Question

Sum of areas of 2 squares is 468m^. If difference of their perimeters is 24m,then find the sides of 2 squares.

Answer 1

Let the side of the larger square be x.
Let the side of the smaller square be y.
According to the question, x² + y² = 468 --------- (1)
Perimeters of the squares = 4x and 4y.
⇒ 4x - 4y = 24
⇒ x– y = 6
⇒ x= 6 + y
x² + y² = 468
Substituting the value of x in equation (1)
⇒ (6+y)² + y² = 468
⇒ 2y² + 12y + 36 = 468
⇒ y²+ 6y + 18 = 234
⇒ y² + 6y - 216 = 0
⇒ (y – 12)(y + 18) = 0
⇒ y = 12 or -18
Since y denotes a length, it cannot be negative.
∴ y = 12
⇒ x = (12 + 6) = 18 m

∴ Sides of the squares are 18 m and 12 m.