Area Problems
A rectangle has 20 cm as its length and 200 sq cm as its area. If the area is increased by 1 1/ 5 time the original area by increase its length only then the perimeter of the rectangle so formed (in cm) is
Area of 4 walls = 2(l+b)h=2(10+7) x 5 = 170 sq mArea of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq marea to be planted = 170 -12 = 158 sq m Cost of painting = Rs. 158 x 3 = Rs. 474
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
EXAMIANS