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PROP. XI. THEOR.

Cones and cylinders of the same altitude, are to one an. See N.

other as their bases.

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Let the cones and cylinders, of which the bases are the circles ABCD, EFGH, and the axes KL, MN, and AC, EG the diameters of their bases be of the same altitude. As the circle ABCD to the circle EFGH, so shall the cone AL be to the cone EN.

If it be not so, the circle ABCD must be to the circle EFGH, as the cone AL to some solid either less than the cone EN, or greater than it. First, let it be to a solid less than EN, viz. to the solid X; and let Z be the solid which is equal to the excess of the cone EN above the solid X; therefore the cone EN is equal to the solids X, Z, together. In the circle EFGH inscribe the square EFGH; therefore this square is greater than the half of the circle: upon the square EFGH erect a pyramid of the same altitude with the cone; this pyramid shall be greater than half of the cone: for, if a square be described about the circle, and a pyramid be erected upon it, having the same vertex with the cone f, the pyramid inscribed in the cone is half of the pyramid circumscribed about it, because they are to one another as their bases *: but the cone is less than the circumscribed pyramid; therefore the pyramid of which the base is the square EFGH, and its vertes the same with that of the cone, is greater than

N half of the cone.

Divide the circumferences EF, FG, GH, HE, each into two

С ЕК G equal parts in the points 0, P, R, S, and join EO, OF, FP; PG, GR, RH, HS, SE: therefore each of the triangles EOF, FPG, GRH, HSE, is greater than half of the segment of the eircle in which it is : upon

* 6. 12.

H
R

А

X

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Vertex is put in place of altitude, which is in the Greek, because the pyramid, in what follows, is supposed to be circumscribed about the cone, and so must have the same vertex. And the same change is made in some places following.

R

* Lemma.

* 1. 12.

* 2. 12.

* 11.5.

each of these triangles erect a pyramid having the same vertex with the cone; each of these pyramids is greater than the half of the segment of the cone in which it is: and thus dividing each of these circumferences into two equal parts, and from the points of division drawing straight lines to the extremities of the circumferences, and upon each of the triangles thus made erecting pyramids having the same vertex with the cone, and so on, there must at length remain some segments of the cone which are together less * than the solid Z: let these be the segments upon EO, OF, FP, PG, GR, RH, HS, SE: therefore the remainder of the cone, viz. the pyramid of which the base is the polygon EOFPGRHS, and its vertex the same with that of the cone, is greater than the solid X. In the circle ABCD inscribe the polygon ATBYCVDQ similar to the polygon EOFPGRHS, and upon it erect a pyramid having the same vertex with the cone AL: and because as the square of AC is to the square of EG, so* is the polygon ATBYCVDQ to the polygon EOFPGRHS; and as the

square of AC to the square of EG, so is * the
circle ABCD to the circle CFGH; therefore the
circle ABCD is * to the circle EFGH, as the polygon
ATBYCVDQ to the poly-
gon EOFPGRHS: but as

N
the circle ABCD to the
circle EFGH, so is the cone
AL to the solid X; and as the

С в
polygon ATBYCVDQ to

T the polygon EOFPGRHS, so is* the pyramid of which the base is the first of these polygons, and vertex L, to the pyramid of which the base is the other polygon, and its vertex N: therefore, as the cone AL to the solid X, so is the pyramid of which the base is the polygon ATBYCVDQ, and vertex L, to the pyramid the base of which is the polygon EOFPGRHS, and vertex N: but the cone AL is greater than the pyramid contained in it; therefore the solid X is greater * than the pyramid in the cone EN: but it is less, as was shewn; which is absurd : therefore the circle ABCD is not to the circle EFGH, as the cone AL to any solid which is less than the cone EN.

V

H
S

R

Al

* 6, 12.

* 14. 5.

In the same manner it may be demonstrated, that the circle EFGH is not to the circle ABCD, as the cone EN to any solid less than the cone AL. Nor can the circle ABCD be to the circle EFGH, as the cone AL, to any solid greater than the cone EN. For, if it be possible, let it be so to the solid I, which is greater than the cone EN: therefore, by inversion, as the circle EFGH to the circle ABCD, so is the solid I to the cone AL: but as the solid I to the cone AL, so is the cone EN to some solid, which must be less * than the cone AL, be- * 14. 5. cause the solid I is greater than the cone EN; therefore, as the circle EFGH is to the circle ABCD, so is the cone EN to a solid less than the cone AL, which was shewn to be impossible: therefore the circle ABCD is not to the circle EFGH, as the cone AL is to any solid greater than the cone EN. And it has been demonstrated, that neither is the circle ABCD to the circle EFGH, as the cone AL to any solid less than the cone EN: therefore the circle ABCD is to the circle EFGH, as the cone AL to the cone EN: but as the cone is to the cone, so * is the cylinder to the * 15. 5. cylinder, because the cylinders are triple* of the cones, . 10. 12. . each of each: therefore as the circle ABCD to the circle EFGH, so are the cylinders upon them of the same altitude. Wherefore, cones and cylinders of the same altitude are to one another as their bases. Q. E. D.

