This plane separates two tetraedrons, HBCG and HBFG. The two faces, HBC and HBG, of the tetraedron HBCG, are sections made by the diagonal planes; and the two faces, HCG and BCG, are each half of one side of the prism. The tetraedron HBFG has for one of its faces the base HFG of the prism; for a second face, the triangle HBG, being the section made by the diagonal plane; and, for the other two, the triangles HBF and GBF, each being half of one of the sides of the prism.

Now, consider these two tetraedrons as having their bases BCG and BFG. These are equal triangles lying in one plane. The point H is the common vertex, and therefore they have the same altitude; that is, a perpendicular from H to the plane BCGF.

Next, consider the first and last tetraedrons described, HBCD and BFGH, the former as having BCD for its base, and H for its vertex; the latter as having FGH for its base, and B for its vertex. These bases are equal, being the bases of the given prism. The vertex of each is in the plane of the base of the other. Therefore, the altitudes are equal, being the distance between these two planes.

Lastly, consider the tetraedrons BCDH and BCGH as having their bases CDH and CGH. These are equal triangles lying in one plane. The tetraedrons have the common vertex B, and hence have the same altitude.

677. Corollary.-Any prism may be divided into tetraedrons in several ways; but the methods above explained are the simplest.

678. REMARK.-On account of the importance of the above problem in future demonstrations, the student is advised to make · a model triangular prism, and divide it into tetraedrons. A po tato may be used for this purpose. The student will derive most benefit from those models and diagrams which he makes himself.

EQUAL PRISMS.

679. Theorem.-Two prisms are equal, when a base and two adjacent sides of the one are respectively equal to the corresponding parts of the other, and they are similarly arranged.

For the triedrals formed by the given faces in the two prisms must be equal (599), and may therefore be made to coincide. Then the given faces will also coincide, being equal. These coincident points include all of one base, and several points in the second. But the second bases have their sides respectively equal, and parallel to those of the first. Therefore, they also coincide, and the two prisms having both bases coincident, must coincide throughout.

680. Corollary.-Two right prisms are equal when they have equal bases and the same altitude.

681. The theory of similar prisms presents nothing difficult or peculiar. The same is true of symmetrical prisms, and of symmetrically similar prisms.

AREA OF THE SURFACE.

682. Theorem.-The area of the lateral surface of a prism is equal to the product of one of the lateral edges by the perimeter of a section, made by a plane perpendicular to those edges.

Since the lateral edges are parallel, the plane HN, perpendicular to one, is perpendicular to all of them. Therefore, the sides of the polygon, HK, KL, etc., are severally perpendicular to the edges of the prism which they unite (519).

Then, in order to measure the area of each face of the prism, we take one edge of the prism as the base

of the parallelogram, and one side of the polygon HN as its altitude.

Thus,

area AGAB X HP,

area EB EC × HK, etc. By addition, the sum of the areas of these parallelograms is the lateral surface of the prism, and the sum of the altitudes of the parallelograms is the perimeter of the polygon HN. Then, since the edges are equal, the area of all the sides is equal to the product of one edge, multiplied by the perimeter of the polygon.

683. Corollary.—The area of the lateral surface of a right prism is equal to the product of the altitude by the perimeter of the base.

684. Corollary.—The area of the entire surface of a regular prism is equal to the product of the perimeter of the base by the sum of the altitude of the prism and the apothem of the base.

EXERCISES.

685.—1. A right prism has less surface than any other prism of equal base and equal altitude; and a regular prism has less surface than any other right prism of equivalent base and equal altitude.

2. A regular pyramid and a regular prism have equal hexagonal bases, and altitudes equal to three times the radius of the base; required the ratio of the areas of their lateral surfaces.

3. Demonstrate the principle stated in Article 683, without the aid of Article 682.

MEASURE OF VOLUME.

686. A PARALLELOPIPED is a prism whose bases are parallelograms. Hence, a parallelopiped is a solid inclosed by six parallelograms.

687. Theorem.-The opposite sides of a parallelopiped are equal.

For example, the faces AI and BD are equal.

For IO and DF are equal, being opposite sides of the parallelogram IF. For

a like reason, EI is equal to CD. But, since these equal sides are also parallel, the included angles EIO and CDF are equal. Hence, the parallelograms are equal.

E

B

688. Corollary.-Any two opposite faces of a parallelopiped may be assumed as the bases of the figure.

689. A parallelopiped is called right in the same case as any other prism. When the bases also are rectangles, it is called rectangular. Then all the faces are rectangles.

690. A CUBE is a rectangular parallelopiped whose length, breadth, and altitude are equal. Then a cube is a solid, bounded by six equal squares. All its vertices, being trirectangular triedrals, are equal (602). All its edges are of right diedral angles, and therefore equal (555).

The cube has the simplest form of all geometrical solids. It holds the same rank among them that the square does among plane figures, and the straight line among lines.

The cube is taken, therefore, as the unit of measure of volume. That is, whatever straight line is taken as the unit of length, the cube whose edge is of that length is the unit of volume, as the square whose side. is of that length is the measure of area.

VOLUME OF PARALLELOPIPEDS.

691. Theorem.-The volume of a rectangular parallelopiped is equal to the product of its length, breadth, and altitude.

In the measure of the rectangle, the product of one line by another was explained. Here we have three lines used with a similar meaning. That is, the number of cubical units contained in a rectangular parallelopiped is equal to the product of the numbers

A

B

of linear units in the length, the breadth, and the altitude.

If the altitude AE, the length EI, and the breadth IO, have a common measure, let each be divided by it; and let planes, parallel to the faces of the prism, pass through all the points of division, B, C, D, etc.

By this construction, all the angles formed by these. planes and their intersections are right angles, and each of the intercepted lines is equal to the linear unit used in dividing the edges of the prism. Therefore, the prism is divided into equal cubes. The number of these at the base is equal to the number of rows, multiplied by the number in each row; that is, the product

Geom.-20