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

$$
\begin{gathered}
V_{A B C D K}=\frac{1}{3} S_{A B C D} \cdot K K 1=\frac{1}{3} A B \cdot B C \cdot K K 1=\frac{1}{3} \cdot 2 \cdot 3 \cdot 1=2, \\
V_{B C K M}=\frac{1}{6} B C \cdot K M \cdot d \cdot \sin 90^{\circ}=\frac{1}{6} \cdot 3 \cdot 5 \cdot 1 \cdot 1=\frac{5}{2}
\end{gathered}
$$

Therefore, the volume of the polyhedron $A B C D K M$ is $2+\frac{5}{2}=\frac{9}{2}$.

## Answer

$\frac{9}{2}$
3 [ Volume of a Tetrahedron and Pyramid $\quad$]

Consider a rectangle $A B C D$, where $A B=2, B C=3$. Segment $K M$ is parallel to $A B$ (see figure), located 1 unit away from the plane $A B C D$ and $K M=5$. Find the volume of the polyhedron $A B C D K M$.

#