be the smaller and suppose Aa to be the altitude of a prism, which having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, &c. each less than Aa, and let k be one of those parts; through the points of division pass planes parallel to the plane of the bases; the corresponding sections formed by these planes in the two pyramids will be respectively equivalent, namely DEF to def, GHI to ghi, &c. (Prop. III. Cor. 2.). This being granted, upon the triangles ABC, DEF, GHI, &c. taken as bases, construct exterior prisms having for edges the parts AD, DG, GK, &c. of the edge SA; in like manner, on bases def, ghi, klm, &c. in the second pyramid, construct interior prisms, having for edges the corresponding parts of Sa. It is plain that the sum of all the exterior prisms of the pyramid S-ABC will be greater than this pyramid; and also that the sum of all the interior prisms of the pyramid S-abc will be less than this pyramid. Hence the difference, between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves. : Now, beginning with the bases ABC, abc, the second exterior prism DEF-G is equivalent to the first interior prism def-a, because they have the same altitude k, and their bases DEF, def, are equivalent; for like reasons, the third exterior prism GHI-K and the second interior prism ghi-d are equivalent; the fourth exterior and the third interior; and so on, to the last in each series. Hence all the exterior prisms of the pyramid S-ABC, excepting the first prism ABC-D, have equivalent corresponding ones in the interior prisms of the pyramid S-abc hence the prism ABC-D, is the difference between the sum of all the exterior prisms of the pyramid S-ABC, and the sum of the interior prisms of the pyramid S-abc. But the difference between these two sets of prisms has already been proved to be greater than that of the two pyramids; which latter difference we supposed to be equal to the prism a-ABC: hence the prism ABC-D, must be greater than the prism a-ABC. But in reality it is less; for they have the same base ABC, and the altitude Ax of the first is less than Aa the altitude of the second. Hence the supposed inequality between the two pyramids cannot exist; hence the two pyramids S-ABC, S-abc, having equal altitudes and equivalent bases, are themselves equivalent. PROPOSITION XVI. THEOREM. Every triangular pyramid is a third part of the triangular prism having the same base and the same altitude. Let F-ABC be a triangular pyramid, ABC-DEF a triangular prism of the same base and the same altitude; the pyramid will be equal to a third of the prism. B Cut off the pyramid F-ABC from the prism, by the plane FAC; there will remain the solid F-ACDE, which may be considered as a quadrangular pyramid, whose vertex is F, and whose base is the parallelogram ACDE. Draw the diagonal CE; and pass the plane FCE, which will cut the quadrangular pyramid into two triangular ones F-ACE, F-CDE. These two triangular pyramids have for their common altitude the perpendicular let fall from F on the plane ACDE; they have equal bases, the triangles ACE, CDE being halves of the same parallelogram; hence the two pyramids F-ACE, F-CDE, are equivalent (Prop. XV.). But the pyramid F-CDE and the pyramid F-ABC have equal bases ABC, DEF; they have also the same altitude, namely, the distance between the parallel planes ABC, DEF; hence the two pyramids are equivalent. Now the pyramid F-CDE has already been proved equivalent to F-ACE; hence the three pyramids F-ABC, F-CDE, F-ACE, which compose the prism ABC-DEF are all equivalent. Hence the pyramid F-ABC is the third part of the prism ABC-DEF, which has the same base and the same altitude. A E Cor. The solidity of a triangular pyramid is equal to a third part of the product of its base by its altitude. PROPOSITION XVII. THEOREM. The solidity of every pyramid is equal to the base multiplied by a third of the altitude. 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 pyra- A mids, or the polygonal pyramid S-ABCDE will be measured by the sum of the triangles ABE, BCE, CDE, or the polygon ABCDE, multiplied by one third of SO; hence every pyramid is measured by a third part of the product of its base by its altitude. E 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 each 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. 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. 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. B 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. F F C H H 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-FYH, 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 Ff, and consequently parallel to the plane of the base: hence these pyramids are equivalent. But 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=f; hence FHK: fgh:: FK × FH: fg × fh (Book IV. Prop. XXIV.); but because of the parallels, FK=fg, hence FHK fgh FH fh. X We have also, FHG FHK:: FG: FK or fg. 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 therr homologous sides. Let CBD-P, cbd-p, be two similar triangular prisms, of which BC, bc, are homologous sides: then will the prism CBD-P be to the prism cbd-p, as BC3 to bc33. P D a C 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 |