 mints later than the schedule time and in order to reach its destination 500 Km away in time it has to increase its speed by 50km/hr from its usual speed. Find its usual speed. [4] [4] [4] 6

9 CBSE TEST PAPER-0 CLASS X Mathematics (Quadratic Equation) [ANSWERS] Ans. (b) Ans. (c) Ans. (a) Acs4. (c) Ans5. abx a x b x ab = 0 ax( bx x) b( bx a) ( 4bx a)( ax + b) = = 0 4bx a = 0 a x = 4b ax + b = 0 b x = a Ans6. ( ) p x + p q x q = 0 a = p, b = p q, c = q D = b ac 4 ( p q ) 4 p ( q ) = = p + q p q + 4 p q 4 ( p q ) = + b ± D x = 9 ( p q ) ( p q ) ± + = p p + q + p + q p + q p =, x = p p q p x =, x = p p q x =, x = p 7

10 Ans7. Kx ( ) + x + = 0 a = k, b =, c = b = b 4ac = 4 k = 4 4k For real and distinct root D > 0 4 4k > 0 4k > 4 k < Ans8. -4 is root of x + px 4 = 0 ( ) p ( ) = p 4 = 0 4 p = p = x px k + + = 0(Given) x + x + k = 0 = 4 D b ac 0 = ( ) 4 k [For equal root D = 0 4k = 9 k = 9 4 Ans9. x+ x = 6 x x = 6 x = 6 5x Put 5 x = y 5y 5 + = 6 y y y = 0 8

11 y y y + = y ( y ) ( y ) ( y 5)( 5y ) = = 0 y = 5 y = 5 = x = 5 x x = 5 = 5 x x = Ans0. Distance = km Let speed = xkm/hr New speed = ( x + ) km/hr Time taken by normal speed = hr x Time taken by new speed = ATQ = 0 x x + 60 x + x = x + x 6 x + x = x + x = 0 x x x + 4 = 0 xx ( x ) ( x + )( x ) = = 0 x = x = Km/hr x + hr Ans. Let B takes xdays to finish the work then A alone can finish it in ( x 6) days ATQ + = x x 6 4 9

12 x 6 + x = x 6x 4 x 6 = x 6x 4 6 = 8 4 x x x x 4x + 4 = = 0 x x x x( x ) ( x ) ( x )( x ) = 0 = 0 x = x = (Neglect) x = Ans. Let usual speed = xkm/hr New speed ( x 50) = + km/hr Total distance = 500 km Time taken by usual speed = 500 hr x Time taken by new speed = ATQ = x x x x = x + 50x x + 50x = x + 50x = x + 50x = 0 x x x = 0 ( ) ( x ) x x = 0 x = 750, x = 000 x = hr x

13 CBSE TEST PAPER-0 CLASS X Mathematics (Quadratic Equation). Which of the following have real root. (a) x + x = 0 (b) x (c) x + x + = 0 6x + 6 = 0 (d) x + 5x + 0 = 0 Solve for x x = x (a) x = (b) x = (c) x = (d) x =. If one root of the equations x + ax + = 0is find the value of a. (a) = 4 (b) = 5 (c) = (d) = 4. Find k for which the quadratic equations 4x kx + = 0has equal root. 4 (a) = ± (b) = (c) = ± (d) = ± Determine the nature of the roots of the Quadratic equation a b x abcdx c d = 0 6. Find the discriminant of ( x )( x ) 7. Find the value of K so that ( x ) is a factor of = 0 [] k x [] [] [] kx [] 8. = + + p + q + x p q x Solve for xby factorization method. [] 9. 5x 6x = 0, Solve for xby the method of completing the square. [] 0. A motor boat, whose speed is 5km/hr in still water, goes 0km down stream and comes back in a total time of 4 hr 0 mint find the speed of the stream.. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hr. faster than the first pipe and four hr. slower than the third pipe. Find the time required by each pipe to fill the pool separately.. A two digit no. is such that the product of its digits is 8. When 6 is subtracted from the no. the digit interchange their places. Find the no. [4] [4] [4]

14 CBSE TEST PAPER-0 CLASS X Mathematics (Quadratic Equation) Ans. Ans. Ans. Ans4. (a) (c) (b) (c) [ANSWERS] Ans5. = D b 4ac ( ) 4abcd 4 9a b 6c d = = a b c a b c d = Ans6. ( x )( x ) = 0 x x x + = 0 x x + = 0 a =, c =, c = D = b 4ac ( ) = 4 = 9 8 = P x = K x x Ans7. Let ( ) ( ) ( ) ( ) P = K K 0 = K k K k k + ( ) ( k ) k k + = 0 ( k )( k ) + = 0 k =, k =

15 Ans8. = + + p + q + x p q x = + p + q + x x p q x p q x p + q = + + x px qx pq ( ) p + q p + q = + + x px qx pq = + + x px qx pq x + px + qx = pq x px qx pq = 0 x( x p) q ( x p) ( x + p)( x + q) = = 0 x = p, x = q Ans9. x x 5 6 = 0 x x 6 x = x + = x = x = x = x = ± x = ± x = or x = 5 5

16 Ans0. Speed of motor boat in still water = 5 km/hr Speed of stream = x km/hr Speed in down word direction = 5 + x Speed in down word direction = 5 x 0 0 ATQ + = x 5 x ( x) + ( + x) ( + x)( x) = x x 9 = x 5 ( x ) 9 5 = x = 00 x = 5 Speed of stream = 5 km/hr Ans. Let x be the no. of hr required by the second pipe alone to till the pool and first pipe ( x + 5) hr while third pipe ( x 4) hr + = x + 5 x x 4 x + x + 5 = x + 5x x 4 x 8x 0 = = 0 x x x x( x ) ( x ) ( x 0)( x + ) = 0 x = = 0 x = (Neglect) 4

17 Ans. Let digit on unit place = x Digit on ten s place = y xy =8 Number = 0. y + x 8 = 0 + x ATQ 0 + x ± 6 = 0x + x x x x + + = x x x x = x x x x x = 0 ( x x x ) = 0 ( ) ( x ) x x = 0 x = x = 9 8 No. = 0 + = 9 5

18 CBSE TEST PAPER-04 CLASS X Mathematics (Quadratic Equation). Solve by factorization (a) 7 x =, (b) x x = 0 7 x =, (c). The quadratic equation whose root are and - is (a) (c) x 9 = 0 (b) x x + = 0 (d) x x = 0 x + = 9 0 x =, (d) ±. The product of two Consecutive positive integers is 06. Representation is quadratic equations (a) (c) x x + x 06 = 0 (b) x + x 06 = 0 (d) x 4. Which is quadratic equations 5. (a) (c) x x + x + = 0 (b) x 4 x + 06 = 0 x 06 = 0 + x + = 0 + x + = 0 (d) x + = 0 The sum of two no. is 6. The sum of their reciprocals is. Find the no. [] 6. Solve for x 7 x = x 7 [] 7. Solve for x + = 0 [] a b x b x a x 8. Using quadratic formula solve for x ( ) ( ) 9x 9 a + b x + a + 5ab + b = 0 [] [] [] 9. A factory kept increasing its output by the same percent ago every year. Find the percentage if it is known that the output is doubled in the last two years. 0. Two pipes running together can fill a cistern in minute. if one pipe takes minute more than the other to fill fit, find the time in which each pipe would fill the cistern.. If the roots of the equations ( a b) x ( b c) x ( c a) + + = 0 are equal prove that a = b + c. Two circles touch internally. The sum of their areas is 6π cm and the distance between their centres is 6 cm. find the ratio of the circles. [4] [4] [4] [4] 6

19 CBSE TEST PAPER-04 CLASS X Mathematics (Quadratic Equation) Ans. Ans. Ans. Ans4. (a) (a) (a) (a) [ANSWERS] Ans5. Let no. be x ATQ + = x 6 x 6 6x x = x 6x + 48 = 0 x x x = 0 x = or x = 4 Ans6. 7 x = ( x 7) ( 7 x) = ( x 7) 7 x = x x x x x x = 0 x 68 = 0 x x x = 0 x = or x = 8 Ans7. a b x b x a x + + = 0 b x( a x ) ( a x ) ( a x + )( b x ) = 0 x =, x = a b + + = 0 7

20 Ans8. = D b 4ac ( 9( a b) ) 4 9 ( a 5ab ab ) = ( a b) ( a ab b ) ( a b ab a ab b ) = = = 9 a + b ab ( a b) = 9 b ± D x = a ( a + b) ± ( a b) 9 9 = 9 x = ( a + b) ± ( a b) 9 ( a + b) ± ( a b) x = 6 a + b + a b a + b + a b x =, x = 6 6 4a + b 4a + b =, x = 6 6 a + b a + b =, x = Ans9. ATQ P r = p + 00 r = + 00 r = 00 ( ) r = 00 8

21 Ans0. Let the faster pipe takes xmint to fill the cistern and the slower pipe will take ( x + ) mint. ATQ x + x + = 40 + = x x + 40 x + + x = x + x 40 x 4x 0 = 0 x x x = 0 ( ) ( x ) x x = 0 4 x = 5, x = x = 5 Ans. ( a b) x ( b c) x ( c a) + + = 0 D = b ac 4 ( b c) 4 ( a b) ( c a) = ( ) ( ) ( ) ( b c a) ( ) = b c bc ac a bc ab = b c bc 4ac 4a 4bc 4ab = b + c + a + bc 4ac 4ab = + For equal root D = 0 ( b c a) ( b c a) + = 0 + = 0 b + c = a 9

22 Ans. Let r and r be the radius of two circles ATQ π r + π r = 6π 6... ( ) r + r = i r r = (Given) 6 r = 6 + r Put the value of r in eq. (i) r ( r ) = 6 r ( r ) ( r ) ( r + 0)( r 4) = 0 r r + r = 6 r + r 80 = 0 r + 6r 40 = 0 r + 0r 4r 40 = = 0 r r r = 0(Neglect) r = 4cm r = 6 + r = r = 0cm 0

23 CBSE TEST PAPER-05 CLASS X Mathematics (Quadratic Equation). Discriminant of x + x + = 0is (a), (b),. For equal root, kx ( x ) + 6 = 0 value of k is (c), (d), (a) k = 6 (b) k = (c) k = (d) k = 8. Quadratic equations whose roots are + s, sis (a) x 4x = 0 (b) x (c) x ( x ) x ( ) = 0 (d) x + 4x + = 0 4x + = 0 4. If αand βare root of the equations x + 5x 7 = 0, then αβ equal to (a) (b) (c) (d) 5. Solve by factorization x + = x 6. Find the ratio of the sum one produce of the roots of 7x x + 8 = 0 [] 7. If αand βare the roots of the equation find K. x + kx + = 0, such that α β = then 8. In a cricket match Kapil took one wicket less than twice the number of wickets taken by Ravi. If the product of the no. of wickets taken by these two is 5, find the no. of wickets taken by each. [] [] [] 9. The sum of a no. and its reciprocal is 7 4. Find the no. [] 0. A piece of cloth costs Rs00. If the piece was 5m longer and each metre of cloth costs Rs less the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre? [4]. ax + bx + c = 0, a 0 Solve by quadratic formula. [4]. The length of the hypotenuse of a right exceeds the length of the base by cm and exceeds twice the length of the altitude by cm. find the length of each side of the. [4]

24 CBSE TEST PAPER-05 CLASS - X Mathematics (Quadratic Equation) Ans. Ans. Ans. Ans4. (a) (a) (a) (b) [ANSWERS] Ans5. x + = x x x + = 0 x x x + = 0 ( ) ( x ) x x = 0 x = x = Ans6. x x = 0 b α + β = = a 7 c 8 αβ = = a 7 α + β = = = αβ Ans7. k α + β = α β = αβ =

25 ( ) ( ) α + β = α β + 4αβ ( k ) ( ) = + 4. k = 49 k = ± 7 Ans8. Let no. of wicket taken by Ravi = x No. of wicket taken by Kapil = x ATQ ( x ). x = 5 x x 5 = 0 5 x =, x = (Neglects) x = Ans9. Let no. be x x 7 ATQ + = x 4 x + 7 = x 4 x x + = x( x ) ( x ) ( x 4)( 4x ) = = 0 x x + = x x x + = x = 4, x = 4 Ans0. Let the length of piece = xm Rate per metre = 00 x New length = ( x + 5) New rate per metre = 00 x + 5

26 ATQ x x + 5 = ( x + ) ( x + ) x = 5 00x x = x + 5x x + 5x = 500 x + 5x 500 = 0 x x x = 0 x( x ) ( x ) ( x + 5)( x 0) = = 0 x = 5, x = 0 Rate per metre =0 Ans. ax bx c + + = 0, x x b c + x + = 0 a a b b b c + x + = 0 a a a a b b c x + = 4 a a a b b 4ac x + = a 4a b b ac 4 x + = ± a 4a b b ac 4 x = ± a a b ± b 4ac x = a b b 4ac b b 4ac + x =, x = a a 4

27 Ans. Let base = x Altitude = y Hypotenuse = h ATQ h = x + h = y + x + = y + x + = y x = y x = y x + y h x y x x + = + x x x ( x ) = 0 x x x = 0 ( x )( x ) 5 + = 0 x = 5, x = Base = 5cm Altitude = x + = 8 cm Hypotenuse = 7cm 5

28 CBSE TEST PAPER-0 CLASS X Mathematics (Arithmetic Progression). The next term of the AP in,5,7,7...is (a) 97 (b) 9 (c) 99 (d) 95. The 0 th term of the AP in, 7,, is (a) 45 (b) 47 (c) 48 (d) 50. Find the first term and the common difference 5 9,,,... [] 4. Is, 6, 9,...form AP [] 5. Which is the next term of the AP, 8, 8,,... [] 6. Find the th term from the lest term of the AP 0, 7, 4,,-6 [] 7. For what value of n are the n th term of the following two AP s are same, 9, 5,.. and 69, 68, 67, [] 8. Check whether 0 is a term of the list of no. 5,, 7,,.. [] 9. Which term of the sequence 0, 9, 8, 7,... is the first negative term? 4 4 [4] 0. The p th term of an AP is q and q th term is p. Find its ( ) th p + q term. [4]. If m times the m th term of an A.P is equal to n times its n th term, Show that the [4] ( m n) th + term of the AP is zero.. The sum of the 4 th and 8 th terms of an AP is 4 and the sum of the 6 th and 0 th terms is 44. Find the first three terms of the AP. [4] 6

29 CBSE TEST PAPER-0 CLASS X Mathematics (Arithmetic Progression) [ANSWERS] Ans. Ans. (a) (b) Ans. 5 9,,,... a d = 5 4 = = Ans4. a =, a = 6, a = a d = 6 ( ) = d = 9 6 = 6 Since d dhence not A.P Ans5., 8, 8,,... d = 8 = = Next term is 50 Ans6. a d ( ) = 6, = 7 0 = a a d = + 0 ( ) = = 7

30 Ans7. n th term of, 9, 5, = nth term of 69, 68, 67, +(n-) 6 = 69+(n-)(-) N = 9 Ans8. d = 5 = 6 a = 5 n n = 5 ( ) ( n ) a = a + n + d 0 = 5 + d So, 0 is not a term of the given list Ans9. For first negative term a n < ( n ). < 0 [ d = 0 = n + < n + < n < 0 n < 8 n > 8 8 th term is first negative term. Ans0. q = a + ( p ) d... ( i) ( ) ( ) _ p = _ a ± q d... q p = p q + d ( ii) q p = d p q d = Put the value of d in eq...(i) 8

31 ( )( ) q = a + p q = a p + a = q + p p+ q ( ) a = a + p + q d ( q p ) ( p q )( ) = = q + p p q + = 0 Ans. mam = n an ( + ( ) ) = ( + ( ) ) m a m d n a n d ma + m d md = na + n d nd ( ) ( ) a m n + m n d md + nd = 0 ( ) ( )( ) ( ) ( m n) a + ( m + n ) d = 0 a + ( m + n ) d = 0 a m n + m n m + n d m n d = 0 = 0 Hence prove a m + n Ans. a4 + a8 = 4(Given) a + d + a + 7d = 4 a + 0d = ( ) a + sd =... i a + a = 44 a + sd + a + ad = 44 a + 4d = 44 ( ) a + 7d =... ii On solving eq. (i) and (ii) d = 5, a = First three terms are -,-8,- 9

32 CBSE TEST PAPER-0 CLASS X Mathematics (Arithmetic Progression). If in any AP, A = -8.9, d =.5, an =.5 then the value of n is (a) = 5 (b) =7 (c).0 (d) =9.,, 6 the missing term in AP is (a) (b) (c) 4 (d) 8. Find the missing term in the boxes of AP,, 6 [] 4. The 7 th term of an AP exceeds its 0 th term by 7. Find the common difference [] 5. Find the sum of first n positive integers. [] 6. In AP, given l = 8, S = 44 And there are total 9 terms. Find a [] 7. In a flower bed, there are rose plants in the first row, in the second, 9 in the third and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed? [] 8. In the AP, find the missing terms in the boxes.,,, [] 9. In an A.P, the sum of first n terms is n 5 n. + find its 5 th term. [4] 0. Which term of the arithmetic progression 8,4,0,6, will be 7 more than its 4 st term.. The ratio of the sum of n terms of two AP s is ( n ) ( n ) : 9 + 6, find the ratio of [4] [4] their 8 th term.. In an AP, the first term is and the sum of the first five terms is one-fourth of the sum of next five terms. Show that 0 th term is. [4] 0

33 CBSE TEST PAPER-0 CLASS X Mathematics (Arithmetic Progression) Ans. Ans. (c) (c) [ANSWERS] Ans.,, 6 a = a = 6 a + d = 6 + d = 6 d = 4 d = a = a + d = + 4 Ans4. ATQ a7 = a0 + 7 a + 6d = a + ad + 7 7d = 7 d = Ans5.,,,...n a =, l = n n Sn = ( a + l ) n n = + = + ( n) ( n ) Ans6. In AP given l = 8, S=44, n = 9

34 n S = ( a + l ) 9 44 = + 8 a = 4 ( a ) Ans7.,, 9,, 5 a =, d = =, a n = 5 ( n )( ) 5 = + n = 0 Ans8. a =, a4 = a + d =... i ( ) ( ) a + d =... ii On solving d = 5 Put the value of d in eq. (i) a 5 = a = 8 a = 8 a = 8 = 5 Ans9. n 5n Sn = + Put n =,,... 5 S = + = S = + = = a = s = 4 a = S S = 4 = 7 d = a a, = 7 4 = a = 5 = a + 4d = = = 46

35 Ans0. a = 8, d = 4 8 = 6 Let n th term of the given AP be 7 more than its 4 st term a = 7 + a n ( n ) = n = 8 8 n = = 5 6 Ans. ATQ n ( ) a + n d 5n + 4 = n a ( n n ) d + Put n = 5on both side ( ) ( ) a + 5 d = a + 5 d a + 7d 79 = a + 7d Hence ratio of 8 th term is 79 Ans. a = S = 4 S S 4S = S S 5S ( ) = S ( 5 ) ( 0 ) a + d = a + d 5 0 [ 4 + 4d ] = [ 4 + ad ] 0 + 0d = 8 + 8d d = 6 a = a + 9d ( ) = =

36 CBSE TEST PAPER-0 CLASS X Mathematics (Arithmetic Progression). th term of the AP,,,... is (a) 8 (b) (c) -8 (d) 48. If 7 th term of an AP exceeds its 0 th term by 7. The common difference is (a) (b)- (c) (d). If x +, x, and 4x + are in A.P, Find the value of x [] 4. Find the sum of first n odd natural no. [] 5. Find the th term of the AP,,5... [] 6. Find the sum of first terms of AP, 6, 0 [] 7. Determine the AP whose third term is 6 and the 7 th term exceeds the 5 th term by []. 8. Find the sum of AP in 5 + ( 8) + ( ) ( 0) [] 9. If the sum of n terms of an AP is n 5 + nand its m th term is 64, find the value of m [4] 0. If the sum of three no. in AP, be 4 and their product is 440, find the no. [4]. If a, b, c are in A.P then prove that,, b + c c + a a + b are in A.P [4]. If S, S, Sbe the sum of n, n and n terms respectively of an AP, prove that [4] ( ) S = S S 4

37 CBSE TEST PAPER-0 CLASS X Mathematics (Arithmetic Progression) Ans. Ans. (b) (d) [ANSWERS] Ans. Since x +, xand 4x + are in A.P ( ) x = x + + 4x + 6x = 5x + 6x 5x = x = Ans4.,, 5, 7,... a = d = = n Sn = ( ) a + n d n = + ( n ). n = + = n [ n ] Ans5. a =, d = = a a d = + ( ) = + = + = 5

38 Ans6., 6, 0,... a = d = 6 = 4 S = ( ) 4 + = [ ] = = 4 Ans7. a = ( ) a + d = 6... i a = a + a + 6d = a + 4d + d = d = 6 Put the valve of d in eq. (i) a + 6 = 6 a = 6 a = 4 4,0,6... Ans8. a = 5 ( ) d = 8 5 = = a n n = 0 ( ) ( n )( ) a = a + n d 0 = = 5 n = n 6

39 Ans9. Sn = n + 5n Put n =,,,... S = 8, S = = a = S = 8 a = S S = 8 = 4 d = a a = 4 8 = 6 a m = 64 ( ) ( m )( ) a + m d = = m 6 = 64 6m = 64 6m = 6 6 m 6 = 7 Ans0. Let no. be a-d, a, a+d ( a d ) + a + ( a + d ) = 4 (Given) = 8 ( a d )( a)( a d ) ( d ) ( d ) + = = d = d = ± a = 8 a = 8 d = + d = Then AP 5,8,, Then AP,8,5,. 7

40 Ans. Given a, b, c are in AP Then b = a + c...( i) If,, are in AP then b + c c + a a + b = c + a b + c a + b c + a b + c c + a c + a a + b = c + a b + c a + b c + a ( ) ( ) ( )( ) b a c b = b + c a + b b c a ii = +... ( ) ( ) ( )( ) ( ) From (i) and (ii) proved n S a n d Ans. = + ( ) n S = ( ) a + n d n S = a + ( n ) d R.H.S = ( S S ) n n = + + n = [ 4a + 4nd d a nd + d ] n = ( a + nd d ) n = a + ( n ) d ( a ( n ) d ) ( a ( n ) d ) 8 = n n = S76 = [ 5 0] 76 = [ 5] = 8( 5) = 890 8

41 CBSE TEST PAPER-4 CLASS X Mathematics (Arithmetic Progression). Which of the following list of no. does form an AP? (a), 4, 8, 6, (c) 0., 0., 0.,.. (b), 5,, 7,. (d),, 9, 7,.. The n th term of the AP in, 5, 8,.. is (a) n- (b) n- (c) n- (d) n-. Find the sum of first hundred even natural no. divisible by 5. [] 4. Find a0 a0for the A.P 9, 4, 9, 4,... [] 5. Find the common difference and write the next two terms of the AP,5,7,7,... [] 6. Show that sequence defined by a = + nis an AP [] n 7. In a n AP, an = 4, d =, Sn = 4find nand a [] 8. Find a0 a0for the A.P in -9,-4,-9,-4,.. [] 9. The ratio of the sums of m and n terms of an AP is m th and n th term is ( m ) ( n ) :. m : n show that the ratio of the [4] 0. If the sum of first p terms of an AP is the same as the sum of its first q term, show that the sum of the first ( p q) + term is zero. [4]. a4 a6 For the A.P a, a, a,...if = find. a, a 7 8 [4]. In an AP p th, q th and r th terms are respectively a, b and c. Prove that p ( b c) + q( c a) + r ( a b) = 0 [4] 9

42 CBSE TEST PAPER-4 CLASS X Mathematics (Arithmetic Progression) Ans. Ans. (b) (a) [ANSWERS] Ans. Even natural no. divisible by 5 are 0, 0, 0. a = 0, d = 0 n = AP = ( 0 ) ( 00 ).0 + = = [ ] Ans4. a = 9 ( 4) ( 9) d = = = ( ) ( 9 ) a a = a + d = a + 9d a 9d = 0d = 0 5 = 50 Ans5., 5, 7, 7,...,5,49,7,... d = a a = 5 = 4 d = 49 5 = 4 d = 7 49 = 4 Hence it is AP a a 5 6 = 7+ 4 = 97 = = 40

