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
Perimeter of rectangle = Circumference of circle = 2?r=2 x ( 22/7 ) x 42= 264 cm Now perimeter of rectangle = 2 x ( 6a + 5a )? 2 x (6a + 5a) = 264? a = 12Smaller side of rectangle = 5a = 60 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 %
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 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