Area Problems
The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
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.
Let original length = x metres and original breadth = y metres. Original area = xy sq.m Increased length = 120 100 and Increased breadth = 120 100 New area = 120 100 x * 120 100 y = 36 25 x y m 2 The difference between the Original area and New area is: 36 25 x y - x y 11 25 x y Increase % = 11 25 x y x y * 100 = 44%
speed = 12 km/h = 12 × 5 18 = 10 3 m / s distance covered = 20 × 2 × 22 7 × 50 = 44000 7 m time taken = distance /speed = 44000 7 × 3 10 s e c = 4400 × 3 7 × 1 60 m i n = 220 7 m i n
We know that area of triangle = ( base x height ) / 2So area of triangle = (BC x AD) / 2 = (AC x BE) / 2 ? (10 x 4.4) / 2 = (11 x h) / 2? h= (10 x 4.4)/11 = 4 cm
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
EXAMIANS