Question

Explain how this formula that gives the number of diagonals of an n-gon can be derived; n(n-3)/2

Answer 1

Well from a single vertex on the n-gon you can draw n-3 diagonals.

Answer 2

So if you labeled the vertices from 1 to n, and started drawing all the diagonals you could starting at vertex 1, you would draw \(n-3, n-3, \;n-4, \;n-5, ..., \;n-(n-1)\) diagonals. Or, \(1+2+3+...+n-4+n-3+n-3\) diagonals. If you add these up, you get \[{(n-3)(n-2) \over 2} +n-3\]

Answer 3

If you simplify this, you get the formula you desired.

Answer 4

\[{(n-3)(n-2) \over 2} + (n-3)={n^2-5n+6 \over 2}+{2n-6 \over 2}={n^2-3n \over 2}={n(n-3) \over 2}\]

Answer 5

In case you were wondering, you can draw \(n-3\) diagonals from vertex two after you've drawn them from vertex 1 since you didn't draw any diagonals between the two.