Question

There are two rectangles AEFD and EBCF in the square ABCD, The area of the square is 400$c{m}^2$. The areas of those rectangles are in the ratio 2:3 ,then find the perimeter of bigger rectangle

• A. 240 cm
• B. 64 cm
• C. 56 cm
• D. 40 cm

Answer 1

The correct option is B 64 cm

Area of square ABCD = $sid{e}^2$
= $A{B}^2$
Given Area = 400 $c{m}^2$
Side of the square = $\sqrt{4}00$ = 20 cm
Let breadth of rectangle BCFE be y cm

The length of both the rectangles will be equal to the side of square.

According to condition
$\frac{\text{area of ADFE}}{\text{area of BCFE}}$ = $\frac{3}{2}$
= $\frac{AD\times (20-y)}{BC\times y}$ = $\frac{3}{2}$
= $\frac{20-y}{y}$ = $\frac{3}{2}$ ...(AD=BC)
= 2$\times$(20 - y) = 3$\times$y
= 40 - 2y = 3y
= 40 = 3y +2y = 5y
= y = $\frac{40}{5}$ = 8 cm = FC = EB

Hence AE = DF = 20-y = 12 cm
Hence perimeter of rectangle AEFD = 2(DF+AD)
= 2(12+20) = 2$\times$32
= 64 cm