43 Ans6. a = + n n So a a a a4 = 5, = 7, = 9, = 7 5 = 9 7 = 9 = Hence it is A.P Ans7. = + ( ) a a n d n ( n ) ( ) ( ) 4 = a +. a + n = 6... i n Sn = n 4 = ( a + 4) 8 = n[6 n + 4) [ a = 6 n n 5n 4 = 0 n = 7, n = a = 8 Ans8. a = ( ) ( ) d = 4 9 = = 5 = 0d = 0 5 = 50 ( 9 ) ( 9 ) a a = a + d a + d Ans9. m a ( m ) d Sm + = Sn n a + ( n ) d ( ) ( ) m m a + m d = n n a + n d m a + md d = n a + nd d 4

44 am + mnd md = an + mnd nd am an md + nd = 0 ( ) ( ) a m n m n d = 0 ( m n)( a d ) = 0 a = d + ( ). + ( + ). [ + m ] [ + n ] m = n Hence Proved ( ) ( ) am a + m d ATQ = an a + n d a m a = a n a a = a Ans0. Sp = Sq p q ( ) ( ) a + p d = a q d + p a pd d q a qd d [ + ] = [ + ] ap + p d pd = aq + q d qd ( ) ( ) ( ) a p q + p q d p q d = 0 ( ) ( )( ) ( ) ( ) ( ) a p q + p q p q d p q d = 0 4 Ans. ( Given ) p q a + p + q d = 0 Sp + q = 0 a a = 7 a + d = a + 6d a + 9d = a + d a = d 9d a = d a a =? a6 a + 5d = a a + 6d 4

45 d + 5d = d + 6d 8d = 9d = 8 : 9 Ans. ( ) A + p D = a...( i) ( )...( ) ( )...( ) A + q D = b ii A + r D = c iii ( ii) ( iii) ( ) ( ) ( ) ( ) ( )...( ) b c = q D r D b c = D q r p b c = p D q r iv Similarly ( ) = ( )...( ) ( ) = ( )...( ) q c a q D r p v r a b r D p q vi Adding (iv), (v) and (vi) p ( b c) + q( c a) + r ( a b) = 0 4

46 CBSE TEST PAPER-05 CLASS X Mathematics (Arithmetic Progression). If a, ( a ) and a are in AP than value of a is (a) - (b) - (c) (d). The Sum of first n positive integers is given by (a) ( ) n n, (b) ( ) n n +, ( ) n n + (c) (d) none of these. The first term of an AP is -7 and common difference 5. Find its general term [] 4. How many terms are there in A.P? 8,5,,..., 47 [] 5. In an A.P the sum of first nterms is n + find its nd term. n. [] 6. Show that the progression 4,7,0,,7,... is an AP 4 4 [] 7. Find the sum to n term of the AP in 5,, -, -4, -7 [] 8. Find the sum of first 4 terms of the list of no. whose n th term is given by an = + n 9. If ( p + ) th term of an A.P is twice the ( q + ) th term show that ( p + ) th term is twice [] [4] the ( p q ) + + term. th 0. The sum of four numbers in A.P is 50 and the greatest number four times the least. Find the numbers. [4]. Find the sum of all integers between 84 and 79 which are multiples of 5. [4]. If m th term of an A.P is n and the nth term is, show that the sum of mnterm is m ( ). mn + [4] 44

47 CBSE TEST PAPER-05 CLASS - X Mathematics (Arithmetic Progression) Ans. Ans. (b) (c) [ANSWERS] Ans. a = 7, d = 5 n ( ) ( n )( ) a = a + n d = = 7 + 5n 5 a n = 5n Ans4. 8 a = 8, d = 5 = a = 47 n n ( ) a = a + n d 5 47 = 8 + ( n ) = n + n = 7 Ans5. n Sn = + n Put n =,,... 6 S = = 8 S = 9 a = s = 8 a = S S = 9 8 = 45

48 Ans = = = = 4 4 Hence AP Ans7. a = 5, d = 5 = n Sn = ( ) a + n d n Sn = 5 ( )( ) + n n = [ 0 n + ] n [ n] Ans8. a = + n n Put n =,,,... a = 5, a = 7, a = a,... a = 5, d = 7 5 = 4 S 4 = 5 ( 4 ) + = = 67 [ ] Ans9. a p+ = aq+ ( ) ( ) [ ] a + p + d = a + q + d a + pd = a + qd a + pd = a + qd pd qd = a ( ) p q d = a 46

49 p+ q+ + ( + ) ( ) ( ) + + ( + ) a p+ a p d = a a + p + q + d p q d pd = = p q p q d Ans0. Let no. be ( a d ),( a d ),( a + d ),( a + d ) ( a d ) ( a d ) ( a d ) ( a d ) = 50 4a = 50 a = = ATQ 4 ( a + d ) = ( a d ) a + d = 4a d a = 5d a = 5d 5 = 5d 5 = d Numbers be 5, 0, 5, 0 Ans. 85, 90, 95, a = 85, d = 5, a = 75 ( ) ( n ) a + n d = a = 75 n = 7 7 S7 = = n n ( ) n = + + Ans. a ( m ) d... ( i) a n d ii m = + + ( )...( ) On solving (i) and (ii) a =, d = mn mn mn Smn = ( ) a + mn d mn =. ( mn ). + mn mn = ( mn + ) 47

50 CBSE TEST PAPER-0 CLASS-X Mathematics (Circles). The perimeter of a sector of a circle of radius 8 cm is 5 what is area of sector? (a) 50cm (b) 4cm (c) 5cm (d) none of these. In fig given below PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q,. If PA = 0cm and DQ = cm. what is length of PC? (a) 8cm (c) 4 cm (b) 7 cm (d) none of these. Tangent of circle intersect the circle (a) Only one point (c) Three points (b) Two points (d) None of these 4. From a point Q the length of the tangent to a circle is 4 cm and the distance of Q from the centre is 5cm. the radius of the circle is (a) 7 cm (c) 5 cm (b) cm (d) 4.5 cm 5. In two con centric circle. Prove that all chords of the outer circle which touch the [] inner are of equal length 6. PA and PB are tangents from P to the circle with centre O. At the Point M a tangent [] is drawn cutting PA at K and PB at N. Prove that KN=AK+BN 48

51 7. In the given Fig. O is the centre of the circle with radius 5 cm [] AB CD. AB CD. AB = 6cm Find OP. 8. Prove that the tangents at the end of a chord of a circle make equal angles with the [] chord 9. Two tangents TP and TQ are drawn from an external point T with centre O. as [] shown in Fig. If they are inclined to each other at an angle of 00 0, then what is the value of POQ? 0. Two concentric circles are of radii 5 cm and cm. find the length of the chord of [] the larger circle which touches the smaller circle. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC []. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q [] intersect at point T. Find the length TP. In the given fig XY and X Y are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and X Y at B. Prove that AOB = 90 X X' O P Q A 4 B Y C Y' [5] 49

52 CBSE TEST PAPER-0 Class X - Mathematics (Circles) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5. Ans6. (D) (A) (A) (A) AB and CD are two chords of the circle which touch the inner circle at M and N Respectively OM = ON AB = CD [ AB and CD are two chords of larger circle] We know that the lengths of the tangents drawn form an external point to a circle are equal PA = PB...( i) A K KA = KM...( ii) M O NB = NM...( iii) (ii) + (iii) P N B KA + NB = KM + NM AK + BN = KM + MN = KN AK + BN = KN Ans7. OP AB OP bisects AB AP = AB = 6 = cm From right OAP OA = OP + AP 5 = OP + OP = 4cm Ans8. In ADBand ADC BD = DC And ADB = ADC = 90 AD = AD [Common] ADB ADC [SAS] ABD = ACD [By C.P.C.T.] 50

53 Ans9. Ans0. Ans. Ans. TP and TQ Are tangents and O Is the centre of the circle OP PT, OQ QT TPO + TQO = 80 Quadrilateral OPTQ is cyclic? PTQ + POQ = POQ = 80 POQ = = 80 PQ be the chord of the larger circle which touches the smaller circle at the point L. since PQ is tangent at the point L to the smaller circle with centre O. OL PQ PQ is a chord of the bigger circle and OL PQ OLbisects PQ PQ = PL In = OPL PL = OP OL 5 = 5 9 = 4 chord PQ = PL =8 length of chord PQ = 8 cm AP, AS are tangents from a point A (Outside the circle) to the circle. AP = AS Similarly BP = BQ CQ = CR DR = DS Now AB + CD = AP + PB + CR + RD = AS +BQ + CQ +DS = (AS +DS) + (BQ + CQ) = AD + BC AB + CD = AD + BC Join OT TP = PQ [tangents from T upon the circle] OT PQ And OT bisects PQ PR = PQ = 4cm Now OR = OP PR OR = 5 4 = cm Now TPR + RPO = 90 [ TPO=90 ] = TPR + PTR RPO = PTR TRP TRQ [By AA similarity] 5

54 TP RP = PO RO TP 4 = 5 0 TP = cm Ans. Join OC In OAPand AOC we have AP = AC [ tangents from A to the circle are equal] AO = AO OP = OC [radius] OAP AOC[By C.P.C.T] = PAC = Similarly CBQ = 4 Now PAC + CBQ = 80 [ XY X Y } + 4 = = 90 But in AOB AOB + OAB + ABO = 80 AOB = 80 AOB + 90 = 80 AOB = 90 5

55 CBSE TEST PAPER-0 Class-X Mathematics (circles). How many tangents can a circle have? (a) (b) (c) 0 (d) infinite. If PA and PB are tangents from a point P lying outside the circle such that PA = 0 cm and APB = 60. Find length of chord AB (a) 0cm (b) 0cm (c) 0cm (d) 40cm. A tangent PQ at a point P to a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = cm the length of PQ. (a) cm (b) cm (c) 0cm (d) None of these 4. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80 then POA is equal to (a) 50 (b) 60 (c) 70 (d) Find the locus of the centre of circles which touch a given line at a given point [] 6. In the given fig, find the perimeter of ABC, if AP = 0 cm [] 7. If PA and PB are tangents drawn from external point P such that PA = 0cm and APB = 60 find the length of chord AB 8. If AB, AC, PQ are tangents in the given fig and AB = 5cm find the perimeter of APQ 9. A circle is touching the side BC of ABCat P and touching AB and AC produced at [] [] [] Q and R respectively. Prove that AQ = (perimeter of ABC ) 0. If PA and PB are two tangents drawn from a point P to a circle with centre o touching it at A and B. prove that OP is the perpendicular bisector of AB.. In the given fig. PQ is tangent at point R of the circle with centre O. if TRQ = 0 find m PRS. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC [] [] [] [5] 5

56 CBSE TEST PAPER-0 Class X - Mathematics (circles) [ANSWERS] Ans. (D) Ans. (A) Ans. (B) Ans4. (A) Ans5. Let APB be the given line and let a circle with centre O touch APB at P. then OPB = 90 let there be another circle with centre O which touches the line APB at P. Thus O ' PB = 90 OPB = O ' PB = 90 Ans6. BC touches the circle at R tangents drawn from external point to the circle are equal AP = AQ, BR = BP And CR = CQ Perimeter of ABC = AB + BC + AC = AB + ( BR + RC) + AC = AB + BP + CQ + AC = AP + AQ = AP = 0 = 0cm P B A R C Q Ans7. APB = 60 AOB = 0 [O is centre of circle] OAB = OBA = 0 each PAB = 60, PBA = 60 PABis equilateral AB = PA = 0cm Ans8. Perimeter of APQ = AP + AQ + PQ = AP + AQ + PX + XQ ( AP PB) ( AQ QC) = = AB + AC = AB = 5 = 50cm B P A R Q C 54

57 Ans9. We know that the two tangents drawn to a circle from an external point are equal AQ = AR, BP = BQ, CP = CR perimeter of ABC = AB + BC + AC = AB + BP + PC + AC = AB + BQ + CR + AC [ BP = BQ, PC = CQ] = AQ + AR = AQ = AR [ AQ = AR] AQ = AR = [perimetre of ABC ] Ans0. Let OP intersect AB at a point C we have to prove that AC = CB and ACP = BCP = 90 PA, PBare two tangents from a point P to the circle with centre O APO = BPO [ O lies on the bisector of APB] In two S ACP and BCP, we have AP = BP [ tangents from P to the circle are equal] PC = PC [Common] APO = BPO [proved] ACP BCP [By SAS rule] AC = CB [C.P.C.T] ACP = BCP C. P. C. t And [ ] But ACP + BCP = 80 ACP = BCP = 90 Hence OP is perpendicular bisector of AB Ans. Given PQ is tangent at point R and TRQ = 0 PRQ = 80 QRT = 0 TRS = 90 [ Tangent of a circle is perpendicular to Radius] PRS = 80 0 = 60 PRS = 80 0 = 60 Ans. Let the circle touch the sides AB, BC, CD and DA at the points P, Q, R, and S respectively. Join OP, OQ, OR and OS. Join OA, OB, OC and OD since the two tangents drawn from an external point subtend equal angles at the centre =, = 4, 5 = 6, 7 = 8 But = 60 [ sum of all angles around a point = 60 ] = 60 [ ] And ( ) = 60 ( + ) + ( 6 + 7) = 80 And ( 4 + 5) + ( 8 + ) = 80 AOB + COD = 80 And BOC + AOD = 80 55

58 Ans. Let the sides BC, CA, AB of ABCtouch the in circle at D, E, F respectively. Join the centre O of the circle with A, B, C, D, E, F Since tangents to a circle from an external point are equal CE = CD = 6cm BF = BD = 8cm AE = AF = xcm ( say) OE = OF = OD = 4 cm [Radii of the circle] Area of OAB = ( 8 + x) x cm...( i) =( ) Or of OBC = 4 4 = 8 cm...( ii ) Or OCA = ( 6 + x) 4 = + x...( iii ) or ABC = 6 + x + + x + 8 = ( 4x + 56 ) cm...( iv) Again perimeter of ABC = AC + AB + BC = 6 + x x ( ) ( ) x ( x) cm ( + x) = 8 + = S = = 4 + x Area of AOC = S ( s a)( s b)( s c) = ( 4 + x)( 4 + x 4)( 4 + x 6 x)( 4 + x 8 x) ( ) = = 4 + x 48x = + 67x 48 x...( v) ( ) ( ) 4x + 56 = 67x + 48x[By 4 and 5] 4x + 56 = 67x + 48x ( x ) ( x x ) = ( ) ( x )( x ) x + 4 = 4x + x = + 4 x x x x = 0 x = 7, x = 4 But x = 4is not possible x = 7 AB = 8 + x = = 5cm AC = 6 + x = = cm 56

59 CBSE TEST PAPER-0 CLASS-X Mathematics (circle). The length of tangent drawn to a circle with radius cm from a point 5 cm from the centre of the circle is (a) 6 cm (b) 8 cm (c) 4 cm (d) 7 cm. A circle touches all the four sides of a quadrilateral ABCD whose sides AB = 6 cm, BC = 7 cm, CD = 4 cm Then AD = --- (a) cm (b) cm (c) 5 cm (d) 6cm. If a point lies on a circle, then what will be the number of tangents drawn from that point to the circle? (a) (b) (c) (d) infinite 4. A quadrilateral ABCD is drawn to circumscribe a circle IF AB = 4 cm, CD = 7 cm, BC = cm, Then length of AD? (a) 7 cm (b) cm (c) 8 cm (d) none of these 5. Find the unknown length x [] 8cm 6. In the given fig OD is perpendicular the chord AB of a circle CA whose centre is O. if BC is a diameter. Find? OD = [] 57

60 7. In the given fig XP and XQ are tangents from X to the circle with centre O. R is a point on the circle such that ARB is a tangent to the circle prove that XA + AR = XB + BR 8. Prove that the segment, joining the points of contact of two parallel tangents, passes thro the centre. [] [] 9. Prove that parallelogram circumscribing a circle is a rhombus [] 0. If two tangents are drawn to a circle from an external point then (i) They subtend equal angles at the centre (ii) They are equally inclined to the segment joining the centre to that point. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. prove that PTQ = OPQ [] []. The lengths of two tangents drawn from an external point to a circle are equal []. In the given fig, PT is tangent and PAB in a secant If PT = 6 cm, AB = 5 cm. Find the length PA [5] 58

61 CBSE TEST PAPER-0 Class X - Mathematics (circle) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5. Ans6. Ans7. Ans8. Ans9. (C) (B) (A) (C) PT is tangent to a circle and PAB is a secant PA. PB = PT ( x) = x = 64 9 x = = 7.8cm 8 Since BC is a diameter CAB = 90 Also OD AB ODB = 90 ACB DOB [ CAB = ODB = 90 ] ABC = DBO [common] CA CB r = = = OD OB r In the given fig XP and XQ are tangents from external point XP = XQ...( i) AR = AP...( ii) BR = BQ...( iii) [ length of tangents are equal from external point] Xp = XQ XA + AP = XB + BQ [By (ii) and (iii)] XA + AR = XB + BR [By (ii) and (iii)] Given two parallel tangents PQ and RS of a circle with centre O Draw line OC RS i.e. PQ. PAO + COA = 80 = 90 + COA = 80 COA = 90 Given ABCD is a parallelogram in which all the sides touch a given circle 59

62 To prove:- ABCD is a rhombus Proof:- ABCD is a parallelogram AB=DC and AD = BC Again AP, AQ are tangents to the circle from the point A AP = AQ Similarly BR = BQ CR = CS DP = DS AP + DP + BR + CR = AQ + DS + BQ + CS = AQ + BQ + CS + DS ( ) ( ) ( ) ( ) AD + BC = AB + DC BC + BC = AB + AB [ AB = DC, AD = BC] BC = AB BC = AB Hence parallelogram ABCD is a rhombus Ans0. Ans. Given on a circle C (o,r), two tangents AP and AQ are drawn from an external point A. To prove (i) AOP = AOQ (ii) OAP = OAP Construction:- Join AO, PO, QO Proof: In APQ and AQO AP = AQ [Length of the tangents drawn from an external point] PO = QO [Radii of the same circle] AO = AO [common] APO AQO [By sss theorem of congruence] (i) AOP = AOQ [C.P.C.T] (ii) OAP = QAO [By C.P.C.t.] Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact To Prove: PTQ = OPQ Proof: Let PTQ = θ Since TP, TQ are tangents drawn from point T to the circle. TP = TQ TPQ is an isocler triangle TPQ = TQP = ( 80 θ ) θ θ = 90 Since TP is a tangent to the circle at point of contact P OPT = 90 OPQ = OPT TPQ 60

63 θ = θ = = PTQ Thus PTQ = OPQ Ans. Ans. Given P is an external point to the circle C(o,r). PQ and PR are two tangents from P to the circle To Prove PQ = PR Construction Join OP Proof: a tangent to a circle is perpendicular to the radius through the point of contact OQP = 90 = ORP Now in right triangles POQ and POR OQ = OR [Each radius r] Hyp. OP = Hyp. OP [common] POQ POR [By RHS rule] PQ = PR Join OT, OA, OP. Draw OM AB Let radius of the circle = r OT PT [ OT is radius and PT is a tangent] OP = PT + OT [from right OPT ] = 6 + r OP r = 6 OP OA = 6...( i) [ OA = OT = r] Also from right OMA OA = OM + AM OP 6 = OM + AM OP OM AM = 6 PM AM = 6 ( PM AM )( PM AM ) ( PM + AM ) PA = 6 ( PM MB) PA 6 ( PB)( AP) = 6 PA( PA + AB) = 6 + = 6 + = [ AM = MB, OMbisects AB] PA + PA = ( PA )( PA ) = 0 PA = 4. or PA = 9[it cannot be -ve] 6

64 CBSE TEST PAPER-04 CLASS-X Mathematics (circle). How many normal s can a circle have? (a) 0 (b) (c) (d) Infinite. A straight line can meet a circle in not more than points (a) one (b) two (c) Tree (d) none of these. A tangent PQ at point P of a circle of radius cm meets a line thro the centre O to a point Q so that OQ = 0 cm Length of PQ is (a) 4 cm (b) 5 cm (C) 6 cm (d) 0 cm 4. A line intersecting a circle in two points is called (a) Tangent (b) secant (c) diameter (d) none of these 5. In fig if OL = 5 cm, OA = cm, Find AB? [] 6. In the given fig ABCD is a cyclic quadrilateral and PQ is a tangent to the circle at C. If BD is a diameter, OCQ = 40 and ABD = 60, find BCP [] 7. Two chords AB and CD of a circle Intersect each other at P outside the circle. If AB = 5 cm, BP = cm and PD = cm, find CD [] 6

65 8. In the adjoining fig, if AD, AE and BC are tangents to the circle at D, E and F respectively than AD=AB+BC+CA [] 9. The circle of ABCtouches the sides BC, CA and AB at D,E,F, respectively. If AB = AC. Prove that BD = CD 0. A circle touches the side BC of a ABCat a point P and touches AB and AC when [] [] produced at Q and R respectively show that AQ = [Perimeter of ABC ]. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ = OPQ []. Prove that the parallelogram circumscribing a circle is a rhombus []. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6cm respectively. Find the sides AB and AC [5] 6

66 CBSE TEST PAPER-04 CLASS X - Mathematics (Circle) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5. Ans6. Ans7. Ans8. (D) (B) (C) (B) AB = AL = OA OL = 5 = 69 5 = 44 = = 4cm BD is a diameter BCD = 90 [Angle in the semi circle] BCP = = 50 Two chords AB and CD of a circle intersect each other at P PA. PB = PC. PD [Bach = length of tangent from P] AB + PB. PB = ( PD + PC) PD ( ) ( 5 )() ( x) + = + 4 = ( + x) x = 0 CD = 0cm CD = CF, BE = BF CD + BE = CF + BF = BC AD = AB + BC + AC Now AD = AC + CD = AC + CF AE = AB + BE = AB + BF AD + AE = AB + AC + BC 64

67 Ans9. Tangents drawn from an external point to a circle are equal in length AF = AE [Tangents from A] (i) BF = BD [Tangents from B] (ii) CD = CE [Tangents from C].(iii) Adding (i)(ii)and(iii) we get AF + BF + CD = AE + BD + CE AB + CD = AC + BD But AB = AC (given) CD = BD Ans0. Since the tangents from an external point to a circle are equal in length, therefore BP = BQ (i)[from point B] CP = CR.(ii)[from point C] And AQ = AR..(iii)[From point A] From () we have AQ = AR AB + BQ = AC + CR AB + BP = AC + CP...( iv) [using(i) and (ii)] Now perimeter of ABC AB + BC +AC = AB +(BP + PC) + AC = (AB + BP) + (AC +PC) = (AB + BP) [using 4] = (AB + BQ) [using ] = AQ AQ = (perimeter of ABC) Ans. Given A circle with centre O. an external point T and two tangents TP and TQ to the circle, where P and Q the points of contact To prove PTQ = OPQ Proof let PTQ = θ In TPQ, we have TP = TQ [length of the tangents drawn from an external point to a circle are equal] So. TPQ is an isosceles triangle TPQ = TOP...( i) In TPQ, we have TPQ + TQP + PTQ = 80 [ sum of three angles of a is 80 ] 65

68 TPQ + Q = 80...( i) TPQ = 80 θ TPQ = (80 θ ) = 90 θ...( ii) But OPT = 90...( iii) [Angle between the tangent and radius of a circle is 90 ] Now OPQ = OPT TPQ = θ = θ = PTQ OPQ = PTQ PTQ = OPQ Ans. Ans. Given ABCD be the parallelogram circumscribing a circle with centre O such that the sides AB, BC, CD and DA touch a circle at P, Q, R and S respectively. To prove : gm ABCD is a rhombus. Proof : AP = AS..(i) BP = BQ.(ii) CR = CQ.(iii) DR = DS (iv) [Tangent drawn from an external point to a circle are equal] Adding (i), (ii), (iii) and (iv), we get Ap + BP + CR + DR = AS + BQ + CQ + DS ( AP + BP) + ( CR + DR) = ( AS + DS ) + ( BQ + CQ) AB + CD = AD + BC Let the sides BC, CA, AB of ABCtouch the in circle at D, E, F respectively. Join the centre O of the circle with A, B, C, D, E, F Since tangents to a circle from an external point are equal CE = CD = 6cm BF = BD = 8cm AE = AF = xcm ( say) OE = OF = OD = 4 cm [Radii of the circle] = Area of OAB ( x) =( + ) 6...( ) x cm i 66

