continue the process till this difference is as small as we please; but this difference will always be greater than the difference of the two pyramids, which must consequently be infinitely small, that is, nothing. We say, therefore--Two triangular pyramids which have the same height and equivalent bases, are equal in volume.

252. As every pyramid of more than three lateral faces, may be cut into triangular pyramids of the same height, by planes passing through the summit and the diagonals of the base, this law of triangular pyramids will be true of all pyramids composed of triangular pyramids. We say, therefore, that-Pyramids of the same height and equivalent bases, are equal in volume.

253. We have seen that polyedral bodies may be cut into tetraedrons (or triangular pyramids) by planes passing through them diagonally (237). Suppose we have Fig.136. the triangular prism ABC-DEF (fig. 136). First, pass a plane through the vertex E and the edge AC of the base; it will cut off the tetraedron E-ABC, whose base and altitude are the same as those of the prism. Suppose the remainder ACDEF, to be cut through the vertices C, D, E, it will be divided into two tetraedrons, one of which, CDEF, may be considered as having C for its summit, and DEF, one of the bases of the prism, for its base; this tetraedron has also its base and height equal to those of the prisın. The remaining seg ment EДCD, may be considered as a tetraedron whose summit is at E, and whose base is ACD; but as the edge BE is parallel to the opposite face of the prism, the point B is at the same distance from the plane of the opposite face, as the point F is; we may therefore substitute for the tetraedron EACD, the equivalent tetraedron BACD, having the same base ACD, and the same height. But in the tetraedron BACD, we may take ABC for the base, and the point D for the summit; this tetraedron has also the same base and the same height with the prism. We have, therefore, three equivalent tetraedrons composing this prism. We, therefore say—A tetraedron is equivalent to one third of a triangular prism of the same base and height.

254. By reviewing the analysis of the last article, it will be perceived that the body ABC-DEF is equivalent to three tetraedrons which have ABC for a common base, and their vertices respectively in the three points D, E,

and F. If, therefore, the body were a truncated prism, that is, if the plane DEF were inclined to the base, we should say--A truncated triangular prism is equivalent to three pyramids whose common base is the base of the prism, and whose heights are respectively equal to the distances of the opposite vertices from the base.

255. As all pyramids of more than three lateral faces may be cut into tetraedrons of the same height, the sum of whose bases will be the base of the pyramid; and as the prisms constructed upon the bases of these tetraedrons taken together, will constitute a prism of the same height, constructed upon the base of the pyramid in question; and as each of the tetraedrons is one third of the corresponding triangular prism, the sum of all the tetraedrons will be one third of the sum of all the triangular prisms. We therefore say (249)—Any pyramid is one third of a prism of the same base and height, or one third of a rectangular parallelopiped of the same height and an equivalent base.

256. Let us now compare together bodies of different heights. Suppose the two rectangular parallelopipeds AG and AL (fig. 137) to have the same base ABCD; Fig.137. and AE and AI, for their heights. Supppose these heights to be to each other in any definite ratio, for instance in the ratio of 9 to 5. Divide AE into 9 equal parts, AI will contain 5 of them; through these points of division pass planes parallel to the base; they will divide the parallelopiped AG into 9 rectangular parallelopipeds of equal bases and equal heights; and as the parallelopiped AL contains 5 of these, the two parallelopipeds will be to each other as 9 to 5; that is, as AE to AI, We should have the same result if their heights were in any other commensurable ratio; the bodies would be in the ratio of their heights.

Suppose their heights to be incommensurable; we may divide one of the lines into equal parts so small that the difference between one of these parts and the remainder of the division of the other line by such part, is less than any given difference; we thus find an approximate ratio of the two bodies, which is indefinitely near to the ratio of the two lines; but which ratio we never pass, on which side soever we commence the approximation; we therefore infer that the ratio of the heights is, in this case also, the ratio of the two bodies. We therefore say-Two

rectangular parallelopipeds of the same base, are to each other in their ratio of the heights.

