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TZW, UZX, VOY be made parallel to the base FGH. The section NQL is equal to the section WZT (12. 3. Sup.); as also ORI to XEU, and PSM to YoV; and therefore, also the prisms that stand upon the equal sections are equal (1. Cor. 8. 3. Sup ), that is, the prism which stands on the base BCD, and which is between the planes BCD and NQL is equal to the prism which stands on the base FGH, and which is between the planes FGH and WZT ; and so of the rest, because they have the same altitude : wherefore, the sum of all the prisms described about the pyramid ABCD is equal to the sum of all those described about the pyramid EFGH. But the excess of the prisms described about the pyramid ABCD above the pyramid ABCD is less than Z (13. 3. Sup.); and therefore, the excess of the prisms described about the pyramid EFGH above the pyramid ABCD is also less than Z. But the excess of the pyramid EFGH above the pyramid ADCD is equal to Z, by hypothesis ; therefore, the pyramid EFGH exceeds the pyramid ABCD, more than the prisms described about EFGH exceed the same pyramid ABCD. The pyramid EFGH is therefore greater than the sum of the prisms described about it, which is impossible. The pyramids ABCD, EFGH, therefore, are not unequal, that is, they are equal to one another. Therefore, pyramids, &c. Q. E. D.

PROP. XV. THEOR.

Every prism having a triangular base may be divided into three pyra:

mids that have triangular bases, and that are equal to another.

Let there be a prism of which the base is the triangle ABC, and let DEF be the triangle opposite the base : The prism ABCDEF may be divided into three equal pyramids having triangular bases.

Join AE, EC, CD; and because ABED is a parallelogram, of which AE is the diameter, the triangle ADE is equal (34. 1.) to the triangle ABE : therefore the pyramid of which the base is the triangle ADE, and vertex the

F point C, is equal (14. 3. Sup.) to the pyramid, of which the base is the triangle ABE, and vertex the point C. But the pyramid D. of which the base is the triangle ABE, and vertex the point C, that is, the pyramid ABCE is equal to the pyramid DEFC (14. 3. Sup.), for they have equal bases, viz. the triangles ABC, DEF, and the same altitude, viz. the altitude of the prism ABCDEF. Therefore the three pyramids ADEC, ABEC, DFEC are equal to one another. But the pyramids ADEC, ABEC, DFEC make up the whole prism ABCDEF; therefore, the prism ABCDEF is divided into three equal pyra

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B mids. Wherefore, &c. Q. E. D.

Cor. 1. From this it is manifest, that every pyramid is the third part of a prism which has the same base, and the same altitude with it; for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases.

Cor. 2. Pyramids of equal altitudes are to one another as their bases ; because the prisms upon the same bases, and of the same altitude, are (1. Cor. 38. Sup.) to one another as their bases.

PROP. XVI. THEOR.

If from any point in the circumference of the base of a cylinder, a straight

line be drawn perpendicular to the plane of the base, it will be wholly in the cylindric superficies.

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Let ABCD be a cylinder of which the base is the circle AEB, DFC the circle opposite to the base, and GH the axis ; from E, any point in the circumference AEB, let EF be drawn perpendicular to the plane of the circle AEB ; the straight line EF is in the superficies of the cylinder.

Let F be the point in which EF meets the plane DFC opposite to the base ; join EG

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Н. and FH ; and let AGHD be the rectangle (14. def. 3. Sup.) by the revolution of which

E the cylinder ABCD is described.

Now, because GH is at right angles to GA, the straight line which by its revolution describes the circle AEB, it is at right angles to all the straight lines in the plane of that circle which meet it in G, and it is therefore at right angles to the plane of the circle AEB. But EF is at right angles to the same plane ; therefore, EF and GH are parallel A

B (6. 2. Sup.), and in the same plane. And since the plane through GH and EF cuts the parallel planes AEB, DFC, in the straight lines EG and FH, EG is parallel to FH (14. 2. Sup.). The figure EGHF is therefore a para lelogram,and it has the angle EGH a right angle, therefore it is a rectangle, and is equal to the rectangle AH, because EG is equal to AG. Therefore, when in the revolution of the rectangle AH, the straight line AG coincides with EG, the two rectangles AH and EH will coincide, and the straight line AD will coincide with the straight line EF. But AD is always in the superficies of the cylinder, for it describes that superficies; therefore, EF is also in the superficies of the cylinder. Therefore, &c. Q. E. D.

PROP. XVII. THEOR.

A cylinder and a parallelepiped having equal bases and altitudes, are

equal to one another.

Let ABCD be a cylinder, and EF a parallelepiped having equal bases, viz. the circle AGB and the parallelogram EH, and having also equal altitudes; the cylinder ABCD is equal to the parallelepiped EF.

