Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the sides of two squares be 'a' m and 'b' m.

As we know that area of square = (side)² and perimeter of square = 4 x side

It is given that sum of the areas of two squares is 260 m².

⇒ a² + b² = 260 ………. (1)

And, the difference of their perimeters is 24 m.

⇒ 4a - 4b = 24

⇒ 4(a - b) = 24

⇒ a - b = $\dfrac{24}{4}$

⇒ a - b = 6

⇒ a = 6 + b ………. (2)

Substituting the value of a in equation (1), we get

⇒ (6 + b)² + b² = 260

⇒ 36 + b² + 12b + b² = 260

⇒ 36 + 2b² + 12b - 260 = 0

⇒ 2b² + 12b - 224 = 0

⇒ 2(b² + 6b - 112) = 0

⇒ b² + 6b - 112 = 0

⇒ b² + 14b - 8b - 112 = 0

⇒ b(b + 14) - 8(b + 14) = 0

⇒ (b + 14)(b - 8) = 0

⇒ (b + 14) = 0 or (b - 8) = 0

⇒ b = -14 or b = 8

Since, side of square cannot be negative.

Side of one square = 8 m

Side of other square = 6 + 8 = 14 m

Thus, side of squares = 8 m and 14 m.