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Book XII. solid BGML is sextuple the pyramid ABCG, and the solid

EHPO sextuple the pyramid DEFH; therefore the solid a I. Ax. 5. BGML is equal to the solid EHPO: But the bases and alti

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A
B

DE tudes of equal solid parallelepipeds are reciprocally proporb 34. 11. tional; therefore, as the base BM to the base EP, so is the alti

tude of the solid EHPO to the altitude of the solid BGML: c 15. 5. But, as the base BM to the base EP, so is the triangle ABC

to the triangle DEF; therefore, as the triangle A BC to the triangle DEF, so is the altitude of the solid EHPO to the altitude of the solid BGML: But the altitude of the solid EHPO is the same with the altitude of the pyramid DEFH; and the altitude of the solid BGML is the same with the altitude of the pyramid ABCG: Therefore, as the base ABC to the base DEF, so is the altitude of the pyramid DEFH to the altitude of the pyramid ABCG: Wherefore the bases and altitudes of the pyramids ABCG, DEFH are reciprocally proportional.

Again, Let the bases and altitudes of the pyramids A BCG, DEFH be reciprocally proportional, viz. the base ABC to the base DEF, as the altitude of the pyramid DEFH to the altitude of the pyramid ABCG: The pyramid ABCG is equal to the pyramid DEFH.

The same construction being made, because as the base ABC to the base DEF, so is the altitude of the pyramid DEFH to the altitude of the pyramid ABCG. And as the base ABC to the base DEF, so is the parallelogram BM to the parallelogram EP; therefore the parallelogram BM is to EP, as the altitude of the pyramid DEFH to the altitude of the pyramid ABCG: But the altitude of the pyramid DEFH is the same with the altitude of the solid parallelepiped EHPO; and the altitude of the pyramid ABCG is the same with the

altitude of the solid parallelepiped BGML: As, therefore, the Book XII. base BM to the base EP, so is the altitude of the solid parallelepiped EHPO to the altitude of the solid parallelepiped BGML. But solid parallelepipeds having their bases and altitudes reciprocally proportional, are equal d to one another. d 34. 11. Therefore the solid parallelepiped BGML is equal to the solid parallelepiped EHPO. And the pyramid ABCG is the sixth part of the solid BGML, and the pyramid DEFH is the sixth part of the solid EHPO. Therefore the pyramid ABCG is equal to the pyramid DEFH. Therefore, the bases,&c. Q. E. D.

PROP. X. THEOR.

Every cone is the third part of a cylinder, which has the same base, and is of an equal altitude with it.

Let a cone have the same base with a cylinder, viz. the circle ABCD, and the same altitude: The cone is the third part of the cylinder; that is, the cylinder is triple the cone.

If the cylinder be not triple the cone, it must either be greater than the triple or less than it. First, Let it be greater than the triple; and describe the square ABCD in the circle ; this square is greater than the half of the circle ABCD * Upon the square ABCD erect a prism of the same altitude with the cylinder; this prism is greater than half the cylinder; because if a square be described about the circle, and a prism erected upon the square, of the same altitude with the cylinder, the inscribed square is half the circumscribed square ; and upon these square bases are erected solid parallelepipeds, viz. the prisms of the same altitude; therefore the prism upon the square ABCD is the half of the prism upon the square described about the circle: Because they are to one another as their bases a : And the cylinder is less than the prism upon a 32. 11. the square described about the circle ABCD: Therefore the prism upon the square ABCD of the same altitude with the cylinder is greater than half the cylinder. Bisect the circumferences AB, BC, CD, DA in the points E, F, G, H; and join AE, EB, BF, FC, CG, GD, DH, HA: Then, each of the triangles A EB, BFC, CGD, DHA is greater than the half of the segment of the circle in which it stands, as was shown in

As was shown in Prop. 2. of this book.

12.

Book XII. Prop. 2. of this Book. Erect prisms upon each of these 'tri

angles of the same altitude with the cylinder;. each of these
prisms is greater than half the segment of the cylinder in wbich
it is; because if, through the points
E, F, G, H, parallels be drawn to А
AB, BC, CD, DA, and parallelo-
grams be completed upon the same E

H
AB, BC, CD, DA, and solid paral-
lelepipeds be erected upon the pa- B

D rallelograms; the prisms upon the triangles A EB, BFC, CGD, DHA

are the halves of the solid parallele- F b 2. Cor. 7. pipeds b. And the segments of the cylinder which are upon the segments

