altitude, because their bases are in the same plane AD, and their vertices at the same point F; hence, they are equal in volume (P. XV.). The pyramids ABC-F and DEF-C, have their bases ABC and DEF, equal because they are the bases of the given prism, and their altitudes are equal because each is equal to the altitude of the prism; they are, therefore, equal in volume: hence, the three pyramids into which the prism is divided, are all equal in volume; which was to be proved.

Cor. 1. A triangular pyramid is one-third of a prism, having an equal base and an equal altitude.

Cor. 2. The volume of a triangular pyramid is equal to one-third of the product of its base and altitude.

The volume of any pyramid is equal to one-third of the product of its base and altitude.

Let S-ABCDE, be any pyramid: then is its volume equal to one-third of the product of its base and altitude. For, through any lateral edge, as SE,

pass the planes SEB, SEC, dividing the pyramid into triangular pyramids. The altitudes of these pyramids will be equal to each other, because each is equal to that of the given pyramid. Now, the volume of each triangular pyramid is equal to onethird of the product of its base and altitude (P. XVI., C. 2); hence, the sum of

the volumes of the triangular pyramids, is

E

B

C

equal to one-third of the product of the sum of their bases

by their common altitude.

But the sum of the triangular pyramids is equal to the given pyramid, and the sum of 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 a mean proportional between their bases.

For, through the edge FH, pass the plane FHg, and through the edge fg, pass the plane fgH, dividing the frustum into three pyramids. The pyra

mid g-FGH, has for its base the lower base FGH of the frustum, and its altitude is equal to that of the frustum, because its vertex g, is in the plane of the upper base. The pyramid H-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.

9

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-FfH and g-FfH, be equal; for they have a common base, and their altitudes are equal, because their vertices K 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 GH; 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, FKH is a mean proportional between. FKL and FGH (B. IV., P. XXIV., C.), or between fgh and FGH. The pyramid f-FKH, 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-FKH is equal in volume to the pyramid g-FfH: 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

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, cbd-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 b are equal, and the faces which bound them are similar (D. 16): hence, these triedral angles may be

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 BH: then will ah also be perpendicular to the base BDC (B. VI., P. XVII.); and AH, 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,

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

base CBD × AĦ : base cbd × ah :: CB3 : cb3.

But, base CBD × AH 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.