A polygon has n sides the number of diagonals in this polygon is:
(a) 1/2n(n-1)
(b) 1/2×n(n-2)
(c) 1/2×n(n-/3)
(d) 1/2×n(n-4)

To find the number of diagonals in a polygon with \( n \) sides, we use the following formula:

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

This formula works because:

1. Each vertex of the polygon can be connected to \( n-3 \) other vertices to form a diagonal (excluding itself and its two adjacent vertices).
2. Since the formula counts each diagonal twice (once from each endpoint), we divide by 2.

Thus, the correct answer is:

**(c) \( \frac{1}{2} \times n \times (n-3) \)**

Would you like more details or have any questions?

Here are some related questions:

1. How do you derive the formula for the number of diagonals in a polygon?
2. How many diagonals are in a hexagon (6-sided polygon)?
3. What is the number of diagonals in a 10-sided polygon (decagon)?
4. How does the number of sides of a polygon affect the number of diagonals?
5. If a polygon has 20 diagonals, how many sides does it have?
6. How would you prove the diagonal formula geometrically?
7. What happens to the number of diagonals as \( n \) increases?
8. Can the number of diagonals in a polygon ever be a prime number?

**Tip:** When working with polygons, remember that every pair of non-adjacent vertices forms a diagonal.

Discover how to calculate the number of diagonals in a polygon with n sides using the formula: (n × (n-3)) / 2. This formula counts each diagonal once and excludes edges of the polygon. Suitable for high school students learning polygon geometry and combinatorics.

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