PROPOSITION 11.
Cones and cylinders which are of the same height are to one another as their bases.
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, [
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. [
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, [
XII. 1] while, as the square on
AC is to the square on
EG, so is the circle
ABCD to the circle
EFGH, [
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. [
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; [
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. [
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; [
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.