2.15. Find the lateral surface of a regular triangular prism with height $h$, if a line passing through the center of the upper base and the midpoint of the side of the lower base is inclined to the plane of the base at an angle of $60^{\circ}$.
2.15. Let the side of the base be denoted by $a$. Then $S_{\text {6o }}=3 a h$. Draw the height $O_{1} O$ (Fig. 2.14). In $\triangle D O O_{1}$, we have $\angle O O_{1} D=30^{\circ}$, so $O_{1} D=2 O D$ and $4 O D^{2}-O D^{2}=h^{2}$, from which $O D=\frac{h \sqrt{3}}{3}$. On the other hand, $O D=\frac{a \sqrt{3}}{6}$, and therefore $a=2 h$.

Thus, $S_{\text {бок }}=3 \cdot 2 h \cdot h=6 h^{2}$.

Answer: $6 h^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-463.jpg?height=416&width=496&top_left_y=1186&top_left_x=44)

Fig. 2.15