[The volume of a body is equal to the sum of the volumes of its parts] Volume of a tetrahedron and a pyramid

A convex polyhedron $A B C D F E$ has five faces: $C D F, A B E, B C F E, A D F E$ and $A B C D$. Edge $A B$ is parallel to edge $C D$. Points $K$ and $L$ are located on edges $A D$ and $B C$ respectively, such that segment $K L$ divides the area of face $A B C D$ in half. Point $M$ is the midpoint of edge $E F$ and is the vertex of pyramid $M A B C D$, the volume of which is 6. Find the volume of pyramid $E K L F$, given that the volume of polyhedron $A B C D F E$ is 19.
Let $E 1, M 1$ and $F 1$ be the orthogonal projections of points $E, M$ and $F$ onto the plane of parallel lines $A B$ and $C D$. Then $E E 1$ is the height of the quadrilateral pyramid $E A K L B$ with vertex $E, M M 1$ is the height of the quadrilateral pyramid $M A B C D$ with vertex $M, F F 1$ is the height of the quadrilateral pyramid $F C D K L$ with vertex $F$. Let $E E 1=a, M M 1=b, F F 1=c, S_{A B C D}=S$. From the problem statement, it follows that $V_{A B C D}=\frac{1}{3} \cdot \cdot b=6$. Since $M$ is the midpoint of segment $E F, M 1$ is the midpoint of segment $E 1 F 1$. Therefore,

$$
b=M M 1=\frac{1}{2}(E E 1+F F 1)=\frac{++\kappa}{2} .
$$

Then

$$
\begin{aligned}
& V_{E K L F}=V_{A B C D F E}-V_{E A K L B}-V_{F C D K L}=19-\frac{1}{-\frac{1}{3}} S_{A B L K} \cdot a^{\frac{1}{3} S_{3}} S_{C D K L} \cdot c= \\
& =19-\frac{1}{3}\left(\frac{1}{2} S \cdot a+\frac{1}{2} S \cdot c\right)=19-\frac{1}{3} S \cdot \frac{a+6}{2}=19-\frac{1}{3} S \cdot b= \\
& =19-V_{M A B C D}=19-6=13 .
\end{aligned}
$$

## Answer

13.00