Sum of the areas of two squares is $640 m^{2}$ , if the difference of their perimeters is $64 m$ , find the sides of two squares.

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Solution

Let the side of the first square be $x$ and that of another be $y$ .
Case(i)
Area of square $= (S i d e)^{2}$
$\Rightarrow x^{2} + y^{2} = 640 m^{2} \dots E q (i)$
Case(ii)
Difference of their perimeter $= 64 m$
So,
$\Rightarrow 4 x - 4 y = 64$
$\Rightarrow x - y = 16$
$\Rightarrow x = 16 + y$
Put the value of $x$ in $E q (i)$ we get,
$\Rightarrow (16 + y)^{2} + y^{2} = 640$
$\Rightarrow 256 + y^{2} + 32 y + y^{2} = 640$
$\Rightarrow 2 y^{2} + 32 y = 384$
$\Rightarrow y^{2} + 16^{y} = 192$
$\Rightarrow y^{2} + 24 y - 8 y - 192 = 0$
$\Rightarrow y (y + 24) - 8 (y + 24) = 0$
$∴ y = 8$
Hence, $x = 16 + 8 = 24$
So, the length of the required sides are $8 m$ and $24 m$ .

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