Question:
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.
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.