11. A pyramid is a solid contained by several triangular planes proceeding from the same point S, and terminating in the different sides of the same polygonal plane ABCDE.

The polygon ABCDE is called the base of the pyramid, the point S its vertex; and the triangles ASB, BSC, &c., A form its convex or lateral surface.

12. If a pyramid SABCDE, be cut by

S

E

C

B

a plane abcde parallel to its base, that portion cut off by this plane and lying next to its base, is called a frustum, or trunk of a pyramid.

13. A pyramid is triangular, quadrangular, &c., according as its base is a triangle, a quadrilateral, &c.

14. A pyramid is regular, when its base is a regular polygon, and when, at the same time, the perpendicular let fall from the vertex to the plane of the base, passes through the centre of this base. That perpendicular is then called the

axis of the pyramid.

15. The altitude or height of a pyramid is the perpendicular let fall from the vertex upon the plane of the base; as SO in the preceding figure. In a regular pyramid it is the length of the axis.

16. The slant-height of a regular pyramid, is the distance from the vertex to the middle of one of the sides of the base; as SH.

17. The diagonal of a polyedron is a straight line joining the vertices of two solid angles which are not adjacent to each other.

18. Similar polyedrons are such as are contained by the same number of simi

S

B

D

E

A

Ус

H

B

lar planes, similarly situated, and having like inclinations to each other.

Note. By the vertices of a polyedron, is meant the points situated at the vertices of its different solid angles.

PROPOSITION I. THEOREM.

Two prisms are equal when the three planes containing a solid angle of the one, are respectively equal to the three planes containing a solid angle of the other, and are similarly situated.

Let the base ABCDE be equal to the base abcde, the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to bchg; then will the prism ABCI be equal to the prism abci.

For, lay the base ABCDE upon its equal abcde; these two bases will coincide. But the three plane angles, which form the solid angle B, are respectively equal to the three plane angles, which form the solid angle b, viz., ABC=abc, ABG abg, and GBC=gbc; they are also similarly situated: hence the solid angles B and b are equal; (21. 6. Sch. ;) therefore the side BG will fall on its equal bg. It is likewise evident, that by reason of the equal parallelograms ABGF,

abgf, the side GF will fall on its equal gf, and in the same manner GH on gh; hence the upper base FGHIK will exactly coincide with its equal fghik. But the upper base FGHIK, is equal and similar to the lower base ABCDE; (Def. 4. 7;) also the upper base fghik is equal and similar to the lower base abcde; hence ABCDE=abcde, and will exactly coincide with it; consequently the parallelogram AG will coincide with its equal ag, BH with bh, and CI with ci: hence the two prisms coincide throughout, and are therefore equal. (Ax. 13.) Therefore, Two prisms are equal when, &c.

Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF will be equal to abgf; so also will the rectangle BGHC be equal to bghc; and thus the three planes, which form the solid angle B, will be equal to the three which form the solid angle b. Hence the two prisms are equal.

PROPOSITION II. THEOREM.

In every parallelopipedon the opposite planes are equal and

parallel.

E

By the definition of this solid, the bases ABCD, EFGH are equal parallelograms, and their sides are parallel : it remains only to show, that the same is true of any two opposite lateral faces, such as AEHD, BFGC. Now B

A

AD is equal and parallel to BC, because the figure ABCD is a parallelogram; for a like reason, AE is parallel to BF: hence the angle DAE is equal to the angle CBF, and the planes DAE, CBF are parallel; (13. 6;) hence also the parallelogram DAEH is equal to the parallelogram CBFG. In

the same way, it might be shown that the opposite parallelograms ABFE, DCGH are equal and parallel. Therefore, In every parallelopipedon, the opposite planes, &c.

Cor. 1. Since the parallelopipedon is a solid bounded by six planes, whereof those lying opposite to each other are equal and parallel, it follows that any face and the one opposite to it, may be assumed as the bases of the parallelopipedon.

Cor. 2. A plane BDHF passing through the opposite diagonal edges BF, DH of a parallelopipedon, will divide it into two triangular prisms, equivalent to each other. For, since the plane cuts the parallelopipedon through its diagonal edges, it will divide the upper base EG, into two equal triangles FEH, FHG; (30. 1. Cor. 1;) for the same reason it will also divide the lower base AC into two equal triangles BAD, BDC. And since the parallelogram AF is equal and parallel to DG, and ED to FC, hence the planes which contain the prisms BEH, HCB, are equal and similar to one another; and being parallel they are also equally inclined to each other; therefore the prism BEH must be equivalent to HBC. (1. 7.)

Cor. 3. Every triangular prism ABDHEF is half of the parallelopipedon AG described on the same solid angle A, and the same edges AB, AD, AE.

Scholium. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelopipedon may be formed on those lines. For this purpose, a plane must be extended through the extremity of each line, and parallel to the plane of the other two; that is, through the point B a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of those planes will form the parallelopipedon required.

In every parallelopipedon, the opposite solid angles are symmetrical; and the diagonals drawn through the vertices of those angles bisect each other.

First. Compare the solid angle A E with its opposite one G. The angle EAB, equal to EFB, is also equal to HGC, the angle DAE DHE CGF; A and the angle DAB DCB=HGF; therefore the three plane angles, which

=

B

H

G

form the solid angle A, are equal to the three which form the solid angle G, each to each. It is easy, moreover, to see that their arrangement in the one is different from their arrangement in the other: hence the two solid angles A and G are symmetrical. (21. 6. Sch.)

Second. Imagine two diagonals EC, AG to be drawn. both through opposite vertices: since AE is equal and parallel to CG, the figure AEGC is a parallelogram; hence the diagonals EC, AG will mutually bisect each other. (33. 1.) In the same manner, it may be shown that the diagonal EC and another DF bisect each other; hence the four diagonals will mutually bisect each other, in a point which may be regarded as the centre of the parallelopipedon. Hence, In every parallelopipedon, &c.

PROPOSITION IV. THEOREM.

In every prism ABCI, the sections NOPQR, STVXY, formed by parallel planes, are equal polygons.

For the sides ST, NO, are parallel, being the intersections of two parallel planes with a third plane ABGF;