by their common altitude.

But the sum of the triangular given pyramid, and the sum of

pyramids is equal to the their bases is equal to the base of the given pyramid: hence, the volume of the given pyramid is equal to onethird of the product of its base and altitude; which was to be proved.

Cor. 1. The volume of a pyramid is equal to one-third of the volume of a prism having an equal base and an equal altitude.

Cor. 2. Any two pyramids are to each other as the products of their bases and altitudes. Pyramids having equal bases are to each other as their altitudes. Pyramids having equal altitudes are to each other as their bases.

Scholium. The volume of a polyedron may be found by dividing it into triangular pyramids, and computing their volumes separately. The sum of these volumes will be equal to the volume of the polyedron.

The volume of a frustum of any triangular pyramid is equal to the sum of the volumes of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base of the frustum, and a mean proportional between the two bases.

Let FGH-h be a fi ustum of any triangular pyramid: then will its volume be equal to that of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base FGH, the upper base fgh, and mean proportional between their bases.

For, through the edge FII, pass the plane FIIg, and through the edge fg, pass the plane jgII, dividing the frustum into three pyramids. The pyra

mid g-FGII, has for its base the lower base FGH of the frustum, and its alitude is equal to that of the frustum, >ecause its vertex g, is in the plane of he upper base. The pyramid II-fgh, has for its base the upper base fgh of the frustum, and its altitude is equal to that of the frustum, because its vertex lies in the plane of the lower base.

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H

The remaining pyramid may be regarded as having the triangle FfH for its base, and the point g for its vertex. From g, draw gK parallel to fF, and draw also KH and Kf. Then will the pyramids K-FJH and g-FfII, be equal; for they have a common base, and their altitudes are equal, because their vertices and g are in a line parallel to the base (B. VI., P. XII., C. 2).

Now, the pyramid K-FfH may be regarded as having FKH for its base and f for its vertex. From K, draw KL parallel to GII; it will be parallel to gh then will the triangle FKL be equal to fgh, for the side FK is equal to fg, the angle F to the angle f, and the angle K to the angle g. But, FKII is a mean proportional between FKL and FGH (B. IV., P. XXIV., C.), or between fgh and FGH. The pyramid f-FKII, has, therefore, for its base a mean proportional between the upper and lower bases of the frustum, and its altitude is equal to that of the frustum; but the pyramid f-FKII is equal in volume to the pyramid g-FII: hence, the volume of the given frustum is equal to that of three pyramids whose common altitude is equal to that of the frustum, and whose bases are the upper base, the lower base, and a mean proportional between them ; which was to be proved.

Cor. The volume of the frustum of any pyramid is equal to the sum of the volumes of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base of the frustum, and a mean proportional between them.

A

S

D

For, let ABCDE-e be a frustum of any pyramid. Through any lateral edge, as eE, pass the planes eEBb, eECc, dividing it into triangular frustums. Now, the sum of the volumes of the triangular frustums is equal to the sum of three sets of pyramids, whose common altitude is that of the given frustum. The bases of the first set make up the lower base of the given frustum, the bases of the second set make up the upper base of the given frustum, and the bases of the third set make up a mean proportional between the upper and lower base of the given frustum : hence, the sum of the volumes of the first set is equal to that of a pyramid whose altitude is that of the frustum, and whose base is the lower base of of the frustum; the sum of the volumes of the second set is equal to that of a pyramid whose altitude is that of the frustum, and whose base is the upper base of the frustum ; and, the sum of the third set is equal to that of a pyramid whose altitude is that of the frustum, and whose base is a mean proportional between the two bases.

PROPOSITION XIX. THEOREM.

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

Let CBD-P, obd-p, be two similar triangular prisms, and let BC, bc, be any two homologous edges: then will the prism CBD-P be to the prism cbd-p, as BC3 to be3

For, the homologous angles B and are equal, and the faces which bound them are similar (D. 16): hence, these triedral angles may be

P

applied, one to the other, so that the angle cbd will coincide with CBD, the edge ba with BA. In this case, the

Ρ

prism cbd-p will take the position Bcd-p. From A draw AH perpendicular to

H

the common base of the prisms: then will the plane BAH be perpendicular to the plane of the common base (B. VI., P. XVI.). From a, in the plane BAH, draw ah perpendicular to BII: then will ah also be perpendicular to the base BDC (B. VI., P. XVII.); and AII, ah, will be the altitudes of the two prisms.

Since the bases CBD, cbd, are similar, we have (B. IV., P. XXV.),

base CBD : base cbd :: CB2: cb2.

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

АП : ah :: CB: cb;

hence, multiplying these proportions term by term, we have,

base CBD x AH: base cbd × ah ::

CB3 : cb3.

But, base CBD AII is equal to the volume of the prism CDB-A, and base cbd x ah is equal to the volume of the prism cbd-p; hence,

prism CDB-P : prism cbd-p ::

CB3 : cb3;

which was to be proved.

Cor. 1. Any two similar prisms are to each other as the cubes of their homologous edges..

For, since the prisms are similar, their bases are similar polygons (D. 16); and these similar polygons may each be divided into the same number of similar triangles, similarly placed (B. IV., P. XXVI.); therefore, each prism may be divided into the same number of triangular prisms, having their faces similar and like placed; consequently, the triangular prisms are similar (D. 16). But these triangular prisms are to each other as the cubes of their homologous edges, and being like parts of the polygonal prisms, the polygonal prisms themselves are to each other as the cubes of their homologous edges.

Cor. 2. Similar prisms are to each other as the cubes of their altitudes, or as the cubes of any other homologous lines.

PROPOSITION XX. THEOREM.

Similar pyramids are to each other as the cubes of their homologous edges.

Let S-ABCDE, and S-abcde, be two similar pyramids, so placed that their homologous angles at the vertex shall coincide, and let AB and ab be any two homologous edges: then will the pyramids be to each other as the cubes of AB and ab.

For, the face SAB, being similar to Sab, the edge AB is parallel to the edge ab, and the face SBO being similar to Sbc, the edge BC is parallel to bc; hence, the planes of the bases are parallel (B. VI., P. XIII.).

E

B

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