12.051. At the base of a right triangular prism lies an isosceles triangle $ABC$; where $AB = BC = a$ and $\angle BAC = \alpha$. A plane is drawn through the side $AC$ at an angle $\varphi (\varphi < \pi / 2)$ to the base. Find the area of the section, given that a triangle is obtained in the section.
## Solution.

Let $A L C$ be the given section (Fig. 12.53), $L K$ be the perpendicular to $A C$, so $K B \perp A C$ (by the theorem of the $3$ perpendiculars), $K$ is the midpoint of $A C$. From $\triangle A B K \quad A K=A B \cos \alpha=a \cos \alpha . A C=2 A K=2 a \cos \alpha$. The area of $\triangle A B C \quad S_{A B C}=\frac{1}{2} A C \cdot A B=a^{2} \cos \alpha \sin \alpha=\frac{a^{2} \sin 2 \alpha}{2}$. Since $\triangle A B C$ is the projection of $\triangle A L C$, then $S_{A L C}=\frac{S_{A B C}}{\cos \varphi}=\frac{a^{2} \sin 2 \alpha}{2 \cos \varphi}$. Answer: $\frac{a^{2} \sin 2 \alpha}{2 \cos \varphi}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-752.jpg?height=588&width=434&top_left_y=94&top_left_x=173)

Fig. 12.53

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-752.jpg?height=446&width=502&top_left_y=235&top_left_x=701)

Fig. 12.54