69 Or of OBC = 4 4 = 8 cm...( ii ) Or OCA = ( 6 + x) 4 = + x...( iii ) ABC = 6 + x + + x + 8 = 4x + 56 cm...( iv) or ( ) Again perimeter of ABC = AC + AB + BC x ( x) ( ) x ( x) cm ( + x) = = 8 + = S = = 4 + x Area of AOC = S ( s a)( s b)( s c) ( 4 x)( 4 x 4)( 4 x 6 x)( 4 x 8 x) = ( ) = = 4 + x 48x = + 67x 48 x...( v) ( ) ( ) 4x + 56 = 67x + 48x[By 4 and 5] 4x + 56 = 67x + 48x ( x ) ( x x ) = ( ) ( x )( x ) x + 4 = 4x + x = + 4 x x x x = 0 x = 7, x = 4 But x = 4is not possible x = 7 AB = 8 + x = = 5cm AC = 6 + x = = cm 6cm xcm xcm 8cm 67

70 CBSE TEST PAPER-05 CLASS-X Mathematics (Circle). The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. what will be the radius of circle? (a) cm (c) cm (b) cm (d) none of these. In the fig given below, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = cm and CQ = cm, what is length of PC (a) 9 cm (c) cm (b) 0 cm (d) cm. The tangent of a circle makes angle with radius at point of contact. (a) 60 (b) 0 (c) 90 (d) none of these 4. If tangent PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80, then what is the value of POA? (a) 0 (b) 50 (c) 70 (d) In fig PA and PB are tangents from P to the circle with centre O. R is a point on the circle Prove that PC + CR = PD + DR [] 6. The length of tangents from a point A at distance of 6 cm from the centre of the circle is 0cm what will be the radius of the circle? [] 68

71 7. In the given fig If TP and TQ are the two [] tangents to a circle with centre O so that POQ = 0 then find PTO 8. In the fig given below PA and PB are tangents to the circle drawn from an external point P. CD is a third C A [] tangent touching the circle at Q. If PB = 0 cm and P Q CQ = cm what is length of PC? D B 9. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord [] 0. In the given fig. if AB = AC prove that BE = EC []. Find the locus of centre of circle which two intersecting lines []. In the given fig. a circle is inscribed in a quadrilateral ABCD in which B = 90. If AD = cm AB = 9 cm and DS = 5 cm, find the radius of the circle []. From a point P two tangents are drawn to a circle with centre O. if OP = diameter of the circle, show that APB is equilateral [6] 69

72 CBSE TEST PAPER-05 CLASS X- Mathematics (Circle) Ans. Ans. Ans. Ans4. (C) (A) (C) (B) [ANSWERS] Ans5. And6. Since length of tangents from an external point to a circle are equal in length PA = PB CA = CR.(i) And DB = DR Now PA = PB PC + CA = PD + DB PC + CR = PD + DR By (i) Since tangents to a circle is perpendicular to radius through the point of contact OTA = 90 In right OTA = 90, we have T OA = OT + AT (6) = OT + (0) OT OT = = 576 OT = 4 A 0cm 6cm O Hence radius of circle is 4 cm Ans7. Since POQ + PTO = 80 [ OPT = 90, OQT = PTQ = 80 PTQ = 80 0 = 70 Ans8. Ans9. PA=PB=0 cm CQ = CA = cm PC = PA CA = 0 = 8 cm Let NM be chord of circle with centre C Let tangents at M.N meet at the point O. Since OM is a tangent OM CM i.e. OMC=90 ON is a tangent ON CN i.e. ONC = 90 70

73 Again in CMN, CM = CN = r CMN = CNM OMC CMN = ONC CNM OML = ONL Thus tangents make equal angle with the chard Ans0. Since tangents from an exterior point A to a circle are equal in length AD = AF...( i) Similarly tangents from an exterior point B to a circle are equal in length BD = BE...() Similarly for C CE = CF.() Now AB = AC AB AD = AC AD AB AD = AC AF...( by ) BD = CF BE = CF...( by ) BE = CE [ BD = BE, CE = CF] (by ) Ans. Let l, l be two intersection lines. Let a circle with centre P touch the two lines l and l at M and N respectively. PM = PN [Radii of same circle] P is equi distance from the lines l and l similarly centre of any other circle. Which touch the two intersecting lines l, l will be equidistant from l and l P lies on la bisector of the angle between l and l [ the locus of points equidistant from two intersecting lines is the pair of bisectors of the angle between the lines] Hence locus of centre or circles which touch two intersecting lines is the pair of bisectors of the angles between the two lines. Ans. In the given fig OP BCand OQ + BA Also, OP = OQ =r OPBQis a square BP = BQ =r But DR = DS = 5 cm..(i) 7

74 AR = AD DR = 5 = 8cm AQ = AR = 8cm BQ = AB AQ = 0 8 = cm r = cm Ans. Join OP Suppose OP meets the circle at Q Join AQ we have i.e. OP = diameter OQ + PQ = diameter PQ = Diameter radius [ OQ = r] PQ = radius Thus, OQ = PQ = radius Thus OP is the hypotenuse of right triangle OAP and Q is the mid point of OP OA = AQ = OQ [ midpoint of hypotenuse of a right triangle is equidistant from the vertices] OAQis equilateral AOQ = 60 So APO = 0 APB = APO = 60 Also PA = PB PAB = PBA But APB = 60 PAB = PBA = 60 Hence APBis equilateral c 7

75 CBSE TEST PAPER-0 CLASS-X Mathematics (Construction). Construct a ABC in which AB = 6.5 cm, B = 60 and BC = 5.5cm Also construct a [6] triangle ABC similar to the ABC ABC whose each side is times the corresponding side of. Draw a circle of radius 4cm. from a point P,7cm from the centre of the circle, draw a pair of tangents to the circle measure the length of each tangent segment. Draw a right triangle in which the sides containing the right angle are 5cm and [6] [6] 4. 4cm construct a similar triangle whose sides are 5 times the sides of the above triangle Construct a ABC in which CA=6cm AB=5cm and BAC = then construct a [6] triangle similar to the given triangle whose sides are 6 of the corresponding sides 5 of the ABC 5. Construct a circle whose radius is equal to 4cm. Let P be a point whose distance from its centre is 6cm. construct two tangents to it form P. 6. Draw a triangle ABC with sides BC = 6.cm, AB = 5.cm and ABC = 60. Then [6] [6] construct a triangle whose sides are 4 times the corresponding sides of ABC 7

76 CBSE TEST PAPER-0 CLASS - Mathematics (Construction) Ans. Steps of construction. Draw a line segment AB = 6.5cm. At B construct ABX = 60. With B as centre and radius BC = 5.5 cm draw an arc intersecting BX at C 4. Join AC. Triangle so obtained is the required triangle [ANSWERS] 5. Construct an acute angle BAY at A on opposite side of vertex C of ABC 6. Locate points A, A, Aon AY such that AA = A A = A A 7. Join Ato B and draw the line through AParallel to A B intersecting the extended line segment AB at B 8. Draw a line through B parallel to BC intersecting the extended line segment AC at C 9. AB C so obtained is the required triangle Ans. Steps of construction. Take a point O in the plane of a paper and draw a circle of the radius 4 cm.. Make a point P at a distance of 7cm from the centre O and Join OP. Bisect the line segment OP. Let M be the mid point of OP. 4. Taking M as a centre and OM as radius draw a circle to intersect the given circle at the points T and T 5. Join PT and PT Then PT and PT are required Tangents PT = PT = 5.75cm 74

77 Ans. Steps of construction. Draw a line segment BC = 5cm. At B construct CBX = 90. Which B as centre and radius = 4cm draw an arc intersecting the line BX at A. 4. Join AC to obtain the required ABC 5. Draw any ray BY making an acute angle with BC on the opposite side to the vertex A 6. Locate 5 points B, B, B, B4,and B5on BY so that BB = B B = BB = BB4 = B4B5 7. Join Bto C and draw a line through B5parallel to BC intersecting the extended line segment BC at C 8. Draw a line through C parallel to CA intersecting the extended line segment BA at A Thus A BC is the required right triangle Ans4. Steps of construction. Draw a line segment AB = 5cm. At A make BAY = 45. Take A as centre and radius = AC = 6cm draw an arc cutting AY at C. 4. Join BC to obtain the ABC 5. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C. 6. Locate 6 points A, A, A, A4, A5 and A6on AX so that AA = A A = A A = A A4 = B4B5 = B5B6 7. Join A5to B and draw a line through A6 parallel to A5 B intersecting the extended line segment AB at B 8. Draw a line through B parallel to BC intersecting the extended line segment AC at C. Then AB C is required triangle 75

78 Ans.5 Steps of construction. Take a point O in the plane of the paper and draw a circle of radius 4cm.. Make a point P at a distance of 6cm from the centre O and join OP. Bisect the line segment OP. Let the point of bisection be M. 4. Taking M as centre and OM as radius, draw a circle to intersect the given circle at the point T and T 5. Join PT and PT to get the required tangents, Ans.6 Steps of construction. Draw a line segment BC = 6.cm. At B make CBX = 60. With B as centre and radius equal to 5.cm, draw an are intersecting BX at A 4. Join AC, Then ABC is the required triangle 5. Draw any ray by making an acute angle with BC on the opposite side to the vertex A 6. Locate le points B, B, B, B4, on by so that BB = B B = BB = BB4 7. Join Bto C and draw al line through B4parallel to BC intersection the extended line segment BC at C 8. Draw a line through C parallel to CA intersecting the extended line segment BA at A Thus A BC is the required triangle. 76

79 CBSE TEST PAPER-0 CLASS-X Mathematics (Construction). Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at [6] 60. Draw the tangents at the extremities of a diameter AB of a circle of radius cm. Are [6] these tangents parallel. Given reasons?. Construct the pair of tangents form a point cm away from the centre of a circle of [6] radius cm and measures their tangents. 4. Construct a ABCin which AB = 4cm, BC = 5cm and AC = 6cm. Now construct a [6] triangle similar to ABCsuch that each of its sides is two third of the corresponding sides of ABC Also prove your assertion. 5. Construct a ABCin which AB = 6.5cm B = 60 and BC = 5.5cm. Also construct a [6] AB ' C ' similar to ABCwhose each side is times the corresponding sides of the ABC 6. Construct a ABCwhose sides are 7.5cm, 7cm, and 6.5cm. construct another [6] similar to ABCand with sides rd of the corresponding sides of ABC 77

80 CBSE TEST PAPER-0 CLASS X- Mathematics (Construction) [ANSWERS] Ans. Steps of construction. Draw a circle with center O and radius 5cm. Draw any diameter AOC. Construct BOC = 60 meeting the circle at B 4. At A and B draw perpendiculars to OA and OB intersecting at P. 5. PA and PB are required tangents and APB = 60 Ans. Steps of construction. Draw a circle of radius cm. Draw any diametre AOB. Draw AT AB and BM AB 4. AT and BM are tangents extremities of a dia AB 5. = 90, = 90, = are alternate angles AT BM Ans. Steps of construction. Draw a circle with centre O and radius cm. Draw line segment OP = 5cm Q. Bisect OP let point of bisection be M 4. Taking M as a centre and OM radius draw a circle P 5cm M O which intersect the given circle at Q and R 5. Join PQ and PR R 6. Length of tangents PQ = PR = 4.6cm 78

81 Ans.4 Steps of construction. Draw ABCwith sides BC = 5cm, AB = 4cm and AC = 6cm. Below BC make acute CBX. Along BX mark off three points B, B, B, such that BB = B B = BB 4. Join BC 5. Draw BC ' BC also C ' A' CA 6. BC ' A' is required Ans.5 Steps of construction X. Construct a ABCis which AB = 6.5cm, B = 60 and BC = 5.5cm. Draw a ray AX making any acute angle with AB A F on the apposite side of the vertex C.. Cut three equal parts from AX say AX = XX = X X 4. Join X to B. 6cm B O' D D E O8cm P C 5. From X draw X B ' X Bat B' 6. At B draw B' C ' BCintersecting AY at C 7. AB ' C ' is required triangle similar to ABC Ans.6 Steps of construction. Draw ABCwith sides BC = 7.5cm AB = 7cm and AC = 6.5cm. Make acute angle CBX. Along BX, cut off BX = XX = X X 4. Join X C 5. Draw X C ' X C 6. Draw C ' A' CAThen A' BC ' is required triangle 79

82 CBSE TEST PAPER-0 CLASS-X Mathematics (Construction). Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at [6] 60. Draw a circle of radius 4cm with centre O. Draw a diameter POQ. Through P or Q [6] draw tangent to the circle. Draw a circle of radius.cm take a point P on it. Without using the centre of the [6] circle. Draw tangent to it at P. 4. Construct a triangle with sides 5cm, 6cm and 7cm and then another triangle whose [6] sides are 7 of the corresponding sides of first triangle 5 5. Let ABC be a right triangle in which AB = 6cm, BC = 8cm and B = 90 BD is the [6] perpendicular from B on AC. The circle through B,C and D is drawn construct the tangents form A to this circle. 6. Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it [6] whose sides are of the corresponding sides of the first triangle 80

83 CBSE TEST PAPER-0 CLASS - Mathematics (Construction) [ANSWERS] Ans. Steps of construction. Draw a circle with centre O and radius 5cm. Draw any radius OT.. Construct TOT ' = = 0 4. Draw TP OTand TP OTThen PT ' and PT are the two required tangents such that TPT ' = 60 Here PT = PT ' Ans. Steps of construction. Draw a circle of radius 4cm. Draw diameter POQ. Construct PQT = Produce PQ to T ' Then TQT ' is the required tangent at the point Q. Ans. Steps of construction. Draw a circle of radius.cm and take a point P on it. R Q. Draw chord PQ. Mark a point R in the major arc QP 4. Join PR and RQ T P T' 5. Draw QPT = PRQ 6. Produce TPto T ' as shown in fig then TPT ' is the required tangent at the point P. 8

84 Ans.4 Steps of construction. Draw a line segment AB = 5cm C'. With A as centre and radius 6cm draw an arc. C. Again B as centre and radius 7cm draw another are cutting the previous are at C. Join 7cm AC and BC then ABC is required triangle 6cm 4. Draw any ray AX making acute angle 5. Locate 7 points A, A, A, A4, A5, A6, A7on AX so A 5cm B B' that A AA = A A = A A = A A4 = A4 A5 = A5 A6 = A6 A7 A A 6. Join A5to B and draw a line through A7 Parallel A 4 A 5 to A5 B intersecting the extended line segment AB at B ' 7. Draw B ' C ' BCthen AB ' C ' is the required triangle A 6 A 7 Ans.5 Steps of construction. Draw ABCwith BC = 8cm, AB = 6cm and B = 90. Draw perpendicular BD from B to AC. Let O be the mid point of BC. Draw a circle with centre O and radius OB = OC. This circle will pass through the point D 4. Join AO and bisect AO 5. Draw a circle with centre O ' and O ' Aas radius cuts the previous circle at B and P 6. Join AP, AP and AB are required tangents drawn from A to the circle passing through B,C and D. 6cm A B X O' D D E O8cm P F C 8

85 Ans.6 Steps of construction. Draw ABCwith AB = 4cm, BC = 6cm and AC = 5cm.. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.. Locate points B, Band Bon BX so that BB = B B = BB 4. Join BCand draw BC ' BC 5. Draw a line through C ' such that C ' A' CAThen A' BC ' is the required triangle 8

86 CBSE TEST PAPER-05 CLASS-X Mathematics (Construction). Construct a ABCin which AB = 6.5cm B = 60 and BC = 5.5cm. Also construct a [6] triangle AB C similar to ABCwhose each side is times the corresponding side of the ABC. Draw a triangle ABC with sides BC = 6.cm. AB = 5.cm and ABC = 60 Then [6] construct a triangle whose sides are 4 times the corresponding sides of ABC. Draw a right ABC with AB = 4.5cm, AC = 7.5cm and B = 90 construct a similar [6] triangle whose sides are 4 5 times the sides of ABC 4. Draw a circle of radius 6cm form a point 0cm away form its centre construct the [6] pair of tangents to the circle and measure their length 5. Draw a ABCwith side BC = 7cm, B = 45 A = 05 then construct a triangle [6] whose sides are 4 times the corresponding sides of ABC 6. Draw a circle of radius 4cm take a point P outside the circle. Without using the [6] centre of the circle draw two tangents to the circle form point P. 84

87 CBSE TEST PAPER-05 CLASS - Mathematics (Construction) [ANSWERS] Ans. Steps of construction. Draw ABCin which AB = 6.5cm, B = 60 and BC = 5.5cm. Construct BAY. Locate points A, A, Aon AY such that AA = A A = A A 4. Join Ato B and draw a line through A parallel to A Bintersecting the extended line segment AB at B ' 5. Draw B ' C ' BC 6. AB ' C ' is the required triangle Ans. Steps of construction X. Draw ABC is which BC = 6.cm, AB = 5.cm A' ABC = 60 A. Draw any ray BY making an acute angle with BC. Locate 4 Points B, B, B, and B4on BY 4. Join BCand draw B4C ' BC 5. Draw C ' A' CA B B 6.cm C C' 6. A' BC ' is required triangle B B B 4 Y 85

88 Ans. Steps of Construction. Draw ABCin which AB = 4.5cm, B = 90 and AC = 7.5cm. Draw any ray BY making an acute angle ABY 7.5cm C C'. Locate 5 points B, B, B, B4, B5on BY so that BB = B B = BB = BB4 = B4B5 A A' 4.5cm B 4. Join B5to A and draw B4 A' B5 A B B 5. Now draw A' C ' AC 6. A' BC ' is the required right triangle Y B 5 B 4 B Ans.4 Steps of construction. Draw a circle of radius 6cm. Mark a point P at a distance 0cm from centre O and Join OP. Bisect the line segment OP Let M be the mid point of OP 4. Taking M as centre and OM radius draw a circle to intersect the given circle at point T and T' 5. Join PT and PT' 6. PT and PT'are required tangents and PT = PT'=8cm Ans.5 Steps of construction. Draw line segment BC = 7cm construct B = 45 and C = 80 ( ) = 0. Suppose BX and CY intersect at A then ABC is the given triangle. Construct an acute angle CBZ 4. Locate 4 points B, B, B, B4on BZ such that BB = B B = BB = BB4 5. Join Bto C and draw B4C ' BC 6. A' BC ' is the required triangle 86

89 Ans.6 Steps of construction. Draw a circle of radius 4cm. Take a point P outside the circle and draw a secant PAB meeting the circle at A and B.. Produce AP to C such that AP = CP bisect CB at Q 4. Draw a semicircle with CB as diameter 5. Draw PD CBmeeting the semi circle at D 6. With P as centre and PD as radius draw an are to meet the given circle at T and T' 7. Join PT and PT' then PT and PT'are the requi9red tangents 87

90 CBSE TEST PAPER-0 CLASS-X Mathematics (Area related to circle). The circumference of a circular field is 58cm. Then its radius is (a) 4cm (c) 7cm (b) 84cm (d) 56cm. The circumference of a circle exceeds its diameter by 80cm. Then its radius is (a) cm (c) 40cm (b) 6cm (d) 4cm. Area of the sector of angle 60 of a circle with radius 0cm is 5 8 (a) 5 cm (b) 5 cm 4 (c) 5 cm (d) none of there 4. Area of a sector of angle P of a circle with radius R is P (a) π R (b) 80 P 80 π R P (c) π R (d) 60 P π R Find the circumference of a circle of diameter 4cm [] 6. The diameter of a circular pond is 7.5m. It is surrounded by a path of width.5m. Find the area of the path. [] 7. Find the area of a sector of a circle with radius 6cm. if angle of the sector is 60 [] 88

91 8. An arc of a circle is of length 5π cmand the sector is bounds has an area of find the radius of the circle. 0π cm [] 9. The cost of fencing a circular field at the rate of Rs. 4 per metre is Rs The field is to be ploughed at the rate of Rs Per m. Find the cast of ploughing the field. π = 7 [] 0. The length of the minute hand of a clock is 4cm Find the area swept by the minute hand in 5 minutes []. An umbrella has 8 ribs which are equally spaced as given in the figure. Assuming umbrella to be a flat circle of radius 45cm. find the area between two consecutive ribs of the umbrella. []. Find the area of the shaded region if PQ = 4cm PR = 7cm and O is the centre of the circle. []. The area of an equilateral ABCis 7.5cm with each centre of the triangle as vertex a circle is drawn with radius equal to half the length of the side of triangle [5] find the area of the shaded region. π =.4 =

92 CBSE TEST PAPER-0 CLASS - Mathematics (Area related to circle) [ANSWERS] Ans. Ans. Ans. Ans4. (B) (D) (B) (D) Ans5 d = 4cm Ans6 r = 7cm Circumference = π r = 7 = 44cm 7 Diameter of pond = 7.5cm R = 8.75m Width =.5m Outer radius = R = =.5m Now Area of path = π R r [ R r)( R r) = π + = =.50 = 7 = 7 = m [ )( ) Ans7 We know that Area of sector θ = 60, r = 6 θ = π r 60 90

93 Required area = ( ) 6 6 = = = cm Ans8 Let r be the radius of the circle by the given condition, we have Area of sector x 5 π r = x = 00 5 OAPB = of the area of the circle 8 Ans9 since for Rs. 4 The length of fencing = metre for Rs 580, the length of fencing = 580 = 0m 4 Perimeter i.e. circumference of the field = 0m Let r be radius of the field. π r = r = 0 r = 7 40 = 5 7 = 5m Area of the field = ( ) π r = π r = π 5 = 5π m Rate = Rs 0.50 Per m Total cast of plareghing the field Rs 5 = Rs 7 ( 5π 0.50) [ ) = Rs 75 = Rs95 Ans0 Angle covered by minute hand in 60 minutes = Angle covered in minutes = = 6 60 Angle covered in 5 minutes = 6 5 = 0 9

94 We know that area swept by the minute hand during this period =Area of sector with sector angle 0 θ 0 ( π r ) = = 8 = 7 = 5 cm Ans Area between two consecutive ribs of the umbrella = Area of the sector with angle 60 made at centre 8 60 / 8 = π ( 45) 60 = ( 45) = 05cm = cm nearly. Ans Area of the shaded region = Area of the semi-circle with O as centre and OQ as radius- ar of PQR...( i) Since QR is a diameter passing thro the centre O of the circle RPQ = 90 [Angle of semi-circle] QR = PR + PQ = = = 65 = 5 QR = 5 diameter of the circle = 5cm r = 5 cm area of the semi circle = π r 5 5 = 7 65 = cm 8 9

95 Also area of PQR = PR PQ = 7 4 = 7 = 84cm Hence area of the shaded region = 84 = = = 8 8 = 6.644cm Ans Area of shaded region = area of ABC (Area of sector BPR) Let a be the side of the equilateral a = 00 = 00 a = = a = a = = π Required area = ( ) = = = = 60.5cm 9

96 CBSE TEST PAPER-0 CLASS-X Mathematics (Area related to circle). If the sum of the circumferences of two circles with radii R and R is equal to the circumference of a circle of Radius R then (a) R + R = R (b) R + R > R (c) R + R < R (d) None of these. If the perimeter of a circle is equal to that of a square, then the ratio of their area is (a) :7 (b) 4: (c) 7: (d) :4. Area of a sector of angle 0 p of a circle with radius R is P (a) π R (b) 80 P (c) π R (d) 60 P 80 π R P π R Area of the sector of angle 60 of a circle with radius 0cm is 5 8 (a) 5 cm (b) 5 cm 4 (c) 5 cm (d) none of there 5. Find the area of the shaded region where ABCD is a square of side 4cm [] 6. The radius of a radius of a circle is 0cm Three more concentric circles are drawn inside it in such a manner that it is divided into four parts of equal area. Find the radius of the largest of the three concentric circle [] 94

97 7. OACB is a quadrant of a circle with centre O and radius 7 cm. it OD = 4cm, then find area of shaded region. [] 8. A pendulum swings through on angle of 0 and describes an arc 8.8cm in length. Find the length of pendulum. 9. The perimeter of a sector of a circle of radius 5.7m is 7.m calculate. (i) The length of arc of the sector in cm. (ii) The area of the sector in cm correct to the nearest cm [] [] 0. Find the area of shaded region in the given fig. where ABCD A B [] is a square of side 0cm and semi-circles are drawn with each side of square as diameter. [ π =.4] IV I III II D 0cm C. Find the area of the shaded region where a circular are of [] radius 6cm has been drawn with vertex O of an equilateral O triangle OAB of side cm as centre P Q A B. In the given fig. ABC is an equilateral triangle inscribed in a A [] circle of radius 4 cm and centre O. Show that the area of the 4 4 π cm shaded region is ( ) O D B P C. The given fig depicts a racing track whose left and [5] right ends are semi circular the difference between the two inner parallel line segments is 60m and they are each 06m long. If the track is 0m wide find. (i) The distance around the track along its inner edge (ii) The area of the track. 95

