Question

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

(OR) 

If the sum of the areas of two squares is 468 m² and the difference of their perimeters is 24m, then find the measurements of their sides.

Answer 1

Let the side of first square = x m say 

Then perimeter of the first square = 4x [∵ P = 4 . side] 

By problem, perimeter of the second square = 4x + 24 (or) 4x – 24 

∴ Side of the second square = 

Now sum of the areas of the two squares is given as 468 m² 

x² + (x + 6)² = 468

⇒ x² + x² + 12x + 36 = 468 

⇒ 2x² + 12x + 36 – 468 = 0 

⇒ 2x² + 12x – 432 = 0 

⇒ x² + 6x – 216 = 0 

⇒ x² + 18x – 12x – 216 = 0 

⇒ x(x + 18)- 12(x + 18) = 0 

⇒ (x + 18) (x – 12) = 0 

⇒ x + 18 = 0 (or) x – 12 = 0 

⇒ x = -18 (or) 12

But x can’t be negative. 

∴ x = 12 

i.e., side of the first square = 12 

∴ Perimeter = 4 × 12 = 48 

∴ Perimeter of the second square = 48 + 24 = 72 

∴ Side of the second square = 72/4 = 18 m.

(or) 

x² + (x – 6)² = 468 

⇒ x² + x² – 12x + 36 = 468 

⇒ 2x² – 12x – 432 – 0 

⇒ x² – 6x – 216 = 0

⇒ x² – 18x + 12x – 216 = 0 

⇒ x(x-18) + 12(x-18) = 0 

⇒ (x – 18) (x + 12) = 0 

⇒ x – 18 = 0 (or) x + 12 = 0 

⇒ x = 18 (or) – 12 

But x can’t be negative.

∴ x = 18 

i.e., side of the first square = 18 m 

∴ Perimeter = 4 × 18 = 72 

Perimeter of the second square = 72 – 24 = 48 

∴ Side of the second square = 48/4 = 12 m. 

i.e., In any way, the sides of the squares are 12m, 18m.