Area Problems
Radhika runs along the boundary of a rectangular park at the rate of 12 km/hr and completes one full round in 15 minutes. If the length of the park is 4 times its breadth , the area of the park is?
Speed = 12 x (5/18) m/sec =10/3 m/sec there4; perimeter = (10/3) x 15 x 60 m=3000 m? 2( a + 4a) = 3000 m? a = 300 mSo, length = 1200 m and breadth = 300 m ? Area = (1200 x 300 ) m2 = 360000m2
Let original radius be r.Then, according to the questions,? (r + 1)2 - ?r2 = 22? ? x [(r + 1)2 - r2] = 22? (22/7) x (r + 1 + r ) x (r + 1 - r) = 22? 2r + 1 = 7 ? 2r = 6 ? r = 6/2 = 3 cm
Circumference = No.of revolutions * Distance covered Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.Number of revolutions per min. =(1100/4.4) = 250.
Let original length = x metres and original breadth = y metres. Original area = xy sq.m Increased length = 120 100 and Increased breadth = 120 100 New area = 120 100 x * 120 100 y = 36 25 x y m 2 The difference between the Original area and New area is: 36 25 x y - x y 11 25 x y Increase % = 11 25 x y x y * 100 = 44%