Solutionsÿ to. Surender Verma. M.Sc. (Maths), B.Ed Delhi Public School, Dwarka, New Delhi

Solutionsÿ to Mathematics PULLOUT WORKSHEETS FOR CLASS X First Term By Surender Verma P.W - 88 pages / 0-06-06 M.Sc. (Maths), B.Ed Delhi Public School, Dwarka, New Delhi New Saraswati House (India) Pvt. Ltd. New Delhi-000 (INDIA)

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CONTENTS. Real Numbers Worksheets ( to 6)... 6 Assessment Sheets ( and )... Chapter Test.... Polynomials Worksheets (0 to 5)...5 Assessment Sheets ( and 4)... Chapter Test... 6. Pair of Linear Equations in Two Variables Worksheets (8 to 9)...9 Assessment Sheets (5 and 6)...45 Chapter Test... 48 4. Triangles Worksheets ( to 45)...5 Assessment Sheets (7 and 8)...68 Chapter Test... 7 5. Introduction to Trigonometry Worksheets (50 to 58)...7 Assessment Sheets (9 and 0)... 84 Chapter Test... 86

6. Statistics Worksheets (6 to 65)...88 Assessment Sheet... 9 Chapter Test... 97 PRACTICE PAPERS ( to 5)... 99 4

Solutions to PULLOUT WORKSHEETS AND PRACTICE PAPERS [Summative Assessments] [FIRST TERM]

Chapter REAL NUMBERS WORKSHEET. (C) We know that the factors of a prime are and the prime itself only. Therefore, the common factor of p and q will be only. Hence, HCF (p, q).. (A) As prime factors of 005 are: 005 5 67. 7 is not a prime factor of 005. 5 5. 0.065 5 4. 5 termi- 4. 5 6 5 6 Clearly, the decimal form of nates after four places. 4. Terminating 4 9 Hint: 0.9. 5 000 First number Second number 5. LCM HCF 96 404 4 404 9696. 4 6. (i) 660 (ii) 0 Hint: Going in opposite direction to the factor tree, we obtain 65 0 (ii) and 0 660 (i). 7. HCF ; LCM 40 Hint: ; 5 5; 7. 8. (i) Terminating 54 54 Hint: 50 5. (ii) Non-terminating repeating 9 Hint: 08. 9. Hint: Let 5 a b ; b 0 5 b a b As RHS of this equation is rational, but LHS is irrational so a contradiction. 0. Let a be any odd positive integer and b 4. By Euclid s lemma there exist integers q and r such that a 4q + r, 0 r < 4... a 4q or 4q + or 4q + or 4q +. Therefore, for a to be odd, we have to take a 4q + or 4q +.. The maximum capacity (in kg) of a bag will be the HCF of 490, 588 and 88. Let us find out the required HCF by prime factorisation method. 490 5 7 588 7 88 7 HCF 7 98 Thus, the maximum capacity of a bag is 98 kg. WORKSHEET. (A) HCF (p, q) p and q are coprime. If p and q are coprime with q 0 and p q is a rational number, then q has only and 5 as prime factors. i.e., q m 5 n where, m and n are nonnegative integers.. (B) Going to opposite direction to the factor tree, we obtain 7 (ii) and 4 (i)...44... and.7... Therefore, we can take.5 as < <. 6 M A T H E M A T I C S X

4. Required number 449 6 449 7 07. 5. Hint: As 576 > 405... 576 405 + 40 Further 405 40 9 + 7 Further 40 7 + 48 Further 7 48 + 4 Further 48 4 + 4 Further 4 4 5 + 4 Further 4 4 6 + 0. In the last equation, remainder is zero. Hence, the required HCF 4. 6. First given number is composite as 5 + (5 + ) 6 8 But second given number is prime as 5 7 + 7 + 5 + + 59. 7. No. Prime factors of 6 n will be of type n n. As it doesn't have 5 as a prime factor, so 6 n can't end with the digit 5. 8. Hint: Let a be any positive integer... a q or q + or q +... a 9q m; m q or a (q +) m +, m q (q + ) or a (q + ) m +, m q + 4q +. 9. We represent 6, 7 and 0 in their prime factors. 6 7 0 5 Now, HCF 6 And LCM 5 60. 0. Hint: Let 5 x, a rational number x + 5 Squaring both sides, we get x +5+x 5 x 5 x RHS of this last equation is rational, but LHS is irrational which is a contradiction.. Length 6 m 0 cm 60 cm Breadth 5 m 85 cm 585 cm Height m 60 cm 60 cm The required length (in cm) of the tape will be the Highest Common Factor (HCF) of the numbers 60, 585 and 60. Let us find out the HCF. 60 5 7 585 5 60 5 HCF 5 45 Hence, the length of the tape will be 45 cm. WORKSHEET 4 4 5 5. (C) 4 4 4 5 ( 5) 0 0.05 Hence, the number terminates after four places of decimal.. (A) É É É É. is a rational number.. 8 7 ; 40 4 5. Now, HCF (8, 40) 4 6. 4. HCF LCM Hint: First number Second number. 5. No. Hint: Prime factors of 5 n does not contain p 5 q in factor, p, q being positive integers. 6. Rational number 0.7 Irrational number 0.6000000.... 45 9 8 7. (i) 0.. 65 5 8 000 7 5 875 (ii) 0.0875. 80 5 0000 8. Let us assume, to the contrary that is rational. We can take integers a and b 0 such that a, where a and b are coprime. b b a a is divisible by R E A L N U M B E R S 7

a is divisible by...(i) We can write a c for some integer c a 9c b 9c (... a b ) b c b is divisible by b is divisible by...(ii) From (i) and (ii), we observe that a and b have atleast as a common factor. But this contradicts the fact that a and b are coprime. This means that our assumption is not correct. Hence, is an irrational number. 9. As: 0 408 + 6...(i) 408 6 + 9...(ii) 6 9 + 4...(iii) 9 4 8 + 0...(iv) HCF 4 From (iii) 4 6 9 6 [408 6] [Use (ii)] 6 408 [0 408] 408 [Use (i)] 4 0 5 408 m. 0. Hint: Let x be any positive integer. Then it is of the form q or q + or q +. If x q, then x (q) 9m; m q If x q +, then x (q + ) 9m + ; m q(q + q + ). If x q +, then x (q +) 9m + 8; m q (q +6q + 4).. The maximum number of columns must be the highest common factor (HCF) of 66 and. Let us find out the HCF by the method of Euclid's division lemma. Since 66 >, we apply division lemma to 66 and, to get 66 9 + 8 Since the remainder 8 ¹ 0, we apply the division lemma to and 8, to get 8 4 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8, the HCF of 66 and is 8 Hence, the maximum number of columns is 8. WORKSHEET 4. (B) Non-terminating repeating. Hint: Denominator is not in the exact form of m 5 n, where m, n are non-negative integers.. (C) 0 r < b.. É 6 5 É 6 5 É 6 É 5 6 5 Rational number. 4. Terminating decimal form as denominator 4 of 07 4 is of the form n 5 m. Here n, m 0 5. (i) 00 (ii) 9 Hint: 7 (ii) and (ii) (i). 6. Let us represent each of the numbers 0, 7 and 4 as a product of primes. 0 5 7 4 4 Now, HCF 6 and LCM 4 5 60. 7. Here, 96 > 8. 96 8 4 + 68 Further 8 68 + 4 Further 68 4 4 + Further 4 + Further 6 + 0 In the last equation, the remainder is zero and the divisor is. Hence, the required HCF. 8. Hint: Let + 5 a b ; b 0 ab 5 Rational b 8 M A T H E M A T I C S X

R Which is a contradiction as 5 is an irrational number. 9. (i) The given fraction can be written as 4 4 5 0.05 4 4 5 0 Hence, the given number terminates after four places of decimal. (ii) The given fraction can be written as E 59 59 5 5 5 5 5 A L N 4 U M 5744 00000 0.05744 Hence, the given number terminates after five places of decimal. 0. The required number of students will be the highest common factor (HCF) of, 60 and 56. Let us find out the HCF by the method of prime factorisation. 60 5 56 HCF 5 Number of buses required Total number of students Number of students in one bus + 60 + 56 4 5 Thus, the maximum number of students in a bus and number of buses required are 5 and 4 respectively.. Hint: Let x any positive integer x 5m, 5m +, 5m +, 5m + or 5m +4 Now take square of all these form. WORKSHEET 5. (C) Let the quotient is m when n is divided by 8. n 8 m n 0, 8, 6, 4,,.... n, 9, 7, 5,,... B E R S n ±, ±, ± 7 ± 5, ±,... n An odd integer is the right Answer.. Hint: HCF (65, 7) Now, 65m 7. m will satisfy this equation.. Hint: LCM of 8, 4, 0, 4 50 Required number 50 + 5. 4. Prime factors of numbers to 0 are: ; ; ; 4 5 5; 6 ; 7 7; 8 ; 9 ; 0 5 Now, LCM 5 7 8 9 5 7 50 is required number. 5.. Hint: 5 5 x 5 4 5 x 5 x, which is a rational number. 6. Hint: Any odd positive integer will be type of 4q + or 4q + (4q + ) 6q + 8q + 8 (q + q) + 8n + Also, (4q +) 6q +4q +9 8 (q +q + ) + 8n +. 7. 5 cm Hint: Find HCF. 8. Hint: Let 5 a b where a, b are integers and b 0 Squaring on both sides, a 5 8 6 0 b a 6 0 b b a 0... a contradiction. 6b 9

9. (i) Terminating (ii) Terminating. 0. The required number of burfis will be the highest common factor of 40 and 0. Let us find out the HCF using Euclid's division lemma. It is clear that 40 > 0. We apply Division lemma to 40 and 0, to get 40 0 + 0 Since the remainder 0 ¹ 0, so we apply Division lemma to 0 and 0, to get 0 0 4 + 0 Again the remainder 0 ¹ 0, so we apply Division lemma to 0 and 0, to get 0 0 + 0 Now, the remainder is zero. So the HCF of 40 and 0 is the divisor at the last stage that is 0. Hence, the required number of burfis is 0.. Let n q, q + or q +. Case I. If n q, then n q divisible by n + q + Not divisible by n +4 q + 4 (q +) + Not divisible by. Case II. If n q + then only n + q ++ q + (q + ) is divisible by. Case III. If n q + then only n +4 q + 6 (q +) is divisible by. WORKSHEET 6. (C) 85 5 7 So, is not a prime factor of 85.. (C) As p and p + are two consecutive natural numbers, HCF and LCM p (p +). 5 7. Hint: The given number is or 500 500 Denominator 500 5 Clearly, the denominator is exactly in the form m 5 n, where m and n are nonnegative integers; so the given number has a terminating decimal expansion. 4. 800 Hint:... 8 ;9 ; 5 5 HCF (8, 9, 5) LCM (8, 9, 5) 800. 5. 9 Hint: HCF (0, 55) 5 0 5 + 55y 5 55y 5 050 045 y 9. 55 6. Irrational Hint: x 7 4 x 7 x Irrational. 7. Rational Number 0.55 Irrational number 0.5477477747.... 8. 5 Hint: HCF (80, 455, 60) 5. 9. (i) 0.05. (ii) 5.85. 0. We know that any positive integer is either of the form q, q + or q + for some integer q. Now, three cases arise. Case I. When p q, p + q + and p + 4 q + 4 Here, p q is exactly divisible by p + q + leaves as remainder when it is divided by p + 4 q + 4 or (q + ) + leaves as remainder when it is divided by. Case II. When p q +, p + q + and p + 4 q + 5 Here, p q + leaves as remainder when it is divided by p + q + or (q + ) is exactly divisible by p + 4 q + 5 or (q + ) + leaves as remainder when it is divided by. 0 M A T H E M A T I C S X

Case III. When p q +, p + q + 4 and p + 4 q + 6 Here, p q + leaves as remainder when it is divided by. p + q + 4 or (q + ) + leaves as remainder when it is divided by p + 4 q + 6 or (q + ) is exactly divisible by. Hence, in all the cases, one and only one number out of p, p + and p + 4 is divisible by, where p is any positive integer. OR Any positive odd integer is type of q + where q is a whole number. (q +) 4q + 4q + 4q (q +)+...(i) Now, q(q + ) is either 0 or even So it is m, where m is some number. From (i) (q +) 8m +.. Since, height of each stack is the same, therefore, the number of books in each stack is equal to the HCF of 96, 40 and 6. Let us find their HCF 96 4 40 4 5 6 4 7 So, HCF 4 48. Now, number of stacks of English books 96 48 Number of stacks of Hindi books 40 48 5 Number of stacks of Mathematics books 6 48 7. ASSESSMENT SHEET. (D) The denominator of each fraction in the options (A), (B) and (C) can be expressed in the form n 5 m, where m, n being nonnegative integer.. (A) Let x be any positive integer then it is of the form q or q + or q +. So, x can be written in the form m or m +.. HCF LCM Product of the two numbers 40 5 p 50 6600 p 50 6600 650. 40 5 4. No; because HCF must divide LCM and here HCF 8 which doesn't divide LCM which is 80. 5. True, If the number n ends with the digit 0, then its prime factorisation contains the primes and 5. But by the Fundamental Theorem of Arithmetic, there is no prime other than in the factorisation of n. 6. The required number would be the HCF of 967 7 960 and 060 048. Let us find the HCF of 960 and 048 by using Euclid s algorithm. Since 048 > 960 048 960 + 8 960 8 7 + 64 8 64 + 0 Since the remainder becomes zero and the divisor at this stage is 64, the HCF of 960 and 048 is 64. Hence, the required number is 64. 7. Clearly, 456 9 and 60 5 HCF 4 Hence LCM 456 60 6840. 4 8. (i) Time taken by Ram to complete one cycle 80 seconds. Time taken by Shyam to complete one cycle 50 seconds. Consider LCM of 80 and 50. R E A L N U M B E R S

80 5 50 5 LCM of 80 and 50 5 4 5 9 900 seconds 900 5 minutes 60 They both will again meet after 5 minutes. (ii) Since they started at 6 a.m. and they will be meeting again after 5 minutes. The time will be 6:5 a.m. (iii) L.C.M. of real numbers. (iv) Since Ram and Shyam go for morning walk daily. So, it depicts their discipline and health consciousness. 9. Let a be any odd positive integer. Then, it is of the form 6p +, 6p + or 6p + 5. Here, three cases arise. Case I: When a 6p +, a 6p + p + 6p(6p + ) + 6q +, where q p(6p + ). Case II: When a 6p +, a 6p + 6p + 9 6p + 6p + 6 + 6(6p + 6p + ) + 6q +, where q 6p + 6p +. Case III: When a 6p + 5, a 6p + 60p + 5 6p + 60p + 4 + 6(6p + 0p + 4) + 6q +, where q 6p + 0p + 4. Hence, a is of the form 6q + or 6q +. ASSESSMENT SHEET. (D) 4587 50.6696. Clearly, the decimal expansion terminates after four decimal places.. (C) LCM (p, q) x y z.. HCF LCM Product of the two numbers. 9 LCM 06 657 LCM 06 657 8. 9 4. As given number can be written as 55 which is product of prime numbers: 5 5 0. Hence it is a composite number. 5. The maximum number out of, 5, 5, 5, 75 is 75. Therefore, the HCF of 55 and 000 is 75. 6. The denominator of 57 5000 or 54 is 0000. 0000 0000 0 4 4 5 4 57 Further, 5000 54 4 0 0.054. 7. Let x p + and y q + x + y (p + ) + (q + ) 4p + 4p + + 4q + 4q + 4(p + p + q + q) + S + T where S 4(p + p + q + q) and T S is divisible by 4 and so an even integer. T is not divisible by 4 but an even integer. Therefore, S + T is even, as sum of any two even number is even, and not divisible by 4. 8. Let us assume the contrary that 5 is a rational number. We can take coprime a and b (say) such that 5 a b ; b 0 b 5 a Square both the sides to get 5b a a is divisible by 5 a is divisible by 5 because if square of a number is divisible by a prime, then the number is divisible by the prime. Let us take some integer c such that a 5c Squaring both sides, we get a 5c M A T H E M A T I C S X

Substitute a 5c in 5b a to get 5b 5c b 5c b is divisible by 5 b is divisible by 5 Therefore, both a and b are divisible by 5. This contradicts the fact that a and b are coprime that is a and b have no common factor. Our assumption is false. So, we conclude that 5 is an irrational number. 9. (i) First we will find HCF of 7 and 94 by using Euclid s lemma: 94 7 + 0 7 0 + 0 0 0 + 6 0 6 7 + 0 The last divisor 6 HCF 6 The least number of students in which 7 apples can be distributed such that each student will get 6 apples 7 6. Similarly, the least number of students in which 94 oranges can be distributed such that each student will get 6 oranges 94 57. 6 Total least number of students required + 57 79. (ii) HCF of two real numbers. (iii) Harmony and love. CHAPTER TEST. (D) Since 844 7 7 So, is not prime factor of 844.. (C) As 8q is even and 6 is even, 8q + 6 is even..... 06 4 LCM 8. 8 4. Yes. 5 7+ ( 5 7 + ) 5 a composite number. 5. É 9 8 9 8 irrational. 6. No. Hint: Prime factors of 9 n will be type of n, i.e.,... Even no. of times. 7.... 0.565 565 00000 449 800 449 s 5... n 5 m 5 5 449 5 s 5 n 5,m. 8. 0 5 05 5 7 50 5 HCF 5 5 And LCM 5 7 8 5 7 400. 9. Hint: Let x, where x is rational. É x +7 6 6 x 9 x 6 6 9 x 6. 6 Since 6 is not a perfect square. So always irrational. It's a contradiction. 6 is R E A L N U M B E R S

0. We know that any positive integer is of the form q or q + or q +. Case I: n q n (q) 9 q 9m n +9m +, where m q. Case II: n q + n (q +) 7q + + 7q + 9q 9q (q + q + ) + 9m + n + 9m +, where m q(q +q + ). Case III: n q + n (q +) 7q + 8 + 54q + 6q n + 7q +54q +6q +9 9(q + 6q + 4q + ) 9m, where m q +6q + 4q +. Hence, n + can be expressed in the form 9m, 9m + or 9m +, for some integer m.. (i) We will find HCF of 96 and by using Euclid s lemma: 96 + 6 and 96 6 6 + 0 the last divisor 6 HCF 6 The minimum number of boxes required for apples 96 6 6 and the minimum number of boxes required for oranges 6 7 Total minimum number of boxes required 7 + 6. (ii) Concept used is HCF of two real numbers using Euclid s lemma. (iii) By distributing fruits in orphanage his kindness and concern towards the needful has been reflected. qq 4 M A T H E M A T I C S X

Chapter POLYNOMIALS WORKSHEET 0. (C) Since the given graph of y p(x) cuts x-axis at three points, so the number of zeroes of p(x) are.. Sum of zeroes b ( 5) 5 a c Product of zeroes a. Hint: BC B C BC. 4. Let one zero be α, then the other one will be B. α. B k 5 8. Solving α + β and α β, we get α, β Polynomial is x (α + β) x + αβ p(x)x x +. 9. According to the division algorithm, p(x) g(x) q(x) + r(x) x x + x + g(x) (x ) + ( x + 4) (As given in question) g(x) x x + x x To find g(x), we proceed as given below. P k 5. 5. Sum of zeroes (S) O L Y N O M I A 4 8 4 5 4 Product of zeroes (P) 4 Now, required polynomial will be x Sx + P, i.e., x + 5 4 x or 4 x + 5x. 6. Let f (x) x + ax + 5x + 0 If x + a is a factor of f (x), then f ( a) 0 Therefore, a a 5a + 0 0 a. L S 7. x 4x + x +6 Hint: If the zeroes are a, b and γ of a cubical polynomial, then the polynomial will be (xα) (x β) (x γ) (x )(x )(x +)x 4x + x +6. Thus, g(x) x x +. 0. ; Hint: 6x 7x 6x 9x +x x(x ) + (x ) (x ) (x +) x 0 gives x x + 0 gives x α + β + 7 6 b a α. β. c a. 5

. Let p(x) x 4 + x 4x 4x + 0 Given zeroes of p(x) are and (x ) (x + ) x 4 4 is a factor of p(x). We divide p(x) by x 4, x 4 x + x 0 x 4 + x 4 x 4 x+ 0 x 4x + x 0 x 4 x +0 4 x 4x + 0 x + 0 0 x + 0 + 0 p(x) (x 4) (x + x0) Other zeroes of p(x) are given by x + x 0 0 x + 6x 5x 0 0 x(x + 6) 5(x + 6) 0 (x 5) (x + 6) 0 x 5, 6 Hence, all the zeroes are,, 5 and 6. WORKSHEET 5. p Hint: () () + () p 0 8 + 6 p 0 p 0 p. 6. Let a and b be the two zeroes of f(x) ax + x + a Then, a + b a and ab a a According to the question, a a. 7. Let the third zero be a, then coefficient of x sum of the zeroes coefficient of x ++a 6 a Hence, the third zero is. 8. Let us divide 6x 4 +8x +7x +x +7 by x +4x +.. (D) Let us take option (D) p(x) (x )(x +x) x This is a linear polynomial.. Q p(x) x x + Sum of zeroes Product of zeroes.. Let a 5 and b 5, then the quadratic polynomial will be x (a + b)x + ab or x 5. 4. p(x) 4x 4x + 4x x x + x (x ) (x ) (x ) (x ) For zeroes, x 0 and x 0 x,. Clearly, the remainder is x +. Now, ax + b x + Comparing the coefficients of like powers of x both the sides, we obtain a, b. 9. We know: Dividend (Divisor Quotient) + Remainder 4x 8x +8x + g(x) (x )+x + g(x) (x ) 4x 8x +7x g(x) 4 x 8x 7 x x 6 M A T H E M A T I C S X

Now, x x x + 4 x 8 x + 7 x 4 x x + 6 x +7 x 6 x +x + 4 x 4 x + 0 Hence, g(x) x x +. 0. and Hint: x x x + (x ) (x ) For zeroes, x 0 and x 0 x, Now, sum of zeroes + And product of zeroes Coefficient of x Coefficient of x Constant term Coefficient of x.. (i) To find the number of sweets which was distributed among the slum children, we divide the total number of sweets by number of children Mr. Vinod has. Remainder thus obtained is the required number of sweets. x 4 x + x + 6x + 8 x 4 +x x x+ x4 4 x+ x + 6 x 6 x x + 6 x 4 x+ 8x + 8 x 0 x+ 8 x x+ 4 + x Hence, the number of sweets which was distributed amoung the slum children was x. (ii) Helping one another, fair division. WORKSHEET. (C) Sum of zeroes (5) 5 µ Product of zeroes 9. c. (A) Let the zeroes be α, β, γ. Then αβγ If γ, then αβ c...(i) Further, ( ) + a ( ) + b ( ) + c 0 + a b + c 0 c b a +...(ii) From equations (i) and (ii), we have αβ b a +.. Sum of zeroes 6 6 k k 6. 4. Let one zero be α, then the other one will be B. So, α. B 4a a 4 a 4a + 4 0 (a ) 0 a. 5. Given polynomial is: f(x)x px p c α+β p and α.βp c (α + ) (β + ) αβ + (α + β) + 4 p c + p + 4 (4 c). P O L Y N O M I A L S 7

6. λ 6 Hint: (α + β) (α β) + 4αβ. 7. x or ; f(x) x x Hint: x or, Sum of zeroes Product of zeroes p(x) x (α + β)x + αβ x x. 8. x x 47 4 Hint: f(x) {x (sum of zeroes) x + (product of zeroes)} 9. The number which to be subtracted is the remainder when 4x 4 +x 8x +x 7 is divided by x + x. To find the remainder, we proceed as following. x x +x 4 x4 + x 8 x + x 7 4 4 x + x 4x + 4x + x 7 4x x + 4 + + 5x Hence, 5x must be subtracted from 4x 4 +x 8x +x 7 so that it becomes exactly divisible by x + x. 0. g(x) x + x + Hint: p(x) g(x) q(x)+r(x) g(x) p( x) r( x) qx ( ) where, p(x) x + x + x + 5 q(x) x 5 and r(x) 9x + 0. 5. Since x and x 5 are zeroes of p(x) x 4 + 6x x 0x 5, so p(x) is 5 5 divisible by + ž Ÿ x Ÿ ž x, i.e., x 5. Here, other two zeroes of p(x) are the two zeroes of quotient x +6x + Put x +6x + 0 (x +) 0 x and x Hence, all the zeroes of p(x) are 5, and. WORKSHEET 5,. (B) Sum of zeroes 99 ve Product of zeroes 7 +ve If the sum of both zeroes is negative, then the zeroes would be either both negative or one negative and other one positive. If the product of both the zeroes is positive, then the zeroes would be either both positive or both negative. Consequently, we obtain that both the zeroes are negative.., Hint: Given polynomial can be written as: p(x) x +x Sum of zeroes b a Product of zeroes c a.. We know that the degree of the remainder is less than the degree of divisor or does't exist. 8 M A T H E M A T I C S X

Here, degree of the divisor is, therefore, the possible degree of the remainder according to the options can be any out of 0, and. 4. k 0. Hint: Substitute x in x + x + k 0. 5. Since a, b are the zeroes of x + px + q, then a + b p; ab q α+β Now, + p α β αβ q and α β αβ q So the polynomial having zeroes will be q(x) x x + žÿ B C žÿ B C p x + x + q q or q(x) qx + px +. 6. g(x) x + x + 7. Hint: Divide x + x 4 by x. 7. One example is: p(x) x x +. g(x) x x + 4 q(x) r(x) 0. 8., 7 7 Hint: For zeroes: x 0 B and C a a (a 4)(a +) For other zeroes, put a 4 0 and a + 0 a, 4 Thus, the other two zeroes are and 4. 0. g(x) x +. Hint: Applying division algorithm, we get x 4 + g(x) (x x + x )+ 4 x g(x) x x + x x+ x x + x x + x +.. b ac c Hint: B C B C B C É BC BC b ac. B C c OR Let us divide x 4 +x +8x +x + 8 by x +5. x 7 x p. 7 9. Since a is a zero of a a 0a + 4, therefore a a 0a + 4 is divisible by a. Further the obtained quotient will provide the other two zeroes. P O L Y N O M I A L S 9

Clearly, the remainder is x +. Now, px + q x + Comparing the coefficients of like powers of x both the sides, we get p, q. WORKSHEET 4. (C) Let each of two equal roots be α. Then α + α b a and α.α c a b a µ c a a(4ac b ) 0 But a 0 So, b 4ac 0 b 4ac This last equation holds iff a and c have same sign.. (A) a + b, ab (a b) (a + b) 4ab 9 4 4 a b ± a, b or a, b a+ 5, b+ or a+, b + 5. Hence, the required polynomial can be x 5 µ x + 5, i.e., x x + 5.. p(x) x (sum of zeroes) x + product of zeroes x ( + )x + x 5 x + 4. Let zeroes be a and b. a + b 6, ab 4 Using (a b) (a + b) 4ab, we get (a - b) 6 4 4 0 a b, ± 5 Thus, the difference of zeroes is ± 5. 5. B C B C C B BC É BC BC BC 5 6 6. 6. x (x + ) (x ) x or, both will satisfy with the given polynomial. We get, p + q + r + s + t 0...(i) and p q + r s + t 0...(ii) From (ii), p + r + t q + s (q + s) 0 q + s 0 [From (i)] p + r + t q + s 0. 7. No. Hint: Divide q(x) by g(x). If the remainder obtained is zero, then the g(x) is a factor of q(x) otherwise not. 8. a, b 7 Hint: Put remainder 0 and equate coefficient of x in the remainder and constant term with zero. 9. According to division algorithm, p(x)g(x) q(x) + r(x) (i) p(x) 6x +x +, g(x) q(x) x + x, r(x) (ii) p(x) 8x + 6x x + 7, g(x) x + q(x) 4x +, r(x) 5x +4 (iii) p(x) 9x + 6x + 5, g(x) x +, q(x) x, r(x) 5. 0. Given quadratic polynomial is 5 5 x + 0x + 8 5 5 5 x + 0x + 8 5 5 5 x + 0x + 0x + 8 5 5x( 5 4) 5x + 4 x + + 5 ( ) ( 5x + 5) ( 5x + 4) 0 M A T H E M A T I C S X

