Let $M N \| A B$. Planes $A M E$ and $B N D$ divide the polyhedron $A B C D E F M N$ into two triangular pyramids $A F E M$ and $B C D N$ with heights $M M 1=N N 1=h$ and a triangular prism $A M E B N D$ with bases $A M E$ and $B N D$. Further, we have:

$$
\begin{gathered}
S_{\triangle A F E}=S_{\triangle B C D}=\frac{1}{2} B C \cdot C D \cdot \sin 120^{\circ}=\frac{1}{4} a^2 \sqrt{3}, \\
S_{A B D E}=A B \cdot B D=a \cdot a \sqrt{3}=a^2 \sqrt{3}, \\
V_{A F E M}=V_{B C D N}=\frac{1}{3} S_{\triangle B C D} \cdot N N 1=\frac{1}{3} \cdot \frac{1}{4} a^2 \sqrt{3} \cdot h=\frac{a^2 h \sqrt{3}}{12}, \\
V_{A M E B N D}=\frac{1}{2} S_{A B D E} \cdot h=\frac{1}{2} a^2 \sqrt{3} \cdot h=\frac{1}{2} a^2 h \sqrt{3} .
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
V_{A B C D E F M N}=V_{A F E M}+V_{B C D N}+V_{A M E B N D}= \\
=\frac{2 a^2 h \sqrt{3}}{12}+\frac{1}{2} a^2 h \sqrt{3}=\frac{2}{3} a^2 h \sqrt{3}
\end{gathered}
$$

## Answer

$\frac{2}{3} a^2 h \sqrt{3}$.
3 [ Volume of a Tetrahedron and a Pyramid $\quad$]

Given a regular hexagon $A B C D E F$ with side length $a$. Segment $M N$ is parallel to one of the sides of the hexagon, equal to its side, and located at a distance $h$ from its plane. Find the volume of the polyhedron $A B C D E F M N$.