The sum of areas of two squares is 468m2 If the difference of their pe

1.	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» 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 Therefore, sides are 18m & 12m.

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»

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
Therefore, sides are 18m & 12m.