Consider the section of the given pyramid by a plane passing through the vertex $S$ and the midpoints $A$ and $K$ of the sides $R T$ and $P Q$ of the square $P Q R T$, respectively. Let $S O$ be the height of the pyramid. Then $O$ is the center of the base $P Q R T$. From the right triangles $P K S$ and $K O S$, we find that

$$
\begin{gathered}
S K=\sqrt{S P^{2}-P K^{2}}=\sqrt{100-36}=8 \\
S O=\sqrt{S K^{2}-O K^{2}}=\sqrt{64-36}=\sqrt{28}=2 \sqrt{7}
\end{gathered}
$$

Let $A M$ be the height of the isosceles triangle $A K S$, dropped to the side $K S$. Then the line $A M$ is perpendicular to the plane of the face $P Q C$ and

$$
A M=\frac{A K \cdot S O}{K S}=\frac{12 \cdot 2 \sqrt{7}}{B}=3 \sqrt{7}
$$

Let $C$ and $D$ be the orthogonal projections of points $T$ and $B$ respectively onto the plane of the face $P Q S$. Since the line $R T$ is parallel to the plane of the face $P Q S$, the distances from all points of the line $R T$ to this plane are equal, so $T C=A M=3 \sqrt{7}$, and since the lines $T C$ and $B D$ are perpendicular to the plane of the face $P Q S$, $T C \| B D$. Therefore, the points $S, T, B, C$ and $D$ lie in the same plane. From the similarity of triangles $S T C$ and $S B D$, we find that

$$
B S=S T \cdot \frac{B D}{C t^{\prime}}=10 \cdot \frac{\frac{9 \sqrt{7}}{}}{3 \sqrt{7}}=15
$$

Therefore, $B T=B S-A S=15-10=5$.

## Answer

5.00
[ Regular pyramid ] [ Sections, unfoldings, and other frameworks. ]

On the extension of edge $S T$ beyond point $T$ of the regular quadrilateral pyramid $S P Q R T$ with vertex $S$, a point $B$ is taken such that the distance from it to the plane $S P Q$ is $\frac{9 \sqrt{7}}{2}$. Find the segment $B T$, if $Q R=12$, and $S R=10$.