247 and x – y = 3 .............................. (ii) Adding equation (i) and (ii) × 11 2x + 11y = 84 11x – 11y = 33 13x x = = 117 9 Putting the value of x in equation (ii) 9 – y = 3 or, 9 – 3 = y y = 6 (a) Hence, the number is 10x + y = 10 × 9 + 6 = 96 (b) And, reversed number = 59 Original number is increased by = 96 – 69 69 × 100% = 2700 69 % = 39.13% (ii) Let 10x + y be required two digits number with digits x and y at tenth place and unit place respectively. then, y – x = 2 or, y = x + 2 ....................... (i) and 10x + y + 3(x + y) = 10y + x or, 10x + y + 3x + 3y – 10y – x = 0 or, 12x – 6y = 0 or, 2x – y = 0 ...................... (ii) Substituting the value of y from equation (i) to (ii) 2x – (x + 2) = 0 or, 2x – x – 2 = 0 x = 2 Putting the value of x in equation (i) y = 2 + 2 = 4 (a) Hence, the number is 10x + y = 10 × 2 + 4 = 24 (b) Reversed number is 42 and original number is decreased to reversed number by = 42 – 24 42 × 100% = 1800 42 % = 42.86% 14. (i) 5 kg of rice and 3 kg of flour cost Rs. 145; and 3 kg of rice and 5 kg of flour cost Rs.151. (ii) 5 kg of apples and 15 kg of mangoes cost Rs. 2350 and 3 kg of apples and 7 kg of mangoes cost Rs. 1190. (a) Find the cost of each item. (b) What is the increased or decreased percent of the first item than the second item? Solution: (i) Let x and y are unit cost of rice and flour respectively then, 5x + 3y = 145 ...................... (i) and 3x + 5y = 151 ...................... (ii) Subtracting equation (ii) × 3 from (i) × 5 25x + 15y = 725 9x + 15y = 453 – – – 16 x x = = 272 17
248 Putting the value of x in equation (i) 5 × 17 + 3y = 145 or, 3y = 145 – 85 or, y = 60 3 y = 20 (a) Hence, cost of rice (x) = Rs. 17 cost of flour (y) = Rs. 20 (b) The cost of rice is decreased to cost of flour by = 20 – 17 20 × 100% = 15% (ii) Let the unit cost of apple and mango are x and y respectively Then, 5x + 15y = 2350 or, x + 3y = 470 x = 470 – 3y ..................... (i) and 3x + 7y = 1190 ................ (ii) Substituting the value of x from equation (i) to (ii) 3(470 – 3y) + 7y = 1190 or, 1410 – 9y + 7y = 1190 or, – 2y = – 220 y = 110 Putting the value of y in equation (i) x = 470 – 3 × 110 or, x = 470 – 330 x = 140 (a) Hence, the cost of apple (x) = Rs. 140 the cost of mango (y) = Rs. 110 (b) The cost of apple is increased to the cost of mango by = 140 – 110 110 × 100% = 3000 110 % = 27.27% 15. (i) If the numerator of a fraction is multiplied by 4 and a denominator is reduced by 2, the result is 2. If the numerator of the fraction is increased by 15 and 2 is subtracted from the double of the denominator, the result is 9 7 . Find the fraction. (ii) If 1 is added to both numerator and denominator of a fraction; the fraction becomes 4 5 . If 1 is subtracted from numerator and denominator both, the new fraction becomes 3 4 . What is that fraction ? Solution: (i) The x y be required fraction such that 4x y – 2 = 2 or, 2x y – 2 = 1 or, 2x = y –2 or, y = 2x + 2 ..................... (i) and x + 15 2y – 2 = 9 4 ...................... (ii)
249 Substituting the value of y from equation (i) to equation (ii) x + 15 2(2x + 2) – 2 = 9 7 or, 7x + 105 = 9(4x + 4 – 2) or, 7x + 105 = 9(4x + 2) or, 7x + 105 = 36x + 18 or, 105 – 18 = 36x – 7x or, 87 = 29x x = 3 Putting the value of x in equation (i) y = 2 × 3 + 2 = 8 Hence, required fraction x y = 3 8 . (ii) Let x y be the required fraction such that x + 1 y + 1 = 4 5 or, 5x + 5 = 4y + 4 or, 5x – 4y = – 1 ..................... (i) and x – 1 y – 1 = 3 4 or, 4x - 4 = 3y – 3 or, 4x – 3y = 1 .............. (ii) Subtracting equation (ii) × 4 from (i) × 3 16x – 12y = – 3 16x – 12y = 4 – + – – x x = = – 7 7 Putting the value of x in equation (i) 5 × 7 – 4y = – 1 or, – 4y = – 1 – 35 or, y = – 36 – 4 y = 9 Hence, required fraction x y = 7 9 . 16. (i) A mother says to her daughter “5 years ago, I was 5 times as old as you were but 10 years hence I shall be only two times as old as you will.” Find their present ages. (ii) Dharmendra said to his son Sanni “I was thrice as old as you were when I was as old as you are.” Also, said that if the sum of our ages is 120 years, find what our ages now would be. Solution: (i) Let x and y are present ages of mother and daughter respectively. Then, x – 5 = 5(y – 5) or, x – 5 = 5y – 25 or, x = 5y – 20 .............................. (i) and x + 10 = 2(y + 10) or, x + 10 = 2y + 20 or, x = 2y + 10 ........................... (ii)
250 From equation (i) and (ii) 5y – 20 = 2y + 10 or, 3y = 30 y = 10 Putting the value of y in equation (i) x = 5 × 10 – 20 = 30 Hence, present age of mother (x) = 30 yrs. present age of daughter (y) = 10 yrs. (ii) Let present age of Dharmantra and Sanni are x and y respectively. then, {x – (x – y)} = 3{y – (x – y)} or, y = 3(2y – x) or, y = 6y – 3x or, 3x = 5y or, x = 5y 3 ....................... (i) and x + y = 120 .............. (ii) Putting the value of x from equation (i) to (ii) 5y 3 + y = 120 or, 5y + 3y 3 = 120 or, 8y = 360 y = 45 Putting the value of y in equation (i) x = 5 × 45 3 = 75 Hence, present age of Dharmendra (x) = 75 yrs. present age of Sanni (y) = 45 yrs. 17. (i) Katrina started a car for the journey of 70 km from Beni to Jomsom at 7 AM with an average speed of 20km per hour. After one hour, Ranvir started his motorbike on the same journey with average speed of 30 km/hr. At what time would they meet? (ii) A bus departed at 9 PM for the journey of 200km from Kathmandu to Pokhara with an uniform speed of 40km per hour. After one and half hour, Sansar started his motorbike on the same journey with uniform speed of 60km/hr. At what time would they meet? Solution: (i) Let Katrina and Ranvir meet x hours after 7 AM so that Katrina travels for x hours, but Ranvir travels only (x – 1) hours. Then, 20x = 30(x – 1) or, 20x = 30x – 30 or, 30 = 3x – 20x or, 30 = 10x x = 3 hours Hence, they meet at 7 AM + 3 hours = 10 AM. (ii) Let bus and bike meet x hours after 9 PM so that bus travels for x hours and bike travels for (x – 1.5) hours. Then, 40x = 60(x – 1.5) or, 40x = 60x – 90 or, 90 = 60x – 40x or, 90 = 20x x = 4.5 hours. Hence, bus and bike meet at 9 PM + 4.5 hours = 1:30 AM.
251 CHAPTER INDICES 12 12.1 Simplification of Indices By Using Their Laws PRACTICE 12.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) What is an index or exponent ? (b) Express m a into exponential form. (c) What is the index of x such that its value becomes 1? Solution: (a) A quantity representing the power to which a given number or expression is to be raised usually meant to multiply the number or expression for that times. (b) m a = a 1 m (c) x0 = 1 2. Write the following in the exponential form (i.e., by using index notation). (a) 3 × 3 × 3 × 3 (b) 81 (c) a × a × a × a (d) 32 Solution: (a) Here, 3 × 3 × 3 × 3 = 34 (b) 81 = 3 × 3 × 3 × 3 = 34 (c) a × a × a × a = a4 (d) 32 = 2 × 2 × 2 × 2 × 2 = 25 3. Write the following in the product form: (a) 8a3 (b) 64 (c) x2 y3 (d) 16 Solution: (a) 8a3 = 2 × 2 × 2 × a × a × a (b) 64 = 6 × 6 × 6 × 6 (c) x2 y3 = x × x × y × y × y (d) 16 = 2 × 2 × 2 × 4. What is the value of: (a) 8– 2 3 (b) 27– 2 3 (c) 8 27 – 2 3 (d) 64– 2 3 (e) 33 × 32 (f) 23 .22 .24 (g) 37 ÷ 24 × 22 (h) 3 2 2 Solution: (a) 8– 2 3 = 1 8 2 3 = 1 (23 ) 2 3 = 1 23 × 2 3 = 1 22 = 1 4
