---
book: 12
number: 11
id: "XII.11"
kind: "theorem"
uses: ["[[book-12/proposition-6]]", "[[book-10/proposition-1]]", "[[book-12/proposition-1]]", "[[book-12/proposition-2]]", "[[book-5/proposition-11]]", "[[book-5/proposition-16]]", "[[book-12/proposition-10]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.11"
license: "CC-BY-SA-4.0"
---

# XII.11

*Cones and cylinders which are of the same height are to one another as their bases.*

## Proof

Let there be cones and cylinders of the same height, let the circles *ABCD*, *EFGH* be their bases, *KL*, *MN* their axes and *AC*, *EG* the diameters of their bases; I say that, as the circle *ABCD* is to the circle *EFGH*, so is the cone *AL* to the cone *EN*.

For, if not, then, as the circle *ABCD* is to the circle *EFGH*, so will the cone *AL* be either to some solid less than the cone *EN* or to a greater.

First, let it be in that ratio to a less solid *O*, and let the solid *X* be equal to that by which the solid *O* is less than the cone *EN*; therefore the cone *EN* is equal to the solids *O*, *X*.

Let the square *EFGH* be inscribed in the circle *EFGH*; therefore the square is greater than the half of the circle.

Let there be set up from the square *EFGH* a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [[book-12/proposition-6|XII. 6]] while the cone is less than the circumscribed pyramid.

Let the circumferences *EF*, *FG*, *GH*, *HE* be bisected at the points *P*, *Q*, *R*, *S*, and let *HP*, *PE*, *EQ*, *QF*, *FR*, *RG*, *GS*, *SH* be joined.

Therefore each of the triangles *HPE*, *EQF*, *FRG*, *GSH* is greater than the half of that segment of the circle which is about it.

On each of the triangles *HPE*, *EQF*, *FRG*, *GSH* let there be set up a pyramid of equal height with the cone; therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it.

Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the solid *X*. [[book-10/proposition-1|X. 1]]

Let such be left, and let them be the segments on *HP*, *PE*, *EQ*, *QF*, *FR*, *RG*, *GS*, *SH*; therefore the remainder, the pyramid of which the polygon *HPEQFRGS* is the base and the height the same with that of the cone, is greater than the solid *O*.

Let there also be inscribed in the circle *ABCD* the polygon *DTAUBVCW* similar and similarly situated to the polygon *HPEQFRGS*, and on it let a pyramid be set up of equal height with the cone *AL*.

Since then, as the square on *AC* is to the square on *EG*, so is the polygon *DTAUBVCW* to the polygon *HPEQFRGS*, [[book-12/proposition-1|XII. 1]] while, as the square on *AC* is to the square on *EG*, so is the circle *ABCD* to the circle *EFGH*, [[book-12/proposition-2|XII. 2]] therefore also, as the circle *ABCD* is to the circle *EFGH*, so is the polygon *DTAUBVCW* to the polygon *HPEQFRGS*.

But, as the circle *ABCD* is to the circle *EFGH*, so is the cone *AL* to the solid *O*, and, as the polygon *DTAUBVCW* is to the polygon *HPEQFRGS*, so is the pyramid of which the polygon *DTAUBVCW* is the base and the point *L* the vertex to the pyramid of which the polygon *HPEQFRGS* is the base and the point *N* the vertex. [[book-12/proposition-6|XII. 6]]

Therefore also, as the cone *AL* is to the solid *O*, so is the pyramid of which the polygon *DTAUBVCW* is the base and the point *L* the vertex to the pyramid of which the polygon *HPEQFRGS* is the base and the point *N* the vertex; [[book-5/proposition-11|V. 11]] therefore, alternately, as the cone *AL* is to the pyramid in it, so is the solid *O* to the pyramid in the cone *EN*. [[book-5/proposition-16|V. 16]]

But the cone *AL* is greater than the pyramid in it; therefore the solid *O* is also greater than the pyramid in the cone *EN*.

But it is also less: which is absurd.

Therefore the cone *AL* is not to any solid less than the cone *EN* as the circle *ABCD* is to the circle *EFGH*.

Similarly we can prove that neither is the cone *EN* to any solid less than the cone *AL* as the circle *EFGH* is to the circle *ABCD*.

I say next that neither is the cone *AL* to any solid greater than the cone *EN* as the circle *ABCD* is to the circle *EFGH*.

For, if possible, let it be in that ratio to a greater solid *O*; therefore, inversely, as the circle *EFGH* is to the circle *ABCD*, so is the solid *O* to the cone *AL*.

But, as the solid *O* is to the cone *AL*, so is the cone *EN* to some solid less than the cone *AL*; therefore also, as the circle *EFGH* is to the circle *ABCD*, so is the cone *EN* to some solid less than the cone *AL*: which was proved impossible.

Therefore the cone *AL* is not to any solid greater than the cone *EN* as the circle *ABCD* is to the circle *EFGH*.

But it was proved that neither is it in this ratio to a less solid; therefore, as the circle *ABCD* is to the circle *EFGH*, so is the cone *AL* to the cone *EN*.

But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [[book-12/proposition-10|XII. 10]]

Therefore also, as the circle *ABCD* is to the circle *EFGH*, so are the cylinders on them which are of equal height.

Therefore etc. Q. E. D.