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 their areas will be x² and y² respectively.

It is given that 4x − 4y = 24 or x − y = 6

x = y + 6

Also, x² + y² = 468

= (6+y)² +y² = 468

= 36 + y² +12y + y² = 468

= 2y² +12y – 432 = 0

= y² + 6y – 216 = 0

= y² +18y – 12y – 216 = 0

= y(y+18)(y – 12) = 0

= 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