Answer
$\frac{n(n-3)}{2}$
Work Step by Step
Given a convex polygon with n sides.
Since the number of vertices is equal to number of sides.
Therefore, $\mathbf{Number}$ $\mathbf{of}$ $\mathbf{sides =n}$.
We want to find the number of diagonals.
A diagonal joins two vertices of the polygon, but they must be vertices that are not already joined by a side of the polygon. Thus there are n - 3 diagonals emanating from each vertex of the polygon (we've excluded two of the n - 1 other vertices that are side of polygon).
$\mathbf{Number}$ $\mathbf{of}$ $\mathbf{diagonals}$ $\mathbf{per}$ $\mathbf{vertex}$ $\mathbf{=n-3}$
$\mathbf{Number}$ $\mathbf{of}$ $\mathbf{diagonals}$ =$\mathbf{n(n-3)}$ (by product rule ).
Since a diagonal connects two vertices, each diagonal was counted twice, once for each vertex.
Thus,
$\mathbf{Number}$ $\mathbf{of}$ $\mathbf{diagonals}$ =$\mathbf{\frac{n(n-3)}{2}}$.
(Note that the convexity of the polygon had nothing to do with the problem-we were counting the diagonals, whether or not we could be sure that they all lay inside the polygon.)
