Sum of the area of two squares is $468 \mathrm{~m}^{2}$.

Question. Sum of the area of two squares is $468 \mathrm{~m}^{2}$. If the difference of their perimeters is $24 \mathrm{~m}$, find the sides of the two squares.

Solution:

Let the sides of the two squares be $x \mathrm{~m}$ and $y \mathrm{~m}$. Therefore, their perimeter will be $4 x$ and $4 y$ respectively and their areas will be $x^{2}$ and $y^{2}$ respectively.

It is given that $4 x-4 y=24$

$x-y=6$

$x=y+6$

Also, $x^{2}+y^{2}=468$

$\Rightarrow(6+y)^{2}+y^{2}=468$

$\Rightarrow 36+y^{2}+12 y+y^{2}=468$

$\Rightarrow 2 y^{2}+12 y-432=0$

$\Rightarrow y^{2}+6 y-216=0$

$\Rightarrow y^{2}+18 y-12 y-216=0$

$\Rightarrow y(y+18)-12(y+18)=0$

$\Rightarrow(y+18)(y-12)=0$

$\Rightarrow y=-18$ or 12

However, side of a square cannot be negative.

Hence, the sides of the squares are 12 m and

$(12+6) m=18 m$