How do you figure how many diagonals are in a pentagon?

Answer 1

A polygon with n sides has n(n-3)/2 diagonals.

How do you figure that for a polygon with n sides:

Start with any vertex. You get a diagonal by joining that vertex to another vertex - but not to all of them. Obviously you don't get a diagonal by joining it to itself! Also, you don't get a diagonal by joining it to the one immediately next to it on either side - you just get a side of the polygon. Thus three of the vertices are no good so that you are left with (n-3) diagonals.

Now, this same argument applies to each of the n vertices in the n-gon so yo would think you'd have n*(n-3) diagonals. But what you have to remember is that you'll be counting each diagonal twice - once from each end. So you need to divide the answer by 2.

So, for a pentagon, n = 5 and number of diagonals = 5*(5-3)/2 = 5*2/2 = 5