# Volume of truncated prism

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.

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.

You see this as follows:

First we can reduce to the case where $h_1=0$ by observing that volume of the piece up to height $h_m=\min(h_1,h_2,h_3)$ is exactly $h_m \cdot A$.

Assume now $h_0=0$ and $h_1, h_2 \geq 0$. This denote the corners of the triangle $D,E,F$ where $D$ is the one with no vertical edge attached on it.

We denote the face bordered by the edges with lengths $h_2$ and $h_3$ by $B$. We observe, that the remaining truncated prism is actually a pyramid over $B$ with apex $D$.

That's the main observation, the rest is relatively simple calculation.

Denote now the height of triangle $A$ on vertex $D$ by $h_D$ and the triangle side opposite of $D$ by $d$. We get the area of $A = \tfrac{1}{2} d h_D$. Note that $h_D$ is perpendicular to $B$

For the volume $V$ we have $$V = \tfrac{1}{3} B h_D = \tfrac{1}{3} \frac{d (h_2 + h_3)}{2} h_D = \tfrac{1}{3} \frac{d h_D}{2} (h_2 + h_3) = \tfrac{1}{3} A (h_2 + h_3)$$

Observe that if you put, on top of your truncated prism, another truncated prism with the same base and vertical edges $x-h_1$, $x-h_2$, $x-h_3$ (where $x\ge\max(h_1,h_2,h_3)$), then you get a prism of base $A$ and height $x$. We have then the relation: $$V(h_1,h_2,h_3)+V(x-h_1, x-h_2, x-h_3)=Ax.$$ If $h_3=\max(h_1,h_2,h_3)$, set $x=h_3$ in the above formula, to get: $$V(h_1,h_2,h_3)=Ah_3-V(h_3-h_1, h_3-h_2, 0).$$ But $V(a,b,0)={1\over3}A(a+b)$ is easy to compute, because in that case the truncated prism is just a pyramid (see wonko's answer), so we have: $$V(h_1,h_2,h_3)=Ah_3-{1\over3}A(h_3-h_1+h_3-h_2)={1\over3}A(h_1+h_2+h_3).$$

The book (Solid mensuration, by Willis F. Kern and James R. Bland. ) that you are looking for can be read on the site https://babel.hathitrust.org/cgi/pt?id=mdp.39015064340733