5. Find the area of the cross-section of a regular hexagonal pyramid $S A B C D E F$ by a plane passing through the vertex $F$ of the base $A B C D E F$ and parallel to the median $C M$ of the lateral face $S C D$ and the apothem $S N$ of the lateral face $S A F$, if the side of the base of the pyramid is $4 \sqrt{7}$, and the distance from the vertex $S$ to the cutting plane is $\sqrt{7}$. (20 points)
Solution. We will construct the section of the pyramid. In the plane $SCD$, through point $S$, draw a line $SQ$ parallel to $CM$, where $Q$ lies on line $CD$, and $CM$ is the midline of triangle $SQD$, $QC=a$, where $a$ is the side of the base of the pyramid.

The plane $SNQ$ is parallel to the plane of the section. Through point $F$, draw a line $FV$ parallel to $NQ$, where $V \in CQ$, $QV=VC=a/2$. $R$ is the intersection point of lines $FV$ and $BC$. The plane of the section intersects the base of the pyramid along segment $FR$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-04.jpg?height=1514&width=1498&top_left_y=845&top_left_x=356)

$O_{1}$ is the intersection point of $NQ$ and $AD$, triangles $AO_{1}N$ and $DO_{1}Q$ are similar. $AO_{1}: O_{1}D=1: 4$, $AO_{1}=2a/5$, $RC=AO_{1}=2a/5$.

Point $Z$ is the intersection point of lines $RF$ and $DE$. Point $T$ is the intersection point of lines $BC$ and $DE$. Triangles $FZE$ and $RZT$ are similar, and $FE: RT=ZE: ZT=5: 7$, $ZE=5a$.

In the plane $SCD$, through point $V$, draw a line $VK$ parallel to $CM$, where $K$ lies on edge $SD$, and $L$ is the intersection point of line $VK$ with edge $SC$.

By Thales' theorem, we have: $SK: KM: MD=1: 1: 2$, $SL: LC=1: 1$.

In the plane $SDE$, point $P$ is the intersection point of lines $ZK$ and $SE$.

By Thales' theorem, we have $SP: PE=2: 5$.

The desired section is $KLRFP$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-05.jpg?height=505&width=551&top_left_y=387&top_left_x=1392)

We will calculate the area of the section using the formula

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-05.jpg?height=403&width=1093&top_left_y=958&top_left_x=867)

$S_{c e 4}=\frac{S_{n p}}{\cos \varphi}$, where $S_{n p}$ is the area

of the projection of the section onto the base plane, $\varphi$ is the angle between the plane of the section and the base plane. We will find the area of the projection of the section onto the base plane.

The projection is a pentagon $TXHJR$. The area of the projection of the section is calculated using the formula

$S_{\text {pr }}=S_{F R G}+S_{O G H}+S_{O H X}+S_{O X F}=\frac{a^{2} \sqrt{3}}{5}-\frac{2}{5} \cdot \frac{1}{2} \cdot \frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot$

$\frac{1}{4} \cdot \frac{a^{2} \sqrt{3}}{4}+\frac{1}{4} \cdot \frac{2}{7} \cdot \frac{a^{2} \sqrt{3}}{4}+\frac{2}{7} \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{303 a^{2} \sqrt{3}}{35 \cdot 32}, \quad S_{\text {pr }}=\frac{303 \sqrt{3}}{10}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-05.jpg?height=434&width=497&top_left_y=1825&top_left_x=1316)

Let the distance from point $S$ to the plane of the section be $d$, $d=\sqrt{7}$. The distance from point $C$ to the section is $d$. In triangle $RCV$, draw the height $CU$, and denote its length by $h$. Then $\sin \varphi=\frac{d}{h}$, $\cos \varphi=\sqrt{1-\left(\frac{d}{h}\right)^{2}}$. We will find $RV$ using the cosine theorem:

$$
R V^{2}=\frac{4 a^{2}}{25}+\frac{a^{2}}{4}-\frac{a^{2}}{5}=\frac{21 a^{2}}{100}, R V=\frac{\sqrt{21} a}{10}
$$

Using different formulas for finding the area of triangle $RCV$, we have

$\frac{\sqrt{21} a h}{10}=\frac{a^{2}}{5} \cdot \frac{\sqrt{3}}{2}, h=\frac{a}{\sqrt{7}}$. Then $\cos \varphi=\sqrt{1-\frac{7 d^{2}}{a^{2}}}=\frac{\sqrt{a^{2}-7 d^{2}}}{a}=\frac{3}{4}$.

Finally, we have

$S=\frac{303 a^{3} \sqrt{3}}{35 \cdot 32 \sqrt{a^{2}-7 d^{2}}}, \quad S=\frac{202 \sqrt{3}}{5}$.

Answer: $\frac{202 \sqrt{3}}{5}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-06.jpg?height=192&width=740&top_left_y=560&top_left_x=1089)