11.177. A cone is formed by rotating a right triangle with area $S$ around one of its legs. Find the volume of the cone if the length of the circumference described by the point of intersection of the triangle's medians during its rotation is $L$.


Solution.

The required volume $V=\frac{1}{3} \pi r^{2} h$. Let the cone be formed by rotating $\triangle A B C$ around the leg $B C$ (Fig. 11.70); then $A C=r, B C=h$. By the condition $\frac{1}{2} r h=S$; then $V=\frac{2}{3} \pi r S$. Further, by the condition, $2 \pi \cdot D N=L$, where $D-$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0879.jpg?height=535&width=377&top_left_y=108&top_left_x=187)

Fig. 11.70

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0879.jpg?height=563&width=567&top_left_y=78&top_left_x=677)

Fig. 11.71

is the point of intersection of the medians, $D N \perp B C$. But $D N: A C=D M: A M=1: 3, \Rightarrow$ $D N=\frac{r}{3} ;$ therefore, $\frac{2}{3} \pi r=L, r=\frac{3 L}{2 \pi}$. We have $V=\frac{2}{3} \pi S \cdot \frac{3 L}{2 \pi}=S L$.

Answer: $S L$.