### 11.211 The side of the base of a regular quadrilateral pyramid

is equal to $a$, the lateral edge forms an angle of $30^{\circ}$ with the height. A plane is drawn through the vertex of the base of the pyramid, perpendicular to the opposite lateral edge. This plane divides the pyramid into two parts. Determine the volume of the part of the pyramid adjacent to the vertex.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-266.jpg?height=376&width=315&top_left_y=712&top_left_x=118)

Let $F A B C D$ be the given regular pyramid, and the section $D M K N$ is perpendicular to $B F$.

By the condition $\angle B F D=60^{\circ} \Rightarrow \triangle B F D$ is an equilateral triangle with side $a \sqrt{2} \Rightarrow K$ is the midpoint of $B F$,

$$
\begin{aligned}
& K F=\frac{a}{\sqrt{2}}, K D=\frac{a \sqrt{6}}{2} . \text { Further, } \triangle F M K=\Delta F N K \Rightarrow \\
& \Rightarrow F M=F N \Rightarrow M N \| A C \Rightarrow M N \perp F D \\
& (\text { since } A C \perp F D) \Rightarrow M N \perp K D \text { (Theorem 12) } \Rightarrow
\end{aligned}
$$

$\Rightarrow S_{D M K N}=\frac{1}{2} M N \cdot K D$. Let $\alpha=\angle B F A$.

By the cosine rule for $\triangle F A B \cos \alpha=\frac{2 B F^{2}-A B^{2}}{2 B F^{2}}=\frac{3}{4} \quad(A B=a, B F=a \sqrt{2})$.

Further, $\triangle F N M$ is equilateral $\Rightarrow M N=F M=\frac{K F}{\cos \alpha}=\frac{4 a}{3 \sqrt{2}}$.

From this, we have: $V_{F D M K N}=\frac{1}{3} K F \cdot \frac{M N \cdot K D}{2}=\frac{\sqrt{6} a^{3}}{18}$.

Answer: $\frac{\sqrt{6} a^{3}}{18}$.