The sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m, find the sides of the square.

The sum of the areas of two squares is 640 m². If the difference of their perimeters is 64 m, find the sides of the square.

Answers (1)

Given $\rightarrow$ sum of areas of two squares = 640 m²difference of other perimeter = 64 m
Let square 1 has side S₁ and square 2 has S₂
Now $S_{1}^{2}+S_{2}^{2}= 640---(i)$
$4S_{1}-4S_{2}= 64$
$\Rightarrow S_{1}= 16+S_{2}---(ii)$
Put value of S₁ in (i)
$\left ( 16+S_{2} \right )^{2}+S_{2}^{2}= 640$
$16^{2}+S_{2}^{2}+2\times 16\times S_{2}+S_{2}^{2}= 640$
$\Rightarrow 2S_{2}^{2}+2\times 16S_{2}+256-640= 0$
$\Rightarrow S_{2}^{2}+16S_{2}+128-320= 0$ (divide by 2)
$\Rightarrow S_{2}^{2}+16S_{2}-192= 0$
$\Rightarrow S_{2}^{2}+\left ( 24-8 \right )S_{2}-192= 0\; \; \left ( 192= 24\times 8 \right )$
$\Rightarrow S_{2}^{2}+24S_{2}-8S_{2}-192= 0$
$\Rightarrow S_{2}\left ( S_{2} +24\right )-8\left ( S_{2}+24 \right )= 0$
$\Rightarrow \left ( S_{2} -8\right )\left ( S_{2}+24 \right )$
$\Rightarrow S_{2} = 8,-24$
Since $S_{2}$ can not be negative
$\Rightarrow S_{2} = 8$
$S_{1} = 16+8$ (from (ii))
$S_{1} = 24$