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.

Let the side of the large square and small square be x cm and y cm respectively.

According to the question,

x²+y² = 400...(1)

and,

4(x-y) = 16

⇒ x-y = 4

⇒ x = y+4 ...(2)

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

(y+4)²+y² = 400

⇒ 2y²+8y+16 = 400

⇒ 2y²+8y-384 = 0

⇒ y²+4y-192 = 0

⇒ y²+16y-12y-192 = 0

⇒ y(y+16)-12(y+16) = 0

⇒ (y-12)(y+16) = 0

⇒ (y-12) = 0 or (y+16) = 0

∴ y = 12 or y = -16

the value of y = -16 is invalid because the length of the side can't be negative.

Thus, Side of small square = y = 12 cm

and Side of large square = x = 4 + y [from (2)]

= 16 cm