A right-angled triangle with legs 6 cm and 8 cm is spun once about its HYPOTENUSE. Find the volume of the solid formed, in ${\text{cm}}^3$, in terms of π.

The hypotenuse is √$(36+64)=10$ cm. Spinning about it sweeps two cones stuck base to base; their shared base radius is the altitude drawn to the hypotenuse, which is $(6\times 8)/10=4.8$ cm, and their two heights add up to the whole hypotenuse, 10 cm. So the volume = (1/3)π$(4.8^2)(10)=(1/3)$π × 230.4 = 76.8π ${\text{cm}}^3$. (96π and 128π are the single cones you get by spinning about the 8 cm and the 6 cm leg instead; 230.4π forgets the 1/3.)