To find its zeroes, put 5x 5 5 x + 4 0. x 5 i.e., x 5 5 and x 4 5 and x 4 5 5 0 and So, sum of zeroes 5 4 5 6 5 5 5 5 And product of zeroes 5 4 5 8 5 5 5. Coefficient of x Also, sum of zeroes Coefficient of x 0 5 5 6 5 5 Constant term And product of zeroes Coefficient of x 8 5 5 5 8 5. OR q(x) x x + Hint: Let S P B B C C C C B µ µ B Hence verified. Required polynomial q(x) x Sx + P.. As and are the zeroes of the given quadratic polynomial, so x and x + will be the factors of that. Conse- quently, x x + x must be the factor of that. Let us divide x 4 0x +5x +5x by x., i.e., µ x x 0 x+8 x4 0 x +5 x +5 x 4 x x + 0 x+8 x+5x 0 x +5x + 8x 8x + 0 Now, x 4 0x +5x +5x x µ (x 0x +8) By splitting 0x, we factorise x 0x + 8 as (x 4) (x ). So, its zeroes are given by x 4 and x. Therefore, all zeroes of the given polynomial are,, and 4. WORKSHEET 5. (D) Let zeroes be a and b, then (a b) 44 a b ±...(i) a + b p...(ii) ab 45...(iii) Also, we have (a b) (a + b) 4ab 44 p 80 p ± 8.. c Hint: f(x) x px (p + c) (α + ) (β +) αβ +(α + β)+.. p r Hint: B CH BC CH HB BCH 4. Let the given linear polynomial be y ax + b...(i) This passes through points (, ), (, ) and,0 µ. P O L Y N O M I A L S

\ a + b...(ii) a + b...(iii) 0 a + b...(iv) Solving equations (ii) and (iii), we get a, b which satisfy to equation (iv). Consequently, using equation (i), we get y x Polynomial is p(x) x Since p(x) 0 if x x is zero of p(x). 5. Let us divide ax + bx c by x + bx + c by the long division method. x + bx + c ax ab ax + bx c ax + abx + acx abx +( b ac) x c abx abx abc + + + ( ab + b ac) x + abc c Put remainder 0 (ab + b ac)x + (abc c) 0 ab + b ac 0 and abc c 0 Consider abc c 0 (ab ) c 0 ab or c 0. Hence, ab. 6. Hint: Let f (x) x mx npx + np (x p) is a factor of p(x) f (x) 0 at x p. p p m p n 0 p [(p (m + n)] 0 p m + n since p 0. 7. x 4x + x + 6 Hint: The required cubic polynomial is given by (x )(x ) (x + ) or x 4x + x + 6 This is the required polynomial. 8.,, 4 Hint: α + β + γ 5 αβ + βγ + αγ αβγ 4 Let αβ γ α + β 7 (α β) α β ± α β or α β Solving α + β 7 and α β, we get α 4, β And solving α + β 7 and α β we get α, β 4. 9. f(x) would become exactly divisible by g(x) if the remainder is subtracted from f(x). Let us divide f(x) by g(x) to get the remainder. x + 6x + 8 x 4 +x x x 4 x + x + x4 4 x+x + 6 x 6 x x + 6 x 4 x + 8x + 8 x 0 x + 8 x x + 4 + x Hence, we should subtract x from f(x). 0. If ± are zeroes of p(x), then x ( + ) and x ( ) are factors of p(x). Consequently x ( + ) x ( ) { } { } i.e., (x ), i.e., x 4x + is factor of p(x). Further, x 4 x+ x x 5 x46 x6 x+8 x5 4 x 4 x + x + x 7 x +8 x 5 x+8 x x + + 5 x +40 x5 5 x +40 x5 + + 0 M A T H E M A T I C S X

Clearly x x 5 is a factor of p(x) (x 7)(x + 5) is a factor of p(x) x 7 and x + 5 are factors of p(x) x 7 0 and x + 5 0 give other zeroes of p(x) x7 and x 5 are other zeroes of p(x). Hence, 7 and 5 are required zeroes.. Hint: 4 4 α β α + β + β α α β {(α + β) αβ} α β. αβ OR Given polynomial is: f(x) pqx + (q pr)x qr pqx + (q pr)x qr pqx + q x prx qr qx(px + q) r(px + q) (px + q)(qx r) px + q 0 and qx r 0 provide the zeroes of f (x). So zeroes are q p and r q. Sum of zeroes q p + r q pr q pq Coefficient of x Coefficient of x q r Product of zeroes p q qr pq Constant term Coefficient of x. So, b 4ac 0 b 4ac This last equation holds iff a and c have same sign.. Required quadratic polynomial x (sum of zeroes)x + product of zeroes x x 5.. p(x) x ax (a +) At x, p(x) () a( ) (a +) + a a 0 q(x) ax x (a +) At x, q(x) a( ) ( ) (a +) a + a 0 Therefore, x + is the common factor of p(x) and q(x). 4. Correct, f (x) x p(x +) c x px (c + p) α + β p; αβ (c + p) Now, (α + ) (β + ) αβ + (α + β) + (c + p) + p + c p + p + c. 5. Let f(x) 6x + x 0x 4 As is a zero of f(x), (x ) is a factor of f(x). Let us divide f(x) by ( x ). ASSESSMENT SHEET. (C) Let each of two equal roots be α. Then α + α b a and α.α c a b a But a 0. c a a(4ac b ) 0 f (x) ( ) ( ) ( ) x É6x 7 x 4 x É6x x 4 x 4 x 4 x x É É P O L Y N O M I A L S

Hence, É x4 É x gives x or x Therefore, other two zeroes are. 6. p(y) y + 5 y 5 (y + 5 y 0) (y +4 5 y 5y 0) and [y(y + 5 ) 5(y + 5 )] (y + 5 )(y 5 ) y + 5 0 and y 5 0 give the required 5 zeroes, that are 5 and. 7. α and β are zeroes of f (x) x x Sum of zeroes α + β...(i) Product of zeroes αβ...(ii) (α+) + (β+) (α + β)+ () + [Using (i)] 4...(iii) And (α+) (β +) 4αβ + α + β + 4αβ + (α + β)+ 4 ( ) + () + [Using (i) and (ii)] 5...(iv) Now, required polynomial can be given by x {(α+) + (β + )}x +(α+)(β +) i.e., x 4x 5. [Using (iii) and (iv)] 8. Let us divide p(x) by x 5. x 5 x + 4 x + 5 4 6 x + 8x 5x + ax + b 6 x4 5x + 8 x+ 0 x+ ax + b 8 x 0x + 0 x + (0 + ax+b ) 0 x 5 + (0 + ax+b ) + 5 Here, remainder is (0 + a)x + b + 5. If the polynomial p(x) is exactly divisible by x 5, the remainder must be zero. (0 + a)x +(b + 5) 0 Comparing the coefficients of like powers of x between both the sides, we have 0 + a 0 and 5 + b 0 a 0 and b 5. ASSESSMENT SHEET 4. (C) Sum of zeroes Product of zeroes.. At x, p(x) 0, i.e., p() 0 a () (a ) 0 4a 6a + 6 0 a 5.. Sum of zeroes α + β 5 Product of zeroes αβ 4 Now, α + β αβ α+β αβ αβ 5 4 4 7. 4 4. Using division algorithm, we have g(x) (x ) x + 4 x x + x + 4 M A T H E M A T I C S X

g(x) x x x x Here, at x, x x +x 8 + 6 0 x x + x (x ) (x x +) g(x) ( x)( x x ) ( x ) g(x) x x +. 5. Given s and p The required polynomial is given by k[x sx + p] i.e., k µ x x, where k is any real number. Let us find zeroes of this polynomial. k(x x ) k (x x ) k ( x )( x +) x 0 and x + 0 provides the zeroes. Thus, the zeroes are α and β 4 Sum of zeroes α + β 5 4 + 4 5 5 4 Coefficient of x Coefficient of x Product of zeroes αβ. 4 4 Constant term Coefficient of x. Hence verified. 7. (i) Let y p(x) y x + x + 6 The table for some values of x and their corresponding values of y is given by x 0 y 0 4 6 6 4 0 Let us draw the graph of p(x) using this table. Hence zeroes. and are the required 6. Let f(x) 4 x 5 x 4 x 8 x x 4 x( x ) ( x ) ( x )(4 x ) To find zeroes of f (x), put x 0 and 4 x 0 x and x 4 P O L Y N O M I A L S 5

From the graph, it is clear that the zeroes of p(x) are and. (ii) Let y p(x) y x 4x The table for some values of x and their corresponding values of y is given by x 0 y 0 0 0 Let us draw the graph of p(x) by using this table. Since, x 4 + x + 8x + ax + b is divisible by x +, therefore remainder 0 i.e.,(a )x + (b 7) 0 or (a )x + (b 7) 0.x + 0 Equating the corresponding terms, we have a 0 and b 7 0 i.e., a and b 7 (ii) Common good, Social responsibility. CHAPTER TEST. (B) Let p(x) x 7 µ x Zeroes are given by 7 µ x 0 and x 0. Hence zeroes are 0 and 7. From the graph, it is clear that the zeroes of p(x) are, 0 and. 8. (i) First we divide x 4 + x + 8x +ax + b by x + as follows: x x7 4 x x x 8x axb x 4 x x 7x axb x x 7 x ( a) xb 7x 7 ( a) x( b7). Q α + β 5, αβ α + β + αβ.. p(x) x (α + β)x + αβ x + x. 4. Let p(x) x + 4x + 5x + 7 Now, p(x) g(x) x + (7 5x) p( x)(75 x) g(x) x x 4x 5x 7 7 5x x x + x + 5. 5. 6 5, 9 5 4 Coefficient of x Hint: α + β Coefficient of x Constant term αβ Coefficient of x 6. Hint: α β α + β 0 b a 0. 6 M A T H E M A T I C S X

7. f (x) ax + bx + cx + d g(x) ax + bx + c q(x) x r(x) d. 8. If α, β and γ are the zeroes of a cubic polynomial f(x), then f(x) x (α + β + γ) x + (αβ + βγ + γα) xαβγ Here, α + β + γ 4, αβ + βγ + γα and αβγ 6 f(x) x 4x + x + 6. 9. We have + ( x+ )( 4 x ) 4 x 5 x So, the value of 4 x 5 x is zero when, x + 0 or 4 x 0, i.e., when x or x 4. Therefore, the zeroes of 4 x + 5 x are and 4. Now, sum of zeroes + 4 5 4 Product of zeroes Coefficient of x Coefficient of x 4 µ µ t Constant term Coefficient of x. 0. (i) Let p(x) Total Relief Fund 4 g(x) Number of families who received Relief Fund q(x) Amount each family received r(x) Amount left after distribution When the polynomial p(x) is divided by a polynomial g(x) such that q(x) and r(x) are respectively the quotient and the remainder, the division algorithm is p(x) g(x). q(x) + r(x)...(i) According to the question, p(x) x + x + x + 5 q(x) x 5 r(x) 9x + 0 Substituting these values of p(x), q(x) and r(x) in the equation (i), we get x + x + x + 5 g(x). (x 5) + 9x + 0 (x 5) g(x) x + x + x + 5 9x 0 x + x 7x 5 x x 7x5 g(x) x 5 To find g(x), we proceed as following: x x x5 x x 7x5 x 5x 6x 7x5 6x 0x x 5 x 5 0 Thus, g(x) x + x +. (ii) Common good, Accountability, social responsibility.. Since and are zeroes. Therefore, x x+ will be a factor of p(x), i.e., x is a factor of p(x). P O L Y N O M I A L S 7

x x 5 x + 8 4 x 5 x + 7 x + 5 x 6 x 4 x + 5 x +8 x +5 x 6 5 x +5x + 8 x 6 8 x 6 + 0 Here, x 5x + 8 x 5x + 6 (x ) (x ) Other zeroes are given by x 0 and x 0. So, other zeroes are and. Hence all the zeroes are,, and. ÿqq 8 M A T H E M A T I C S X

Chapter PAIR OF LINEAR EQUATIONS IN TWO VARIABLES WORKSHEET 8. (B) Since (, a) lies on the line x y 5, therefore, (, a) must satisfy this equation. () (a) 5 a a.. k 6 Hint:. k 9. k 6 Hint: The condition of inconsistency of two equations a x + b y c and a x + b y c is a given by b c x a b c. 4. Adding the given equations, we get 80x + 80y 40 or x + y...(i) Subtracting given first equation from other one, we get 6x 6y 6 or x y...(ii) Solving equations (i) and (ii), we obtain x, y. 5. x, y Hint: Let x y u, v. x y Given equations become 0u + v 4 and 5u 5v. 6. False. Let us substitute c 40, The given equations become x y 8 or 5x 0y 40 Here, 5 0 8 40 The equations represent a pair of coincident lines. The equations have infinitely many solutions at c 40 and no solutions at c 40. For no value of c, the given pair has a unique solution. 7. The given equations are 4(x + y) 9 + 7y and x + y 4 or 8x + 5y 9 0 x + y 4 0 By cross-multiplication, we have x 0 + 8 y + 7 x y 5 6 5 x and y 5 Hence, x, y 5 is the solution of the given system of equations. 8. To draw a line, we need atleast two solutions of its corresponding equation. x +y 6; at x 0, y and x, y. So, two solutions of x + y 6 are: x 0 y x y ; at x 0, y 4 and at x 6, y 0 So, two solutions of x y are: x 0 6 y 4 0 Now, we draw the graph of given system of equations by using their corresponding solutions obtained in the above tables. P A I R O F L I N E A R E Q U A T I O N S... 9

From the graph, the two lines intersect the y-axis at (0, ) and (0, 4). 9. Let the fixed charges and change per km be ` x and ` y respectively. x + 0y 05...(i) x + 5y 55...(ii) Subtracting equation (i) from equation (ii), we get 5y 50 y 0...(iii) From equations (i) and (iii), we get x 5 Now, the fare for travelling a distance of 5 km x + 5y 5 + 5 0 ` 55. Fixed charge ` 5 Charge per km ` 0 Total charge for 5 km ` 55. WORKSHEET 9. (D) As y and y 7, both represent straight lines parallel to x-axis y and y 7 are parallel lines. Hence, the given pair of equations has no solution.. x 5y 5. (, k) lies on it. 5(k) 5 5(k) k 5.. Condition for parallel lines is 0 M A T H E M A T I C S a a b b k c c k 6. 4. The given lines to be coincident, if k I ( k ) k k II III Taking I and II, we have k 6 k ± 6....(i) Taking II and III, we have k k k k(k 6) 0 k 0 or 6... (ii) Using (i) and (ii), we obtain k 6. 5. x 5, y Hint: Adding the given equations, we get x + y...(i) Subtracting the given equations, we get x + y 7...(ii) 6. Yes. Applying the condition a a b c b c, we have 6 9 That is true. Therefore, the pair of equations is consistent with infinitely many solutions. X

7. x 55 5, y 65 04 Hint: The system: 9x 0y + 5 0 5x + 6y 60 0 By cross-multiplication, we have x y. 600 90 540 75 54 + 50 8. For equation x + y 0, x 0 y For equation x y 5 0, x y As the lines corresponding to the given equations intersect each other at (, ), the required solution is x, y. 9. Let the man's starting salary and fixed annually increment be x and y respectively. According to the question, x + 4 y 5000...(i) x + 0 y 8000...(ii) Equations (i) and (ii) form the required pair of linear equations. Let us solve this pair. Subtracting equation (i) from equation (ii), we get 6y 000 y 500 Substituting y 500 in equation (ii), we get x 000 Hence, starting salary was ` 000 and annual increment was ` 500. The value imbibe by the man are: consistency, hard work and sincerity WORKSHEET 0. (D) Line x a is parallel to y-axis and the line y b is parallel to x-axis. These lines intersect each other at (a, b).. As the lines are intersecting each other, a a.. x y 5 0 and 6x y k 0 have no solution These equations represent a pair of parallel lines. 6 5 k k 0. 4. No. For infinitely many solutions, the following condition must be satisfied. λ 6 7 4 But, here 6 7 4 as Hence, no value of λ provides the pair of infinitely many solutions. 5. The given system of equations can be written as ax + by (a b) 0 bx ay (a + b) 0 By cross-multiplication, x ba ( + b) a( a b) ( i) a b ( iii) y aa ( + b) + b( a b) ( ii) P A I R O F L I N E A R E Q U A T I O N S...

Taking (i) and (iii), we get ab b a + ab x a b Taking (ii) and (iii), we get a ab+ ab b y a b Hence x, y is the solution of the given system of equations. 6. x 6, y 4, m 0 Hint: Take x u and y v. 7. No; (6, 0), (4, 0) Hint: For x + y 6 For x + 9y x 0 y x 4 y 0 Let us draw the graph of lines using the tables obtained above. In the graph, lines are parallel. So, the pair of equations is not consistent. The lines intersect the xaxis at (4, 0) and (6, 0). 8. (i) Let l length of the rectangle b breadth of the rectangle According to question, (l + 7)(b ) lb...(i) (l 7)(b + 5) lb...(ii) From equation (i), lb + 7b l lb 7b l...(iii) From equation (ii), lb 7b + 5l 5 lb 7b + 5l 5...(iv) Adding equations (iii) and (iv), we get l 56 l 8 m Putting the value of l in equation (iii), we get b 5 m. l 8 m, b 5 m. (ii) Solution of system of linear equations in two variables. (iii) Love for environment and human beings. WORKSHEET. (D) Let unit's and ten's digit be x and y respectively. x + y 9...(i) 0y + x + 7 0x + y...(ii) Solving equations (i) and (ii), we have x 6, y Hence, the required number is 0 + 6, that is, 6.. Given equation is 5(x y) 5x 5y 0 Let a 0, b 0 and c 6 a b c For coincident; a b c a 5 b 5 Hence ; a 0 b 0 c and c 6 a b c a b c So required equation which can coincide is 0x 0y 6 0. p 6 Hint: 5 p 6 p 0 M A T H E M A T I C S X

Note: At p 6, the given system has both zero and non-zero solutions. 4. a 5, b Hint: According to the condition of infinitely many solutions, we reaches at a + b a b. 7 5. x, y Hint: Simplifying the given linear equations, we have 7 8 7 5, + 5 y x y x Now take x u, v; and solve. y 6. x 4 a b, y a +4 b 5a 5b x Hint: b ( a + b ) + b ( a + b ) y a( a+ b) + a( a+ b) at batb Take first and third terms as well as second and third terms and solve. 7. a 7, b Hint: For infinitely many solutions, 4 ( a 4 ) b ( a ) 5b Take a 4 and a b 5b. 8. Table for values of x and y as regarding equation x + y 5 0 is x 0 y 5 Similarly table for equation x y 5 0 is x 0 y 5 Let us draw the graph of lines using the tables obtained above. The lines intersect y-axis at (0, 5) and (0, 5). 9. (i) Total distance 00 km Let speed of train x km/h and speed of bus y km/h As speed distance time time distance speed According to question, and 00 x From equation (i), 60 x + 40 y + 00 y 4 hours...(i) 4 hours and 0 minutes....(ii) 5 x + 60 y...(iii) From equation (ii), 00 x + 00 y 4 x + 8 y 6 5 6 hours x + y...(iv) 4 We will solve equations (iii) and (iv) by elimination method. Applying (iii) 5 (iv), we get: P A I R O F L I N E A R E Q U A T I O N S...

5 x + 60 y 5 x + 0 y 5 4 0 y 5 4 0 9 y 4 0 s 4 y 0 8 80 km/h 9 Putting the value of y in equation (iv), we get, x + 80 4 x 4 40 4 6 40 4 s 40 4 s 40 x 60 km/h 60 (ii) Solution of system of linear equations in two variables. (iii) By opting for public transport it depicts that she is a responsible citizen, so her responsibility and rationality have been depicted here. WORKSHEET. (D) The condition to be coincident for lines ax + by + c 0 and dx + ey + f 0 is given by a d b e c f ae bd; bf ce. Note: Two lines are coincident if both the equations follow the condition of infinitely many solutions.. (A) For no solutions, k ( k ) k ± 6 k k If k 6 6 6 4 x (True) 6 6 6 If k 6 6 8 4 x 6 6 (True) Required value of k, can be 6 or 6.. Let the required equation be ax + by + c 0. a b c Then, 5 a b k (say) a k, b k, k 5 c any real number Then, kx ky + c 0 x y + c k 0 Putting k c, we have x y. 4. For infinite number of solutions, we have p+ q 7 ( p q) (4 p q) On solving, p+ q ( p q) 4 M A T H E M A T I C S and 7 ( p+ q) (4 p+ q ), we obtain p 5, q. 5. x, y Hint: Adding and subtracting the given two equations, we have x + y...(i) and x y...(ii) Now, solve equations (i) and (ii). 6. x a, y b Hint: Given system of linear equations may be written as bx + ay ab (a + b) 0 b x + a y a b 0 Solve these two equations by the method of cross-multiplication. 7. Let the two digits number be 0x + y. Since ten's digit exceeds twice the unit's digit by X

x y + x y 0...(i) Since the number obtained by interchanging the digits, i.e., 0y + x is 5 more than three times the sum of the digits.... 0y + x (x + y) + 5 x 7y + 5 0...(ii) On solving equations (i) and (ii), we obtain x 8 and y... 0x + y 8 Hence, the required two-digit number is 8. 8. Tables for equations x + y 0 and x y 0 are respectively and x 4 y x 0 4 y Let us draw the graph. From the graph, it is clear that the lines intersect each other at a point A(, ). So the solution is x, y. The line x + y 0 intersects the y-axis at B(0, ) and the line x y 0 intersects the y-axis at C (0, ). Draw the perpendicular AM from A on the y-axis to intersect it at M. Now, in Δ ABC, base BC + units, height AM units.... ar(δabc) base height 8 sq. units Hence, x, y ; area 8 sq. units. 9. Speed of boat 6 km/hr, Speed of stream km/hr Hint: Let the speed of boat in still water x km/h and the speed of stream y km/h 40 + 8 x y x + y...(i) Distance Using Time Speed ± 6 + 8 x y x + y...(ii) Put x y u, x + y v and solve further to find x and y. OR Let each boy receives ` x and the number of boys be y. Then sum of money which is distributed is ` xy. Had there been 0 boys more, each would have received a rupee less,... (y + 0) (x ) xy 0x y 0...(i) Had there been 5 boys fewer, each would have received ` more,... (y 5) (x + ) xy 5x y 5...(ii) Solving (i) and (ii), we get x 5 and y 40... xy 00 Hence, sum of money ` 00 And number of boys 40. P A I R O F L I N E A R E Q U A T I O N S... 5

WORKSHEET. (B) As the given equations are homogeneous so only solution will be x 0, y 0.. (C) x 4, y 9 Hint: Put x u, v and solve. y 9. Here, x 6 9 5... Lines are parallel. 4. The given equations have a unique solution a b l m am bl. 5. The given equation can be written as 6ax + 6by a + b...(i) and 6bx 6ay b a...(ii) Multiplying equation (i) by a and (ii) by b and adding the results, we have 6(a + b )x (a + b ) x Substituting x in equation (i), we have 6a + 6by a + b Þ 6by b Þ y Thus, the solution is x, y. 6. a 5, b Hint: Two linear equations a x + b y + c 0 and a x + b y + c 0 have infinite number of solutions if a b c a b c. 7. Given system of linear equations can be written as (a + b) x + (a b) y 0 (a b) x + (a + b) y 0 By cross-multiplication, x ( a b) + ( a+ b) ( i) y ( a + ) b + ( a ) b ( ii) ( a + b ) ( a + b ) ( a b ) ( a b ) ( iii) Taking (i) and (iii), Again taking (ii) and (iii), 5 b a x 0 ab a+0b y 0 ab 5 b a a+0b Thus, x, y is the 0 ab 0 ab solution of the given system of equations. 8. Speed of rowing 6 km/hr, Speed of current 4 km/hr Hint: (x + y) 0 [Time Speed Distance] (x y) 4 Where, x speed of rowing and, y speed of current. OR Let fare from A to B and from A to C be ` x and ` y respectively. According to the given conditions, x + y 795...(i) x + 5y 00...(ii) Solving eqn. (i) and (ii), we obtain x 75, y 5 Hence, fares from A to B is ` 75 and from A to C is ` 5. 9. Let us make the table for the values of x and corresponding values of y to the equation x + y 8 0 x 4 y 4 0 6 M A T H E M A T I C S X

Similarly, for the equation x y 0 x 4 y Let us draw the graph. From the graph, the lines intersect each other at the point A(, ). Therefore, the solution is x, y. The lines intersect the y-axis at B(0, 8) and C(0, ). To find the area of the shaded portion, that is, ΔABC, draw perpendicular AM from A on the y-axis to intersect it in M. Now, AM units and BC 8 + 9 units.... ar(δabc) BC AM 9 7 sq. units Hence, x, y ; area.5 sq.units. WORKSHEET 4. (A) In the case of no solution, 6 5 k 0. k. (D) x 80, y 0 Hint: x + y 40, x + 4y 60.. For unique solution, 4 p p 4. 4. True. According to the conditions of consistency, either 5 5 x or 5 5 Clearly, the first condition holds. Hence, the system of equations is consistent with a unique solution. 5. For infinitely many solutions, p+ q ( p q ) (5 p ) 4 4p + 4q 6p 6q and p q 5p + p 0q 0 and p q p 5, q. 6. x, y, Hint: Take u, v x y x y Given equation can be written as: u + v 4 4u + 4v and u v 8 4u 4v. 7. x, y Hint: Put x u and y v. P A I R O F L I N E A R E Q U A T I O N S... 7

8. Table for values of x and y corresponding to equation 4x 5y 0 0 is x 5 0 y 0 4 Similarly for the equation x +5y 5 0 x 5 0 y 0 Let us draw the graphs for the two equations. Distance travelled by the car that starts from A AC 5 x Similarly distance for other car BC 5 y... AC BC 5x 5y 5x 5y 00 x y 0...(i) Case II: When two cars travel in opposite directions. Let the cars meet at D Distance travelled by the car that starts from A AD x Similarly distance for other car BD y... AD + BD x + y x + y 00...(ii) Solving equations (i) and (ii), we get x 60 and y 40 Hence, speeds of two cars that start from places A and B are 60 km/h and 40 km/h respectively. As the graphs of the two lines intersect each other at the point A(5, 0), the required solution is x 5, y 0. The graphs intersect the y-axis at B (0, ) and C(0, 4). Therefore, the coordinates of vertices of the triangle ABC are A(5, 0), B(0, ) and C(0, 4). Hence the answer: x 5, y 0 and (5, 0), (0, ), (0, 4). 9. Let speeds of two cars that start from places A and B be x km/hr and y km/hr respectively. Case I: When two cars travel in same direction. Let the cars meet at C WORKSHEET 5. (B) x y 0...(i) x y...(ii) + x (Subtracting)... x. Further y x.. The given equations represent to be parallel lines if ( k ) k k. 8 M A T H E M A T I C S X

