2.47. The base of the prism $A B C A_{1} B_{1} C_{1}$ is an equilateral triangle $A B C$ with side $a$. The vertex $A_{1}$ is projected onto the center of the lower base, and the edge $A A_{1}$ is inclined to the base plane at an angle of $60^{\circ}$. Determine the lateral surface area of the prism.
2.47. Let's draw a section through $B C$ perpendicular to $A A_{1}$ (Fig. 2.41); then $S_{\text {side }}=(B C+M C+$ $+M B) A A_{1}=(B C+2 M C) A A_{1}$. Since $\angle A_{1} A O=60^{\circ}$ (by the problem statement), then

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-480.jpg?height=417&width=479&top_left_y=1441&top_left_x=672)

Fig. 2.41
$A A_{1}=2 A O=\frac{2 a \sqrt{3}}{3}$. Further, in $\triangle A M D$ we have $\angle A D M=30^{\circ}$ and, consequently, $A M=\frac{a \sqrt{3}}{4}$. Then from $\triangle A M C$ we find $M C=\sqrt{A C^{2}-A M^{2}}=\sqrt{a^{2}-\frac{3 a^{2}}{16}}=\frac{a \sqrt{13}}{4}$. Thus,

$$
S_{\text {side }}=\left(a+\frac{a \sqrt{13}}{2}\right) \frac{2 a \sqrt{3}}{3}=\frac{a^{2} \sqrt{3}(\sqrt{13}+2)}{3}
$$

Answer: $\frac{a^{2} \sqrt{3}(\sqrt{13}+2)}{3}$.