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circle; for [by 32. 11.) they are to one another as their bases; and so also are the third parts of them: Wherefore the pyramid whose base is the fquare ABCD is one half of the pyramid erected upon the square described about the circle. But the pyramid erected upon the square described about the circle, is greater than the cone, for this is contained in that. Therefore the pyramid whose bafe is the square ABCD, and vertex the fame as that of the cone, is greater than one half the cone. Bifect the parts AB, BC, CD, DA of the circumference in the points E, F, G, H, and join AE, EB, BF, FC, CG, GD, DH, HA; then each of the triangles AEB, BFC, CGD, DHA is greater than one half of the respective segment of the circle ABCD: Upon each of the triangles AEB, BFC, CGD, DHA erect pyramids of the fame altitude as that of the cone; then every one of these pyramids thus erected, is greater than one half the respective fegment of the cone: And fo bisecting the remaining parts of the circumference, and joining right lines, and erecting a pyramid upon every one of the triangles having the fame vertex as that of the cone, and doing this continually, there will at laft [by 1. 10.] be left fome portions of the cone that will be less than the excess of the cone above one third part of the cylinder. Let there be such left, which let be those upon AE, E B, BF, FC, CG, GD, DH, HA. Then the remaining pyramid, whose base is the polygon AEBFCGDH, and vertex the fame as that of the cone, is greater than a third part of the cylinder. But [by cor. 7. 12.] the pyramid whose base is the polygon AEBFCGD EBFCGDH, and vertex the fame as that of the cone, is one third part of the prism whose base is the polygon AEBFCGDH, and altitude the fame as that of the cylinder: Therefore the prism whose base is the polygon A EBFCGDH, and altitude the fame as that of the cylinder, is greater than the cylinder whose base is the circle ABCD. But it is also less: For it is comprehended by it; which is impoffible. Therefore the cylinder is not less than thrice the cone. It has alfo been proved not to be greater than thrice the cone: Therefore the cylinder is thrice the cone; and accordingly the cone is one third part of the cylinder.

1

i

;

1

1

Therefore

Therefore every cone is one third part of the cylinder having the fame base as it, and an equal altitude. Which was to be demonstrated.

d Some demonstrate this theorem thus: Every cone may be confidered as a pyramid with a polygonous base, of an infinite number of exceeding small equal fides; and every cylinder as a prism with such a polygonous base. And therefore fince [by cor. prop. 7.] every pyramid is a third part of a prism of the fame base and altitude, every cone, thus taken for a pyramid, will also be one third of a cylinder, of the same altitude, thus taken for a prifm.

PROP. XI. THEOR.

Cones and cylinders having the same altitude, are to one another as their bases.

Let the cones and cylinders whose bases are the circles ABCD, EFGH, axes KL, MN, and diameters of the bases are A C, E G, have the same altitude: I say as the circle ABCD is to the circle EFGH, so is the cone A L, to the cone EN.

For if it be not so, it will be as the circle ABCD is to the circle EFGH, so is the cone AL to some solid either less or greater than the cone EN. First let it be so to the solid x, which is less. Let the folid y be equal to the excess of the cone EN above the folid x. Then the cone