. m 4 m Hint: x. 4. For the point of intersection of any line with x-axis, put y 0... x + 7 (0) x So the required point is (, 0). 5. For inconsistency, k + 6 (k ) 4 k + 4 and (k + ) 8 k and k ( ± ) Hence, k. 6. Given system of equations can be written as x + y 8 0...(i) x y 0...(ii) Now, x Hence the system has unique solution. Now, by cross-multiplication on (i) and (ii), we get x y 6 6 4 + 8 4 x 6, y Thus, the solution of given system is x 6, y. 7. x 5, y Hint: Take u, v and solve. x+ y x y 8. Let Meena received x notes of ` 50 and y notes of ` 00 Since total number of notes is 5... x + y 5...(i) Since the value of both types of notes is ` 000.... 50x + 00y 000 x +y 40...(ii) Solving equations (i) and (ii), we get x 0, y 5 Hence, Meena received 0 notes of ` 50 and 5 notes of ` 00. OR Let the length and breadth of rectangle be x units and y units respectively. Then area of rectangle xy sq. units Case I. The length is increased and breadth is reduced by units.... (x + ) (y ) xy 8 xy 4 x + y xy 8 x y...(i) Case II. The length is reduced by unit and breadth increased by units. (x ) (y + ) xy + xy y + x xy + x y 5...(ii) Solving equations (i) and (ii), we get x and y Hence, the length of the rectangle is units and the breadth is units. 9. The given linear equations are 4x y 8 0...(i) and x y + 6 0...(ii) To draw the graphs of the equations (i) and (ii), we need atleast two solutions of each of the equations. These solutions are given below: x 0 y 4x 8 8 0 x 0 y x 6 0 Plot the points A(0, 8), B(, 0), C(0, ) and D(, 0) on graph paper and join them to form the lines AB and CD as shown in figure. P A I R O F L I N E A R E Q U A T I O N S... 9

4. False. a As... a 4, a b b 5 0, c c a b b c c They are parallel. 6 5. a, b 5 The graphs of these lines intersect each other at P(, 4). So, unique solution: x, y 4. Also, the graphs meet the x-axis at D(, 0) and B(, 0). Hence, the triangle formed by the lines and the x-axis is PBD with vertices P(, 4), B(, 0) and D(, 0). WORKSHEET 6. (C) For coincident lines, k 7 4 k 4.. (A) The given system of equations can be written as x + y 40, x + 4y 60 Solving this system, we obtain x 80, y 0.. Adding the given equations, we have x 0 x 0 Substituting x 0 in any of the given equations, we get y 0 Hence, the required solution is x 0, y 0. ( a + 5) 5 Hint:. b + 9 5 6. Put x u and v in given system of y equations. u + v 7 0...(i) u + v 7 0...(ii) By cross-multiplication, u v 7 + 7 + 4 u 4, v x 4, y Hence, x 4, y is the solution of the given system of equations. 7. x, y 5 and m Hint: x + y x y Substitute this value of y in x 4y 4 and solve for x. 8. The given system of linear equations is x y 5 0...(i) x + y 5 0...(ii) To draw the graph of equations (i) and (ii), we need atleast two solutions of each of the equations, which are given below: x 0 4 y x 5 5 x 0 y x +5 5 4 40 M A T H E M A T I C S X

Using these points, we are drawing the graphs of lines as shown below: From the graph, the lines intersect each other at the point P(, ). Therefore, the solution is x, y. The lines meet the y-axis at the points Q(0, 5) and R(0, 5). 9. Let the fixed charge and additional charge for each day be ` x and ` y respectively. Since Saritha paid ` 7 for a book kept for 7 days... x + 4y 7...(i) Also, Susy paid ` for the book kept for 5 days... x + y...(ii) Subtracting equation (ii) from (i), we get y 6 y Again substituting y in equation (ii), we get x 5 Hence, the fixed charge is ` 5 and additional charge for each day is `. OR Son's age 0 years, father's age 40 years. Hint: Let the present age of father and son be x and y years respectively.... x + 5 (y + 5) And x 5 7 (y 5). WORKSHEET 7. (A) A 70, B 5, C 0, D 7. Hint: In a cyclic quadrilateral ABCD, A + C 80 and B + D 80.. (C) x 0, y 0 Hint: Both lines are passing through the origin.. For infinite number of solutions, p + q p q 7 p + q 6 and p q 9 p 5, q. 4. False. Hint: As a + 5b 0. 5. False, x 4, y does not satisfy the second equation. 6. No solution Hint: x + y 7, 6x + 9y Here, x 7 Parallel lines. 6 9 7. The given system of linear equations can be written as px + qy (p q) 0 qx py (p + q) 0 To solve the system for x and y, using the method of cross-multiplication, we have x y qp ( + q) pp ( q) pp ( + q) + qp ( q) p q x p q y p q p q x, y. 8. The given system of equations can be written as x 4y 0...(i) 6x 8y + 5 0...(ii) To draw the graph, we need atleast two solutions for each of the equations (i) and (ii), which are respectively given below: P A I R O F L I N E A R E Q U A T I O N S... 4

x 7 y x 4 5 x y 6 x 5 6 8 Let us draw the graph by using these points. x, y 9 Hence, the required fraction is 9. OR ` 6000, ` 550 Hint: Let incomes of X and Y be 8x and 7x respectively; and expenditures of them be 9y and 6y respectively. 8x 9y 50...(i) 7x 6y 50...(ii) WORKSHEET 8. (C) The condition that the given system of equations has unique solution is represented by a a b b.. (A) Multiplying first equation by and the other one by and adding, we get.8x 0.9 x From the graph, it is clear that the lines are parallel. Hence, the given system of equations is inconsistent. 9. Let the fraction be x y On adding to each of numerator and denominator, the fraction becomes 6 5... x + 6 y + 5 5x + 5 6y + 6 5x 6y...(i) Further, on subtracting 5 from each of numerator and denominator, the fraction becomes... x y 5 5 x 0 y 5 x y 5...(ii) Solving equations (i) and (ii), we get Substituting x in any of the given equations, we have y. x, y.. k 6 k k Hint:. k k 4. The condition that the given system of equations represents parallel lines is p + p 5 p + 5p 5 p. 5. True. The condition for parallel lines is 6 6 5 5 The condition holds. The lines are parallel. 4 M A T H E M A T I C S X

6. x a, y b Hint: Put x u and y v. 7. Given system of linear equations can be written as: (a b) x + (a + b) y (a ab b ) 0 (a + b) x + (a + b) y (a + b ) 0 By cross - multiplication, x ( a+ b)( a + b ) + ( a+ b) ( a ab b ) y ( )( + ) + ( + )( a b a b a b a ab b ) ( a b) ( a+ b) ( a+ b) ( a+ b) x y b ( a + b) 4ab b ( a + b ) Hence, the solution of given system of equations is x a + b, y ab a+ b. 8. To draw graph of the equation, we need atleast two solutions. Two solutions of the equation 4x +y 4 0 are mentioned in the following table: x 0 6 y 8 0 Similarly, two solutions of each of the equations x y 0 and y + 4 0 are respectively x 0 y 0 and x 0 4 y 4 4 Using the tables obtained above, let us draw the graph. Observing the graph, we get the lines meet each other pairwise in A(, 4), B(, 4) and C(9, 4). Hence, the vertices of the triangle ABC so obtained are A(, 4), B(, 4) and C(9, 4). Area of ΔABC base height 0 8 40 sq. units. 9. ` 600, ` 700 Hint: Let cost price of trouser be ` x and that of shirt ` y. Then 5 0 x+ y 50 ² 00 00» 0 5 x+ y 55 00 00 ¼ 5 x+ y 0400²» x+ 5 y 0700 ¼ OR 6 l of 50% and 4 l of 5%. Hint: Let x litres of 50% acid and y litres of 5% acid should be mixed. P A I R O F L I N E A R E Q U A T I O N S... 4

50 5 40 x + y ( x+ y) ² 00 00 00» x+ y0 ¼ x y²» x+ y 0 ¼ WORKSHEET 9. (C) x 9, y 6 Hint: x y and x +y 6.. (A) Solving x y 4 and x + y 5, we have x, y. Now, substituting these values of x and y in y x + m, we have + m m.. p 6 8 4 50 75 p p. 4. For inconsistency, α α α α α 6 and α α α α ± 6 and α 0 or α 6 α 6. 5. x b, y a Hint: a xb y ab(a +b), ax by ab Solving the equations, we get x b, y a. 6. x a 6b, y 5 5 Hint: 4bx + ay ab 0 bx + ay 8ab 0. 7. x + y 800, x + 8y 000; Not possible Hint: Let cost of chair be ` x and that of table be ` y. x +y 800, x + 8y 000. 8. Let the actual prices of tea-set and lemonset be ` x and ` y respectively According to the question, Case I. Selling price Cost price Profit 0.95x +.5y (x + y) 7 0.05x + 0.5y 7...(i) Case II. Selling price Cost price Profit.05x +.0y (x + y) 0.05x + 0.0y...(ii) Solving equations (i) and (ii), we get x 00, y 80 Hence, actual prices of tea-set and lemonset are ` 00 and ` 80 respectively. OR The person invested ` 500 at the rate of % per year and ` 700 at the rate of 0% per year. Hint: Let the person invested ` x at the rate of % per year and ` y at the rate of 0 % per year \ x 0y + 0 00 00 6x + 5y 6500...(i) and 0 + 00 00 5x + 6y 6700...(ii) Adding and subtracting (i) and (ii), we get x + y 00...(iii) x y 00...(iv) 9. Two solutions of 6y 5x + 0 are: x 4 y 0 5 Two solutions of y 5x 5 are x y 0 5 Now, we draw the graph of the system on the same coordinate axes. 44 M A T H E M A T I C S X

. For coincident lines a b c a b c a b a b 7 4a b (i) From the graph, we look that the two lines intersect each other at A(4, 5). (ii) The vertices of the triangle: A(4, 5); B(, 0); C(, 0). Height of ΔABC corresponding to the base BC, h 5 units and base, b BC 5 units Now, ar(δabc) b h 5 5.5 square units. ASSESSMENT SHEET 5. (C) Let us check option (C). x + 5 y ( ) + 5 () 5 7 a 5b 0. 4. False, because the given pair of equations has infinitely many solutions at k 40 and no solutions at k 40. 5. Given equations are y x.(x + y) x + y y x...(i) and (x + y) x y...(ii) Substituting the value of x + y from equation (i) in equation (ii), we get x y y x ( x y ) x y (x y) x y ± x y...(iii) or x y...(iv) Substituting x y and x y in equation (ii), we get respectively x + y...(v) and x + y...(vi) Solving equations (iii) and (v), we have x ; y. x y ( ) ( ) 9 + 8.. 4 x y 4 x y...(i) x y 8...(ii) + y 6 (Subtracting) From (i) x 4 x + y 0. P A I R O F L I N E Therefore, xy 4 Solving equations (iv) and (vi), we have x 4 ; y 4 Therefore, xy 6. Hence, xy 4 or 6. A R E Q U A T I O N S... 45

6. Given equations can be written as x a + y (a + b) 0 b x y + 0 a b Let us apply cross-multiplication method to solve these equations. x y a + + b b b b a a a ab ba bx b + a ay a + b ab a b Taking and bx b + a a b a b ay a + b ab a b ab ( a b) ( a b) a b x and y b ( a b) a ( a b) x a and y b. 7. Given equations of lines are: x + y + 4 0...(i) and 6x y + 4 0...(ii) To draw the graphs of lines (i) and (ii), we need atleast two solutions of each equation. For equation (i), two solutions are: x 0 y 4 5 For equation (ii), two solutions are: x 0 y 8 Let us draw the graphs of the lines (i) and (ii). From the graph it is clear that the two lines intersect each other at a point, P(, ), therefore, the pair of equations consistent. The solution is x, y. 8. Let the cost price of the saree and the list price of the sweater be ` x and ` y respectively. Now two cases arise. Case I. 8 Sale price of the saree x + x 00 08 00 x Sale price of the sweater y y 0 00 90 00 y 08 00 x + 90 00 y 008 08x + 90y 00800...(i) Case II. Sale price of the saree x + x 0 00 0 x 00 Sale price of the sweater y y 8 00 46 M A T H E M A T I C S X

9 00 y 0 00 x + 9 00 y 08 0x + 9y 0800...(ii) Adding equations (i) and (ii), we get 8x + 8y 0600...(iii) Subtracting equation (i) from (ii), we get x + y 000 or 8x + 8y 8000...(iv) (Multiplying by 09) Solving equations (iii) and (iv), we have x 600 and y 400 Hence, the cost price of the saree is ` 600 and the list price of the sweater is ` 400. ASSESSMENT SHEET 6. (B) For infinitely many solutions, 9 k k 6 k + 4 Taking 9 k k. 6. For no solution, 7 k k 4k + 7 k 4. k. x a and y k must satisfy both the given equations. Let us substitute these values of x and y in bx ay + ab 0 b(a) ak + ab 0 ak + ab 0 k b. 4. Yes, because a a b b c c, i.e., 6 9. 5. Given equations are: 4x +5y (p +q)x +(p +8q)y q p + Here, a 4, b 5, c, a p + q, b p + 8q, c q + p For infinitely many solutions, a b c a b c Taking 4 p + q 5 p +8q q + p 4 5 p + q p + 8q 0p + 0q 4p + q 6p q p q...(i) Also, taking 4 p + q q + p 4p 4q 8q +4p 4 q p Substitute q p in equation p q to get p 9. Hence, p 9, q 9. 6. Given system of equations is 4x +67y 4...(i) 67x + 4y 4...(ii) Adding (i) and (ii); and subtracting (i) from (ii), we get respectively 0x + 0y 0 and 4x 4y 48 i.e. x + y 0...(iii) and x y...(iv) Adding equations (iii) and (iv); and subtracting equation (iv) from (iii), we get respectively x and y. 7. One of the given equations is x + y 4...(i) Here, at x 0, y 4 and at x, y 0 P A I R O F L I N E A R E Q U A T I O N S... 47

Two solutions of equation(i) are given in the following table: x 0 y 4 0 Another given equation is x y 4...(ii) Two solutions of equation (ii) are given by the following table: x 0 y 4 0 Let us draw the graph of the two equations (i) and (ii) by using their corresponding tables. (4 + 4) 8 square units. Thus, area of the triangle is 8 square units. 8. Let Aman had a total of x oranges; and he made lot A of p oranges and lot B of remaining x p oranges. There are two cases now. Case I. Selling price of lot A ` p Selling price of lot B ` (xp) p + x p 400 x p 00...(i) Case II. Selling price of lot A ` p Selling price of lot B ` 4 5 (x p) From the graph, vertices of the triangle ABC are A(0, 4), B(0, 4) and C(, 0). ar(δabc) base height AB OC p + 4 (x p) 460 5 4x + p 00 Add equations (i) and (ii) to get 48 M A T H E M A T I C S...(ii) 7x 500 x 500 Hence Aman had a total number of 500 oranges. CHAPTER TEST. (C) A pair of linear equations is said to be consistent, if the lines either intersect each other at a point or coincide.. (C) 6, 6 Hint: Let the son's age x, And father's age y y 6x and y + 4 4(x + 4) Solve yourself.. The lines are coincident 6 k 8 6 k. X

4. Yes. For consistency, either a 4a b b or a 4a b b a a Here only the relation a 4a b b a a, The pair is consistent. i.e., holds. 5. x + 47y 0 47x + y 6 68x + 68y 7 (Adding) x + y 4...(i) Subtracting the given equations from one another, we get 6x + 6y 5 x y...(ii) Solve equations (i) and (ii) to get x, y. 6. We are given xy x+ y...(i) and xy x y 0...(ii) Taking equation (i), xy x+ y x + y 4xy...(iii) Now, taking equation (ii), xy x y 0 6x + y 0xy...(iv) Multiplying equation (iii) by and adding its result to (iv), we get 9y 8xy x Putting x in equation (iv), we get + y 5y y Thus, x and y. 7. The given system of equations will have infinite number of solutions if a b a+ b a+ b a b a+ b and a+ b a+ b a + b a b and a + b 4 a + b a + b a + b and a + b 4 b and a Hence, the given system of equations will have infinite number of solutions, if a, b. 8. (i) Let fixed charge ` x and charges for a distance of km ` y Now, According to question, x + y 89...(i) x + 0y 45...(ii) We will solve equations (i) and (ii) by elimination method. Subtract equation (ii) from equation (i): 8y 56 y 56 8 7 Putting value of y in equation (i), we get x + (7) 89 x 89 84 5 x 5; y 7 For a journey of 0 km charge paid x + 0y 5 + 0(7) 5 + 0 ` 5. (ii) Solution of pair of linear equations in two variables. (iii) Love towards environment. 9. To draw the graph of a line, we are required atleast two solutions of its corresponding equation. At x 0, 5x y 5 gives y 5 P A I R O F L I N E A R E Q U A T I O N S... 49

At x, 5x y 5 gives y 0 Thus, two solutions of 5x y 5 are given in the following table: x 0 Similarly, we can find the solution of each remaining equation as given in the following tables: x + y : y 5 0 x 9 y 4 0 6x + y 7: x 4 y 5 7 Now, we will draw the graphs of the three lines on the same coordinate axes. From the graph, it is clear that the lines form a triangle ABC with vertices A(, 0), B(, ) and C(, 5) qq 50 M A T H E M A T I C S X

Chapter 4 TRIANGLES WORKSHEET. (D) Observing the figure, we obtain A R, B Q, C P ΔABC ~ ΔRQP.. Q ΔABC ~ ΔDEF AB DE BC EF AC DF ar Δ ar Δ ( ABC) ( DEF) QR QS QT PQ [Using (i)] and (Common) ΔPQS ~ ΔTQR (SAS criterion) Hence proved. 6. : 4 Hint: ar(δdef) 4 ar(δabc) A Taking BC EF ar Δ ar Δ BC (5.4) 64 ( ABC) ( DEF) ar( Δ DEF) ar( Δ ABC) 4. 7. m Hint: Distance between tops AD B D E F C BC BC. cm. 64 5.4 5.4. Yes. Here, 6 4 + 0 676 AC AB + BC ΔABC is a right triangle. 4. In ΔABC, LM CB AM AL (i) AB AC (Basic Proportionality Theorem) Similarly in ΔADC, AN AL (ii) AD AC Comparing equations (i) and (ii), we have AM AN. Hence proved. AB AD 5. As PQ PR (i) QR QS QT PR (Given) AD AE + DE (5) () m. 8. Hint: Use Pythagoras Theorem. 9. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio. Proof: ABC is a given triangle in which DE BC. DE intersects AB and AC at D and E respectively. We have to prove AD BD AE CE Let us draw EM AB and DN AC. Join BE and CD. T R I A N G L E S 5

Now, ar(δade) base height. DC BC + BD BC + AB AD EM...(i) BC + 4 (AC BC ) Also, ar(δade) ar(δbde) AE DN...(ii) BD EM...(iii) 9 + (5 9) 9 + 4 4 DC cm. ar(δcde) CE DN...(iv) Dividing equation (i) by equation (iii) and equation (ii) by equation (iv), we have and ar ( ΔADE) ar( ΔBDE) ar( ΔADE) ar( ΔCDE) AD BD AE CE...(v)...(vi) But ar(δbde) ar(δcde)...(vii) (Triangles are on the same base DE and between the same parallels BC and DE) Comparing equations (v), (vi) and (vii), we have AD BD AE CE. nd Part: As B C AB AC AD + DB AE + EC AD AE (... BD EC) AD DB AE EC By converse of Basic Proportionality Theorem, DE BC. Hence proved. WORKSHEET 4. (A) ΔBAC ~ ΔADC BC AC AC DC y 6 4 y 8 cm.. x 8cm Hint: As DE BC AD DB AE EC x x x+ 5. x 4. DE BC and DB is transversal EDA ABC (Alternate interior angles) Similarly, AED ACB Consequently, ΔADE ~ ACB (AA similarity) AD AB ar Δ ar Δ AD 9AD ar Δ ( ADE) ( ABC) ( ADE) 5 ar(δade) 7cm. 5. No. Here, And Q DP PE 5 0 DQ QF 6 8 DP PE DQ QF Therefore, PQ is not parallel to EF. 6. Hint: Use Basic Proportionality Theorem. 5 M A T H E M A T I C S X

7. As AB BC AC AD BC BD BC Let us construct a ΔPQR with Q 90 such that PQ A B and QR B C...(ii) Using Pythagoras Theorem AB AD + BD A AD AB BC AB B C D 4 4AD AB. Hence proved OR Let ABCD be a rhombus Since, diagonals of a rhombus bisect each other at right angles, AO CO, BO DO, AOD DOC COB BOA 90 Now, in ΔAOD AD AO + OD...(i) Similarly, DC DO + OC...(ii) CB CO + BO...(iii) and BA BO + AO...(iv) Adding equations (i), (ii), (iii) and (iv), we have AD + DC + CB + BA (DO + CO + BO + AO ) BD AC BD CA + + + 4 4 4 4 BD + CA. Hence proved 8. Hint: BD DE EC BC Use Pythagoras Theorem. 9. Statement: In a triangle, if square of the largest side is equal to the sum of the squares of the other two sides, then the angle opposite to the largest side is a right angle. Proof: We are given a triangle ABC with A C A B + B C...(i) We have to prove that B 90 In ΔPQR, PR PQ + QR (Pythagoras Theorem) A B + B C...(iii) [From (ii)] But A C A B + B C...(iv) [From (i)] From equations (iii) and (iv), we have PR A C PR A C...(v) Now, in ΔA B C and ΔPQR, A B PQ [From (ii)] B C QR [From (ii)] A C PR [From (v)] Therefore, ΔA B C ΔPQR (SSS congruence rule) B Q (CPCT) But Q 90... B 90. Hence proved. nd Part In ΔADC, D 90... AC AD + DC 6 + 8 6 + 64 00 In ΔABC, AB + AC 4 + 00 676 and BC 6 676 Clearly, BC AB + AC Hence, by converse of Pythagoras Theorem, in ΔABC, BAC 90 ΔABC is a right triangle. T R I A N G L E S 5

. (B) WORKSHEET 5 ar( ΔADE) DE ar( ΔABC) BC BC ar(δade) BC. ΔOAB ~ ΔOCD OA OC OB OD OB 4 6 cm. 8 6 cm.. In ΔABC, to make DE AB, we have to take AD DC BE EC x + 9 x + x + 4 x + 9x x + 4x + 9x + 6x x. 4. No, ' ΔFED ~ ΔSTU Corresponding sides of the similar triangles are in equal ratio. DE TU EF ST DE ST EF TU. 5. AB PQ AP AO BQ BO AC PR AP AO CR CO x (i) (ii) From (i) and (ii), BQ BO CR CO BC QR. (By converse of BPT) 6. :. Hint: Let AB BC a... AC a... ar( ΔABE) ar( ΔACD) AB AC. 7. In Δ ABC and Δ AMP, A A (Common) ABC AMP 90 (i) ΔABC ~ ΔAMP, (AA criterion) (ii) CA PA BC MP. (... Corresponding sides of similar triangles are proportional.) 8. Hint: ar(δ AXY) ar(bxyc).ar(δaxy) ar(bxyc) + ar(δ AXY) ar(δabc) A B ar( ABC) ar( ΔAXY) 54 M A T H E M A T I C S As ar ( ΔABC) ar ( Δ AXY) ΔABC ~ Δ AXY AB AX Δ X AB AX BX AB 9. Hint: Prove converse of Pythagoras Theorem.. (A) In triangle ABC, WORKSHEET 6 AD DC 6 7 BE 8 4 EC 4 7 AD DC BE DE AB. EC. (B) ΔABD ~ ΔBCD AB BC BD CD 5.4 BC.6 5. 5.4 5. BC.6 7.8 cm. A D Y B C E. X C

ar ( ΔDEF) EF. ar ( ΔABC) BC ar(δdef) 54 6 9 96 cm. 4. In Δ ABC and Δ ADE, BAC DAE (Common angle) ACB AED (Each 90 ) ΔABC ~ Δ ADE (AA criterion) AQ BP AQ 9 AO BO 0 6 AQ 0 9 6 AQ 5 cm. OR Let the height of the tower be h metres AB AC + BC 5 + 44 cm Now, AB AD BC DE AC AE DE 5 AE DE 6 5 cm and AE cm. 5. No. Ratio of areas of two similar triangles Square of ratio of their corresponding altitudes 9 5 5 6 5. Hence, it is not correct to say that ratio of areas of the triangles is 6 5. 6. AE AC + EC (i) BD DC + BC (ii) Adding (i) and (ii), we get AE + BD AC + EC + DC + BC (AC + BC ) + (EC + DC ) AE + BD AB + DE. Hence proved. 7. In ΔAQO and ΔBPO, QAO PBO (Each 90 ) AOQ BOP (Vertical opposite angles) So, by AA rule of similarity, ΔAQO ~ ΔBPO ΔABC ~ ΔPQR. AB PQ BC QR h 8 40 40 h 60 metres. 8 8. Hint: As ΔAOB ~ ΔCOD ar( ΔAOB) AB ar( ΔCOD) CD É CD CD 4. 9. Hint: Prove Pythagoras Theorem. For nd Part: AB AD + BD (i) Also AC AD + CD (ii) From (i) and (ii), AB AC BD CD AB + CD AC + BD. Hence proved. WORKSHEET 7. (A) M 80 ( L + N) (ASP) 80 (50 + 60 ) 70 Q ΔLMN ~ ΔPQR M Q Q 70. T R I A N G L E S 55

. (C) In ΔKMN, as PQ MN, KP PM KQ QN KP PM KQ KN KQ KN KQ PM KP 0.4 KQ 4 0.4 KQ + 4 7 4 KQ 0.4 4 7 KQ 4.8 cm.. ΔABC ~ ΔDEF. 4. Q ΔABC ~ ΔPQR ar( Δ PRQ) QR ar( ΔBCA) BC 9 9 :. 5. True Hint: Use Basic Proportionality Theorem 6. Hint: Use: 4. In right ΔAOE, AO AE + OE...(ii) From equations (i) and (ii), we have AF + OF AE + OE...(iii) Similarly, we can find out that BD + OD BF + OF...(iv) and CE + OE CD + OD...(v) Adding equations (iii), (iv) and (v), we arrive AF +BD +CE AE +BF +CD. Hence the result. 8. ΔABC ~ ΔPQR AB PQ BC and B Q QR 7. Draw EOF AD A E B O OB EO + EB OD OF + DF OB + OD EO + EB + OF + DF EO + CF + OF + AE ['DF AE, EB CF] (EO + AE ) + (CF + OF ) OB + OD OA + OC. OR Join OA, OB and OC In right ΔAOF, AO AF + OF...(i) D F C BC AB PQ QR and B Q AB PQ BP and B Q QM (Q BD DC and QM MR) ΔABD ~ ΔPQM AB PQ AD PM. Hence proved. 9. Let the two given triangles be ABC and PQR such that ΔABC ~ ΔPQR AB PQ BC...(i) QR 56 M A T H E M A T I C S X

