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

Explanation

To solve the problem, we need to find the sides of two squares given that the sum of their areas is 640 m² and the difference in their perimeters is 64 m. Let's denote the sides of the two squares as \( A \) and \( B \).

### Step-by-Step Solution:

1. **Set Up the Equations**:
- The area of the first square is \( A^2 \).
- The area of the second square is \( B^2 \).
- According to the problem, we have the equation for the sum of the areas:
\[ A^2 + B^2 = 640 \quad \text{(1)} \]
- The perimeter of a square is given by \( 4 \times \text{side} \), so the perimeters of the two squares are \( 4A \) and \( 4B \). The difference in their perimeters gives us:
\[ |4A - 4B| = 64 \quad \text{(2)} \]
- This simplifies to:
\[ |A - B| = 16 \quad \text{(3)} \]

2. **Express One Variable in Terms of the Other**:
- From equation (3), we can express \( A \) in terms of \( B \):
\[ A - B = 16 \quad \text{or} \quad B - A = 16 \]
- Let's take \( A - B = 16 \):
\[ A = B + 16 \quad \text{(4)} \]

3. **Substitute into the Area Equation**:
- Substitute equation (4) into equation (1):
\[ (B + 16)^2 + B^2 = 640 \]
- Expanding this gives:
\[ B^2 + 32B + 256 + B^2 = 640 \]
- Combine like terms:
\[ 2B^2 + 32B + 256 = 640 \]
- Rearranging gives:
\[ 2B^2 + 32B - 384 = 0 \]
- Dividing the entire equation by 2 simplifies it:
\[ B^2 + 16B - 192 = 0 \quad \text{(5)} \]

4. **Solve the Quadratic Equation**:
- We can use the quadratic formula \( B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 16, c = -192 \):
\[ B = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-192)}}{2 \cdot 1} \]
\[ B = \frac{-16 \pm \sqrt{256 + 768}}{2} \]
\[ B = \frac{-16 \pm \sqrt{1024}}{2} \]
\[ B = \frac{-16 \pm 32}{2} \]
- This gives us two possible solutions for \( B \):
\[ B = \frac{16}{2} = 8 \quad \text{(valid)} \]
\[ B = \frac{-48}{2} = -24 \quad \text{(not valid)} \]

5. **Find the Value of A**:
- Using equation (4):
\[ A = B + 16 = 8 + 16 = 24 \]

6. **Final Answer**:
- The sides of the two squares are:
\[ A = 24 \, \text{m} \quad \text{and} \quad B = 8 \, \text{m} \]