Let S-ABCDE be a pyramid.

Pass the planes SEB, SEC, through the diagonals EB, EC; the polygonal pyramid S-ABCDE will be divided into several triangular pyramids all having the same altitude SO. But each of these pyramids is measured by multiplying its base ABE, BCE, or CDE, by the third part of its altitude SO (Prop. XVI. Cor.); hence the sum of these triangular pyramids, or the polygonal pyramid S-ABCDE will be measured by the sum of the triangles ABE, BCE, CDE, or the polygon ABCDE, inultiplied by one third of SO; hence every pyramid is measured by a third part of the product of its base by its altitude.

B

Cor. 1. Every pyramid is the third part of the prism which has the same base and the same altitude.

Cor. 2. Two pyramids having the same altitude are to cach other as their bases.

Cor. 3. Two pyramids having equivalent bases are to each other as their altitudes.

Cor. 4. Pyramids are to each other as the products of their · bases by their altitudes.

Scholium. The solidity of any polyedral body may be computed, by dividing the body into pyramids; and this division may be accomplished in various ways. One of the simplest is to make all the planes of division pass through the vertex of one solid angle; in that case, there will be formed as many partial pyramids as the polyedron has faces, minus those faces which form the solid angle whence the planes of division proceed.

PROPOSITION XVIII. THEOREM.

If a pyramid be cut by a plane parallel to its base, the frustum that remains when the small pyramid is taken away, is equivalent to the sum of three pyramids having for their common altitude the altitude of the frustum, and for bases the lower base of the frustum, the upper base, and a mean proportional

between the two bases.

0*

Let S-ABCDE be a pyramid cut by the plane abcde, parallel to its base; let T-FGH be a triangular pyramid having the same altitude and an equivalent base with the pyramid S-ABCDE. The two bases may be regarded as situated in the same plane; in which case, the plane abcd, if produced, will form in the triangular pyramid a section fgh situated at the same distance above the common plane of the bases; and therefore the section fgh will be to the section abcde as the base FGH is to the base ABD (Prop. III.), and since the bases are equivalent, the sections will be so likewise. Hence the pyramids S-abcde, T-fgh are equivalent, for their altitude is the same and their bases are equivalent. The whole pyramids S-ABCDE, T-FGH are equivalent for the same reason; hence the frustums ABD-dab, FGH-hfg are equivalent; hence if the proposition can be proved in the single case of the frustum of a triangular pyramid, it will be true of every other.

h

B

F

F

f

K

h

Let FGH-hfg be the frustum of a triangular pyramid, having parallel bases: through the three points F, g, H, pass the plane FgH; it will cut off from the frustum the triangular pyramid g-FGH. This pyramid has for its base the lower base FGH of the frustum; its altitude likewise is that of the frustum, because the vertex g lies in the plane of the upper base fgh.

GRUN

This pyramid being cut off, there will remain the quadrangular pyramid g-fhHF, whose vertex is g, and base fhHF. Pass the plane fgH through the three points f, g, H; it will divide the quadrangular pyramid into two triangular pyramids g-FfH, g-fhH. The latter has for its base the upper base gfh of the frustum; and for its altitude, the altitude of the frustum, because its vertex H lies in the lower base. Thus we already know two of the three pyramids which compose the frustum.

It remains to examine the third g-FfH. Now, if gK be drawn parallel to fF, and if we conceive a new pyramid K-FƒH, having K for its vertex and FfH for its base, these two pyramids will have the same base FfH; they will also have the same altitude, because their vertices g and K lie in the line gK, parallel to Fƒ, and consequently parallel to the

H

H

But

plane of the base: hence these pyramids are equivalent. the pyramid K-FƒH may be regarded as having its vertex in f, and thus its altitude will be the same as that of the frustum: as to its base FKH, we are now to show that this is a mean proportional between the bases FGH and fgh. Now, the triangles FHK, fgh, have each an equal angle F=ƒ; hence

FHK: fgh: FKx FH: fgx fh (Book IV. Prop. XXIV.); but because of the parallels, FK=fg, hence FHK fgh: FH fh.

We have also,

FHG : FHK :: FG: FK or fg.
But the similar triangles FGH, fgh give
FG: fg FH : fh ;

hence,

FGH: FHK:: FHK : fgh;

or the base FHK is a mean proportional between the two bases FGH, fgh. Hence the frustum of a triangular pyramid is equivalent to three pyramids whose common altitude is that of the frustum and whose bases are the lower base of the frustum, the upper base, and a mean proportional between the two bases.

