Elementary Geometry for College Students (6th Edition)

(a) The hexagon $ABCB'A'D$ has 6 vertices. We can find the number of diagonals $D$: $D = \frac{n~(n-3)}{2} = \frac{(6)(6-3)}{2} = 9$ (b) When we draw the nine diagonals. we can see that two diagonals lie outside the hexagon. The diagonal $AA'$ and the diagonal $BB'$ lie outside the hexagon.