11.092. A triangle with sides 10, 17, and 21 cm is rotated around the larger side. Calculate the volume and surface area of the resulting solid of revolution.
## Solution.

From $\triangle A O B(O B=21-x) R^{2}=17^{2}-(21-x)^{2}$ (Fig. 11.80). From $\triangle A O C$ $(O C=x) R^{2}=10^{2}-x^{2}$. Solving the system: $\left\{\begin{array}{l}R^{2}=17^{2}-(21-x)^{2}, \\ R^{2}=10^{2}-x^{2},\end{array}\right.$ we get $O C=x=6 \text{ cm}$, and $O B=21-6=15 \text{ cm}$. Then $R=\sqrt{10^{2}-6^{2}}=8 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-705.jpg?height=58&width=1148&top_left_y=1567&top_left_x=98)
The lateral surface area of the smaller cone $S_{6 \text{ok}_2}=\pi R l_{2}=\pi 8 \cdot 10=80 \pi \text{ cm}^{2}$. Then the surface area of the desired figure is: $S=S_{\text{bok}_1}+S_{\text{bok}_2}=216 \pi$ cm $^{2}$. The volume of the larger cone $V_{1}=\frac{1}{3} \pi R^{2} H_{1}=\frac{1}{3} \pi 64 \cdot 15=320 \pi$ cm $^{3}$. The volume of the smaller cone $V_{2}=\frac{1}{3} \pi R^{2} H_{2}=\frac{1}{3} \pi 64 \cdot 6=128 \pi \text{ cm}^{3}$. Then the volume of the desired figure is: $V=V_{1}+V_{2}=448 \pi \text{ cm}^{3}$.

Fig. 11.79

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-706.jpg?height=322&width=494&top_left_y=383&top_left_x=193)

Fig. 11.80

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-706.jpg?height=626&width=354&top_left_y=79&top_left_x=815)

Answer: $448 \pi \text{ cm}^{3} ; 216 \pi \text{ cm}^{2}$.