The sum of the areas of two squares is `157 m^(2)` . If the sum of their perimeters is 68 m , find the sides of the two squares .

Answer

To solve the problem, we need to find the sides of two squares given the sum of their areas and the sum of their perimeters. Let's denote the side lengths of the two squares as \( A \) and \( B \).

### Step 1: Set up the equations based on the problem statement.
1. The sum of the areas of the two squares is given as:
\[ A^2 + B^2 = 157 \quad \text{(Equation 1)} \]
2. The sum of the perimeters of the two squares is given as:
\[ 4A + 4B = 68 \quad \text{(Equation 2)} \]
We can simplify this equation by dividing everything by 4:
\[ A + B = 17 \quad \text{(Equation 2 simplified)} \]

### Step 2: Express one variable in terms of the other.
From Equation 2 simplified, we can express \( B \) in terms of \( A \):
\[ B = 17 - A \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 3 into Equation 1.
Now, we substitute Equation 3 into Equation 1:
\[ A^2 + (17 - A)^2 = 157 \]
Expanding \( (17 - A)^2 \):
\[ A^2 + (289 - 34A + A^2) = 157 \]
Combining like terms:
\[ 2A^2 - 34A + 289 = 157 \]

### Step 4: Rearrange the equation into standard quadratic form.
Subtract 157 from both sides:
\[ 2A^2 - 34A + 132 = 0 \]
Now, divide the entire equation by 2 to simplify:
\[ A^2 - 17A + 66 = 0 \quad \text{(Equation 4)} \]

### Step 5: Factor the quadratic equation.
We need to factor Equation 4. We are looking for two numbers that multiply to \( 66 \) and add up to \( -17 \). The numbers \( -11 \) and \( -6 \) work:
\[ (A - 11)(A - 6) = 0 \]

### Step 6: Solve for \( A \).
Setting each factor to zero gives us:
\[ A - 11 = 0 \quad \Rightarrow \quad A = 11 \]
\[ A - 6 = 0 \quad \Rightarrow \quad A = 6 \]

### Step 7: Find the corresponding values of \( B \).
Using Equation 3, we can find \( B \) for both values of \( A \):
1. If \( A = 11 \):
\[ B = 17 - 11 = 6 \]
2. If \( A = 6 \):
\[ B = 17 - 6 = 11 \]

### Conclusion:
The sides of the two squares are \( 11 \, m \) and \( 6 \, m \).