Area Problems
The length and breadth of a square are increased by 40% and 30% respectively. The area of a resulting rectangle exceeds the area of the square by?
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%
let ABCD be the given parallelogram area of parallelogram ABCD = 2 x (area of triangle ABC) now a = 30m, b = 14m and c = 40m s=1/2 x (30+14+40) = 42 Area of triangle ABC = s s - a s - b s - c = 42 12 28 2 = 168sq m area of parallelogram ABCD = 2 x 168 = 336 sq m
Let original radius be r.Then, according to the questions,? (r + 1)2 - ?r2 = 22? ? x [(r + 1)2 - r2] = 22? (22/7) x (r + 1 + r ) x (r + 1 - r) = 22? 2r + 1 = 7 ? 2r = 6 ? r = 6/2 = 3 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
Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%
EXAMIANS