Sum of the areas of two squares is 640 m2.


Sum of the areas of two squares is $640 \mathrm{~m}^{2}$. If the difference of their perimeters is $64 \mathrm{~m}$. Find the sides of the two squares.


Let the sides of the squares are $x \mathrm{~m}$ and $=y \mathrm{~m}$. Then

According to question,

Sum of the difference of their perimeter $=64 \mathrm{~m}$

$4 x-4 y=64$


$y=x-16 \ldots .(1)$

And sum of the areas of square $=640 \mathrm{~m}^{2}$


Putting the value of x in equation (2) from equation (1)



$x^{2}+x^{2}-32 x+256=640$

$2 x^{2}-32 x+256-640=0$


$2 x^{2}-32 x-384=0$

$2\left(x^{2}-16 x-192\right)=0$

$x^{2}-16 x-192=0$

$x^{2}-24 x+8 x-192=0$









Sides of the square never are negative.

Therefore, putting the value of x in equation (1)




Hence, sides of the square be $24 \mathrm{~m}$ and $8 \mathrm{~m}$ respectively.

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