Question

The sum of the areas of two squares is 640 m². If the difference in their perimeter be 64 m, find the sides of the two squares.

The sum of the areas of two squares is 640 m². If the difference of their perimeters is 64 m, find the sides of the squares.

Answer 1

Let the length of the side of the first and the second square be x and y, respectively.

According to the question: 

x² + y² = 640   ...(1)

Also, 

4x – 4y = 64 

⇒ x – y = 16 

⇒ x = 16 + y  

Putting the value of x in (1), we get:  

x² + y² = 640 

⇒ (16 + y)² + y² = 640 

⇒ 256 + 32y + y² + y² = 640 

⇒ 2y² + 32y – 384 = 0 

⇒ y² + 16y – 192 = 0 

⇒ y² + (24 – 8)y – 192 = 0 

⇒ y² + 24y – 8y – 192 = 0 

⇒ y(y + 24) – 8(y + 24) = 0 

⇒ (y + 24)(y – 8) = 0 

⇒ y = –24 or y = 8 

∴ y = 8   ...(∵ Side cannot be negative) 

∴ x = 16 + y

= 16 + 8 

= 24 m 

Thus, the sides of the squares are 8 m and 24 m.