252 (b) 27– 2 3 = 1 272 3 = 1 (33 ) 2 3 = 1 33 × 2 3 = 1 32 = 1 9 (c) 8 27 – 2 3 = 27 8 2 3 = 33 23 2 3 = 3 3 × 2 3 23 × 2 3 = 32 22 = 9 4 (d) 64– 2 3 = (26 ) – 2 3 = 26 × ( ) – 2 3 = 2–4 = 1 24 = 1 16 (e) 33 × 32 = 33 + 2 = 35 = 243 (f) 23 . 22 . 24 = 23 + 2 + 4 = 29 = 512 (g) 37 ÷ 24 × 22 = 37 24 × 22 = 37 24 – 2 = 37 22 = 2187 4 (h) 3 2 2 = 32 22 = 9 4 5. Write in simplest exponential form: (a) (i) 22 .23 (ii) (x + y)2 (x + y)–3 (iii) a2x.a3 . ax .a3 (iv) (2x + 3y)2m (2x + 3y)3n (b) (ii) (3a)2 (ii) (xy)3 (iii) (ab)2x (iv) [(x + y) (x – y)]4 (c) (i) 5 3 2 (ii) a b 3 (iii) a b c (iv) a + y x – y 3 (d) (i) x4 y–2 (ii) x4 (iii) 3 xy2 × 3 x2 y (iv) ( ) x– 2 3 ÷ y– 1 2 6 Solution: (a) (i) 22 . 23 = 22 + 3 = 25 (ii) (x + y)2 (x + y)–3 = (x + y)2 – 3 = (x + y)–1 (iii) a2x . a3 . ax . a3 = a2x + 3 + x + 3 = a3x + 6 (iv) (2x + 3y)2m (2n + 3y)3n = (2x + 3y)2m + 3n (b) (i) (3a2 ) 2 = 32 . a2 = 9a2 (ii) (xy)3 = x3 y3 (iii) (ab)2x = a2x . b2x (iv) [(x + y) (x – y)]4 = (x2 – y2 ) 4 (c) (i) 5 3 2 = 52 32 (ii) a b 3 = a 3 b3 (iii) a b c = a c bc (iv) a + y x – y 3 = (a + y)3 (x – y)3 (d) (i) x4 y-2 = (x4 y–2) 1 2 = x4 × 1 2 y– 2 × 1 2 = x2 y–1 (ii) x4 = ( x4 ) 1 2 = {(x4 ) 1 2} 1 2 = x4 × 1 2 × 1 2 = x (iii) 3 xy2 × 3 x2 y = 3 xy2 × x2 y = 3 x1 + 2 y2 + 1 = (x3 y3 ) 1 3 = x3 × 1 3 × y3 × 1 3= xy (iv) ( ) x – 2 3 ÷ y– 1 2 6 = x – 2 3 y– 1 2 6 = y 1 2 x 2 3 6 = 1 2 × 6 x 2 3 × 6 = y3 x4
253 Check Your Performance 6. Simplify: (a) (81) 1 4 × 1 27 2 3 (b) 625 81 – 3 4 (c) 154 × 65 105 × 92 (d) 284 × 64 213 × 182 (e) 81 16 3 4 ÷ 32 243 – 3 5 of 27 8 – 2 3 (f) 2 3 – 1 5 10 3 – 3 2 Solution: (a) Here, (81) 1 4 × 1 27 2 3 = (34 ) 1 4 × 1 33 2 3 = 34 × 1 4 × 1 2 3 33 × 2 3 = 3 × 1 32 = 1 32 – 1 = 1 3 (b) Here, 625 81 – 3 4 = 81 625 3 4 = 34 54 3 4 = 3 4 × 3 4 5 4 × 3 4 = 33 53 = 27 125 (c) Here, 154 × 65 105 × 92 = (3 × 5)4 × (2 × 3)5 (2 × 5)5 × (32 ) 2 = 34 × 54 × 25 × 35 25 × 55 × 34 = 34 + 5 – 4 × 54 – 5 × 25 – 5 = 35 × 5–1 × 20 = 243 5 = 48 3 5 (d) 284 × 64 213 × 182 = (2 × 2 × 7)4 × (2 × 3)4 (3 × 7)3 × (2 × 3 × 3)2 = 24 × 24 × 74 × 24 × 34 33 × 73 × 22 × 32 × 32 = 24 + 4 + 4 – 2 × 34 – 3 – 2 – 2 × 74 – 3 = 210 × 3–3 × 7 = 1024 × 7 27 = 7168 27 = 265 13 27 (e) Here, 81 16 3 4 ÷ 32 243 – 3 5 of 27 8 – 2 3 = 81 16 3 4 ÷ 243 32 3 5 of 8 27 2 3 = 34 24 3 4 ÷ 35 25 3 5 of 23 33 2 3 = 3 4 × 3 4 2 4 × 3 4 ÷ 3 5 × 3 5 25 × 3 5 of 2 3 × 2 3 33 × 2 3 = 33 23 ÷ 33 23 of 22 32 = 33 23 ÷ 33 × 22 23 × 32 = 33 23 × 23 × 32 33 × 22 = 33 + 2 – 3 23 + 2 – 3 = 32 22 = 9 4 (f) Here, 2 3 – 1 5 10 3 – 3 2 = 3 2 1 5 10 3 – 3 2 = 3 2 1 5 × 10 3 – 3 2 = 3 2 2 3 – 3 2 = 3 2 2 3 × ( ) – 3 2 = 3 2 – 1 = 2 3 7. Simplify: (a) 1 1 + xa – b + 1 1 + xb – a (b) (1 – xm – n) –1 + (1 – xn – m) –1 (c) a a + b–1 + 1 ab + 1 (d) {(a – b)–2} –m × {(a – b)m} 2 Solution: (a) 1 1 + xa – b + 1 1 + xb – a = 1 1 + xa xb + 1 1 + xb xa = 1 xb + xa a b + 1 xa + xb xa = xb xb + xa + xa xa + xb = xb xb + xa + xa xb + xa = (xb + xa ) (xb + xa ) = 1 (b) (1 – xm – n) –1 + (1 – xn – m) –1 = 1 – xm xn –1 + 1 – xn xm –1 = xn – xm xn –1 + xm – xn xm –1 = xn xn – xm + xm xm – xn = xn xn – xm – xm xn – xm = xm – xm xn – xm = 1
254 (c) a a + b–1 + 1 ab + 1 = a a + 1 b + 1 a b + 1 = a ab + 1 b + 1 ab + 1 = a b ab + 1 + 1 ab + 1 = (ab + 1) (ab + 1) = 1 (d) {(a – b)–2}–m × {(a – b)m}2 = (a – b)(– 2) × (– m) × {(a – b)m × 2} = (a – b)2m × (a – b)2m = (a – b)2m + 2m = (a – b)4m 8. Simplify: (a) 27x3 ÷ 243x5 (b) 8xy3 × 2x3 y (c) 4 625x8 y12 (d) 3 –216a6 b9 c 0 (e) 3 –8x3 y–9 × 4 81x4 y4 (f) 121a4 b–8c 8 ÷ 3 1331a–6b9 c 3 (g) 3 8x–9y12z 6 ÷ 4 1296x8 y–4 (h) 4 81a12b–8c 4 ÷ 3 125a–6b6 c 9 (i) 4 256x–8y16 × 3 27x6 y–6 × 4 x4 y–8 (j) 5 16x4 y6 z 2 × 5 2x6 y4 z –7 ÷ 16x8 y12 Solution: (a) 27x3 ÷ 243x5 = 27x3 243x5 = 33 x3 35 x5 = 1 35 – 3 x5 – 2 = 1 32 x2 = 1 3x (b) 8xy3 × 2x3 y = 8xy3 × 2x3 y = 23 xy3 × 2x3 y = 23 + 1x1 + 3y3 + 1 = 24 x4 y4 = 22 x2 y2 = 4x2 y2 (c) 4 625x8 y12 = (54 x8 y12) 1 4 = 54 × 1 4 x8 × 1 4 y12 × 1 4 = 5x2 y3 (d) 3 –216a6 b9 c 0 = (–63 a 6 b9 . 1)1 3 = (–6)3 × 1 3 a6 × 1 3 b9 × 1 3 = – 6a2 b3 (e) 3 –8x3 y–9 × 4 81x4 y4 = {(–2)3 x3 y–9) 1 3 × (34 x4 y4 ) 1 4 = (–2) 3 × 1 3 x3 × 1 3 y–9 × 1 3 × 34 × 1 4 x4 × 1 4 y4 × 1 4 = –2 × x × y–3 × 3 × x × y = – 6x2 y–2 = – 6x2 y2 (f) 121a4 b–8c 8 ÷ 3 1331a–6b9 c 3 = (112 a 4 b–8c 8 ) 1 2 ÷ (113 a –6b9 ) 1 3 = 112 × 1 2 a4 × 1 2 b–8 × 1 2 c8 × 1 2 ÷ 113 × 1 3 a–6 × 1 3 b9 × 1 3 = 11a2 b–4c 4 ÷ 11a–2b3 = 11a2 b–4c 4 11a–2b3 = a 2 + 2c 4 b3 + 4 = a 4 c 4 b7 (g) 3 8x–9y12z 6 ÷ 4 1296x8 y–4 = 3 23 x–9y1226 ÷ 4 64 x8 y–4 = 2x–3y4 22 ÷ 6x2 y–1 = 2x–3y4 z 2 6x2 y–1 = y4 + 1z 2 3x2 + 3 = y5 z 2 3x5 (h) 4 81a12b–8c 4 ÷ 3 125a–6b6 c 9 = (34 a 12b–8c 4 ) 1 4 ÷ (53 a –6b6 c 9 ) 1 3 = 3a3 b–2c ÷ 5a–2b2 c 3 = 3a3 b–2c 5a–2b2 c 3 = 3a3 + 2 5b2 + 2c 3 – 1 = 3a5 5b4 c 2 (i) 4 256x–8y16 × 3 27x6 y–6 × 4 x4 y–8 = 4 44 (x–2) 4 (y4 ) 4 × 3 33 (x2 ) 3 (y–2) 3 × 4 x4 (y–2) 4 = 4 x–2 y4 × 3x2 y–2 xy–2 = 12x–2 + 2 + 1 y4 – 2 – 2 = 12x1 y0 = 12x
255 (j) 5 16x4 y6 z 2 × 5 2x6 y4 z –7 ÷ 16x8 y12 = 5 16 × 2x4 + 6 y6 + 4 z2 – 7 × 16x8 y12 = 5 32x10y10z -5 ÷ 16x8 y12 = 5 25 (x2 ) 5 (y2 ) 5 (2–1) 5 ÷ 42 (x4 ) 2 (y6 ) 2 = 2x2 y2 z –1 ÷ 4x4 y6 = 2x2 y2 z –1 4x4 y6 = 1 2x4 – 2 y6 – 2 z1 = 1 2x2 y4 z 9. Simplify: (a) 3n + 1 × 32n – 3 33n – 4 (b) 2n + 2 × {2(n – 1)} (n + 1) 2n(n – 1) (c) 5x + 5x + 1 5 . 5x (d) 4a + 22a – 1 3 × 22a – 1 (e) 52x + 1 + 25 × 52x – 1 10 × 52x (f) 4m + 2 – 4m + 1 4m + 1 + 2 . 4m + 1 Solution: (a) 3n + 1 × 32n – 3 33n – 4 = 3n + 1 + 2n – 3 – 3n + 4 = 32 = 9 (b) 2n + 2 × {2(n – 1)}(n + 1) 2n(n – 1) = 2n + 2 × 2(n – 1) (n + 1) 2n(n – 1) × 1 4n + 1 = 2n + 2 × 2n2 – 1 2n2 – n × 1 (22 ) n + 1 = 2n + 2 + n2 – 1 – n2 + n 22n + 2 = 22n + 1 22n + 2 = 22n + 1 – 2n – 2 = 2–1 = 1 2 (c) 5x + 5x + 1 5 . 5x = 5x + 5x . 51 5 . 5x = 5x (1 + 5) 5 . 5x = 6 5 = 1 1 5 (d) 4a + 22a – 1 3 × 22a – 1 = 22a + 22a – 1 . 2 3 × 22a . 2–1 = 22a(1 + 2–1) 3 × 22a . 2–1 = 1 + 1 2 3 2 = 2 + 1 2 3 2 = 3 2 3 2 = 1 (e) 52x + 1 + 25 × 52x – 1 10 × 52x =52x . 51 + 25 × 52x . 5–1 10 × 52x = 52x(5 + 25/5) 10 × 52x = 5 + 5 10 = 10 10 = 1 (f) 4m + 2 – 4m + 1 4m + 1 + 2 . 4m + 1 = 4m . 42 – 4m . 41 4m . 41 + 2 × 4m × 416 = 4m(42 – 4) 4m(4 + 2 × 4) = 16 – 4 4 + 8 = 12 12 = 1 10. Simplify: (a) 1 1 + xa – b + xc – b + 1 1 + xb – c + xa – c + 1 1 + xc – a + xb – a (b) 1 1 + yr – p + yr – q + 1 1 + yq – 1 + yq – p + 1 1 + yp – q + yp – r Solution: (a) Here, 1 1 + xa – b + xc – b + 1 1 + xb – c + xa – c + 1 1 + xc – a + xb – a = 1 1 + xa xb + xc xb + 1 1 + xb xc + xa xc + 1 1 + xc xa + xb xa = 1 xb + xa + xc xb + 1 xc + xb + xa xc + 1 xa + xc + xb xa = xb xb + xc + xa + xc xb + xc + xa + xa xb + xc + xa = xb + xc + xa xb + xc + xa = 1
