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the cone EFGHN to some solid, which must be less i than the core Book XIE ABCDL, because the folid Z is greater than the cone EFGHN. m therefore the cone EFGHN has to a solid which is less than the cone i. 14. 5. ABCDL, the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible. therefore the cone ABCDL has not to any folid 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 folid 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 is to the cone, fok k. 15. 5. the cylinder to the cylinder, for every cone is the third part of the cylinder upon the same base, and of the fame 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.

PROP. XIII. THEOR. Fa cylinder be cut by a plane parallel to its opposité See N.

planes, or bascs; 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.

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Let the cylinder AD be cut by the plane GH parallel to the opposite planes AB,CD, meeting the axis EF in the point K, and let the line GH be the common section of the plane GH and the surface

R of the cylinder AD. let' A'EFC be the parallelogram, in any position of it, by the revolution of which about the straight A line EF the cylinder AD is described ; and let GK be the common fection of the plane GH, and the plane AEFC. and because the parallel planes AB, GH are cut by the plane AEKG, AE, KG, their common sections with it, are parallel *; wherefore AK is a parallelogram, T and GK equal to E A the straight line from the center of the circle A B. for V the same reason, each of the straight lines

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P.

Book XII. drawn from the point to the line GH may be proved to be Wequal to those which are drawn from the center of the circle

AB to its circumference, and are therefore all equal to one anob.15.Def.s. ther, therefore the line GH is the circumference of a circle 5 of

which the center is the point K. therefore the 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 Mewn that the cylinder AH is to the cylinder HC, as the axis EK to the axis Kr.

Produce the axis EF both ways ; and take any number of straight lines EN, NL, each equal to EK; and any oumber FX, XM, each equal to FK ; and let planes parallel to AB, CD pass through the 0 L points L, N, X, M. therefore the common sections of these planes with the cylinder produced are circles the centers

R N S of which are the points L, N, X, M, as was proved of the plane GH; and these planes cut off the cylinders PR,

E B RB, DT, TQ. and because the axes LN, NE, EK are all equal, therefore the cylinders PR, RB, BG are to one another as their bases. but their bases

G

K K are equal, and therefore the cylinders

d F D PR, RB, BG are equal. and because the axes LN, NE, EK are equal to one T X Y another, as also the cylinders PR, RB, BG, and that there are as many axes as V M e cylinders; therefore whatever multiple the axis KL is of the axis KE, the fame 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 multiple 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; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ ; and if less, less. since therefore there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD, and that of the axis EK and cylinder EG there has been taken any equimultiples whatever, viz. the axis KL and cylinder PG; and of the axis

A

C. II. 12.

AA

KF and cylinder GD, any equimultiples whatever, viz. the axis Book XII.
KM and cylinder GQ; and it has been demonstrated if the axis m
KL be greater than the axis KM, the cylinder PG is greater than .
the cylinder GQ; and if equal, equal; and if less, less. therefore d d. 5. Der si
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. TMEOR.

.

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 7

the cylinder EB to the cylinder FD, fo 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 . but their bases 2. ti.it are equal, therefore also the cylinders EB, CM are equal. and be

F K cause the cylinder FM is cut by the plane CD parallel to its oppofite planes, as the cylinder CM to E the cylinder FD, fo is o the axis

b. 13. 725 LN to the axis KL. but the cy1 linder CM is equal to the cylin

der EB, and the axis LN to the axis GH. therefore as the cylin-A

H

B der EB to the cylinder FD, so is

M

N the axis GH to the axis KL. and as the cylinder EB to the cylinder FD, fo is the cone ABG to the cone CDK, because the c.is. cylinders are triple of the cones. therefore also the axis GH is d. 10. 120 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:

IL

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PRO P. XV. THE O R.
THE bases and altitudes of cqual cones and cylin-

ders are reciprocally proportional; and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another.

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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 alti2. 11. 11. tude being to one another as their bases, therefore the base ABCD b. A. S.

is equal to the bale EFGH; and as the base ABCD is to the base
EFGH, fo is the al-
titude MN to the al.

N

R
titude KL. but let the
altitudes KL, MN be
unequal, and MN the

І.

X Y
greater of the two,
and from MN take
MPequalto KL, and,
thro' the point P, cut A

C E
the cylinder EO by K

M the plane TYS paral

E

B lel to the opposite planes of the circles EFGH, RO; therefore the common section of the plane TYS and the cylinder EO is a circle, and confequently ES is a cylinder, the base of which is the circle EFGH,

and altitude MP. and because the cylinder AX is equal to the cho 6. 7. S. linder EO, as AX is to the cylinder ES, fo is the cylinder EO to

the faine ES. but as the cylinder AX to the cylinder ES, fo '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 cylinder ES, d. 13. 12. fod is the altitude MN to the altitude MP, because the cylinder

G

EO is cut by the plane TYS parallel to its opposite planes. there- Book XII. fore 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, fo is the altitude MN to the altitude KL ; MN is equal is to KL, and therefore the b. A. 5. 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 altirude MN to the altitude KL, therefore MN is greater

b than KL; then, the fame construction being made as before, because as the base ABCD to the base EFGH, fo is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD is to the base EFGH, as the cyhinder AX to the cylinder ES; and as the altitude MN to the altitude MP or KL, fo is the cylinder EO to the cylinder ES. therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the same ES. whence the cylinder AX is equal to the cylinder EO. and the same reasoning holds in cones.

2. 11, 12

Q. E. D.

PRO P. XVI. PROB.

0

the same center, a polygon of an even number of equal fides, that shall not meet the lesser circle.

T describe in the greater of two circles that have

Let ABCD, EFGH be two given circles having the same center K. it is required to inscribe in the greater circle ABCD a polygon of an even number of equal sides, that shall not meet the leser circle.

Thro' the center K draw the straight line BD, and from the poiat G, where it meets the circumference of the lefler circle, draw

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