Area Problems
A rectangle has 20 cm as its length and 200 sq cm as its area. If the area is increased by 1 1/ 5 time the original area by increase its length only then the perimeter of the rectangle so formed (in cm) is
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%
Distance travelled in 1 revolution = circumference of the wheel = ?d= ( 22/7 ) x 63= 198 cmSo the distance travelled in 100 revolutions = 100 x distance travelled in 1 revolution = 100 x 198 cm= 19800 cm = 198 m
Let area 100 m2Then, side = 10 m New side = 125 % of 10= (125/100) x 10= 12.5 m New area = 12.5 x 12.5 m2=(12.5)2 sq. m? Increase in area = (12.5)2 - (10)2 m2= 22.5 x 2.5 m2=56.25 m2% Increase = 56.25 %
EXAMIANS