# Parallelepipeds of Same Height have Volume Proportional to Bases

Jump to navigation
Jump to search

## 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 $AB$.

- 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 $FH$.

- 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:

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