Question

A polygon of $ n $ sides has 105 diagonals, then $ n $ is equal to

• A. 20
• B. 21
• C. 15
• D. -14

Hint:

To find the number of diagonals in a polygon, use the formula \( D = \frac{n(n - 3)}{2} \), where \( n \) is the number of sides.

Answer 1

The Correct Option is C

A polygon with \( n \) sides can have diagonals that are calculated using the formula:

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

We know the polygon has 105 diagonals. Thus, we set up the equation:

\[ \frac{n(n-3)}{2} = 105 \]

Solve for \( n \) by first multiplying both sides by 2 to clear the fraction:

\[ n(n-3) = 210 \]

Now, expand and rearrange to form a standard quadratic equation:

\[ n^2 - 3n - 210 = 0 \]

Next, we factorize the quadratic equation:

\[ (n-15)(n+14) = 0 \]

The solutions for \( n \) are obtained by setting each factor to zero:

• \( n-15 = 0 \) leads to \( n = 15 \)
• \( n+14 = 0 \) leads to \( n = -14 \)

We disregard \( n = -14 \) as a polygon cannot have negative sides. 

Therefore, the solution is: \( n = 15 \)

The polygon has 15 sides.

Answer 2

To determine the number of sides in a polygon given the total number of lines connecting its vertices, we follow these steps:

1. Problem Analysis:
The number of lines connecting any two vertices (sides and diagonals) of an n-sided polygon is given by the combination formula:
$ \binom{n}{2} = \frac{n(n-1)}{2} $

2. Given Condition:
We are told the total number of lines is 105:
$ \frac{n(n-1)}{2} = 105 $

3. Solving the Equation:
Multiply both sides by 2:
$ n(n-1) = 210 $
Rearrange to standard quadratic form:
$ n^2 - n - 210 = 0 $

4. Factoring the Quadratic:
Find factors of -210 that sum to -1:
$ n^2 - 15n + 14n - 210 = 0 $
Factor by grouping:
$ (n - 15)(n + 14) = 0 $

5. Valid Solution:
The solutions are n = 15 or n = -14.
Since a polygon can't have negative sides, we take:
$ n = 15 $

Final Answer:
The polygon has $15$ sides.