11.4. The lateral surface of a right circular cone is a circle of radius $R$, from which a sector corresponding to the central angle $(2-\sqrt{3}) \pi$ has been removed. Find the maximum value of the area of a plane section of the cone passing through its vertex.
Answer: $\frac{1}{2} R^{2}$

Solution. The arc of the circle of development is $2 \pi-(2-\sqrt{3}) \pi=\pi \sqrt{3}$. From the ratio $2 \pi r=2 \pi R \cdot \frac{\pi \sqrt{3}}{2 \pi}$, where $r$ is the radius of the base, we get $r=\frac{R \sqrt{3}}{2}$. If $\alpha$ is the angle at the vertex of the axial section, then $\sin \frac{\alpha}{2}=\frac{r}{l}$, where $l$ is the length of the generatrix of the cone. In our case, $l=R$, so $\sin \frac{\alpha}{2}=\frac{\sqrt{3}}{2}, \frac{\alpha}{2}=60^{\circ}, \alpha=$ $120^{\circ}$.

The area of the section passing through the vertex of the cone is $S=$ $\frac{1}{2} R^{2} \sin \varphi$, where $\varphi$ is the angle between the generatrices of the section. The maximum area will be for the section where $\varphi=90^{\circ}$ (this will not be the axial section, for which $\varphi=\alpha=120^{\circ}$). Such a section always exists, as the angle changes from $120^{\circ}$ to $0^{\circ}$.

Thus, the area of the maximum section will be

$$
S_{\max }=\frac{1}{2} R^{2}
$$