11.017. The side of the base of a regular triangular pyramid is $1 \mathrm{~cm}$, and its lateral surface area is $3 \mathrm{~cm}^{2}$. Find the volume of the pyramid.
Solution.

Let the side of the base of a regular triangular pyramid be denoted by $a$, i.e., $a=1(\mathrm{cm}) ; S_{\text {side }}=3\left(\mathrm{~cm}^{2}\right)$. Since $S_{\text {side }}=p l$, where $p=\frac{3 a}{2}$ is the semi-perimeter of the base, and $l$ is the apothem, then $l=\frac{S_{\text {side }}}{p}=\frac{3}{3 / 2}=2$ (cm). The radius of the circle inscribed in a regular triangle is $r=\frac{a \sqrt{3}}{6}=$ $=\frac{\sqrt{3}}{6}$ (cm). Then we can find $h=\sqrt{l^{2}-r^{2}}=\sqrt{4-\frac{3}{36}}=\frac{\sqrt{47}}{2 \sqrt{3}}$ (since the base of the height in a regular pyramid is the center of the circle inscribed in the triangular base). The area of the base

\[
\begin{aligned}
S_{\mathrm{base}} & =\frac{a^{2} \sqrt{3}}{4}=\frac{\sqrt{3}}{4}\left(\mathrm{~cm}^{2}\right) . \text { Then the volume } \\
& V=\frac{1}{3} S_{\mathrm{base}} \cdot h=\frac{1}{3} \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{47}}{2 \sqrt{3}}=\frac{\sqrt{47}}{24}\left(\mathrm{~cm}^{3}\right) .
\end{aligned}
\]

Answer: $\frac{\sqrt{47}}{24} \mathrm{~cm}^{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-660.jpg?height=478&width=402&top_left_y=173&top_left_x=187)

Fig. 11.14

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-660.jpg?height=624&width=498&top_left_y=28&top_left_x=701)

Fig. 11.15