11.015. Find the total surface area of a regular triangular pyramid, the side of the base of which is equal to $a$, and the dihedral angle at the base is $60^{\circ}$.
## Solution.

The radius of the circle inscribed in the equilateral

$\triangle A B C$ is $r=\frac{a \sqrt{3}}{6}$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-658.jpg?height=450&width=420&top_left_y=815&top_left_x=750)

Fig. 11.13

(Fig. 11.13); $S H$ is the height of the pyramid, $S K$ is the apothem, $K$ is the midpoint of $A C, \angle S K H=60^{\circ}$. The apothem $l=\frac{r}{\cos 60^{\circ}}=\frac{a \sqrt{3}}{3}$. The lateral surface area $S_{\text {lateral }}=p \cdot l$, where $p$ is the semiperimeter of the base $p=\frac{3 a}{2}$. Then $S_{\text {lateral }}=\frac{a \sqrt{3}}{3} \cdot \frac{3 a}{2}=\frac{a^{2} \sqrt{3}}{2}, S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. The total surface area $S=S_{\text {base }}+S_{\text {lateral }}=\frac{a^{2} \sqrt{3}}{4}+\frac{a^{2} \sqrt{3}}{2}=\frac{3 a^{2} \sqrt{3}}{4}$. Answer: $\frac{3 a^{2} \sqrt{3}}{4}$.