Question

The difference of areas of two squares is A. The difference of their perimeters is P. The sum of their perimeters is

• A. $\frac{2A}{P}$
• B. $\frac{4A}{P}$
• C. $\frac{8A}{P}$
• D. $\frac{16A}{P}$

Answer 1

The correct option is D

$\frac{16A}{P}$

Let the sides of squares be $x$ and $y$

Then their areas are $x^2,y^2$

Their perimeters are $4x,4y$

Sum of their perimeters = $4x+4y$ = $4(x+y)$

Given,
$x^2$ – $y^2=A$

$\Rightarrow (x-y)(x+y)=A$ .....(1)

Also, $4x-4y=P$

$\Rightarrow 4(x-y)=P$ $\Rightarrow x-y=\frac{P}{4}$

Substituting value of $x-y$ in (1)

$(x-y)(x+y)=A$

$\frac{P}{4}(x+y)=A$

$(x+y)=\frac{4A}{P}$

$4(x+y)=\frac{16A}{P}$ [Multiplying both sides by 4, as we need sum of perimeters]

Therefore, the sum of their perimeters is $\frac{16A}{P}$.