Area Problems
The length of a class room floor exceeds its breadth by 25 m. The area of the floor remains unchanged when the length is decreased by 10 m but the breadth is increased by 8 m. The area of the floor is
Let the breadth of floor be 'b' m. Then, length of the floor is 'l = (b + 25)' Area of the rectangular floor = l x b = (b + 25) × b According to the question, (b + 15) (b + 8) = (b + 25) × b b 2 + 8 b + 15 b + 120 = b 2 + 25 b 2b = 120 b = 60 m. l = b + 25 = 60 + 25 = 85 m. Area of the floor = 85 × 60 = 5100 sq.m.
Let the side of the square = 100 m So area of square = 100 x 100 = 10000.New length = 140 m, New breadth = 130 mIncrease in area = [(140 x 130) - (100 x 100)] m2= 8200 m2? Required increase percent = (8200/ 10000) x 100 % = 82%
Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares 13 . 5 * 10000 m 2 = 135000 m 2 Let altitude = x metres and base = 3x metres.Then, 1 2 * 3 x * x = 135000 ? x 2 = 90000 ? x = 300 Base = 900 m and Altitude = 300 m.
EXAMIANS