Sum of the areas of two squares is $468 \, m^2$. If the difference of their perimeters is $24 \, m$,find the sides of the two squares.

(A) Let the sides of the two squares be $x \, m$ and $y \, m$. Therefore,their perimeters are $4x$ and $4y$ respectively,and their areas are $x^2$ and $y^2$ respectively.
It is given that the difference of their perimeters is $24 \, m$:
$4x - 4y = 24$
$x - y = 6$
$x = y + 6$
It is also given that the sum of their areas is $468 \, m^2$:
$x^2 + y^2 = 468$
Substituting $x = y + 6$ into the area equation:
$(y + 6)^2 + y^2 = 468$
$y^2 + 12y + 36 + y^2 = 468$
$2y^2 + 12y - 432 = 0$
Dividing by $2$:
$y^2 + 6y - 216 = 0$
Factoring the quadratic equation:
$y^2 + 18y - 12y - 216 = 0$
$y(y + 18) - 12(y + 18) = 0$
$(y + 18)(y - 12) = 0$
This gives $y = -18$ or $y = 12$. Since the side of a square cannot be negative,we take $y = 12 \, m$.
Then,$x = 12 + 6 = 18 \, m$.
Thus,the sides of the two squares are $18 \, m$ and $12 \, m$.