11.100. The height of the cone is divided into three equal segments, and planes parallel to the base are drawn through the division points, dividing the cone into three parts. Find the volume of the middle truncated cone if the volume of the given cone is $V$.
Solution.

$$
V_{\text {truncated cone }}=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right) ; \quad S_{1}=\pi R_{1}^{2}, S_{2}=\pi R_{2}^{2} \text {, then we get, }
$$

that $V_{\text {truncated cone }}=\frac{1}{3} h \pi\left(R_{1}^{2}+R_{2}^{2}+R_{1} R_{2}\right)$ (Fig. 11.87). According to the condition, we have
$h=\frac{H}{3}$, where $H=\frac{3 V}{S}=\frac{3 V}{\pi R^{2}}$, therefore, $h=\frac{V}{\pi R^{2}}$. We get that $V_{\text {truncated cone }}=\frac{1}{3} \frac{V}{R^{2}}\left(R_{1}^{2}+R_{2}^{2}+R_{1} R_{2}\right)=\frac{V}{3}\left(\left(\frac{R_{1}}{R}\right)^{2}+\left(\frac{R_{2}}{R}\right)^{2}+\frac{R_{1} R_{2}}{R^{2}}\right)$. Since

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-711.jpg?height=488&width=460&top_left_y=346&top_left_x=154)

Fig. 11.87 $\triangle B O_{1} E \sim \triangle B O C$, then we have $\frac{O_{1} E}{O C}=\frac{B O_{1}}{B O} ; \frac{R_{1}}{R}=\frac{\frac{1}{3} H}{H} ; \frac{R_{1}}{R}=\frac{1}{3} ; \operatorname{since}$ $\triangle B O_{2} F \sim \triangle B O C$, then $\frac{R_{2}}{R}=\frac{2 / 3 \cdot H}{H} ; \frac{R_{2}}{R}=\frac{2}{3}$, then $V_{\text {truncated cone }}=\frac{V}{3}\left(\left(\frac{1}{3}\right)^{2}+\left(\frac{2}{3}\right)^{2}+\frac{2}{9}\right)=\frac{7 V}{27}$. Answer: $\frac{7 V}{27}$.