Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of the two squares.

Given:

Sum of the areas of two squares = 400 cm².

The difference of their perimeters = 16 cm

Let a and b be the sides of 2 square.

Sum of Areas,

⇒ a² + b² = 400 ……………(1)

Difference of Perimeters,

⇒ 4a - 4b = 16

⇒ 4(a - b) = 16

⇒ a - b = $\dfrac{16}{4}$

⇒ a - b = 4

⇒ a = 4 + b

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

⇒ (4 + b)² + b² = 400

⇒ 4² + b² + 2 x 4 x b + b² = 400

⇒ 16 + b² + 8b + b² = 400

⇒ 16 + 2b² + 8b - 400 = 0

⇒ 2b² + 8b - 384 = 0

⇒ b² + 4b - 192 = 0

⇒ b² + 16b - 12b - 192 = 0

⇒ b(b + 16) - 12(b + 16) = 0

⇒ (b + 16)(b - 12) = 0

⇒ b = - 16 or 12

Since the side of a square cannot be negative, b = 12 cm.

So, a = 4 + b = 16 cm

Hence, the sides of the squares are 16 cm and 12 cm.