2.21. The height of the cylinder is $H$, and the radius of its base is $R$. Inside the cylinder, a pyramid is placed, the height of which coincides with the generatrix $A A_{1}$ of the cylinder, and the base is an isosceles triangle $A B C(A B=A C)$ inscribed in the base of the cylinder. Find the lateral surface area of the pyramid if $\angle A=120^{\circ}$.
2.21. Draw $A D \perp B C$ and connect points $A_{1}$ and $D$ (Fig. 2.19). According to the theorem of three perpendiculars, we have $A_{1} D \perp B C$. Since arc $C A B$ contains $120^{\circ}$, and arcs $A C$ and $A B$ each contain $60^{\circ}$, then $B C=R \sqrt{3}$, $A B=R$. In $\triangle A B D$ we have $A D=\frac{R}{2}$. Further, from $\triangle A A_{1} D$ we get $A_{1} D=\sqrt{H^{2}+\frac{R^{2}}{4}}=\frac{1}{2} \sqrt{R^{2}+4 H^{2}}$. There-

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-466.jpg?height=469&width=388&top_left_y=327&top_left_x=752)

Fig. 2.19

fore,

$$
\begin{gathered}
S_{\triangle A_{1} A B}=\frac{1}{2} A B \cdot A A_{1}=\frac{1}{2} R H \\
S_{\triangle A_{1} B C}=\frac{1}{2} B C \cdot A_{1} D=\frac{1}{2} R \sqrt{3} \cdot \frac{1}{2} \sqrt{R^{2}+4 H^{2}}=\frac{1}{4} \sqrt{3 R^{2}+12 H^{2}}
\end{gathered}
$$

Finally, we find

$$
\begin{gathered}
S_{60 \mathrm{~K}}=2 S_{\triangle \mathrm{A}_{1} A B}+S_{\triangle \mathrm{A}_{1} B C}=R H+\frac{1}{4} R \sqrt{3 R^{2}+12 H^{2}}= \\
=\frac{R}{4}\left(4 H+\sqrt{3 R^{2}+12 H^{2}}\right)
\end{gathered}
$$

Answer: $0.25 R\left(4 H+\sqrt{3 R^{2}+12 H^{2}}\right)$.