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


Two squares have sides $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$. The sum of their areas is $656 \mathrm{~cm}^{2}$. Find the sides of the squares.


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 $=656 \mathrm{~cm}^{2}$




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

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

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


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

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

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









Sides of the square never are negative.

Therefore the side of the other square is




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

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