volumes of two prisms of equal altitudes are to each other as their bases. The same is true of pyramids.

704. Corollary.—Symmetrical prisms are equivalent. The same is true of symmetrical pyramids.

705. The volume of a frustum of a pyramid is found by subtracting the volume of the pyramid cut off from the volume of the whole. When the altitude of the whole is not given, it may be found by this proportion: the area of the lower base of the frustum is to the area of its upper base, which is the base of the part cut off, as the square of the whole altitude is to the square of the altitude of the part cut off.

EXERCISES.

706.-1. What is the ratio of the volumes of a pyramid and prism having the same base and altitude?

2. If two tetraedrons have a triedral vertex in each equal, their volumes are in the ratio of the products of the edges which contain the equal vertices.

3. The plane which bisects a diedral angle of a tetraedron, divides the opposite edge in the ratio of the areas of the adjacent faces.

SIMILAR POLYEDRONS.

707. The propositions (640 to 643) upon the ratios of the areas of the surfaces of similar tetraedrons, may be applied by the student to any similar polyedrons. These propositions and the following are equally applicable to polyedrons that are symmetrically similar.

708. Problem.-Any two similar polyedrons may be divided into the same number of similar tetraedrons, which shall be respectively similar, and similarly arranged.

For, after dividing one into tetraedrons, the construc

tion of the homologous lines in the other will divide it in the same manner. Then the similarity of the respective tetraedrons follows from the proportionality of the lines.

709. Theorem.-The volumes of similar polyedrons are proportional to the cubes of homologous lines.

First, suppose the figures to be tetraedrons. Let AH and BG be the altitudes.

E

H

I

B

G

Then (641), EIO: CDF :: EI2: CF2 :: AH2: BG2. By the proportionality of homologous lines, (634), AH: BG :: EI : CF :: AH : BG.

Multiplying these proportions (701), we have

AEIO BCFD:: EI3: CF3:: AH3: BG3,

or, as the cubes of any other homologous lines.

Next, let any two similar polyedrons be divided into the same number of tetraedrons. Then, as just proved, the volumes of the homologous parts are proportional to the cubes of the homologous lines. By arranging these in a continued proportion, as in Article 436, we may show that the volume of either polyedron is to the volume of the other as the cube of any line of the first is to the cube of the homologous line of the second.

710. Notice that in the measure of every area there are two linear dimensions; and in the measure of every volume, three linear, or one linear and one superficial.

EXERCISE.

711. What is the ratio between the edges of two cubes, one of which has twice the volume of the other?

This problem of the duplication of the cube was one of the celebrated problems of ancient times. It is said that the oracle of Apollo at Delphos, demanded of the Athenians a new altar, of the same shape, but of twice the volume of the old one. The efforts of the Greek geometers were chiefly aimed at a graphic solution; that is, the edge of one cube being given, to draw a line equal to the edge of the other, using no instruments but the rule and compasses. In this they failed. The student will find no difficulty in making an arithmetical solution, within any desired degree of approximation.

REGULAR POLYEDRONS.

712. A REGULAR POLYEDRON is one whose faces are equal and regular polygons, and whose vertices are equal polyedrals.

The regular tetraedron and the cube, or regular hexaedron, have been described.

The regular octaedron has eight, the dodecaedron twelve, and the icosaedron twenty faces.

Geom.-21

The class of figures here defined must not be confounded with regular pyramids or prisms.

713. Problem.—It is not possible to make more than five regular polyedrons.

First, consider those whose faces are triangles. Each angle of a regular triangle is one-third of two right angles. Either three, four, or five of these may be joined to form one polyedral vertex, the sum being, in each case, less than four right angles (612). But the sum of six such angles is not less than four right angles. Therefore, there can not be more than three kinds of regular polyedrons whose faces are triangles, viz. the tetraedron, where three plane angles form a vertex; the octaedron, where four, and the icosaedron, where five angles form a vertex.

The same kind of reasoning shows that only one regular polyedron is possible with square faces, the cube; and only one with pentagonal faces, the dodecaedron.

Regular hexagons can not form the faces of a regular polyedron, for three of the angles of a regular hexagon are together not less than four right angles; and therefore they can not form a vertex.

So much the more, if the polygon has a greater number of sides, it will be impossible for its angles to be the faces of a polyedral. Therefore, no polyedron is possible, except the five that have been described.

MODEL REGULAR POLYEDRONS.

714. The possibility of regular polyedrons of eight, of twelve, and of twenty sides is here assumed, as the demonstration would occupy more space than the principle is worth. However, the student may construct models of these as follows. Plans for the regular tetraedron and the cube have already been given.

For the octaedron, draw eight equal regular triangles, as in the diagram.

For the dodecaedron, draw twelve equal regular pentagons, as in the diagram.

For the icosaedron, draw twenty equal regular triangles, as in the diagram.

There are many crystals, which, though not regular, in the geometrical rigor of the word, yet present a certain regularity of shape.

EXERCISES.

715.—1. How many edges and how many vertices has each of the regular polyedrons?

2. Calling that point the center of a triangle which is the intersection of straight lines from each vertex to the center of the opposite side; then, demonstrate that the four lines which join the vertices of a tetraedron to the centers of the opposite faces, intersect each other in one point.

3. In what ratio do the lines just described in the tetraedron divide each other?

4. The opposite vertices of a parallelopiped are symmetrical triedrals.

5. The diagonals of a parallelopiped bisect each other; the lines which join the centers of the opposite edges bisect each other; the lines which join the centers of the opposite faces bi