Question

Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of the two squares ?

Answer 1

Let the sides of the two squares be x cm and y cm where x > y.

Then, their areas are x² and y² and their perimeters are 4x and 4y.

By the given condition, x² + y²= 400 and 4x − 4y= 16

4x − 4y= 16 ⇒ 4(x − y)= 16 ⇒ x − y= 4

⇒ x = y+4 … (1)

Substituting the value of y from (1) in x² + y²= 400, we get that (y+4)² + y²= 400

⇒ y²+16 + 8y + y²= 400

⇒ y²+4y − 192 = 0

⇒ y²+16y −12y − 192 = 0

⇒ y(y+16) −12(y+16) = 0

⇒ (y+16) (y −12) = 0

⇒ y= −16 or y = 12

Since the value of y cannot be negative, the value of y =12.

So, x = y+4 = 12+4 = 16

Thus, the sides of the two squares are 16 cm and 12 cm.