# Sum of the areas of two squares is 640 m2.

Question:

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.

Solution:

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.