С of the circle cut off by AB, BC, CD, DA, are less than the solid parallelepipeds which contain them; therefore the prisms upon the triangles A EB, BFC, CGD, DHA, are greater than half the segments of the cylinder in which they are; therefore, if each of the circumferences be divided into two equal parts, and straight lines be drawn from the points of division to the extremities of the circumferences, and upon the triangles thus made, prisms be erected of the same

altitude with the cylinder, and so on, there must at length rec Lemma. main some segments of the cylinder which together are less o

than the excess of the cylinder above the triple of the cone. Let them be those upon the segments of the circle A E, EB, BF, FC, CG, GD, DH, HA. Therefore the rest of the cylinder, that is, the prism of which the base is the polygon AEBFCGDH, and of which the altitude is the same with that

of the cylinder, is greater than the triple of the cone: But this d 1. Cor. 7. prism is triple d of the pyramid upon the same base, of which

the vertex is the same with the vertex of the cone; therefore the pyramid upon the base A EBFCGDH, having the same vertex with the cone, is greater than the cone, of which the base is the circle ABCD: But it is also less, for the pyramid is contained within the cone; which is impossible. Nor can the cylinder be less than the triple of the cone. Let it be less, if possible: Therefore, inversely, the cone is greater than the third part of the cylinder. In the circle ABC describe a

this square is greater than the half of the circle: And upon the square ABCD erect a pyramid, having the same vertex with the cone: this pyramid is greater than the half of the cone; because, as was before demonstrated, if a square be described about the circle, the square ABCD is the half of it; and if upon these squares there be erected solid parallelepi

12.

square; this

peds of the same altitude with the cone, which are also prisms, Book XII. the prism upon the square ABCD will be the half of that which is upon the square described about the circle ; for they are to one another as their bases e; as

H

e 32. 11. are also the third parts of them : A A Therefore the pyramid, the base of which is the square ABCD, is half the pyramid upon the square E

G described about the circle: But this last pyramid is greater than the cone which it contains; there

B fore the pyramid upon the square ABCD, baving the same vertex

F with the cone, is greater than the half of the cone. Bisect the circumferences AB, BC, CD, DA in the points E, F, G, H, and join AE, EB, BF, FC, CG, GD, DH, HA: Therefore each of the triangles, AEB, BFC, CGD, DHA is greater than half the segment of the circle in which it is : Upon each of these triangles erect pyramids having the same vertex with the cone. Therefore each of these pyramids is greater than the half of the segment of the cone in which it is, as before was demonstrated of the prisms and segments of the cylinder; and thus dividing each of the circumferences into two equal parts, and joining the points of division and their extremities by straight lines, and upon the triangles erecting pyramids having their vertices the same with that of the cone, and so on, there must at length remain some segments of the cone, which together shall be less than the excess of the cone, above the third part of the cylinder. .Let these be the segments upon AE, EB, BF, FC, CG, GD, DH, HA. Therefore the rest of the cone, that is, the pyramid, of which the base is the polygon AEBFCGDH, and of which the vertex is the same with that of the cone, is greater than the third part of the cylinder. But

this pyramid is the third part of the prism upon the same base AEBFCGDH, and of the same altitude with the cylinder. Therefore this prism is greater than the cylinder of which the base is the circle ABCD. But it is also less, for it is contained within the cylinder, which is impossible. Therefore the cylinder is not less than the triple of the cone. And it has been demonstrated that neither is it greater than the triple. Therefore the cylinder is triple of the cone, or, the cone is the third part of the cylinder. Wherefore, “every cone,” &c. Q. E, D.

Book XII.

PROP. XI. THEOR.

See N.

Cones and cylinders of the same altitude, are to one another as their bases.

Let the cones and cylinders, of which the bases are the circles ABCD, EFGH, and the axes KL, MN, and AC, EG the diameters of their bases, be of the same altitude. As the circle ABCD to the circle EFGH, so is the cone AL to the cone EN.

If it be not so, let the circle ABCD be to the circle EFGH, as the cone AL to some solid either less than the cone EN, or greater than it. First, Let it be to a solid less than EN, viz. to the solid X; and let Z be the solid which is equal to the excess of the cone EN above the solid X: therefore the cone EN is equal to the solids X, Z together. In the circle EFGH describe the square EFGH, therefore this square is greater than the balf of the circle: Upon the square EFGH erect a pyramid of the same altitude with the cone : this pyramid is greater than half the cone. For, if a square be described about the circle, and a pyramid be erected upon it, having the same vertex with the cone *, the pyramid inscribed in the cone is half the pyramid circumscribed about it, because they are to one another as their bases a : But the cone is less than the circumscribed pyramid; therefore the pyramid of wbich the base is the square EFGH, and its vertex the same with that of the cone, is greater than half the cone : Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points O, P, R, S, and join EO, OF, FP, PG, GR, RH, HS, SE: Therefore each of the triangles EOF, FPG, GRH, HSE, is greater than half the segment of the circle in which it is : Upon each of these triangles erect a pyramid having the same vertex with the cone; each of these pyramids is greater than the half of the segment of the cone in which it is: And thus dividing each of these circumferences into two equal parts, and from the points of division drawing straight lines to the extremities of the circumferences, and upon each of the triangles thus made erecting pyramids, having the same vertex with the cone, and so on, there must at length

a 6. 12.

* Vertex is put in place of altitude which is in the Greek, because the pyra. mid, in what follows, is supposed to be circumscribed about the cone, and so must have the same vertex. And the same change is made in some places following.

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