Area Problems
The area of a rectangle is thrice that of square. Length of the rectangle is 40 cm and the breadth of the rectangle is ( 3 / 2 ) times that of the side of the square. The side of the square in cm is?
Let the side of the square = y cmThen, breadth of the rectangle = 3y/2 cm ? Area of rectangle = (40 x 3y/2) cm2= 60y cm2? 60y = 3y2? y = 20Hence, the side of the square = 20 cm
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 the side of the square = 100 m So area of square = 100 x 100 = 10000.New length = 140 m, New breadth = 130 mIncrease in area = [(140 x 130) - (100 x 100)] m2= 8200 m2? Required increase percent = (8200/ 10000) x 100 % = 82%
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%
EXAMIANS