Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y, respectively
And the area of the squares will be $x^{2}$ and $y^{2}$, respectively.
Given,
4x – 4y = 24
x – y = 6
x = y + 6
Also, $x^{2}+ y^{2}$ = 468
⇒$ (6 + y^{2}) + y^{2} $= 468
⇒ $36 + y^{2} + 12y + y^{2}$ = 468
⇒ $2y^{2} + 12y + 432$ = 0
⇒$y^{2} + 6y – 216$ = 0
⇒$y^{2} + 18y – 12y – 216$ = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = -18, 12
As we know, the side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.