256 (b) Here, 1 1 + yr – p + yr – q + 1 1 + yq – 1 + yq – p + 1 1 + yp – q + yp – r = 1 1 + y–p y–r + y–q y–r + 1 1 + y–r y–q + y–p y–q + 1 1 + y–q y–p + y–r y–p = 1 y–r + y–p + y–q y–r + 1 y–q + y–r + y–p y–q + 1 y–p + y–q + y–r y–p = y–r y–r + y–q + y–p + y–q y–r + y–q + y–p + y–p y–r + y–q + y–p = y–r + y–q + y–p y–r + y–q + y–p = 1 11. Simplify: (a) xa xb c × xb xc a × xc xa b (b) xp xq r × x–r x–q p × x–p x–r q Solution: (a) xa xb c × xb xc a × xc xa b = xca xbc × xab xca × xbc xab = xca + ab + bc – bc – ca – ab = x0 = 1 (b) xp xq r × x–r x–q p × x–p x–r q = xrp xqr × r –rp x–pq × x–pq x–qr = xrp – rp – pq – qr + pq + qr = x0 = 1 12. Simplify: (a) pq ap/q aq/p × qr aq/r ar/q × rp ar/p ap/r (b) c x1/c a x1/a × a x1/a b x1/b × b x1/b c x1/c (c) xa + b xc a – b × xb + c xa a – b × xc + a xb c – a (d) xa xb a + b – c × xb xc b + c – a × xc xa c + a – b Solution: (a) Here, pq a p/q a q/p × qr a q/r a r/q × rp a r/p a p/r = a p/a a q/p 1/rq × a q/r a r/q 1/qr × a r/p qp/r 1/rp = a p q × 1 pa a q p × 1 pq × a q r × 1 qr a r q × 1 qr × a r p × 1 rp a p r × 1 rp = a 1 q2 a 1 p2 × a 1 r 2 a 1 q2 × a 1 p2 a 1 r 2 = a 1 q2 + 1 r 2 + 1 p2 – 1 p2 – 1 q2 – 1 r 2 = a0 = 1 (b) c x1/c a x1/a × a x1/a b x1/b × b x1/b c x1/c = x 1 c × 1 c x 1 a × 1 a × x 1 a × 1 a x 1 b × 1 b × x 1 b × 1 b x 1 c × 1 c = x 1 c2 x 1 a2 × x 1 a2 x 1 b2 × x 1 b2 x 1 c2 = x 1 c2 + 1 a2 + 1 b2 – 1 a2 – 1 b2 – 1 c2 = x0 = 1 (c) xa + b xc a – b × xb + c xa a – b × xc + a xb c – a = (xa + b) a – b (xc ) a – b × (xb + c) b – c (xa ) b – c × (xc + a) c – a (xb ) c – a = xa2 – b2 xca – bc × xb2 – c2 xab – ca × xc2 – a2 xbc – ab = xa2 – b2 + b2 – c2 + c2 – a2 xca – bc + ab – ca + bc – ab = x0 x0 = 1
257 (d) xa xb a + b – c × xb xc b + c – a × xc xa c + a – b = xa2 + ab – ca xab + b2 – bc × xb2 + bc – ab xbc + c2 – ca × xc2 + ca – bc xca + a2 – ab = xa2 + ab – ca + b2 + bc – ab + c2 + ca – bc xab + b2 – bc + bc + c2 – ca + ca + a2 – ab = xa2 + b2 + c2 xa2 + b2 + c2 = xa2 + b2 + c2 – a2 – b2 – c2 = x0 = 1 13. Prove that: (a) x–1y y–1z z –1x = 1 (b) 3 ab2 3 a2 b4 3 a–3b–6 = 1 (c) xa xb 2 xb xc 2 xc xa 2 = 1 Solution: (a) Here, x–1y y–1z z –1x = 1 LHS = x–1y y–1z z –1x = x0 y0 z 0 = 1 × 1 × 1 = 1 = 1 = RHS, proved. (b) Here, 3 ab2 3 a 2 b4 3 a –3b–6 = 1 LHS = 3 ab2 3 a 2 b4 3 a –3b–6 = 3 a 1 + 2 – 3 b2 + 4 – 6 = 3 a 0 b0 = 3 1 . 1 = 1 = RHS, proved. (c) Here, xa xb 2 xb xc 2 xc xa 2 = 1 LHS = xa xb 2 xb xc 2 xc xa 2 = xa xb × xb xc × xc xa 2 = (xa + b + c – a – c – a) 2 = (x0 ) 2 = 12 = 1 = RHS, proved. 14. Show that: (a) a(x – y)z a(y – z)x a(z – x)y = 1 (b) (xa – b) a + b (xb – c) b + c (xc – a) c + a = 1 (c) (xa – b)a2 + ab + b2 . (xb – c)b2 + bc + c2 . (xc – a)c2 + ca + a2 = 1 Solution: (a) Here, a(x – y)z a(y – z)x a(z – x)y = 1 LHS = a(x – y)z a(y – z)x a(z – x)y = azx – yz . axy – xz . ayz – xy = aax – yz + xy – zx + yz – xy = a0 = 1 = HRS, proved. (b) Here, (xa – b) a + b (xb – c) b + c (xc – a) c + a = 1 LHS = (xa – b) a + b (xb – c) b + c (xc – a) c + a = xa2 – b2 . xb2 – c2 . xc2 – a2 = xa2 – b2 + b2 – c2 + c2 – a2 = x0 = 1 = RHS, proved. (c) Here, (xa – b)a2 + ab + b2 . (xb – c)b2 + bc + c2 . (xc – a)c2 + ca + a2 = 1 LHS = (xa – b)a2 + ab + b2 . (xb – c)b2 + bc + c2 . (xc – a)c2 + ca + a2 = xa3 – b3 . xb3 – c3 . xc3 – a3 = xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1 = RHS, proved. 15. Simplify: (a) ax ay z ay az x az ax y (b) xa xb a + b xb xc b + c xc xa c + a (c) m + (mn2 ) 1/3 + (m2 n)1/3 m – n × 1 – n1/3 m1/3 Solution: (a) Here, a x ay z ay a z x a z a x y = a zx ayz × a xy a zx × a yz a xy = azx + xy + yz – yz – zx – xy = a0 = 1 (b) Here, xa xb a + b xb xc b + c xc xa c + a = (xa – b) a + b . (xb – c) b + c (xc – a) c + a = xa2 – b2 . xb2 – c2 . xc2 – a2 = xa2 – b2 + b2 – c2 + c2 – a2 = x0 = 1 (c) Here, m + (mn2 ) 1/3 + (m2 n)1/3 m – n × 1 – n1/3 m1/3 = (m1/3) 3 + m1/3(n1/3) 2 + (m1/3) 2 n1/3 (m1/3) 3 – (n1/3) 3 × m1/3 – n1/3 m1/3
258 = m1/3{(m1/3) 2 + m1/3 n1/3 + (n1/3) 2 } (m1/3 – n1/3) {(m1/3) 2 + m1/3 n1/3 + (n1/3) 2 × m1/3 – n1/3 m1/3 = 1 16. Show that: (a) bc xb – c . ca xc – a . ab xa – b = 1 (b) x– (a2 + b2 ) xab b – a × xbc x– (b2 + c2 ) b – c × x(c2 + a2 ) x– ca c – a = 1 Solution: (a) Here, bc xb – c . ca xc – a . ab xa – b = 1 LHS = bc xb – c . ca xc – a . ab xa – b = xb – c bc . xc – a ca . xa – b ab = xb – c bc + c – a ca + a – b ab = xab – ca + bc – ab + ca – bc abc = x 0 abc = x0 = 1 = RHS, proved. (b) Here, x– (a2 + b2 ) xab b – a × xbc x– (b2 + c2 ) b – c × x(c2 + a2 ) x– ca c – a = 1 LHS = x– (a2 + b2 ) xab b – a × xbc x– (b2 + c2 ) b – c × x(c2 + a2 ) x– ca c – a = (x– a2 – b2 – ab)(– (a – b)) × (xbc + b2 + c2 )b – c × (xc2 + a2 + ca)c – a = x– (a2 + b2 + ab) × (– (a – b)) × x(b2 + bc + c2 ) (b – c) × x(c2 + a2 + ca) (c – a) = x(a – b) (a2 + b2 + ab) × x(b – c) (b2 + bc + c2 ) × x(c – a) (c2 + a2 + ca) = xa3 – b3 × xb3 – c3 × xc3 – a3 = xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1 = RHS, proved. 17. Simplify: (a) x + 1 y m x – 1 y n y + 1 x m y – 1 x n (b) 1 – n m n m – n 1 + m n m m – n m n – 1 n m – n n m + 1 m m – n Solution: (a) x + 1 y m x – 1 y n y + 1 x m y – 1 x n = xy + 1 y m xy – 1 y n xy + 1 x m xy – 1 x n = (xy + 1)m ym × (xy – 1)n yn (xy +1)m xm × (xy – 1)n xn = (xy + 1)m ym × (xy – 1)n yn × xm (xy + 1)m × xn (xy – 1)n = xm + n ym + n = x y m + n (b) 1 – n m n m – n 1 + m n m m – n m n – 1 n m – n n m + 1 m m – n = 1 – n m n m – n m n – 1 n m – n × 1 + m n m m – n n m + 1 m m – n = m – n m n m – n m – n n n m – n × m + n n m m – n n + m m m m – n = n m n m – n × m n m m – n = m n – n m – n × m n m m – n = m n – n m – n + m m – n = m n – n + m m – n = m n m – n m – n = m n
259 Additional Practice – IV 1. An arithmetic sequence is given: 5, 9, 13, 17, … (a) Find the common differences of the given sequence. (b) Write the general term of the given arithmetic sequence. (c) Which term of the given sequence is 49? Find it. Solution: Here, 5, 9, 13, 17 (a) d = t2 – t1 = t3 – t2 or, d = 9 – 5 = 13 – 9 = 4 Common difference (d) = 4. (b) Since first term (a) = 5, Common difference (d) = 4 The general term is given by tn = a + (n – 1)d = 5 + (n – 1) × 4 = 5 + 4n – 4 = 4n + 1 (c) Let nth term is 49, i.e. tn = 49 4n + 1 = 49 or, 4n = 48 or, n = 12 The 12th term is 49. 2. The first five terms of the geometric sequence are given: 5, 10, 20, 40, 80…………… (a) Write the formula to find the general term of a geometric sequence. (b) Find the common ratio of the given sequence. (c) Find the 10th term of the given sequence. (d) Which term of the given sequence is 5120? Find it. Solution: Here, geometric sequence 5, 10, 20, 40, 80, ......... (a) The general term of geometric sequence is tn = arn – 1 (b) Common ratio (r) = t2 t1 = 10 5 = 2 (c) Since a = 5, r = 2, tn = arn – 1 = 5 × 2n – 1 = 5 × 2n – 1 when n = 10, t10 5 × 210 – 1 = 5 × 29 = 5 × 512 = 2560 (d) Let, nth term be 5120. i.e. tn = 5120 or, arn – 1 = 5120 or, 5 × 2n – 1 = 5120 or, 2n – 1 = 1024 or, 2n – 1 = 210 or, n = – 1 = 10 or, n = 11 The 11th term is 5120. 3. The second term and sixth term of a geometric sequence are 192 and 12 respectively. (a) Find the common ratio of the given sequence. (b) Which term of the given sequence is 3? Find it. (c) Find the 8th term of the sequence.
