10. The base of the right prism $A B C A_{1} B_{1} C_{1}$ is a triangle $A B C$ with angle $B$ equal to $90^{\circ}$ and angle $C$ equal to $30^{\circ}$. Find the area of the section of the prism by a plane passing through the center of the lateral face $A A_{1} C_{1} C$ and vertex $B$ and parallel to the diagonal of the lateral face $A B_{1}$, if the distance from vertex $C$ to the cutting plane is 2, and the hypotenuse of the base of the prism is $\sqrt{14}$?
Solution: Construction of the section. Through point $O$ - the center of the lateral face $A A_{1} C_{1} C$ - draw $O S \| A B_{1}, S \in(A B C), O S=A B_{1} / 2$ and $O H \perp A A_{1}, H \in A C$. Then $S H \| A B, S H=A B / 2$. Connect points $B$ and $S, D=B S \cap A C$. Extend $D O$ until it intersects the extension of edge $C C_{1} ; T=(D O) \cap\left(C C_{1}\right), E=D T \cap A_{1} C_{1}, F=B T \cap B_{1} C_{1}$. The trapezoid $B D E F-$ is the desired section.

Draw $C K \perp B S$, then $T K \perp B S$ and the plane of triangle $T K C$ is perpendicular to the cutting plane. Draw $C P \perp T K, P \in T K$; the length of $C P$ is equal to the given distance from vertex $C$ to the cutting plane. Introduce the notation: $A C=a, C P=d$.

Since $\triangle H D S \sim \triangle A D B$ and $H S=A B / 2, H D=A D / 2=a / 6 ; E C_{1}=A D=a / 3$; $D C=2 E C_{1} ; T D=2 T E, B D=2 F E$, and the area of triangle $B D T$ is four times the area of triangle FET. Accordingly, the area of the section $S_{\text {BDEF }}=3 / 4 \cdot S_{\triangle B D T}$.

In the plane of the base, draw $C K \perp B S$ and $B M \perp A C$.

$B S=\sqrt{B F^{2}+F S^{2}}=\sqrt{\left(\frac{a \sqrt{3}}{2 \cdot 2}\right)^{2}+\left(\frac{a}{2}\right)^{2}}=\frac{a \sqrt{7}}{4}, B D=\frac{2}{3} B S=\frac{a \sqrt{7}}{6}$. In triangle $B D C$

$C D=\frac{2}{3} a, B M=\frac{a \sqrt{3}}{2 \cdot 2}, C K \cdot B D=B M \cdot C D$. Therefore, $C K=\frac{B M \cdot C D}{B D}=\frac{2 a}{3} \cdot \frac{a \sqrt{3}}{4} \cdot \frac{6}{a \sqrt{7}}=\frac{a \sqrt{3}}{\sqrt{7}}$.

In $\triangle C K T K P=\sqrt{C K^{2}-C P^{2}}=\sqrt{\frac{3}{7} a^{2}-d^{2}}=\frac{\sqrt{3 a^{2}-7 d^{2}}}{\sqrt{7}}$;

$K T=\frac{C K^{2}}{K P}=\frac{3 a^{2}}{7} \frac{\sqrt{7}}{\sqrt{3 a^{2}-7 d^{2}}}=\frac{3 a^{2}}{\sqrt{7} \sqrt{3 a^{2}-7 d^{2}}}$.

The area of triangle $B D T \quad S_{\triangle B D T}=\frac{1}{2} B D \cdot K T=\frac{1}{2} \cdot \frac{a \sqrt{7}}{6} \cdot \frac{3 a^{2}}{\sqrt{7} \sqrt{3 a^{2}-7 d^{2}}}=\frac{a^{3}}{4 \sqrt{3 a^{2}-7 d^{2}}}$.

The area of the section $S_{\text {BDEF }}=\frac{3}{4} \cdot S_{\triangle B D T}=\frac{3 a^{3}}{16 \sqrt{3 a^{2}-7 d^{2}}}=\frac{3 a^{2}}{16 \sqrt{3-7(d / a)^{2}}}$.

For reference: the height of the pyramid $h=\frac{\sqrt{3} d}{2 \sqrt{3-7(d / a)^{2}}}$.

Answers: $a=\sqrt{14}, d=2, S=\frac{3 \cdot 14}{16 \sqrt{3-7(4 / 14)}}=\frac{21}{8} \quad(h=\sqrt{3})$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-14.jpg?height=2116&width=1590&top_left_y=400&top_left_x=290)

## Solution for Variant 12