Area Problems
The length of hall is (4/3) times its breadth. If the area of hall be 300 square meters, the difference between the length and the breadth is?
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
Original area = ?(d/2)2= (?d2) / 4New area = ?(2d/2)2= ?d2Increase in area = (?d2 - ?d2/4)= 3?d2/4? Required increase percent = [(3?d2)/4 x 4/(?d2) x 100]%= 300%
let original radius = r and new radius = (50/100) r = r/2 original area = ?r 2 and new area = ? r / 2 2 decrease in area = 3 ?r 2 / 4 * 1 ?r 2 *100 = 75%
EXAMIANS