Question

The sum of areas of two squares is 640 m\(^2\). If the difference in perimeters is 64 m, find the sides.

Hint:

In speed-distance problems, the equation is usually written as: \(\text{Time}_{\text{slow}} - \text{Time}_{\text{fast}} = \text{Difference}\).

Answer 1

Step 1: Solving OR (B):
1. Let sides be \(a\) and \(b\). \(4a - 4b = 64 \implies a - b = 16 \implies a = b + 16\). 2. Sum of areas: \((b+16)^2 + b^2 = 640\). 3. \(b^2 + 32b + 256 + b^2 = 640 \implies 2b^2 + 32b - 384 = 0\). 4. \(b^2 + 16b - 192 = 0 \implies (b+24)(b-8) = 0\). 5. Side \(b = 8\) m, \(a = 8 + 16 = 24\) m.
Step 2: Final Answer (OR):
The sides of the squares are 24 m and 8 m.