Which one of the following polygon has as many diagonals as the number of sides in it?

Options

A. 4
B. 6
C. 5
D. 7

Answer

To solve the problem of finding which polygon has as many diagonals as the number of sides, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find a polygon where the number of diagonals is equal to the number of sides.

2. **Recall the Formula for Diagonals**: The formula to calculate the number of diagonals \(D\) in a polygon with \(n\) sides is given by:
\[ D = \frac{n(n-3)}{2} \]

3. **Set Up the Equation**: According to the problem, we want the number of diagonals \(D\) to be equal to the number of sides \(n\). Therefore, we can set up the equation:
\[ n = \frac{n(n-3)}{2} \]

4. **Clear the Fraction**: Multiply both sides of the equation by 2 to eliminate the fraction:
\[ 2n = n(n-3) \]

5. **Rearrange the Equation**: Rearranging gives us:
\[ n^2 - 3n - 2n = 0 \]
This simplifies to:
\[ n^2 - 5n = 0 \]

6. **Factor the Equation**: We can factor out \(n\):
\[ n(n - 5) = 0 \]

7. **Solve for \(n\)**: This gives us two solutions:
\[ n = 0 \quad \text{or} \quad n = 5 \]
Since \(n = 0\) does not represent a polygon, we take \(n = 5\).

8. **Identify the Polygon**: A polygon with 5 sides is called a pentagon. Therefore, the polygon that has as many diagonals as the number of sides is a pentagon.

9. **Conclusion**: The correct answer is option C, which corresponds to a pentagon.