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axis KL and cylinder PG; and of the axis KF and cylinder GD, B. XII. 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: therefore d the axis EK is to d 5. def. the axis KF, as the cylinder BG to the cylinder GD. Where- 5. fore, if a cylinder, &c. R. 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 equat 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: but a 11. 12. their bases are equal, therefore also the cylinders EB, CM are equal. And because the cylinder FM is cut by the plane

F

K CD parallel to its opposite planes, as the cylinder CM to the cylinder FD, so is b the axis LN to the axis KL. But E

C с 1
the cylinder CM is equal to
the cylinder EB, and the axis
LN to the axis GH: therefore
as the cylinder EB to the cylin-
der FD, so is the axis GH to
the axis KL: and as the cylin- А

H
B

N der EB to the cylinder FD, so ise the cone ABG to the cone CDK, because the cylinders are c 15. 5. triple d of the cones: therefore also the axis GH is to the axis d 10. 12. KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore, cones, &c. Q. E. D.

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M M

B. XII.

PROP. XV. THEOR.

See N.

THE bases and altitudes of equal cones and cylinders 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 axis, 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 a 11. 12. of the same altitude being to one another as their bases a, thereb A. 5. fore the base ABCD is equal b to the base EFGH; and as the

base ABCD is to the base EFGH, so is the altitude MN to
the altitude KL.

N
But let the alti-

R
tudes KL, MN
be unequal, and
MN the greater

L
of the two, and

X Y

S
from MN take
MP equal to KL,
and, through the

H
point P, cut the

A

к
cylinder EO by

E
M

G
the plane TYS
parallel to the op-

B

F posite planes of the circles EFGH, RO;' therefore the common section of the plane TYS and the cylinder EO is a circle, and consequently ES is a cylinder, the base of which is the circle EFGH, and altitude MP: and because the cylinder AX

is equal to the cylinder EO, as AX is to the cylinder ES, so c7.5. cis the cylinder EO to the same ES. But 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 d 13. 12. cylinder EO to the cylinder Es, so d 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 B. XII. 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 the altitude MN to the altitude KL; MN is equal b to KL, and therefore b. A. 5. the cylinder AX is equal a to the cylinder EO.

a 11. 12. 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 b 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 a to the base EFGH, as the cylinder AX to the cylinder ES; and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES: therefore the cylinder AX is to the cylinder ES, as the cylinder EO to the same ES; whence the cylinder AX is equal to the cylinder EQ: and the same reasoning holds in cones. Q. E, D.

PROP. XVI. PROB.

TO describe in the greater of two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the lesser circle.

Let ABCD, EFGH be two given circles having the same centre 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 lesser circle.

Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser

B. XII. circle, draw GA at right angles to BD, and produce it to C;

therefore AC touches a the circle EFGH: then, if the circuma 16. 3. ference BAD be bisected, and the half of it be again bisected, b Lemand so on, there must at length remain a circumference less ma than AD; let this be LD; and

from the point L draw LM per-
pendicular to BD, and produce

H
it to N; and join LD, DN. There-
fore LD is equal to DN: and be-

È
cause LN is parallel to AC, and that

K GM
B

D
AC touches the circle EFGH ;
therefore LN does not meet the
circle EFGH: and much less shall

N the straight lines LD, DN meet the circle EFGH: so that if straight lines equal to LD be applied in the circle ABCD from the point L around 10 N, there shall be described in the circle a polygon of an even number of equal sides not meeting the lesser circle. Which was to be done.

LEMMA II.

IF two trapeziums ABCD, EFGH be inscribed in the circles, the centres of which are the points K, L; and if the sides AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG be all equal to one another; but the side AB greater than EF, and DC greater than HG: the straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle.

If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less. First, let KA be equal to LE: therefore, because in two equal circles, AD, BC in the

one are equal to EH, FG in the other, the circumferences a 28. 3. AD, BC are equal a to the circumferences EH, FG; but be

cause the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG: therefore the whole circumference ABCD is greater than the whole EFGH; but it is also equal to it, which is

impossible: therefore the straight line KA is not equal to B. xn. LE.

But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM, which are respectively parallel a to, and less than EF, FG, GH, HE: then because EH a 2. 6. is greater than MP, AD is greater than MP; and the circles

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ABCD, MNOP are equal; therefore the circumference AD is greater than MP; for the same reason, the circumference BC is greater than NO; and because the straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN: therefore the circumference AB is greater than MN; and, for the same reason, the circumference DC is greater than PO: therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewise equal to it, which is impossible : therefore KA is not less than LE; nor is it equal to it; the straight line KA must therefore be greater than LE. Q. E. D.

Cor. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the centre to the circumference of the circle de scribed about the triangle.

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