Question

The sum of the areas of two squares is 400 sq.m. If the difference between their perimeters is 16 m, find the sides of two squares.

Answer 1

Let the sides of the two squares be x cm and y cm(x > y).

Then, their areas are x 2 and y 2 and their perimeters are 4x and 4y.

According to the first condition,

sum of the areas of two squares is 400 sq.m

∴ x² + y² = 400 ........(i)

According to the second condition,

difference between the perimeters is 16 m

∴ 4x − 4y = 16

∴ 4(x − y) = 16

∴ x − y = 4

∴ x = y + 4

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

(y + 4)² + y² = 400

∴ y² + 8y + 16 + y² = 400

∴ 2y² + 8y – 384 = 0

∴ y² + 4y – 192 = 0

∴ y² + 16y – 12y – 192 = 0

∴ y(y + 16) – 12(y + 16) = 0

∴ (y + 16) (y – 12) = 0

∴ y + 16 = 0 or y – 12 = 0

∴ y = – 16 or y = 12

But, y ≠ – 16 as the side of a square cannot be negative.

∴ y = 12

∴ x = y + 4 = 12 + 4 = 16

∴ The sides of the two squares are 16 cm and 12 cm.