98 CBSE TEST PAPER-0 CLASS X Mathematics (Area related to circle) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5 (A) (C) (D) (B) Area of square =96cm ABCD = 4 4cm 4 Diameter of each circle = = 7cm radius of each circle = 7 cm Ans6 Let r be the radius of the largest of the three circles Area of largest circle π r = π 4 ( 0) = [area of given circle] 4 r = 00 r = 00 = Ans7 Area of quadrant OACB = π ( ) = = cm Area of shaded region = Area of quadrant OACB ar of OAD = ( 7 4) = 4 = = 4.5cm 96

99 Ans8 Ans9 (ii) Let r be the length of pendulum π π AOB = 0 = 0 = 80 6 l θ = r π 8.8 = 6 r r = = 6.8 cm π Let O be the centre of a circle of radius 5.7m and OACB be the given sector with perimeter 7.m Then OA=OB=5.7m Now OA + OB + arc AB = 7.m 5.7m m + arc AB = 7.m arc AB = 7..4 = 5.8m Length of arc AB = 5.8m Area of sector OACB = radius arc A O 0 0 r B = cm = 45.0cm Area of sector correct to nearest cm = 45cm Ans0 Area of region I + II = area of ABCD area of two semicircles of each radius 5cm = 0 0 π 5 = 00 5π = =.5cm Similarly area of II + ar IV = Area of region I, II, III, and IV Thus area of shaded region.5cm = Area ABCD Area of (I, II, III, IV) = 00 4 = 57cm =.5 = 4cm 97

100 Ans Area of shaded region = Area of major sector OPLQO + Area of equilateral OAB 00 = π ( 6) ( ) = 6 cm Ans In OBD Let BD = a, OB = 4cm BD sin 60 = OB 9 = 4 4 a = = cm BC = a = = 4 cm OD = cos 60 OD = 4 = cm 4 Area of shaded region = Area of sector OBPC Area of OBC 0 = π = π cm 4 4 Ans(i) (ii) The distance around the track along the inner edge ( π π ) = = + 60 = = 7 m The area of the track = π ( )( ) = π 70 0 = = = 40m 98

101 CBSE TEST PAPER-0 CLASS-X Mathematics (Area related to circle). The circumference of a circular field is 58 cm. Then its radius is (a) 4cm (b) 84cm (c) 7cm (d) 56cm. If the perimeter of a circle is equal to that of a square, then the ratio of their area is (a) :7 (b) 4: (c) 7: (d) :4. Area of a sector of angle p of a circle with radius R is P (a) π R (b) 80 P 80 π R (c) OR P π R (d) 60 P π R 70 (a) π RP 80 π R P (b) 80 (c) π RP 60 π R P (d) Area of the sector of angle 60 of a circle with radius 0cm is 5 8 (a) 5 cm (b) 5 cm 4 (c) 5 cm (d) none of these 5. The cost of fencing a circular field at the rate of Rs4 per metre is Rs 580. The field is to be ploughed at the rate of Rs.0.50per m Find the cost of ploughing the field. (Take r = ) 7 6. Find the difference between the area of regular hexagonal plot each of whose side 7m and the area of the circular swimming take in scribed in it. (Take r = ) 7 [] [] 99

102 7. In the given Fig. areas have been drawn of radius cm each with vertices A,B,C, and D of quadrilateral ABCD as centers. Find the area of the shaded region. [] 8. Find the area of a sector of a circle with radius 6cm if angle of the sector is 60. [] 9. The radii of two circles are 8cm and 6cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles. 0. A chord of a circle of radius 0cm subtends a right angle at the centre. Find the area of the corresponding: (Use π =.4) (i) minor sector (ii) major sector (iii) minor segment (iv) major segment. Find the area of the shaded region in the given figure where ABCD is a square of side 0cm and semi-circus are drawn with each side of the square as diameter. ( π =.4). Find the area of the shaded region where a circular are of radius 6cm has been drawn with vertex O of an equilateral triangle OAB of side cm as centre. Three horses are tethered with 7 metre long ropes at the three corners of a triangle field having sides 0m, 4m. Find the area of the plot which can be grazed by the horses. Also find the area of the plot which remains ungrazed. [] [] [] [] [5] 00

103 CBSE TEST PAPER-0 CLASS X- Mathematics (Area related to circle) Ans. Ans. Ans. Ans4. (B) (C) (D) (B) [ANSWERS] Ans5. Since for Rs. 4, the length of fencing = metre for Rs.580, the length fencing = 580 = 0meters 4 Perimeter i.e circumference of the field = 0meters Let r be the radius of the field π r = 0 area of the field r ( ) = π = π 5 = 5π m Rate = Rs.0.50 per m total cost of ploughing the field = Rs. 7 = Rs. ( π ) ( ) = Rs. 75 = Rs.95 Ans6. Side of hexagonal plot = 7m Area of equilateral triangle OAB = ( side) = ( 7 ) = 96 m 4 4 Area of hexagonal plot = 6 Area of triangle OAB ( ) = 6 96 = = 468.0m OC = OA = AC = ( 7) = =

104 Ans7 Required area = Area of the circle with radius = r ( ) = cm 7 = 6 = 86cm Ans8 We know that Area of sector = Here θ = 60, r = required area = ( ) = = 7 6 = = cm θ π 60 r A O 60 0 B Ans9 A = Area of the first circle = ( ) r 8 = 64π cm π 6 = 6π cm A = Area of the second circle = ( ) A + A = Total area = 64π + 6π = 00π cm Let R be the radius of the circle with area A + A π R = 00π R = 00 R = 0cm Hence required radius = 0cm. Ans0 (i) Area of minor sector = 90 = ( 0) 60 7 = (.4 )( 00 ) 4 4 = = = 78.5cm 4 θ π 60 r 0

105 (ii) Area of major sector = Area of circle Area of minor sector 90 = π = 60 4 = = 5.5cm ( 0) π ( 0) (.4)( 00) (.4)( 00) (iii) We know that area of minor segment = Area of minor sector OAB Area of OAB area of OAB = ( OA)( OB) sin AOB = ( OA)( OB)( AOB = 90 ) θ Area of sector = π r 60 = (.4 )( 00 ) 50 = 5 (.4 ) 50 4 = = 8.5cm (iv) Area of major segment = Area of the circle Area of minor segment ( ) ( ) = r = = = 85.5cm Ans Let us mark the four unshaved regions as I, II, III and IV Thus area of I + Area of III = Area of ABCD Area of two semi circle of each of radius 5 cm = π = 00 5π = = =.5cm Similarly area of II + area of IV =.5cm Hence area of the unshaded region i.e I, II, III, IV = (.5)=45cm Thus area of the shaded region = Area ABCD Area of (I,II,III and IV) = 00-4 = 57cm 0

106 Ans Area of the shaded region = Area of major sector OP L QO + Area of equilateral triangle OAB area of equilateral triangle with side 00 = π ( 6) + ( ) 60 4 a = a and POQ = = + cm 7 Ans Let A = θ, B = θ and C = θ Area which can be grazed by three horses = Area of sector with central angle θ and radius 7 cm + Area of sector with central angle θ and radius 7 cm + Area of sector with central angle θ and radius 7cm π r θ π r θ π rθ = π r = Cθ + θ + θ 60 π r = Sum of three angles of a = = = 77m 7 60 Sides of plot ABC are a = 0cm, b = 4m and c = 4m semi perimeter, s = = 48m Are of triangular plot = Area of ABC = s ( s a)( s b)( s c) = = 6m Area grazed by the horses = 77m ungrazed area = (6-77) = 59m 04

107 CBSE TEST PAPER-04 CLASS-X Mathematics (Area related to circle). The area of a circle is 94.4cm. Then the radius of the circle is (a).4cm (c).cm (b).cm (d).cm. If the sum of the areas of two circles with radii R and R is equal to the area of a circle of radius R, then (a) R + R = R (b) R + R = R (c) R + R < R (d) R + R < R. Circumfuse of a sector of angle P of a circle with radius R is P (a) π R (b) 80 P (c) π R (d) 60 P 80 π R P π R If the perimeter and area of circle are numerically equal, then the radius of the circle is (a) units (c) 4 units (b) π units (d) 7 units 5. A wheel has diameter 84cm. find how many complete revolutions it must make to cones 79 meters. [] 6. The given figure is a sector of a circle of radius 0.5cm. Find the perimeter of the sector. (Take π = ) 7 A 60 0 B [] 05

108 7. A car had two wipers which do not overlap. Each wiper has a blade of length 5cm sweeping though an angle of 5. Find the total area cleaned at each sweep of the blades. 8. In the given Fig arcs have been drawn with radii 4 cm each and with centres P, Q and R. find the area of the shaped region [] [] 9. In Akshita s house there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 40 on and the difference of their circumference is 88cm. find the diameter of the circular top and bottom. 0. A chord of a circle of radius 5cm subtends an angle of 60 at the centre. Find the area of the corresponding minor and major [] [] segments of the circle. (use π =.4and =.7). Find the area of the shaded region of the two concentric circles with centre O are 7cm and 4cm respectively and AOC = 40 []. From each corner of a square of side 4cm, a quadrant of a circle of radius cm is cut and also a circle of diameter cm is cut as shown in the figure. Find the area of the remaining portion of the square. []. In given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56m. If the corner of each circular flower bed is the point of enter section O of the diagonals of the square lawn. Find the sum of the areas of the lawn and the flower beds. [5] 06

109 CBSE TEST PAPER-04 CLASS X- Mathematics (Area related to circle) [ANSWERS] Ans. Ans. Ans. Ans4. (C) (A) (C) (A) Ans5 Ans6 Diameter, r = 84cm = = 84 7 Distance covered in one revolution= circumference = π r π ( r) = = 64cm Thus for distance covered 64cm, number of revolution = for distance covered 79 metres = 7900cm number of revolutions = = We know that circumference i.e, perimeter of a sector of angle P of a circle with radius R P = π R + r 60 required perimeter 60 = ( 0.5) ( 0.5) A 60 0 B 44 = = cm Ans7 Radius of each wiper = 5cm Angle = 5 θ = 5 Total area cleaned at each sweep of the blades = 5 θ 5 5 Area = π r

110 = = = = cm 6 = 54.96cm Ans8 Area of shaded region = π r ( 4) = = 4 7 = 4 = 08cm Ans9 Sum of radii of circular top and bottom = 40cm Let radius of top = r cm radius of bottom = ( 40 r) cm Circumference of top = rcm π circumference of bottom = π ( ) 40 r cm Difference of circumference = π π ( 40 ) By the given condition = π r π ( r) r r cm 40 = π [ r 40 + r] = 88 r 40 = = = 4 r 7 54 r = r = = 77 Radius of top = 77cm Diameter of top = 77 = 54cm Radius of bottom = 40-r=40-77=6cm Diameter of bottom = 6 = 6cm Ans0 We know that area of minor segment = Area of minor sector OAB area of OAB 08

111 ( ) ( ) 60.7 = π ( 5) ( 5) = (.4)( 5) [ OAB is an equilaterl triangle] (.7)( 5) = (.57)( 75) 4 = = 0.475cm Again, area of major segment = Area of the circle area of minor segment r ( ) ( )( ) = = = = cm Ans Area of the shaded region = area of sector OAC area of sector OBD 4 40 π = π = = 47 = = = cm = 5 cm ( 4) π ( 7) ( 96 49) Ans Area of the shaded region = Area of square ABCD -4 (area of quadrant of a circle with radius ) -π ( ) = ( ) ( ) π π = π π = π ( radius of the circle = cm) = 6 = = cm Ans Area of the square lawn ABCD = m...( i) Let OA = OB = xmeter By Pythagoras theorem 09

112 x + x = x = x = ii ( ) Again area of sector OAB = 90 π x = π r ( )[ ( )] = 4 7 m iii From ii Also area of OAB m 4 [Here AOB = 90. since square ABCD is divided into 4 right triangles] AB m from& 4...( iv) area of flowerbed ( ) = = 7 = 448 m...( v) Similarly area of the other flowerbed = 448m Total area of the lawn and the flowerbeds = = = 40m 0

113 CBSE TEST PAPER-05 CLASS-X Mathematics (Area related to circle). The radius of a circle is 7 cm them the area of the circle is π (a) 54cm (b) 49 cm π (c) cm (d) 49 cm. Area of a sector of angle P of a circle with radius R is P (a) π R (b) 80 P (c) π R (d) 60 P 80 π R P π R 70. The diameter of a circle whose area is equal to the sum of the area of the two circles of radii 4cm and 7cm is (a) cm (c) 6cm (b) 5cm (d) 50cm 4. The circumference a circle is 58cm. Then its area is (a),76 cm (b),576 cm (c),76 cm (d) 4,576 cm 5. The radii of two circles are 9cm and 9cm respectively. Find the radius of the circle which has its circumference equal to the sum of the circumference of the two circles 6. A car travels 0.99km distance in which each wheel makes 450 complete revolutions. Find the radius of its wheel 7. A sector is cut from a circle of diameter cm. if the angle of the sector is 50 find its area [] [] []

114 8. In the given figure AOBCA represent a quadrant of area [] 9.65 cm. Calculate the area of the shaded portion. 9. The length of the minute hand of clock is 4cm. Find the area swept by the minute hand is 5 minutes 0. A round table cover has six equal designs as shown is the figure. If the radius of the cover is 8cm. find the cost of [] [] making the design at the rate of Rs.0.5 per cm (use =.7). Find the area of the shaded region where ABCD is a square of side 4cm []. ABCD is a flower bed If OA = m and DC = 4m. Find the area of the bed []. An elastic belt is placed round the rein of a pulley of radius 5cm. one point on the belt is pulled directly away from the centre O of the pulley until it is at P, 0cm from O. Find the length of the best that is in contact with the rim of the pulley. Also find the shaded area. [5]

115 CBSE TEST PAPER-05 CLASS X - Mathematics (Area related to circle) [ANSWERS] Ans. Ans. Ans. Ans4. (D) (A) (A) (A) Ans5 C = circumference of the st circle = π (9) = 8π cm C = circumference of the nd circle = π (9) = 8π cm C C = 8π + 8π = 56π cm Ans6 Distance travelled by a wheel in 450 complete revolutions = 0.99 kms = 990m. Distance travelled in one revolution = Let r be the radius of the wheel π r = 5 r = r = m = 0 = 5m = 5 m Ans7 We have = cm radius = cm Angle of sector = 50

116 θ 50 Area of the sector = A = π r = = = = 44.8 cm 7 4 Ans8 Required are 9 = Area of quadrant OAB-Area of OAB = π = = = = = 6.5cm 8 (.5) ( )(.5) Ans9 Angle covered by minute hand in 60 minutes = 60 Angle covered in minute = = Angle covered in 5 minutes = 6 5 = 0 We know that area swept by the minute hand during this period = Area of the sector with sector angle 0 θ = = = ( π r ) [ r 4 cm] 7 = 8 = 54 = = 5 cm Ans0 Area of one design = ( ) π - area of OAB π = ( 8) - area of equilateral OAB 6 π = 6 4 π = ( 8) 6 4 ( 8) ( 8) 4

117 [Area of equilateral triangle with side a =.7 = ( 8) 7 4 Total cost = ( ) a ] ( ) = ( 8) = = = Rs.6.68cm Ans Area of square ABCD = 4 4cm 96cm Diameter of each circle = 4 = 7cm radius of each circle = 7 cm area of one circle = π r = cm = = cm area of the four circles = 4 cm 54cm = Ans Here OA = R = m and OC = r = 4m Area of the flower bed (i.e. shaded portion) = area of quadrant of a circle of radius R of the quadrant of a circle of radius r. π = π R π r = ( R r ) = = = 4 7 ( ) ( 4 ) m [ R m and r 4m] = = m 4 7 = 9.5m ( 4)( 4) m 5

118 Ans OA 5 cosθ = = = OP 0 θ = 60 AOB = 0 = 0 Arc 0 π 5 AB = cm 60 0π θ = cm l π r = 60 Length of the belt that is in contact with the rim of the pulley = Circumference of the rim length of arc AB = 0 π 5cm πcm 0 = πcm 0 5 θ Now area of sector OAQB = π 5 cm = π cm since Area π r 60 = 60 = Area of OAP = 5 cm Area of quadrilateral OAPB ( ) AP = 00 5 = 75 = 5 cm 5 Hence shaded area = 5 π ( ) cm 5 = π 6

119 CBSE TEST PAPER-0 CLASS X Mathematics (Height and Distance). A man standing on the deck of a ship, which is 0m above the water level, observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of the hill as 0. Calculate the distance of the hill from the ship and the height of the hill. A boy is standing on the ground and flying a kite with 00m of string at an elevation of 0 Another boy is standing on the roof of a 0m high building and is flying his kite at an elevation of 45. Both the boys are on the opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60. At a point Y 40m vertically above X, the angle of elevation is 45 find the height of the tower PQ and the distance XQ. 4. At the foot of a mountain the elevation of its summit is 45. After ascending 000m towards the mountain up a slope of 0 inclination, the elevation is found to be 60. Find the height of the mountain. 5. A round balloon of radius r subtends an angle α at the eye of the observer while [] the angle of the elevation of its centre is β. Prove that the height of the centre the balloon is r sin β cosec α 7

120 6. An areoplane flying horizontally km above the ground is observed at an [] elevation of 60. After 0 seconds, its elevation is observed to0. Find the speed of the areoplane in km/hr. 7. From a window (h m high above the ground) of a house in a street, the angles of [] elevation and depression of the top and foot of another house on the opposite side of the street are α and β respectively show that the height of the opposite house is ( + α β ) h tan.cot m. 8. A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away [] from the wall through a distance a so that it slides a distance b down the wall making an angle β with the horizontal. Show that a cosα cos β = b sin β sinα 9. A girl who is. m tall, spots a balloon moving with the wind in a horizontal line at [] a height of 88.m from the ground. The angle of elevation of the balloon from the eye of the girl at any instant is 60. After some time, the angle of elevation reduces to 0. Find the distance travelled by the balloon during the interval. 0. From the top of a tower 96m high, the angles of depression of two cars on a road at [] the same level as the base of the tower and on same side of it are θ and φ, where tanθ = and 4 tanφ =. Find the distance between the two cars. 8

121 CBSE TEST PAPER-0 CLASS X Mathematics (Height and Distance) [ANSWERS] Ans. Let AB be the ship and CD be the hill. Let BD = xmin right ABD 0 tan 0 x = 0 = x x = 0 m In right CEA h tan 60 AE = 0m A B x C h E D 0 h = 0 h = 0m Height of the hill = h + 0 =0+0 =40m. Ans. AE= 00m (Given) and DC= 0m (Given) Let EF = hm In right ABE h + 0 = sin 0 00 h + 0 = 00 h = 0m Now in right EFD h sin 45 ED = 0 = ED ED = 0 m m A B C F E h 0m 45 0 D 0m 9

122 Ans. In right QRY QR tan 45 x = QR x = QR = x In right QPX x + 40 = tan 60 x x + 40 = x x = x + 40 x = x = + ( ) x = 0 + = X Y 45 0 x 60 0 x Q x R 40 P PQ = x + 40 = = 94.64m In right QPX x + 40 = sin 60 XQ = XQ XQ = 09.m Ans4. In right OEC CE = cos0 000 OE = 000 OE = 500m 0

123 In right ADO h 500 = tan 60 OD A h 500 = OD h 500 OD = h-500 h In right ABC h h = tan 45 C m E B D 500 h = h 500 h = h 00 h = h = 69.86m Ans5. In right OBD OD α = cos ec r α OD = r cos ec In right OCD A O r OC = Sinβ OD OC sin β = α r cos ec α OC = r sin β.cos ec D C B ß

124 Ans6. In right ACE AC x = tan 60 = x x = m In right BDE 60 0 A km B km = tan 0 x + y = + y E 0 0 x C y D = + y + y = y = km Speed = distance time = = km / hr. A Ans7. Let DE = hm H DC = xm In right EDC h = tan β x h = x...( i) tan β In right ABE E hm D x x B C h

125 H = tanα x H = tan α [from (i) h tan β H tan β = h tanα H = h tan α.cot β AC = H + h = h tan α. cotβ + h ( tan α.cot β ) = h + Hence Proved Ans8. In right ACD b + h x = sin α and = cosα l l Similarly in right BCE A b h a + x Sin β =,cos β = l l cosα cos β R. H. S = sin β sinα x a + x x a x l l l a = = = h b + h h b h b l l l E a β l D α l x B C h Hence Proved Ans9. In right HEF 87 tan 60 x = 87 = x 87 x = 87 x =

126 = 87 = 9 m In right GDF H 87 G = tan 0 x + y 87 = 9 + y F.m x E y 88.m D.m y = 87 9 A B C = 58 m Ans0. In right ABC 96 = tanθ x 96 = x x = [tanθ = 4 x = 4 x = 8m Ø A 96m In right ABD 96 = tanφ x + y 96 = [ tanφ = x + y 96 = 8 + y 8 + y = 8 y = 88 8 y = 60m D Ø Y C X B 4

127 CBSE TEST PAPER-0 Class X Mathematics (Height and Distance). If the angles of elevation of the top of a tower from two points at distances a and b from the base and in the same straight line with it are complementary then the height of the tower is (a) ab (b) ab (c) a b (d) a b. If the height of tower is half the height of the flagstaff on it and the angle of elevation of the top of the tower as seen from a point on the ground is 0 then the angle of elevation of the top of the flagstaff as seen from the same point is (a) 0 (b)45 (c) 60 (d) 90. As observed from the top of a 75m tall lighthouse the angles of depression of two ships are 0 and 45. If one ship is exactly behind the other on the same side of the light hours, then distance between the two ships is (a) 75( ) (b) 75 (c) 75 (d) The angle of elevation of the top a building from the foot of the tower is 0 and the angle of elevation of the top of the tower from the foot of the building is 60 if the tower is 50m high then the height of the building is 5

128 (a) 50 (b) 50 (c) 50 (d) A vertical tower stands on a horizontal plane and surmounted by vertical flagstaff [] of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are αand βrespectively prove that the height of the tower is h tanα [ ] tan β tanα 6. The angle of elevation of the top of a tower at a point on the level ground is 0 [] after walking a distance of 00m towards the foot of the tower along the horizontal line through the foot of the tower on the same level ground the angle of elevation to the top of the tower is 60 find the height of the tower 7. As observed from the top of light house 00m high above sea lever the angle of [] depression of a ship sailing directly towards it changes from 0 to 60 Determine the distance travelled by the ship during the period of observation (use =.7) 8. The angles of elevation of the top of a tower from two points P and Q at distances [] of a and b respectively from the base and in the same straight line with are complementary prove that the height of the tower is ab where a>b 6

129 CBSE TEST PAPER-0 Class - X Mathematics (Height and Distance) [ANSWERS] Ans. (A) Ans. (C) Ans (A) Ans4. (B) Ans5. Let AB = Height of tower = H =? Let BC = Height of flagstaff = h In right angle triangle OAB and OAC x = cotα H x = H cotα x And = cot β x = ( H + h) cot β H + h Equating value of x we get α = ( + ) ( ) H cot H h cot β H cotα cot β = hcot β h cot β H = cotα cot β O β α X C Flag Staff h B H Tower A h Ans6. In BCD, tan 60 x = = h = x...( i) h In ACD, = tan 0 = 00 + x h = 00 + x h = 00 + h h = 00 h = = h = = 50 = 50.7 = 86.6m D (Top of Tower) h=? A 00m B C Ground 7

130 Ans7. Let PQ be the light house such that PQ = 00m. Let A.B be the positions of ship when the angle of depression changes from 0 to 60 respectively Let AB = xmand BP = y From right angled APQ 00 tan 0 x y 00...( i) x + y = = + = From right angled triangle BPQ 00 = tan 0 = y y = =...( ii ) From (i) and (ii) 00 x + = 00 cm Ans8. Let AB be tower of height h. let P and Q be the given points in the same straight line with the foot B of the tower Let BP = a, BQ = b APB = Q, AQB = 90 θ From right angle APB h = tanθ a h = a tan θ...( i) From right angle AQB h = tan ( 90 θ ) b h = cotθ b b h = bcot θ =...( ii) tanθ Multiplying (i) and (ii) we get h h = = ab ab A P A B P X y θ a Q 90- θ b Q h B Q h=? 8

