Sidebilder
PDF
ePub

therefore also the cylinders EB, CM are equal. And because the cylinder FM is cut by the plane CD parallel to its opposite planes, as the

-F

K cylinder CM to the cylinder FD, SO is (13. 12.) the axis LN to the axis KL. But the cylinder CM is equal to the cylinder EB, and the axis LN E

С

D to the axis GH: therefore as the cylinder EB to the cylinder FD, so is the axis GH to the axis KL: and as the cylinder EB to the cylinder FD, so is (15. 5.) the cone ABG to the cone CDK, because the cylinders

M are triple (10. 12.) of the cones: A

B 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.

[ocr errors]

H

PROP. XV. THEOR.

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 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, (11. 12.) therefore the base ABCD

R

N is equal (A. 5.) to the base

0 EFGH; and as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL.

I X But let the altitudes KL, MN

Y be unequal, and MN the greater of the two, and from MN take MP, equal to KL, and, through

H the point P, cut the cylinder EO A

NC by the plane TYS, parallel to the к

E

IG

M opposite planes of the circles EFGH, RO; therefore the common section of the plane TYS and the cylinder EO is a circle, and

[ocr errors]

B

F

* See Note.

consequently ES is a cylinder, the base of which is the circle EFHG, and altitude MP: and because the cylinder AX is equal to the cylinder EO, as AX is to the cylinder ES, so (7. 5.) is the cylinder EO to the same ES. But as the cylinder AX to the cylinder ES, so (11. 12.) 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, so (13. 12.) is the altitude MF 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 the altitude MN to the altitude KL; MN is equal (A. 5.) to KL, and therefore the cylinder AX is equal (11. 12.) 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 (A. 5.) 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 (11. 12.) 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 is to the same ES; whence the cylinder AX is equal to the cylinder EO; and the same reasoning holds in cones. Q. E. D.

PROP. XVI. PROB.

To describe in the greater of the 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 circle, draw GA at right angles to BD, and produce it to C; therefore AC touches (16. 3.) the circle EFGH: then, if the circumference BAD be bisected, and the half of it be again bisected, and so on, there must at length remain a circumference less (Lemma.) than AD: let this be

LD; and from the point L draw LM per

A pendicular to BD, and produce it to N;

H

I. and join LD, DN. Therefore LD is equal to DN ; and because LN is paral

E K GM lel to AC, and that AC touches the circle B!

D EFGH, therefore LN does not meet the circle EFGH: and much less shall the straight lines LD, DN meet the circle

N EFGH, so that if straight lines equal to LD be applied in the circle ABCD from the point L around to 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.

Ir 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 EC greater than HG; the straight line KĂ 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, DC in the one, are equal to EH, FG in the other, the circumferences AD, BC are equal (28. 3.)

to the circumferences EH, FG; but because 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 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 (2. 6.) to and less than EF, FG, GH, HE: then because EH is greater than MP, AD is greater than MP; and the circles 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.

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

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 described about the triangle.

PROP. XVII. PROB.

To describe in the greater of two spheres which have the some centre, a solid polyhedron, the superficies of which shall not meet the lesser sphere.*

Let there be two spheres about the same centre A; it is required to describe in the greater a solid polyhedron, the superficies of which shall not meet the lesser sphere.

Let the spheres be cut by a plane passing through the centre; the common sections of it with the spheres shall be circles: because the sphere is described by the revolution of a semicircle about the diameter remaining unmoveable: so that in whatever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle ;. and this is a great circle of the sphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater (15. 3.) than any straight line in the circle or sphere: let then the circle made by the section of the plane with the greater sphere be BCDE, and with the lesser sphere be FGH; and draw the two diameters BD, CE at right angles to one another; and in BCDE, the greater of the two circles, describe (16. 12.) a polygon of an even number of equal sides, not meeting the lesser circle FGH; and let its sides, in BE, the fourth part of the circle, be BK, KL, LM, ME; join KA and produce it to N; and from A draw AX at right angles to the plane of the circle BCDE, meeting the superficies of the sphere in the point

* See Note.

X; and let the planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall produce great circles on the superficies of the sphere; and let BXD, KXN be the semicircles thus made upon the diameters BD, KN: therefore, because XA is at right angles to the plane of the circle BCDE, every plane which passes through XA is at right (18. 11.) angles to the plane of the circle BCDE; wherefore the semicircles BXD, KXN, are at right angles to that plane; and because the semicircles BED, BXD, KXN, upon the equal diameters BD, KN are equal to one another, their halves BE, BX, KX, are equal to one another; therefore, as many sides of the polygon as are in BE, so many there are in BX, KX equal to the sides BK, KL, LM, ME: let these polygons be described, and their sides be BO, OP, PR, RX; KS, ST, TY, YX, and join OS, PT, RY; and from the points O, S, draw OV, SQ perpendiculars to AB, AK: and because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, OV is drawn perpendicular to AB the common section of the planes, therefore OV is perpendicular (4. def. 11.) to the plane BCDE: for the same reason SQ is perpendicular to the same plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ; and because in the equal semicircles BXD, KXN the circumferences BO, KS are equal, and OV,

X

R

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

SQ are perpendicular to their diameters, therefore (26. 1.) OV is equal to SQ, and BV equal to KQ: but the whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder

« ForrigeFortsett »