Area Problems
A person observed that he required 30 s time to cross a circular ground along its diameter than to cover it once along the boundary. If his speed was 30m/min, than the radius of the circular ground is (take ? = 22/7)
Let the radius of circular field = r m.Speed of person in m/s = 30/60 = 1/2m/sAccording to the question,[(2?r) /(1/2)] - [(2r)/(1/2)] = 30? 4?r - 4r = 30? [4 x (22/7) - 4]r =30? (125 - 4)r = 30 ? (8.5)r = 30? r = 30/8.5 = 3.5 m
Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %
Area of park = 100 x 100 = 10000 m2Area of circular lawn = Area of park - area of park excluding circular lawn= 10000 - 8614 = 1386Now again area of circular lawn = (22/7) x r2 = 1386 m2? r2 = (1386 x 7) / 22= 63 x 7= 3 x 3 x 7 x 7? r = 21 m
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