In a regular truncated quadrilateral pyramid with lateral edges $A A 1, B B 1, C C 1, D D 1$, the side of the upper base $A 1 B 1 C 1 D 1$ is 1, and the side of the lower base is 7. A plane passing through the edge $B 1 C 1$ perpendicular to the plane $A D 1 C$ divides the pyramid into two parts of equal volume. Find the volume of the pyramid.
Since the secant plane passes through the line $B 1 C 1$ parallel to the base plane $A B C D$, these planes intersect along a line $l$ parallel to $B 1 C 1$, and thus to the line $B C$. Let the line $l$ intersect the lines $A B$ and $C D$ at points $E$ and $F$ respectively. If points $E$ and $F$ were on the extensions of the edges $A B$ and $C D$, the volume of the part of the truncated pyramid below the secant plane would be greater than the volume of the remaining part of the pyramid, so points $E$ and $F$ lie on the edges $A B$ and $C D$. Let $Q$ be the center of the lower base of the truncated pyramid, and $M$ be the foot of the perpendicular dropped from point $B 1$ to the line $D 1 Q$. The line $B 1 M$ lies in the plane of the diagonal section $B B 1 D 1 D$, and the line $A C$ is perpendicular to this plane, so the line $B 1 M$ is perpendicular to two intersecting lines $D 1 Q$ and $A C$ in the plane $A D 1 C$. Therefore, the line $B 1 M$ is perpendicular to the plane $A D 1 C$, and the plane passing through the intersecting lines $B 1 C 1$ and $B 1 M$ is perpendicular to the plane $A D 1 C$ (if one of two planes passes through a line perpendicular to the other plane, then these planes are perpendicular). Thus, the section in question is a trapezoid $B 1 C 1 F E$. Let the height of the truncated pyramid be $h, C F=B E=t$, and the volume of the truncated pyramid be $V$. On the extension of the segment $C 1 B 1$ beyond point $B 1$, lay off the segment $B 1 G=6$. Then the polyhedron $C C 1 F B G E$ is a triangular prism with lateral edges $G C 1=E F=B C=7$. The distance from the lateral edge $G C 1$ of the prism to the plane of the opposite lateral face $B C F E$ is $h$. If $V 1$ is the volume of the prism, then

$$
V 1=\frac{1}{2} S_{B C F E} \cdot h=\frac{1}{2} \cdot 7 t h=\frac{7}{2} t h .
$$

Let $V 2$ be the volume of the tetrahedron $B E B 1 G$, which complements the part of the truncated pyramid below the secant plane to form the prism. The opposite edges $B E$ and $B 1 G$ of the tetrahedron are perpendicular, and the distance between the lines $B E$ and $B 1 G$ is $h$, so

$$
V 2=\frac{\frac{1}{6}}{B} B E \cdot B 1 G \cdot h \cdot \sin 90^{\circ}=\frac{1}{6} t \cdot 6 \cdot h=t h
$$

and since $V 1-V 2=\frac{1}{2} V$, we have

$$
\frac{7}{2} t h-t h=\frac{1}{2} \cdot \frac{1}{3} h\left(72+12+\sqrt{7^{2} \cdot 1^{2}}\right)
$$

19

from which we find that $t=5$. The diagonal section $B B 1 D 1 D$ is an isosceles trapezoid $B B 1 D 1 D$, and point $Q$ is the midpoint of the larger base $B D$. Let $T$ be the point of intersection of the line $B 1 M$, perpendicular to $D 1 Q$, with the base $B D, K$ and $L$ be the feet of the perpendiculars dropped from the vertices $B 1$ and $D 1$ to $B D$. Point $T$ lies in the planes $B 1 C 1 F E$ and $A B C D$, so it lies on the line $E F$ of intersection of these planes. Then

$$
\begin{gathered}
\frac{B T}{B D}=\frac{B E}{A B}=\frac{\frac{19}{5}}{7}=\frac{19}{35}, B T=\frac{19}{35} B D=\frac{19}{35} \cdot 7 \sqrt{2}=\frac{19}{5} \sqrt{2}, \\
B L=\frac{1}{2}(B D+B 1 D 1)=\frac{1}{2}(7 \sqrt{2}+\sqrt{2})=4 \sqrt{2}, B Q=\frac{1}{2} B D=\frac{7}{2} \sqrt{2}, \\
Q L=B L-B Q=4 \sqrt{2}-\frac{7}{2} \sqrt{2}=\frac{\sqrt{2}}{2}, B K=\frac{1}{2}(7 \sqrt{2}-\sqrt{2})=3 \sqrt{2}, \\
K T=B T-B K=\frac{19}{5} \sqrt{2}-3 \sqrt{2}=\frac{4}{5} \sqrt{2} .
\end{gathered}
$$

The right triangles $B 1 K T$ and $Q L D 1$ are similar by two angles, so $\frac{B_{1} K}{K T}=\frac{Q L}{D_{1} L}$, or $\frac{\frac{h}{5} \sqrt{2}}{\frac{\frac{\sqrt{2}}{2}}{h}}$, from which $h=\frac{2}{\sqrt{5}}$. Therefore,

$$
V=\frac{1}{3} h(49+1+7)=\frac{1}{3} \cdot \frac{2}{\sqrt{5}} \cdot 57=\frac{38}{\sqrt{5}}
$$

## Answer