A regular hexagon with sides of length $4$ has an isosceles triangle attached to each side. Each of these triangles has two sides of length $5$. The isosceles triangles are folded up to make a pyramid with the hexagon as its base. What is the volume of the pyramid?

1. Each triangular face has base $4$ and legs $5$, so its altitude from the apex is $\sqrt{5^2-2^2}=\sqrt{21}$.
2. The hexagon has apothem $2\sqrt3$. The apex is above the center, so the pyramid's height is $\sqrt{(\sqrt{21})^2-(2\sqrt3)^2}=\sqrt{21-12}=3$.
3. The base area is $\frac{3\sqrt3}2\cdot4^2=24\sqrt3$, so $V=\frac13\cdot24\sqrt3\cdot3=\boxed{24\sqrt3}$.