11.125. 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 $A_{1}$ projects 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.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0837.jpg?height=591&width=815&top_left_y=74&top_left_x=273)

Fig. 11.21
## Solution.

Let $O$ be the center of $\triangle ABC$ (Fig. 11.21), $O$ is the projection of $A_{1}$ onto the plane $ABC, \angle OAB = \frac{1}{2} \angle BAC = 30^{\circ}, \angle A_{1}AO = 60^{\circ}$.

Then $\cos \angle A_{1}AB = \cos \angle A_{1}AO \cdot \cos \angle OAB = \frac{\sqrt{3}}{4}$;

$$
\sin \angle A_{1}AB = \sqrt{1 - \cos^2 \angle A_{1}AB} = \sqrt{1 - \frac{3}{16}} = \frac{\sqrt{13}}{4}.
$$

From $\triangle AOA_{1} (\angle AOA_{1} = 90^{\circ}): A A_{1} = \frac{AO}{\cos \angle A_{1}AO} = \frac{a \sqrt{3}}{3 \cos 60^{\circ}} = \frac{2a \sqrt{3}}{3}$. In the face $AA_{1}B_{1}B$, a perpendicular $BK$ is dropped from $B$ to $AA_{1}$. Then the plane $BKC$ is perpendicular to the edge $AA_{1}$ of the prism, and $\triangle BKC$ is the perpendicular section of the prism.

From $\triangle AKB (\angle AKB = 90^{\circ}): BK = AB \sin \angle A_{1}AB = \frac{a \sqrt{13}}{4}$.

Perimeter of $\triangle BKC: P = 2BK + BC = \frac{a \sqrt{13}}{2} + a = \frac{a(2 + \sqrt{13})}{2}$, lateral surface area of the prism $S = P \cdot AA_{1} = \frac{a^2 \sqrt{3}(2 + \sqrt{13})}{3}$.

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