PROP. XII. THEOR.

Similar cones and cylinders have to one another the tripli- See N.

cate ratio of that which the diameters of their bases have.

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Let the cones and cylinders of which the bases are the circles ABCD, EFGH, and the diameters of the bases AC, EG, and KL, MN the axes of the cones or cylinders, be similar: the cone of which the base is the circle ABCD and vertex the point L, shall have to the cone of which the base is the circle EFGH and vertex N, the triplicate ratio of that which AC has to EG.

For if the cone ABCDL has not to the cone EFGHN the triplicate ratio of that which AC has to EG, the cone ABCDL must have the triplicate of that ratio to some solid which is less or greater than the cone EFGHN. First, let it have it to a less, viz, to the solid X. Make

the same construction as in the preceding proposition, and it may be demonstrated the

very same way as in that proposition, that the pyramid of which the base is the polygon EOFPGRHS, and vertex N, is greater than the solid X. Inscribe also in the circle ABCD the polygon ATBYCVDQ similar to the polygon EOFPGRHS, upon which erect a pyramid having the same vertex with the cone; and let LAQ be one of the triangles containing the pyramid upon the polygon ATBÝCVDQ, the vertex of which is L; and let NES be one of the triangles containing the pyramid upon the polygon

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EOFPGRHS of which the vertex is N; and join KQ, MS. Then, because the cone ABCDL is similar to the cone EFGHN, AC is * to EG as the axis KL to the axis MN; and as AC to EG, so* is AK to EM; therefore as AK to EM, so is KL to MN; and alternately, AK to KL, as EM to MN: and the right angles AKL, EMN, are equal: therefore, the sides about these equal angles' being proportionals, the triangle AKL is similar* to the triangle EMN. Again, because AK is to KQ, as EM to MS, and that these sides are about equal angles AKQ, EMS, because these angles are, each of them, the same part of four right angles at the centres K, M; therefore the triangle AKQ is similar* to the triangle EMS. And because it has been shewn that as AK to KL, so is EM to MN, and that AK is equal to KQ, and EM to MS: therefore as QK to KL, so is SM to MN: and therefore, the sides about the right angles QKL, SMN being proportionals, the triangle LKQ is similar to the triangle NMS. And because of the similarity of the triangles AKL, EMN, as LA is to AK, so is NE to EM; and by the similarity of the triangles AKQ, EMS, as KA to AQ, so ME to ES: therefore ex æquali is to AQ, as NE to ES. Again, because of the simi

* 24 Def. 11.

* 15.5.

* 6. 6.

* 6. 6.

* 22.5.

* LA

* 22.5.

larity of the triangles LQK, NSM, as L'Q to QK, so NS to SM; and from the similarity of the triangles KAQ, MES, as KQ to QA, so MS to SE: therefore ex æquali*, LQ is to QA, as NS to SE: and it was proved that QA is to AL, as SE to EN: therefore, again, ex æquali, as QL to LA, so is SN to NE: wherefore the triangles LQA, NSE, having the sides about all their angles proportionals

, are equiangular * *5.6. and similar to one another : and therefore the pyramid of which the base is the triangle AKQ, and vertex L, is similar to the pyramid the base of which is the triangle EMS, and vertex N, because their solid angles are equal* to one another, and they are contained by * B. 11. the same number of similar planes : but similar pyramids which have triangular bases have to one another the triplicate* ratio of that which their homologous * 8. 12. sides have; therefore the pyramid AKQL has to the pyramid EMSN the triplicate ratio of that which AK has to EM. In the same manner, if straight lines be drawn from the points D, V, C, Y, B, T, to K, and from the points H, R, G, P, F, O, to M, and pyramids be erected upon the triangles having the same vertices with the cones, it may be demonstrated that each pyramid in the first cone has to each in the other, taking them in the same order, the triplicate ratio of that which the side AK has to the side EN; that is, which AC has to EG: but as one antecedent to its consequent, so are all the antecedents to all the consequents*; therefore as the pyramid AKQL to the py- * 12.5. ramid EMSN, so is the whole pyramid the base of which is the polygon DQATBYCV, and vertex L, to the whole pyramid of which the base is the polygon HSEOFPGR, and vertex N: wherefore also the first of these two last-named pyramids has to the other the triplicate ratio of that which AC has to EG; but, by the hypothesis, the cone of which the base is the circle ABCD, and vertex L, has to the solid X, the triplicate ratio of that which AC has to EG; therefore, as the cone of which the base is the circle ABCD, and vertex L, is to the solid X, so is the pyramid the base of which is the polygon DQATBYCV, and vertex L, to the pyramid the base of which is the polygon HSEOFPGR, and vertex N: but the said cone is greater than the pyramid contained in it; therefore the solid X is greater* than the pyramid, the base of which is the * 14. 5.

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