Further, in the question, ar( % ABC) BC ar( % DEF) EF Let us draw perpendiculars AD and PM from A and P to BC and QR respectively. ADB PMQ 90...(ii) Now, in ΔABD and ΔPQM, B Q (QΔABC~ΔPQR) ADB PMQ [From (ii)] So, by AA rule of similarity, we have ΔABD ~ ΔPQM AB PQ AD...(iii) PM From equations (i) and (iii), we get BC QR AD PM...(iv) ar ( ΔABC) BC AD Now, ar( ΔPQR) QR PM BC BC QR QR [Using (iv)] BC...(v) QR Similarly, we can prove that ar ( ΔABC) ar( ΔPQR) AB PQ...(vi) ar( % ABC) and AC...(vii) ar( % PQR) PR From equations (v), (vi) and (vii), we obtain ar( % ABC) ar( % PQR) AB PQ BC QR AC PR. Hence, the theorem. BC 64 BC 5.4 5.4 64 5.4 5.4 8 WORKSHEET 8 5.4. cm.. (B) Ratio of areas of two similar triangles Ratio of squares of their corresponding sides. 4 : 9 6 : 8.. (C) M Q 5 (Corresponding angles) PQ ML QR MN (Ratio of corresponding sides) MN 5 6. In ΔABC, DE BC AB AC AD AE 4. Yes. Here, AB 5 7 AP AQ 5 7.5 BP BR 4 6 AP AQ BP BR 0 cm. 5 cm. Hence, due to the converse of Basic Proportionality Theorem, AB QR. T R I A N G L E S 57

5. Q DB BC and AC BC DB AC Now, DBA BAC (Alternate angles) And, DEB ACB (Each 90 ) ΔBDE ~ ΔABC (AA similarity) 6. AX AB BE AC DE BC BE DE AC BC. (Corresponding sides) Hint: See Worksheet 5, Sol. 8. 7. Hint: AM AB; AL AC Use Pythagoras Theorem. Hence proved. 8. Hint: Join AC and use Basic Proportionality Theorem. D E A X 9. As D and F are mid-points of AB and AC respectively. DF P BC and DF BC. DF BC Also, as ΔADF ~ ΔABC ar (% ADF) (DF)...(i) ar( % ABC) (BC) 4 As ar(δadf) ar(δbde) ar(δcfe) ar(δdef) ar (% DEF) (i) ar (% ABC) 4 A D F B F C (ii) Yes, as DE P AC and DE AC (Using mid-point theorem) But as AB AC BC DE BC. (iii) Concept of similarity of two triangles and mid-point theorem. (iv) His ability to think rationally and taking unbiased decision. WORKSHEET 9. (A) Let the length of shadow is x metres. BE. 4 4.8 m ΔABC ~ ΔDEC AB DE BC EC.6 0.9 4.8 + x x.6x 4. + 0.9x. x 4..6 m..7. (B) Here, (a) + ( ) a a + a. 4a (a) According to the converse of Pythagoras Theorem, the angle opposite to longest side is of measure 90. AD DB AB AD AD AB AD AB AD 5 DE BC ΔABC ~ ΔADE BC DE AB AD 5. 4. No. In ΔPQD and ΔRPD, PDQ PDR 90 B E C 58 M A T H E M A T I C S X

But neither PQD RPD nor PQD PRD Therefore, ΔPQD is not similar to ΔRPD. 5. Hint: ΔBAC ~ ΔADC BA AD AC DC BC AC CA BC CD. 6. AD DB 5 4 AD AB AD 5 4 5AB 5AD 4AD AD AB 5 9...(i) As DE BC, ΔADE ~ ΔABC DE BC AD AB 5...(ii) 9 [Using (i)] Q DE BC and DC is a transversal EDC ~ BCD (Alternate interior angles) i.e., EDF BCF...(iii) Similarly, DEF CBF... (iv) From equations (iii) and (iv), we have ΔDEF ~ ΔCBF (AA similarity) ar ( ΔDEF) DE 5 ar ( ΔCFB) BC 8. [Using equation (ii)] 7. A B D A D C ar( ΔABC) AL ar( ΔDEF) DM AL DM 4 5 Ratio of corresponding heights is 4 : 5. OR Proof: Draw a ray DZ parallel to the ray XY. In ΔADZ, XY DZ AY AX YZ XD YZ AY In ΔYBC, BY DZ YZ BD ZC DC YZ ZC From (i) and (ii), ZC AY Now, AC AY + YZ + ZC... (i) (... BD DC) AY + AY + AY 8 AY... (ii)... (iii) 4AY Therefore, AC : AY 4 :. Hence proved. 8. 5 cm Hint: BD BC; EB AB Use Pythagoras Theorem. 9. Statement: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Proof: We are given a right triangle ABC right angled at B. T R B I A L AB AC; DE DF AB AC DE DF C N G L E S E M F AB DE AC also A D DF ΔABC ~ ΔDEF We need to prove that AC AB + BC Let us draw BD AC. Now, ΔADB ~ ΔABC 59

AD So, AB AB AC or AD. AC AB (Sides are proportional) Also, ΔBDC ~ ΔABC CD So, BC BC AC or CD. AC BC Adding equations (i) and (ii), we get AD. AC + CD. AC AB + BC AC(AD + CD) AB + BC AC. AC AB + BC...(i)...(ii) AC AB + BC. Hence proved. nd Part: AB AD + BD AD + (CD) AD + 9CD AD + CD + 8CD AC + 8CD µ AC + 8 BC 4 ' CD BC ª 4 ¹ AB AC + BC. Hence proved.. (A) WORKSHEET 40 AB PQ BC QR AC PR (Q Δ ABC ~ Δ PQR) 9 7 x 0 y x 7 9 and y 9 0. Required ratio 4 5. 6 5 4 5 4 : 5.. 7 m Hint: P S Use Pythagoras Theorem and find OP. N 8m W 5 m O 4. Hint: Let AB c AC b BC a... a b + c Also, ar(δabe) 4 c ar(δbcf) 4 a ar(δacd) 4 b. 5. See Worksheet- 6, Sol. 6. 6. Let ABCD be a quadrilateral of which diagonals intersect each other at O. It is given that AO CO BO DO AO or BO CO...(i) DO In ΔAOB and ΔCOD, AOB COD (Vertically opposite angles) AO BO CO [From (i)] DO Hence, by SAS rule of similarity, we obtain ΔAOB ~ ΔCOD BAO DCO i.e. BAC DCA These are alternate angles. Therefore, AB CD and AC is transversal ABCD is a trapezium. Hence proved E 60 M A T H E M A T I C S X

OR Hint: As BAC EFG ; ABC FEG and ACB FGE ACB FGE ACD FGH and DCB HGE ΔDCA ~ ΔHGF Similarly ΔDCB ~ ΔHGE. 7. Hint: B A Prove that ΔAEB ~ ΔDEC. 8. See worksheet, Sol. 9 (st part). nd Part Join EF and join BD to intersect EF at O. E D C DE + EF + DF AB + BC + CA (iv) 4 Perimeter of ΔDEF ++.5 [Taking (ii) and (iv)] Perimeter of ΔDEF 5 cm.. DE BC x x x + x x 4 x x x 4.. ΔKNP ~ ΔKML x a c x ac b+ c b+ c. 4. Hint: A B L P Q AB DC, and EF AB, AB DC EF In ΔABD, EO AB, DE AE DO...(viii) BO (Basic Proportionality Theorem) Similarly, in ΔBCD, DO BO CF...(ix) BF Using equations (viii) and (ix), we obtain the required result, i.e., AE ED BF FC. D Prove that ΔADL ~ ΔCPD. 5. Hint: AP PC AP AC C Similarly, BQ BC Use Pythagoras Theorem. 6. PQ BC AP PB AQ QC ΔAPQ ~ ΔABC ar( ΔABC) AB ar ( ΔAPQ) AP () 9 WORKSHEET 4 ' AB ª AP ¹. (B) DE AB EF BC DF AC (i) (ii) (iii) ar( ΔABC) 8 ar ( ΔAPQ) T R I A N G L E S 6

ar( BPQC) 8 ar (% APQ) ar( % APQ) ar ( BPQC) 8. Ratio of areas of ΔAPQ and trapezium BPQC is : 8. OR Let the given square be ABCD. Let us draw an equilateral triangle APB and another equilateral triangle AQC on the side AB and on the diagonal AC respectively. We need to prove ar(δapb) ar (ΔAQC) In right ΔABC, Now, ar(δapb) And ar(δaqc) AC AB + BC AB...(i) (Q AB BC) 4 AB...(ii) 4 AC ( AB 4 ) AB...(iii) Dividing equation (ii) by equation (iii), we obtain ar ( ΔAPB) ar ( ΔAQC) AB 4 AB ar(δapb) ar(δaqc). Hence proved. 7. Hint: Extend AD till E such that AD DE and similarly, PM MN Prove that ΔACE ~ ΔPRN...(i) But 5, (CPCT) 4 and 4 6... 5 6...(ii) Adding (i) and (ii), + 5 + 6 BAC QPR... ΔABC ~ ΔPQR. (By SAS) 8. Let us take two similar triangles ABC and PQR such that ΔABC ~ ΔPQR... AB. PQ BC QR CA...(i) RP We need to prove ar ( ΔABC) AB ar( ΔPQR) PQ BC QR CA RP Let us draw AM BC and PN QR. Q ΔABC ~ ΔPQR B Q...(ii) In ΔABM and ΔPQN, B Q [From (ii)] and M N (Each 90 ) ΔABM ~ ΔPQN (AA criterion) AB PQ AM PN...(iii) 6 M A T H E M A T I C S X

From equations (i) and (iii), we have Now, AM PN BC QR ar(δabc)...(iv) base height BC AM And ar(δpqr) QR PN Therefore, ar( ΔABC) BC AM ar( ΔPQR) QR PN BC...(v) QR [Using (iv)] From results (i) and (v), we arrive ar( ΔABC) ar( ΔPQR) AB PQ BC QR CA RP. Hence the result. Further, consider the question in the following figure. ABO CDO and BAO DCO (Alternate angles) ΔAOB ~ ΔCOD (AA rule) ar ( Δ AOB) AB ar ( ΔCOD) CD ar(δcod) 84 ' CD AB cm. WORKSHEET 4. (A) ΔOBC ~ ΔODA (AA criterion) OB OD OC OA BC DA BC DA 4 8 cm.. (D) Let the given areas be x and x. Required ratio x : x :.. x or 8 Hint: Use OD OB OC OA. 4. True. Geometrical figures which are equiangular i.e., if corresponding angles in two geometrical figure are same, are similar. 5. In right ΔADC, AD AC CD (CD) CD [Q AC BC CD] CD. 6. Hint: BMDN is a rectangle. DBMD ~ DDMC DN DM DM MC DM DN MC Also, DBND ~ DDNA. DM DN DN AN 7. Let BE x and EC 4x. In DBCD, GE DC DBGE ~ DBDC BE BC GE DC x x + 4x GE AB DN DM AN. ( DC AB) GE 6 AB...(i) 7 Similarly, DDGF ~ DDBA FG AB 4 7 FG 4 7 AB...(ii) Adding equations (i) and (ii), we get GE + FG 6 7 AB + 4 7 AB T R I A N G L E S 6

EF 0 7 AB 7 EF 0AB. Hence proved. 8. Hint: Let AB BC AC a Draw AE BC and BE EC a BD BC a Using Pythagoras Theorem AD AE + DE AE + (BE BD) AD AE + BE + BD.BE.BD AD AE + EC + a AC + a a a a 9 a a a 7a + 9 9 9AD 7AB. 9. See Worksheet 9, Sol. 9 (st part). Hint: nd Part: AC AD + DC AD + (BD) AD + 9BD AD + BD + 8BD AB + 8 BC µ 4 AC AB + BC. B D A E C. Hint: Draw AM BC and DN BC As ΔAOM ~ ΔDON ar( Δ ABC) ar( Δ DBC) s BC s AM s BC s DN AM AO DN OD. 4. Hint: Use concept of similarity. 5. Draw AP BC AB AP + BP AP + (BD + DP) AB AP + BD + DP + BD. DP AD + BD (BD + DP) AB AD BD CD. [... BP PC] Hence proved. 6. Hint: ab From figure, show x a + b. 7. In ΔMDE and ΔMCB, MDE MCB (Alternate angles) MD MC (M is mid-point of CD) DME CMB (Vertically opposite angles) WORKSHEET 4. (C) 94 cm Hint: Prove that ΔOBP ~ ΔOAQ.. 6 cm Hint: Use AA-similarity to prove ΔAOB ~ ΔCOD. ΔMDE ΔMCB, (ASA criterion) DE CB (CPCT) 64 M A T H E M A T I C S X

AE AD BC AE BC...(i) (Q BC AD) Now, in ΔLAE and ΔLCB, LAE LCB (Alternate angles) ALE CLB (Vertically opposite angles) ΔLAE ~ ΔLCB (AA criterion) AE BC LE (Corresponding sides) BL BC BC EL BL [Using equation (i)] EL BL. Hence proved. OR Hint: As AD is median so, AB + AC (AD + BD ) AB + AC «BC º AD» 4 ¼ (AB +AC ) 4AD + BC...(i) Similarly, (AB + BC ) 4BE + AC...(ii) (AC +BC ) 4CF + AB...(iii) Add (i), (ii) and (iii), (AB + AC + BC ) 4(AD +BE +CF ). 8. See Worksheet, Sol. 9 (st part). nd Part: Draw EM AB M is a point on CB EM AB Also in ΔBCD, DE EB CM MB From (i) and (ii), CE AE DE EB. WORKSHEET 44. (C) BE 4 a a 4 BE AB + BC + CA a + a + a a...(ii) 4 BE 4 BE.. (B) 60 cm Hint: Use Pythagoras Theorem.. x 4 Hint: Use Basic Proportionality Theorem. 4. Hint: In ΔACD and ΔABC, A A ADC ACB 90 ΔACD ~ ΔABC AC AB. AD (i) ΔBCD ~ ΔBAC BC BA. BD (ii) Applying (ii) (i) gives the result. 5. Let the given parallelogram be ABCD We need to prove that AC + BD AB + BC + CD + DA Let us draw perpendiculars DN on AB and CM on AB produced as shown in figure. In ΔABC, CE AE CM MB...(i) T R I A N G L E S 65

In ΔBMC and ΔAND, BC AD (Opposite sides of a gm ) BMC AND (Each 90 ) CM DN (Distance between same parallels) ΔBMC ΔAND (RHS criterion) BM AN...(i) (CPCT) In right triangle ACM, AC AM + CM (AB + BM) +BC BM AB + AB. BM + BM + BC BM AB + BC + AB. BM...(ii) In right triangle BDN, BD BN + DN (AB AN) + (AD AN ) AB AB. AN + AN + AD AN BD AB +DA AB. AN BD CD +DA AB. BM...(iii) [Using (i) and AB CD] Adding equations (ii) and (iii), we have AC + BD AB +BC +CD +DA. Hence proved. 6. Hint: AP QB RC Use Basic Proportionality Theorem. 7. (i) PQ PR + QR (6) (x) + {(x + 7)} 676 4x + 4(x + 49 + 4x) 676 4x + 4x + 96 + 56x 8x + 56x 480 0 x + 7x 60 0 x + x 5x 60 0 x(x + ) 5(x + ) 0 (x 5)(x + ) 0 x 5 or x (reject it) x5 PR 5 0 km QR (5 + 7) 4 km Before construction of the highway the distance travelled 0 + 4 4 km After construction of the highway the distance travelled 6 km Distance saved 4 6 8 km. (ii) Pythagoras theorem (iii) Yes: as it will save time and fuel. Ravi is innovative in his thoughts, so his rationality and social responsibility is reflected here. 8. Let us produce AD to J and PM to K so that DJ AD and MK PM. Join CJ and RK. In ΔADB and ΔJDC, AD JD, ADB JDC, BD CD ΔADB ΔJDC (SAS criterion of congruence) AB JC...(i) (CPCT) Similarly, we can prove that PQ KR...(ii) According to the given conditions, we have AB PQ AD PM AC PR JC KR AJ PK AC PR [Using (i) and (ii)] JC KR AJ PK AC PR ΔAJC ~ ΔPKR (SSS criterion of similarity) JAC KPR (Corresponding angles) i.e., DAC MPR...(iii) Similarly, we can prove that DAB MPQ...(iv) Adding equations (iii) and (iv), we obtain 66 M A T H E M A T I C S X

BAC QPR Thus, in ΔABC and ΔPQR, we have...(v) AB PQ AC (Given) PR and BAC QPR [From (v)] Therefore, ΔABC ~ ΔPQR. (SAS criterion of similarity) Hence proved. OR ΔABE, ΔACE and ΔADE are right angled triangles right angle at E each. AB AE +BE...(i) AC AE +CE...(ii) and AD AE +DE...(iii) Adding equations (i) and (ii), we get AB +AC AE + BE + CE AE + (BD DE) + (CD + DE) AE +BD BD DE + DE +CD + CD DE + DE AE +BD BD DE + DE +BD + BD DE + DE (... BD CD) AE + DE + BD (AE +DE BC )+ µ (D is a mid-point of BC) AD + BC [Using (iii)] WORKSHEET 45. (D) ΔABC ~ ΔPQR 0 h 0 50 Hence proved.. (A) The ratio of similar triangles is equal to the ratio of squares of their corresponding altitudes. 00 49 5 h h 5 49 h h 00. Altitude AM divides base BC in two equal parts. That is BM MC 7 cm. Using Pythagoras Theorem In right ΔABM, 5 49 00 5 7 0.5 cm. AM 5 7 (5 + 7) (5 7) 8 4 cm. 4. (i) We know that diagonal of a square side In square AEFG, AF AG...(i) In square ABCD, AC AD...(ii) Using equations (i) and (ii), we obtain AF AG AC AD....(iii) (ii) GAF DAC (Each 45 ) GAF GAC DAC GAC CAF DAG...(iv) From equations (iii) and (iv), we have ΔACF ~ ΔADG. (SAS criterion) 5. Hint: Q PQ PR QR QS QT PQ. 6. Hint: Draw AM BC and DN BC. 7. Hint: Fig. A b p c h 50 0 0 00 m. C a B T R I A N G L E S 67

8. Hint: For st part: Prove Pythagoras Theorem. For nd part: AC AB 9. Hint: Let the DC AB x Then (AD + CD ) (AD + BD ) QC 4 5 x and AP 5 x ΔQRC ~ ΔPRA. OR See Worksheet 44, Sol. 5. ASSESSMENT SHEET 7. (C) In ΔABC, PQ BC. AP BP AQ QC.4 BP BP.6 cm AB AP + BP.4 +.6 6 cm. ar( ΔABC) BC ar( ΔDEF) EF 9 4 BC EF BC EF.. Draw AM BC and DN BC Q AMO DNO 90 and AOM DON ΔAMO ~ ΔDNO (AA similarity) AM DN AO DO...(i) Now, Therefore ar( ΔABC) ar( ΔDBC) ar ar É% É% ABC DBC BC AM BC DN AO DO. [Using (i)] AO DO Hence proved. 4. True, because ΔBCD ~ ΔCAD CD BD. AD. 5. PQ BC and AB is transversal APQ ABC...(i) (Corresponding angles) In ΔABC and ΔAPQ, BAC PAQ (Common) ABC APQ [From (i)] so, by AA criterion of similarity, ΔABC ~ ΔAPQ ar( ΔABC) AB ar( ΔAPQ) AP Subtracting unity from both the sides, we have ar( ΔABC) ar( ΔAPQ) AB ar( Δ APQ) AP ar(trapezium BPQC) AB ar( ΔAPQ) AP...(ii) It is given that AP PB PB AP PB AP + + PB + AP AB AP AP AB 9...(iii) AP From equations (ii) and (iii), we have 68 M A T H E M A T I C S X

ar( ΔAPQ) ar(trapezium BPQC) 8 ar(δapq) : ar(trapezium BPQC) : 8. 6. See Worksheet 6, Sol. 6. 7. See Worksheet, Sol. 9 (st part). 8. We are given two triangles ABC and PQR such that ΔABC ~ ΔPQR. Draw perpendiculars AD and PM on BC and QR respectively. We need to prove ar( ΔABC) AD ar( ΔPQR) PM In ΔABD and ΔPQM, ADB PMQ 90 ABD PQM (... ΔABC ~ ΔPQR) ΔABD ~ ΔPQM (AA criterion of similarity) AB PQ AD...(i) PM (Corresponding sides) We know that the ratio of areas of two similar triangles is equal to ratio of squares of their corresponding sides ar( ΔABC) AB...(ii) ar( ΔPQR) PQ From equations (i) and (ii), we have ar( ΔABC) AD ar( ΔPQR) PM. Hence proved. ASSESSMENT SHEET 8. (A) ΔPQR ~ ΔCAB, PQ CA PR BC QR AB or AB QR BC PR CA PQ.. x 6 5 6+ x 0 8, x 5 4 cm.. In ΔABC, LM AB. Using Basic Proportionality Theorem, we have AC AL BC BM x x + x x x 4x x + x 6x 9 x 9 x 9. 4. False, because QPR 90. 5. Let the given right angled triangle be ABC with C 90 such that AC b, BC a and AB c. Using Pythagoras Theorem, we have AB AC + BC c b + a...(i) Area of equilateral triangle drawn on side BC 4 a...(ii) Similarly, areas of equilateral triangles drawn on side AC and side AB are respectively 4 b...(iii) and 4 c...(iv) Sum of areas of equilateral triangles drawn on the sides BC and AC 4 a + 4 b [Adding (ii) and (iii)] T R I A N G L E S 69

4 (a + b ) 6. Q 4 c [Using (i)] Area of equilateral triangle drawn on hypotenuse AB. Hence proved. ΔABC ~ ΔPQR AB PQ BC QR BC QR AB PQ BD QM...(i) (a) In ΔABD and ΔPQM, AB PQ BD QM [From (i)] ABD PQM (QΔABC ~ ΔPQR) So, by SAS criterion of similarity, we have ΔABD ~ ΔPQM AB PQ AD PM (b) Q ΔABD ~ ΔPQM, [From part (a)] ADB PMQ 80 ADC 80 PMR [From figure] ADC PMR. Hence proved. 7. Converse of Pythagoras Theorem: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Proof: We are given a ΔABC in which AC AB + BC...(i) We need to prove ABC 90. Let us construct a ΔPQR such that PQR 90 and PQ AB...(ii) QR BC...(iii) Using Pythagoras Theorem in ΔPQR, we have PR PQ + QR PR AB + BC...(iv) [Using equations (ii) and (iii)] From equations (i) and (iv), we have AC PR...(v) Now, in ΔABC and ΔPQR, AB PQ (From (ii)) BC QR [From (iii)] AC PR [(From (v))] So, ΔABC! ΔPQR (SSS congruence) ABC PQR (CPCT) But PQR 90 (By construction) ABC 90. Hence proved 8. BD BE...(i) (Given) In ΔOBD, AF OB and BD OB AF BD ΔOAF ~ ΔOBD OA OB AF BD OA OB AF BE...(ii) [Using (i)] In ΔAFC and ΔBEC, FAC EBC (Each 90 ) FCA ECB (Vertically opposite angles) 70 M A T H E M A T I C S X

So by, AA criterion of similarity, ΔAFC ~ ΔBEC AC BC AF...(iii) BE Comparing equations (ii) and (iii), we have OA OB AC BC OA OB OC OA OB OC OA OB OA OC OB OC OB OA (OA + OB) OC OA OB Dividing both sides by OA OB OC, we get OA + OB OC. CHAPTER TEST Hence proved. 4. Yes. MQ PQ PM 5. 5.7 9.5 cm NR PR PN.8 4.8 8 cm Now, and Clearly, PM MQ 5.7 9.5 0.6 PN NR 4.8 8 0.6 PM MQ PN NR MN QR. 5. ΔAOB ~ ΔCOD (AAA criterion of similarity) AO CO BO DO (Corresponding sides). (B) BC 5 + cm 7 x 9 x 9 x 8 x x 7x 8x 6x 9x + 8 x x 8 0 (x ) (x + 4) T ΔABD ~ ΔCBA AB BC AD AC AD 5 R I A 60 cm.. (D) Δ P Δ P 40 6 50 5 Δ : Δ 6 : 5. AD. DB AE EC.5 EC EC.5 cm. N G L E S x or x 4 x. (Negative value rejected) 6. ΔABE ΔACD AB AC and AE AD (CPCT) Consider AB AC AD + DB AE + EC DB EC...(i) (... AE AD) Also AD AE...(ii) (Proved above) Dividing equation (ii) by equation (i), we have AD DB AE EC...(iii) 7

Hence, in ΔABC AD DB AE EC DE BC (Converse of Basic 7. Hint: Proportionality Theorem) ADE ABC and AED ACB ΔADE ~ ΔABC. ΔPAC ~ ΔQBC x z AC BC ΔRCA ~ ΔQBA y z AC AB. 8. Hint: Draw MN AD, passing through O to intersect AB at M and DC at N. (Corresponding angles) A A (Common) ΔADE ΔABC (AA-criterion) (ii) As ΔADE ΔABC ar É% ADE ar É% ABC É AD É AB É AD ÉAB [' ar(δade) ar(decb) ar(δade) ar(decb) + ar(δade) ar(δade) ar( % ADE) ar(δabc) ar É % ABC ] AD AB AD AB AD AB AB Use Pythagoras Theorem for Δ AOM, ΔBOM, Δ CON and ΔDON. 9. (i) As DE & BC. AB AD AB µ BD AB BD s AB Hence proved. (iii) Concept of similarity of two triangles. (iv) Honesty and rationality to divide his land equally between his two children. qq 7 M A T H E M A T I C S X

