have equivalent corresponding ones in the interior prisms of the pyramid s-abc ; therefore, the prism ABCD is the difference between the sum of 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 greater than the difference between the two pyramids, which latter difference we suppose to be ABCx; hence, the prism ABCD, must be greater than the prism ABCx; but in reality it is less ; for they have the same base, ABC, and the altitude, k, of the first, is less than the altitude, A.x, of the second. Hence, the supposed inequality between the two pyramids cannot exist. Hence, the two pyramids, S-ABC, s-abc, having equivalent bases and equal altitudes, are equivalent. PROPOSITION XXIV. THEOREM. E D Every triangular pyramid is the third part of the triangular prism having the same base and the same altitude. Let S-ABC be a triangular pyramid, ABCDES a triangular prism, having the same base and the same altitude; then will the pyramid be equal to a third of the prism. Cut from the prism the pyramid S-ABC by the plane SAC; there will remain the solid S-ACDE, which may be considered as a quadrangular pyramid whose vertex is S, and whose base is the parallelogram ACDE: draw the diagonal CE, and pass the plane SCE, which will divide the quadrangular pyramid into two triangular pyramids, S-ACE, S-DCE. These two triangular pyramids have for their common altitude the perpendicular let fall from S on the plane ACDE; they have equal bases, the triangles ACE, DCE being halves of the same parallelogram ; hence, the two pyramids, S-ACE, S-DCE, are equivalent (Prop. XXIII.). But, the pyramid S-DCE, and the pyramid S-ABC, have equal bases, ABC, DES; they have also the same altitude, for this altitude is the distance of the parallel planes, ABC, DES. Hence, the two pyramids, S-ABC, А B S-DCE, are equivalent. Now, the pyramid S-DCE has already been proved equivalent to the pyramid S-ACE ; hence, the three pyramids, S-ABC, S-DCE, S-ACE, which compose the prism ABCD, are equivalent. Hence, the pyramid S-ABC is the third of the prism ABCD, which has the same base and the same altitude. Cor. The volume of a triangular pyramid is equal to one-third of the product of its base by its altitude. PROPOSITION XXV. THEOREM. Every pyramid, S-ABCDE, is measured by the third part of the product of its base, ABCDE, by its altitude, SO. For, by passing the planes SEB, SEC through the diagonals EB, EC, the polygonal pyramid S-ABCDE will be divided into several triangular pyramids, having the same altitude, SO. But, by the preceding theorem, each of these pyramids is measured by multiplying its base, ABE, BCE, or CDE, by the third part of its altitude, SO; 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, multiplied by one-third of SO; hence, every pyramid is measured by the third of the product of its base by its altitude. E B COR. 1. Every pyramid is the third part of the prism having the same base and the same altitude. Cor. 2. First.—Any two pyramids are to each other as the products of their bases by their altiludes. Secondly.—Two pyramids having the same altitude are to each other as their bases. Thirdly.-Two pyramids having equivalent bases are to each other as their altitudes. SCHOLIUM I. In order to find the volume of a polyedron, we divide it into pyramids, compute the volumes of these pyramids, and add together the numbers thus obtained. In order to divide the polyedron into pyramids, we can assume any point whatever in space and join it to all the vertices of the polyedron. The bases of the different pyramids thus formed will be the faces of the polyedron, and the altitudes of the pyramids the perpendiculars let fall from the point taken upon the planes of these faces. The volume of the polyedron will be the arithmetical or the algebraic sum of the volumes of the pyramids, according as their common vertex (the point taken) is within or without the polyedron. Sometimes the division of the polyedron into pyramids is effected by taking one of the vertices, and from that drawing diagonals to all the other vertices not adjacent to this. SCHOLIUM 2. When a point can be found in the interior of a polyedron, equidistant from all its faces, the pyramids composing the polyedron, which have this point for a common vertex, will have as a common altitude the perpendicular let fall from this point on any one of the faces, and the volume of the polyedron will have for its measure the third of the product of its surface by this perpendicular. As any polygonal pyramid may be divided into triangular pyramids (tetraedrons), it is evident that any polyedron may be divided also into tetraedrons. PROPOSITION XXVI. THEOREM. T If a polygonal pyramid, S-ABCDE, and a triangular pyramid, T-FGH, having equivalent bases lying in the same plane and the same altitude, be cut by a plane, abd, parallel to the plane of the bases, the frustums, ABD-abd, FGH-fgh, thus cut off, will be equivalent. For the plane abd, produced, forms in the triangular pyramid a section, fgh, situated at the same height above the common plane of the bases; and therefore, since the base ABCDE is equivalent to FGH, the section abcde will be equivalent to the section fgh (Prop. XXII., Cor.). Hence, the pyramids S-abcde and 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-abd, FGH-fgh, which remain after taking the small pyramids from the wholes respectively, are equivalent. a F E H A B G PROPOSITION XXVII. THEOREM. D F E C B Every frustum of a pyramid is equal to the sum of three pyramids, having for their common altilude 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. From the preceding theorem it follows that if the proposition can be proved in the single case of the frustum of a triangular pyramid, it will be true of any other frustum. Let ABC-DEF be the frustum of a triangular pyramid. The planes AEC, DEC divide it into three triangular pyramids, E-ABC, E-DCF, E-DCA. The first, E-ABC, has for its base the lower base, ABC, of the frustum ; its altitude is likewise that of the frustum, since its vertex E lies in the plane of the upper base, EDF. If we take the point C for its vertex, the second pyramid, E-DCF, has for its base DEF, the upper base of the frustum ; its altitude is also the altitude of the frustum, since its vertex, C, lies in the lower base, ABC. Thus we know two of the pyramids which compose the frustum. It remains to consider the third pyramid, E-DCA. To measure this, we compare it with the second, E-DCF. These two pyramids having the same altitude (considered with reference to the common vertex, E), are to each other as their bases, CDA, CDF. But CDA : CDF : : AC : DF, since the triangles have the same altitude. Hence E-DCA : E-DCF : : AC : DF. But since the bases of the frustum, ABC and DEF, are similar, AC : DF :: ABC : VDEF (Book III., Prop. XXVII.). Therefore E-DCA : E-DCF :: VABC : VDEF, : or E-DCA VABC = E-DCF ✓ DEF Now E-DCF – DEF X of the altitude of frustum. 3 Hence I . Х E-DCA = VABC . DEF. X of the altitude of the frustum. 3 Therefore, the third pyramid, E-DCA, is equivalent to a pyramid having for its base a mean proportional between the two bases of the frustum, and for its altitude the altitude of the frustum. Hence, the frustum of a pyramid is equivalent to three pyramids whose common altitude is the altitude of the frustum, and whose bases are respectively the lower and upper bases of the frustum, and a mean proportional between these two bases. SCHOLIUM. Let V be the volume of a frustum of a pyramid, B and B' its bases, and H its altitude. The volumes of the three pyramids being H H 3 Вх we have In two similar tetraedrons, S-ABC, T-DEF, the homologous faces are similar, and the homologous triedrals are equal. |