Question

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.

Answer 1

Let the sides of the two squares be x m and ym. Therefore, their perimeter will be 4x and 4yrespectively and their areas will be x2 and y2 respectively.
It is given that
4x - 4y = 24
x - y = 6
x = y + 6
Also, x2 + y2 = 468
⇒ (6 + y2) + y2 = 468
⇒ 36 + y2 + 12y + 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, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Answer 2

Let the side of the larger square be x.
Let the side of the smaller square be y.
According to the question, x2  + y2 = 468 ---- (mark as 1)

Perimeters of the squares  4x and 4y.
⇒ 4x - 4y = 24
⇒ x– y = 6
⇒ x= 6 + y
x2  + y2 = 468

Substituting the value of x in equation (1)
 (6+y)^2 + y^2 = 468
 2y^2 + 12y + 36 = 468
 y^2 + 6y + 18 = 234
 y^2 + 6y - 216 = 0
 (y – 12)(y + 18) = 0
 y = 12 or -18

 y says a length, it cannot be negative.
so,
∴ y = 12
⇒ x = (12 + 6) = 18 m

Sides of the squares are 18 m and 12 m.