Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the side of larger square = x m
let the side of smaller square = y m
According to question, x² + y² = 468 …(i)
Difference between parimeters, 4x – 4y = 24
⇒ x – y = 6
⇒ x = 6 + y …(ii)
Putting the value of x in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468
⇒ 2y² + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18) (y – 12) = 0
⇒ (y + 18) = 0 or (y – 12) = 0
Either y = – 18 or y = 12
But, y ≠ – 18, as x is the side of square, which can’t be negative. So, y = 12
Hence, the side of smaller square = 12 m
Putting the value of y in equation (ii). we get
side of large square = x = y + 6 = 12 + 6 = 18 m.