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as the cone, the pyramid upon the inscribed square is one half that upon the circumfcribing square; for (by 6. 12.] they are to one another as their bafes; and the cone is less than the circumfcribed pyramid: Therefore the pyramid whose base is the square EFGH, and vertex the fame as that of the cone, is greater than one half the cone. Bisect the parts EF, FG, GH, HE of the circumference in the points O, P, R, s; and join Ho, OF, EP, TF, FR, RG, GS, SH. Then each of the triangles HOE, EPF, FRG, GSH is greater than one half of the respective segment of the circle. Upon every one of the triangles HOE, EPF, FRG, GSH erect a pyramid of the fame altitude with that of the cone. Then each of these erected pyramids is greater than one half the respective segment of the cone; and fo bisecting the remaining parts of the circumferenice, joining right lines, and upon every one of the triangles erecting pyramids of the fame altitude as that of the cone, and always doing this, there will at laft [by 1. 10.] be left some fegments of the cone that will be less than the solid v. Let such be left; and let them be those that stand on Ho, OE, EP, PF, FR, RG, GS, SH, and the remaining pyra mid, whose base is the polygon HOEPFRGS, and altitude the fame as that of the cone, is greater than the folid x. In the circle ABCD describe the polygon DTAVBQCZ, similar and alike fituate to the polygon HOEPFRGS, and upon it raise a pyramid of the fame altitude as the cone AL. Then because [by 20. 6. and 1. 12.] as the square of A C is to the square of EG, so is the polygon DTA YBQCZ to the polygon HOEPFRGS, and as the square of AC is to the square of GH, so is [by 2. 12.] the circle ABCD to the circle EFGH; [by 1. 5.] as the circle ABCD is to the circle EFGH, so is the polygon DTAVBQCZ to the polygon HOEPFRGS. But [by conftr.] as the circle ABCD is to the circle EFGH, so is the cone AL to the folid x; and [by 6. 12.] as the polygon DTAVBOCZ is to the polygon HOEPFRGS, so is the pyramid whose base is the polygon DTAVBQCZ and vertex the point L, to the pyramid whose bafe is the polygon HOEPFRGS, and vertex the point w: Therefore as the cone AL is to the folid x, so is the pyramid whose base is the polygon DTAVBQCZ, and vertex the point 1, to the pyramid whose base is the polygon

HOEPFRGS

HOEPFRGS, and vertex the point N: Wherefore alternately as the cone AL is to the pyramid which is in it, so is the folid x to the pyramid which is in the cone EN. But the cone AL is greater than the pyramid which is in it: Therefore the folid x is greater than the pyramid which is in the cone EN: But it is less too; which is abfurd. Therefore it is not as the circle ABCD to the circle EFGH, so is the cone AL to some solid less than the cone EN. In like manner we demonstrate that it is not as the circle EFGH is to the circle ABCD, so is the cone En to some solid less than the cone AL: I say also that it is not as the circle ABCD is to the circle EFGH, so is the cone AL to some folid greater than the cone EN. For if possible, let it be so to some solid x, which is greater. Then inversely, as the circle EFGH is to the circle ABCD, so will the solid x be to the cone AL. But as the solid x is to the cone AL, so is the cone EN to some solid less than the cone AL: And therefore as the circle EFGH is to the circle ABCD, so is the cone EN to some solid less than the cone AL; which has been demonstrated to be impossible; therefore it is not as the circle A B C D is to the circle EFGH, so is the cone AL to some solid greater than the cone EN. It has also been proved that it is not so to any folid which is less. Therefore as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN. But as cone is to cone, so is cylinder to cylinder; for [by 10. 12.] the one is thrice the other; and therefore as the circle ABCD is to the circle EFGH, so are the cones and cylinders of the fame altitude standing upon them.

Wherefore cones and cylinders of the fame altitude, are to one another as their bases. Which was to be demonftrated.

e This may be demonstrated shorter, by the method of indivisibles, and from the notes upon the fixth and tenth theorems.

PROP. XII. THEOR. Similar cones and cylinders are to one another in the triplicate ratio of the diameters of their bases.

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Let there be similar cones and cylinders, whose bases are the circles ABCD, EFGH, diameters of the bases BD, FH, and axes of the cones or cylinders KL, MN; I say the cone whose base is the circle A B CD, and vertex the point L, will be to the cone whose base is the circle EFGH, and vertex the point L, in the triplicate ratio of

BD to FH.

For if the cone ABCDL be not to the cone EFGHN in the triplicate ratio of BD to FH, the cone ABCDL will have a triplicate ratio to some folid either less than the cone

EFGH. Again, upon the square EFGH erect a pyramid of the fame altitude as that of

the cone and so this

crected pyramid is greater than one half the cone. Divide the parts EF, FG, GH, HE of the circumference into equal parts in the points O, P, R, s, and join E O, OF, FP, PG, GR, RH, HS, SE: Then each of the triangles É OF, FPG, GRH, OSE is greater than one half of the refpective fegment of the circle EFGH. Again, upon each of the triangles EOF, FPG, GRH, HSE erect a pyramid, having the same vertex as the cone, and each of the erected pyramids is greater than one half of the respective segment of the cone. Therefore bisecting the remaining fegments of the circumference, and joining right lines, and crecting pyramids upon every one of the triangles having the fame vertex as the cone, and always doing this, at last [by