131 CBSE TEST PAPER-0 Class X Mathematics (Height and Distance). A.6m tall girl stands at a distance of.m from a lamp post and casts a shadow of 4.8m on the ground then the height of the lamp post is (a) 8 (b) (c) 8 (d) 8 8. A tree breaks due to storm and broken part bends so that the top of the tree touches the ground making an angle of 0 with ground. If the distance between the foot of the tree to the point where the top touches the ground is 8m then the height of the tree is (a) 8 (b) 8 (c) 8 (d) 8. The angle of elevation of the sum when the length of the shadow of a pole is times the height of the pole is (a) 0 (b) 45 (c) 60 (d) From the fig find the angle of elevation θ (a) 0 (b) 60 (c) 45 (d) 75 O Ø Shadow B A B h An aeroplane flying horizontally at a height of.5 km above the ground is observed at a certain point on the earth to subtend an angle of 60. After 5 seconds., its angle of elevation is observed to be 0 calculate the speed of aero plane in km/hr. 6. A man is standing on the deck of a ship which is 5m above water level. He observes the angle of elevation of the top of a lighthouse as 60 and the angle of depression of the base of the light house as 45 calculate the height of the lighthouse 7. An aeroplane when flying at a height of 5 m form the ground passes vertically below another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 0 and 60 respectively. Find the distance between the two planes at that instant 8. A boy standing on a horizontal plane finds a bird flying at a distance of 00 m from him at an elevation of 0. A girl standing on the roof of 0m high building finds the angle of elevation of the same bird to be 45. Both the boy and the girl are on opposite side of the bird. Find the distance of the bird from the girl. O Ø 50 A [] [] [] [] 9

132 CBSE TEST PAPER-0 Class - X Mathematics (Height and Distance) [ANSWERS] Ans. (A) Ans. (C) Ans. (A) Ans4. (B) Ans5. Let O be the observation point let A be the position of aero plane such that AOC = 60 and AC =.5km = BD Let B the position of aero plane after 5 seconds. BOD = 0, OC = xkm, CD = ykm In right.5 x = x OCA = cot 60 =.5 5 x = =...( i ) 0 In right ADB x + y = cot 0 =.5 x + y = ( ).5 x + y =...( ii) eq.( ii) eq.( i) O A C.5km B D y = = ( ) = Distance covered in 5 seconds y = km Distance covered in second = 5 km Distance covered in 600 seconds = 600 = 40 km 5 0

133 Ans6. H = Height of lighthouse = h + 5.(i) x In right ADC = cot 45 = 5 x = 5m x In right ADE = cot 60 = h 5 = h = 5 h Now H = h + 5 = = 5( + ) m B x x h C E D H=height of lighthouse Ans7. Let Aand Abe the positions of the two aero planes Let A A = x? And OM = y y = cot 0 = 5 y = 5...( i) Also ( ) y cot x = = ( ) 5 = 5 + x 5 + x = 5 ( )( ) ( ) x = 5 = 5 = 650m O y A A x 5m M Ans8. Positions of bird at A, boy at P and girl at are as shown in figure In ABP AB = sin 0 = AD = = 50m Also BC = DQ = 0m AC = AB BC = 50 0 = 0m AC In ACQ = sin 45 AQ 0 = AQ = 0 m AQ Hence the bird is 0 m away from the girl. P (Boy) 00m A (bird) 45 0 Q 0m 0m B D

134 CBSE TEST PAPER-04 Class X Mathematics (Height and Distance). A person walking 0m towards a chimney in a horizontal line through its bare observer that its angle of elevation changes from 0 to 45 then height of chimney is (a) 0 + (b) 0 (c) 0( ) (d) 0( + ). Find height of the tower if length of shadow = 0m sun s altitude = 45 (a) 0m (b) m (c) 0m (d) none of these. If the angles of elevation of a tower from two points at distances a and b where a>b From its foot and in the same straight line from it are 0 and 60 then height of the tower is a (a) a + b (b) ab (c) a b (d) b 4. The angle of elevation of the top of a tower from a point on the ground which is 0m away from the foot of the tower is 0, the height of tower is (a) 0 (b) 0 (c) 0 (d) 0 5. At a point on level ground, the angle of a elevation of a vertical tower is found to be 5 such that its tangent is, on walking 9.m to wards the tower, the tangent of the angle of elevation is [] 4. Find the height of tower 6. From the top of a building 60m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 0 and 60 respectively. Find (i) the horizontal distance between the building and the lamp post (ii) The height of the lamp post [Take =.7] 7. A man standing on the deck of a ship which is 0m above the water level observes the angle of elevation of the top of a hill as 60 and the angle of depression of the base of the hill as 0 calculate the distance of the hill from the ship and the height of the hill 8. The angle of elevation of a jet-plane from a point P on the ground is 60. After a flight of 5 seconds the angle of elevation (change to 0 ). If the jet plane is flying at a constant height of500 find the speed of the jet plane is km/hour [] [] []

135 CBSE TEST PAPER-04 Class - X Mathematics (Height and Distance) [ANSWERS] Ans. (B) Ans. (A) Ans. (B) Ans4. (B) Ans5. Suppose height of tower is h mtr In ABD h tanβ = =...( i ) x 4 In ABC h tanα = 9 + x h 5 =...( ii ) 9 + x ( i) ( ii) h x = 4 x = h x C α 9 D β x B A h Ans6. Suppose height of lamppost h mtr. In DEB x = cot 0 = 60 h ( ) x = 60 h...( i) x In right CAB = cot 60 = 60 h D h (60-h) B 60m 60 x = = 0 x = 0.7 = 4.64 m...( ii) ( ) By eq. (i) and (ii) (60-h) =0 60 h = 0 h = 40m C 60 0 x A

136 Ans7. Let H= Height of hill CE = CD + DE = 0 + h...( i) In right ADE x = 0 h x = x In right ADC = cot 0 0 x = 0 Equating the values of x, we wet h = 0 h = 0cm From H = 0 + h = 0+0 = 40m And x = distance of hill from ship = 0 A 0 B E h Hill = H D 0 C Ans8. Let A be the point on the ground E is the position of aero plane such that EAD = 60 and ED = 500 = CB C is the position of plane after 5 sec. CAB = 0, AB = x, AD = y x In right ABC. = cot 0 E 500 C x = 500 x = 4500m y 500 m In ADE = cot y A B = y D 500 X y = 500m Distance BD = x y = = 000m D S = T Speed = 000 = 00 m / sec = = 00 km / hr

137 CBSE TEST PAPER-05 Class X Mathematics (Height and Distance). If height of the tower = shadow of the tower then angle of elevation (a) 0 (b) 45 (c) 60 (d) 90. If length of shadow = 0m and angle of elevation = 60. Then height of tower. (a) 0m (b) 0 m (c) 0 m (d) 0 m. A little boy is flying a kite. The string of kite makes an angle of 0 with the ground. If length of the kite is h = m then length of string (a) 6m (b) 4m (c)5m (d) m 4. Line joining on eye and the object to be viewed is called (a) Horizontal (b) line of sight (c) Vertical line (d) None of these 5. A pole 5m high is fixed on the top of a tower. The angle of elevation of the top of pole observed from a point A on the ground is 60 and the angle of depression of the point A from the top of the tower is 45. Find the height of the tower. (Take =.7) 6. Form a window 5m high above the ground in a street. The angles of elevation and depression of the top and foot of another house on the apposite side of the street are 0 and 45 respectively. Show that the height of the apposite house is.66m [Take =.7] 7. The angle of elevation of the top of a tower as observed from a point on the ground is ' α ' and on moving a metre towards the tower. The angle of elevation is a tanα tan β ' β ' prove that height of tower is tan β tanα 8. A TV tower stands vertically on a bank of a canal. From a point on the other bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60 from a point 0m away from this point on the same back the angle of elevation of the top of the tower is 0. Find the height of the tower and the width of the canal. [] [] [] [] 5

138 CBSE TEST PAPER-05 Class - X Mathematics (Height and Distance) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5. (B) (D) (B) (B) In ADB AD h AD h AD = cot 45 = = h In ADC AD = cot 60 = h + 5 Horizontal C 45 0 B Pole h = = h + 5 h + 5 = h ( ) 5 = h [ AD h] ( ) h = = + = 6.8m A h D Tower Ans6. Suppose AE = hm and BC = xm Then AB = ( 5 + h) m In AED 6

139 tan 0 = = h x h x A h x = h...( i) In DEB D x E 5 tan 45 = x 5 = x x = 5m 5m C x 5m B Then 5 h = by...( i ) 5 h = = 5 m Height of another house =5 + 5 = Ans7. = =.66m Let AB b tower and height of tower = hm In ABC A h tan β = x h x =...( i) tan β h Tower In ABD α β D a C x B 7

140 h tanα = x + a h = x + a tanα ( ) h h = + a tan α by...( i) tan β h tanα h = + a tanα tan β h tan β = h tanα + a tanα tan β ( ) h tan β tanα = a tanα tan β a tanα tan β h = tan β tan β Ans8. Let h be the height of tower and x be the width of the river h In ABC = tan 60 x h = x h In ABD, + 0 = tan 0 x h x + 0 = = h =...( ii ) A Equating (i) and (ii) x + 0 x = x = x + 0 x = 0 x = 0m D C 0 x B Put x = 0in (i) h = x h = 0 m 8

141 CBSE TEST PAPER-0 CLASS X Mathematics (Coordinate Geometry). The distance between the point ( a, b),( a, b) is ( ) ( ) ( ) ( ) a a + b b a b c a + b d a + b. The area of whose vertices are (, ),( 4,6 ) and (, 5) is ( a) ( b) ( c) 4 ( d ) 5. The point ( 5, ) lies in (a) st quadrant (b) nd quadrant (c) rd quadrant (d) 4 th quadrant 4. The distance between the points ( Cosθ Sinθ ) ( Sinθ Cosθ ), and, is ( a) ( b) ( c) ( d ) 5. Find the distance between the point ( ) (,, ) A at at B at at [] 6. Determine if the points (,5 ),(,) and (, ) collinear [] 7. If A(, ), B( a, b) and C (,4 ) are the vertices of a isosceles show that [] a + b =, if AB = BC 8. Find the value of P if the point A( 0,) is equidistant from (, p) and (,) p [] 9. Find the centroid of the whose vertices are ( 4, 8)( 9, 7) and ( 8, ) [] 0. If the points ( x, y) is equidistant from the points ( a + b, b a) and ( a b, a + b) [4] prove that bx = ay. (, ),( x,8) and ( 6, y) are three concyclic points whose centre is ( ),5. Find the [4] possible value of xand y.. Find the vertices of the the mid points of whose sides are (, ),( 5,6) and (,) [4] 9

142 CBSE TEST PAPER-0 CLASS X Mathematics (Coordinate Geometry) Ans. Ans. Ans. Ans4. (a) (c) (d) (d) [ANSWERS] Ans5. AB = ( at at ) + ( at at ) ( ) ( ) 4a ( t t ) = a t t t + t + ( ) ( ) 4a ( t t ) ( )(( ) 4) = a t t t + t + = + + a t t t t ( ) ( ) = a t t t + t + 4 Ans6. Let A = (,5 ), B = (,) and C = (, ) ( ) ( ) AB = + 5 = 5 ( ) ( ) BC = + = ( ) ( ) AC = + 5 = 65 AB + BC AC hence not colliner Ans7. AB = BC(Given) AB = BC ( a + ) + ( b ) = ( a) + ( 4 b) a a + b + 4 4b = + a + a b 8b 4a + 4b = 4 a + b = 40

143 Ans8. Let B(, p) and C ( p,) AB = AC (Given) AB = AC ( 0 ) + ( p) = ( p 0) + ( ) p 4 p = p + 4 p = p = Ans9. Let ( x, y) be the coordinate of centroid x + x + x x = = = = y + y + y y = = 0 8 = = 4 Coordinate is (, 4) Ans0. Let P( x, y), A( a + b, b a) and B( a b, a + b) PA = PB PA = PB ( Given ) ( a + b x) + ( b a y) = ( a b x) + ( a + b y) = a b x ab ax ax b a y ab ay by a b x ab bx ax a b y ab by ay 4ab 4bx 4ab = ay ay 4bx = 4ay bx = ay 4

144 Ans. OA = OB = OC = Radius of circle OA = OB = OC OB = OA ( x ) + ( 8 5) = ( + ) + ( 5 ) A(-,) x + 4 4x + 9 = x 4x = 0 x x x 6 + = 0 x( x ) ( x ) ( x 6)( x + ) = 0 x = 6 or x = = 0 B (x,8) O (,5) C (6,y) OC ( 6 ) + ( y 5) = ( + ) + ( 5 ) ( ) = OA 4 + y + 5 0y = y 0y + 6 = 0 y y y = 0 y ( y ) ( y ) ( y 8) y = 0 y = 8 or y = 8 8 = 0 4

145 Ans. Let vertices of ABC be A( x y ), B( x y ) and C ( x, y ) By mid points formula x + x ( ) = x + x = 6... i y + y = y + y =... ( ii) x + x = 5 x + x = 0... ( iii) y + y = 6 y + y =... ( iv) x + x = x + x = 6... v y + y = y + y = 4... ( vi) Adding (i), (iii) and (v) ( x x x ) + + = 0 ( ) x + x + x = 5... vii Adding (ii),(iv)and (vi) ( y y y ) + + = 8 ( ) y + y + y = 9... viii ( ) Subtracting (i),(iii) and (v) from (vii) We get x =, x = 5, x = (x,y ) D (,) B A (x,y) (5,6) E F (-,) (x C,y ) Subtracting (ii),(iv) and (vi) from eq. (viii) We get y = 7, y =, y = 5 4

146 CBSE TEST PAPER-0 CLASS X Mathematics (Coordinate Geometry). If (, ) ( 4, y), ( x,6 ) and (,5) are the vertices of a parallelogram taken in order. Then ( x, y ) is ( a) ( 6, ) ( b) ( 6, ) ( c) ( 6,4 ) ( d ) (,4). The coordinates of the point which divides the join of (, 7 ) and ( 4, ) in the ratio :. is ( a) (, ) ( b) (, ) ( c) (, ) ( d ) (,). The coordinates of a point A, where AB is the diameters of a circle whose centre (, ) and B is (, 4 ) is. ( a) (, 9 ) ( b) (,9 ) ( c) (, 0 ) ( d ) ( 4,5) 4. If the area of a quadrilateral ABCD is zero, then the four points A,B,C,D are (a) Collinear (b) Not collinear (c) Nothing can be said (d) None of these 5. Prove that the points (, 0) (4, 5) (-, 4) and (-,-) taken in order form a rhombus. [] 6. Show that (4,4) (,5) (-.,) are vertices of a right [] 7. Prove that in a right, the mid point of the hypotenuse is equidistant from the vertices. [] 8. Prove that diagonals of a rectangle bisect each other and are equal. [] 9. The line joining the points (, ) and ( 5, 6) is bisected at P. if P lies on the line [] x + 4y + k = 0, Find the value of K. 0. The two opposite vertices of a square are (, 6) and ( 5,4 ). Find the coordinates of [4] the other two vertices.. Find the coordinates of the circum centre of a triangle whose vertices are A( 4,6 ), B ( 0,4) and C ( 6, ) Also find its circum radius.. If two vertices of an equilateral triangle be ( 0,0)(, ), Find the third vertex. [4] [4] 44

147 CBSE TEST PAPER-0 CLASS X Mathematics (Coordinate Geometry) Ans. Ans. Ans. Ans4. (b) (a) (c) (a) [ANSWERS] Ans5. Let A (,0 ), B ( 4,5 ), C (,4 ) and D (, ) ( ) ( ) AB = = 6 ( ) ( ) BC = = 6 ( ) ( ) CD = = 6 ( ) ( ) DA = = 6 Since AB = BC = CD = DA Hence ABCD is rhombus Ans6. Let A (4,4) B (,5) C (-,) AB = (-4) + (5-4) = AC = (--4) + (5-4) = 4 BC = (--) + (-5) = Since AC = AB +BC Hence ABC is right Ans7. Let A( a,0 ), B( 0,b) O ( 0,0) are the vertices of right a b Coordinate of C, B O (0,b) C (a,b) A i.e. (a,b) (0,0) (a,0) 45

148 OC = a + b AC = a + b BC = a + b Hence C is Equidistant from the vertices. Ans8. Let ABCD be a rectangle take A as origin the vertices of a rectangle are A( 0,0 ), B( a,0 ), C ( a, b), D( o, b) ( 0) ( 0) A = a + b = a + b ( 0 ) ( 0 ) BD = a + b = a + b AC = BD Mid point of AC = a + b a b, =, Mid point of BD = a + b a b, =, Mid point of AC = Mid point of BC Hence Proved (0,b) (a,b) C D A (0,0) (a,o) B Ans9. A P (x,y) B (,-) (:) (5,-6) Coordinate of 7 7 =, P =, P lies on eq. x + 4y + k x + 4y + k = K = K = 0 K = 7 46

149 Ans0. AB = BC AB = BC ( x ) + ( y + 6) = ( x 5) + ( y 4) x + x + y y = x + 5 0x + y + 6 8y 8x + 0y 4 = 0 x + 5y = x y = 5 In right ABC AC = ( AB) + ( BC ) ( x ) + ( y + 6) + ( x 5) + ( y 4) = ( 5 ) + ( 4 + 6) ( x y x y) = 8 ( ) x + y x + y = i Put the value of y in eq. (i) x 9x 74x 464 = 0 x x x + 6x + = x 6 = 0 x x x = 0 x( x ) ( x ) ( x 8)( x + ) = 0 x = 8 x = = 0 Now x =, y = And x = 8 y = D A(,-6) (5,4) C B (x,y) Ans. Let P be the circum centre of ABC then PA = PB = PC PA = PB = PC PA = PB ( x 4) + ( y 6) = ( x 0) + ( y 4) (4,6) A P (x,y) 8x + 4y = 6 (0,4) B (6,) C 47

150 ( ) x + y = 9... i PB = PC ( x 0) + ( y 4) = ( x 6) + ( y ) x 4y = 4 ( ) x y = 6... ii On solving eq. (i) and (ii) x =, y = Circum radius (PA) = ( 4 ) + ( 6 ) = 0 Ans. OA = OB = AB OA = OB = AB OA = OB + =... ( ) x y i y A ( ) OB = AB ( ) x + y = 6... ii 6 x y = (0,0) O (x,y)b x Put the value of y in eq. (i) 6 x x + = x = 0 or x = x = 0 y = ( 0, ) x = y = (, ) 48

151 CBSE TEST PAPER-0 CLASS X Mathematics (Coordinate Geometry). The ratio of the points of trisection of the line segment joining the points, B 7, 4 are A( ) and ( ) ( a) :, : ( b) :, : ( c) :, : ( d ) :, :. The valve of K if the points A(, ), B( 4, K ) and ( 6, ) ( a) ( ) ( b) ( ) ( c) ( ) ( d ) ( 0) C are collinear is. The mid-point of the line segment joining ( a,4) and (,b) is (,a + ) values of a and b is a a =, b = b a =, b = c a =, b = d a =, b = ( ) ( ) ( ) ( ) 4. Coordinate of A and B are (,α ) and (, α + 4 ). The mid-point of AB is (,) value of α is a b c d ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). The. The 5. Find the coordinates of the points which divide the line segment joining the points (-,0) and (0,8) in four equal parts. 6. Find the area of the rhombus if its vertices are (,0), (4,5), (-,4) and (-,-) taken in order. 7. Show that the points ( a, b + c),( b, c + a) and ( c, a b) [] [] + are collinear. [] 8. The length of a line segment is 0. If one end point is (, ) and the abscissa of the second end point is 0, show that its ordinate is either or Using section formula show that the points (, )( 5,0) and (,) are collinear. [] 0. If P and Q are two points whose coordinates are ( at, ) a a, t t of t respectively and S is the points (,0 ), at and a show that + is independent SP SQ. Find the area of the quadrilateral whose vertices taken in order are 4,,,5,,, ( ) ( ) ( ) and ( ). The vertices of ABC are A( 4,6 ), B(,5 ) and C ( 7, ). A line is drawn to inter sect AD AE sides AB and AC at D and E respectively such that = =. Calculate the AB AC 4 area of the ADEand compare it with the area of ABC. [] [4] [4] [4] 49

152 CBSE TEST PAPER-0 CLASS - X Mathematics (Coordinate Geometry) [ANSWERS] Ans. Ans. Ans. Ans4. (a) (d) (a) (a) Ans5. Q is the mid point of AB (-,0) (0,8) A P Q R B Coordinate of Q +, + = (-,4) Coordinate of P =, Coordinate of R =,6 Ans6. Let A (,0) B (4,5) C (-,4) and D (-,-) ( ) ( ) AC = = 4 ( ) ( ) BD = = = 6 Area of rhombus = d d AC BD 4 6 = 4 50

153 Ans7. For collinear ( ) + ( ) + ( ) = a( c a a b) b( a b b c) c( b c c a) a( c b) b( a c) c( b a) x y y x y y x y y 0 = = + + = ac ab + ba bc + cb ca = 0 Ans8. Let (,-) 0 (0,y) ( 0 ) + ( y + ) = ( 0) y A B + y + + y = + 6y 7 = 0 y y y = 0 y ( y ) ( y ) ( y + 9)( y + 9) = 0 y = 9 y = = 0 Ans9. If points A(, ), B( 5,0) and ( ), are collinear then one point divides the join of other two in the same ratio let C (,) divides the join of A(, ) and B( 5,0) in the ratio K: A C (x,y) B (-,) (K:) (5,0) 5K 0 + = and = K + K + K + = 5K and K + = K = K = Hence Proved 5

154 Ans0. SP = ( at a) + ( at 0) ( ) = a t + 4t = a t + t + 4t 4 ( t ) ( ) = a + = a t + a at SQ = a + 0 t t a = ( + t ) t a = ( + t ) t ( + t ) ( + ) t + = + SP SQ a t a t = a t = a H. P ( + ) ( + ) Ans. ar ( ABC ) = ( ) ( ) = 0.5 sq. unit ar of = [ x, y y + x y y + x y y ar ( ACD) = ( 8 9 4) ( 6 4 ) + = [ + ] = 7.5 sq.unit ( ) ( ) ( ) ar of quadrilateral = = 8 sq. unit. D(,) (,-)C A(-4-) (--5) B 5

155 Ans. AD AE = = AB AC 4 AB AC 4 = = AD AE AD + DB AE + EC = = 4 AD AE DB EC + = + = 4 AD AE DB EC = = AD AE AD AE = = DB EC AD : DB = AE : EC = : Now coordinate of D and E are, 4 4 and 9,5 4 ar ar ar ar ( ADE) ( ABC ) ( ADE) ( ABC ) = :6 = = = A (4,6) D B (,5) E (7,) C 5

156 CBSE TEST PAPER-04 CLASS X Mathematics (Coordinate Geometry). The ratio n which the points (, ) and ( 5,6 ) divided by the x- axis is : : : : ( a) ( b) ( c) ( d ). The distance between P( a,7) and Q (, ) is 5. The value of a is ( a) ( 4, ) ( b) ( 4, ) ( c) ( 4, ) ( d ) ( 4,). On which axes point ( 4,0) lie ( a) x axis ( b) y axis ( c) ( d ) - both none of these 4. The distance of the point ( 4, 6) from the origin is ( a) 5 ( b) ( c) ( d ) 5. If the coordinates A and B are (-,-) and (,-4) respectively. Find the coordinates of P such that AP = 7 ABand P lies on the line segment AB. [] 6. In what ratio is the line segment joining the points (-,) and (,7) divided by the y-axis? 7. Find the relation between xand y such that the point ( x, y) is equidistant from the points ( 7,) and (,5) 8. Determine the ratio in which the line x + y 4 = 0divides the line segment joining the points A(, ) and B( ),7? 9. Show that the points A( 5,6 ), B(,5 ), C (,) and D( 6,) square. 0. Prove that the points ( a, a),( a, a) equilateral. Calculate the area of this.. A A( 4, 8 ), B(, 6 ) and C ( 5, 4) and ( a, a) are the vertices of a ABC, are the vertices of a are the vertices of an Dis the mid point of AD BC and P is a point on AD joined such that =, find the coordinates of P. PD. The coordinates of the vertices of ABC that three area of ABCis, find the value of K. are A( 4, ), B(, ) and C ( O, K ). Given [] [] [] [] [4] [4] [4] 54

157 CBSE TEST PAPER-04 CLASS - X Mathematics (Coordinate Geometry) Ans. Ans. Ans. Ans4. (b) (c) (a) (b) [ANSWERS] Ans5. Coordinate of P are (-,-) A :4 (,-4) B x = = = ( 4) ( ) 8 0 y = = = Ans6. Let A (-,-) and B (,7) P (0,y) and ratio be K: (-,-) A P(o,y) K: (,7) B k 7k Coordinate of P are, k + k + k = 0 k + k = or : 55

