#### Answer

(a) There are 9 diagonals.
(b) Two diagonals lie in the exterior of the hexagon.

#### Work Step by Step

(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.