parallelopipedon is equal to the product of its base by its altitude; hence, that of the triangular prism is also equal to the product of its base, or half that of the parallelopipedon, by its altitude.

Third. Any prism may be divided into as many triangular prisms of the same altitude as there are triangles in the polygon taken for a base. But the volume of each triangular prism is equal to the product of its base by its altitude; and, since the altitude is the same in each, it follows that the sum of all these prisms is equal to the sum of all the triangles taken as bases multiplied by the common altitude.

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

THEOREM XVI.

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

KO, to the bases ABC, F HI. Take A K' equal to FK, and join A N. Draw K' O' perpendicular to A N in the plane AND, and K' O' will be perpendicular to the plane ABC, and equal to K O, the altitude of the prism FHI-M. For, conceive the triedral angles A and F to be applied the one to the other; the planes containing them, and therefore the perpendiculars K' O', KO, will coincide.

Now, since the bases ABC, FHI are similar, we have

(Theo. XIX. Bk. IV.),

2

2

Base ABC: Base FHI :: A B2 : F' H2 ;

and, because of the similar triangles DAN, KFO, and of the similar parallelograms D B, KH, we have

DN:KO :: DA : KF'::AB: F' H.

Hence, multiplying together the corresponding terms of these proportions, we have

3

Base ABC × D N : Base FHI × KO : A B3 : FH3. But the product of the base by the altitude is equal to the solidity of a prism (Theo. XV.); hence

Prism ABC-E : Prism F'HI-M : : A B3 : FH3.

THEOREM XVII.

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

Let ABCDE-S be a right pyra

mid, and SM its slant height; then the convex surface is equal to the perimeter AB+BC+CD+DE+E A multiplied by S M.

S

tances from a perpendicular let fall from the vertex S to the

CD, &c. multiplied by half the common altitude, SM; that is, the convex surface of the pyramid is equal to the perimeter of the base multiplied by half the slant height.

305. Cor. The lateral faces of a right pyramid are equal isosceles triangles, having for their bases the sides of the base of the pyramid.

THEOREM XVIII

306. If a pyramid be cut by a plane parallel to its base,

1st. The edges and the altitude will be divided proportionally.

2d. The section will be a polygon similar to the base.

Let the pyramid A B C D E – S, whose altitude is SO, be cut by a plane, GHIKL, parallel to its base; then will the edges SA, SB, SC, &c., with the altitude SO, be divided proportionally; and the section G HIKL will be similar to the base A B C D E.

First. Since the planes ABC, GHI A

are parallel, their intersections AB, GH,

by the third plane SAB, are parallel

(Theo. III.); hence the triangles SA B,

G

S

LK
PI

H

E D

B

0

C

SG Hare similar (Theo. XVI. Bk. IV.), and we have

SA:SG::SB: SH.

For the same reason, we have

SB:SH::SC:SI;

and so on. Hence all the edges, SA, SB, SC, &c., are cut proportionally in G, H, I, &c. The altitude S O is likewise cut in the same proportion, at the point P; for BO and HP are parallel; therefore we have

SO:SP: : S B : SH.

Secondly. Since G His parallel to AB, HI to BC, IK to CD, &c., the angle GHI is equal to ABC, the angle

HIK to BCD, and so on (Theo. VI.). Also, by reason of

the similar triangles SAB, SG H, we have

AB:GH::SB:SH;

and by reason of the similar triangles SBC, SHI, we have

SB:SH::BC : HI;

hence, on account of the common ratio SB: S H,

AB:GH ::BC :Η Ι.

For a like reason, we have

BC:HI :: CD : IK,

and so on. Hence the polygons ABCDE, GHIKL have their angles equal, each to each, and their homologous sides proportional; hence they are similar.

307. Cor. 1. If two pyramids have the same altitude, and their bases in the same plane, their sections made by a plane parallel to the plane of their bases are to each other as their bases.

For, the two polygons ABCDE, G H I K L being similar, their surfaces are as the squares of the homologous sides

AB, G H (Theo. XXI. Bk. IV.). But

AB:GH::SA:SG.

2

Hence, ABCDE:GHIKL : : SA2 : SG2.

For the same reason,

MNO:PQR : : SM2 : SP2.

But since G H I K L and PQR are in the same plane, we have also (Theo. VIII.),

hence,

SA:SG::SM:SP;

ABCDE: G H I K L ::MNO:PQR, therefore the sections GHIKL, P Q R are to each other as the bases A B C D E, M Ν Ο.

308. Cor. 2. If the bases ABCDE, MNO are equivalent, any sections, GHIKL, PQR, made at equal distances from those bases, are likewise equivalent.

THEOREM ХІХ.

309. The convex surface of a frustum of a right pyramid is equal to half the sum of the perimeters of its two bases, multiplied by its slant hight.

Let ABCDE-L be the frustum of a right pyramid, and MN its slant hight; then the convex surface is equal to the sum of the perimeters of the two bases ABCDE, GIHKL, multiplied by half of MN.

L

G

K

M

I

H

E

A

D

N

B

C

For the upper base G H I K L is similar to the base ABCDE (Theo. XVIII.), and ABCDE is a regular polygon (267); hence the sides GH, HI, IK, KL, and LG are all equal to each other. The angles GA B, A BH, HBC, &c. are equal (Theo. XVII. Cor.), and the edges AG, BH, CI, &c. are also equal (Theo. XVIII.); therefore the faces AH, BI, CK, &c. are all equal trapezoids (23), having a common altitude, MN, the slant hight of the frustum. But the area of either trapezoid, as A H, is equal to (AB+GH) × MN (Theo. VII. Bk. IV.); hence the areas of all the trapezoids, or the convex surface of frustum, are equal to half the sum of the perimeters of the two bases multiplied by the slant hight.