260 (d) If the first three terms make an arithmetic sequence, what should be the value of the third term? Justify. Solution: (a) Here, tn = arn – 1 t2 = 192 or, ar2 – 1 = 192 or, ar = 192 ......................... (i) and t6 = 12 or, ar6 – 1 = 12 or, ar5 = 12 .......................... (ii) Dividing equation (ii) by equation (i) ar5 ar = 12 192 or, r4 = 1 16 = 1 24 r = 1 2 (b) Let, nth term of sequence be 3. i.e. tn = 3 we know that, r = 1 2 ar = 192 or, 1 2 = 192 or, a = 384 Now, tn = 3 or, arn – 1 = 3 or, 1 2 n – 1 = 3 384 = 1 128 = 1 2 7 or, n – 1 = 7 n = 8 Hence, 8th term of sequence is 3. (c) The 8th term, t8 = ar8 – 1 = 384 × 1 2 7 = 384 × 1 128 = 3 (d) Since t1 = a = 384 t2 = 192 d = t2 – t1 = 192 – 384 = – 192 Hence, t3 = a + (3 – 1)d = 384 + 2 × (–192) = 0 i.e. To be arithmetic sequences the third term is 0. 4. The first four terms of the arithmetic sequence are given: 5, 11, 17, 23, … (a) Find the common difference of the given sequence. (b) Find the 8th term of the sequence. (c) Which term of the given sequence is 131? Find it. (d) If the first three terms make a geometric sequence, what should be the value of the third term? Justify. Solution: Here, arithmetic sequence, 5, 11, 17, 23 (a) d = t2 – t1 = 11 – 5 = 6
261 (b) Since a = 5, d = 6 tn = a + (n – 1)d or, t8 = 5 + (8 – 1) × 6 = 5 + 7 × 6 = 47 (c) Let, nth term be 131 i.e. tn = 131 or, a + (n – 1)d = 131 or, 5 + (n – 1) × 6 = 131 or, 6(n – 1) = 126 or, (n – 1) = 21 or, n = 22 Hence, 22nd term is 131. (d) Here, t1 = a = 5 t2 = 11 The to be geometric series we need the common ratio as t2 t1 = 11 5 . Then, tn = arn – 1 or, t3 = 5 × 11 5 3 – 1 = 5 × 11 5 2 = 121 5 To be geometric series the third term should be 121 5 . 5. (a) Find the HCF of : x3 + 27 and x4 + 9x2 + 81 (b) Simplify: xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 (c) Simplify: 273n + 1 × (243)–4n/5 9n + 5 × 33n – 7 Solution: (a) Here, x3 + 27 and x4 + 9x2 + 81 1st expression = x3 + 27 = x3 + 33 = (x + 3) (x2 – 3x + 32 ) = (x + 3) (x2 – 3x + 9) 2nd expression = x4 + 9x2 + 81 = (x2 ) 2 + 92 + 9x2 = (x2 + 9)2 – 2 . x2 . 9 + 9x2 = (x2 + 9)2 – 9x2 = (x2 + 9)2 – (3x)2 = (x2 + 9 + 3x) (x2 + 9 – 3x) = (x2 + 3x + 9) (x2 – 3x + 9) HCF = x2 – 3x + 9 (b) Here, xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 = (xa – b)a2 + ab + b2 × (xb – c)b2 + bc + c2 × (xc – a)c2 + ca + a2 = xa3 – b3 × xb3 –c3 × xc3 –a3 = xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1 (c) Here, 273n + 1 × (243)–4n/5 9n + 5 × 33n – 7 = (33 ) 3n + 1 × (35 ) –4n/5 (32 ) n + 5 × 33n – 7 = 39n + 3 × 3–4n 32n + 10 × 33n – 7 = 39n + 3 – 4n – 2n – 10 – 3n + 7 = 30 = 1 6. (a) Simplify: a + b xa2 xb2 × b + c xb2 xc2 × c + a xc2 xa2 (b) Factorize: p3 q3 – q3 p3 (c) Factorize: y4 x4 – 21 y2 x2 + 100 Solution:
262 (a) Here, a + b xa2 xb2 × b + c xb2 xc2 × c + a xc2 xa2 = (xa2 – b2 ) 1 a + b × (xb2 – c2 ) 1 b + c × (xc2 – a2 ) 1 c + a = x (a + b) (a – b) (c + a) × x (b + c) (b – c) (b + c) × x (c + a) (c – a) (c + a) = xa – b × xb – c × xc – a = aa – b + b – c + c – a = x0 = 1 (b) Here, p3 q3 – q3 p3 = p q 3 – q p 3 = p q – q p p q 2 + p q . q p + q p 2 = p2 – q2 pq p2 q2 + 1 + q2 p2 = (p2 – q2 ) pq p4 + p2 q2 + q4 p2 q2 = (p + q) (p – 1) {(p2 ) 2 + (q2 ) 2 + p2 q2 } p3 q3 = (p + q) (p – q) {(p2 + q2 ) 2 – 2p2 q2 + p2 q2 } p3 q3 = (p + q) (p – q) {(p2 + q2 ) 2 – p2 q2 } p3 q3 = (p + q) (p – q) (p2 + q2 + pq) (p2 + q2 – pq) p3 q3 = (p + q) (p – q) (p2 + pq + q2 ) (p2 – pq + q2 ) p3 q3 Alternative Method p3 q3 – q3 p3 = p6 – q6 p3 q3 = (p3 ) 2 – (q3 ) 2 p3 q3 = (p3 + q3 ) (p3 – q3 ) p3 q3 = (p + q) (p2 – pq + q2 ) (p – q) (p2 + pq + q2 ) p3 q3 = (p + q) (p – q) (p2 + pq + q2 ) (p2 – pq + q2 ) p3 q3 (c) Here, y4 x4 – 21 y2 x2 + 100 = y4 – 21x2 y2 + 100x4 x4 = 1 x4 {(y2 ) 2 + (10x2 ) 2 – 21x2 y2 } = 1 x4 {(y2 – 10x2 ) 2 + 2 . y2 . 10x2 – 21x2 y2 } = 1 x4 {(y2 – 10x2 ) 2 – x2 y2 } = 1 x4 {(y2 – 10x2 ) 2 – (xy)2 } = 1 x4 {(y2 – 10x2 + xy) (y2 – 10x2 – xy)} = 1 x4 (y2 + xy – 10x2 ) (y2 – xy – 10x2 ) 7. (a) If xyz = 1, prove that: 1 1 + x + y–1 + 1 1 + y + z–1 + 1 1 + z + x–1 = 1 (b) Find the LCM of: x2 – x + 1 and x4 + x Solution: (a) Here, xyz = 1 xz = 1 y = y–1 i.e. y–1 = zx ........................ (i)
263 Again, xyz = 1 y = 1 xz = ........................ (ii) Now, LHS = 1 1 + x + y–1 + 1 1 + y + z–1 + 1 1 + z + x–1 = 1 1 + x + zx + 1 1 + 1 zx + 1 z + 1 1 + z + 1 x = 1 1 + x + zx + 1 zx + 1 + x zx + 1 x + zx + 1 x = 1 1 + x + zx + zx zx + 1 + x + x x + 2x + 1 = 1 1 + zx + x + zx 1 + zx + x + x 1 + zx + x = 1 = RHS, proved. (b) Here, x2 – x + 1 and x4 + x 1st expression = x2 – x + 1 = (x2 – x + 1) 2nd expression = x4 + x = x(x3 + 1) = x(x + 1) (x2 – x + 1) LCM = (x2 – x + 1) x(x + 1) = x(x + 1) (x2 – x + 1) 8. There is a two-digit number between 10 and 100. The number is 8 times the sum of its digits. If 45 is subtracted from the number, the places of the digits are interchanged. (a) Write a two-digit number in terms of x and y. (b) Write two equations in terms of x and y. (c) Find the two-digit number. Solution: Let x and y be two digit to form a number between 10 to 100. Then, (a) The required number be 10x + y. (b) The equations in terms of x and y are 10x + y = 8(x + y) .................... (i) 10x + y – 45 = 10y + x ............. (ii) Solving equation (i) and (ii) From equation (ii), 10x + y – 45 = 10y + x or, 9x – 9y – 45 = 0 or, x – y – 5 = 0 or, x = y + 5 ..................................... (iii) From equation (i) 10x + y = 8(x + y) or, 10x + y = 8x + 8y or, 2x – 7y = 0 Substituting for x = y + 5 from (iii), 2(y + 5) – 7y = 0 or, 2y + 10 – 7y = 0 or, 10 – 5y = 0 or, 10 = 5y or, y = 2 Substituting for y = 2, x = y + 5 = 2 + 5 = 7 Finally the required number is 10x + y = 10 × 7 + 2 = 72
264 9. According to the price list of Nepal Oil Corporation dated 2080/01/20, the price of one-liter petrol was Rs 10 more than one-liter diesel. On that day, Ramesh had bought 5-liter petrol and 7-liter diesel on Rs. 2030. (a) Present the given statement in the form of linear equation. (b) What is the price of petrol and diesel per liter? Find it. (c) If the prices of petrol and diesel are decreased by 10%, how much equal quantity of petrol and diesel can be bought by Ramesh in Rs 1224? Calculate it. Solution: Let the price of diesel be Rs. x per liter Then the price of petrol will be Rs. (x + 10). (a) According to question, 5 × Rs. (x + 10) + 7 × Rs. x = Rs. 2030 or, 5x + 50 + 7x = 2030 or, 12x = 1980 ............................ (i) which is required linear equation (b) From (i), 12x = 1980 or, x = 165 The price of diesel = Rs. 165/l So, The price of petrol = Rs. (165 + 10) = Rs. 175/l (c) After decreasing the price by 10% the new rates would be for diesel, Rs 165 – 10% of Rs. 165 = Rs. 148.50 for petrol, Rs. 175 – 10% of Rs. 175 = Rs. 157.50 Let Ramesh bought liters of both, then by question, Rs. 148.50 × k + Rs. 157.50 × k = Rs. 1224 or, Rs. 306k = Rs. 1224 or, k = 4 Hence Ramesh bought 4/4 l of both diesel and petrol. 10. In a Dangol Furniture Store, the cost of 6 chairs and 5 tables is Rs. 15000 and the cost of 9 tables and 7 chairs is Rs. 21300. (a) Consider the price of chair to be x and table to be y and form two equations. (b) Find the cost of single chair and table. (c) If the cost of chair is increased by 20% and cost of table is decrease by 25%, how many equal number of chairs and tables can be bought in Rs. 21600? Solution: Let the rate of chairs and tables are Rs. x and Rs. y respectively. (a) According to question, 6x + 5y = 15000 ................. (i) 9x + 7y = 21300 ................. (ii)
265 (b) Applying equation (i) × 7 - equation (ii) × 5 42x + 35y = 105000 45x + 35y = 106500 – – – – 3x x = = – 1500 500 Substituting for x on equation (i) 6 × 500 + 5y = 15000 or, 3000 + 5y = 15000 or, 5y = 12000 y = 24000 Hence, the cost of single chair is Rs. 500 and a table is Rs. 2400. (c) After the change in rates price of a chair = Rs. 500 + 20% of Rs. 500 = Rs. 600 price of a table = Rs. 2400 – 25% of Rs. 2400 = Rs. 1800. Let x be the equation number of chair and tables to be bought in Rs. 21600 i.e. Rs. 600 × x + Rs. 1800 × x = Rs. 21600 or, 2400x = 21600 or, x = 9 Hence 9/9 chairs and table can be bought. 11. Shusil started her car for the journey from Kathmandu to Pokhara at 11 AM with an average speed of 30 km/hr. After two hours, Ashok started his car on the same journey with average speed of 40 km/hr. (a) After what time will they meet? Write in terms of x. (b) After what time would they meet? Calculate it. (c) At what time would they meet? Calculate it. Solution: Here, the speed of Shusil = 30 km/hr. the speed of Ashok = 40 km/hr. Let they meet after x hours Shusil started. (a) Then the distance covered by Shusil = 30x km the distance covered by Ashok = 40(x – 2) km (b) According to question, 30x = 40(x – 2) or, 30x = 40x – 80 or, 10x = 80 or, x = 8 Hence, they meet after 8 hours Shusil started. (c) Since Shusil started at 11 AM they will meet at 11 + 8 = 19 = 7 PM.