Chapter 5 INTRODUCTION TO TRIGONOMETRY WORKSHEET 50. (B) sin A cos (A 6 ) sin A sin {90 (A 6 )} A 90 A + 6 A 6 4 A 9. x x. sin sin sin x sin 6 Q x π x 60. x 6 Q. sin θ 4 5 sin θ sin θ cos θ 7 5 4 5 4 5 cos θ 7 5 Now, tan θ + sec θ sin θ cos θ + cosθ 4 5 7 5 + 7 5 4 7 + 5 7 49 7 7. sin R 4. tan θ. sin R 5. cot 5 + tan 4. 6. True, cos 80 LHS + cos 59 cosec sin0 cos(90 0 ) + cos 59 cosec (90 59 ) sin0 7. sin 0 + cos 59 sec 59 sin 0 cos 59 + cos 59 + Hence, the given equation is valid. 4 cot 0 + sin 60 cos 45 4 ( ) + 4 + 4 8 8 6 6. 8. sin (x + y) and cos (x y) sin (x + y) sin 90 and cos (x y) cos 0 x + y 90 and x y 0 Adding and subtracting, we get respectively x 0 and y 60 i.e., x 60 and y 0. 9. cosec A 0 sin A cosec A 0 cos A sin A tan A cot A sin A cos A tan A 0 0 I N T R O D U C T I O N T O T R I G O N O M E T R Y 7

sec A 0. Hint: RHS cos A 0. Hint: LHS sin A cos A cos A cos A ( cos A) ( + cos A) cos A OR sin θá( sin θ). cos θ (cos θ ) WORKSHEET 5. (A) sin (θ + 6 ) cos θ sin (θ + 6 ) sin (90 θ) θ + 6 90 θ θ 54 θ 7.. (C) Hint: Divide numerator and denominator by cos θ.. sec θ 5 4 sec θ 5 6 sec θ 5 6 tan 9 θ tan θ 6 4. 4. Hint: A 0º, B 90º, C 60º. 5. sec θ + tan θ 6. sin A 7 5 sin R ( sin R) cos R cos R cos R R b b a R a b a sin sin ba ba. b+a b a b a, cos A 4 5, a b 7. sin C 4 5 and cos C 7 5. cos 60 + sin 0 cot 0 tan 60 + sec 45 cosec 45 + s. 8. Given expression cot R tan (90 R) sec (90 R) cosec R sin R cos (90 R) + cos Rsin (90 R) cot R cot R cosec R cosec R sin R sin R + cos Rcos R cot Rcosec R sin R + cos R cosec θ cosec θ (Q sin q + cos q ). 9. Draw Δ ABC with AB BC AC a (say) Draw AD BC BAD DAC θ 0 and BD DC a/ sin θ Þ sin 0. BD a/ AB a tan R tan R 0. LHS tan R tan R tan R tan R tan RÉ tan R tan R Étan R ª tan R¹ 74 M A T H E M A T I C S X

I N T R R tan tan R tan R ª ¹ (tanθ )(tan θ+ + tan θ) tan θ tan θ LHS tan R tan R O D U C T sec R tan R + sec θ cosec θ RHS OR cot A cos A cot A + cos A cos A µ sin A cos A µ sin A sin A + sin A cos A cos A sin A cos A cos A sin A cosec A cosec A + RHS WORKSHEET 5. (A) Hint: tan 5º cot 85º; tan 5º cot 65º.. (D) Hint: ( + sin q) ( sin q) cos q I sec R.. 8 tan x 5 tan x 5 64 sec x 5 64 sec x 7 8 sec x 89 64 cos x 8 7 Now, sin x cos x cos x cos x 64 8 89 7 58 7 7 7. O 4. Hint: sec q cosec 60 cos q sin 60 q 0... cos 0 4 5. Hint: sec 4A cosec (90º 4A). 6. Hint: cos (90º q) sin q, sin (90º q) cos q. 7. (+ sin θ)( sin θ) (+ cos θ)( cos θ ) sin θ cos θ cos sin cosec cosec θ θ cot θ θ + cot θ sec N T O T R I G O N O M E T R Y θ θ 7 8 49 64. OR R R cot cot R É R cot tan + cot θ cot θ tan θ s 7 8 8. 8. sin 0 + sin 45 + sin 60 + sin 90 9. Hint: + + + + 4 4 + LHS Hint: LHS +++4 0 4 4 5. sin R sin R + sin R sin R (sec Rtan R) OR + () cos A sin A tana sin A cos A 75

cos A sin A cos A sin A cos A sin A cos A sin A cos A sin A cos A + sin A. 0. Hint: sin q cos q cos q sin q. WORKSHEET 5. (A) tan q 4 Perpendicular Base BC 4 cos θ 5 5 Base Hypotenuse 4 5 4 cos R 5. + cos R 4 9 + 5. (C) Hint: + tan θ + sec θ + + sec R cot R + cot R+ cosec R cot R. Hint: A + B 90º; A B 0º. 4. tan θ cot (θ + 9 ) tan θ tan [90 (θ + 9 )] θ 90 θ 9 θ 8 θ 7. cos R 5. cot R. cos R 6. True Hint: A 6 + B 6 (A + B ) [(A + B ) A B ]. 7. Hint: sin θ cos θá+ + sinθ cosθ LHS. (sin θá+ cos θ)(sin θ Á cos θ) 8. LHS cosa sin A cosa sin A Dividing numerator and denominator by sin A, we get cot A + coseca cot A + cosec A ( cot A + coseca ) ( cosec A cot A ) cot A + cosec A (coseca + cot A) [ coseca + cot A] cot A coseca + cosec A + cot A RHS. 9. Given expression sin68 o cot5 coso 5tan75 tan 45 tan 0 tan 40 tan 50 tan 70 5 sin (90 ) cot (90 75 ) cos o 5 tan 75 ( ) ( ) tan 90 70 tan 90 50 tan 50 tan 70 5 cos tan75 cos 5tan75 cot 70 cot 50 tan 50 tan 70 5 5 tan 50 tan 70 tan70 tan 50 5 5 5 0 5 5 5. 0. Given expression 8 cosec 0. sin 60. cos 60. cos 45. sin 45. tan 0. cosec 45. 76 M A T H E M A T I C S X

8. sin (90 0 ). sin 0 cos (90 0 ) cos (90 45 ). sin 45. sin 0 cos 0 sin 45 8 sin 0 cos 0. sin 0. sin 45. sin 45. sin 0 cos 0 sin 45 sin 0. sin 0 cos 0 8 sin 0 cos 0 8 8. sin 45 sin 45 sin 45 tan0. (C) + tan 0 +. Hint: tan x 5 8 4. sin x 5 7, cos x 8 7 4 sin x cos x 5 89 64 89 6 89. sin 0 + tan 45 cosec 60 sec 0 + cos 60 + cot 45 + + 4 + + 4+ +. Hint: sec θ sec θ tan θ OR x 6x x 6x x µ 4x 4 + 4 4 4 7 + 6 4 4 4. 7 6 5. Q A + B + C 80º C + A 80º B LHS cot cot cot (90º B ) tan B tan θ ± x 4x µ WORKSHEET 54. (B) As sin A 4, let BC x and CA 4x AB ( ) 4 ( ) x x 7 x Now, tan A BC AB 7 xx 7.. 6. Yes. RHS. Hint: Both sides 7 5. 7. LHS (cosec A sin A) (sec A cos A) sina cosa sin A cos A sin A sin A cos A cos A cos A sin A sin A cos A sin A cos A... (i) I N T R O D U C T I O N T O T R I G O N O M E T R Y 77

RHS tan A + cot A sin Acos A sin A + cos A sin A cos A + cos A sin A sin A cos A...(ii) (Q sin A + cos A ) From equations (i) and (ii), we obtain LHS RHS. 8. 7 sin θ + ( sin θ) 4 Let sin θ x 7x + x 4. (C) sin θ cos θ 0.6 0.8 And tan θ sin θ cos θ 0.8 0.6 4 Now, 5 sin θ tan θ 5 0.8 4 0. Hint: Divide numerator and denominator by sin A. +cota + 4. cota 4 4. sec A sec A sec 0 A 0 4x x 4 x ± sin θ or sin θ sin θ is not possible as θ is acute. cosec θ cos θ sec θ + cosec θ. Hence proved 9. Hint: cos (40º + θ) sin{90 (40 + θ)} sin (50 θ) and cos 40 sin 50. 0. LHS m n (tan θ + sin θ) (tan θ sin θ) 4sin θ tan θ...(i) RHS 4 sin Áθ mn 4 sin θ cos Áθ 4sin θ sec θ 4 sin θ tan θ... (ii) From (i) and (ii), LHS RHS. WORKSHEET 55 64 6 8. (A) Required value 5 + 00 00 6 5 (9 + 6 400) 00 8. A + B 90 B 90 0 60 Now, cosec B cosec 60 5. Given expression 78 M A T H E M A T I C S ( ) + 4 + + 5 0 + ( ) + + 4 9. 4 6. False Hint: ÐA 0º, ÐB 60º. 7. sec θ cosec θ sec θ + cosec θ cos θ sin θ + cos θ sin θ É sin É sin cos R Rs + cos R R s 4 + 4 sin sin sin R cos R sin R + cos R R R cot θ + cot θ 7. Hence proved. X

8. cos R sin R cos R + sin R cot θ. cot θ + (Dividing numerator and denominator by sin θ) p q p q p q + p+ q. 9. Given expression sin 5 cos 55 + cos 60 cos 55 sin 5 sin 5 cos ( 90 5 + ) cos 55 sin ( 90 55 ) sin 5 sin 5 + cos 55 cos 60 cos 55 + cos 60 s OR Given expression cos 58 sin + sin cos 68 cos 60 cos 8 cosec 5 tan 8 tan 5 tan 60 tan 7 tan 55 cos 58 sin ( 90 58 ) + sin cos ( 90 ) ( ) ( ) tan ( 90 5 ) cos 8 cosec 90 8 tan 8 tan 5 tan 60 tan 90 8 cos 58 cos 58 + sin sin cos 8 sec 8 tan 8 tan 5 tan 60 cot 8 cot 5 cos 8 tan8 tan 5 cos 8 tan 8 tan 5 tan 60 tan 60 6. 0. tan A n tan B n cot B and sin A m sin B tana sin B sin A m m cosec B sin A... cosec B cot B m sin A n tan A m n cos A sin A m (n ) cos A m cos A. n Hence proved OR Consider an equilateral triangle PQR in which PS QR. Since PS QR so PS bisects P as well as base QR. We observe that ΔPQS is a right triangle, right-angled at S with QPS 0 and PQS 60. For finding the trigonometric ratios, we need to know the length of the sides of the triangle. So, let us suppose PQ x Then, QS QR x and (PS) (PQ) (QS) PS x x x 4 x 4 I N T R O D U C T I O N T O T R I G O N O M E T R Y 79

x (i) cos 60 QS PQ x (ii) sin 60 PS PQ x x x (iii) tan 0 QS PS x. WORKSHEET 56. (B) b x + a y b a cos θ + a b sin θ a b.. (A) A 90 60 0 cosec A cosec 0.. (C) tan θ 5 + tan θ + sec θ 5 + sin R Now, + sin θ sin θ cos R sec θ+ tan θ sin R sec θ tan θ cos R sin 9 4. cos 6 + sin 9 cos É90 9 + 4 0. 5 5 5 5 5 cos 7 sin 6 5 5 5 4cos 45 5. cos 7 + sin ( 90 7 ) 4 5. Given expression 6. 4 4 + 4 + 6 4 4 4 + + 4 7 4 9 4 8 4. tan A + 7. Q sin θ 4 «º É µ» ¼ sin A + cos A cos A sin A + sin A É + cos A sin A + cos A cos A + cos A + sin A + cos A sin A( + cos A ). cosec θ 4 Q cos θ sin θ 9 6 sec θ 4 7 and cot θ 7 Now, LHS cosec θ cot sec θ 7 9 7 θ 6 7 9 9 6 7 cosec A 7 4 9 9 9 7 RHS. Hence proved. 80 M A T H E M A T I C S X

8. Hint: LHS cos A sin A sin A 9. sin A cos A sin A sin A + cos A sin A cos A ( ) sin A cos A + cos A sin A cos A ( ) ( ) ( ) cos A cos A cot A. sin A cos A OR Using a + b (a + b ab) (a + b), we get sin θ+ cos θ Á + sin θ. cos θ sin θ+ Á cosθ. (sin θ+ cos θ)(sin θ+ cos θ sin θ cos θ) sin θ+ cos θ + sin q cos q sin θ. cos θ + sin θ. cos θ. ( + sin θ)( sin θ) ( + cos θ)( cos θ) ( θ)( θ) ( θ)( θ) + sin sin + cos cos sin cos 5 µ 8 θ θ cos θ sin θ cot θ 5 64. 0. Hint: p sec θ + tan θ + sec θ tan θ + tan θ + tan θ + sec θ tan θ tan θ (tan θ + sec θ) Similarly p + sec θ (tan θ + sec θ).. (B) Hint: WORKSHEET 57 x + y cot A x y cos A x y sin A x+ y and x y cos A sin A + cos A.. (A) 5 Hint: (x + ) x + 5. (A) tan A tan 0 A 0 C 80 A B 80 0 60 Now, sin A cos C + cos A sin C sin 0 cos 60 + cos 0 sin 60 +. 4. cos α cos α cos 60 α 60 tan β tan β tan 0 β 0. Now, sin (α + β) sin (60 + 0 ) sin 90. 5. tan tan... tan 4 tan 44 tan 45 tan 46 tan 47... tan 88 tan 89 (tan tan 89 )(tan tan 88 )...(tan 4 tan 47 )(tan 44 tan 46 ) tan 45 (tan cot )(tan cot )...(tan 4 cot 4 )(tan 44 cot 44 ) tan 45 () ()... () () tan 45 (... ) tan 45. 6. Given expression tan 50 + sec 50 + cos 40 cosec 50 cot 40 + cosec 40 I N T R O D U C T I O N T O T R I G O N O M E T R Y 8

tan 50 + sec 50 + cot (90 50 ) + cosec (90 50 ) cos 40 cosec (90 40 ) tan 50 + sec 50 tan 50 + sec 50 + cos 40. cos 40 +. 7. LHS tan (A B) tan (60 0 ) tan 0. tan A tan B RHS + tan A tan B +. 8. RHS + tan 60 tan 0 tan 60 tan 0 LHS. Hence verified. 6 sin Áθ sin Áθ +. + 6 cos Áθ cos Áθ cos 6 6 R R R R 6 R sin + cos + sin cos cos sin Rcos R 6 cos R 6 R R R sec LHS. [' sin +cos ] OR Hint: Numerator of LHS tan θ + sec θ (sec θ tan θ) (tan θ + sec θ) (tan θ + sec θ) (sec θ tan θ) (tan θ + sec θ) ( sec θ + tan θ). 9. cos θ + sin θ cos θ Squaring both sides, we get cos θ + cos θ sin θ + sin θ cos θ cos θ cos θ cos θ sin θ sin θ cos θ cos θ sin θ sin θ Adding sin θ to both sides, we have sin θ + cos θ cos θ sin θ sin θ + sin θ (cos θ sin θ) sin θ cos θ sin θ sin θ Hence proved. 0. Hint: l tan θ + m sec θ n...(i) l l tan θ m sec θ n...(ii) l ll tan θ + ml sec θ nl l l tan θ m l sec θ n l + (m l + ml ) sec θ nl n l sec θ nl n l ml ml Similarly, tan θ nm mn. lm ml WORKSHEET 58. (D) Given expression cos (90 70 ) + cos 70 sec (90 40 ) cot 40 + {cosec 58 cot 58 tan (90 58 )} sin 70 + cos 70 cosec 40 cot 40 + (cosec 58 cot 58 ) + () +.. (A) sec 5A cosec (A 6 ) sec 5A sec {90 (A 6 )} 5A A + 6 A.. Given expression sin 5 + sin 0... + sin 40 + sin 45 + sin 50 +... + sin 80 + sin 85 + sin 90 cos 85 + cos 80 +... + cos 50 + + sin 50 +... + sin 80 + sin 85 + () (cos 85 + sin 85 ) + (cos 80 + sin 80 ) +... + (cos 50 + sin 50 ) + + ( + +... 8 terms) + + 8 + + 9. 8 M A T H E M A T I C S X

4. tan x. + + tan x tan 45 x 45 o 5. 5. cosec A sin A cos A sin A tan A, cot A Now, sin A + cot A 4 tan A cos A ( ) + 4 µ 4. 6. True Hint: a cos θ + b sin θ 4...(i) sin θ a sin θ b cos θ...(ii) cos θ a cos θ sin θ + b sin θ 4 sin θ a sin θ cos θ b cos θ cos θ + b 4 sin θ cos θ Similarly, a 4 cos θ + sin θ a + b 6 sin θ + 9 cos θ sin θ cos θ + 6 cos θ + 9 sin θ + sin θ cos θ 6 + 9 5. 7. (a b ) sin θ + ab. cos θ a + b Divide by cos θ (a b ) tan θ + ab a b cos R (a b ) tan θ + ab (a + b ). sec θ (a + b ). tan R Squaring both sides: (a b ) tan θ + 4 a b + 4 ab (a b ) tan θ (a + b ) ( + tan θ) (a + b ) +(a + b ) tan θ [(a b ) (a + b ) ] tan θ + 4 a b + 4 ab (a b ) tan θ (a + b ) 0 4a b tan θ + 4ab (a b ) tan θ a 4 b 4 + a b 0 4a b tan θ + 4ab (a b ) tan θ (a b ) 0 4a b tan θ 4ab (a b )tan θ + (a b ) 0 [ab tan θ (a b )] 0 ab tan θ a b tan θ a b ab 8. Hint: Use (a + b ) a 6 + b 6 +a b (a + b ). 9. LHS ( + cot A + tan A )( sin A cos A ) sec A cosec A cos A sin A + + sin A cos A sin A cos A cos A sin A ( ) (sin A cos A + cos A + sin A)(sin A cos A) sin A cos A (sin A cos A)(sin A + cos A + sin Acos A) sin A cos A sin A. cos A RHS. Hence proved. 0. m cosec θ sin θ n sec θ cos θ sin θ sin θ sin θ sinθ. cos cosr cos θ cos cos R Now, LHS ( ) ( mn + mn ) cos sin 4 θ sin θ cos θ sin + θ cos θ sin θ cos R θ sin θ sin R cos R 4 θ θ I N T R O D U C T I O N T O T R I G O N O M E T R Y 8

( ) cos θ + ( ) sin θ cos θ + sin θ RHS. OR LHS ( + cot A cosec A)( + tan A + sec A) cos A sin A sin A sin A cos A cos A µ µ sin A + cos A sin A ( sin A + cos A ) sin A cos A cos A + sin A + cos A sin A + sin A cos A + cos A sin A cos A sin A cos A sin A cos A RHS. Hence proved. ASSESSMENT SHEET 9. (A) cos 0 BC AC 0 x x 0 x 0 sin 0 AB AC y 0 cm y 0 cm.. 5 Hint: tan θ cot (θ + 5 ) tan θ tan [90 (θ + 5 )].. We know: sin θ cos (90 θ) so, given expression cos(90o 5 o) «cos 55o º + \ cos 55 ^» o cos(90o5 o) ¼ cos 60 cos 55 cos 55 + cos 55 cos 55 +. 4. True, because LHS tan 60 and tan 0 RHS tan 0. 5. (sin 6 A + cos 6 A) (cos 4 A + sin 4 A) + {(sin A + cos A) sin A cos A (sin A + cos A)} {(sin A + cos A) sin A cos A} + [Q (a + b) a + b + ab(a + b) and (a + b) a + b + ab] ( sin A cos A) ( sin A cos A) + [Q sin A + cos A ] 6 sin A cos A + 6 sin A cos A + 0 6. sin x + cos y (Given) sin x + cos y 0 sin x + cos y 0... (i) 6 sin x cos y sin x cos y É 4 sin x cos y 4 (Given) 4 É 4 4 (sin x + cos y)... (ii) Substituting cos y sin x from (i) in (ii), we get (sin x + sin x) sin x 4 sin x ± When sin x, cos y When sin x, cos y Hence, sin x, cos y or sin x, cos y. 7. We know that sin (90 θ) cos θ, tan (90 θ) cot θ, sec(90 θ) cosec θ 84 M A T H E M A T I C S X

Now, sec (90 θ) cot θ (sin 5 + sin 65 ) + cos 60 tan 8 tan 6 (sec 4 cot 47 ) sec (90 θ) cot θ sin 5 sin 90 5 { + ( )} É + cos 60 tan 8 tan (90 8 ) sec 4 cot 90 4 { ( )} Rcot R o cosec sin 5 cos 5 + o cos 60 tan 8 cot 8 (sec 4 tan 4 ) 4 + + 6. 8. Given equations are: sin θ + cos θ p... (i) sec θ + cosec θ q... (ii) Squaring both the sides of equation (i), we get sin θ + cos θ + sin θ cos θ p Subtract unity from both the sides to get p sin θ cos θ... (iii) Equation (ii) can be written as q cos θ + sin θ sin RcosR q... (iv) sin R cos R From equations (iii) and (iv), we get q (p ) sin R cos R sin θ cos θ sin R cos R q (p ) (sin θ + cos θ) q (p ) p. Hence proved. ASSESSMENT SHEET 0. (A) Given expression sin 5 cos (90 5 ) + cos 5 sin (90 5 ) sin 5 + cos 5.. sin θcosθ sinθ+ cosθ tan θ tanθ+ cos 45o. sec 0ocosec 0 É sinr cosr cosr cosr sinr cosr cos R cosr 4 4 ( ) ( + )( ) + 5. ( + ) 4 6. 8 4. False, because cos sin 67 0, 0 is not a positive value. 5. LHS cos A + sin A + + sina cos A É cos A + + sina É +sina cosa É cos A + + sin A + sin A É +sin A cosa sina sina cos A É sina sina cos A cos A sec A RHS. Hence proved. É I N T R O D U C T I O N T O T R I G O N O M E T R Y 85

6. Let us construct a triangle ABC in which AB BC AC a (say). Draw AD BC. AD bisects BC a BD DC AD bisects BAC θ 0 In right angled ΔABD. AD AB BD a a a 4 a 4 a AD a Now, in Δ ABD, a tan θ BD AD tan 0 a tan 0. 7. ( a b ) sin θ + ab cos θ a + b (Given) Divide both sides by cos θ to get ( a b ) tan θ + ab (a + b ) sec θ Squaring both sides, we get (a b ) tan θ + 4a b + 4ab(a b ) tan θ (a + b ) sec θ (a b ) tan θ (a + b ) tan θ + 4ab (a b ) tan θ (a + b ) + 4a b 0 (Q sec θ + tan θ) 4a b tan θ + 4ab (a b ) tan θ (a b ) 0 4a b x + 4ab (a b ) x (a b ) 0 where x tan θ This is a quadratic equation in x. Here, discriminant, D 6ab Éa b 4 s4ab Éa b 0 x tan θ 4 ab( a b ) 0 a b s( 4 ab) ab a b ab. Hence proved. 8. Since ABC is a acute angled triangle so, A < 90, B < 90 and C < 90. Also A + B + C 80... (i) sin (A + B C) (Given) sin (A + B C) sin 0 A + B C 0... (ii) Similarly, B + C A 45... (iii ) Add equations (ii) and (iii) to get B 75 B 7 o Subtract equation (ii) from equation (i) to get C 50 C 75 Subtract equation (iii) from equation (i) to get A 5 A Thus, A 67 o o o 67, B 7 and C 75. CHAPTER TEST. (A) x sec θ and x tan θ. É x µ sec R tan x sec R tan R µ 4 cos 0 + cos 70 sin 59 sin o o o É o k sin 70 cos 70o sin 59 cos 59. o R µ 4 4 k k 4. k. sin 4 θ + cos 4 θ + 4k sin θ cos θ (sin θ + cos θ) sin θ cos θ + 4k sin θ cos θ sin θ cos θ ( k) 0 k 0 k. 86 M A T H E M A T I C S X

4. tan θ 4 tan θ + 4 + sec θ 7 0 (tan θ + sec θ) (6 + 7) 0 5. 5. False. Suppose A 0 and B 60 Then, LHS tan (A + B) tan (0 + 60 ) tan 90 LHS undefined... (i) and RHS tan A + tan B tan 0 + tan 60 + + RHS a real number... (ii) From results (i) and (ii), it is clear that the given identity is false. 6. 7 Hint: cos 55 cos (90 5 ) sin 5 cos 70 sin 0 and tan 5 cot 85. 7. 4. Hint: sin 0 cos 60, sin 60, cos 45 sin 45, sin 90. 8. sin θ + cos θ a Squaring both sides. sin θ + cos θ + sin θ cos θ a sin θ cos θ a sin θ cos θ a 4 Now, sin 6 θ + cos 6 θ ( ) sin θ+ cos θ... (i) sin θ cos θ (sin θ + cos θ) a () [Using equation (i)] 4 a 4 ( a ). 4 Hence proved. ( ) 9. LHS ( θ+ θ ) ( ) sec tan secθ+ tanθ + sec θ + tan θ + sec θ tan θ sec θ + tan θ + sec θ tan θ + ( ) ( ) sec θ + tan θ+ sec θtan θ sec θ + tan θ + + sec θ tan θ tan R + tan R + sec R tan R sec R + sec R + sec R tan R tan θ (tan θ + sec θ) sec θ (sec θ + tan θ) tan θ tan θ cos θ sec θ sin θ. cos θ sin θ RHS. cos θ Hence proved. OR sin A sin B cos A cos B + cos A + cos B sin A + sin B Ésin A sin BÉsin A + sin B cos A cos B cos A cos B Écos A + cos B Ésin A + sin B É É sin A sin B cos A cos B Écos A + cos BÉsin A + sin B sin A É cos A) (sin B cos B É É cos A + cos B sin A + sin B ( cos A + cos B )( sin A + sin B ) 0 which is an integer. qq I N T R O D U C T I O N T O T R I G O N O M E T R Y 87

Chapter 6 STATISTICS WORKSHEET 6. (B). Hint: Median Mode + Mean.. Since the mode is 7 k 7 k 4.. In such case, mean will increase by. New mean 8 +. 4. x 6 4fx i i Hint: Mean. 4fi 5. Class Frequency ( f ) Cumulative interval Frequency (cf) 0-8 8 8 8-6 0 8 6-4 6 4 4-4 58-40 5 7 40-48 7 80 N 80 For median class, N 80 40 In the cumulative frequency column, 58 is just greater than 40. So, 4- is the median class. Here, l 4, cf 4, f 4, N 40, h 8 Using formula: N cf Median l + h f žÿ ž 4 + 404 ž 8 6 žÿ 4 Hence, median of the given distribution is 6. 6. 6.5 Hint: Here maximum class frequency is. So, the modal class is 0-40. Now, l 0, f, f 0, f 0, h 0 Use the formula: f f0 Mode l + ž žÿ f f0 f h. 7. Hint: Production No. of Production yield yield farms (in kg/ha) (in kg/ha) more than or cf equal to 50-55 50 00 55-60 8 55 98 60-65 60 90 65-70 4 65 78 70-75 8 70 54 75-80 6 75 6 00 80 0 For more than ogive, plot following points: (50, 00), (55, 98), (60, 90), (65, 78), (70, 54), (75, 6), (80, 0). WORKSHEET 6. (B) median. 4. The given distribution can be represented as: Marks obtained No. of students 0-0 5 0-0 0-0 4 0-40 40-50 6 More than 50 4 Clearly, the frequency of the class 0-40 is. 4. Let us rewrite the given table with cumulative frequencies. Class interval f cf 0-5 0 0 5-0 5 5 0-5 7 5-0 0 57 0-5 9 66 N 66 88 M A T H E M A T I C S X

