100. 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.
100. Let the plane passing through $B_{1} C_{1}$ intersect $A B$ and $D C$ at points $K$ and $L$ (Fig. 12). By the condition, the volumes of the polyhedra $A K L D A_{1} B_{1} C_{1} D_{1}$ and $K B C L B_{1} C_{1}$ are equal. Applying the Simpson's formula (Problem 15) to them, denoting $|A K|=|D L|=a$. Since the heights of these polyhedra are equal, we get the equation for $a$:

$$
7 a+1+4 \frac{(a+1)}{2} \cdot \frac{(7+1)}{2}=(7-a) 7+4 \frac{(7-a)}{2} \cdot \frac{(7+1)}{2}
$$

from which $a=\frac{16}{5}$.

Let the height of the pyramid be $h$. Introduce a coordinate system, taking its origin at the center of $A B C D$, with the $x$ and $y$ axes parallel to $A B$ and $B C$. The points $A, C$, and $D_{1}$ will have coordinates $\left(-\frac{7}{2},-\frac{7}{2}, 0\right),\left(\frac{7}{2}, \frac{7}{2}, 0\right)$,

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-066.jpg?height=350&width=366&top_left_y=523&top_left_x=167)

Fig. 11.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-066.jpg?height=287&width=415&top_left_y=597&top_left_x=657)

Fig. 12.

$\left(-\frac{1}{2}, \frac{1}{2}, h\right)$. It is not difficult to find the equation of the plane $A C D_{1} ; h x - h y + z = 0$. The plane $K L C_{1} B_{1}$ will have the equation $10 h x - 8 z + 3 h = 0$. The normal vector to the first plane is $\boldsymbol{n}(h, -h, 1)$, and to the second is $\boldsymbol{m}(10 h, 0, -8)$. The condition of their perpendicularity gives $10 h^{2} - 8 = 0, h = \frac{2 \sqrt{5}}{5}$. The volume of the pyramid is $\frac{38 \sqrt{5}}{5}$.