If a=2b=8c and a+b+c=13,then the value of `(sqrt(a^(2)+b^(2)+c^(2)))/(2c)` is:

Options

A. `(9)/(2)`
B. `(5)/(6)`
C. `-(9)/(2)`
D. `-(5)/(6)`

Answer

To solve the problem step-by-step, we start with the given equations:

1. **Given Relationships**:
- \( a = 2b = 8c \)
- \( a + b + c = 13 \)

2. **Express Variables in Terms of a Common Variable**:
Let \( a = K \). Then from the relationships:
- \( 2b = K \) implies \( b = \frac{K}{2} \)
- \( 8c = K \) implies \( c = \frac{K}{8} \)

3. **Substitute into the Sum Equation**:
Substitute \( a \), \( b \), and \( c \) into the equation \( a + b + c = 13 \):
\[ K + \frac{K}{2} + \frac{K}{8} = 13 \]

4. **Find a Common Denominator**:
The common denominator for the fractions is 8. Rewrite the equation:
\[ \frac{8K}{8} + \frac{4K}{8} + \frac{K}{8} = 13 \]
Combine the fractions:
\[ \frac{8K + 4K + K}{8} = 13 \]
This simplifies to:
\[ \frac{13K}{8} = 13 \]

5. **Solve for K**:
Multiply both sides by 8:
\[ 13K = 104 \]
Divide by 13:
\[ K = 8 \]

6. **Find Values of a, b, and c**:
Now substitute back to find \( a \), \( b \), and \( c \):
- \( a = K = 8 \)
- \( b = \frac{K}{2} = \frac{8}{2} = 4 \)
- \( c = \frac{K}{8} = \frac{8}{8} = 1 \)

7. **Calculate \( \sqrt{a^2 + b^2 + c^2} \)**:
Now we need to find \( \sqrt{a^2 + b^2 + c^2} \):
\[ a^2 = 8^2 = 64, \quad b^2 = 4^2 = 16, \quad c^2 = 1^2 = 1 \]
Therefore:
\[ a^2 + b^2 + c^2 = 64 + 16 + 1 = 81 \]
So:
\[ \sqrt{a^2 + b^2 + c^2} = \sqrt{81} = 9 \]

8. **Calculate the Final Expression**:
We need to find:
\[ \frac{\sqrt{a^2 + b^2 + c^2}}{2c} \]
Substitute \( c = 1 \):
\[ \frac{9}{2 \cdot 1} = \frac{9}{2} \]

Thus, the final answer is:
\[ \frac{9}{2} \]