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
According to the question, Area of semi- circle = 77 m(1/2) x ? x r2 = 77? r2 = (77 x 2 x 7)/22? r = 7m ?Circumference of semi- circle =?r + 2r= ( ? + 2)r = [(22/7) + 2] x 7 = 36 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