If each side of a regular dodecagon is 1 cm, find the circumradius of the dodecagon.

Options

A. `sqrt(2+sqrt3) cm`
B. `3sqrt3 cm`
C. `1+2sqrt3 cm`
D. none of these

Answer

To find the circumradius of a regular dodecagon (12-sided polygon) with each side measuring 1 cm, we can use the formula for the circumradius \( R \):

\[ R = \frac{s}{2 \sin\left(\frac{180}{n}\right)} \]

where \( s \) is the length of each side and \( n \) is the number of sides.

### Step-by-Step Solution:

1. **Identify the values**:
- The number of sides \( n = 12 \) (since it is a dodecagon).
- The length of each side \( s = 1 \) cm.

2. **Substitute the values into the formula**:
\[ R = \frac{1}{2 \sin\left(\frac{180}{12}\right)} \]

3. **Calculate \( \frac{180}{12} \)**:
\[ \frac{180}{12} = 15 \text{ degrees} \]

4. **Find \( \sin(15^\circ) \)**:
Using the sine subtraction formula:
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) \]
We know:
- \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Substituting these values:
\[ \sin(15^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \]

5. **Substitute \( \sin(15^\circ) \) back into the formula for \( R \)**:
\[ R = \frac{1}{2 \cdot \frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{4}{2(\sqrt{6} - \sqrt{2})} = \frac{2}{\sqrt{6} - \sqrt{2}} \]

6. **Rationalize the denominator**:
Multiply the numerator and denominator by \( \sqrt{6} + \sqrt{2} \):
\[ R = \frac{2(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} = \frac{2(\sqrt{6} + \sqrt{2})}{6 - 2} = \frac{2(\sqrt{6} + \sqrt{2})}{4} = \frac{\sqrt{6} + \sqrt{2}}{2} \]

7. **Final Result**:
The circumradius \( R \) of the regular dodecagon is:
\[ R = \frac{\sqrt{6} + \sqrt{2}}{2} \text{ cm} \]