Sidebilder
PDF
ePub

Book XII. But as the solid Z is to the cone ABCDL, so is the cone

EFGHN to some solid, which must be less a than the cone a 14. 5.

ABCDL, because the solid Z is greater than the cone EFGHN: Therefore the cone EFGIN has to a solid which is less than the cone ABCDL, the triplicate ratio of that

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

which EG has to AC, which was demonstrated to be impossible: Therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ratio of that which AC has to EG; and it was demonstrated, that it could not have that ratio to any solid less than the cone EFGHN: Therefore the cone ABCDL has to the cone EFGHN, the

triplicate ratio of that which AC has to EG: But as the cone b 15. 5. is to the cone, so is the cylinder to the cylinder; for every

cone is the third part of the cylinder upon the same base, and of the same altitude: Therefore also the cylinder has to the cylinder, the triplicate ratio of that which AC has to EG: Wherefore, " similar cones," &c. Q. E. D.

Book XII.

PROP. XIII. THEOR.

If a cylinder be cut by a plane parallel to its oppo. See N. site planes, or bases ; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other.

Let the cylinder AD be cut by the plane GH, parallel to the opposite O

P
planes AB, CD, meeting the axis
EF in the point K, and let the line
GH be the common section of the

R plane GH and the surface of the cy

N

S linder AD: Let AEFC be the parallelogram, in any position of it, by the revolution of which about the straight A E B line EF the cylinder AD is described; and let GK be the common section of the plane GH, and the plane AEFC: And because the parallel G К.

H Н planes AB, GH are cut by the plane AEKG, AE, KG, their common C

F D sections with it, are parallel a ; where

a 16. 11. fore AK is a parallelogram, and GK T X equal to EA the straight line from the centre of the circle AB: For the V M Q same reason each of the straight lines drawn from the point K to the line GH may be proved to be equal to those which are drawn from the centre of the circle AB to its circumference, and are therefore all equal to one another. Therefore the line GH is the circumference of a circle b, of which the centre is the point K : Therefore the b 15 def.1. plane GH divides the cylinder AD into the cylinders AH, GD; for they are the same which would be described by the revolution of the parallelograms AK, GF, about the straight lines EK, KF: and it is to be shown, that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF.

Produce the axis EF both ways; and take any number of straight lines EN, NL, each equal to EK; and any number FX, XM, each equal to FK; and let planes parallel to AB, CD pass through the points L, N, X, M : Therefore the common sections of these planes, with the cylinder produced, are circles the centres of which are the points L, N, X, M, as was proved of the plane GH; and these planes cut off the cylinders,

Book XII. PR, RB, DT, TQ: And because the axes LN, NE, EK are all

equal; therefore the cylinders PR, RB, BG are o to one anc 11. 12. other as their bases; but their bases are equal, and therefore

the cylinders PR, RB, BG are equal : And because the axes
LN, NE, EK are equal to one an-
other, as also the cylinders PR, RB, O L P
BG, and there are as many axes as
cylinders; therefore, whatever multi-
ple the axis KL is of the axis KE,

R N S
the same multiple is the cylinder PG
of the cylinder GB: For the same
reason, whatever multiple the axis
MK is of the axis KF, the same mul- A E

B tiple is the cylinder QG of the cylinder GD: And if the axis Kl be equal to the axis KM, the cylinder PG is equal to the cylinder GQ;

G K H Η and if the axis KL be greater than the axis KM, the cylinder PG is C F D greater than the cylinder GQ; and if less, less : Since therefore there T X are four magnitudes, viz. the axes EK, KF, and the cylinders BG, V M Q GD, and of the axis EK and cylinder BG there have been taken any equimultiples whatever, viz. the axis KL and cylinder PG; and of the axis KF and cylinder GD, any equimultiples whatever, viz. the axis KM and cylinder GQ: and it has been demonstrated, if the axis KL be greater than the axis KM, the cylinder PG is greater

than the cylinder GQ; and if equal, equal; and if less, less : d 5. def. 5. Therefored the axis EK is to the axis KF, as the cylinder BG

to the cylinder GD. Wherefore, if a cylinder,” &c. Q. E. D.

