Question

Sum of the areas of two squares is 640 m². If the difference of their perimeters is 64 m. Find the sides of the two squares.

Answer 1

Let the side of the larger square be x and the side of the smaller square be y.

The sum of the areas is 640 m².

x² + y² = 640   ...(1)

The difference of their perimeters (4 × side) is 64 m.

4x − 4y = 64

4(x − y) = 64

x − y = `64/4`

x − y = 16 

x = y + 16   ...(2)

Putting the value of x in equation (2) from equation (1).

(y + 16)² + y² = 640

(y² + 32y + 256) + y² = 640

2y² + 32y + 256 − 640 = 0

2y² + 32y − 384 = 0

Divide the entire equation by 2 to simplify:

y² + 16y − 192 = 0

y² + 24y − 8y − 192 = 0

y² + 24y − 8y − 192 = 0

y(y + 24) − 8(y + 24) = 0

(y + 24) (y − 8) = 0

y + 24 = 0 or y − 8 = 0

y = −24 or y = 8

Sides of the square are never negative.

∴ y = 8

∴ Side of the smaller square y = 8 m

Side of the larger square y + 16 = 8 + 16

= 24 m