5. Inside a regular quadrilateral pyramid $S A B C D$ with base $A B C D$, there is a regular quadrilateral prism $K L M N K_{1} L_{1} M_{1} N_{1}$, the base $K L M N$ of which lies in the plane $A B C$. The center of the base $K L M N$ of the prism is located on the segment $A C$, $K L\|A C$, $K N\|$ $B D$ (points $K$ and $B$ lie on the same side of $A C$), the side of the base of the prism is 2, and the lateral edge $K K_{1}$ of the prism is 3. The vertices $L_{1}$ and $M_{1}$ of the upper base of the prism $K L M N K_{1} L_{1} M_{1} N_{1}$ belong to the lateral faces $S B C$ and $S C D$ of the pyramid $S A B C D$ respectively. The plane $\gamma$ passes through the line $B D$ and the point $L_{1}$. Find the area of the section of the pyramid $S A B C D$ by the plane $\gamma$, if the side of the pyramid is $8 \sqrt{2}$, and its height is 12. (20 points)
Solution. According to the problem, we have $AB = a = 8\sqrt{2}$, $SO = h = 12$, $KL = b = 2$, $KK_1 = h_0 = 3$.

$PR \parallel AC \parallel KL$, $RF \parallel SO$, $RF = h_0$.

$\triangle RCF \sim \triangle SCO$, $\frac{CF}{CO} = \frac{RF}{SO} = \frac{h_0}{h} = \frac{1}{4}$, $CF = \frac{CO}{4}$.

The plane $PQR$ is parallel to $ABC$.

$GR \perp M_1L_1$, $GR = \frac{b}{2} = 1$. $KE \perp BD$,

$LE = CO - CF - GR = \frac{3CO}{4} - \frac{b}{2} = 6 - 1 = 5$.

Let $\alpha$ be the angle of inclination of the plane $\gamma$ to the base plane $ABC$. Then $\tan \alpha = \frac{h_0}{LE} = \frac{3}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-14.jpg?height=688&width=780&top_left_y=770&top_left_x=1135)

Let point $H$ be the intersection of the plane $\gamma$ and the edge $SC$. Then $\angle HOC = \alpha$. If $HU$ is the perpendicular dropped to the plane $ABC$, then $\frac{HU}{OU} = \tan \alpha = \frac{3}{5}$.

$\triangle HCU \sim \triangle SCO$, $\frac{HU}{SO} = \frac{CU}{CO}$, $\frac{HU}{h} = \frac{CO - OU}{CO}$, $\frac{HU}{h} = \frac{CO - 3HU}{CO}$, $\frac{HU}{12} = \frac{24 - 5HU}{24}$, $7HU = 24$, $HU = \frac{24}{7}$.

By the Pythagorean theorem, we have $OH^2 = OU^2 + HU^2$, $OH = \frac{HU \sqrt{34}}{3} = \frac{8 \sqrt{34}}{7}$.

The area of the section of the pyramid by the plane $\gamma$:

$$
S_{\text{ce}} = \frac{BD \cdot OH}{2} = \frac{64 \sqrt{34}}{7}.
$$

Answer: $\frac{64 \sqrt{34}}{7}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-14.jpg?height=674&width=743&top_left_y=1608&top_left_x=1116)