Two squares have sides x cm and (x + 4) cm.

Solution:

Given that sides of the squares are $=x \mathrm{~cm}$ and $=(x+4) \mathrm{cm}$. Then

According to question,

Sum of the areas of square $=656 \mathrm{~cm}^{2}$

So,

$x^{2}+(x+4)^{2}=656$

 

$x^{2}+x^{2}+8 x+16=656$

$2 x^{2}+8 x+16-656=0$

$2 x^{2}+8 x-640=0$

 

$2\left(x^{2}+4 x-320\right)=0$

$x^{2}+4 x-320=0$

$x^{2}-16 x+20 x-320=0$

$x(x-16)+20(x-16)=0$

 

$(x-16)(x+20)=0$

$(x-16)=0$

$x=16$

or

$(x+20)=0$

$x=-20$

Sides of the square never are negative.

Therefore the side of the other square is

$=(x+4)$

$=16+4$

$=20$

Hence, sides of the square be $16 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively.