Example 66. In the base of the right prism $A B C A_{1} B_{1} C_{1}$ lies a triangle, where $|A B|=|A C|$ and $\widehat{A B C}=\alpha$. It is also known that point $D$ is the midpoint of edge $A A_{1}, \widehat{D C A}=$ $=\beta$ and $|C D|=b$. Find the lateral surface area of the prism.

Construction of the image. Let the figure $A B C A_{1} B_{1} C_{1} \quad$ (Fig. 75) be the image of the given prism. This image is complete. Its parametric number (calculate it yourself) $p=5$. On this image, construct point $D$ - the midpoint of edge $A A_{1}$ - and consider $|C D|=$ $=b$ (without spending new parameters).

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-114.jpg?height=731&width=691&top_left_y=228&top_left_x=1002)

Fig. 75
Solution. It is clear that this problem belongs to type $F$. Let's solve it by direct calculation. Since $A B C A_{1} B_{1} C_{1}$ is a right prism, then

$$
S_{\text {side }}=P \cdot H, \text { where } P=2|A C|+|B C|, \text { and } H=\left|A A_{1}\right| .
$$

Since $\triangle A C D$ is a right triangle, then $|A C|=b \cos \beta,|A D|=$ $=b \sin \beta$, and therefore $\left|A A_{1}\right|=2 b \sin \beta$.

Construct $[A E]$ - the median of $\triangle A B C, [A E] \perp[B C]$. From the right $\triangle A C E$ we find: $|C E|=|A C| \cos \alpha=b \cos \beta \cos \alpha$, i.e., $|B C|=2 b \cos \alpha \cos \beta$.

Thus,

$$
\begin{aligned}
& S_{\text {side }}=(2 b \cos \beta+2 b \cos \alpha \cos \beta) \cdot 2 b \sin \beta= \\
= & 4 b^{2} \cos \beta(1+\cos \alpha) \sin \beta=4 b^{2} \sin 2 \beta \cos ^{2} \frac{\alpha}{2}
\end{aligned}
$$

Following from the problem statement, $00$.

Therefore, $S_{6 \text { side }}=4 b^{2} \sin 2 \beta \cos ^{2} \frac{\alpha}{2}$.