2.51. A plane passing through the vertex of a cone intersects the base along a chord, the length of which is equal to the radius of this base. Determine the ratio of the volumes of the resulting parts of the cone.
2.51. Let $A O=r, S O=h$ (Fig. 2.43), $V$ - the volume of the cone, $V_{1}$ and $V_{2}$ the volumes of its parts. We will find $V_{1}$ as the difference between the volumes of the part of the cone whose base is the sector $A O B$, and the pyramid whose base is $\triangle A O B$. According to the condition, $A B=r$, i.e., $A B$ is a side of the inscribed regular hexagon, and thus, $V_{1}=\frac{1}{6} \cdot \frac{1}{3} \pi r^{2} h-\frac{1}{3} \cdot \frac{r^{2} \sqrt{3}}{4} h=\frac{1}{3} S_{1} h$ 16 Skanavi

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-482.jpg?height=635&width=490&top_left_y=1221&top_left_x=659)

Fig. 2.43
where $S_{1}$ is the area of the segment $A m B$. Then $V_{2}=V-V_{1}=\frac{1}{3} S_{2} h$, where $S_{2}=\pi r^{2}-S_{1}$. Thus, $V_{1}: V_{2}=S_{1}: S_{2}$. Next, we find $S_{1}=\frac{1}{6} \pi r^{2}-\frac{r^{2} \sqrt{3}}{4}=\frac{r^{2}(2 \pi-3 \sqrt{3})}{12}, S_{2}=\pi r^{2}-\frac{r^{2}(2 \pi-3 \sqrt{3})}{12}=$
$=\frac{r^{2}(10 \pi+3 \sqrt{3})}{12}$. Therefore, $V_{1}: V_{2}=\frac{2 \pi-3 \sqrt{3}}{10 \pi+3 \sqrt{3}}$.

Answer: $(2 \pi-3 \sqrt{3}):(10 \pi+3 \sqrt{3})$.