Problem: When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{cm}^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{cm}^3$. What is the length (in cm) of the hypotenuse of the triangle? Solution: For this problem, let $h$ be the height of the triangle and $w$ be the width. Assuming $w > h$, we have that $\frac{w^2h \pi}{3} = 1920\pi$ and $\frac{wh^2\pi}{3} = 800\pi$. Diving the first \expression by the second, we find that $\frac{12}{5}h=w,$ which gives us that $h = 10$ and $w = 24$. After Pythagoras, we get that $\sqrt{h^2+w^2} = 26.$