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.

Answer 1

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.