Question

The sum of the areas of two squares is 468 metre square. If the difference of their perimeters is 24m, find the sides of the squares. (solve it by square completing method) ​

Answer 1

Let the side of the squares be x and y respectively.

According to the question:

☛ 4x - 4y = 24

➜ 4(x - y) = 24

➜ x - y = 6

➜ x = y + 6 ---------(i)

Also,

☛ x² + y² = 468

➜ (y + 6)² + y² = 468 { from (i) }

➜ y² + y² + 36 + 12y = 468

➜ 2y² + 12y = 432

Divide by 2 both sides

➜ y² + 6y = 216

➜ y² + 2 × 3 × y + 3² = 216 + 3²

➜ (y + 3)² = 225

➜ √(y + 3)² = √225

➜ y + 3 = ± 15

Since, Length can not be negative so -15 is neglected.

☛ y + 3 = +15

➜ y = 15 - 3

➜ y = 12

In eq.(i),

☛ x = y + 6

☛ x = 12 + 6

☛ x = 18

Hence, Side of first square is 18 meters and side of the other square is 12 meters.