Increase in circumference of circle = 5%? Increase in radius is also 5%.Now, increase in area of circle = 2a + (a2/100) %Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%
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 of the field =1215/135 = 9 hec= 90000 m2 [1 hec =10000 m2]? Side of the field = ?90000 = 300 mPerimeter of the field = 4 x 300 = 1200 mNow, cost of putting a fence around field = (1200 x 75)/100 = ? 900
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
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