The sum of areas of two squares is 468m^{2} 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 x^{2}+y^{2} = 468
Cond. II 4x-4y = 24
⇒ x - y = 6
⇒ x = 6 + y
x^{2} + y^{2} = 468
⇒ (6+y)^{2} +y^{2} = 468
on solving we get y = 12
⇒ x = (12+6) = 18 m
∴ sides are 18m & 12m.