Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

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 x² and y^2, respectively.

Given,

4x – 4y = 24

x – y = 6

x = y + 6

Also, x² + y² = 468

⇒ (6 + y²) + y² = 468

⇒ 36 + y² + 12y + y² = 468

⇒ 2y² + 12y  - 432 = 0

⇒ y² + 6y – 216 = 0

⇒ y² + 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.