PROPOSITION XIX. THEOREM.

Similar triangular prisms are to each other as the cubes of their homologous sides.

D

a

租

a

с

B

For, since the prisms are similar, the planes which contain the homologous solid angles B and b, are similar, like placed, and equally inclined to each other (Def. 17.): hence the solid angles B and b, are equal (Book VI. Prop. XXI. Sch.). If these solid angles be applied to each other, the angle cbd will coincide with CBD, the side ba with BA, and the prism cbd-p will take the position Bcd-p. From A draw AH perpendicular to the common base of the prisms: then will the plane BAH be perpendicular to the plane of the com

mon base (Book VI. Prop. XVI.). Through a, in the plane BAH,

P

draw ah perpendicular to BH: then will ah also be perpendicular to the base BDC (Book VI. Prop. XVII.); and AH, ah will be the altitudes of the two prisms.

Now, because of the similar triangles ABH,aBh, and of the similar parallelograms AC, ac, we have

D

H

B

AH ah AB: ab :: BC: bc. But since the bases are similar, we have base BCD base bcd: hence,

a

BC2: bc (Book IV. Prop. XXV.);

base BCD base bcd: AH2: ah2.

Multiplying the antecedents by AH, and the consequents by ah, and we have

base BCD × AH: base bcdx ah :: AH3 ah3.

But the solidity of a prism is equal to the base multiplied by the altitude (Prop. XIV.); hence, the

prism BCD-P: prism bcd-p :: AH3 : ah3 : : BC3 : bc3, or as the cubes of any other of their homologous sides.

Cor. Whatever be the bases of similar prisms, the prisms will be to each other as the cubes of their homologous sides.

For, since the prisms are similar, their bases will be similar polygons (Def. 17.); and these similar polygons may be di vided into an equal number of similar triangles, similarly placed (Book IV. Prop. XXVI.): therefore the two prisms may be divided into an equal number of triangular prisms, having their faces similar and like placed; and therefore, equally inclined (Book VI. Prop. XXE); hence the prisms will be similar. But these triangular prisms will be to each other as the cubes of their homologous sides, which sides being proportional, the sums of the triangular prisms, that is, the polygonal prisms, will be to each other as the cubes of their homologous sides.

PROPOSITION XX. THEOREM.

Two similar pyramids are to each other as the cubes of their homologous sides.

For, since the pyramids are similar, the solid angles at the vertices will be contained by the same number of similar planes, like placed, and equally inclined to each other (Def. 17.). Hence, the solid angles at the vertices may be made to coincide, or the two pyramids may be so placed as to have the solid angle S common.

SO: So and consequently,

B

In that position, the bases ABCDE, abcde, A will be parallel; because, since the homologous faces are similar, the angle Sab is equal to SAB, and Sbc to SBC; hence the plane ABC is parallel to the plane abc (Book VI. Prop. XIII.). This being proved, let SO be the perpendicular drawn from the vertex S to the plane ABC, and o the point where this perpendicular meets the plane abc: from what has already been shown, we shall have

C

SA: Sa: AB: ab (Prop. III.);

ISO: So AB : ab.

But the bases ABCDE, abcde, being similar figures, we have ABCDE: abcde :: AB2: ab2 (Book IV. Prop. XXVII.). Multiply the corresponding terms of these two proportions; there results the proportion,

ABCDESO : abcde × So: : AB3 : ab3.

Now ABCDESO is the solidity of the pyramid S-ABCDE, and abcde So is that of the pyramid S-abcde (Prop. XVII.) ; hence two similar pyramids are to each other as the cubes of their homologous sides.

1

General Scholium.

The chief propositions of this Book relating to the solidity of polyedrons, may be exhibited in algebraical terms, and so recapitulated in the briefest manner possible.

Let B represent the base of a prism; H its altitude: the solidity of the prism will be B× H, or BH.

Let B represent the base of a pyramid; H its altitude: the solidity of the pyramid will be BxH, or H×B, or BH.

Let H represent the altitude of the frustum of a pyramid, having parallel bases A and B; AB will be the mean proportional between those bases; and the solidity of the frustum will be Hx (A+B+ √ AB).

In fine, let P and p represent the solidities of two similar prisms or pyramids; A and a, two homologous edges: then we shall have

Pp: A3: a3.