Area Problems
Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is
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%
Let a = 13, b = 14 and c = 15. Then, s = 1 2 a + b + c =21 (s- a) = 8, (s - b) = 7 and (s - c) = 6. Area = s s - a s - b s - c = 21 × 8 × 7 × 6 = 84 sq.cm
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