Q N 66 N Median class 0-5 Modal class 5-0 Required sum 0 + 5 5. 5. In the given distribution, maximum class frequency is 0, so the modal class is 40-50. Here, lower limit of modal class: l 40 Frequency of the modal class: f 0 Frequency of the class preceding the modal class: f 0 Frequency of the class succeeding the modal class: f Size of class: h 0 Using the formula: f f0 Mode l + ž žÿ f f f h 0 0 40 + ž 0 žÿ 0 40 + 4.70 44.70. Hence, mode of the given data is about 45 cars. 6. Let us rewritten the table with class intervals. Class interval f cf 6-8 0 0 8-40 40-4 5 4-44 4 9 44-46 5 4 46-48 4 8 48-50 4 50-5 5 N 5 We mark the upper class limits on x-axis and cumulative frequencies on y-axis with a suitable scale. We plot the points (8, 0); (40, ); (4, 5); (44, 9); (46, 4); (48, 8); (50, ) and (5, 5). These points are joined by a free hand smooth curve to obtain a less than type ogive as shown in the figure. Figure: Less than type ogive To obtain median from the graph: We first locate the point corresponding to N 5 7.5 students on the y-axis. From this point, draw a line parallel to the x-axis to cut the curve at P. From the point P, draw a perpendicular PQ on the x-axis to meet it at Q. The x-coordinate of Q is 46.5. Hence, the median is 46.5 kg. Let us verify this median using the formula. Median l + 46 + N cf h f 7.5 4 µ 4 46 + 7 46 + 0.5 4 46.5 kg. Thus, the median is the same in both methods. S T A T I S T LI C S 89

7. (i) By making the given data continuous, we get: a 57, h. No. of mangoes No. of boxes (f i ) Mid-points (x i ) xi a ui h f i u i 49.5-5.5 5 5 0 5.5-55.5 0 54 0 55.5-58.5 5 a 57 0 0 58.5-6.5 5 60 5 6.5-64.5 5 6 50 400 fu 5 f i µ fu i i Mean a + h µ f 57 + 5 75 57 + i 400 400! 57.9. (ii) Step devitation method (iii) Vikram Singh believes in quality serving, fruits will remian fresh and free from germs and flies. i i WORKSHEET 64. (C) mid-points of the classes.. (D) We have Mode Median Mean 45 Median 7 Median.. 5-0 Hint: N 5 + 8+ + 9. 4. Required number of athletes + 4 + 5 + 7 8. 5. Let us assumed mean be a 5 Here, h 0 Using the formula: 4f i ui Mean a + h 4fi 50 5 +.8. f + 0.9 f 0 q0. f 0.9 f 9....(i) But 68 + f + f 0 f + f 5...(ii) Solving (i) and (ii), we obtain f 8 and f 4. 6. Let us convert the given data into less than type distribution. Class interval f Lifetimes (in hrs.) 0-0 0 less than 0 0 0-40 5 less than 40 45 40-60 5 less than 60 97 60-80 6 less than 80 58 80-00 8 less than 00 96 00-0 9 less than 0 5 cf 90 M A T H E M A T I C S X

S We mark the upper class limits along the x-axis with a suitable scale and the cumulative frequencies along the y-axis with a suitable scale. For this, we plot the points A(0, 0), B(40, 45), C(60, 97), D(80, 58), E(00, 96) and F(0, 5) on a graph paper. These points are joined by a free hand smooth curve to obtain a less than type ogive as shown in the given figure. T A Figure: Less than type ogive 7. The given distribution can be again represented with the cumulative frequencies as given below: Class interval f i x i cf f i x i 00-0 0 0 0-40 4 0 6 80 40-60 8 50 4 00 60-80 6 70 40 00 80-00 0 90 50 900 50 760 T I S T LI C S Σfx i i Mean: Mean Σfi Q Σf i 50 and Σf i x i 760 Mean 760 50 45.0. Hence, the mean is ` 45.0 Median: N cf Median l + h f Q N 50, N 5, f 4, cf, l 0 and h 0 5 Median 0 + µ 0 4 0 + 8.57 8.57 Hence, the median is ` 8.57. Mode: f f Mode l + f f f Q l 0, f 4, f 0 f 8 and h 0 0 µ 0 h 4 Mode 0 + µ s4 8 0 0 + 40 8 5 Hence, the mode is ` 5. 8. C.I. No. of consumers (f i ) (c.f.) 65-85 4 4 85-05 5 9 05-5 c.f. 5-45 0 f 4 45-65 4 56 65-85 8 64 85-05 4 68 N 68 N 68 4 c.f. just greater than 4 is 4. Median class is 5-45. 9

Median l + N cf.. h f 4 5 + 0 0 5 + 7. (ii) 0 + 4 + 8 + 4 46 families. (iii) Since, Mr. Sharma is saving electricity so his consumption is less, which means his monthly bill will also be less. So, he believes in saving and hence is responsible also.. (B) 0-40 Hint: WORKSHEET 65 Class interval Frequency Cumulative (C.I.) ( f ) Frequency 0-0 4 4 0-0 4 8 0-0 8 6 0-40 0 6 40-50 8 50-60 8 46 60-70 4 50. 45 Hint: Draw a line parallel to the x-axis at the point y 40 0. This line cuts the curve at a point. From this point, draw a perpendicular to the x-axis. The abscissa of the point of intersection of this perpendicular with the x-axis determines the median of the data.. The given distribution can also be represented as follows: Class interval Frequency 0-0 0-0 9 0-0 5 0-40 0 40-50 8 50-60 5 As the maximum frequency is 0, the modal class is 0-40. 4. C.I. f i x i f i x i - 9 8-5 4 88 5-7 7 6 6 7-0 7 8.5 44.5 Σf i 75 Σf i x i 4.5 Σfx 4.5 Mean i i 5.5. Σfi 75 5. In the given distribution, the classes are in the inclusive form. Let us convert them into exclusive form by subtracting 6 6, i.e., 0.5 from lower limit and adding the same to upper limit of each class. Class interval f 59.5-6.5 5 6.5-65.5 8 65.5-68.5 4 68.5-7.5 7 7.5-74.5 8 Here, the maximum frequency is 4. l 65.5, f l 4, f 0 8, f 7, h Now, f f0 mode l + f f0 f h 4 8 65.5 + µ 84 8 7 65.5 +.85 67.5 Hence, the modal height of the students is 67.5 cm. 6. The given data may be re-tabulated by the following manner with corresponding cumulative frequencies. 9 M A T H E M A T I C S X

S Heights (in cm.) No. of girls Cumulative C.I. ( f ) frequency (cf ) T Below 40 4 4 40-45 7 45-50 8 9 50-55 40 55-60 6 46 60-65 5 5 N 5 Now, N 5. So, N 5.5. This observation lies in the class 45-50. Then l 45, cf, f 8, h 5 N cf Now, median l + h f žÿ 5.5 45 + ž 5 žÿ 8 49.0. Hence, the median height of the girls is 49.0 cm. 7. C.I. f i x i f i x i 0-7 77-4 56 4-6 8 5 70 6-8 7 Σ f i 50 Σ f i x i 74 Mean mileage A T I S T 4 i i 4 fi LI fx C 74 50 4.48 km/l. (ii) No, the manufacturer is claiming mileage.5 km/l more than average mileage. (iii) The manufacturer should be honest with his customer. 8. 69.5. Hint: Change the given distribution into less than type and more than type distributions. For drawing the less than type S ogive, take upper class limits and corresponding cumulative frequencies; and for drawing the more than type ogive take lower class limits and corresponding cumulative frequencies. ASSESSMENT SHEET. (A) Here, a 5, h 0. x a + h 5 + 0 Σfu i i Σfi µ 7. 0 00. Sum of numbers 5 85 Sum of first 6 numbers 6 9 Sum of last 6 numbers 6 7 6 th number 9 + 85 9.. The modal class is 0-40. h 0, f, f 0, f 0, l 0. Mode l + f f0 f f f 0 h 0 + µ 0 64 0 0 + 6.5 6.5. 4. False, Hint: N 5+5+0+8+60 N 0 5. x i f i f i x i 5 5 6 7 4 8 p + 4 4p + 8 6 48 Σf i Σf i x i 4p + 6 9

Σfx i i Mean Σf 6 4 p 6 4p + 6 4p 6 p 4. i 6. Class interval Frequency ( f ) 0-0 4 0-40 6 40-60 8 60-80 8 80-00 4 The class corresponding to the maximum frequency is 40-60. So, 40-60 is the modal class. f f0 Mode l + f f0 f h Here, l 40, f 8, f 0 6, f 8 and h 0 Mode 40 + 8 6 8 6 8 0 s 0 40 + 50.9. 7. We notice classes are continuous. We form cumulative frequency table by less than method. Marks (C.I.) Number Marks of stu- less cf Point dents than 0-0 5 0 5 (0, 5) 0-0 8 0 (0, ) 0-0 0 0 (0, ) 0-40 9 40 (40, ) 40-50 6 50 8 (50, 8) 50-60 7 60 45 (60, 45) On plotting these points on a graph paper and joining them by a free hand smooth curve, we get a curve called less than ogive. 8. The cumulative frequency table for the given data is given below: Marks (C.I.) No. of Cumulative students frequency ( f ) (cf ) 0-0 0 0 0-0 f 0 + f 0-0 5 5 + f 0-40 0 65 + f 40-50 f 65 + f + f 50-60 0 7 + f + f N 75 + f + f Clearly, N 75 + f + f But N 00 f + f 5... (i) N 50. The median is which lies in the class 0-40. So, l 0, f 0, cf 5 + f, h 0. Using the formula: Median l + N cf f µ µ h µ 0 50 5 f 0 + 0 94 M A T H E M A T I C S X

0 5 f 0 75 5f 0 75 0 f 5 f 9 Substituting f 9 in equation (i), we get 9 + f 5 f 6 Hence, f 9 and f 6. ASSESSMENT SHEET. (D) x + x +... + x n n x x k x +... + k xn x x... xn k k k x n k Required mean x k. k n x k (Dividing throughout by k) (Dividing throughout by n). The first ten prime numbers are:,, 5, 7,,, 7, 9,, 9. Median 4.. Mean 5 4fx i 4f 445 0k 5 7 k k 8. i i 5 6 + 0 k + 5 6 + 0 0+5 5 6+ k +6+0+5 4. False, because the values of these three measures depend upon the type of data, so it can be the same. 5. Let us use the assumed mean method to find the mean of the given data. Marks No. of Class d i (C.I.) students mark xi 5 f i d i ( f i ) (x i ) 0-0 4 5 0 0 0-0 6 5 0 0 0-0 8 5 0 80 0-40 0 5 0 0 40-50 45 0 0 50-60 0 55 0 600 Σf i 70 Σf i d i 400 Here, assumed mean, a 5 Σfd i i Now, required mean a + Σfi 5 + 400 5 + 5.7 40.7. 70 6. Since mode 6, which lies in the class interval 0-40, so the modal class is 0-40. f 6, f 0 f, f, l 0 and h 0. Now, mode l + 6 0 + 6 0 6 0 f f 0 f f f µ 0 6 f f µ h 0 0 6f 60 0f 4f 40 f 0. 7..5 marks. Hint: Classes No. of Cumulative students frequency 0-0 5 5 0-0 8 0-0 6 9 0-40 0 9 40-50 6 5 50-60 6 4 Draw the ogive by plotting the points: (0, 5), (0, ), (0, 9), (40, 9), (50, 5) f f S T A T I S T LI C S 95

and (60, 4). Here N 0.5. Locate the point on the ogive whose ordinate is 0.5. The x-coordinate of this point will be the median.. 8. We prepare the cumulative frequency table by less than method as given below: Cumu- Fre- Score lative Scores quency less fre- Point ( f ) than quency ( f ) 00-50 0 50 0 (50, 0) 50-00 5 00 45 (00, 45) 00-50 45 50 90 (50, 90) 50-400 0 400 0 (400, 0) 400-450 5 450 5 (450, 5) 450-500 40 500 75 (500, 75) 500-550 0 550 85 (550, 85) 550-600 5 600 00 (600, 00) We plot the points given in above table on a graph paper and then join them by free hand smooth curve to draw the cumulative frequency curve by less than method. Similarly for the cumulative frequency curve by more than method, we prepare the corresponding frequency table. Cumu- Fre- Score lative Scores quency more fre- Point ( f ) than quency ( cf) 00-50 0 00 00 (00, 00) 50-00 5 50 70 (50, 70) 00-50 45 00 55 (00, 55) 50-400 0 50 0 (50, 0) 400-450 5 400 90 (400, 90) 450-500 40 450 65 (450, 65) 500-550 0 500 5 (500, 5) 550-600 5 550 5 (550, 5) We plot the points given in this last table on the same graph and join them by free hand smooth curve to draw the cumulative frequency curve by more than method (see figure). Median: The two curves intersect each other at a point. From this point, we draw a perpendicular on the x-axis. The foot of this perpendicular is P(75, 0). The abscissa of the point P, i.e., 75 is the required median. Hence, the median is 75. Figure: Less than and more than type cumulative frequency curves 96 M A T H E M A T I C S X

CHAPTER TEST. (A) Let Σf i N Σ(f i x i x ) Σf i x i Nx N fx 4 i i x µ N (x x ) 0.. Let us rewrite the given distribution in the other manner. Marks No. of students 0-0 0-0 9 0-0 5 0-40 0 40-50 8 50-60 5 N Clearly, the modal class is 0-40.. 7.5 Hint: First, transform the given class-intervals into exclusive form and then find the cumulative frequency table. Here, N + 0 + 5 + 8 + 57 N 8.5. 4. Monthly income No. of (in `) families 0000-000 5 000-6000 6 6000-9000 9 9000-000 7 000-5000 8 5000 or more 5 Hence, required number of families is 9. 5. No, because an ogive is a graphical representation of a cumulative frequency distribution. 6. Yes; as we know mode median mean median mode + mean Median mode + mean mode mode + mean mode+ (mean mode). 7. u i C.I. x f i i A 800-80 80 7 80-840 80 4 840-860 850 9 860-880 870 5 880-900 890 9 x i h f i u i 40 4 0 0 4 0 0 0 0 0 0 0 5 40 0 8 Σf i 64 Σf i u i 5 Let assumed mean be A 850 h 0 Σuf i i Mean A + f h Σ i 5 850 + µ 64 0 850 +.565 85.565. Hence, the required mean is 85.565. 8. Mode l + f f0 f f f 0 h Here, l 0, f 45, f 0 0, f, h 0 45 0 Mode 0 + 90 0 0 0 +.5.5 marks. S T A T I S T LI C S 97

9. (i) Class intervals Frequency Mid-points f i x i (in daily pocket (No. of children) of C.I. allowances) (in `) ( f i ) (x i ) - 7 84-5 6 4 84 5-7 9 6 44 7-9 8 4 9- x 0 0-5 0-5 4 4 96 Σf i 44 + x Σf i x i 75 + 0x Mean 4 4 fx i f i i 75 0x 44 x 75 0x As Mean ` 8 (given) 8 44 x 79 + 8x 75 + 0x 40 x x 0. (ii) Arithmetic mean of grouped data. (iii) One shouldn't be spend thrift, but should save his money for future use. qq 98 M A T H E M A T I C S X

PRACTICE PAPERS PRACTICE PAPER Practice Paper- SECTIONA. (D) a 6 a, b b, c c 0 9 a Þÿi.e., a b c ¹ b c ÞÿThe given lines are parallel.. (B) ÐD ÐQ and ÐE ÐR Þ DDEF ~ DQRP (AA rule of similarity) Þ DE PQ ¹ EF RP..... x t y LCM (x, y) 800 HCF( x, y ) 50. 4. 9aÿ< 90 Þ aÿ< 0 Þÿÿaÿis an acute angle. cos 9a sin a Q µ Þ cos 9a cos B Q Þ 9a B Þÿÿa π 0 \ tan 5a tan π 4. SECTION B 5. True, because out of any two consecutive positive integers, one is even and the other one is odd; and the product of an even and an odd is even. 6. (i) We know that the factors of a prime are and the prime itself only. Therefore, the common factor of p and q will be only. Hence, HCF (p, q). (ii) LCM (p, q) pq 7. No, if two zeroes are a and b of polynomial x + kx + k, then a + b k and a. b k Þ a k and a k (when a b) Þ a k and a k. k Þ k (Comparing both) 4 Þ k 4k ÿþ k 4k 0 k( k 4) 0 ÿ\ k 4, 0. 8. For infinitely many solutions, 9 k k 6 k +4 Þÿ 9 k 6 and 9 k k +4 Þÿk and k k + 4 Þÿk and k, i.e., k. 9. Yes. Here, 6 4 + 0 676 Þ AC AB + BC \ÿdabc is a right triangle. OR No, ' DFED ~ DSTU Corresponding sides of the similar triangles are in equal ratio. \ DE TU EF ST \ DE ST ¹ EF TU. 0. x i f i f i x i 6 8 5 8 40 7 5 05 9 p 9p 8 88 4 5 Sf i p + 4 Sf i x i 9p + 0 99

Mean fx i i f i Þ 7.5 9p + 0 p + 4 Þ 7.5p + 07.5 9p + 0 ÿþ.5p 4.5 \ p. SECTION C. On the contrary let us assume that is a rational number. Then, we can take coprime a and b such that a b a Þ + 8 6 b (Squaring both sides) a Þ 6 0 b ÿþÿ 0 b a 6. b Since, a and b are integers, therefore, RHS of this last equation is rational and so LHS must be rational. But this contradicts the fact that 6 is irrational. This contradiction has arisen due to incorrect assumption that is a rational number. So, we conclude that is an irrational number.. Let a be any odd positive integer. Then it is of the form 6m +, 6m + or 6m +5, where m is an integer. Here, cases arise. Case I. When a 6m +, a (6m + ) 6m + m + m (m +)+ 6q +, where q m (m + ). Case II. When a 6m +, a (6m + ) 6m + 6m + 9 6m + 6m + 6 + 6(6m + 6m + ) + 6q +, where q 6m +6m +. Case III. When a 6m + 5, a (6m + 5) 6m + 60m +5 6m + 60m + 4 + (m + 5m + ) + 6q +, where q (m + 5m + ). Hence, a is of the form 6q + or 6q +. OR As: 0 408 + 6...(i) 408 6 + 9...(ii) 6 9 + 4...(iii) 9 4 8 + 0...(iv) Þ HCF 4 \ From (iii) Þ 4 6 9 6 [408 6] [Using (ii)] 6 408 [0 408] 408 [Using (i)] Þ 4 0 5 408 Þ m.. Since x 5 is a factor of f(x) x 5 x + x 5, so as f(x) may be rewritten. f(x) x 5 x + x 5 x 5 x 5 x + 0x + x 5 x 5 x( x 5) + ( x 5) x ( x 5 x + ) x É 5 ( 5) To find zeroes of f (x), put f (x) 0. ( x 5) ( 5 + ) x x 0 Þ x 5 0 or x 5 + 0 Þ x 5 or x 5 ± 04 Þ x 5 ± 5 or x Þ x 5 or x 5 + or 5 00 MATHEMATICS X

Hence all the zeroes of f(x) are 5, 5 + and 5. 4. The given pair of equations may be rewritten as x+ y xy ; x y 0 xy i.e, x + y 4 ; x + y 0 Adding this last pair, we get y Þÿ y Substituting y in the first equation of last pair, we get x + 4 ÿþÿ x 4 Þÿ x \ x Hence, x, y is the required solution. 5. Let the length of each side of the given equilateral triangle be a, then AB BC CA a...(i) \ a BD...(ii) Draw AP ^ BC to meet BC at P. P will be the mid-point of BC, that is a BP...(iii) a a a \ DP BP BD...(iv) 6 [Using (ii) and (iii)] Now, in right-angled triangle APB, AP AB BP Þÿ AP a a 4 PRACTICE PAPER Þ AP a 4 Also, in right-angled triangle APD, AD AP + DP...(v) Þ AD a a 7a a + 4 6 6 [From (iv) and (v)] Þ 6AD 8a Þ 9AD 7a Þ 9AD 7AB. Hence proved. 6. ÐBAD 90 ÐCAD (QÿÐBAC 90 ) 90 (90 ÐACD) (QÿÐADC 90 ) ÞÐBAD ÐACD...(i) ÐBDA ÐADC 90...(ii) Using equations (i) and (ii) in DABD and DCAD, we have DABD ~ DCAD (AA rule of similarity) Þ BD AD AD CD (Corresponding parts) Þ BD. CD AD. OR Given: AB P CD; AB CD To find: ar (% AOB) ar (% COD) As Ð Ð [Q Alternate interior angles] Ð ÿð4 Þ DAOB ~ DCOD (AA Criteria) ar ( \ % AOB) ar (% COD) 4. AB CD AB µ CD µ CD CD (Q AB CD) 0

7. cos θ sin θ cos θ + sin θ + cos R sin R cos R cos R Þÿ cos R sin R cos R cos R Þ tan R tan R + Þ tan q ÿ \ÿÿÿqÿ 60. 8. In DABC, tan A BC AB \ AC () ( ) 8 sin A, cos A, sin C, cos C Now, sin A. cos C + cos A. sin C + 9. True, LHS 9 + 8 9. cos 80 + cos 59 cosec sin0 cos(90 0 ) sin0 + cos 59 cosec (90 59 ) sin 0 + cos 59 sec 59 sin 0 cos 59 + cos 59 + Hence, the given equation is valid. 0. The given data is C.I. 0-5 5-0 0-5 5-0 0-5 5-0 0-5 f 0 5 0 80 40 0 5 From the table, maximum occuring frequency is 80. So, modal class is 5-0. Q Mode l + f 0 µ f f0 f f h Here, l 5, f 80, f 0 0, f 40, h 5 \ Mode 5 + 80 0 5 60 0 40 5 + 50 90 7.78 Hence, modal size is 7.78 hectares. OR Class Frequency Cumulative interval ( f) Frequency (cf) 0-8 8 8 8-6 0 8 6-4 6 4 4-4 58-40 5 7 40-48 7 80 N 80 For median class, N 80 40 In the cumulative frequency column, 58 is just greater than 40. So, 4- is the median class. Here, l 4, cf 4, f 4, N 40, h 8 Using formula: N cf Median l + h f žÿ ž 4 + 40 4 ž 8 6 žÿ 4 Hence, median of the given distribution is 6. 0 MATHEMATICS X

PRACTICE PAPER SECTION D. Let f (x) x +x + 6 x +9x +4x +6 (Spliting middle term) x( x +)+ ( x +) (x + )( x +) To find the zeroes of f (x), we have x + 0 or x +0 Þ x or \ Zeroes of the given polynomial are and. Now,sum of zeroes ÿ + Coefficient of x Coefficient of x Product of zeroes 6 Constant term Coefficient of x. Hence proved.. (i) Let speed of the train be x km/hr and that of the bus be y km/hr. Distance Time Speed Case I. According to question, we get Þ 60 x + 00 60 y 4 Þ 60 x + 40 y 4 5 Þ x + 60 y Þ x + 4 y 5...(i) Case II. According to the given conditions, we get 00 00 00 x y ¹ 4 + 0 60 Þ 00 x + 00 y 5 6 Þ 4 x + 8 y 6 Dividing by 4, we get x + y 4 Use equation (i) equation (ii), 4 y y 5 4 y 85 0 0 \ y 0 80 km/hr Put y 80 in equation (i), we get... (ii) x + 4 80 5 Þ x 5 0 4 60 60 \ x 60 km/hr. Hence, speed of the train 60 km/hr and speed of the bus 80 km/hr. (ii) Solution of system of linear equations in two variables. (iii) By opting for public transport it depicts that she is a responsibile citizen, so her responsibility and rationality have been depicted here. OR (i) Let l length of the rectangle b breadth of the rectangle According to question, (l + 7)(b ) lb (i) (l 7)(b + 5) lb (ii) From equation (i), lb + 7b l lb Þ 7b l (iii) From equation (ii), lb 7b + 5l 5 lb Þ 7b + 5l 5 (iv) Adding equations (iii) and (iv), we get l 56 Þÿ l 8 m Putting the value of l in equation (iii), we get b 5 m. \ l 8 m, b 5 m. (ii) Solution of system of linear equations in two variables. (iii) Love for environment and human beings. 0

. Table for values of x and y as regarding equation x + y 5 0 is x 0 y 5 Similarly table for equation x y 5 0 is x 0 y 5 Let us draw the graph of lines using the tables obtained above. Comparing equations (i) and (ii), we get AP MQ AQ NP Þ AP AQ NP MQ Further, ar(dbpq)...(iii) PB MQ...(iv) And ar(dcqp) QC NP...(v) But triangles BPQ and CQP are on the same base PQ and between the same parallels PQ and BC, so their areas must be equal. i.e., ar(dbpq) ar(dcqp) Þ PB MQ QC NP The lines intersect y-axis at (0, 5) and (0, 5). 4. We are given a triangle ABC in which a line PQ parallel to BC is drawn to intersect the sides AB and AC at P and Q respectively. We need to prove, AP PB AQ QC Join PC and QB. Draw QM ^ AB and PN ^ AC. Þ [Using equations (iv) and (v)] PB QC NP MQ From, equations (iii) and (vi), we get...(vi) AP AQ PB QC Þ AP PB AQ QC. Hence proved. 5. We are given two triangles ABC and PQR such that DABC ~ DPQR. Draw perpendiculars AD and PM on BC and QR respectively. Now, area of a triangle base height \ ar(dapq) AP MQ...(i) Also ar(dapq) AQ NP...(ii) We need to prove ar( ΔABC) AD ar( ΔPQR) PM In DABD and DPQM, ÐADB ÐPMQ 90 ÐABD ÐPQM (... ÿdabc ~ DPQR) \ DABD ~ DPQM (AA criterion of similarity) Þ AB PQ AD...(i) (Corresponding sides) PM 04 MATHEMATICS X

\ We know that the ratio of areas of two similar triangles is equal to ratio of squares of their corresponding sides ar( ΔABC) AB ÿ...(ii) ar( ΔPQR) PQ From equations (i) and (ii), we have ar( ΔABC) ar( ΔPQR) AD ÿ PM. Hence proved. 6. sin q + cos q (Given) Þ sin q + cos q + sin q cos q (Squaring both the sides) Þ + sin qÿcos q Þ sin qÿcos q...(i) Now, tan qÿ+ cot q sin R cos R cos R sin R R R sin cos [Using (i)] sin Rcos R. Hence proved. 7. Let us take LHS of the given identity. tan R cos ec R tan R sec R cosec R sin R cos R sin R + sin R cos R cos R sin R sin R cos R sin R R R R R R R R sin cos sin cos cos sin.cos sin R cos R R R R R sin cos sin cos sin Rcos R sin R cos R sin R cos R [... ÿsin q + cos q ] RHS. Hence proved. OR tan A n tan B n Þ cot B and sin A m sin B tan A Þ sin B m sin A ÿþÿcosec B m sin A... cosec B cot B m n m n cos A Þ Þÿ sin A tan A sin A Þ m (n ) cos A m Þ cos A.Hence proved n 8. m n (tan q + sin q) (tan q sin q) (tan q + sin q + tan qÿ sin q) (tan qÿ+ sin qÿ tan q + sin q) [Q A B (A + B) (A B)] tan qÿ. sin q 4sin qÿ tan q...(i) 4 mn 4 (tan R sin R)(tan R sin R) 4 tan Rsin R [Q (A + B) (A B) A B ] sin R 4 sin cos R R 4 sin q cos R 4sin θ sec θ 4 sin q tan q... (ii) From (i) and (ii), m n 4 mn. 9. Let us prepare the cumulative frequency table by more than method as given below: Production Production yeild f i yeild (in kg/ha) cf Point (in kg/ha) more than or equal to 50-55 50 50 (50, 50) 55-60 6 55 48 (55, 48) 60-65 8 60 4 (60, 4) 65-70 4 65 4 (65, 4) 70-75 5 70 0 (70, 0) 75-80 5 75 5 (75, 5) Total 50 PRACTICE PAPER 05