266 50° 60° e UNIT V GEOMETRY CHAPTER TRIANGLES 13 13.1 Angles of Triangle Exercise 13.1 1. Observe the given figure and answer the following questions. (a) What is the supplementary angle of 4 in the given figure? (b) Which is the co-interior angle of 5 in the given figure? (c) What is the alternate angle of 2 in the given figure? (d) What is the sum of 1, 2 and 3 in ΔABC? (e) What is the sum of 1, 5 and 6 in the given figure? (f) Which angle is equal to the sum 1 and 2 in the figure? Solution: a. Supplementary of 4 is 3 and 5. b. Co-interior of 5 is 4. c. Alternate angle of 2 is 6. d. 1 + 2 + 3 = 180°. e. 1 + 5 + 6 = 180°. d. 1 + 2 = 4. 2. Observe the given figure and answer the following questions. (a) What is the value of (a + b + c) in the given figure? (b) Which angle is equal to a + b in the given figure? Solution: a. a + b + c = 180° b. a + b = d 3. (a) What is the value of 'e' in the adjoining triangle? (b) What is the measure of the angle 'x' in the given figure? 6 5 1 2 3 4 A B C D E 6 5 1 2 3 4 A B C D E a b c d a b c d 50° x 120°
267 Solution: a. e = 50° + 60° = 110° b. x + 50° = 120° x = 70° 4. Observe the following figures and answer the given questions. (i) State the exterior angle property of a triangle. (ii) Find the unknown angles (denoted by letters such as a, b, c, x, y, z, etc.) by using exterior angle property of the triangle. (a) (b) (c) (d) (e) (f) (g) (h) Solution: i. The exterior angle of triangle producing a side of triangle is equal to the sum of two non-adjacent interior angles of triangle. a. x + 55° + 60° = 180° [‡ Sum of interior angles of is 180°] or, x = 180° – 115° x = 65° And, y = 55° + 60° [‡ An exterior angle equals to sum of opposite interior angles of ] 50° 60° e 50° x 120° 60° 55° x y A B D C 80° 50° a P Q T R S c b 60° x 35° y A D C X Y B 120° 110° x y B D S T z A C P Q R 2a a 3a L M K N x L B N A M C 120° x 130° 130° F E D C 55° x E 110° B D F A C P Q x 60° 55° x y A B D C
268 b. a + 50° + 80° = 180° [‡ Sum of interior angles of is 180°] or, a = 180° – 130° a = 50° And, b = 50° + 80° [‡ An exterior angle equals to sum of opposite interior angles of ] Again, c + 50° = 180° [‡ Linear pair] c = 130° c. (60° + x) + 35° = 180° [‡ Co-interior angles] or, x = 180° – 95° x = 85° y + 60° = 180° [‡ Co-interior angles] y = 120° d. PQB + ABQ = 180° [‡ Co-interior angles] or, PQB + 120° = 180° PQB = 60° x = PQB [‡ Vertically opposite angles] x = 60° BCR + DCR = 180° [‡ Linear pair] or, BCR + 110° = 180° BCR = 70° y = BCR [‡ Corresponding angles] y = 70° x + y + z = 180° [‡ Sum of interior angles of ] or, 60° + 70° + z = 180° z = 50° e. Here, a + 2a + 3a = 180° [‡ Sum of interior angles of ] or, 6a = 180° a = 30° And, x + a = 180° [‡ Linear pair] or, x + 30° = 180° x = 150° f. Here, x + 130° + 120° = 360° [‡ Sum of exterior angles of ] or, x = 360° – 250° x = 110° g. Here, x + 55° = 130° [‡ An exterior angle equals to sum of opposite interior angles of ] x = 75° 80° 50° a P Q T R S c b 60° x 35° y A D C X Y B 120° 110° x y B D S T z A C P Q R 2a a 3a L M K N x L B N A M C 120° x 130° 130° 55° x E D C F
269 h. Here, DQF + EPB = 180° [‡ Co-exterior angles] or, DQF + 110° = 180° DQF = 70° And, x + 30° = DQF [‡ An exterior angle equals to sum of opposite exterior angles of ] or, x + 30° = 70° x = 40° 13.2 Properties of Isosceles Triangle Exercise 13.2 1. (a) Which angles are equal in ΔABC alongside? (i) a and b (ii) a and c (iii) b and c (iv) None (b) If Q = R in ΔPQR, which sides are equal? (i) PQ and PR (ii) PQ and QR (iii) PR and QR (iv) None (c) In ΔLMN, LM = LN and LK MN. What is the relation between MK and NK? (i) MN = MK (ii) MK = KN (iii) MK > KN (iv) MK < MN (d) In ΔABC, AB = AC and ∠BDA = ∠CDA. What is the relation between BAD = CAD? (i) BAD > CAD (ii) BAD < CAD (iii) BAD CAD (iv) BAD = CAD Solution: a. (iii) b and c b. (i) PQ and PR c. (ii) MK = KN d. (iv) BAD = CAD 2. (a) Write down two properties of an isosceles triangle. (b) What is the measure of each acute angle of an isosceles right-angled triangle? (c) What is the relation between the bisector of the vertical angle of an isosceles triangle and its base? (d) Write the relation between the perpendiculars draw from the base points to the equal sides of an isosceles triangle. Solution: a. Property I: two sides are equal. Property II: opposite angles of equal sides are equal. a b c A B C P Q R L M N K A B C D a b c A P B C Q R L M N K A B C E 110° B D F A C P Q x
270 b. Each acute angle of an isosceles right angled triangle is 45°. c. Bisector of the vertical angle of an isosceles triangle is perpendicular on its base. d. The length of perpendiculars drawn from the base points to the equal sides of an isosceles triangle are equal. 3. (a) In BAC, A = 90°, AB = AC. Find the base angles. (b) In ΔPQR, Q = 120°, PQ = QR. Find the base angles. Solution: a. If A = 90° and AB = AC, then B = C = 45°. b. In PQR, Q = 120°, PQ = QR, then P = R = 30°. 4. In the ΔPQR, the measures of some angles are shown. (a) Which angles are equal in an isosceles triangle? (b) Find the values of PQR and PRQ. (c) Is ΔPQR isosceles? Why? Solution: a. Opposite angles of equal sides are equal b. PRS + PRQ = 180° [‡ Linear pair] or, 115° + PRQ = 180° PRQ = 65° And, PQR + PRQ + RPQ = 180° [‡ Sum of interior angles of ] or, PQR + 65° + 50° = 180° PQR = 65° c. PQR + RPQ = PQS [⸪ Exterior angle theorem] PQR + 110° = 150° PQR = 40° Again, PRQ + PRS = 180° [Linear pair] PRQ + 150° = 180° or, PRQ = 30° As PQR PRQ, PQR is not an isosceles triangle. 5. Observe the equal sides marked in the following figures and answer the given questions. (i) (ii) (iii) (iv) (a) Find the values of a, b, c, d, e, wherever applicable. 50° 115° S R P Q 58° b a 3c 4c a 48° a b 120° a 50° 115° S R P Q C A B R P Q 120° 110° 150° P Q S R 110° 150° P Q S R
271 (b) Measure the remaining angles. Solution: i. Remaining b = a Then, a + a + 58° = 180° or, 2a = 122° a = 61° ii. Remaining a = 3c Then, 3c + 3c + 4c = 180° or, 10c = 180° c = 18° iii. Remaining angle of isosceles with 48° is a. Adjacent angle of 48° is b Then, a + a + 48° = 180° or, 2a = 132 a = 66° Again a + b + (48 + b) = 180° or, 66° + 2b + 48° = 180° or, 2b = 66° b = 33° iv. Linear pair of 120° is 60° Remaining of triangle is 60° Then, a + 60° + 60° = 180° a = 60° 6. (i) Find the angles of an isosceles triangle in which each of the base angles is half of the vertical angle. (ii) Find the angles of an isosceles triangle in which the vertical angle is half of the base angle. Solution: i. Alongside figure represent the question. Then, a + a + 2a = 180° or, 4a = 180° a = 45° then 2a = 90° So, 45°, 45° and 90° are angles of ii. Alongside figure represent the question, Then, a + 2a + 2a = 180° or, 5a = 180° a = 36° & 2a = 72° Hence, angles of are 36°, 72° and 72°. 48° a b a b 120° a a 2a a 2a a 2a 58° b a 3c 4c a
272 7. (i) (a) What is an isosceles triangle? Define it. (b) In the adjoining triangle, LM = LN. Show that LMN = MNL. (c) If MLN = 34°, find the measure of LMN and MNL. (ii) Prove that if two angles are equal in a triangle, the sides opposite to the equal angles are also equal. Solution: i. a. An isosceles triangle is the triangle with two equal sides. b. If LMN is drawn with LM = LN, LMN = MNL [The angles opposite to equal sides of isosceles triangle are equal.] c. If MLN = 34°, MLN + LMN + MNL = 180° or, 34° + LMN + LMN = 180° or, 2LMN = 146° LMN = 73° Also, MNL = 73° ii. Given: ABC = ACB To-prove: AB = AC Construction: Let draw AD BC Proofs: S.N. Statements S.N. Reasons 1. In ABC and ACD 1. i. ABD = ACD (A) i. From given ii. ADB = ADC (A) ii. AD BC iii. AD = AD (S) iii. Common sides 2. ABD –ACD 2. By A.A.S axiom 3. AB = AC 3. Being corresponding sides of –Triangles Proved 8. (i) Prove that the perpendicular drawn from the vertex of an isosceles triangle to the base bisects the base. (ii) In the isosceles triangle PR = PQ and QS = RS. (a) Prove that: QPS = RPS (b) Prove that: PSQR (c) If QPS = 30o, find the value of PQR. Solution: i. Given: AB = AC and AD BC To-prove: BD = DC Proofs: S.N. Statements S.N. Reasons 1. In ABC and ACD 1. ................. i. ADB = ADC (R) i. Being AD BC ii. AB = AC (H) ii. From given iii. AD = AD (S) iii. Being common sides 2. ABD –ACD 2. By R.H.S axiom 3. BD = DC 3. Being corresponding sides of –Triangles M N L M N L A B C D P Q R S A B C D
273 ii. Given: In PQR, PR = PQ, QS = RS To-prove: i. QPS = RPS ii. PS QR Proofs: S.N. Statements S.N. Reasons 1. In PQS and PRS 1. ................. i. PQ = PR (S) i. From given ii. QS = RS (S) ii. From given iii. PS = PS (S) iii. Common sides 2. PQS –PRS 2. By S.S.S axiom 3. QPS = RPS, PSQ = PSR 3. Being corresponding angles of –Triangles 4. PS QR 4. PSQ = PSR are Linear pair Proved iii. If QPS = 30° PQR = 90° – 30° [‡ PSQ = 90°] = 60° 9. (i) Prove that the base angles of an isosceles right triangle are equal and each measures half of right angle. (ii) If a triangle is an equilateral, show that it is equiangular triangle and each angle measures 60°. (iii) If a triangle is an equiangular, show that it is equilateral triangle. Solution: i. Given: In ABC, A = 90°, AB = AC To-prove: B = C = 45° We know, base angles of are equal So, B = C [⸪ Base angles] And, A + B + C = 180° [⸪ Angle sum property] or, 90° + B + B = 180° or, 2B = 90° B = 45° Also, C = 45° ii. Given, In ABC, AB = BC = AC To-prove: A = B = C = 60° B = C [‡ AB = AC] A = B [‡ AC = BC] And, A + B + C = 180° or, B + B + B = 180° or, 3B = 180° P Q R S A B C A B C
