=> -X 10 X h = 24 2 .. 24 X 2 h =---------= 4.8 cm 10 19. (a) According to the question, •h 2 V243 — a =------- 4 4 2 _ t/81 X 3 _ 9 a/3 a/3 a/3 a = V9 = 3 cm 20. (6) Let sides of isosceles triangle are 3x, 3x and 4x. ™ ,,n • ,. a + b + c Then, half-perimeter (s) =------------- 2 3x + 3x + 4x ,. =---------------------= 5x 2 Given, area of isosceles triangle = 18-n/I squnits ,/s(s - a) (s - i) {s-c) = 18^5 ^5x (5x - 3x) (5x - 3x) (5x - 4x) = 18^5 ^5x X 2x X 2x X x = 18a/5 => 2a/5x2 = 18a/5 => x2 =9
=> x = 3 .". Third side of isosceles triangle = 4X3 = 12 units 21. (a) Since, AC2 = ABZ + BC2 A \ 15 ^\25 e 20 c
Quadrilateral 24. (a) Required area = Length x Breadth = 15x8 = 120 sqm 25. (d) Required area = Length X Breadth = 100 X40 =4000 sqm = (4 X 103 ) sqm 26. (a) According to the question, Lx B = Area => LX15 = 300 [L = length and B = breadth] L = ^ = 20m 15 27. (c) Let length of rectangle = 5x and breadth of rectangle = 3x According to the question, 5x - 3x = 8 => 2x = 8 x = 4 Length = 5x = 5x4 = 20m Breadth = 3x = 3x4 = 12m .•. Required area = Length X Breadth = 20 X 12 = 240 sqm 28. (G) Let length = 2x and breadth = x According to the question, 2x — x = 5 => x = 5 .'. Required perimeter = 2(2x + x) = 2 x 3x = 2X3X5 = 30 cm
32. (a) Length of rectangle = 40 m Let breadth of = x Then, according of the question, (40 + x)15 = 40 Xx => 600 + 15x = 40x=> 25x = 600 x = 24 m 33. (a) Let the length of rectangle = L m .". Breadth of rectangle = B m Using conditions from the question,
L - B = 23 ...(i) 2(L + B) = 206 L + B = 103 ...(ii) On adding Eqs. (i) and (ii), we get 2L = 126 => L = 63 m => B = 103 -63 = 40 m Then, area of rectangle = L X. B = 63 X 40 = 2520 m2 . 34. (c) Let length and breadth of a rectangle is x and y. Then, as per first condition, (x - 5) (y + 3) = xy - 9 => xy-5y+3x-15 = xy-9 => 3x-5y = 6 ...(i) As per second condition, (x + 3) (y + 2) = xy + 67 => xy+3y+2x+6 = xy+67 => 2x+3y = 61 ...(ii) On multiplying Eq. (i) by 3 and Eq. (ii) by 5, then adding, we get Hence, the length of rectangle is 17 m. 35. (a) Let the width of the rectangle = x units Length = (2 x + 5) units According to the question, Width cannot be negative.
.'. Width = 5 units ∴ Length = 2x+5 =2x5+5 = 15units .". Perimeter of the rectangle = 2(15+5) =40 units 36. (d) Given that, 2 Area of rectangle = 2a = 1 X b => ix jb = 2a2 = lxa => i = 2a Now, In AACD AC2 = AD2 + CD2 = a2 + 4a2 = 5a2 Side of square (AC) - av5 Hence, area of square = (av5) = 5a 37. (c) According to the formula, Percentage increase in diagonals = 5% 38. (6) According to the formula, ■/ Percentage increase in sides = 10%
.•. Percentage increase in diagonals = 10% 39. (d) Let length of the rectangular field = 7x m and breadth of the rectangular field = Zx m According to the question, Area of rectangular field = Length x Breadth => 3584 = 7x X 2x => 14x2 = 3584 => x2 =5^5i = 256 14 => x2 = 256 => x = 16m /. Length of rectangular field = 7x 16 = 112m and breadth of rectangular field = 2X 16 = 32m /. Perimeter of rectangle = 2 (Length + Breadth) = 2(112+ 32) = 2 X 144 = 288 m
47. (e) Distance travelled by A in 15 s
57. (c)
Circle
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