We plot the points mentioning in the table such that lower class limits are on the x-axis and the cumulative frequencies are on the y-axis. By joining these points by free hand smooth curve, We obtain more than type ogive as shown in the adjoining graph. N 60 ÿÿÿÿþÿ x + y + 45 60 Þ x + y 5... (i) N cf Median l + ž q h Ÿ f 0 5 + x Þ 8.5 0 + 0 0 ± On simplifying, x 8...(ii) From equations (i) and (ii), we have x 8 and y 7.. We will use the step-deviation method. Marks No. of Class- d i u i f i u i di (C.I.) students mark x i 45 (f 0 i ) x i To obtain median from graph: Draw a line parallel to x-axis passing through y 50 5. This line meets the ogive at (68., 5). \ÿ Median 68.. 0. Given that median is 8.5. It lies in the class interval 0-0, so 0-0 is the median class. Let us prepare frequency distribution table. C. I. f cf 0-0 5 5 0-0 x 5+x 0-0 0 5 + x 0-40 5 40 + x 40-50 y 40 + x + y 50-60 5 45 + x + y x + y +45 0-0 5 5 40 4 0 0-0 4 5 0 0-0 8 5 0 6 0-40 5 0 40-50 6 45 0 0 0 50-60 5 55 0 5 60-70 0 65 0 0 70-80 8 75 0 4 80-90 5 85 40 4 0 90-00 95 50 5 0 Sf i 85 ÿsf i u i 9 Let a 45. Here h 0 Mean a + fu i i µ s fi 45 + 9 85 0 45 +.4 48.4. h 06 MATHEMATICS X

. (C)... 4 5 7 PRACTICE PAPER Practice Paper- SECTIONA 4 5 4 000 0.00.. (D) The condition for the line parallel is:. As 4. a a b c 5 4. c b c c 5 ar( ΔABC) BC ar( ΔDEF) EF (Result on areas of similar triangles) D B A C 54 ar( ΔDEF) 4 o o o cos (90 70 ) sin 70 sin 70 sin 70 o o E + k F ar(δdef) 96 cm. + cos θ cos θ k k SECTIONB k 6. 5. False, because or is an irrational and sum or difference of a rational number and an irrational number is an irrational number. 6. No, because prime factorisation of any number of type n can not have 5 as one of its prime factor. 7. False, since the discriminant is zero for k ±. 8. The condition for infinitely many solutions a is a b b c c. ( a +) b 5 a b ( a +) and 5 a 5 b b 6a 5a 5 and 9b 0b 5 a 8 and b 5. 9. Yes, because ΔPBC and ΔPDE are similar by SAS rule. As BP DP PC PE and BPC DPE. OR In ΔAQO and ΔBPO, QAO PBO (Each 90 ) AOQ BOP (Vertically opposite angles) So, by AA rule of similarity, ΔAQO ~ ΔBPO AQ BP AQ 9 AO BO 0 6 AQ 5 cm. AQ 0 9 6 0. The empirical relationship among the three measures of central tendency is: Mode Median Mean 55 50 65 00 65. SECTIONC. To prove that the pair of numbers (847, 60) is coprime by using Euclid s algorithm, we have to prove that the highest common factor of the pair is. Since 60 > 847 60 847 + 466 Since the remainder 466 0. 847 466 + 8 Since the new remainder 8 0. 07

466 8 + 85 Since the new remainder 85 0. 8 85 4 + 4 Since the new remainder 4 0. 85 4 + Since the new remainder 0. 4 + Since the new remainder 0. + Since the new remainder 0. + 0 Since, the remainder has now become zero, the divisor at this stage is, the HCF of 847 and 60 is. OR Let us assume, to the contrary, that + 5 is rational. So we can find coprimes a and b such that + 5 a b Rearranging, 5 a b b a and b are integers a b is an integer a b b is rational number 5 should be rational. But we know that 5 is irrational. So our assumption that + 5 is rational is wrong. Hence + 5 is irrational.. Let x be any positive integer. Then it is either of the form q or q + or q +. Case I. x q Cubing both sides, we get x (q) 7q 9m (i) m q Case II. x q + Cubing both sides, we get x (q + ) 7q + + (q + ) q 7q + 7q + 9q + 9q (q + q + ) + 9m + (ii) m q(q + q + ). Case III. x q + Cubing both sides, we get x (q +) 7q + 8 + (q + ) 6q 7q + 54q + 6q + 8 9q (q + 6q + 4) + 8 9m + 8; (iii) m q (q +6q + 4). Thus, from equations (i), (ii) and (iii), it is clear that cube of any positive integer is either of the form 9m, or 9m + or 9m + 8.. We have time distance speed Time taken by Abhay to cover one complete round 60 0 hours Time taken by Ravi to cover one complete round 60 4 hours 5 Abhay and Ravi reach the starting point respectively after 0 hours and 4 hours, and their respective multiples. Therefore, they will meet again at the starting point after the time given by least common multiple of 0 hours and 4 hours Let us determine the LCM of 0 hours and 4 hours. 0 5, 4 LCM 5 0 Hence, the required time is 0 hours. 4. Let α and β be the zeroes of 6x + x + k. α + β (α + β) αβ k 6 6 k 6 But it is given that α + β 5 6 5 6 6 k k 4 k. 6 08 MATHEMATICS X

OR Since a is a zero of a a 0a + 4, therefore a a 0a + 4 is divisible by a. Further the obtained quotient will provide the other two zeroes. Add equations (i) and (ii) to get PR +QR 4PS SM(PS SQ) 4 PQ PR + QR PQ. OR (Q PS SQ) Hence proved. a a (a 4)(a +) For other zeroes, put a 4 0 and a + 0 a, 4 Thus, the other two zeroes are and 4. 5. In right-angled triangle PQS, PS PQ + QS QS PS PQ QR 4 PS PQ QR 4PS 4PQ (i) In right-angled triangle PQR, PR PQ + QR PQ + 4PS 4PQ [Using (i)] PR 4PS PQ. Hence proved. 6. Draw RM PQ. In ΔPRS, PSR > 90 PR PS + RS + PS. SM...(i) [Using result on obtuse-angled triangle] In ΔQRS, QSR < 90 QR RS + SQ SQ.SM...(ii) [Using result on acute-angled triangle] AB Given: PQ AD, PM are medians. To show: ΔABC ~ ΔPQR. AC PR AD PM and Construction: Extend AD up to D' such that AD DD' and extend PM up to M' such that PM MM' Join BD', D'C; QM', M' R. Proof: As AD DD' and BD DC and PM MM' and QM MR ABD' C and PQM' R are Pgm AB D' C and PQ RM' {Q Opposite sides of a Pgm are equal} As it is given that: AB PQ AC PR AD PM DC AC AD AD RM PR PM PM ΔACD' ~ ΔPRM' (SSS criteria) (CPCT) PRACTICE PAPER 09

Similarly 4 + + 4 BAC QPR Now: In ΔABC and ΔPQR AB PQ AC PR (i) and BAC QPR (ii) ΔABC ~ ΔPQR (SAS) Hence proved. 7. sin θ 4 Q sin θ + cos θ cos θ sin θ cos θ 4... (i) Given [Using (i)] cos 7 θ 6 sec θ 6... (ii) 7 Let us take LHS of the given equation. LHS cosec R cot R sec R sec θ [... cosec θ cot θ ] 7 6 9 7 [Using (ii)] 7 RHS. Hence proved. 8. We have sin 60, cos 60, sec 0, cosec 0 +sin 60 +cosec 0 5 + cos 60 + sec 0 + + () 5+ + Multiplying Num. and Deno. by, + +4 4 4 5+ + 4 6 + 9 + 48 60 + + 6 9 79. 9. Consider, left hand side of the given equation, tan θ cot θ + cotθ tanθ tan θ tan θ + tanθ tan θ tan θ + tan θ tan θ tan θ tan R tan R Étan R ( ) ( ) tan θ ( tan θ+ + tan θ) ( θ ) tan θ tan [Q a b (a b) (a + b + ab] sin θ cosθ tan θ + + + + tan θ cos θ sin θ sin θ+ cos θ + sin θ cosθ sec θ. cosec θ + RHS. Hence proved. 0. Let us convert the more than type distribution to the normal distribution. Marks No. of students 0-0 0-40 7 40-60 0 60-80 5 80-00 5 We observe from the table that the value 0 is the maximum frequency. So, the modal class is 40-60. Now, mode l + f f0 h f f f ¹ 0 Here, l 40, f 0, f 0 7, f 5, h 0 0 7 Mode 40 + 0 40 7 5 0 MATHEMATICS X

40 + 60 8 40 + 4.44 54.44 marks. SECTION D. To solve a system of equations graphically we need atleast two solutions of each equation. Two solutions of the equation x y are given in the following table: x 0 y 4 Two solutions of the equation 4x y 8 are given in the following table: x y 0 4 Let us draw the graph of the two given equations. X' Y' From the graph, it is clear that the two lines intersect each other at the point (, 4). Hence, the solution is x, y 4.. (i) First we divide x 4 + x + 8x + ax + b by x + as follows: x x7 4 x x x 8x axb x 4 x x 7x axb x x 7 x ( a) xb 7x 7 ( a) x( b7) Since, x 4 + x + 8x + ax + b is divisible by x +, therefore remainder 0 i.e., (a )x + (b 7) 0 or (a )x + (b 7) 0.x + 0 Equating the corresponding terms, we have a 0 and b 7 0 i.e., a and b 7. (ii) Common good, Social responsibility.. Let the speed of rowing in still water and the speed of the current be u km/hr and v km/hr respectively. The speed of rowing in downstream (u + v) km/hr. The speed of rowing upstream (u v) km/hr. Using the formula: Distance Speed Time According to first condition of the question, 8 u+ v +... (i) u v According to second condition of the question, 6 u+ v + 40 8... (ii) u v Let us put, u+ v x and u v y such that equations (i) and (ii) reduce to 8x + y... (iii) And 6x + 40y 8... (iv) PRACTICE PAPER

Equations (iii) and (iv) form a pair of linear equations. Multiply equation (iii) by and subtract the result from equation (iv) to get 6y y 8 Substitute y in equation (iv) to get 8 6x 8 5 x Q x u+ v u + v Q y u v u v 8 This last system gives u 0 and v Hence, the speed of the rowing in still water 0 km/hr and the speed of the current km/hr. 4. ar(δaxy) ar(bxyc).ar(δaxy) ar(bxyc) + ar(δaxy) ar(δabc) B X ar ( ΔABC) ar ( Δ AXY) As ΔABC ~ ΔAXY AB ar( ΔABC) AX ar( ΔAXY) AB AX AX AB AB BX AB BX AB BX AB. 5. Hint: A Y C ΔABC ~ ΔPQR AB PQ AB PQ AB PQ BC and B Q QR BD and B Q QM BD and B Q QM ΔABD ~ ΔPQM AB AD PQ PM 6. m cosec θ sin θ (Given) m sin θ sin θ sin θ sin θ cos θ m sin θ... (i) (Q sin θ cos θ) 4 m cos θ sin θ... (ii) (Squaring) Further, n sec θ cos θ (Given) cos θ n cosθ n cos θ cos θ 4 sin θ n cos θ... (iii) sin θ n... (iv) cos θ Multiplying equations (ii) and (iii), we get 4 m cos θ sin θ n cos θ sin θ cos θ (m n) / cos θ... (v) Multiplying equations (i) and (iv), we get 4 n sin θ m cos θ cos θ sin θ sin θ (n m) / sin θ... (vi) Adding equations (v) and (vi), we get (m n) / + (n m) / cos θ + sin θ i.e., (m n) / + (n m) /. Hence proved. 7. Given equations are: sin θ + cos θ p... (i) and sec θ + cosec θ q... (ii) Squaring both the sides of equation (i), we get sin θ + cos θ + sin θ cos θ p Subtract unity from both the sides to get p sin θ cos θ... (iii) MATHEMATICS X

Equation (ii) can be written as q cos θ + sin θ q sin RcosR sin R cos R From equations (iii) and (iv), we get... (iv) q (p ) sin R cos R sin θ cos θ sin R cos R q (p ) (sin θ + cos θ) q (p ) p. Hence proved. OR As x a sin θ a x sin θ and y b tan θ b y tan θ LHS a x b y sin θ cos sin sin θ tan θ cos θ sin θ sin θ sin θ RHS Hence proved. 8. As A + B + C 80 B + C 80 A B + C 80 A 90 A sin B + C sin A 90 cos A B C sin + A cos (i) Now, sin A + B + C sin sin A + cos A [Using (i)]. Hence Proved. 9. The cumulative frequency table for the given data is given below: θ θ Marks No. of students Cumulative (C.I.) ( f ) frequency (cf ) 0-0 0 0 0-0 f 0 + f 0-0 5 5 + f 0-40 0 65 + f 40-50 f 65 + f + f 50-60 0 7 + f + f N 75 + f + f Clearly, N 75 + f + f But N 00 f + f 5... (i) N 50. The median is which lies in the class 0-40. So, l 0, f 0, cf 5 + f, h 0. Using the formula: N µ cf Median l + h f 50 5 f µ 0 + 0 0 0 5 f 0 75 5f 0 75 0 f 5 f 9 Substituting f 9 in equation (i), we get 9 + f 5 f 6 Hence, f 9 and f 6. 0. Class-interval Frequency (C.I.) ( f) 0-0 0-0 9 0-0 5 0-40 0 40-50 8 50-60 5 PRACTICE PAPER

The class corresponding to the maximum frequency is 0-40. So, 0-40 is the modal class. Mode l + f f 0 µ f f0 f h Here, l 0, f 0, f 0 5, f 8 and h 0 \ Mode 0 + 0 + 0 + 0 5 µ t058 0 5 µ 60 5 8 0 5 µ 60 0 0 + 5 50 0 0 + 7 7 0 + 5.5 5.5 OR Σfiui Mean A + h 60 + 0 4 Σfi 50 60 + 5.6 65.6 Hence, the required arithmetic mean is 65.6.. Let us convert the given data into less than type distribution. Class f Lifetimes cf interval (in hrs.) 0-0 0 less than 0 0 0-40 5 less than 40 45 40-60 5 less than 60 97 60-80 6 less than 80 58 80-00 8 less than 00 96 00-0 9 less than 0 5 Class Mid- Freq- u i f i u i interval values uency x i A h (x i ) (f i ) 0-0 0 5 0-50 40 8 50-70 60 A 70-90 80 0 90-0 00 0-0 0 40 0 0 0 0 8 0 0 0 0 0 0 0 40 0 6 60 0 6 Σf i 50 Σf i u i 4 Let assumed mean be A 60 Here, h 0 4 MATHEMATICS X

We mark the upper class limits along the x-axis with a suitable scale and the cumulative frequencies along the y-axis with a suitable scale. For this, we plot the points A(0, 0), B(40, 45), C(60, 97), D(80, 58), E(00, 96) and F(0, 5) on a graph paper. These points are joined by a free hand smooth curve to obtain a less than type ogive as shown in the given figure. Practice Paper- SECTIONA. (D) There are infinitely many real numbers of both types rational and irrational between and 5.. (B) For infinite number of solutions: k 5, i.e., k 6. 0. DABC ~ DDEF 6. LCM First number Second number HCF 96 404 4 404 9696. 4 7. True, because we find the remainder zero when x 4 + 5x 7x + x + is divided by x + x +. 8. Infinite number of solutions because the system obeys the following condition: a b c a b c, i.e., 9 9. 9. In DAOD and DCOB, AO OC DO OB and ÿðaod ÿðcob Þ DAOD ~ DCOB AD \ BC ÿ ÿÿþ ÿ 4 BC ÿ Þ BC 8 cm. OR Let the height of the tower be h metres DABC ~ DPQR ÐB ÐE 80 (40 + 65 ) 75. 4. AC BC AB Þ AC AC cosec C BC AB. SECTIONB 5. 8 n can be rewritten as n. Clearly, the prime factor of 8 n is only. To end with the digit 0, one of the prime factors of 8 n must be 5. Hence, 8 n cannot end with the digit zero for any n ÎN. PRACTICE PAPER AB Þ PQ BC QR Þÿÿ h 8 40 40 Þ h 60 metres. 8 0. Since the maximum frequency is 8, so modal class is 4-8. \ l 4, f 8, f 0 4, f 5, h 4 Now, mode l + f f0 f f f 0 h 84 4 + 4 4 +.9 6 4 5 6.9. 5

SECTIONC. Let us assume on the contrary that is a rational number. Then can be written as a, where a and b are coprime b and b ¹ 0. a b (Squaring) Þ a b... (i) Þÿ a is divisible by...(ii) Þ a is divisible by...(iii) [If a prime (here ) divides d, then the same prime divides d, where d is a positive integer.] Þ a c Þ a 4c... (iv) From (i) and (iv), we get 4c b Þ b c Þ b is divisible by Þ b is divisible by...(v) From results (ii) and (v), we have a and b both are divisible by. But this contradict the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that is a rational number. Thus, we conclude that is an irrational number. OR (i) The given fraction can be written as 4 4 5 0.05 4 4 5 0 Hence, the given number terminates after four places of decimal. (ii) The given fraction can be written as 59 59 5 5 5 5 5 4 5744 00000 0.05744 Hence, the given number terminates after five places of decimal.. (i) LCM of,, 4, 5, 6 60 \ Required number 60p + ; p is positive integer (7 8 + 4)p + (7 8p) + (4p + ) Now, this number is to be divisible by 7. Since 7 8p is always divisible by 7; so, we must choose the least value of p which will make 4p + divisible by 7. Putting p,,, 4, 5 etc. in succession we find that p 5. \ Required number 60p + 60 5 + 0 (ii) LCM of two real numbers. (iii) Kindness and love towards society.. p(t) t 5 To obtain zeroes of p(t), put p(t) 0 i.e., t 5 0 t ( 5 ) 0 Þ ( + 5 )( 5 ) Þ t 5, 5 t t 0 So, zeroes of p(t) are 5 and 5 Sum of zeroes 5 + 5 0 0 Coefficient of t Coefficient of t Product of zeroes 5 5 5 5 Constant term Coefficient of t. Hence verified. 4. The given system of equations can be rewritten as 4x + y 48 0 40x 6y 9 0 Applying the method of cross multiplication to solve the system. x Þÿ 576 88 x Þ 864 y 768 + 90 y 5 44 4 0 Þ x 864 5 and y 44 44 Þ x 6 and y 8. 5. We are given a square of side length a. Then length of its diagonal will be a. We know that area of an equilateral triangle of side length x is 4 x. 6 MATHEMATICS X

\ Area of the equilateral triangle described of the side of the square. A side 4 a... (i)ÿÿþÿÿa 4 A side And area of the equilateral triangle described on the diagonal of the square A diagonal 4 (a ) a 4 Aside ÞA diagonal [Using (i)] Þ A side A diagonal. Hence proved. 6. In the figure drawn, AB DC and DAED ~ DBEC. DADC and DBDC both are on the same base DC and lie between same parallels AB and DC. So, ar(dadc) ar(dbdc) Þÿ ar(daed) + ar(ddec) ar(dbec) + ar(ddec) Þ ar(daed) ar(dbec) (i) ar( % AED) (AD) Now, ar( % BEC) (BC) (Q DAED ~ DBEC) Þ ( AD) ( BC) [From (i)] Þ AD BC. Hence proved. 7. Given expression 4 4 «º µ µ» ¼ + 6 6 { } ( ) + () ( + ) + 4 + 4 4 + 4 4. 8. Consider left hand side of the given equation. LHS (cosec A sin A) (sec A cos A) sina cosa sin A cos A sin A cos A. sin A cos A cos A sin A. sin A cos A sin A cos A Also, taking right hand side, RHS tan A + cot A sin A cos A + cosa sin A sin A cos A sin A cos A sin A + cos A Hence, LHS RHS. OR Draw DÿABC with AB BC AC a (say) Draw AD ^ BC \ ÐBAD ÐDAC qÿ 0 and BD DC a/ BD a/ \ sin qÿÿ AB a ÿþ sin 0 ÿÿ. 9. Given equations are: sin qÿ + cos qÿ p... (i) and sec qÿ + cosec qÿ q... (ii) Squaring both the sides of equation (i), we get sin qÿ+ cos qÿ+ÿsin qÿcos qÿ p Subtract unity from both the sides to get p sin qÿcos q... (iii) Equation (ii) can be written as q cos θ + sin θ sin RcosR Þ q... (iv) sin R cos R From equations (iii) and (iv), we get q (p ) sin R cos R sin q cos q sin R cos R Þ q (p ) (sin q + cos q) Þ q (p ) p. Hence proved. PRACTICE PAPER 7

0. Let us use step-deviation method to obtain the mean. C.I. f i x i d i d u i i h x i a d i u i 0-0 7 0 40 4 0-40 p 0 0 p 40-60 50 0 0 0 60-80 4 70 0 4 80-00 9 90 40 8 Sf i ÿÿÿÿÿsf i u i 9+p 8 p Here a 50, h 0 fu i i Mean a +ÿ h fi Þ 50 50 + 8 p 9 + p Þ 8 p 0 \ p 8. OR Since mode 6, which lies in the class interval 0-40, so the modal class is 0-40. \ÿf 6, f 0 f, f, l 0 and h 0. Now, mode l + Þ 6 0 + Þ 6 0 6 0 Þ 0 6f 60 0f Þ 4f 40 Þÿÿÿÿÿf 0. f 0 f f f µ 0 f 6 f f µ SECTION D. We have x sec A cos A cosa cos A f f 0 h cos A cos A \ x sin A cos A Similarly, y cosec A sin A sin A sina sin A sin A y cos A sin A LHS x y (x + y ) x 4 y + x y 4 4 ª º ª º «sin A» «cos A» «cos A» «sin A +» «¹» «¹» ¼ ¼ sin A cos A cos A sin A 4 sin A cos A cos A sin A + 4 4 sin A cos A cos A sin A 4 8 8 4 sin A cos A 4 cos A sin A + sin A cos A 4 cos A sin A 6 6 ( sin A) ( cos A) + sin A + cos A RHS. Hence proved.. Let the original fraction be x y. On adding to both the numerator and the denominator of x y, it becomes 4 5 x + i.e., 4, i.e., 5x + 5 4y + 4 y + 5 i.e., 5x 4y...(i) On subtracting 5 from both the numerator and the denominator of x y, it becomes 8 MATHEMATICS X

i.e., x 5 y 5, i.e., x 0 y 5 i.e., 8x 4y 0...(ii) Subtracting equation (i) from equation (ii), we get x Þ x 7 Substituting x 7 in equation (i), we get 5 7 4y Þ y 9 Hence, the required fraction is 7 9. OR Let incomes of X and Y be 8x and 7x respectively; and their expenditures be 9y and 6y respectively. We know that: Income Expenditure Savings... 8x 9y 50 (i) and 7x 6y 50 (ii) Comparing equations (i) and (ii), we have 8x 9y 7x 6y or x y (iii) Substituting this value of x in equation (i), we get 4y 9y 50 Þ 5y 50 Þ y 50 Substituting this value of y in equation (iii), we get... x 50 750. Now, 8x 8 750 6000 and 7x 7 750 550 hence X s income is 6000 and Y s income is 550.. Let us make the table for the values of x and corresponding values of y to the equation x + y 8 0 x 4 y 4 0 Similarly, for the equation x y 0 x 4 y Let us draw the graph. From the graph, the lines intersect each other at the point A(, ). Therefore, the solution is x, y. The lines intersect the y-axis at B(0, 8) and C(0, ). To find the area of the shaded portion, that is, DABC, draw perpendicular AM from A on the y-axis to intersect it in M. Now, AM units and BC 8 + 9 units.... ar(dabc) BC AM 9 7 sq. units Hence, x, y ; area.5 sq.units. 4. Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Proof: We are given, a DABC in which ÐA 90 We need to prove BC AB +AC. Draw AD ^ BC (see figure) In DABC and DDBA, ÐABC ÐDBA (Common) PRACTICE PAPER 9

ÐBAC ÐBDA (Each 90 ) So, DABC ~ DDBA (A A criterion of similarity) AB Therefore, BD BC AB or AB BD. BC...(i) Similarly, DABC ~ DDAC Therefore, AC DC BC AC AB DC. BC...(ii) Adding equations (i) and (ii), we get BD. BC + DC. BC AB + AC or (BD + DC). BC AB + AC or BC. BC AB + AC (Q BC BD + DC) or BC AB + AC. Hence proved. OR Converse of Pythagoras Theorem: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Proof: We are given a DABC in which AC AB + BC...(i) We need to prove ÐABC 90. Let us construct a DPQR such that ÐPQR 90 and PQ AB...(ii) QR BC...(iii) Using Pythagoras Theorem in DPQR, we have PR PQ + QR \ PR AB + BC...(iv) [Using equations (ii) and (iii)] From equations (i) and (iv), we have AC PR...(v) Now, in DABC and DPQR, AB PQ (From (ii)] BC QR [From (iii)] AC PR [(From (v)] So, DABC @ DPQR (SSS congruence) \ ÐABC ÐPQR (CPCT) But ÐPQR 90 (By construction) \ ÐABC 90. Hence proved. 5. Let the given parallelogram be ABCD We need to prove that AC + BD AB + BC + CD + DA Let us draw perpendiculars DN on AB and CM on AB produced as shown in figure. In DBMC and DAND, BC AD (Opposite sides of a gm ) ÐBMC ÐAND (Each 90 ) CM DN (Distance between same parallels) \ DBMC @ DAND (RHS criterion) Þ BM AN...(i) (CPCT) In right triangle ACM, AC AM + CM (AB + BM) +BC BM AB + AB. BM + BM + BC BM AB + BC + AB. BM...(ii) In right triangle BDN, BD BN + DN (AB AN) + (AD AN ) AB AB. AN + AN + AD AN BD AB +DA AB. AN Þ BD CD +DA AB. BM...(iii) [Using (i) and AB CD] Adding equations (ii) and (iii), we have AC + BD AB +BC +CD +DA. Hence proved. 6. Asÿa, b are zeros of p(x) x px + q. Þÿa + b p; ab q 0 MATHEMATICS X

\ Consider B C C B B C B C ( ) B C ( B C) BC q ª ¹ q 4 4 B C B C B C ( ) q q ( p q) q q 4 4 p 4q 4p qq p q 4p q q q p q 4 4p q RHS. Hence proved. 7. sec q x + 4x or sec q x + 4x or + tan q x + + 6x or tan q x + 6x or tan q x 4 x or tan q + x 4 x Add equations (i) and (ii) to get sec qÿ+ tan q x + or x + 4x + x 4x... (i) (Given) (Squaring)... (ii) 4x x + x or 4x x. Hence proved. 8. LHS ( + cot A cosec A) cos A sin A sin A ( + tan A + sec A) sin A cos A cos A µ µ sin A + cos A sin A cos A + sin A + cos A ( sin A + cos A ) sin A cos A sin A + sin A cos A + cos A sin A cos A sin A cos A RHS. Hence proved. sin A cos A 9. We prepare the cumulative frequency table by less than method as given below: Marks Freq. Marks cf Point less than 0-0 4 0 04 (0, 4) 0-0 0 0 4 (0, 4) 0-0 6 0 0 (0, 0) 0-40 40 5 (40, 5) 40-50 0 50 7 (50, 7) 50-60 8 60 90 (60, 90) 60-70 8 70 98 (70, 98) 70-80 80 00 (80, 00) We take marks on the x-axis and cumulative frequency on the y-axis and then plot the points mentioned in the table. On joining these points by free hand smooth curve, we get less than ogive. Further, we prepare the cumulative frequency table by more than method as given below: Marks Freq. Marks more than cf Point or equal to 0-0 4 0 00 (0, 00) 0-0 0 0 96 (0, 96) 0-0 6 0 86 (0, 86) 0-40 0 70 (0, 70) 40-50 0 40 48 (40, 48) 50-60 8 50 8 (50, 8) 60-70 8 60 0 (60, 0) 70-80 70 (70, ) We will plot the points mentioned in this table on the same graph. On joining these points by free hand smooth curve, we get more than ogive. PRACTICE PAPER

