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%
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 one diagonal be k.Then, other diagonal = (60k/100) = 3k/5 cmArea of rhombus =(1/2) x k x (3k/5) = (3/10) = 3/10 (square of longer diagonal)Hence, area of rhombus is 3/10 times.
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