Area Problems
The two parallel sides of a trapezium are 1 meters and 2 meters respectively . The perpendicular distance between them is 6 meters. The area of the trapezium is?
perimeter = total cost / cost per m = 10080 /20 = 504mside of the square = 504/4 = 126mbreadth of the pavement = 3mside of inner square = 126 - 6 = 120marea of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq mcost of pavement = 246*6*50 = Rs. 73800
Let original length = x and original breadth = y. Decrease in area = xy - ❨ 80 x x 90 y ❩ 100 100 = ❨ xy - 18 xy ❩ 25 = 7 xy. 25 ∴ Decrease % = ❨ 7 xy x 1 x 100 ❩% = 28%
let the side of the square be x meterslength of two sides = 2x metersdiagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0 . 59 x 2 x * 100 % = 30% (approx)
Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = √2x = (1.41x) m. Saving on 2x metres = (0.59x) m. Saving % = ❨ 0.59x x 100 ❩% = 30%
EXAMIANS