274 B = 60° Also, A = C = 60° iii. Given: In ABC, A = B = C To-prove: ABC is equilateral or, AB = BC = AC Here, AB = BC [‡ A = C] BC = AC [‡ A = B] Then, AB = BC = AC Hence, ABC is an equilateral . 10. (i) ABC is an isosceles triangle in which AB = AC and AD bisects CAE. (a) Prove that AD//BC. (b) If EAD = 48°, ACB. (ii) In a triangle ABC, AB = AC and O is an interior point of ΔABC. OB and OC are the bisectors of angles B and C. Prove that OBC is an isosceles triangle. (iii) PQR is an isosceles triangle in which PQ = PR. M and N are the mid-points of PR and PQ. Prove that QM = NR. Solution: i. (a) Given: In ABC, AB = AC, CAD = EAD To-prove: AD//BC Here, B = C [‡ AB = AC] And, B + C = EAC [‡ An exterior equals to sum of opposite interior] or, B + B = 2EAD [‡ CAD = EAD angles of ] or, 2B = 2EAD B = EAD AD//BC [‡ B = EAD are corresponding angles] (b) If EAD = 48° CAD = 48° And, ACB = CAD [‡ Alternate angles] ACB = 48° ii. Given: In ABC, AB = AC, OB and OC are bisector of ABC and ACB To-prove: OB = OC or OBC is an isosceles triangle. Here, ABC = ACB [‡ Being based angles of ] or, 1 2 ABC = 1 2 ACB or, OBC = OCB [‡ Being OB and OC are bisector of ABC = ACB] And, OB = OC [‡ Being opposite sides of OBC = OCB] Hence, OBC is an isosceles triangle. A B C A D B C E A D B C E A B C O
275 iii. Given: In PQR, PQ = PR and, M and N are mid-points of PR and PQ To-prove: QM = NR Proofs: S.N. Statements S.N. Reasons 1. In QMR and RNQ 1. ................. i. MR = NQ (S) i. 1 2 PQ = 1 2 PR ii. MRQ = NQR (A) ii. Base angles of PQR iii. QR = QR (S) iii. Being common sides 2. QMR –RNQ 2. By S.A.S axiom Proved 11. (i) In the given figure, ΔPQR is an isosceles triangle, QS = TR and SPT = 40°. (a) Prove that ΔPST is an isosceles triangle. (b) Find the measure of PST. (ii) In the adjoining figure, AB = AC, AD = AE and ADC = 75°. (a) Prove that BE = CD. (b) Find the measure of DAE. Solution: i. Given: In PQR, PQ = PR, QS = TR To-prove: PST is an isosceles triangle. Proofs: S.N. Statements S.N. Reasons 1. In PQS and PRT 1. ................. i. PQ = PR (S) i. From given ii. PQS = PRT ii. Being base angles of iii. QS = TR (S) iii. From given 2. PQS –PRT 2. By S.A.S axiom 3. PS = PT Hence PST is an isosceles triangle 3. Being corresponding sides of –Triangles Proved Again, SPT = 40° Since, PST = PTS Now, SPT + PST + PTS = 180° or, 40° + PST + PST = 180° or, 2PST = 140° PST = 70° M P Q R N T P Q R S 40° E A B C D 75° T P Q R S 40°
276 ii. Given: In ABC, AB = AC, AD = AE To-prove: BE = CD Proofs: S.N. Statements S.N. Reasons 1. In ABE and ACD 1. ................. i. AEB = ADC (A) i. Being base angles of ADE ii. ABE = ACE (A) ii. Being base angles of ABC iii. AB = AC (S) iii. From given 2. ABE –ACD 2. By A.A.S axiom 3. BE = CD 3. Being corresponding sides of –Triangles Proved Again, ADE = ADC = 75° And, AED = ADE = 75° Now, DAE + ADE + AED = 180° or, DAE + 75° + 75° = 180° or, DAE = 180° – 150° DAE = 30° 12. (i) In the adjoining figure, ΔABC and ΔCDE are both isosceles triangles and ACB = DCE. Prove that: ADC = BEC. (ii) In the given figure, ΔPQR and ΔRST are both equilateral triangles. Prove that: PS = QT (iii) In the given figure, D, E and F are the mid-points of the sides AB, BC and CA of the equilateral triangle ABC respectively. Prove that ΔDEF is also an equilateral triangle. Solution: i. Given: ABC and CDE are isosceles triangles and ACB = DCE To-prove: ADC = BEC Proofs: S.N. Statements S.N. Reasons 1. In ADC and BEC 1. ................. i. AC = BC (S) i. From given ii. ACD = BCE (A) ii. Adding BCD both sides on ACB = DCE iii. CD = CE (S) iii. From given 2. ADC –BEC 2. By S.A.S axiom 3. ADC = BEC 3. Being corresponding sides of –Triangles Proved E A B C D 75° C A D B E R P S Q T F A B C D E C A D B E
277 ii. Given: PQR and PST are equilateral triangles To-prove: PS = QT Proofs: S.N. Statements S.N. Reasons 1. In PSR and QTR 1. ................. i. PR = QR (A) i. From given ii. PRS = QRT (S) ii. Adding ARS both sides on PRQ = SRT = 60° iii. SR = RT (A) iii. From given 2. PSR –QTR 2. By A.S.A axiom 3. PS = QT 3. Being corresponding sides of –Triangles Proved iii. Given: In ABC, D, E and F are mid-points of AB, BC and CA To-prove: DEF is an equilateral triangle Proofs: S.N. Statements S.N. Reasons 1. In ADF and BED 1. ................. i. AD = BD (S) i. D is mid-point of AB ii. DAF = DBE (A) ii. A = B = C iii. AF = BE (S) iii. 1 2 AC = 1 2 BC 2. ADF –BED 2. By S.A.S axiom 3. DF = DE 3. Being corresponding sides of –Triangles 4. DE = EF 4. Proceed same as 3 5. DE = EF = DF 5. From (3) & (4) 6. DEF is an equilateral triangle. 6. Being all equal sides. Proved 13.3 Sides and Angles of a Triangle Exercise 13.3 1. (a) What is the relation of the sum of any two sides of triangle and its third side? (b) What is the relation between b + c and a in the given triangle? (c) Which is the longest side in the given triangle? (d) Which is the shortest side in the given triangle PQR? (e) Which is the greatest angle in the given triangle LMN? R P S Q T F A B C D E a b c B C A 88° 55° 37° Q R P 80° 60° M N L 7 cm
278 Solution: a. Here, Sum of any two sides of triangle is greater than its third side. b. b + c > a c. BC is longest side as it is opposite to greatest angle A = 88°. d. P + Q + R = 180° or, 80° + 60° + R = 180° R = 40° PQ is shortest side as it is opposite to smallest R = 40°. e. MLN is greatest angle as it is opposite to longest side MN = 7 cm. 2. (a) In ΔABC A = 72° and B = 32°. Find the longest and shortest sides. (b) In ΔPQR Q = 65° and R = 42°. Find the longest and shortest sides. (c) In ΔPQR Q = 90° and R = 25°. Find the longest and shortest sides. Solution: a. In ABC, A = 72°, B = 32° Then, A + B + C = 180° or, 72° + 32° + C = 180° C = 76° Hence, longest side = AB (opposite to C) Shortest side = AC (opposite to B) b. In PQR, Q = 65° and R = 42° Then, P + Q + R = 180° or, P + 65° + 42° = 180° P = 73° Hence, Longest side = QR (opposite to P) Shortest side = PQ (opposite to R) c. In PQR, Q = 90° and R = 25° Then, P + Q + R = 180° or, P + 90° + 25° = 180° P = 65° Hence, Longest side = PR (opposite to Q) Shortest side = PQ (opposite to R) 3. (a) In ΔPQR PQ = 6 cm and QR = 8 cm and RP = 7 cm. Determine the greatest and the smallest angles. (b) In ΔXYZ, XY = 5 cm, XZ = 13 cm and YZ = 12 cm. Determine the greatest and the smallest angles. (c) In ΔABC, AB = 15 cm, AC = 5 cm and BC = 18 cm. Determine the greatest and the smallest angles. M N L 7 cm Q R P 80° 60° a b c B C A 88° 55° 37°
279 Solution: a. In PQR, PQ = 6 cm, QR = 8 cm and RP = 7 cm Then, Greatest angle = P (opposite to QR) Smallest angle = R (opposite to PQ) b. In XYZ, XY = 5 cm, XZ = 13 cm and YZ = 12 cm Then, Greatest angle = Y (opposite to XZ) Smallest angle = Z (opposite to XY) c. In ABC, AB = 15 cm, AC = 5 cm, BC = 8 cm Then, Greatest angle = C (opposite to AB) Smallest angle = B (opposite to AC) 4. (a) In an equilateral triangle, any two sides are together greater than the third side. Verify it by experimental method. (b) In a right-angled triangle, the hypotenuse is the greatest side. Verify it experimentally. Solution: a. Given: ABC is an equilateral triangle. To-verity: AB + BC > AC Experiments: Fig. AB BC AC Result i. 3 cm 3 cm 3 cm AB + BC > AC ii. 4 cm 4 cm 4 cm AB + BC > AC iii. 5 cm 5 cm 5 cm AB + BC > AC Conclusion: The sum of length of two sides is greater than length of third side. b. Given: ABC is right angle triangle. To-verify: Hypotenuse AC is longest side. A B C Fig. (i) A B C Fig. (ii) A B C Fig. (iii) A B C Fig. (i) A B C Fig. (ii) A B C Fig. (iii)
280 Experiments: Fig. AB BC AC Result i. 3 cm 4 cm 5 cm AC is longest ii. 6 cm 8 cm 10 cm iii. 5 cm 12 cm 13 cm Conclusion: Hypotenuse is longest side of right angle triangle. 5. (a) Show by measurement that the sum of the medians of a triangle is less than the perimeter of the triangle. (b) Show by measurement that the perimeter of a quadrilateral is greater than the sum of its diagonals. Solution: a. Given: AQ, BR and CP are medians of ABC. To-verify: AQ + BR + CP AB + BC + AC. Experiments: Fig. AQ + BR + CP AB + BC + AC Result i. 5 + 4 + 4 + 13 6 + 4 + 6 = 16 AQ + BR + CP < AB + BC + AC ii. 6 + 5 + 3 = 14 5 + 3 + 6 = 14 iii. 7.5 + 3 + 5.5 = 16 6 + 6 + 8 = 20 Conclusion: The sum of the medians of a triangle is less than the perimeter of the triangle. b. Given: AC and BD are diagonals of Quadrilateral ABCD. To-verify: AC + BD AB + BC + CD + AD Experiments: Fig. AC + BD AB + BC + CD + AD Result i. 5 + 5 = 10 cm 3 + 4 + 3 + 5 = 15 cm AB + BC + CD + AD > AC + BD ii. 7 + 6 = 13 cm 2 + 4 + 3 + 6 = 15 cm iii. 8 + 5 = 13 cm 6 + 3 + 3 + 5 = 17 cm Conclusion: The perimeter of a quadrilateral is greater than the sum of its diagonals. A B C Q P R Fig. (i) A B C P R Q Fig. (ii) A B C P R Q Fig. (iii) A B C D Fig. (i) Fig. (ii) Fig. (iii) A B C D A B C D