257. Let us now compare the two rectangular paralFig.138. lelopipeds AG and IP (fig. 138) of different bases and different heights. Upon the edge IN take II' = AE, and through this point suppose a plane section I'L' parallel to the base; and upon BC take BC'—IM, and through the point C' suppose the plane section C'H' parallel to the face AF. We shall then have the parallelopipeds AG and IL' whose bases are the equal rectangles AH' and IM', and whose heights are the edges AB and

IK; we shall therefore have (256)

vol. AG'

vol. IL/

=

AB

IK

and by comparing the parallelopipeds AG and AG', considered as having for their common base the rectangle vol. AG AD

AD

;

AF, we obtain
By multiplying these
vol. AG'
two proportions in order, omitting the factor AG' com-
mon to the numerator and denominator, and substituting
for AD' its equal IM, we make a comparison of the par-
vol. AG ABX AD
allelopeds AG and IL'; this gives vol. ILIK X IM
Comparing now the parallelopipeds IL and IP, hav-
ing the same base IKLM, we obtain the proportion
vol. IL
vol. IPIN

II'

Multiplying together these two equa

tions, omitting the common factor II' in the two terms of the first ratio, and substituting in the second AE for vol. AG obtain the following: vol. IP

its equal II', we

AB X AD X AE

IK X IM X IN

We therefore say-Any two rectangular parallelopipeds are to each other as the products of the three edges which meet at one of the vertices; that is, as the products of their three dimensions.

258. To illustrate this result, we take as the unit of the measure of volume, the cube whose side or edge is a unit in the measure of lines. Let a g be this cube (fig. Fig.139. 139). If we compare this small cube whose side is 1, with the parallelopiped AG, we shall have

vol. AG
vol, a b

ABX AD × AE
1

; that is, the par

allelopiped AG contains the cube whose side is a b as Fig.139 many times as the product of the lines AB, AD, AE, referred to the common measure ab, contains unity. And this is what is meant, when it is said that— The measure of volume in a rectangular parallelopiped, is the product of its three contiguous edges.

259. We observe that the product AB × AD expresses the number of squares, whose side is unity [a b], contained in the base AC; that is, expresses the area of the base; we say therefore, that-The volume of a rectangular parallelopiped has for its measure the product of its base by its height.

260. As any parallelopiped may be changed into a rectangular parallelopiped of the same height and an equivalent base (248), it follows that-Any parallelopiped has for the measure of its volume, the product of its base by its height.

261. As a rectangular parallelopiped may be divided into two equivalent triangular prisms of the same height with the parallelopiped, whose bases are each half the base of the parallelopiped (249); it follows, that— The volume of a triangular prism has for its measure the product of its base by its height.

262. As every prism of more than three lateral faces may be considered as composed of triangular prisms of the same height with itself, the sum of whose bases make the base of the prism (250), we say-The volume of every prism has for its measure the product of its base by its height. And consequently (255)—The volume of any pyramid has for its measure, the product of one third of the base by the height.

Remark. To obtain the volume of any frustum of a pyramid, find the volume of the entire pyramid, and subtract from it the volume of the partial pyramid cut off.

263. Let P represent any prism, B its base and H its height; we shall have the formula P = B × H; let P' denote a similar prism, B' its base, and H' its height; this gives P'B' × H'. If we compare these, we P B × H But in similar polyedrons, hoP B' × H' mologous faces are proportional to the squares of their homologous measures (237); the bases are then as the

have

H2
H2

P

H2 × H

this will give us

P

H2 × H

H3

; that is-Similar prisms are to each other as the H3 third powers of their heights, or as the cubes of their homologous measures generally.

264. Let p represent a pyramid, b its base and h its height; p' another similar pyramid, b' its base, and h' its height; for the first of these pyramids, we have the expression p b xh ; and for the second, p' By comparing these, we obtain the proporbx h

Ρ

3

; and by a process of reasoning sim

tion

p'

ilar to the above, we obtain

pyramids are to each other as the cubes of their heights; or, as the third powers of their homologous dimensions generally.

265. As all polyedral bodies may be considered as composed of pyramids; we have from the above rule, a method of ascertaining the measure of their volume. And as similar polyedral bodies are composed of the same number of similar tetraedrons or triangular pyramids, we infer that-Similar polyedrons are to each other as the cubes described upon their homologous meas

ures.