158 Ans7. Let P( x, y) be equidistant from the points A( 7,) and B (,5) AP = BP(Given) AP = BP ( x 7) + ( y ) = ( x ) + ( y 5) x 49 4x y y x 9 6x y 5 0y = x y = Ans8. Let the ratio be K : K + 7K Coordinate of P are, K + K + P lies on the line x + y 4 = 0 K + 7K 4 + = 0 K + K + 6K K 4K 4 = 0 9K = 0 K = 9 : 9 AB = = 7 Ans9. ( ) ( ) ( ) ( ) BC = + 5 = 7 ( ) ( ) CD = 6 + = 7 ( ) ( ) DA = = 7 AC = = 4 Diagonal ( ) ( ) BD = = 4 Diagonal ( ) ( ) Hence proved 56

159 Ans0. Let A( a, a), B ( a, a) C ( a, a) ( ) ( ) AB = a a + a a = 8a = a ( ) ( ) AB = a + a + a + a = a ( ) ( ) AC = a a + a a = a AB = BC = AC = a ar ABC = 4 = 4 = a ( a) ( side) Ans. Let A( 4, 8 ), B(,6) and ( 5, 4) C are the vertices of ABC, Dis the mid point of BC AP = PD AP : PD = : (4,-8) A Coordinate of D, i.e ( 4,) (,6) B P D (5,-4) C ( ) Coordinate of P are, i. e, i. e 4, ( ) 57

160 Ans. A( 4, ), B (, ) and C ( 0, k ) ar ABC = x y y + x y y + x y y = 4 ( k ) + ( )( k ) + 0 ( ) = [ k k ] = 7 but ar = [ k] 7 k = ( 7 k ) = ± 7k = 4 k = 7 7k = 4 k = 5 Value of k ( ) ( ) ( ),5 7 58

161 CBSE TEST PAPER-05 CLASS X Mathematics (Coordinate Geometry). The coordinates of the mid point of the line segment joining ( 5,4) and ( 7, 8) ( a) (, ) ( b) (, ) ( c) (, ) ( d ) (, ). Two vertices of a ABC are A(, ) and ( ) 5 be,, then the coordinates of the third vertex C is is B 5,. If the coordinates of its centroid ( a) (, ) ( b) (, ) ( c) (, ) ( d ) (, ). The abscissa of every point on y-axis is ( a) 0 ( b) ( c) ( d ) - 4. The ordinate of every point on x-axis is ( a) ( b) ( c) 0 ( d ) - 5. For what value of P are the points (,) (p,-) and (-) collinear? [] 6. Find the third vertex of a if two of its vertices are at (,) and (,5) and the centroid is at the origin. [] 7. In a seating arrangement of desks in a classroom three students are seated at A(, ), B( 6,4) and ( 8,6) C respectively. Seated in line? 8. Show that (, ),(, ),(, ) are the vertices of an equilateral triangle [] [] 9. If the point P( x, y) is equidistant from the points A( 5,) and B (,5 ) x = y, prove that 0. Find the lengths of the medians of the triangle whose vertices are (, ),( 0, 4) and ( 5,).. The area of a is 5. Two of its vertices are (,) and (, ) y = x +. Find the third vertex.. The third vertex lies on [] [4] [4]. Prove that the point ( a, o),( a, b) and ( ), are collinear if + = a b [4] 59

162 CBSE TEST PAPER-05 CLASS - X Mathematics (Coordinate Geometry) Ans. Ans. Ans. Ans4. (a) (c) (a) (c) [ANSWERS] Ans5. For collinear ( ) ( ) ( ) 0 x y y + x y y + x y y = ( ) p ( ) ( )( ) = 5 + p = 0 p = 5 Ans6. Let third vertex of the be ( x, y) x + + y = 0, = 0 x = 4, y = 7 Hence ( 4, 7) Ans7. ( ) ( ) AB = = 8 = ( ) ( ) BC = = = ( ) ( ) AC = = = 5 AB + BC = AC Hence they seated in a line. 60

163 Ans8. Let A(, ), B (, ), C (, ) ( ) ( ) AB = + = 8 ( ) ( ) BC = = 8 ( ) ( ) CA = + + = 8 Since AB = BC = CA, the ABC is equilateral Ans9. PA = PB(Given) PA = PB ( 5 x) + ( y) = ( x) + ( 5 y) = x x y y x x y y 8x = 0y + y 8x = 8y x = y Ans0. Coordinates of points D, E and F are ,,, and, 5 7 i. e,, (,) and, Length of the median AD F (,-) A E 5 7 = + + = 0 B (0,4) D (-5,) C Length of the median BE ( 0) ( 4) = + = = And length of the median CF 0 CF = = 6

164 Ans. Let the third vertex be A( x, y ). Other two vertices of the are B (,) C (, ) ar of ABC = 5 ( ) ( ) ( ) 5 x y y + x y y + x y y = ± x( ) ( y) ( y ) = ± x + y 7 = ± 0 x + y = 7 x + y = (, ) x y lies on eq. y = x + On solving eq. x + y = 7and y = x + We get 7 x =, y = Similarly on solving eq. x + y = and y = x + We get, and Ans. Since ( a,0 ),( 0, b ) and(,) are collinear Area = 0 ( ) ( ) ( ) 0 x y y + x y y + x y y = a ( b ) + 0( 0) + ( 0 b) = 0 ab a b = 0 ab = a + b Dividing by ab ab a b = + ab ab ab + = a b 6

165 CBSE TEST PAPER-0 Class-X Mathematics (Surface area and Volume). The number of solid spheres each of diameter 6cm that could be molded to form a solid metes cylinder of height 45cm and diameter 4cm is (a) (b) 4 (c) 5 (d) 6. A metallic sphere of radius 0.5 cm is melted and then recast into small cones each of radius.5 cm and height cm the number of such cone is (a) 6 (b) 6 (c) (d) 0. A circular test is cylindrical to a height of 4cm and conical above it. If its diameter is 05 m and its slant height is 40m, then the total area of canvas required is (a) 760 m (b) 640 m (c) 960 m (d) 790 m 4. If the radii of the ends of a bucket are 5cm and 5cm and it is 4cm high, then its surface area is. (a) 85. cm (b) 7. cm (c) 05. cm (d) 60 cm 5. A 0m deep well with diameter 7m is dug up and the earth from digging is evenly [] Spread out to form a platform m 4m Find the height of the platform. 6

166 6. Find the maximum volume of a cone that can be carved out of solid hemisphere of [] radius r. 7. Find the volume of the largest right circular cone that can be cut out of the cube [] whose edge is 7 cm 8. A hollow cylindrical pipe is 40cm long. It s outer and inner diameter are 8 cm and [] 6cm respectively. Find the volume of the copper used in making the pipe. 9. A plate of metal cm thick, 9 cm broad and 8cm long is melted into a cube. Find [] the difference in the surface area of the two solids. 0. The difference between outside and inside surfaces of a cylindrical metallic pipe [] 4cm long is 44 cm. If the pipe is made of 99cu.cm of metal. Find the outer and inner radii of the pipe.. If h, C and V respectively are the height. The curved surface and volume of cone. [] Prove that πvh c h + v = 9 0. A toy is in the form of a cone mounted on a hemisphere of common base radius [] 7cm. The total height of the toy is cm. find the total surface area of the toy use π = 7. The radius of a solid iron sphere is 8cm. eight rings of iron plate of external radius [5] 6 cmand thickness cm are made by melting this sphere. Find internal radius of this rings. 64

167 CBSE TEST PAPER-0 Class X - Mathematics (Surface area and Volume) Ans. Ans. Ans. Ans4. (C) (B) (D) (D) [ANSWERS] Ans5. Volume of earth dug out = 7 7 = 0 r = 7 = 7 5 = 770cm Area of the platform Height of the platform = = = = =.5m 08 8 πr h = 4 = 08m Volume of the earth dug out Area of the platform Ans6. Clearly radius of the base of the cone = radius of hemisphere = r Height of cone = Radius of the hemisphere Volume of cone = πr x r = π r Ans7. For Largest right circular cone to be cut, clearly the circle will be inscribed in a face of the cube and its height will be equal to an edge of the cube. 7 Radius of the cone r = cm Volume of the cone = π r h 7 = = 88.8cm 7 65

168 Ans8. We have H= Height of the cylindrical pipe = 40 cm R= External radius = 8 cm 4cm = R= internal radius = 6 cm cm = Volume of the copper used in making the pipe ( ) = π π = π R h r h R r h = ( 4 ) 40cm = 7 40cm = 940cm 7 7 Ans9. Volume of the plate = Let a be the edge of the cubes = = = a 79 9 a 9cm 9 8 = 79cm Now surface area of the cube = a ( ) Surface area of the plate = [ lb + bh + lh] cm = [ ] = 68cm 6 = 6 9 = 486cm difference in the surface area of the two solids ( ) cm 5cm = = Ans0. Let R cm and r cm be external and internal radii of the metallic pipe We have h = Length of the pipe = 4cm ( ) π Rh π rh = 44 i. e.π R r h = 44 ( R r) 4 = 44 7 R r =...( i ) Also π π π ( ) R h r h = 99 = R r h = 99 ( R + r)( R r) 4 =

169 9 R + r =...( ii ) Solving (i) and (ii) we get R =.5cm r= Hence outer radius =.5cm Inner radius = cm Ans. C.S.A. of cone C = π rl...( i) Volume of cone V = π r h...( ii) L.H.S. = πvh c h + 9v π h π r h π rl h + 9 π r h = { } = π r h π r h l + π r h ( ) = π r h π r h r + h + π r h l = r + h 4 4 π r h π r h π r h π r h = + = 0 Ans T.S.A. of toy= C.S.A of hemisphere + C.S.A. of cone r = 7 cm, h = 4cm = + = + = l r h 7 4 5cm = = = = 858cm = π + π r rl Ans We have volume of solid iron sphere = = 048 cm 4 π 8 cm 67

170 External radius of each iron ring = 0 cm = 6 cm Let internal radius of each ring be r cm 0 Volume of each ring = π r cm Volume of 8 such rings 400 = 8π r cm = 4π r cm 9 Volume of 8 rings = volume of sphere π r = π r = π 9 4π = = = 6 = r r cm 68

171 CBSE TEST PAPER-0 CLASS-X Mathematics (Surface area and Volume). A metallic sphere of radius 0.5cm is meted and then recast into small cones each of radius.5cm and height cm the number of such cones is (a) 6 (b) 6 (c) (d)0. A solid sphere of radius r is melted and cast into the shape of a solid cone of height r, the radius of the base of the cone is (a) r (c) r (b) r (d) 4r. During conversion of a solid from one shape to another, the volume of new shape will (a) increase (c) remain unaltered (b) decrease (d) be doubled 4. A right circular cylinder of radius r cm and height h cm (h>r) just encloses a sphere of diameter (a) r cm (c) h cm (b) r cm (d) h cm 5. Determine the ratio of the volume of a cube to that of a sphere which with exactly [] fit inside the cube 69

172 6. Find the maximum volume of a cone that can be carved out of a solid hemisphere [] of radius r 7. The height of a right circular cone is cm and the radius of its base is 4.5 cm. find [] slant height. 8. How many balls, each of radius cm, can be made from a solid sphere of lead of [] radius 8 cm 9. The diameter of metallic sphere in 6cm. the sphere in melted and drawn into a [] wire of uniform cross section. It the length of wire is 6cm. Find its radius. 0. Water flows at the rate of 0 metre per minute through a cylindrical pipe having [] its diameter at 5mm. How much tine will it take to fill a conical vessel where diameter of base is 40 cm and depth 4 cm?. The radius of the base and the height of solid right cylinder are in the ratio : and [] its volume is 67 cu.cm. Find the total surface area of the cylinder. π = 7. A toy is in form of a cone mounted on a hemisphere of common base radius 7cm. [] The total height of the toy is cm find the total surface area of the toy.. A bucket made up of metal sheet is in the form of frustum of a cone. Its depth is 4 [5] cm and the diameters of the top and bottom are 0 cm and 0 cm respectively. Find the cast of milk which win completely fill the bucket at the rate Rs. 0 Per litre and cast of metal sheet used. If it casts Rs. 0 per 00cm ( use π =.4) 70

173 CBSE TEST PAPER-0 CLASS - Mathematics (Surface area and Volume) [ANSWERS] Ans. Ans. Ans. Ans4. (B) (A) (C) (B) Ans5. Let the radius of the sphere which fits exactly into a cube be r units Then length of each edge of cube = r units Let Vand V be the volumes of the cube and sphere Then V = ( r) V 4 = π r V 8r 6 = = V 4 π r π V : V = 6 : π Ans6. Radius of cone = radius of hemisphere = r Height of cone = radius of hemisphere Volume of cone = πr r = π r Ans7. h = cm, r = 4.5cm l = r + h = Slant height ( ) = = 64.5 ( ) =.86 approx 7

174 Ans8. Ans9. Ans0 Number of balls = 4 π ( 8) = = 5 4 π ( ) Diametre of sphere = 6cm r = cm Volume ( ) Volume of sphere of radius 8cm Volume of sphere of radius cm = π r = π r = π = 6π cm Let r be radius of wire 4 4 π = π = 6π cm Volume of wire = r ( ) π r 6 = 6 π r = r = cm we have volume of the water that flows out in one minute = Volume of cylinder of diametre 5mm and length 0 m. 5 r = mm = cm 4 h = 000cm Volume of cylinder = 000 cm Volume of conical vessel r = 0 cm and h = 4cm = 7 ( ) 0 4cm Suppose the conical vessel is filled in x minutes Volume of the water flows out in x minutes = Volume of conical vessel = x = x = = minutes 00 0 x= 5minutes seconds 7

175 Ans Let r be the radius of the base and h be the height of the solid right circular cylinder. r h = r = h 4 Volume of the cylinder = π r h = π h. h = h = 7 = 4 4 h = Surface area of cylinder = π rh + π r h π 4h = π h + 9 π h 0π h = [ + 8] = 9 9 0π = = 5 7 = 770 cm 9 Ans Total surface area of toy = C.A.S. of hemisphere C.S.A. of cone π π r + rl Here r = 7 cm, h = 4cm = + = + = l r h 7 4 5cm T.S.A. of toy = = = 858cm 0 0 Ans h = 4 cm, r = = 5 cm, r = = 5cm ( ) ( ) l = h + r r = = = 676 = 6cm h r r r r (i) Volume of bucket = π ( + + ) 7

176 =.4 4 ( ) = = 864cm ( ) (ii) Quantity of milk = = litres Cast of litre of milk = Rs.0 Cost of 8.64 litre milk = =Rs. 6.8 T.S.A. of bucket (excluding the upper end) ( ) π ( ) = πl r + r + r = = 7.cm Cost of 00 cm metal sheet = Rs.0 cost of 7. cm metal sheet = 7. 0 = Rs

177 CBSE TEST PAPER-0 Class-X Mathematics (Surface area and Volume). A solid sphere of radius r is melted and cast into the shape of a solid cone of height r the radius of the base of the cone is (a) r (c) r (b) r (d) 4r. A reseuoir is in the shape of a frustum of a right circular cone. It is 8m across at the top and 4m across at the bottom. If it is 6m deep, then its capacity is (a) 76 m (b) 96 m (c) 00 m (d) 0 m. A cone of height 4 cm and radius of base 6 cm is made up of modeling clay. A child reshapes it in the farm of a sphere the radius of the sphere is. (a) 5cm (c) 8cm (b) 6cm (d) cm 4. A circular tent is cylindrical to a height of 4 m and conical alone it. If its diameter is 0 m and its slant height is 40m. The total area of the canvas required in m is (a) 760 (b) 5840 (c) 960 (d) A drinking glass is in the shape of a frustum of a cone of height 4 cm. The [] diameter of its two circular ends one 4cm and cm. Find the capacity of the glass π = 7 75

178 6. The diameter of a sphere is 6cm. It is melted and drawn into a wire of diameter [] cm. what is the length of wire, 7. An iron pipe 0cm long has exterior diameter equal to 5cm. If the thickness of the [] pipe is cm. Find the whole surface area of the pipe. 8. Find the ratio of the volumes of two circular cones. If r : r = : 5and h : h = : [] 9. A well.5m in diameter and 0m deep into be dug in rectangular field 0m by [] 4m. The earth taken out is spread evenly on the field. Find the level of the earth raised in the field. 0. A solid sphere of radius 6cm is melted into a hollow cylinder of uniform thickness. [] If the external radius of the base of cylinder is 5cm and its height is cm, Find the uniform thickness of the cylinder.. A medicine capsule is in the shape of a cylinder with two hemispheres. Stuck t each [] of its ends. The length of the entire capsule is 4 mm and the diameter of the capsule is 5 mm. Find its surface area.. A per stand made of wood is in the shape of a cuboid with four conical depression to hold pens. The dimensions of the cuboid one 5cm by 0cm by.5cm. The radius of each of the dimensions is 0.5cm and the depth is.4cm. Find the volume of the wood in the entire stand.. A solid consisting of a right circular cone. Standing on a hemisphere, is placed upright in a right circular cylinder, full of water and touches the bottom. Find the volume of water left in the cylinder having given that the radius of the cylinder is cm and its height is 6cm. The radius of hemisphere is cm and the height of the cone is 4cm. Give your answer to the nearest cubic centimeter π = 7 [] [5] 76

179 CBSE TEST PAPER-0 CLASS - Mathematics (Surface area and Volume) [ANSWERS] Ans. (A) Ans. (A) Ans. (B) Ans4. (B) Ans5 r = cm r = cm r = 4cm r = cm h = 4cm Capacity of glass π h = r + r + r r 4 = = [ ] = 7 = = 0 cm Ans6 Radius of sphere r = cm 4 Volume of sphere = πr 4 = π = πcm ( ) 6 Let length of the wire = lcm R= Radius of the wire = cm Volume of wire ( ) lπ = 6π l = 6cm Ans7 R =.5cm r =.5 =.5cm h = 0cm = π R h = π l = lπ cm Total surface area of pipe = π ( R + r)( h + R r) ( )( ) = = 68cm 7 77

180 Ans8. Ratio of volumes of Two cones = π r h π r h r h = r h = = = 5 5 Ans9 Radius of well = 7 4 m Depth of well = 0m 7 7 Volume of earth taken out = 0 m = 85 m Area of field = 0m 4m = 80m Area of field excluding well = 6 8 m Level of earth raised = 85 8 = m = 0.79m 6 = 7.9cm 7 7 = m volume of earh taken out Area of field Ans0 4 Volume of sphere = πr 4 = π = πcm ( )

181 Let the internal radius of cylinder r = x cm External radius R = 5cm Volume = π ( R r ) h = π ( x ) 5 Volume of the hollow cylinder = Volume of sphere ( x ) 5 π = 88π 5 = 5 = 9 x x = 6 x = 4cm 88 x Ans Uniform thickness of cylinder = 5-4=cm The length of capsule = 4mm 5 r = mm Length of cylindrical portion = 4 5 of capsule = 9mm Total surface area = = π r + 4 π rh 5 5 = = + = = 0mm π r + π rh + π r Ans Required volume = volume of cuboids 4[V. of one depression] = = = 55.9 = 5.07cm ( ) 79

182 Ans Volume of cylinder = π ( ) Volume of cone π ( ) 6 = 4π = 4 = 6 π = 4 = 6 π Volume of hemisphere π ( ) Volume of water in the cylinder 6 6 = 54π π π = πcm = πcm = cm = 6 cm = 6cm 80

183 CBSE TEST PAPER-04 Class-X Mathematics (Surface area and Volume). The radii of the ends of a bucket 0cm high are cm and 7cm. then its capacity in litres. (a) 9.0 (b) 0.0 (c).0 (d) A solid is hemispherical at the bottom and conical above. It the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is (a) : (b) : (c) : (d) :. The diameter of a sphere is 6cm. It is melted and drawn into a wire of diameter cm. The length of the wire is (a) cm (b) 8cm (c) 6cm (d) 66cm 4. If the radii of the circular ends of a bucket of height 40cn are of length 5cm and 4cm. then volume of the bucket in cubic centimeter is (a) (b) (c) (d) A solid iron pole consists of a cylinder of height 0 cm and of base diameter [] 4cm. Which is surmounted by a cone 9cm high, find the mass of the pole. Given that cm of iron has 8g mass approx. 55 π = 6. cubes each of volume 64cm are joined end to end. Find the surface area of the [] resulting cuboid. 7. Kuldeep mode a bird bath for his garden in the shape of a cylinder with a hemi [] spherical depression at one end. The height of the cylinder is.45cm and its radius 8

184 is 0cm. find the total surface area of the bird-bath π = 7 8. A vessel is in the form of an inverted cone. Its height is 8cm and radius of its top, [] which is open, is 5cm. it is filled with water up to brim. When lead shots. Each of which is a sphere of radius 0.5cm, are dropped into the vessel. One forth of the water flows out. Find the number of lead shots dropped. 9. A medicine capsule is in the shape of a cylinder with two hemispheres. Stuck to [] each of its ends. The length of the entire capsule is 4 mm and the diameter of the capsule is 5 mm. Find its surface area. 0. A spherical glass vessel has a cylindrical neck 8cm long, m in diameter, the [] diameter of the spherical part 8.5 cm. by measuring the amount of water it holds, a child finds its volume to be 45cm check whether she is correct, taking the above as the is side measurements [ π =.4]. Metallic sphere of radii 6cm, 8cm and 0cm respectively are melted to form a [] single solid sphere. Find the radius of the resulting sphere. A shuttle cock used for playing badminton has the shape of a frustum of a cone [] mounted on a hemisphere. The external diameter of the frustum are 5cm and cm. the height of the entire shuttle cock is 7cm. find the external surface area.. A farmer connects a pipe of internal diameter 0cm from a canal into a cylindrical [5] tank in his field which is 0m in diameter and m deep. If water flours through the pipe at the rate of km/h in how much time will the tank be filled? 8

185 CBSE TEST PAPER-04 CLASS - Mathematics (Surface area and Volume) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5 (B) (B) (C) (B) For cylinder r = = 6 cm. h = 0cm For cone r = = 6 cm, h = 9cm Volume of pope = volume of cylindrical portion + volume of cone r h π π = π r h + = π + 55 = 6π ( 0 + ) = 6 = 6 55 = 780cm ( 6) 0 ( 6) ( 9) Ans6 Two cubes joined end to end we get cuboid l = = 8cm b = 4cm h = 4cm Surface area of cuboid = [ lb + bh + lh] [ ] [ + + ] = = 80 = 60cm Ans7 Let h be the height of cylinder and r be the common radius of the cylinder and hemisphere Total surface area of bird bath= C.S.A. of cylinder + C.S.A. of hemisphere 8

186 ( ) = π rh + π r = π r h + r = 0( ) cm 7 = 000cm =.m Ans8. Radius of lead shot = 0.5cm Radius of cone = 5cm Let xbe numbers of lead shots are dropped xyvolume of one lead shot = volume of the cone 4 4 x π ( 0.5) = π ( 5) x = = ( ) 00 lead shots one dropped Ans9. The length of capsule = 4mm 5 r = mm Length of cylindrical portion = 4 5 of capsule = 9mm Total surface area = π r + π rh + π r = π r + 4 π rh 5 5 = = + = = 0mm Ans0. For cylindrical part r = cm = cm h = 8cm For spherical part: Radius (R) = = 4 cm 84

187 Volume of glass solid = Volume of cylindrical part + Volume of the spherical part 4 4 = π r h + π R = π r h R = cm = cm = = cm = 46.5cm Ans. Sum of the volumes of three 4 π Given spheres = ( 6) + ( 8) + ( 0) 4 = π = π 78 = 4π 576 = 04π cm [ ] Let R be the radius of single solid sphere since volume remains the same 4 π R R = 04 π 04 4 = = R = = R = cm ( ) Ans. External surface area = C.S.A. of Frustum of the come + S.A. of hemisphere π r + r l + π r = ( ) [.5]( 6.) π ( ) [ r, r.5, h 7 6cm] = π + = = = = 85

188 5 68 = = ( ) = + l h r r l = + = 6.8 ( ) 6.5 = = 74.6cm Ans. Rate of water flowing = = 5 6 m / sec m / sec In second the water flows = 5 6 m Internal diameter = 0cm = m 5 Volume of the water that flows through the pipe in one second = 5 0 = = m 0 Volume of water in the tank = π r h r = = 5 m, h = m 0 = 7 00 = 7 m Time taken to fill the tank = = = Second 7 = 00 minutes = hr. 40 minutes πr h 86

