3. A skewed prism is given, whose base is an equilateral triangle with a side of $10 \mathrm{dm}$, and the lateral edge is $13 \mathrm{dm}$. One of the vertices of the upper base is normally projected onto the center of the lower base of the prism. Calculate the surface area of the prism.
Solution. Let point $O$ be the orthogonal projection of vertex $A_{1}$ of the prism $A B C A_{1} B_{1} C_{1}$ onto its base $A B C$ (see diagram). We will prove that the face $B C C_{1} B_{1}$ is a rectangle.

Since $O$ is the orthogonal projection of $A_{1}$ onto the plane $A B C$, we get that the plane $A P A_{1}$ is perpendicular to the edge $B C$ ($P$ is the midpoint of edge $B C$). Therefore, from $A A_{1} \parallel B B_{1} \parallel C C_{1}$ it follows that $B B_{1} \perp B C$ and $C C_{1} \perp B C$, which was to be proved.

![](https://cdn.mathpix.com/cropped/2024_06_05_faf4f26986d5dd6dd724g-9.jpg?height=560&width=452&top_left_y=225&top_left_x=1031)

From the right triangles $A_{1} N A$ and $A_{1} M A$ ($N$ is the midpoint of $A B$ and $M$ is the midpoint of $A C$), we get $\overline{N A_{1}}=12$ and $\overline{M A_{1}}=12$. For the surface area $P$ of the prism, we have:

$$
\begin{aligned}
P & =2 B+M=2 \cdot \frac{\overline{A B}^{2}}{4} \sqrt{3}+\overline{A B} \cdot \overline{N A_{1}}+\overline{A C} \cdot \overline{M A_{1}}+\overline{B B_{1}} \cdot \overline{B C} \\
& =50 \sqrt{3}+10 \cdot 12+10 \cdot 12+10 \cdot 13=50 \sqrt{3}+370 .
\end{aligned}
$$