For, in the first place, any parallelopipedon (T. VII.) is equivalent to a rectangular parallelopipedon having the same altitude and an equivalent base. Now the volume of the latter is equal to its base multiplied by its height; hence the volume of the former is, in like manner, equal to the product of its base by its altitude.

In the second place, and for a like reason, any triangular prism is half of the parallelopipedon so constructed as to have the same altitude and a double base; but the volume of the latter is equal to its base multiplied by its altitude; hence that of a triangular prism is also equal to the product of its base multiplied into its altitude.

In the third place, any prism may be divided into as many triangular prisms of the same altitude, as there are triangles capable of being formed in the polygon which constitutes its base; but the volume of each triangular prism is equal to its base multiplied by its altitude; and since the altitude is the same for all, it follows that the sum of all the partial prisms must be equal to the sum of all the partial triangles which constitute their bases, multiplied by the common altitude.

Hence the volume of any polygonal prism is equal to the product of its base by its altitude.

Cor. Comparing two prisms which have the same altitude, the products of their bases by their altitudes will be as the bases simply. Hence two prisms of the same altitude are to each other as their bases. For a like reason, two prisms of the same base are to each other as their altitudes.

THEOREM XIII.

Similar prisms are to one another as the cubes of their homologous sides.

AH, ah perpendicular to the bases BCD, bcd: join BH; take Baba, and in the plane BHA draw ah perpendicular to BH: then ah will be perpendicular to the plane CBD, and equal to ah the altitude of the other prism; for if the solid angles B and b were applied the one to the other, the planes which contain them, and consequently the perpendiculars ah, ah would coincide.

Now because of the similar triangles ABH, abh, and the similar figures P, p, we have

AH: ah:: AB: ab:: BC: bc;

and because of these similar bases, the

base BCD: base bcd:: BC2: be2. (B. III., T. XXVII.) Taking the product of the corresponding terms of these proportions, we have

AH x base BCD: ah × base bcd:: BC3 : bc3.

But AH x base BCD expresses the volume of the prism P, and ah × base bcd expresses the volume of the other prism p; therefore

prism P: prism p:: BC3: bc3.

THEOREM XIV.

If a pyramid be cut by a plane parallel to its base, the section thus formed will be a polygon similar to the base, and the lateral edges and the altitude will be cut into proportional parts.

First, since the planes ABCDEF, abcdef are parallel, it follows that AB and ab, BC and bc, CD and cd, etc., are parallel (B. V., T. X.); consequently the angles ABC and abc, BCD and bcd are respectively equal (B. V., T. XIII.). Moreover the couples of similar triangles SAB and Sab, SBC and Sbc, etc., give the following equal ratios:

SA: Sa:: AB: ab:: SB: Sb,
SB: Sb:: BC: be:: SC: Sc,
SC: Sc:: CD:cd:: SD: Sd,
&c.

&c.

&c.

F

E

S

D

0

C

A

B

Hence, by equality of ratios, we have

AB: ab:: BC: be:: CD: cd:: etc.

Consequently the two polygons ABCDEF, abcdef are similar (B. III., T. X.).

Now, considering only the ratios between the lateral edges, we have

SA: Sa:: SB: Sb:: SC: Sc:: etc.,

from which we see that the edges are cut into proportional parts.

Finally, if we pass a plane through the edge SB and the altitude SO, its intersection bo with the plane abcde will be parallel to BO (B. V., T. X.), and the two similar triangles SBO, Sbo will give

SO: So::SB: Sb:: SA: Sa :: etc.,

which establishes the theorem.

Cor. When two pyramids S- ABCDE, T-MNP have equivalent bases situated in the same plane, and equal altitudes, the sections made by a plane parallel to their bases, are also equivalent.

For, since the polygons ABCDE, abcde are similar, we have the proportion

ABCDE: abcde:: AB2: ab2;

we also have, by reason of the foregoing relations,

ABCDE: abcde:: SO2: So2.

In the same manner the two polygons MNP, mnp give

MNP: mnp:: TQ2 : Tq2,

we have, moreover, by supposition, SOTQ, So= Tq; hence, ABCDE: abcde:: MNP: mnp.

Now, by hypothesis, ABCDE = MNP; consequently, abcde mnp.

THEOREM XV.

The lateral surface of a right pyramid is equal to the perimeter of its base multiplied by half the slant height.

Since the lateral surface is composed of equal isosceles triangles SAB, SBC, SCD, etc., each one of which is measured by its base into one half its altitude, which altitude is the same as SF the slant height of the pyramid, it follows that the lateral surface is equal to the sum of the bases of all these triangles, or the perimeter of the pyramid's base, into one half the slant height.

Cor. I. If two right pyramids have the same slant heights, their lateral surfaces will be to each other as the perimeters of their bases.

Cor. II. The lateral surface of a frustum of a regular pyramid is measured by half the sum of the perimeters of its two bases multiplied by its slant height.

For, since the frustum is formed from a regular pyramid, the two bases are similar and regular polygons (T. XIV.). Consequently the lateral surface of a frustum of a right pyramid is composed of equal trapezoids, the altitude of each being the same as the slant height of the frustum.

The area of each trapezoid is measured by half the sum of its parallel bases multiplied into its altitude (B. III., T. XXIII., C. II.). Hence the sum of all these trapezoids, or the lateral surface of the frustum, is measured by half the sum of the perimeters of its bases into its slant height.

THEOREM XVI.

Two triangular pyramids having equivalent bases and equal altitudes, are equivalent, or equal in volume.

Let S-ABC, S-abc be those two pyramids; let their equivalent bases ABC, abc be situated in the same plane, and let AT be their common altitude. If they are not equivalent, let S abc be the smaller; and suppose Aa to be the altitude of a prism, which, having ABC for its base, is equal to their dif ference.

Divide the altitude AT into equal parts Ax, xy, yz, etc., 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 (T. XIV., C.), namely, DEF to def, GHI to ghi, etc.

This being granted, upon the triangles ABC, DEF, GHI, etc., taken as bases, construct exterior prisms, having for edges the parts AD, DG, GK, etc., of the edge SA. In like manner, on the bases def, ghi, klm, etc., 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