189 CBSE TEST PAPER-05 Class-X Mathematics (Surface area and Volume). The diameter of a metal lice sphere is 6cm. it is melted and drawn into a wire of diameter of the cross-section 0.cm. then the length of wire is (a) m (c) 6m (b) 8m (d) 66m. The ratio between the volumes of two spheres is 8:7. What is the ratio between their surface areas? (a) : (b) 4:5 (c) 5:6 (d) 4:9. A hollow cube of internal edge cm is filled with spherical marbles of diameter 0.5cm and it is assumed that number of marbles that the cube can accommodate is (a) 496 (b) 496 (c) 4496 (d) 4596 space of the cube remains unfilled. Then the 8 4. A solid is hemispherical at the bottom and conical above. It the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is (a) : (b) : (c) : (d) : 5. A cone of height 4 cm and radius of base 6 cm is made up of modeling clay. Find [] the volume of the cone. 6. cubes each of volume 64cm are joined end to end. Find the surface area of the [] resulting cuboid. 87

190 7. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The [] diameter of the hemisphere is 4cm and the total height of the Vessel is cm. find inner surface Area? 8. A spherical ball of diameter cm is melted and recanted into cubes each of side [] cm. find the no. of cubes thus formed π = 7 9. How many silver coins.75cm in diameter and of thickness mm must be melted [] to form a cuboid 5.5cm 0cm.5cm? 0. A container like a right circular having diameter cm and height 5cm is full of [] ice-cream. The ice-cream is to be filled in cones of height cm and diameter 6cm having a hemispherical shape on the top. Find number of such cones which can be filled with ice-cream.. Water flowing at the rate of 5 km per hour through a pipe of diameter 4cm into [] a rectangular tank which is 50m long and 44m wide. Find the time in which the level of water in the tank will rise by cm.. A solid cylinder of diameter cm and height 5cm is melted and recast into toys [] with the shape of a right circular cone mounted on a hemisphere of radius cm. if the height of the boy is cm find the number of toys so formed. [5]. A cone of radius 0cm divided into two parts by drawing a plane through the midpoint of its axis, parallel to its base. Compare the volume of the two parts. 88

191 CBSE TEST PAPER-05 Class X - Mathematics (Surface area and Volume) [ANSWERS] Ans. (C) Ans. (D) Ans. (A) Ans4. (B) Ans5 h = 4 cm, r = 6cm Volume of cone = π ( 6) ( 4) = π r h Ans6 6 4 π = 88 πcm Two cubes joined end to end we get cuboid l = = 8cm b = 4cm h = 4cm Surface area of cuboid = [ lb + bh + lh] [ ] [ + + ] = = 80 = 60cm Ans7 Ans8 Inner surface area = π rh + π r [ radius of base of the cylinder = radius of hemisphere] = π r h + = = 44 = 57cm ( r) ( ) 4 Volume of spherical ball = πr = 4 π cm 89

192 Volume of each cube required no. of cubes Volume of ball = Volume of cube 4 = 7 = 485 = cm Ans9 Volume of the cuboid = = 85 cm.75 Radius of the coin = r = = 0.875cm Thickness h = mm = 0.cm Volume of one coin required no. of coins = = = = 0 0 = = π r h = Volume of cuboid Volume of each coin Ans0 Volume of cylinder = π r h = π 5 = 540π Diameter of cone = cm r = 6cm Volume of ice cream = Volume of ice-cream Cone + Volume of hemispherical top of ice-cream cm 90

193 = π r h + π r π = + = 6π + 8π = 54π ( ) ( ) π ( ) No. of ice-cream cones Volume of cylinder = V. of icecream cone + Volume of hemisph top 540π = = 0ice-cream cones 54π Ans Km = 000m 5km = 5000m Volume of cylinder = 4 Radius = = 7cm 7 = 00 cm πr h Volume of water flowing through the cylindrical pipe in an hour at the rate of 5km/hr 7 7 = 5000 = m Volume of cuboid = lbh Volume of required quantity of water in the tank. = cm m 46m 00 = = 00 Since m of water falls into tank in hour 46m of water falls into tank in = 46 = hours Ans 6cm H=5cm cm 9

194 Volume of the cylinder Number of toys = Volume of one toy π R H π R H = = π r + π r h π r r + h = = = 5 ( ) ( ) Ans (i) (ii) Let OAB be the cone and OQ be its axis and P be the midpoint of OQ Let OQ = h cm Then h OP = PQ = cm And QB = 0cm Also OPD OQB OP PD h / PD = = = OQ QB h 0cm PD = 5cm A smaller cone of radius = 5cm and height = h/cm Frustum of a cone in which h R = 0 cm, r = 5 cm, height = cm Volume of smaller cone = h π 5 5 = 5π h cm 6 h = π cm Volume of frustum of the cone ( ) ( ) 75π h = cm 6 5π h 75π h Ratio of required volume = : 6 6 = 5 :75 = : 7 A C O P Q D B 9

195 CBSE TEST PAPER-0 Class-X Mathematics (Probability). Cards each marked with one of the numbers 4,5,6,,0 are placed in a box and mixed. Thoroughly, One card is drawn at random from the box. What is the probability of getting an even prime number? (a) 0 (b) (c) (d). A bag contains 5 red and 4 black balls. A ball is drawn at random from the bag. What is the probability of getting a black ball? (a) (b) 9 (c) 4 9 (d) None of these. A dice is thrown once what is the probability of getting a prime number? (a) (b) (c) (d) 0 4. What is the probability that a number selected from the numbers,,,,5 is a multiple of 4? (a) 5 (b) (c) (d) 5. Why is tossing a coin considered is be way of deciding which team should get the ball at the beginning of a football match? [] 6. An unbiased die is thrown what is the probability of getting an even number [] 7. Two unbiased coins are tossed simultaneously find the probability of getting two heads. 8. One card is drawn from a well shuffled deck of 5 cards. Find the probability of getting a Jack of hearts 9. 8 cards numbered,,,,8 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probabilities that the card bears (i) an even number (ii) a number divisible by or [] [] [] 9

196 0. A bag contains 5 red balls, 4 green balls. and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drown is (a) white (b) neither red nor white. A box contains 0 balls bearing numbers,,,4,,0. A ball is drawn at random from the box what is the probability that the number on the ball is. (i) an odd number (ii) Divisible by or (iii) Prime number. A bag contains 5 red and some blue balls (i) If probability of drawing a blue ball from the bag is twice that of a red ball. Find the number of blue balls in the bag. (ii) If probability of drawing a blue ball from the bag is four times that of a red ball. Find the number of blue balls in the bag.. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is (i) a card of spades of an ace (ii) a red king (iii) neither a king nor a queen (iv) either a king or a queen (v) a face card (vi) cards which is neither king nor a red card. [] [] [] [5] 94

197 CBSE TEST PAPER-0 Class X - Mathematics (Probability) Ans. Ans. Ans. Ans4. Ans5. Ans6. (A) (C) (B) (A) [ANSWERS] Probability of head P( H ) = Probability of tail P( T ) = i.e. P( T ) = P( H ) = Probability of getting head and tail both are same tossing a coin considered to be fair way. Total number of outcomes are,,,4,5,6 which are 6 in number favourable case = [ is the only even prime] required probability = 6 Ans7. Total number of outcomes are HH, HT, TH, TT which are 4 in numbers Favorable outcomes = HH which is only required probability = 4 Ans8. Total number of outcomes = 5 Favorable cares = [There is only one Jack of hearts] required probability = 5 Ans9. Total no. of possible outcomes = 8 (i) Favorable cases are,4,6,8,0,,4,6,8 i.e.9 number Required probability = 9 = 8 95

198 (ii) Favorable cares are,,4,6,8,9,0,,4,5,6,8 i.e. in number. Required probability = = 8 Ans0. Total number of balls in the bag = 5+4+7=6 total number of possible outcomes = 6 (a) Favorable outcomes for a white ball = 7 Required probability = 7 6 (b) Favorable outcomes for neither red nor white ball=number of green balls =4 Required probability = 4 = 6 4 Ans. Total number of outcomes = 0 (i) Favorable outcomes are,,5,7,9,,,5,7,9 i.e. 0 in number. required probability = 0 = 0 (ii) Number divisible by are,4,6,8,0,,4,6,8,0 i.e. 0 in number Numbers divisible by are,6,9,,5,8. i.e. 6 in number Numbers divisible by or are 6,,8 i.e. in number. numbers divisible by or =0+6-= Favorable outcomes = required probability = 0 (iii) Prime numbers are,,5,7,,,7,9 i.e.8 in number Favorable outcomes = 8 8 Required probability = = 0 5 Ans. Let number of blue balls = x Total number of balls = 5 + x 5 Probability of red ball = 5 + x x Probability of blue ball = 5 + x By given condition x 5 =. 5 + x 5 + x x = 0 96

199 No. of blue balls = 0 5 x (ii) Here = x 5 + x x = 0 Hence number of blue balls = 0 Ans. Total possible outcomes = 5 (i) no. of spade = No. of ace = 4 card is common [ace of spade] Favorable outcomes = +4-=6 required probability = 6 = 4 5 (ii) no. of red kings = Favorable outcomes = required probability = = 5 6 (iii) no. of king and queen = 4+4=8 Favorable outcomes = 5-8=44 44 Required probability = = 5 (iv) no. of king and queen =4+4=8 8 Required probability = = 5 (v) no. of face cards = 4+4+4= [Jack, queen and king are face card] Required probability = = 5 (vi) The no. of cards which are neither red cards nor kings = 5-(6+4-) =5-8=4 required probability = 4 =

200 CBSE TEST PAPER-0 Class- X Mathematics (Probability). Cards marked with the numbers to 5 are placed in a box and mixed thoroughly one card is drawn from this box. Find the probability that the number on the card is an even number (a) (b) (c) (d) None of these. The king, queen and jack of clubs are removed from a deck of 5 playing cards and then well shuffled. One card is selected from the remaining cards. Finds the probability of getting a king (a) 49 (b) (c) 7 7 (d) none of these. What is probability of getting a number less than 7 in a single throw of a die (a) (b) 0 (c) (d) none of these 4. One card is drawn from a well shuffled deck of 5 cards. Find the probability of drawing 0 of a black suit. (a) 6 (b) (c) (d) 0 5. A game consists of tossing a one-rupee coin times and noting its outcome each time. Hanif wins it all the tosses give the same result i.e. three heads or three fails and loses otherwise. Calculate the probability that hanif will lose the game 6. Gopy buys a fish a fish from a shop for his aquarium. The shopkeeper taker out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish? 7. A lot consists of 44 ball pens of which 0 are defective and the others are food. Arti will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that [] [] [] 98

201 (i) She will buy it (ii) she will not buy it? 8. Harpreet tosses two different coins simultaneously (say one is of Rs and other is Rs ) what is the probability that she gets at least one head? 9. A box contains blue marble, white marbles, if a marble is taken out at random from the box, what is the probability that it will be a white one? Blue one? Red one? 0. The Integers from to 0 inclusive are written on cards ( one number on one [] [] [] card). These card one put in a box and well mixed. Joseph picked up one card. What is the probability that his card has (i) number 7 (ii) an even number (iii) a prime number. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavored candy? (ii) a lemon flavored candy?. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red, find the number of blue balls in the bag.. Cards marked with numbers,,,..5 are placed in a box and mixed thoroughly and one card is drawn at random from the box what is the probability that the number on the card is (i) a prime number? (ii) a multiple of or 5? (iii) an odd number? (iv) neither divisible by 5 nor by 0? (v) perfect square (vi) a two digit number. [] [] [5] 99

202 CBSE TEST PAPER-0 CLASS - Mathematics (Probability) [ANSWERS] Ans. Ans. Ans. Ans4. Ans5. (A) (A) (C) (A) Since a coin is tossed times possible outcomes are = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} x( s ) = 8 heads and tails {HHH,TTT} x( heads and tail) = P(Hanif will win the game) = = 8 4 P ( Hanif will lose the game) = = 4 4 Ans6. Total no. of fishes = 5+8= No. of male fish = 5 P(male fish) = 5 Ans7. Total no. of ball pens = 44 Number of defective pens = 0 number of good pens = 44-0=4 4 (i) P(she will buy it) = = (ii) P (she will not buy) = = 44 6 Ans8. For tossing coins Total possible outcomes are = {HH,TT,TH,HT} = 4 Favorable outcomes = at least one head = {TH,HT,HH} = Required probability = 4 00

203 Ans9. Total no. of possible outcomes = ++4 = 9 The no. of favorable outcomes for white marbles = Required probability = 9 The no. of favorable outcomes for blue marbles = Required probability = = 9 The no. of favorable outcomes for red marbles =4 Required probability = 4 9 Ans0. Total no. of possible outcomes = 0 (i) P (the no.7) = 0 (ii) Even no. are,4,6,8,0,,4,6,8,0,,4,6,8,0 Favorable outcomes = 5 Required probability = 5 = 0 (iii) Prime numbers from to 0 {,,5,7,,,7,9,,9} No. of favorable outcomes = 0 Required probability = 0 = 0 Ans. the bag has lemon flavored candies only. (i) P(an orange flavored candy) = 0 0 = (ii) P (a lemon flavor ) = = Ans. Suppose no. of blue bolls = x Total no. of balls = ( x + 6) x Probability of blue balls = x Probability of red balls = x + 6 Acc. to. question x 6 =. 6 + x 6 + x x = Hence no. of blue balls = 0

204 Ans. Total no. of possible outcomes = 5 (i) favorable cases are,,5,7,,,7,9, which are 9 in number req. probability = 9 5 (ii) Multiple of or 5 Favorable cases are,5,6,9,0,,5,8,0,,4,5 which are in number required probability = 5 (iii) favorable cases are,,5,7,9,,,5,7,9,,,5 which are in number required probability = 5 (iv) favorable cases are,,,4,6,7,8,9,,,,4,6,7,8,9,,,,4 which are 0 in number Required probability = 0 = (v) Perfect square no. are (,4,9,6,5) Favorable cases are = 5 5 Required probability = = 5 5 (vi) A two digit no. are {0,,,,4,5,6,7,8,9,0,,,,4,5} =6 Required probability = 6 5 0

205 CBSE TEST PAPER-0 CLASS- X Mathematics (Probability). Cards marked with the numbers to are placed in a box and mixed thoroughly one card is drawn from the box. Find the probability that the number on the card is a prime number less than 8. (a) (b) 7 0 (c) (d) none of these. Two dice are thrown simultaneously. Find the probability that the total of the numbers on the dice is (a) (b) (c) (d) 0. A bag contains 5 red balls. 8 white balls 4 green balls and 7 black balls. A ball is drawn at random from the bag. Find the probability that it is not green (a) 5 6 (b) 7 6 (c) (d) tickets of a lottery of a lottery were sold and there are 5 prizes on these tickers. If a man has purchased one lottery ticket. What is the probability of winning a prize (a) 0.00 (b) 0.00 (c) (d) none of these 5. One card is drawn from a well shuffled deck of 5 playing cards. Find the probability of getting (i) a non-face card (ii) a black king or a red queen 6. A child has a die whose six faces show the letters as given below [] [] A B C D E A The die is thrown once what is the probability of getting (i) A? (ii) D? 7. A bag contains 5 balls out of which x are blue (i) if one ball is drown at random, what is the probability that it will be a blue ball? (ii) if 7 more blue balls are put in the bag. The probability of drawing a blue ball will be double than in (i) find x [] 0

206 8. A game of chance consist of spinning an arrow which comes to rest pointing at one of the numbers,,,4,5,6,7,8 and these are 7 8 [] equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than? (iv) a number less than 9? 9. A box contains 90 dicer which are numbered from to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by A box contains 5 red marbles, 8 white marbles and 4 green marbles, one marble is taken out of the box at random what is the probability that the marble out will be (i) red? (ii) white? (iii) not green?. Five cards- the ten, Jack, queen, king and ace of diamonds are well shuffled with their face downwards. One card is than picked up at random (i) What is the probability that the card is the queen? (ii) if the queen is drawn and put aside what is the probability that the second card picked up is (a) an ace? (ii) a queen?. Savita and kavita are friends. What is the probability that both will have (i) different birthdays? (ii) The same birthday? (ignoring a leap year). A die is thrown once find the probability of getting (i) a prime number (ii) a number between and 6 (iii) a number greater than 4 (iv) a number less than or equal to 4 (v) an odd number (vi) a no. divisible by [] [] [] [] [5] 04

207 CBSE TEST PAPER-0 CLASS - Mathematics (Probability) [ANSWERS] Ans. Ans. Ans. Ans4. (B) (D) (A) (C) Ans5. (i) Number of face cards in a dock of 5 playing cards = 4 king + 4 queen +4 Jacks = Number of non face cards in a deck of 5 playing cards = total no. of cards No. of face cards = 5-= probability of getting a non face card = = (ii) 5 As there are two black kings and red queens in a deck of 5 playing cards. No. of a black king or a red queen = +=4 Probability of getting a black king or a red queen = 4 = 5 Ans6. When a die is thrown once The number of all possible outcomes =6 A B C D E A Favorable outcomes = A A = P(A) = = 6 (ii) Favorable outcomes for letter D = P (D) = 6 Ans7. (i) Total no. of balls = 5 Total no. of possible outcomes = 5 No. of blue balls = x Favorable outcomes = x x P (drawing blue balls) = 5 (ii) When 7 more blue balls are put in the bag then total no. of balls = 5+7=4 05

208 No. of blue balls = x+7 P(drawing blue balls) = Acc. to question x + 7 x = 4 5 x = x 5 After solving x=5 x Ans8. arrow comes to rest pointing any numbers,,,4,5,6,7,8 The total number of possible outcomes = 8 (i) Let A be the event the arrow will point at 8 no. of favorable outcomes = P(A) = 8 (ii) Odd numbers are {,,5,7} Favorable outcomes = 4 required probability P(E) = 4 = 8 (iii) The no. greater than are {,4,5,6,7} = 6 Favorable outcomes = 6 6 Required probability = = 8 4 (iv) The no. less than 9 are {,,,4,5,6,7,8,} Favorable outcomes = 8 Required probability = 8 8 = Ans9. Total no. of dices which are numbered from to 90 are 90 Total no. of outcomes = 90 (i) Let A be the event that the disc drawn at random bears a two digit number Two digit numbers are (0,,,..to 90) favorable outcomes = 8 So P(E) = 8 = ,,,4,5,6,7,8,9 (ii) Perfect square numbers are ={ } (iii) Favorable outcomes = 9 9 P (E) = = 90 0 The number divisible by 5 are (5,0,5, 85,90) 06

209 Favorable outcomes = 8 Required probability = 8 = 90 5 Ans0. No. of red marbles = 5 No. of green marbles = 4 No. of white marbles = 8 Total no. of marbles = 7 Total no. of possible outcomes = 7 (i) no. of red marbles = 5 5 P(R) = 7 (ii) No. of white marbles = 8, Favorable outcomes = 8 8 P(W) = 7 (iii) No. of green marbles = 4 4 P(G) = 7 P(Not green marbles ) = - 4 = 7 4 = 7 7 Ans. Total no. of possible outcomes = 5 [Ten, Jack, queen, King and ace] (i) no. of favorable out comes = [queen] P(Queen)= 5 7 (ii) (a) when queen is drawn and put aside then remaining no. of cards =5-=4 [ten + Jack + king + ace] Then total possible outcomes = 4 The second card picked up is an ace Favorable outcomes = P(E) = 4 (b) The second card picked up is queen Queen is already removed No. of favorable outcomes = 0 P(E) = = 07

210 Ans. Savita birthday can be any day of the year now Kavita birthday can also be any day of 65 days in the year. (i) It Kavita birthday is different from Savita s birthday Than number of favorable outcomes = 65-=64 P ( Kavita s birthday different from Savita s birthday )= (ii) P (Savita and Kavita have the same birthday) 64 = = Ans. All possible outcomes = {,,,4,5,6} = 6 A prime number are {,,5} Favorable outcomes = Required probability = = 6 (ii) No. between and 6 {,4,5} favorable outcomes = Required probability = = 6 (iii) No. greater than 4 are {5,6} No. of favorable outcomes = Required probability = = 6 (iv) No. less than or equal to 4 are = {,,,4} Favorable outcomes = 4 Required probability = 4 = 6 (v) Odd number = {,,5} Favorable outcomes = Required probability = = 6 (vi) Number divisible by = {,4,6} Favorable outcomes = Required probability = = 6 08

211 CBSE TEST PAPER-04 CLASS- X Mathematics (Probability). One card is drawn from a well-shuffled deck of 5 cards. What is the probability of getting red face card (a) 6 (b) (c) (d) none of these. Two dice are thrown simultaneously. What is the probability of getting an even number as the sum (a) (b) (c) 4 (d). A bag contains lemon flavored can dies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavored candy? (a) (b) (c) (d)0 4. The maximum value of probability (a) (b) 0 (c) (d) None of these 5. Suppose you drop a die at random on the rectangular region shown in the fig. what is the probability that it will land inside the circle with diameter m? 6. (i) A lot of 0 bulbs contain 4 defective ones. One bulb is drawn at random fro the lot. What is the probability that this bulb is defective? (ii) Suppose the ball drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that the bulb is not defective? 7. A bag contains 5 red balls, 8 white balls 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is (i) black (ii) red [] [] [] 8. A carton consists of 00 shirts of which 88 are good, 8 have minor defects and 4 [] 09

212 have major defects Jimmy a trader. Will only accept the shirts which are good, but sujatha, another trades, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that (i) It is acceptable to Jimmy? (ii) It is acceptable to Sujatha? 9. 7 cards numbered,,, 6,7 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the no. on the card is (i) odd (ii) divisible by (iii) divisible by and both 0. A card is drawn at random from a pack of 5 cards. Find the probability that the card drawn is (i) a black queen (ii) a Jack a queen or a king (iii) neither an ace nor a king. There are 40 students in class X of whom 5 are girls and 5 are boys. The class [] [] [] teacher has to select one student as a clan representative. She writer the name of each student on a separate card. The card being identical and she puts cards in a bag and stirs thoroughly she then draws one card from the bag. What is the probability that the name written on the card is the name of a (i) girl (ii) boy?. A piggy bank contains hundred 50P coins, fifty Rs coins, twenty Rs coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the hand is turned upside down. What is the probability that the coin (i) will be a 50P. coin (ii) will not be a Rs 5 coin?. (a) A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability that the sum of the two numbers that turn up is 8? (b) Find the probability of obtaining (i) a total of 6 (ii) a total of 0 (iii) The same no. on both dice [] [5] 0

213 CBSE TEST PAPER-04 CLASS - Mathematics (Probability) [ANSWERS] Ans. Ans. Ans. Ans4. (A) (A) (D) (A) Ans5. Area of rectangle = = 6m Area of circle = π r = π π = 4 m P (die will land inside the circle) π Area of circle 4 π = = = Area of rectangle 6 4 Ans6. (i) P (The bulb is defective) Number of defective bulbs = Total number of bulbs in a lot 4 = = 0 4 (ii) No. of good bulbs = 0-4=6 One bulb is drawn and not replaced then no. of good bulbs = 6-=5 Total no. of bulbs = 0-=9 Required probability = 5 9 Ans7. Total number of outcomes = =4 (i) Favorable outcomes =7 7 Required probability = 4 (ii) favorable outcomes = 5 required probability = 5 4 Ans8. Total no. of shirts = 00 Total no. of possible outcomes = 00

214 (i) The no of outcomes favorable to jimmy = P (shirt is acceptable to Jimmy ) = = (ii) The no. of outcomes favorable to Sujatha = 88+8=96 Required probability = = Ans9. Total no. of possible outcomes =7 (i) For odd no. favorable cases,,5,7,9,,,5,7 =9 Required probability = 9 7 (ii) Divisible by {,4,6,8,0,,4,6} = 8 Favorable cases = 8 Required probability = 8 7 (iii) Divisible by and both i.e. divisible by 6 {6,} Favorable cases = Required probability = 7 Ans0. Total no. of possible outcomes = 5 (i) No. of black queen = Favorable outcomes = P (a black queen) = = 5 6 (ii) No. of Jack = 4 queen = 4 King =4 No. of favorable outcomes = Required probability = = 5 (iii) Ace = 4, King = 4 Total = 4+4=8 Neither an ace not a king = 5-8=44 P(Neither an ace nor a king) = 44 5 = Ans. Total no. of possible outcomes = 40 (i) No. of outcomes favorable for girls = 5 P(card with name girl) = 5 = (ii) No. of boys = 5 no. of favorable outcomes = 5 P (card with name of a boy) = 5 = 40 8

