10,11 Given a rectangle $A B C D$ and a line $M N$, parallel to $A B$ and at a distance $h$ from the plane of the rectangle (see figure). It is known that $A B=a, B C=b, M N=c$. Find the volume of the polyhedron $A B C D M N$.

#
The plane $B M C$ divides the polyhedron $A B C D M N$ into a quadrilateral pyramid $A B C D M$ with base $A B C D$ and a triangular pyramid $B C M N$. The height $M M 1$ of the pyramid $A B C D M$ is equal to the distance from point $M$ to the plane $A B C D$, i.e., $M M 1=h$. The opposite edges $B C$ and $M N$ of the triangular pyramid $B C M N$ are perpendicular, and the distance between them is equal to the distance from the line $M N$ to the plane $A B C D$, i.e., $h$. Therefore,

$$
\begin{gathered}
V_{A B C D M}=\frac{1}{3} S_{A B C D} \cdot h=\frac{1}{3} a b h \\
V_{B C M N}=\frac{1}{6} B C \cdot M N \cdot h \cdot \sin 90^{\circ}=\frac{1}{6} b c h
\end{gathered}
$$

Therefore,

$$
V_{A B C D M N}=\frac{1}{\overline{3}} a b h+\frac{1}{6} b c h=\frac{1}{6} b h(2 a+c)
$$

## 

$\overline{6} b h(c+2 a)$.