Question

Sum of the areas of two squares is 260m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer 1

Let the sides of two squares be 'a' m and 'b' m.

As we know that area of square = (side)² and perimeter of square = 4 × side

It is given that sum of the areas of two squares is 260 m².

⇒ a² + b² = 260   ..... (1)

And, the difference of their perimeters is 24 m.

⇒ 4a − 4b = 24

⇒ 4(a − b) = 24

⇒ a − b = `24/4​`

⇒ a − b = 6

⇒ a = 6 + b   .....(2)

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

⇒ (6 + b)² + b² = 260

⇒ 36 + b² + 12b + b² = 260

⇒ 36 + 2b² + 12b − 260 = 0

⇒ 2b² + 12b − 224 = 0

⇒ 2(b² + 6b − 112) = 0

⇒ b² + 6b − 112 = 0

⇒ b² + 14b − 8b − 112 = 0

⇒ b(b + 14) − 8(b + 14) = 0

⇒ (b + 14)(b − 8) = 0

⇒ (b + 14) = 0 or (b − 8) = 0

⇒ b = −14 or b = 8

Since, side of square cannot be negative.

Side of one square = 8 m

Side of other square = 6 + 8 = 14 m

Thus, side of squares = 8 m and 14 m.