Parallelepipeds of Same Height have Volume Proportional to Bases

Theorem

In the words of Euclid:

Parallelepidedal solids which are of the same height are to one another as their bases.

(The Elements: Book $\text{XI}$: Proposition $32$)

Proof

Let $AB$ and $CD$ be parallelepipeds of the same height.

Let $AE$ be the base of $AB$ and $CF$ be the base of $AB$.

Using Proposition $45$ of Book $\text{I} $: Construction of Parallelogram in Given Angle equal to Given Polygon:

Let the parallelogram $FH$ equal in area to $AE$ be applied to $FG$.

Let the parallelepiped $GK$ be constructed with the same height as $CD$ on the base $FH$.

Then from Proposition $31$ of Book $\text{XI} $: Parallelepipeds on Equal Bases and Same Height are Equal in Volume:

the parallelepiped $AB$ equals the parallelepiped $GH$.

We have that the parallelepiped $CK$ is cut by the plane $DG$ which is parallel to opposite planes.

Therefore from Proposition $25$ of Book $\text{XI} $: Parallelepiped cut by Plane Parallel to Opposite Planes:

the parallelepiped $CD$ is to the parallelepiped $DH$ as the base $CF$ is to the base $FH$.

But:

the base $FH$ equals the base $AE$

and:

the parallelepiped $GK$ equals the parallelepiped $AB$.

Therefore, as the base $AE$ is to the base $CF$, the parallelepiped $AB$ is to the parallelepiped $CD$.

$\blacksquare$

Historical Note

This proof is Proposition $32$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions