Let the radius of the park be r, then?r + 2r = 288(? + 2)r = 288? [(22/7) + 2]r = 288? r = (288 x 7)/36 = 56 ? Area of the park = (1/2)?r2= (1/2) x (22/7) x 56 x 56= 4928
Distance covered in one revolution = total distance travelled / total number of revolution. = ( 88 x 1000) / 1000 m = 88 mWe know that the distance covered in one revolution = circumference of the wheel.? ?d = 88? 22d / 7 = 88? d = 28 m
Let length of rectangular field = 5y,so width = 4y.From question5y - 4y = 20m? x = 20m ? Length = (5 x 20)m = 100 mBreadth = (4 x 20) m = 80 m ? Perimeter = 2 (100 + 80 ) m = 360 m
Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%
EXAMIANS