the sum of areas of two squares is 468m2 if

Question

The sum of areas of two squares is 468m²  If the difference of their perimeters is 24cm, find the sides of the two squares.

Ans:    Let the side of the larger square be x.

Let the side of the smaller square be y. APQ x²+y² = 468

Cond. II          4x-4y = 24

⇒ x - y = 6

⇒ x = 6 + y

x²  + y² = 468

⇒ (6+y)² +y² = 468

on solving we get y = 12

⇒ x = (12+6) = 18 m

∴ sides are 18m & 12m.