215 Ans. Total no. of coins = =80 (i) No. of 50p coin = 00 Required probability = 00 = 5 Ans (ii) No. of Rs 5 coin = 0 No. of coin which is not Rs.5 = P (Not Rs 5 coin) = = When a black die and a white die are thrown Total no. of possible outcomes ( 6, ),(, ),(, ),(,4 ),(,5 ),(,6 ) (, ),(, ),(, ),(,4 ),(,5 ),(,6) (, ),(, ),(, ),(,4 ),(,5 ),(,6 ) ( 4, ),( 4, ),( 4, ),( 4,4 ),( 4,5 ),( 4,6) ( 5, ),( 5, ),( 5, ),( 5,4 ),( 5,5 ),( 5,6) ( 6, ),( 6, ),( 6, ),( 6,4 ),( 6,5 ),( 6,6) Total no. of possible outcomes = 6 6 = 6 Outcomes in which sum of two numbers is 8 are (,6),(,5),(4,4),(5,),(6,) which are 5 in number Required probability = 5 6 (i) For total 6 Favorable outcomes are (,5),(,4),(,),(4,),(5,) favorable no. of outcomes = 5 Required probability = 5 6 (ii) For a total of 0, we have (4,6),(5,5),(6,4) Favorable outcomes = Required probability = = 6 (iii) For same no. on both dice or no. of doublets (,),(,),(,),(4,4),(5,5),(6,6) Favorable outcomes = 6 6 Required probability = = 6 6

216 CBSE TEST PAPER-05 CLASS- X Mathematics (Probability). If E be any event the value of P(E)lie in between (a) 0 < P( E) < (b) 0 P( E) < (c) 0 P( E) (D) 0 P( E). Maximum and minimum value of probability is (a)(,) (b)(,0) (c)(0,) (d) none of these. An unbiased die is thrown. What is the probability of getting an even number or a multiple of? (a) (c) (b) (d) none of these 4. Let E be any event then the value of P(E) + P (not E) equals to (a) (b) 0 (c) (d) 5. If two dice are thrown one find the probability of getting 9 [] 6. A card is drawn from a well shuffled deck of playing cards. Find the probability of a face card [] 7. What is the probability of having 5 Mondays in a leap year? [] 8. Cards bearing numbers to 0 are placed in a bag and mixed thoroughly. A card is taken out from the bag at random what is the probability that the number on the card taken out is an even number? 9. A bag contains 5 red 4 black and green balls. A ball is taken out of the bag at random find the probability that the selected ball is (i) of red colour (ii) not of green colour [] [] 4

217 0. From a well shuffled pack of 5 cards black aces and black queens are removed. From the remaining cards a card is drawn at random find the probability of drawing a king or a queen. Which of the following experiments have equally likely outcomes? Explain. (i) a driver attempts to start a car. The car starts or does not start. (ii) A player attempts to shoot a basket ball she/he shoots or misses the shot (iii) A baby is born. It is a boy or a girl.. Find the probability that a number selected at random from the numbers,,,..,5 is a (i) Prime number (ii) multiple of 7 (iii) multiple of or 5. From a pack of 5 playing cards, Jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a black Jack (iv) a honorable card [] [] [] [5] 5

218 CBSE TEST PAPER-05 Class X - Mathematics (Probability) [ANSWERS] Ans. Ans. Ans. Ans4. (C) (B) (A) (A) Ans5. Total number of possible outcomes of throwing two dice = 6 6 = 6 Number of outcomes of getting 9 i.e. (+6),(4+5),(5+4),(6+) = 4 The required probability 4 = = 6 4 Ans6. Total number of possible outcomes = 5 Favorable out comes = 4+4+4= [4 Jack, 4 queen, 4 king] required probability = = 5 Ans7. Total number of days in a leap year = 66 This contains 5 weeks and days The remaining two days may be MT, TW,WTh, Th.F, FS, SS, SM Favorable cases are MT, SM i.e. out of 7 cases required probability = 7 Ans8. Total number of outcomes = 0-=7 Cards in the box having even numbers are 4,6,8,0,,4,6,8,0. What are 9 in number favorable outcomes in this care = 9 P (an even number) = 9 7 6

219 Ans9. Total number of balls in the bag = 5+4+ total number of outcomes = (i) No. of red balls = 5 the required probability for a red ball = 5 (ii) favorable cases for non green ball = -=9 the required probability for a non green ball = 9 = 4 Ans0. Total number of cards = 5 Number of black aces = Number of black queens = Cards left = 5--=48 Total number of equally likely cases = 48 Number of kings and queens left in the 48 cards = 4+=6 Favorable cases = 6 required probability = 6 48 = 6 Ans. (i) A driver attempts to start a car the can starts or does not start are not equally likely (ii) A player attempts to shoot a basket ball she/he shoots or misses the shoot are not equally likely (iii) A baby is born It is a boy or a girl is an equally likely event Ans. (i) Prime number are,,5,7,,,7,9,,9, = Total number of outcomes = 5 P (Prime number) = 5 (ii) Multiples of 7 are 7,4,,8,5 =5 (iii) P (a multiple of 7) = 5 = 5 7 Multiple of are,6,9, = Multiple of 5 are 5,0,5,,5 = 7 7

220 Multiple of and 5 are 5 0 = Multiple of or 5 = +7-=6 P (Multiple of or 5) = 6 5 Ans. Total number of outcomes = 5 Cards removed = +++ = 8 [ Jack, queens, kinks and aces of red colour] remaining number of cards = 5-8=44 Total number of outcomes = 44 (i) Favorable outcomes = [There are black queen] required probability = = 44 (ii) Favorable outcomes = number of red cards left = 6-8=8 Probability for a red card = 8 = 9 44 (iii) Favorable outcomes = number of black Jacks = required probability = = 44 (iv) Number of picture cards left =++ [Jacks, queens, Kings are picture (v) cards] required probability = 6 = 44 Honorable cards [ace, Jack, Queen and king] No. of honorable cards left = +++=8 required probability = 8 = 44 8

221 CBSE TEST PAPER- Class X Mathematics Time :-.5. Hrs. M.M. :- 40. If cos θ = 7 ( + sin θ )(-sin θ )., evaluate: 8 ( + cos θ )(-cos θ ). A ladder 0m long reaches a window 8m above the ground. Find the distance of the foot of the ladder from the base of the wall.. Find the co Ordinates of a point A, where AB is the diameter of a circle is (,- ) and B is (, 4). 4. Show that: θ θ θ θ θ θ (sin + cosec ) +(cos + sec ) = 7 + tan + cot. [] 5. Find the length of the medians of the whose vertices are (,-),(0. 4) and (- 5, ). 6 The co ordinates of one and point of a diameter of a circle are (4,-) and the co ordinates of the centre of the circle (,-) Find the co ordinates of the other end of the diameter. [] [] 7. Evaluate: θ + θ cos(40 ) sin(50 ) cos 40 + cos 50 Sin 40 + sin [] 8. Find the co ordinates of the points of trisection of the line joining (4, -) and (-, -). 9. Evaluate without using trigonometric tables. [] [] cotθ tan(90 - θ ) - sec (90 - θ ) cosec θ + sin (tan 5 tan45 tan 85 ). 0. If sec θ = x+. xprove that sec θ+ tan θ = xor 4 x []. Prove the following trigonometric Identities: - cos eca - cot A sin A = sin A - cos ec A + cot A []. The angle s of elevation of the top of a building from the foot of the tower is 0 [] 9

222 and the angle of elevation of the top of the tower from the foot of the building is 60. If the tower is 50m high. Find the height of the building.. Find the area of a triangle whose vertices are (, 8) (-4, ) and (5, - ). [] 4. The angle of elevation of a jet plane from point A on the ground is 60. After a [6] flight of 5 seconds the angle of elevation changes to 0. If the jet plane is flying at a constant height of 500 m, find the speed of the jet plane. 5. If the angle s of the elevation of a cloud from a point h meters above a lake is a [6] and the angle of depression of its reflection in the lake is β, prove that the height of the cloud is h (tan β + tan α) (tan β tan α) 0

223 CBSE TEST PAPER- Class X Mathematics Time :-.5. Hrs. M.M. :- 40. The sum and product of zeros of the polynomial are -/ and - respectively. What is the quadratic polynomial?. Determine the nature of the roots of the equation 4x x -9 = 0.. Find the th term of an A.P with first term ½ and common difference /. 4. For what value of K the numbers K + 4, 7K +, K 5 are in A.P. 5. What type of graph following pair of lines represent x + 7x = - and 6x + y =5. 6 How many two digit numbers are divisible by. [] 7. Solve. 4x + y =.7, 7x y = Which term of the sequence 4, 09, 04 is the first negative term. [] 9. Find the zeros of the polynomial 8x x. [] 0. If two zeroes of the polynomial x 4 + x 0x 6x + 6 are and -. [] Find the other zeroes of the polynomial.. Draw the graph of x + y = 6, x y = Find the area of x = 0, y = o and x y =.. Some students arranged a picnic. The budget for food was Rs. 40. Because 4 students of the group failed to go the cost of food to each students got increased by Rs. 5. How many students went for the picnic?. If the m th term of an A.P is /n and n th term is /m show that sum of (mn) term is ½ (mn + ). 4. Two pipes running together can fill a cistern in / minutes. If one pipe takes minutes more than other to fill it. Find the time in which each pipe would fill the cistern. 5. One fourth of a herd of camel was seen in the forest twice the square root of the herd gone to mountain and the remaining 5 were seen on the bank of a river. [] [] [] [] [6]

224 Find the total number of camels. 6. A boat goes km upstream and 40km down stream in 8 hours. It can go 6km upstream and km down stream in the same term. Find the speed of the boat and speed of the stream. [6] SECTION C 6. Using division algorithm, find the HCF of three numbers 40, 67, and For what values of a and b the following system of linear equations have an infinite number of solutions. x + y = 7, (a b) x + (a + b) y = a + b. 8. For what values of K does the quadratic equation 9x + 8kx + 6 = 0 have equal roots. 9. A two digit number is such that product of its digit is 5 when 8 is added to the number, the digits interchange their places find the number. 0. How many terms of the AP. 7, 69, 66,..make the sum 897 explain the double answer.. As observed form top of a light house, 00 m high above sea level, the angle of depression of a ship, sailing directly towards it changes from 0 o to 45 o Determine distance traveled by ship during period of observation (Take =.7). If the sum of the squares of zeros of polynomial 6x + x + k is 5 6 find value of k.. Equilateral triangles are drawn on the sides of a right triangle. Show that area of the triangle on the hypotenuse is equal to sum of areas of triangles on the other two sides. 4. A pole has to be erected at a point on the boundary of a circular park of diameter metres in such a way that the differences of its distance from two diametrically opposite fixed gates A and B on the boundary is 7 mt. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected.

225 5. Write prime factorization of 048 by using the factorization tree. SECTION - D 6. Prove that in a right angle triangle, the square on the hypotenuse is equal to sum of squares on other two sides using above prove the following in ABC, A = 90 o and AD BC. Prove that AB + CD = BD + AC. 7. If the angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in lake is β. Prove that height of cloud is h(tan β + tan α) tan β - tan β 8. Solve graphically the following system of linear equations. x + y + = o x y + 8 = o Shade the area of region bounded by the lines and x-axis. 9. A passenger train takes hours less for a journey of 00 km, if its speed is increased by 5km/hr from its original speed find the original speed of the train. 0. Find the other zero s of the polynomial x 4 5x + x + 0x 8 if it is given that two of its zero are - and.

226 CBSE TEST PAPER- Class X Mathematics Time :-.5. Hrs. M.M. :- 40. State fundamental theorem of Arithmetic.. In ABC. DE BC =cm. DB = 7 cm. What is the ratio of ABC to area of DEF.. The length of tangent from a point A at a distance of 5 cm from the centre of the circles is 4 cm. Find the radius of the circle. 4. Stare Euclid s Division lemma. 5. Write 4560 as product of product of prime factors. 6 Find the HCF of 96 and 404. Hence, find their LCM. [] 7. Check whether 6 n end with O or not. [] 8. Prove that tangents from an external point to the same circle are equal. [] 9. A vertical stick 0m long casts a shadow 8m long. At the same a tower casts a shadow 0m long. Determine the height of the tower. 0. Construct a circle whose radius is equal to 4 cm. Let P be a point whose distance from its centre is 7 cm. Constructs two tangents from P. [] []. State and prove Basic Proportionality Theorem. []. Show that square of any positive integer is of the form m or m + for some integer m []. O is any point inside a rectangle ABCD. Prove that OB + OD = OA + OC. [] 4. If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus. 5. Prove that the ratio of areas of two similar s is equal to the square of the ratio of their corresponding sides. Use the about theorem in the following. The area of two similar s are 8 cm and 44 cm. If the largest side of the smaller is 7 cm, find the largest side of larger. 6. In an equilateral ABC,D is a point on side BC such the BD = / BC. Prove that 9AD = 7 AB. [] [6] [6] 4

227 CBSE TEST PAPER- Class X Mathematics Time :- Hrs. M.M. : General Instructions.. All questions are compulsory.. The question paper consists of 0 questions divided into four section A, B C and D. Section A comprises ten questions of mark each, Section B comprises of five questions of marks each, Section C comprises of ten questions of marks each and Section D comprises of five questions of 6 marks each.. All question in section-a are to be answered in one sentence or as per the exact requirement of the question. 4. There is no overall choice. However an internal choice has been provided in one question of marks, three question of marks each and two question of 6 marks each. You have attempt only one of the alternatives in all such questions. SECTION A 0 MARKS. If 7 th term of an A.P exceeds the 0 th term by 7. Find the common difference.. The three consecutive terms of an A.P are P + 4, and 5P. Find a term other than.. The perimeters of two similar triangles ABC and PQR are respectively cm. and 4 cm. If PQ = cm. find the length of side AB. 4. ABCD is a trapezium. AB CD,If OA =,OC = x-, OB = x-5, OD = x 9. Find the value of x. 5. Check whether has terminating decimal representation. 6. State Fundamental Theorem of Arithmetic. 5

228 7. Write zeros and number of zeros in fig.,for the polynomial y=p(x) 8. Find the discriminant of the quadratic equation X + x - 0 = 9. For what value of A, trigonometric ratios tan A and cot A are equal? 0. Find the value of θ. If cos θ, O < θ < SECTION B. If an = 5-n. Find the sum of n term of A.P. OR If the 9 th term of an A.P is zero. Prove that its 9 th term is double of 9 th term.. In right angled ABC : B = 90, from the vertex B. BD AC is drown. Prove [] [] that = + BD BC AB. If (, ), (4, y), (x, 6) and (, 5) are the vertices of parallelogram taken in order. Find the value of x and y. 4. The length of a line segment is 0 unit. If one end is at (,-) and abscissa of the second end is 0. Find the ordinate of the other end. 5. Without using the trigonometric table evaluate the following: [] [] [] Sin70 4 cos 5. cosec 7 7Cos0 7 tan5.tan 5.tan55.tan

229 SECTION C 0 MARKS 6. If Band Qare acute angles such that Sin B= Sin Q then prove that B = Q. [] OR If Sin (A B) = and Cos (A + B) = O < A + B 0 90 and A > B. Find A. and < B. 7. Prove that + cosθ sinθ + cosθ =. cosθ + sinθ sinθ [] 8. If the sum of first P terms of an A. P is q and sum of q terms of the A.P is P. Find the sum of (p + q) terms. 9. The line segment joining the points (,) and (5, - 8) is trisected by the points P and Q. If p lies on the line x y + k = 0. Find the possible values of K. OR If A and B are ( -, _) and (, -4) respectively. Find the coordinates of P, if p lies on AB and AP AB = 7 0. If the vertices of ABCare A (8, ). B (K,-4) and C (,-5) are area of ABC is square units. Find the values of K.. In the given fig.. PQRS is a rhombus and QRC = BAC. Prove that PQ = AP. [] [] [] [] QC.. ABCD is a parallelogram. M is the mid-point of CD. The line BM is drawn intersecting AC at L and AD produced to E prove that EL = BL OR In the given figure 4. [] 7

230 r AB Prove that AC, EF = + y x z r AC, DC r AC, AB = x, EF = y, CD = z. Use Euclid Division algorithm to show that the square of any positive integer is either of the form m or m+ for some integer m. [] 4. If two zeros of the polynomial zeros. 4 x x x x are ±. Find other [] 5. Solve the system of equations for x b x a y = ab( a + b) and a b b x - a y = a b SECTION D 0 MARKS 6. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Using above result prove that In ABC. [6] AB = ( + ) BX ; Give that XY BC. Where X and Y are the points on the sides AB and AC respectively. 7. State and prove converse of Pythagoras theorem. Using above theorem. Find area of ABC if BC cm. 0 AOC = 90,AB = cm. OA = cm. and OC = 4 cm. [6] 8

231 8. The students of a class are made to stand in rows. If three students are extra in a row, there would be one row less. If three students one less in a row, would be two rows more. Find the number of students in the class. 9. An aeroplane left 0 minutes later than its scheduled time and in order to reach the destination 500 km away in time, it has to increase its speed by 50 km/hr from the usual speed. Determine its usual speed. OR Speed of a boat in still water is km/hr. It can go km upstream and return downstream to the original points in hours 45 minutes. Find the speed of the stream. 0. The angles of depression of the top and bottom of an 8 meter tall building from the top of a multi-storeyed building are 0 0 and 45 0 respectively. Find the height of the multi-storeyed building and the distance between the two buildings. OR An aeroplane when 000 m high, passes vertically above another aeroplane at an instant, when the angles of elevation of the two aeroplanes from the point, on the ground are 60 0 and 45 0 respectively. Find the vertical distance between the aeroplanes. [6] [6] [6] 9

232 CBSE TEST PAPER- Class X Mathematics Time :-. Hrs. M.M. :- All the questions are compulsory. Section A consists of 0questions of mark each ( to 0) Section B consists of 5 questions of marks each ( to 5) Section C consists of 0 questions of marks each (6 to 5) Section D consists of 5 questions of 6 marks each (6 to 0) Section A. Given that LCM (50, 00) = 00 find HCF (50, 00). Find the number of zero s for p(x).. Is the following equation quadratic? Give reasons : 4x -8 x + = 0 4. The 8 th term of an AP is 5 and its 0 nd term is 7. Find the AP. 5. Find the sum of all odd integers between and 00 divisible b. Is MNP similar to EFG 0

233 6. 7. Evaluate in simplest form. cos 60 o cos 0 o + sin 60 o sin 0 o 8. Prove that cos θ + sin θ = cosec θ sinθ 9. Given the following equations solve for x and y. x 4y = 4x y = 6 0. Without actual division state if the following rational numbers will have terminating or non terminating repeating decimal expansion 50. Solve for x and y Section B x + y x y + + =8 and + = 9. In an AP the sum of its first n terms is n + x Find its 8 th term.. Using Basic proportionality theorem prove that the lines drawn through the points of trisection of one side a triangle parallel to another side trisect the third side.

234 4. Without using trigonometric tables find value of 0 0 tan 50 + sec 50 cot 40 + cosec cos 40 cosec State euclids dirsion lemma. SECTION C 6. Using division algorithm, find the HCF of three numbers 40, 67, and For what values of a and b the following system of linear equations have an infinite number of solutions. x + y = 7 (a b) x + (a + b) y = a + b. 8. For what values of K does the quadratic equation 9x + 8kx + 6 = 0 have equal roots. 9. A two digit number is such that product of its digit is 5 when 8 is added to the number, the digits interchange their places find the number. 0. How many terms of the AP. 7, 69, 66,..make the sum 897 explain the double answer.. As observed form top of a light house, 00 m high above sea level, the angle of depression of a ship, sailing directly towards it changes from 0 o to 45 o Determine distance traveled by ship during period of observation (Take =.7). If the sum of the squares of zeros of the polynomial 6x + x + k is 5 6 k. find value of. Equilateral triangles are drawn on the sides of a right triangle. Show that area of the triangle on the hypotenuse is equal to sum of areas of triangles on the other two sides. 4. A pole has to be erected at a point on the boundary of a circular park of diameter metres in such a way that the differences of its distance from two diametrically

235 opposite fixed gates A and B on the boundary is 7 mt. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected. 5. Write prime factorization of 048 by using the factorization tree. SECTION - D 6. Prove that in a right angle triangle, the square on the hypotenuse is equal to sum of squares on other two sides using above prove the following in ABC, A = 90 o and AD BC. Prove that AB + CD = BD + AC. 7. If the angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in lake is β. Prove that height of cloud is h(tan β + tan α) tan β - tan β 8. Solve graphically the following system of linear equations. x + y + = o x y + 8 = o Shade the area of region bounded by the lines and x-axis. 9. A passenger train takes hours less for a journey of 00 km, if its speed is increased by 5km/hr from its original speed find the original speed of the train. 0. Find the other zero s of the polynomial x 4 5x + x + 0x 8 if it is given that two of its zero are - and.

236 Time : Hours CBSE UNIT TEST PAPER-05 CLASS - X (MATHEMATICS) M.M. GENERAL INSTRUCTIONS :-. All questions are compulsory.. There are four section A,B, C and D Section A contains 0 questions of mark each Section B contains 5 questions of marks each Section C contains 0 questions of marks each and Section D contains 5 questions carrying 6 marks each. There is one internal choice in section B, three in section C and two in section D. 4. Use of calculator and mobile phone is not permitted SECTION A. State the Fundamental Theorem of Arithmetic s.. The graph of y = f (x) is given below. Find the number of zeroes of f (x). Write a rational number between and 4. What is the nature of roots of quadratic equation 4x x+9 = 0? 5. Find perimeter of the given shaded region. 4

237 6. The length of tangent from a point A at a distance of 5cm from centre of a circle is 4cm. What will be radius of the circle? 7. Which measure of central tendency is given by x-coordinate of the point of intersection of the more then ogive and less than ogive? 8. A bag contains 5 red and 5 black balls. A ball is drawn at random from the bag. What is the probability of getting a red ball? 9. What is the distance between two parallel tangents of a circle of radius 4cms? 0. The height of a tower is 0m. Find the altitude of sun if the shadow of the tower is equal to the height of the tower. SECTION B. From pocket money, child saves Re first day, Re second day, Re third day and so on in a month. How much money will the child save in the month of February 008?. Express Sin65 o + Cos75 o in terms of trigonometric ratios of angles between 0 o and 45 o OR B + C If A,B and C are interior angles of a ABC then show that Cos = sin A. in the figure given below, DE ll BC and DE : BC =4:5 then find the ratio of the areas of ADE and ABC 5

238 4 All the Queens, Jacks and diamonds have been removed from a pack of 5 cards. The remaining cards have been reshuffled and a card is drawn at random. Find the probability that it is a ) face card ) black card. 5. Find the value of x for which the distance between the points P (,-) and Q (x,5) is 0 units. SECTION D 6. Prove that - is irrational. 7. Draw the graph of equations x+y=and x+5y=. Does their point of intersection lie on the x-axis? Find the area of the triangle formed by these lines and the line y=o. 8. Solve following for x and y (a-b) x+ (a+b) y=a -ab-b (a+b)(x+y)=a +b OR Solve the pair of linear equations for x and y 47x+5y=4 5x+47y=59 9. The sum of first n terms an A.P is n +n. Find the A.P and also the n th term. OR Find the sum of all three digit number which leave remainder when divided by 5. 0 Prove that SecA SecA + + Sec A+ Sec A- = coseca. Find out if the points A (-,), B (,-) and C (4,) are the vertices of an isosceles right triangle?. In what ratio does the points P (,-5) divide the line segment joining A (-,5) and B (4,-9). Construct a right angled ABC right angled at B with AB =4cms and BC=6cms. Now draw a circle with AB as diameter. From C, draw tangents to this circle. OR Construct a right angled ABC in which AB=4cms,BC=6cms and ABC=60 o such that each side of new triangle is 4/ of the corresponding side of ABC 4. In the given figure, AB, AC and PQ are tangents to a circle and AB=5 cm. Find the perimeter of APQ. 6

239 5. Find the area and perimeter of the shaded region in figure given below, where AP=PQ=QR=4 cm. 6. Find the mean, mode and median of the following data. 7. In a right angled triangle, prove that the square on the hypotenuse is equal to the sum of the squares on other two sides. Prove. Using the above, in the given figure, find the length of PQ if angle QPR = angle POR=90 o OP= 6 cm, OR=8 cm and QR= 6 cm 7