An n sided polygon has 'n' diagonals, then the value of n is :

Options

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

Answer

To find the value of \( n \) for which an \( n \)-sided polygon has \( n \) diagonals, we can use the formula for the number of diagonals in a polygon:

\[ \text{Number of diagonals} = \frac{n(n-3)}{2} \]

Given that the number of diagonals is equal to \( n \), we can set up the equation:

\[ \frac{n(n-3)}{2} = n \]

### Step 1: Eliminate the fraction
To eliminate the fraction, we can multiply both sides of the equation by 2:

\[ n(n-3) = 2n \]

### Step 2: Rearrange the equation
Next, we can rearrange the equation to bring all terms to one side:

\[ n(n-3) - 2n = 0 \]

This simplifies to:

\[ n^2 - 3n - 2n = 0 \]

Combining like terms gives us:

\[ n^2 - 5n = 0 \]

### Step 3: Factor the equation
Now we can factor the equation:

\[ n(n - 5) = 0 \]

### Step 4: Solve for \( n \)
Setting each factor equal to zero gives us:

1. \( n = 0 \)
2. \( n - 5 = 0 \) which leads to \( n = 5 \)

Since \( n \) must be a positive integer representing the number of sides in a polygon, we discard \( n = 0 \).

Thus, the value of \( n \) is:

\[ \boxed{5} \]