281 13.4 Similar Triangles Exercise 13.4 1. (a) By which condition, the given triangles ABC and DEF are similar? (i) ASA (ii) AAS (iii) AAA (iv) SSS (b) Which is the name of similar condition for ΔPQR and ΔLMN where ∠Q = ∠M and PQ QR = LM MN . (i) ASA (ii) SAS (iii) SSS (iv) AAA (c) What is the similar condition for ΔABC and ΔPQR when AB QR = BC RP = AC PQ ? (i) SSS (ii) SSA (iii) AAA (iv) SAS (d) By which condition ΔPQR and ΔPQS are similar? (i) SSA (ii) AAA (iii) SSS (iv) ASA Solution: a. ABC ~ DEF by (iii) AAA axiom. b. PQR ~ LMN by (iii) SAS axiom c. ABC ~ PQR by (i) SSS axiom d. PQR ~ PQS by (ii) AAA axiom 2. (a) By which axiom the triangles ABC and PQR are similar? A B C D E F P Q R L M N A B C P Q R P Q R S A B C P Q R 55° 55° 9 cm 6 cm A B C E F D P Q R M N L A B C Q R P P Q R S
282 (b) By which condition the adjoining triangles are similar? (c) By which condition the given triangles are similar? Solution: a. B = Q = 55° And, AB BC = PQ QR or, 6 9 = 4 6 or, 2 3 = 2 3 ABC ~ PQR by SAS axiom b. In ABC and PQR i. B = Q = 50° ii. C = R = 65° iii. A = P (being remaining angles) So, ABC ~ PQR by A.A.A. axiom c. In PQR and XYZ PQ XY = QR YZ = PR XZ or, 15 10 = 21 14 = 18 12 or, 3 2 = 3 2 = 3 2 Hence, PQR ~ XYZ by S.S.S. axiom. 3. (a) The given triangles ABC and PQR are similar. Write a pair of two corresponding angles, other than give angles. (b) If ΔXYZ ~ ΔPQR, write a pair of two corresponding sides. Solution: a. In ABC and PQR B = Q 50° 65° 50° 65° A B C P Q R 55° 55° 9 cm 6 cm A B C 50° 65° P Q R 50° 65° P Q R X Y Z A B C P Q R X Y Z P Q R
283 And, AB BC = PQ QR or, 4 6 = 2 3 2 3 = 2 3 Hence, ABC ~ PQR by S.A.S. axiom. b. In XYZ and PQR i. X = Q ii. Y = R iii. Z = P Hence, XYZ ~ PQR by A.A.A. axiom. 4. Solve the following numerical problems: (a) In the following figure AB//DE, find the values of x and y. (b) In the adjoining figure, AC//BD, find the values of x and y. (c) In the figure. DE//BC, find the values of x and y. (d) In the figure, DA//BC and ∠DBA = ∠ACB. Find length of DA and DB. Solution: a. In CDE and CBA i. C = C (being common angles) ii. E = B (being 90°) iii. D = A (corresponding angles) CDE ~ CBA by A.A.A. axiom Now, sides of similar triangles are proportion. CD AC = DE AB = CE BC A C B 4 cm 12 cm E D x y 5 cm A C B D O x y A B C 7 cm D E 4 cm x y 2 cm D A B C 3.5 cm A B C P Q R X Y Z P Q R A C B 4 cm 12 cm E D x y 5 cm
284 Taking 1st & 2nd Taking 2nd & 3rd CD AC = DE AB DE AB = CE BC or, x 13 = 4 12 or, 4 12 = y 5 x = 4.33 cm y = 1.67 cm b. In AOC and BOD i. A = B (alternate angles) ii. O = O (V.O.A.) iii. C = D (alternate angles) AOC ~ BOD by A.A.A. axiom Now, OC OD = AC BD = OA OB or, x 10 = 3 5 = 6 y Taking 1st & 2nd Taking 2nd & 3rd x 10 = 3 5 3 5 = 6 y x = 6 cm y = 10 cm c. Here, In ABC and ADE i. A = A (common angle) ii. B = D (corresponding angles) iii. C = E (corresponding angles) ABC ~ ADE by AAA axiom Now, AB AD = BC DE = AC AE or, y + 2 y = 7 4 = x + 5 5 Taking 2nd & 3rd Taking 1st & 2nd 7 4 = x + 5 5 y + 2 y = 7 4 or, 35 4 = x + 5 or, 1 + 2 y = 7 4 or, 8.75 – 5 = x or, 2 y = 3 4 x = 3.75 cm or, y = 8 3 y = 2.66 cm d. In ABD and BCA i. A = B (‡ alternate angles) ii. B = C (‡ given) iii. D = A (‡ remaining angles) A C B D O x y A B C 7 cm D E 4 cm x y 2 cm D A B C 3.5 cm
285 Hence, ABD ~ BCA by AAA axiom Now, AD AB = AB BC = BD AC or, AD 3 = 3 3.5 = BD 4 Taking 1st & 2nd Taking 2nd & 3rd AD 3 = 3 3.5 3 3.5 = BD 4 or, AD = 9 3.5 or, 12 3.5 = BD AD = 2.57 cm BD = 3.43 cm 5. In the following figures, prove that ΔABE ~ ΔCDE: (a) (b) (c) Solution: a. In ABE and CDE AEB = CED And, AE BE = CE ED = 1 ABE ~ CDE by S.A.S. axiom b. In ABE and DCE i. A = D (‡ alternate angles) ii. B = C (‡ alternate angles) iii. AEB = DEC (‡ V.O.A.) ABE ~ DCE by A.A.A. axiom c. In ABE and DCE i. B = D (‡ being 90°) ii. AEB = CED (‡ common angles) iii. BAE = DCE (‡ remaining angles) Hence, ABE ~ DCE by A.A.A. axiom 6. In the following figures, prove that ΔBAC ~ ΔBDA: (a) (b) (c) A B C D E A B C D E A B E D C B A C D B A C D A B C D A B C D E A B C D E A B E D C
286 Solution: a. In BAC and BDA i. BAC = BDA (‡ being 90°) ii. ABC = ABD (‡ common angles) iii. ACB = BAD (‡ remaining angles) Hence, BAC ~ BDA by A.A.A. axiom b. In BAC and BDA i. ABC = BDA (‡ being 90°) ii. BAC = BAD (‡ common angle) iii. ACB = ABD (‡ remaining angles) Hence, BAC ~ BDA by A.A.A. axiom c. In BAC and BDA i. BCA = BAD (‡ given) ii. ABC = ABD (‡ common angle) iii. BAC = ADB (‡ remaining angles) Hence, BAC ~ BDA by A.A.A. axiom 7. (a) In the trapezium ABCD, AB//DC and the diagonal AC and BD intersect each other at E. Prove that AB:BE = CD:DE. (b) In a triangle ABC, X and Y are any points on AB and AC respectively and XY//BC. Prove that AX:AB = AY:AC. Solution: a. Given: AB//DC, diagonals AD and BD intersect at E. To-prove: AB : BE = CD : DE Proofs: S.N. Statements S.N. Reasons 1. In ABE and CDE 1. i. ABE = CDE (A) i. Being alternate angles ii. AEB = CED (A) ii. Being V.O.A. iii. BAE = DCE (A) iii. Being alternate angles 2. ABE ~ CDE 2. By A.A.A axiom 3. AB BE = CD DE AB : BE = CD : DE 3. Being corresponding sides of ~ Triangles Proved b. Given: XY//BC To-prove: AX : AB = AY : AC B A C D B A C D A B C D A B D C E A B D C E A B C X Y
287 Proofs: S.N. Statements S.N. Reasons 1. In AXY and ABC 1. i. AXY = ABC (A) i. Being corresponding angles ii. AYZ = ACB (A) ii. Being corresponding angles iii. XAY = BAC (A) iii. Being common angle 2. AXY ~ ABC 2. By A.A.A axiom 3. AX AB = AY AC AX : AB = AY : AC 3. Being corresponding sides of ~ Triangles Proved 8. (a) In ΔABC, ∠A = 90° and AD ⊥ BC. If BC = a, AC = b, AB = c, BD = x and DC = a – x, prove that a2 = b2 + c2 . (b) In the following figures, Quad. ABCD ~ Quad. EFGH. Find the length of AC, HF, AD and GH. Solution: a. Given: A = 90°, AD BC, BC = a, AC = b, AB = c, BD = x, DC = a – x To-prove: a2 = b2 + c2 Here, In ABC & DAC i. BAC = ADC (‡ being 90°) ii. ACB = ACD (‡ being common angle) iii. ABC = CAD (‡ being reaming angles) Hence, ABC ~ DAC by AAA axiom Now, AB AD = AC CD = BC AC or, c y = b a – x = a b Taking 2nd and 3rd, b a – x = a b or, b2 = a2 – ax or, ax = a2 – b2 ............(1) Similarly, ABC ~ DBA Then, AB BD = AC AD = BC AB or, c x = b y = a c C A B D b c y A B D C 3 cm E F H G C A B D b c y
288 Taking 1st and 3rd, c x = a c or, c2 = ax ..................(2) From (1) and (2) c2 = a2 – b2 i.e. a2 = b2 + c2 proved b. Here, Quad ABCD ~ Quad EFGH AB EF = BC FG = CD GH = AD EH = AC EG = BD FH or, 1 2 = 1.5 3 = 2 GH = AD 5 = AC 5 = 3 HF Taking 1st and 5th Taking 1st and 6th 1 2 = AC 5 1 2 = 3 HF AC = 2.5 cm HF = 6 cm Taking 1st and 4th Taking 1st and 3rd 1 2 = AD 5 1 2 = 2 GH AD = 2.5 cm GH = 4 cm 9. (a) Observe the adjoining picture. Find the height of the window of the tree house. (b) Observe the adjoining picture. Find the height of the building. (c) Observe the adjoining picture at the same time. Find the height of the tree. Solution: a. In ABC and EFC i. ABC = EFC (‡ being 90°) ii. ACB = ECF (‡ being common angle) iii. BAC = FEC (‡ remaining angles) Now, ABC ~ EFC by A.A.A. axiom and, AB EF = BC FC or, AB 5 = 72 40 AB = 9 ft b. In ABC and DEC i. ABC = DEC (‡ being 90°) ii. ACB = DCE (‡ both 50°) iii. BAC = EDC (‡ remaining angles) A B D C 3 cm E F H G A B C F E 5 ft 40 ft 72 ft 50° 50° D E B A 20 ft 200 ft C
289 Now, ABC ~ DEC and, AB DE = BC EC or, AB 5.5 = 200 20 AB = 55 ft c. In ABC and DEF i. ABC = DEF (‡ being 90°) ii. ACB = DFE (‡ shadow at same time) iii. BAC = EDF (‡ being reaming angles) Now, ABC ~ DEF And, AB DE = BC EF or, AB 1.5 = 50 2 or, AB = 75 2 AB = 37.5 m A C B 50 m D F E 2 m 1.5 m
290 CHAPTER PARALLELOGRAM 14 14.1 Properties of Parallelogram Exercise 14.1 1. Define the following: (a) Parallelogram (b) Rectangle (c) Rhombus (d) Square Solution: a. Parallelogram: The quadrilateral having parallel opposite sides is called parallelogram. b. Rectangle: The parallelogram or quadrilateral having all angles equal to 90° is called rectangle. c. Rhombus: The parallelogram having all equal sides is called rhombus. d. Square: The rectangle having all equal sides is called square. 2. Write any two properties of the following quadrilaterals. (a) Parallelogram (b) Rectangle (c) Rhombus (d) Square Solution: a. Parallelogram: Property I: Opposite sides are parallel. Property II: Opposite sides are equal. b. Rectangle: Property I: All angles are equal and 90°. Property II: Diagonals are equal. c. Rhombus: Property I: All sides are equal. Property II: Diagonal bisect each other perpendicular. d. Square: Property I: All sides and angles are equal. Property II: Diagonals are equal. 3. (a) Write two properties of a rectangle distinct from a parallelogram. (b) What are two properties of a square distinct from a rectangle? (c) Write two properties of a square distinct from a rhombus. (d) Write two properties of a rhombus distinct from a rectangle. Solution: a. Rectangle Parallelogram i. All angles are equal ii. Diagonals are equal i. Opposite angles are equal ii. Diagonals bisect each other b. Square Rectangle i. All sides are equal. ii. Diagonals bisect perpendicularly. i. Opposite sides are equal. ii. Diagonals bisect each other.
