Question

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.

Answer 1

Let the side of the first square be 'a' m and that of the second be ′A′ m.

Area of the first square = a² sq m.

Area of the second square = A² sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A - 4a = 24

A - a = 6              ......(1)

A² + a² = 468      ......(2)

From (1), A = a + 6

Substituting for A in (2), we get

(a + 6)² + a² = 468

a² + 12a + 36 + a² = 468

2a² + 12a + 36 = 468

a² + 6a + 18 = 234

a² + 6a - 216 = 0

a² + 18a - 12a - 216 = 0

a(a + 18) - 12(a + 18) = 0

(a - 12)(a + 18) = 0

a = 12, - 18

So, the side of the first square is 12 m. and the side of the second square is 18 m.