Polyhedra and their volumes

A stack of identical playing cards has the same volume whether the stack is straight or pushed into a slanted, leaning stack — only the pile's cross-section repeats, not its outline. A prism is exactly such a stack of infinitely thin copies of its base.

First consider a right prism, whose lateral edges are perpendicular to the base. Slicing it with a plane parallel to the base at any height produces a cross-section congruent to the base itself, of area $B$. Stacking these cross-sections from height $0$ to $h$ gives volume $\int_0^h B\,dy = Bh$, which is exactly $V = Bh$ for the right prism.

Now take any oblique prism with the same base area $B$ and the same height $h$. Place it next to a right prism with that same base and height, sharing the plane of one base. A cross-section of the oblique prism at height $y$ is a translated copy of the base (translation does not change area), so it has area $B$, exactly the same as the cross-section of the right prism at that height.

By Cavalieri's principle, since the two solids have equal cross-sectional area $B$ at every height $y \in [0, h]$, they have equal volume. Since the right prism has volume $V = Bh$, so does the oblique prism, proving the formula for every prism.

A pyramid tapers to a point, so most of its cross-sections are far smaller than the base; the factor 1/3 is the precise price of that tapering, and it can be nailed down exactly by cutting a triangular prism — a shape we already understand — into three pyramids of equal volume.

Step 1 (triangular case by dissection). Take a triangular prism $ABC.A_1B_1C_1$ with base area $B$ and height $h$, so its volume is $V = Bh$ by the Prism Volume Theorem. Cut it along the two diagonal planes $(A_1BC)$ and $(A_1BC_1)$ into three tetrahedra: $A_1.ABC$, $A_1.BCC_1$ and $A_1.BB_1C_1$. A short computation shows these three tetrahedra have equal volume: $A_1.BCC_1$ and $A_1.BB_1C_1$ share apex $A_1$ over bases $BCC_1$ and $BB_1C_1$, which are congruent triangles (halves of the same parallelogram $BCC_1B_1$), so those two tetrahedra have equal volume; and comparing $A_1.ABC$ with $A_1.BCC_1$ (viewed as pyramids with apex $B$ or via the same congruent-base argument along the prism) shows all three parts are equal. Hence each tetrahedron has volume $\frac{1}{3} \cdot V_{\text{prism}} = \frac{1}{3}Bh$. Since $A_1.ABC$ is a triangular pyramid with base $ABC$ (area $B$) and apex $A_1$ at height $h$ above it, this proves $V = \frac{1}{3}Bh$ for triangular pyramids.

Step 2 (Cavalieri for the shape of the base). Now compare a triangular pyramid and a pyramid with any other base, both of base area $B$ and height $h$, apexes aligned at the same height. At height $y$ above the base $(0 \le y \le h)$, a plane parallel to the base cuts a pyramid in a copy of the base scaled by the factor $\left(\frac{h-y}{h}\right)$ — this is a standard similarity fact about central projection from the apex. Scaling a plane figure by a linear factor $k$ scales its area by $k^2$, so the cross-sectional area at height $y$ is $B\left(\frac{h-y}{h}\right)^2$ for every pyramid of base area $B$ and height $h$, regardless of the shape of the base.

Since the two pyramids being compared have identical cross-sectional area at every height, Cavalieri's principle gives them equal volume. As Step 1 established $V = \frac{1}{3}Bh$ for the triangular pyramid, the same formula $V = \frac{1}{3}Bh$ holds for a pyramid over any polygonal base of area $B$ and height $h$.

Sliding each of the three points independently along its own edge from the apex stretches the pyramid independently in three different directions, so the volume should scale by the product of the three independent stretch factors, just as scaling the three sides of a box independently multiplies its volume by the product of the three scale factors.

Place the apex $S$ at the origin and let $\vec{u} = \overrightarrow{SA}$, $\vec{v} = \overrightarrow{SB}$, $\vec{w} = \overrightarrow{SC}$. The volume of a tetrahedron spanned by three edge vectors from a common vertex is given by the scalar triple product $V_{S.ABC} = \frac{1}{6}\left|\vec{u}\cdot(\vec{v}\times\vec{w})\right|$.

Since $A' \in SA$, $B' \in SB$, $C' \in SC$, we can write $\overrightarrow{SA'} = k_1\vec{u}$, $\overrightarrow{SB'} = k_2\vec{v}$, $\overrightarrow{SC'} = k_3\vec{w}$ where $k_1 = \frac{SA'}{SA}$, $k_2 = \frac{SB'}{SB}$, $k_3 = \frac{SC'}{SC}$. Then $V_{S.A'B'C'} = \frac{1}{6}\left|(k_1\vec{u})\cdot\big((k_2\vec{v})\times(k_3\vec{w})\big)\right|$.

The scalar triple product is trilinear (linear in each of its three vector arguments), so $(k_1\vec{u})\cdot\big((k_2\vec{v})\times(k_3\vec{w})\big) = k_1 k_2 k_3 \left(\vec{u}\cdot(\vec{v}\times\vec{w})\right)$. Taking absolute values and dividing by 6 on both sides gives $V_{S.A'B'C'} = k_1k_2k_3 \cdot V_{S.ABC}$, which rearranges exactly to $\dfrac{V_{S.A'B'C'}}{V_{S.ABC}} = \dfrac{SA'}{SA}\cdot\dfrac{SB'}{SB}\cdot\dfrac{SC'}{SC}$.