291 c. Square Rhombus i. All sides and angles are equal ii. Diagonals are equal and bisect perpendicularly i. All sides are equal ii. Diagonals bisect perpendicularly d. Rhombus Rectangle i. All sides are equal ii. Diagonals bisect perpendicularly i. All angles are equal ii. Diagonals are equal 4. (a) In the figure, AB = CD and AB// DC. What is the relation between AD and BC? (b) In the given figure, PQ = RS and PQ// RS. Write the relation between PS and QR. Solution: a. If AB = CD and AB//DC, then AD = BC and AD//BC b. If PQ = RS and PQ//RS, then PO = OS and QO = OR 5. (a) The diagonals of AC and BD in the given parallelogram ABCD ....... to each other. (i) intersect (ii) bisect (iii) trisect (iv) divide (b) In the figure, OS = OQ and OP = OR. The quadrilateral PQRS is a .......... . (i) rectangle (ii) rhombus (iii) parallelogram (iv) square (c) The opposite sides of a parallelogram are ..... (i) equal (ii) parallel (iii) not equal (iv) both (i) and (ii) (d) The diagonals of a square are ..... (i) not equal (ii) parallel (iii) bisected (iv) perpendicularly bisected Solution: a. Diagonal AC and BD (ii) bisect to each other b. In quad PQRS, if OS = OQ and OP = OR, then quad PQRS is (iii) parallelogram. D A B C Q P R S O D A B C Q P R S O D A B C O S P Q R O D A B C O S P Q R O
292 c. The opposite sides of parallelogram are (i) equal. d. The diagonals of square are (iv) perpendicularly bisected. 5. (i) Observe the given parallelogram ABCD in which AB = 2x + 3 cm and CD = x + 7 cm. (a) Write the relation between the opposite sides of a parallelogram. (b) Find the value of x in the given parallelogram. (c) Find the length of AB and CD. (ii) Study the given parallelogram PQRS, the diagonals PR and QS intersect at a point O. Also, PO = 3x + 1 cm and RO = x + 7 cm. (a) Write the relation between the opposite angles of a parallelogram. (b) Find the value of x in the given parallelogram PQRS. (c) Find the length of PR. (iii) In given figure, KL = MN and KL//MN. If KM = 3x + 6 cm and LN = 5x – 2 cm, find the length of KM. (iv) In the given figure, AB = CD and A//CD, AO = 5x and DO = x + 8 cm. Find the length of AD. Solution: i. a. Opposite sides of parallelogram are equal. i.e. AB = CD and AD = BC b. Here, AB = CD or, 2x + 3 = x + 7 x = 4 cm c. AB = 2 × 4 + 3 = 11 cm CD = AB = 11 cm ii. a. Diagonal of parallelogram bisect each other, i.e. PO = OR and SO = OQ b. Here, PO = OR or, 3x + 1 = x + 7 or, 2x = 6 x = 3 cm c. Here, PQ = PO + OR PR = 3x + 1 + x + 7 = 4x + 8 = 4 × 3 + 8 = 20 cm iii. If KL = MN and KL//MN then, KM = LN and KM//LN Here, KM = LN i.e. 3x + 6 = 5x – 2 or, 8 = 2x x = 4 cm And, KM = 3 × 4 + 6 = 18 cm D A B C x + 7 2x + 3 S P Q R O L K M N 5x – 2 3x + 6 B A C D O D A B C x + 7 2x + 3 S P Q R O L K M N 5x – 2 3x + 6
293 iv. If AB = CD and AB//C AO = OD & BO = OC Here, AO = OD i.e. 5x = x + 8 or, 4x = 8 x = 2 cm And, AD = 5x + x + 8 = 6x + 8 = 6 × 2 + 8 = 20 cm 6. Find the value of x, y and z from the following figures: (i) (ii) (iii) (iv) In the parallelogram, if ∠D = (4x + 28°) and ∠B = (6x + 20°). (a) Write the relation between the diagonals of a parallelogram. (b) Find the value of x in the given parallelogram ABCD. (c) Find the measures of ∠B and ∠A of the parallelogram. (v) In the following figure, ABCD is a parallelogram. (a) What is parallelogram? Define it. (b) Find the values of x, y and z. (c) How many degrees added in z will form a rectangle? (vi) Observe the given figure ABCDE. (a) In which condition does a parallelogram form a rectangle? (b) Find the values of x, y and z. (c) What is the relation between the angles x and z? (d) Write the name of the quadrilateral AECB. Solution: i. Here, x = 150° (‡ corresponding angles) And, y = x (‡ alternate angles) y = 150° Also, z + 150° = 180° (‡ Co-interior angles) z = 30° ii. Here, z = 150° = 180° (‡ Co-interior angles) z = 30° And, y + z = 80° (‡ An exterior & equal to sum of two opp. interior angles of ) y = 50° Then, x = y (‡ alternate angles) x = 50° y z x 150° z x y 80° 150° y x 4x z B A C D O y z x 150° z x y 80° 150°
294 iii. Here, y = x (‡ corresponding angles) And, 4x + y = 180° (‡ Co-interior angles) or, 4x + x = 180° or, 5x = 180° x = 36 And, y = 36° z = y (‡ opposite angles of ) z = 36° iv. a. Opposite angles of parallelogram are equal. i.e. D = B and A = C b. Here, D = B 4x + 28 = 6x + 20° or, 28 – 20° = 6x – 4x or, 8° = 2x x = 4° c. B = 6x + 20° = 6 × 4 + 20° = 44° d. A + B = 180° (‡ Co-interior angles) or, A + 44° = 180° A = 136° v. a. Parallelogram is quadrilateral with opposite parallel sides. b. BAC = HAG (‡ V.O.A.) = x ACD = BAC (‡ Alternate angles) = x DCE + ACD + ACB = 180° (‡ Straight angle) or, 58° + x + 88° = 180° or, x = 180° – 146° x = 34° z = 58° (‡ corresponding angles) y = z (‡ alternate angles) y = 58° vi. a. A parallelogram becomes rectangle when diagonals are equal. b. ADE = AED = z (‡ base angles of ) ADE = BCD = x (‡ corresponding angles) So, x = z And, x + z + 40° = 180° (‡ sum of angles of ) or, x + x = 140° or, 2x = 140° x = 70° Also, z = 70° y x 4x z 4x + 28° 6x + 20° A B D C x y G F H A D C B 58° E 88° z A E D C 40° z 40° x y B
295 Again, x + y + 40° = 180° (‡ sum of angles of ) or, 70° + y + 40° = 180° or, y = 180° – 75° y = 70° c. The relation of x and z are equal. d. Quadrilateral AECB is isosceles trapezium. 7. (a) Prove that the opposite sides of parallelogram are equal. (b) In the parallelogram RHLA, RH//AL and RA//HL. Prove that ∠ARH = ∠HLA and ∠RHL = ∠RAL. (c) If the opposite sides of a quadrilateral are equal then the quadrilateral is a parallelogram. (d) If the opposite angles of a quadrilateral are equal then the quadrilateral is a parallelogram. (e) Prove that the diagonals of a parallelogram bisect each other. (f) In the given quadrilateral PQRS, PO = RO and QO = SO. Show that PQRS is a parallelogram. Solution: a. Given: In ABCD, AB//CD and AD//BC To-prove: AB = CD, AD = BC Construction: Let joint B and D Proofs: S.N. Statements S.N. Reasons 1. In ABD and CDB 1. i. ABD = CDB (A) i. Being alternate angles ii. BD = BD (S) ii. Common sides iii. ADB = CBD (A) iii. Alternate angles 2. ABD –CDB 2. By A.S.A axiom 3. AB = CD, AD = BC 3. Being corresponding sides of –Triangles Proved b. Given: In RHLA, RH//AL, RA//HL To-prove: ARH = HLA and RHL = RAL Proofs: S.N. Statements S.N. Reasons 1. ARH + RAL = 180° 1. Being co-interior angles 2. HLA + RAL = 180° 2. Being co-interior angles 3. ARH + RAL = HLA + RAL ARH = HLA 3. From (1) & (2) 4. RHL = RAL 4. Proceed same as 3. Proved c. Given: AB = DC and AD = BC To-prove: ABCD is parallelogram Construction: Let join A and C D A B C A R H L D C A B
296 Proofs: S.N. Statements S.N. Reasons 1. In ABC and CDA 1. ................. i. AB = CD (S) i. From given ii. BC = AD (S) ii. From given iii. AC = AC (S) iii. Common sides 2. ABC –CDA 2. By S.S.S axiom 3. BAC = ACD, ACB = CAD 3. Being corresponding sides of –Triangles 4. AB//CD and BC//AD 4. Being equal alternate angles 5. ABCD is a parm. 5. Being//opposite sides. Proved d. Given: In quad. ABCD A = C and B = D To-prove: ABCD is parallelogram Proofs: S.N. Statements S.N. Reasons 1. A = C & B = D 1. From given 2. A + B + C + D = 360° 2. Being sum of interior angles of quad 3. A + B + A + B = 360° 2(A + B) = 360° A + B = 180° 3. From (1) & (2) 4. AD//BC 4. Being supplementary co-interior angles 5. AB//DC 5. Proceed same as 4. 6. ABCD is parallelogram 6. Being parallel opposite sides Proved e. Given: In ABCD, diagonals AC and BD intersect at O. To-prove: AO = OC, BO = OD Proofs: S.N. Statements S.N. Reasons 1. In AOD and COB 1. ................. i. OAD = OCB (A) i. Being alternate angles ii. AD = BC (S) ii. Being opposite sides of parm. iii. ADO = OBC (A) iii. Being alternate angles 2. AOD –COB 2. By A.S.A axiom 3. AO = OC, DO = OB 3. Being corresponding sides of –Triangles Proved f. Given: In quad PQRS, PO = RO, QO = SO To-prove: PQRS is parallelogram D C A B D C A B O S R P Q O