Area Problems
A circular road runs rounds a circular ground. If the difference between the circumference of the outer circle and the inner circle is 66 meters, the width of the road is?
Let the parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
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 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
EXAMIANS