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

Solution:

Given that the sides of two square be $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$

Then according to question

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

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

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

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

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

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

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

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

$(x+20)=0$

$x=-20$

Or

$(x-16)=0$

$x=16$

Since, sides of the squares being a positive, so x cannot be negative.

Therefore,

When $x=16$ then

$x+4=16+4$

$=20$

Thus, sides of the squares be $16 \mathrm{~cm}, 20 \mathrm{~cm}$