The sum of areas of two squares is 468 sq.m. If the difference between

1.	The sum of areas of two squares is 468 sq.m. If the difference between their perimeter is 24 m, then find sides of both squares.
Answer» Let the side of a square is x m. Perimeter of that square = 4x m Difference in perimeter is 24 m Perimeter of second square = 4x + 24 m Then, side of second square = (4x + 24)/4 = 4(x + 6)/4 = (x + 6)m Area of first square = x² sq m Area of second square = (x + 6)² sq m = x² + 12x + 36 sq m Sum of areas of both squares = 468 sq m x² + (x² + 12x + 36) = 468 ⇒ 2x² + 12x + 36 – 468 = 0 ⇒ 2x²+ 12x – 432 =0 ⇒ 2(x² + 6x – 216) = 0 ⇒ x² + 6x – 216 = 0 ⇒ x² + 18x – 12x – 216 = 0 ⇒ x(x + 18) – 12(x + 18) = 0 ⇒ (x + 18)(x – 12) = 0 when x + 18 = 0, then x = -18 (not possible) or x – 12 = 0, then x = 12 ∴ x = 12 Side of smaller square = 12 m and side of larger square = x + 6 = 12 + 6 = 18 m Thus sides of both squares are 12 m. and 18 m respatively

The sum of areas of two squares is 468 sq.m. If the difference between their perimeter is 24 m, then find sides of both squares.

Answer»

Let the side of a square is x m.

Perimeter of that square = 4x m

Difference in perimeter is 24 m

Perimeter of second square

= 4x + 24 m

Then, side of second square

= (4x + 24)/4 = 4(x + 6)/4 = (x + 6)m

Area of first square = x² sq m

Area of second square

= (x + 6)² sq m = x² + 12x + 36 sq m

Sum of areas of both squares = 468 sq m

x² + (x² + 12x + 36) = 468

⇒ 2x² + 12x + 36 – 468 = 0

⇒ 2x²+ 12x – 432 =0

⇒ 2(x² + 6x – 216) = 0

⇒ x² + 6x – 216 = 0

⇒ x² + 18x – 12x – 216 = 0

⇒ x(x + 18) – 12(x + 18) = 0

⇒ (x + 18)(x – 12) = 0

when x + 18 = 0, then x = -18 (not possible)

or x – 12 = 0, then x = 12

∴ x = 12

Side of smaller square = 12 m

and side of larger square

= x + 6 = 12 + 6

= 18 m

Thus sides of both squares are 12 m. and 18 m respatively