The volume of the cone formed by rotating the right-angled triangle around one of its legs is $800 \pi \mathrm{cm}^{3}$, and the volume of the cone formed by rotating around the other leg is $1920 \pi$ $\mathrm{cm}^{3}$. What is the length of the hypotenuse of this triangle? (in cm)
The volume of a cone with a base radius of $\mathrm{r}$ and height $\mathrm{h}$ is $\frac{\pi}{3} \mathrm{hr}$. Let $\mathrm{a}, \mathrm{b}$ represent the two legs of a right triangle, then we have
$$
\frac{\pi}{3} \mathrm{ba}^{2}=800 \pi \text { and } \frac{\pi}{3} \mathrm{ab}^{2}=1920 \pi \text {. }
$$

By comparing the two equations, we get $\frac{\mathrm{a}}{\mathrm{b}}=\frac{5}{12}$, which means $\mathrm{a}=\frac{5}{12} \mathrm{~b}$.
$$
\begin{array}{l}
\text { Therefore, } \mathrm{a}^{3}=\left(\frac{5}{12} \mathrm{~b}\right) \mathrm{a}^{2}=\frac{5}{12}\left(\mathrm{ba}^{2}\right) \\
=\frac{5}{12}(800)(3)=1000 .
\end{array}
$$

Thus, we find: $a=10, b=24$. Using the Pythagorean theorem, the hypotenuse of the triangle is $\sqrt{10^{2}+24^{2}}=26(\mathrm{~cm})$.