I needed to find the volume of what Wikipedia calls a truncated prism, which is a prism (with triangle base) that is intersected with a halfspace such that the boundary of the halfspace intersects the three vertical edges of the prism at heights $h_1, h_2, h_3$.
I was able to find the formula
$$
V=A\frac{h_1+h_2+h_3}{3},
$$
(where $A$ is the area of the triangle base) online, but without proof.
I was also able to prove this formula myself, but with a really nasty proof. (I integrated the area of the horizontal cross-sections; after passing the first intersection with the hyperplane at height $h_1$ these cross-sections have the form of the base triangle minus a quadratically increasing triangle, then after crossing the first intersection at height $h_2$ they have the form of a quadratically shrinking triangle)
Do you know of an elegant proof of the volume formula?
PS: Wikipedia cites 'William F. Kern, James R Bland,Solid Mensuration with proofs, 1938, p.81' for the name truncated prism, but I cannot find this book.
Best Answer
The existing two answers are both pretty good, so here's my two cents in the form of a proof without words . . . Okay, a few words won't hurt :)
Let $h_1$ be the smallest height (up to purple), $h_3$ be the largest height to the top, and $h_2$ be the intermediate height (up to teal).
$$\mathcal{V}_{\text{whole}} = \mathcal{V}_{\text{purple}} + \mathcal{V}_{\text{teal}} + \mathcal{V}_{\text{top}} $$
The area of the base triangle is denoted $A$ as in the OP.
\begin{align*} \mathcal{V}_{\text{purple}} &= h_1 A & &\text{as a straight up prism} \\ \mathcal{V}_{\text{teal}} &= \frac23 (h_2 - h_1) A & &\text{as a straight up prism with height $h_2-h_1$ minus} \\ &&&\text{a triangular cone upside-down of the same height} \\ \mathcal{V}_{\text{top}} &= \frac13 (h_3 - h_2) A & &\text{as a triangular cone which base is still $A$ by Cavalieri principle} \end{align*} The coefficient associated with each height is easily seen to be all $1/3.\quad$Q.E.D.