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 $=64 \mathrm{~m}$
$4 x-4 y=64$
$x-y=16$
$y=x-16 \ldots .(1)$
And sum of the areas of square $=640 \mathrm{~m}^{2}$
$x^{2}+y^{2}=640$.......(2)
Putting the value of x in equation (2) from equation (1)
$x^{2}+(x-16)^{2}=640$
$x^{2}+x^{2}-32 x+256=640$
$2 x^{2}-32 x+256-640=0$
$2 x^{2}-32 x-384=0$
$2\left(x^{2}-16 x-192\right)=0$
$x^{2}-16 x-192=0$
$x^{2}-24 x+8 x-192=0$
$x(x-24)+8(x-24)=0$
$(x-24)(x+8)=0$
$(x-24)=0$
$x=24$
or
$(x+8)=0$
$x=-8$
Sides of the square never are negative.
Therefore, putting the value of x in equation (1)
$y=(x-16)$
$=24-16$
$=8$
Hence, sides of the square be $24 \mathrm{~m}$ and $8 \mathrm{~m}$ respectively.
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