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If not, let them be unequal ; and first, let the cylinder be less than the parallelepiped EF ; and from the parallelepiped EF let there be cut off a part ÈQ by a plane PQ parallel to NF, equal to the cylinder ABCD. "In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the parallelogram PH (Cor. 1.4. 1. Sup.), and cut off from the parallelogram EH, a part OR equal to the polygon AGKBLM. The point R will fall between P and N. On the polygon AGKBLM let an upright prism AGBCD be constituted of the same altitude with the cylinder, which will therefore be less than the cylinder, because it is within it (16. 3. Sup.); and if through the point R a plane RS parallel to NF be made to pass, it will cut off the parallelepiped ES equal (2. Cor. 8. 3. Sup.) to the prism AGBC, because its base is equal to that of the prism, and its altitude is the same. But the prism AGBC is less than the cylinder ABCD, and the cylinder ABCD is equal to the parallelepiped EQ, by hypothesis ; therefore, ES is less than EQ, and it is also greater, which is impossible. The sylinder ABCD, therefore, is not less than the parallelepiped EF ; and in the same manner, it may be shewn not to be greater than EF. Therefore they are equal. Q. E. D.

PROP. XVIII, THEOR.

If a. cone and a cylinder have the same base and the same altitude, the

cone is the third part of the cylinder.

Let the cone ABCD, and the cylinder BFKG have the same base, viz. the circle BCD, and the same altitude, viz. the perpendicular from the point A upon the plane BCD, the cone ABCD is the third part of the cylinder BFKG.

If not, let the conè ABCD be the third part of another cylinder LMNO, having the same altitude with the cylinder BFKG, but let the bases BCD and LIM be unequal ; and first, let BCD be greater than LIM.

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Then, because the circle BCD is greater than the circle LIM, a polygon may be inscribed in BCD, that shall differ from it less than LIM does (4. 1. Sup.), and which, therefore, will be greater than LIM. Let this be the polygon BECFD; and upon BECFD let there be constituted the pyramid ABECFD, and the prism BCFKHG.

Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG, is greater than the cylinder LMNO, for they have the same altitude, but the prism has the greater base. But the pyramid ABECFD is the third part of the prism (15. 3. Sup.) BCFKHG, therefore it is greater than the third part of the cylinder LMNO. Now, the cone ABECFD is, by hypothesis, the third part of the cylinder LMNO, therefore, the pyramid ABECFD is greater than the cone ABCD, and it is also less, because it is inscribed in the cone, which is impossible. Therefore, the cone ABCD is not less than the third part of the cylinder BFKG: And in the same manner, by circumscribing a polygon about the circle BCD, it may be shewn that the cone ABCD is not greater than the third part of the cylinder BFKG; therefore, it is equal to the third part of that cylinder. Q. E. D.

PROP. XIX. THEOR.

If a hemisphere and a cone have equal bases and altitudes, a series of

cylinders may be inscribed in the hemisphere, and another series may be described about the cone, having all the same altitudes with one ano other, and such that their sum shall differ from the sum of the hemisphere, and the cone, by a solid less than any given solid.

Let ADB be a semicircle, of which the centre is. C, and let CD be at right angles to AB ; let DB and DA be squares described on DC, draw CE, and let the figure thus constructed revolve about DC : then, the sector BCD, which is the half of the semicircle ADB, will describe a hemisphere having C for its centre (7. def. 3. Sup.), and the triangle CDE will describe a cone, having its vertex at C, and having for its base the circle (11. def. 3. Sup.) described by DE, equal to that described by BC, which is the base of the hemisphere. Let W be any given solid. A series of cylinders may be inscribed in the hemisphere ADB, and another described about the cone ECI, so that their sum shall differ from the sum of the hemisphere and the cone, by a solid less than the solid W.

Upon the base of the hemisphere let a cylinder be constituted equal to W, and let its altitude bé cx. Divide CD into such a number of equal parts, that each of them shall be less than CX ; let these be CH, HG, GF, and FD. Through the points F, G, H, draw FN, GO, HP parallel to CB, meeting the circle in the points K, L and M; and the straight line CE in the points Q, R and S. From the points K, L, M draw Kf, Lg, Mh perpendicular to GO, HP and CB; and from Q, R and S, draw Qq, Rr, Ss perpendicular to the same lines.

It is evident, that the figure being thus constucted, if the whole revolve about CD, the rectangles Ff, Gg, Hh will describe cylinders (14. def. 3. Sup.) that will be circumscribed by the hemisphere BDÀ; and that the rectangles DN, F4,

Gr, Hs, will also describe cylinders that will circumscribe the cone ICE. Now, it may be demonstrated, as was done of the prisms inscribed in a pyramid (13. 3. Sup.), that the sum of all the cylinders described within the hemisphere, is exceeded by the hemisphere by a solid less than the cylinder

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