PROP. XIV. THEOR.

Cones and cylinders upon equal bases are to one another as their altitudes.

Let the cylinders EB, FD be upon the equal bases AB, CD: As the cylinder EB to the cylinder FD, so is the axis GH to the axis KL.

Produce the axis KL to the point N, and make LN equal to the axis GH, and let CM be a cylinder of which the base is CD, and axis LN; and because the cylinders EB, CM have

the same altitude, they are to one another as their bases a : Book XII. But their bases are equal, therefore also the cylinders EB, CM are equal. And because

a 11. 12. the cylinder FM is cut by the plane CD parallel to its oppo

F K site planes, as the cylinder CM to the cylinder FD, so ist the

b 13. 12. axis LN to the axis KL. But E

G
C

D

L
the cylinder CM is equal to
the cylinder EB, and the axis
LN to the axis GH: There-
fore as the cylinder EB to the
cylinder FD, so is the axis GH
to the axis KL: And as the A

H
B

M
В

N
cylinder EB to the cylinder
FD, so is c the cone ABG to

c 15. 5. the cone CDK, because the cylinders are tripled the cones: d 10. 12. Therefore also the axis GH is to the axis KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore, “ cones," &c. Q. E. D.

PROP. XV. THEOR.

The bases and altitudes of equal cones and cylin. See N. ders are reciprocally proportional ; and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another.

Let the circles ABCD, EFGH, the diameters of which are AC, EG, be the bases, and KL, MN the axes, as also the altitudes, of equal cones and cylinders; and let ALC, ENG be the cones, and AX, EO the cylinders: The bases and altitudes of the cylinders AX, EO are reciprocally proportional; that is, as the base ABCD to the base EFGH, so is the altitude MN, to the altitude KL.

Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First, Let them be equal ; and the cylinders AX, EO being also equal, and cones and cylinders of the same altitude being to one another as their bases a, a 11. 12. therefore the base ABCD is equal to the base EFGH; and b A. 5. as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. But let the altitudes KL, MN be unequal, and MN the greater of the two, and from MN take

E

Book XII. MP equal to KL, and through the point P cut the cylinder

EO by the plane TYS, parallel to the opposite planes of the
circles EFGH, RO; therefore the common section of the
plane TYS and the
cylinder EO is a cir-

N

R cle, and consequent

0
ly ES is a cylinder
the base of which is
the circle EFGH,

L
X Y

P

S. and altitude MP: And because the cylinder AX is equal

D

H Н to the cylinder EO,

C E

G as AX is to the cy- K

M c 7. 5. linder ES, so c is the cylinder EO to

B the same ES. But a ll. 12. as the cylinder AX to the cylinder ES, so a is the base

ABCD to the base EFGH: for the cylinders AX, ES are

of the same altitude; and as the cylinder EO to the cylind 13. 12. der ES, sod is the altitude MN to the altitude MP, because

the cylinder EO is cut by the plane TYS, parallel to its opposite planes. Therefore as the base ABCD to the base EFGH, so is the altitude MN to the altitude MP: But MP is equal to the altitude KL; wherefore as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; that is, the bases and altitudes of the equal cylinders AX, EO are reciprocally proportional.

But let the bases and altitudes of the cylinders AX, EO, be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL: The cylinder AX is equal to the cylinder EO.

First, Let the base ABCD be equal to the base EFGH:

then because as the base ABCD is to the base EFGH, so is e A. 5. the altitude MN to the altitude KL: MN is equal e to KL, fll. 12. and therefore the cylinder AX is equal' to the cylinder EO.

But let the bases ABCD, EFGH be unequal, and let ABCD be the greater; and because as ABCD is to the base EFGH, so is the altitude MN to the altitude KL; therefore MN is greater e than KL. Then, the same construction being made as before, because as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD is f to the base EFGH, as the cylinder AX to the cylinder ES; and as the altitude MN to the altitude

« ForrigeFortsett »