Question

how many diagonals are in a heptagon?

Answer 1

There's a neat pattern to this, which you can use to construct a formula for a regular \(n\)-gon.
For a regular \(n\)-gon, you can draw \(n-3\) diagonals from the first two vertices. From the third, you have \(n-4\); from the fourth, \(n-5\); and so on, up to the last two vertices, which do not have any remaining partners that you can connect them to.

If \(S\) denotes the total number of diagonals, then
\[S=\underbrace{(n-3)+(n-3)}_{\text{first two vertices}}+(n-4)+(n-5)+\cdots+2+1+\underbrace{0+0}_{\text{last two}}\]
Rewriting \(S\) in a reverse order, you have
\[S=0+0+1+2+\cdots+(n-5)+(n-4)+(n-3)+(n-3)\]
Adding both forms of \(S\) together, and pairing up the first terms, second terms, and so on, you have
\[2S=(n-3+0)+(n-3+0)+(n-4+1)\\
\quad\quad+\cdots+(1+n-4)+(0+n-3)+(0+n-3)\\
2S=n(n-3)\]
which gives the final formula,
\[S=\frac{n(n-3)}{2}\]
For a heptagon, \(n=7\).