Question

The sum of the areas of two squares is 52 cm² and difference of their perimeters is 8 cm. Find the lengths of the sides of the two squares.

Answer 1

Given: Let the side lengths be a cm and b cm. So: a² + b² = 52 and 4a – 4b = 8.

Step-wise calculation:

1. From perimeters:

4a – 4b = 8 

⇒ a – b = 2

⇒ a = b + 2

2. Substitute into the area equation:

(b + 2)² + b² = 52

3. Expand and simplify:

b² + 4b + 4 + b² = 52

⇒ 2b² + 4b – 48 = 0

⇒ b² + 2b – 24 = 0

4. Factor: (b + 6)(b – 4) = 0

⇒ b = 4   ...(Reject b = –6)

5. Then a = b + 2 = 6.

The side lengths of the two squares are 6 cm and 4 cm (larger = 6 cm, smaller = 4 cm).