What is the number of sides of a polygen in which the sum of the degree measure of the interior angles is 4 times the sum of the degree measures of the exterior angles?

Options

A. `10`
B. `12`
C. `14`
D. No such polygon exist

Answer

To solve the problem, we need to find the number of sides \( n \) of a polygon where the sum of the degree measures of the interior angles is 4 times the sum of the degree measures of the exterior angles.

### Step-by-step Solution:

1. **Understand the formulas:**
- The sum of the interior angles of an \( n \)-sided polygon is given by the formula:
\[ \text{Sum of interior angles} = (n - 2) \times 180^\circ \]
- The sum of the exterior angles of any polygon is always:
\[ \text{Sum of exterior angles} = 360^\circ \]

2. **Set up the equation:**
- According to the problem, the sum of the interior angles is 4 times the sum of the exterior angles. Therefore, we can write the equation:
\[ (n - 2) \times 180^\circ = 4 \times 360^\circ \]

3. **Calculate the right side of the equation:**
- Simplifying the right side:
\[ 4 \times 360^\circ = 1440^\circ \]

4. **Substitute and simplify:**
- Now we have:
\[ (n - 2) \times 180^\circ = 1440^\circ \]
- To isolate \( n - 2 \), divide both sides by \( 180^\circ \):
\[ n - 2 = \frac{1440^\circ}{180^\circ} \]
- Calculating the division:
\[ n - 2 = 8 \]

5. **Solve for \( n \):**
- Add 2 to both sides to find \( n \):
\[ n = 8 + 2 = 10 \]

### Conclusion:
The number of sides of the polygon is \( n = 10 \).