05 00 95 90 85 80 75 70 65 60 55 50 45 40 5 0 5 0 5 0 5 0 Y (0, 00) (0, 96) (0, 70) (0, 86) (40, 5) (0, 0) (0, 4) (70, 98) (40, 48) less than ogive (50, 7) (50, 8) (60, 90) more than ogive (60, 0) (80, 00) (0, 4) (9, 0) (70, ) 0 0 0 40 50 60 70 80 90 00 Median: The abscissa of the point of intersection of both the ogives determines the median of the given distribution. To find such abscissa, we draw a perpendicular from the point of intersection of both the ogives to the x-axis, which meet the axis at (9, 0). Hence the required median is 9 marks. 0. The given distribution can be again represented with the cumulative frequencies as given below: Class interval f i x i cf f i x i 00-0 0 0 0-40 4 0 6 80 40-60 8 50 4 00 60-80 6 70 40 00 80-00 0 90 50 900 50 760 X Σfx i i Mean: Mean Σfi Q Sf i 50 and Sf i x i 760 \ Mean 760 50 ` 45.0. N cf Median: Median l + h f Q N 50, N 5, f 4, cf, l 0 and h 0 \ Median 0 + 5 4 µ 0 ` 8.57 f f0 Mode: Mode l + h µ f f0 f Q l 0, f 4, f 0, f 8 and h 0 4 \ Mode 0 + µ 0 `5 s4 8. We prepare cumulative frequency table from the given data. C.I. Frequency Cumulative ( f ) Frequency (cf ) 0-8 8 8 8-6 0 8 6-4 6 4 4-4 58-40 5 7 40-48 7 80 N 80 Here, N 80 \ N 40 Cumulative frequency just more than 40 is 58. So 4- is the median class. \ l 4, cf 4, f 4, h 8 \ Median l + 40 4 4 + 4 N cf h f 8 4 + 48 4 6. MATHEMATICS X

. (D) Practice Paper-4 SECTIONA s 6 6 5 As 6 5 is irrational, it has nonterminating, non-repeating decimal form.. (C) + k and p +, i.e., k and p. So, p + k.. AD BD AE CE Þÿ DE BC k[x (5 + )x + (5)()] k(x + x 0), k being a real number. f(x) is not a unique polynomial as k is any real number. 8. Condition for infinite number of solutions: a a ÿ b b ÿÿ c i.e., k k 4( k ) ÿ, c 4 i.e., k 5. 9. No. In DAOB and DDOC, ÐAOB ÐDOC (Vertically opposite angles) DE AD ÞÿÿDADE~ÿDABC ÿÿ\ÿ BC AB x.5 Þÿÿÿÿ 7.5 4 ÿÿÿþÿ x 5 8 cm 4. If x 0, cos 0 4 cos 0 4 0. 8 SECTIONB 5. As p and p are common factors. n l \ÿhcf (x, y) is pp. 6. The required highest number will be the HCF of 0, 4 and 56. 0 5; 4 5 7; 56 8 Therefore, HCF 8. 7. Let the required polynomial be f(x). Then f(x) k[x (sum of zeroes) x + product of zeroes] PRACTICE PAPER4 AO OB x as 5 x 6, i.e., 5 x DO OC 0 5 Therefore, DAOB is not similar to DDOC. 0. Yes. Let us take left hand side of the given equation. LHS +sin cos R R cos + sin R cos R R sec q + tan qÿ ( + tan q) + tan q + tan qÿ RHS. OR Qÿ A + B + C 80º \ÿ LHS C + A 80º B cot cot cot (90º B ) tan B RHS. SECTIONC. Let us assume to the contrary that 4 5 is rational. Then we can take integers a and b ¹ 0 such that a b 4 5, i.e., 4 5 a b i.e., 5 b a 4b

RHS of this last equation is rational as a and b are integers, but LHS of it is irrational. This is an incorrect statement due to our wrong assumption that 4 5 is rational. So, we conclude that 4 5 is irrational.. Any positive integer is either of the form q, q + or q +. There are cases now: Case I. When n q, n + q + and n + q +. Here, only n is divisible by. Case II. When n q +, n + q + and n + q + (q + ) Here, only n + is divisible by. Case III. When n q +, n + q + (q +) and n +q +5(q + ) + Hence only n + is divisible by. OR Let x be any positive integer. Then x can be of the form: 5m, 5m +, 5m +, 5m +, 5m + 4 \ If x 5m Þ x 5m 5(5m ) 5q. If x 5m + Þÿÿÿx 5m + + 0m 5(5m + m) + 5q + If x 5m + Þÿ x 5m + 4 + 0m 5(5m + 4m) + 4 5q + 4 If x 5m + Þÿ x 5m + 9 + 0m 5(5m + 6m + ) + 4 5q + 4 If x 5m + 4 Þ x 5m + 6 + 40m 5(5m + 8m + ) + 5q + \ Square of x can be of the form: 5q, 5q + or 5q + 4.. By the division algorithm, Dividend Divisor Quotient + Remainder \ 6x +8x x + 8 g(x) (x + 4) + 6x + 0 Þ g(x) 6x 8x 9x x 4 x (x 4) (x4) x 4 (x 4)(x ) Þ g(x) x 4 Þ g(x) x. 4. Asÿa, b are zeros of p(x) x px + q. Þÿa + b p; ab q B C C B B C B C \ Consider ( ) B C ( B C) BC q ª ¹ q 4 4 B C B C B C ( ) q q ( p q) q q 4 4 p 4q 4p qq p q 4p q q q p q 4 4p q RHS. Hence proved. OR Let the present ages of father and his son are x years and y years respectively. According to the given conditions: x + y 65 After 5 years, the father s age (x + 5) years After 5 years, the son s age (y + 5) years Therefore, x + 5 (y + 5) i.e., x y 5 Thus, the required pair of linear equations is x + y 65... (i) and x y 5... (ii) Subtracting equation (ii) from equation (i), we get y 60 Þÿ y 0 Substituting y 0 in equation (i), we get x + 0 65 Þÿx 45 Thus, present age of father 45 years and present age of his son 0 years. 5. Let us draw MN parallel to AB, which passes through P. So, AM BN and DM CN. 4 MATHEMATICS X

From right-angled triangles APM, BPN, CPN, DPM; we have respectively PA PM + AM... (i) PB PN + BN... (ii) PC PN + CN... (iii) PD PM + DM... (iv) From equations (i) and (ii), PA PB PM PN... (v) From equations (iii) and (iv), PC PD PN PM... (vi) Add equations (v) and (vi) to get PA + PC PB + PD. Hence proved. 6. Let median AD passes through the point O on PQ. To prove: PO QO Proof: In DAPO and DABD, ÐPAO ÐBAD (Common angle) APO ABD º» and AOP ADB ¼ (Corresponding angles) \ÿdapo ~ DABD (By AAA similarity) Þ PO BD AO AD...(i) Similarly, in DAQO and DACD, Þ QO CD AO AD...(ii) From equations (i) and (ii), we have PO BD QO CD Þ PO QO (... BD CD) Hence, median AD also bisects PQ. Hence proved. OR Let the given parallelogram be ABCD We need to prove that AC + BD AB + BC + CD + DA Let us draw perpendiculars DN on AB and CM on AB produced as shown in figure. In DBMC and DAND, BC AD (Opposite sides of a gm ) ÐBMC ÐAND (Each 90 ) CM DN (Distance between same parallels) \DBMC @ DAND (RHS criterion) Þ BM AN...(i) (CPCT) In right triangle ACM, AC AM + CM (AB + BM) +BC BM AB + AB. BM + BM + BC BM AB + BC + AB. BM...(ii) In right triangle BDN, BD BN + DN (AB AN) + (AD AN ) AB AB. AN + AN + AD AN BD AB +DA AB. AN Þ BD CD +DA AB. BM...(iii) [Using (i) and AB CD] Adding equations (ii) and (iii), we have AC + BD AB +BC +CD +DA. Hence proved. cos Rsin R 7. cos Rsin R Using componendo and dividendo, we get cos Rsin Rcos Rsin R cos Rsin Rcos Rsin R cosr Þÿ Þÿcot qÿ sinr Þÿ cot qÿ cot 60 Þ ÿqÿ 60. 8. LHS 4 4 cos R cos R sin R sin R sec qÿ sec 4 qÿ cosec qÿ+ cosec 4 q PRACTICE PAPER4 5

( + tan q) ( + tan q) ( + cot q) + ( + cot q) ( + tan q) ( tan q) ( + cot q) ( cot q) tan 4 qÿ ( cot 4 q) cot 4 qÿ tan 4 qÿ RHS. Hence proved. 9. We know that cosec (90 q) sec q cot (90 q) tan q tan (90 q) cot q sec θ.cosec (90 θ) tan θ.cot(90 θ) + sin 55 + sin 5 Now, tan0.tan 0. tan 60.tan 70.tan 80 sec θ. sec θ tan θ. tan θ+ sin 55 + sin (90 55 ) tan0 tan 0..tan(90 0 ) tan(90 0 ) sec R tan R sin 55o cos 55o tan0.tan0. o o cot0o scot0o tan 0 os.tan 0 o. s tan 0o tan 0o 0. We prepare the cumulative frequency table for the given data. So, cf 00, f 5, l 000. Median l + N cf f µ µ h 50 00 000 + µ 5 500 000 + 7.9 7.9 hours. SECTION D. To draw a line, we need atleast two solutions of its corresponding equation. x 0 x 0 y 4 0 y 0 ÿÿÿÿÿ Two solutions of Two solutions of x y 4 x y From the graph, the two lines intersect each other at the point A(, ). \ x and y Shaded region is DABC. \ÿheight of DABC units And its base BC units. Lifetimes Frequency Cumulative (in hrs.) ( f ) frequency (cf ) 500-000 4 4 000-500 86 0 500-000 90 00 000-500 5 5 500-4000 95 40 4000-4500 7 48 4500-5000 8 500 N 500 Here, h 500 Q N 500, \ÿ N 50. \ ar(dabc) 9 square units. 6 MATHEMATICS X

. Given system of linear equations may be written as bx + ay ab (a + b) 0 b x + a y a b 0 Applying cross-multiplication, we get x ab ab ab 4 y 4 ab ab ab abab x y Þ 4 4 abab ab a b abab (i) (ii) (iii) Taking (i) and (iii), 4 abab aba ( b) x a abab ab( a b) Taking (ii) and (iii), 4 ab a b ab ( b a) y b abab ab( a b) Thus, the required solution is x a, y b.. (i) Let p(x) Total Relief Fund g(x) Number of families who received Relief Fund q(x) Amount each family received r(x) Amount left after distribution When the polynomial p(x) is divided by a polynomial g(x) such that q(x) and r(x) are respectively the quotient and the remainder, the division algorithm is p(x) g(x). q(x) + r(x) (i) According to the question, p(x) x + x + x + 5 q(x) x 5 and r(x) 9x + 0 Substituting these values of p(x), q(x) and r(x) in the equation (i), we get x + x + x + 5 g(x) (x 5) + 9x + 0 Þ (x 5) g(x) x + x + x + 5 9x 0 x + x 7x 5 Þ g(x) x x 7x5 x 5 To find g(x), we proceed as following: x x x5 x x 7x5 x 5x 6x 7x5 6x 0x x 5 x 5 0 Thus, g(x) x + x +. (ii) Common good, Accountability, social responsibility. 4. We are given a DABC in which a line PQ BC intersects the sides AB at P and AC at Q. We need to prove AP AQ. PB QC Draw QM ^ÿab and PN ^ÿac. Join PC and BQ. Proof: Area of a triangle Base Height \ ar(dapq) AP QM Þ Þ AQ PN AP QM AQ PN QM PN AQ AP... (i) Since DBPQ and DCQP are on the same base PQ and between the same parallels PQ and BC, therefore, their areas should be equal. i.e., ar(dbpq) ar(dcqp) Þ PB QM QC PN Þ QM PN QC ÿ PB... (ii) PRACTICE PAPER4 7

Form equations (i) and (ii), it is clear that AQ AP QC AP AQ ÿ,ÿi.e., PB PB ÿ QC. Hence proved. 5. Let us produce AD to J and PM to K so that DJ AD and MK PM. Join CJ and RK. In DADB and DJDC, AD JD, ÐADB ÐJDC, BD CD Þ DADB @ DJDC (SAS criterion of congruence) Þ AB JC...(i) (CPCT) Similarly, we can prove that PQ KR...(ii) According to the given conditions, we have Þ AB PQ JC KR AD PM AC PR AJ PK AC PR [Using (i) and (ii)] JC Þ KR AJ PK AC PR Þ DAJC ~ DPKR (SSS criterion of similarity) Þ ÐJAC ÐKPR (Corresponding angles) i.e., ÐDAC ÐMPR...(iii) Similarly, we can prove that ÐDAB ÐMPQ...(iv) Adding equations (iii) and (iv), we obtain ÐBAC ÐQPR...(v) Thus, in DABC and DPQR, we have AB PQ AC PR (Given) and ÐBAC ÐQPR [From (v)] Therefore, DABC ~ DPQR. (SAS criterion of similarity) Hence proved. 6. m cosec q sin qÿ sin θ sin q sin θ cos θ sinθ sin θ n sec q cos qÿ cos q cosr cos R sin R cos R cos R mn + mn 7. Now, LHS ( ) ( ) cos sin 4 θ sin θ cos θ sin + θ cos θ sin θ cos ( ) cos θ + ( ) sin θ 4 θ θ cos qÿ+ sin qÿ RHS. cot x tan x + + sec x. cosec x tanx cotx LHS cos x cot x tan x + tanx cotx sin x sin x cos x + sin x cos x cos x sin x cos x cos x sin x sin x + sin x cos x sin x cos x sin x cos x cos x sin x(cos x sin x ) + sin x cos x(sin x cos x) cos x sin x sin x. cos x(cos x sin x) 8 MATHEMATICS X

(cos x sin x)(cos xcos x. sin xsin x) sin x. cos x(cos x sin x) cosx. sinx sin x. cos x sin x. cos x + + sec x. cosec x RHS. Hence proved. 8. It is given that sin q + cos q a Squaring both sides, we get sin qÿ+ cos qÿ+ sin qÿcos qÿ a But sin qÿ+ cos q... (i) \ + sin qÿcos q a Þ sin qÿcos q a a Þ sin qÿcos q... (ii) Cubing both sides of equation (i), we get sin 6 qÿ+cos 6 q + sin q cos qÿ (sin q + cos q) Þ sin 6 qÿ+ cos 6 qÿ+ (sin q cos q) [Using (i)] PRACTICE PAPER4 a µ Þ sin 6 qÿ+ cos 6 q + Þ sin 6 qÿ+ cos 6 ( a ) q 4 Þÿ sin 6 qÿ+ cos 6 4( a ) q. 4 Hence proved. OR Consider a sin a + b cos a sin a. cos a Þ a sin a. sin a + b cos a. cos a sin a. cos a Þ b cos a. sin a + b cos a. cos a sin a. cos a (... a sin a b cos a) Þ b sin a + b cos a sin a Þ b (sin a + cos a) sin a \ b sin a...(i) Again, a sin a b cos a Þ a. b b cos a [From (i)] \ a cos a...(ii) Now, squaring and adding equations (i) and (ii), we get b + a sin a + cos a \ a + b.hence proved. 9. We prepare a table for less than series with corresponding cumulative frequencies and points. Marks Freq. Marks Cumul- Point less ative than Freq. 0-0 5 0 5 (0, 5) 0-0 8 0 (0, ) 0-0 6 0 9 (0, 9) 0-40 0 40 9 (40, 9) 40-50 6 50 5 (50, 5) 50-60 5 60 40 (60, 40) We take upper limits on the x-axis and cumulative frequencies on the y-axis. Then we plot the points on the graph paper. By joining these points by free hand smooth curve, we obtain less than ogive as shown in the above figure. OR A more than type distribution from the given distribution is given below: 9

Production yield (in kg/ha) Cumulative freq. More than or equal to 50 00 More than or equal to 55 98 More than or equal to 60 90 More than or equal to 65 78 More than or equal to 70 54 More than or equal to 75 6 To draw more than type ogive, we mark lower limits from the given table on x-axis and cumulative frequency from the above table on y-axis with suitable scale(s). We plot the points (50, 00), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 6). Now, we join these points with free hand smooth curve as shown. in the adjoining figure. 0. Class Mid- Freq. ( f i ) u i f i u i interval values x i A h (x i ) 0-0 0 5 40 0 0 0-50 40 8 Fig. : More than type ogive. 0 8 0 50-70 60 A 70-90 80 0 90-0 00 0 0 0 0 0 0 0 40 0 6 0-0 0 60 0 6 Sf i 50 Sf i u i 4 0 MATHEMATICS X

Let assumed mean be A 60 Here, h 0 Σfiui Mean A + h 60 + 0 4 Σfi 50 60 + 5.6 65.6 Hence, the required arithmetic mean is 65.6.. Class-interval Frequency 0-0 5 0-0 0-0 9 0-40 0 40-50 5 The class corresponding to the maximum frequency is 0-40. So, 0-40 is the modal class. Mode l + f f h 0 µ f f0 f Here, l 0, f 0, f 0 9, f 5, h 0 z x y xy which given a contradiction as LHS is irrational but RHS is rational. Þ x + y is an irrational number.. (C) Coincident lines is given by a b a b c c Here, a p, b q, c 4 a 4, b, c 5 p q 4 p q Now, Þÿÿ 0.8 4 5 4 Þ p., q.4 Therefore, p + q.. Consider A 90 ÿÿÿ\ A 0 So, sin 0 cos 0 sin 0 cos 60 4. DE + EF + FD AB + BC + CA 0. EF BC Mode 0 + 0 + 0 9 t0 9 5 µ µ 60 4 0 0 0 + 6 0 0 + 0 6 0 + 4. 4.. Practice Paper-5 SECTIONA. (B) Let us assume that x + y is rational number and let x + y z; when z is rational. Þ x + y + xy z Þ xy z x y Þ Perimeter of DDEF 4 ( + +.5) Þ Perimeter of DDEF 5 cm. SECTIONB 5. True, because product of an even number and an odd number is an even number. 6. Going in opposite direction to the factor tree, we obtain (i) 0 660 and (ii) 65 0 7. For infinite number of solutions, a a b b c i.e., a a + b 8 c 5 9, i.e., a and a + b 0 i.e., a, b 8. PRACTICE PAPER5

AB 8. DF BC EF CA DE Þ DABC ~ DDFE ÞÿÿÐBÿÿÐF. But ÐB 60, so, ÐF 60. 9. Let p(x) ax + bx + c So, y ax + bx + c should be satisfied by (, 0), (0, ) and (4, 0) Therefore, 0 a b + c; c; 0 6a + 4b + c 0. Þ c, a 4, b 9 4. Hence, p(x) 4 x 9 4 x Þ p(x) 4 (x x 4) This is the required expression. x i 5 7 9 0 + p f i 4 5p 6 Sfi5 + 5p f i x i 0 5 76 00p + 5p 8 Sf i x i 95 +00p + 5p Mean fixi 95 + 00p+ 5p ÿþÿÿ0 fi 5 + 5p Þ 00 + 00p 95 + 00p + 5p Þ 5p 5 Þ p ±. But frequency cannot take negative value. So, p ¹ÿ. Hence, p. OR Relation among mean, median and mode is given by the following impirical formula: Mode Median Mean SECTIONC. Let a be any positive integer. We know that any positive integer is either of the form q or q + for some integers q. \ a q or q + Case I. When a q, a + q +. \ a (a + ) q q + q (q +) r, where r q (q + ) So, a (a + ) is divisible by. Case II. When a q +, a + q + (q + ) \ a (a + ) (q + ) (q + ) where r (q + ) (q + ) So, a (a + ) is divisible by. Hence, multiplication of any two consecutive positive integers is divisible by.. Ram, Ravi and Nitin will meet next after the time given by the LCM of 5 days, 4 days and 9 days. Now, we find out the LCM of 5, 4 and 9. 5 5; 4 ; 9 \ LCM 5 60 They met last on Sunday. So, it will be Sunday after 7n days, where n is a natural number. So, it will be Sunday after 57 days. Therefore, it will be Wednesday after 60 days. Hence, they will meet on next Wednesday. OR We represent 6, 7 and 0 in their prime factors. 6, 7, 0 5 Now, HCF 6 And LCM 5 60.. Let zeroes are a, b, g Let ab 8... (i) Also we know a + b + g 9... (ii) ab + ag + bg 6 abg 4 \ 8(g)ÿ 4 Þ gÿ From (ii), ÿþÿa + b 6... (iii) From (i), Þÿb 8 B \ Use if in (iii) we get a + 8 B 6 Þa 6a + 8 0 (a 4) (a ) 0 Þÿa 4 Þÿÿa. If a 4 ÿþÿÿb and if a Þ b 4 \ÿzeroes are, 4 and. 4. Put x+ y u and v in the given x y system of equation, we get MATHEMATICS X

0u + v 4 5u 5v i.e., 5u + v... (i) 5u 5v... (ii) Multiply equation (i) by 5 and add the result to (ii) Substitute 40u 8 or u 5 u in equation (i) 5 v u and v give x + y 5 5 and x y On solving, we get x and y Hence, x, y is the required solution. 5. In DABC, DE BC Q DABC ~ DADE D A E. 6. In right-angled DABC, AB +BC AC...(i) (By Pythagoras Theorem) M A B N C In DABN, AN AB + BN µ AB BC + (... N is the mid-point of BC) B C AB + 4 BC ar \ ( ΔABC ) ar ( ΔADE) AB AD Again, DE BC Q ar (DADE) ar ( ÿbced) \ ar (DABC) ar (DADE) ar ( ΔABC ) Þ ar ( ΔADE)...(i)... (ii) From equations (i) and (ii), we get AB AD Let AB x and AD x, then from the figure, Now, BD x x ( ) x BD ( ) AB x x Þ 4AN 4AB +BC...(ii) Similarly, in DCBM, 4CM AB +4BC...(iii) Adding equations (ii) and (iii), we get 4AN +4CM (4AB +AB ) + (BC + 4BC ) 5AB + 5BC Þ 4(AN + CM ) 5 (AB + BC ) 5 AC [From (i)] Hence proved. cos 58 7. sin ( ) cos 90 sin cos 8 cosec 5 µ p p tan 5p tan 60p tan 75p cos É90 p 5pcosec 5p» ¼ tan É90 p 75p t t tan 75p ½ PRACTICE PAPER5

sin sin sin 5pt sin 5p cot 75pt t cot 75 µ p. OR LHS cos qÿ cos ( 0 ) cos 60 tan θ RHS + tan θ tan 0 + tan 0 4 + + 4 Thus, LHS RHS. Hence proved. tan θ+ sec θ 8. LHS tan θ sec θ+ ( ) tan θ+ sec θ sec θ tan θ tan θ sec θ+ ( tan θ+ sec θ ) ( sec θ+ tan θ )( sec θ tan θ ) tan θ sec θ+ ( tan θ+ sec θ )( sec θ+ tan θ ) tan θ sec θ+ tan q + sec q sin θ cos θ + cos θ + sin θ RHS. Hence proved. cos θ 9. p q (a cot qÿ+ b cosec q) (b cot qÿ+ a cosec q) a cot q + ab cot q cosec q + b cosec q b cot qÿ ab cot qÿcosec qÿ a cosec q a (cosec qÿ cot q) + b (cosec qÿ cot q) a () + b () b a. 0. Since, the maximum frequency is 4, so the modal class is 0000-5000. \ l 0000, f 4, f 0 6, f 6, h 5000 Now, mode l + 0000 + f f0 u h f f f ¹ 0 4 6 8 6 6 µ t 5000 0000 + 75000 40 0000 + 875 875. Thus, the monthly modal income is ` 875. OR Let a 50; h 0 C.I. f i x i u i xi a h f i u i 0-0 4 0 8 0-40 0 0 0 40-60 8 50 0 0 60-80 6 70 6 80-00 50 90 00 8 Sf i u i 8 fiui 8 x a + h f 50 + 0 ¹ 8 50 + 95 6 i 50 + 8.4 68.4. SECTION D µ. Given that p(x) a(x + ) x(a + ) i.e., p(x) ax (a + )x + a To find zeroes of p(x), put p(x) 0. i.e., ax + a a x x 0 i.e., (ax a x) (x a) 0 i.e., ax (x a) (x a) 0 i.e., (x a) (ax ) 0 i.e., x a, a Thus, a and are the zeroes of p(x). a a Sum of zeroes a a a 4 MATHEMATICS X

Éa a Coefficient of x Coefficient of x Product of zeroes a a a a Constant term Coefficient of x. Hence proved. OR Let p(x) x + x + x + 5 q(x) x 5 r(x) 9x + 0 g(x)? We know p(x) q(x). g(x) + r(x) Þ x + x + x + 5 (x 5) g(x) + 9x + 0 Þ x + x + x + 5 9x 0 g(x). (x 5) x x 7x5 Þ g(x) x 5 Consider x x x5 x x 7x5 x 5x 6x 7x5 6x 0x x 5 x 5 t g(x) x + x +.. Let the actual length be x and breadth be y. Then according to question, xy 9 (x 5) (y + )...(i) and xy + 67 (x + ) (y + )...(ii) Simplifying equations (i) and (ii), we have x 5y 6...(iii) and x + y 6...(iv) On solving (iii) and (iv), x 7, y 9 Hence, length of rectangle is 7 units and that of breadth is 9 units.. Table for values of x and y corresponding to equation 4x 5y 0 0 is Similarly for the equation x +5y 5 0 x 5 0 y 0 Let us draw the graphs for the two equations. As the graphs of the two lines intersect each other at the point A(5, 0), the required solution is x 5, y 0. The graphs intersect the y-axis at B (0, ) and C(0, 4). Therefore, the coordinates of vertices of the triangle ABC are A(5, 0), B(0, ) and C(0, 4). Hence the answer: x 5, y 0 and (5, 0), (0, ), (0, 4). 4. Statement: In a triangle, if square of the largest side is equal to the sum of the squares of the other two sides, then the angle opposite to the largest side is a right angle. Proof: We are given a triangle ABC with A C A B + B C...(i) We have to prove that ÐB 90 Let us construct a DPQR with ÐQ 90 such that, PQ A B and QR B C...(ii) x 5 0 y 0 4 PRACTICE PAPER5 5