altitude of the frustum. The remaining portion is the pyramid having C for its vertex, and DBE for its base.

From C draw CG in the plane AF parallel to AD, meeting DF in G; from G draw GH in the plane DEF, parallel to EF or CB; join BH, EG. Then the triangle DGH will be equal to ABC, and DGE will be a mean proportional between ABC and DEF (V. 22). Now, the pyramid C-DBE is equal to the pyramid G-DBE (VIII. 15). But the pyramid G-DBE is the same as the pyramid B-DGE, which has for its altitude the altitude of the frustum, and for its base a mean proportional between the bases of the frustum.

Hence, the three portions into which the frustum has been divided are equal to three pyramids having a common altitude, which is the altitude of the frustum, and having for bases the two bases of the frustum and a mean proportional between them.

2. The same is true for any frustum of a pyramid.

Let a-bcde be a pyramid, and A-BCD a triangular one having an equivalent base and the same altitude. They are equal (VIII. 15.) If we cut sections from them at equal distances from their bases, by planes parallel to their bases, the sections will be equal in area; hence, the

AA

portions cut off toward the vertices will be equal pyramids (VIII. 15); therefore the remaining frusta are equal. But the volume of the triangular frustum is equal to the volumes of three pyramids having for a common altitude the altitude of the frustum, and for bases the two bases of the frustum and a mean proportional between them. Hence, the same rule is true for the other.

Proposition 20.

Theorem. The area of the convex surface of a cylinder is equal to the circumference of the base multiplied by its altitude; and its volume is equal to the area of the base multiplied by its altitude.

Inscribe in the cylinder a prism whose base is a regular polygon. The area of the lateral surface of this is equal (VIII. 10) to the perimeter of the base multiplied by its altitude, and the volume is equal (VIII. 9) to the area of the base multiplied by the altitude. These being true whatever the number of sides, they are true when that number becomes infinite. In that case the perimeter of the polygon coincides (VI. 10, Cor.) with the circumference of the base, and the prism with the cylinder. Hence the truth of the theorem.

Scholium. If R = radius of the base of the cylinder; H, its altitude; S, its convex surface; and V, its volume, then the theorem is expressed by the following formulæ:

S = 2R × H;
V = πR2 × Η.

Proposition 21.

Theorem. The area of the convex surface of a cone is equal to one half the product of the circumference of the base and the slant height; and its volume is equal to one third the product of the area of the base and the altitude.

Inscribe in the cone a pyramid whose base is a regular polygon. The area of the lateral surface of this is equal (VIII. 11) to one half the product of the perimeter of the base and slant height, and its volume is equal (VIII. 17) to one third the product of the area of the base and the altitude. These being true whatever the number of the

sides, they are true when that number becomes infinite; in which case the perimeter of the polygon coincides (VI. 10, Cor.) with the circumference of the base of the cone, and the prism with the cone. Hence the truth of the theorem.

Scholium. Let R = radius of base, H = altitude, H' = slant height, S = convex surface, V= volume. Then, S = 2R×H' = 2«R×+VR2 + H2 = «RVR2 + H2, and

V = πR2 × H.

Corollary 1. The convex surface of a cone is equal to the hypotenuse of the triangle which forms it, multiplied by the circumference of the circle described by its middle

point.

Let AB be the axis, AC the hypotenuse,
Dits middle point. Then (last Scholium),
S=AC×2CB = AC×2π(CB) = A C×2=DE.

Corollary 2.-A cone is one third of a cylinder on the same base and of the same altitude.

Proposition 22.

Theorem. The area of the convex surface of the frustum of a cone is equal to the sum of the circumferences of the two bases multiplied by one half the slant height; and its volume is equal to the volume of three cones, whose common altitude is the altitude of the frustum, and whose bases are the two bases of the frustum and a mean proportional between them.

Inscribe in the frustum of the cone a frustum of a pyramid whose bases are similar regular polygons. The area of the lateral surface of this is equal (VIII. 11, Sch.) to the sum of the perimeters of the two ends multiplied by one half the slant height; and its volume is equal (VIII. 19) to the volume of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the two bases of the frustum and a mean proportional between them. These being true

whatever the number of sides, they are true when that number becomes infinite. In that case the perimeters of the polygons coincide (VI. 10, Cor.) with the circumferences of the bases, the frustum of the pyramid with the frustum of the cone, and the three pyramids into which the frustum is divided with three cones. Hence the truth of the theorem.

Scholium. If R, R' be the radii of the bases of the frustum, H its altitude, H' its slant height, S its surface, and V its volume, then

S= (2πR+2πR')+H' =ᅲH'(R+R'),

V=1H(πR2+R'2+ VR2×πR'

= πΗ(R2+ R2 + RR').

2

Corollary. The surface of a frustum is equal to the slant

height, multiplied by the circumference of a circle

described by its middle point.

Let AB be a portion of the axis, CD the

C

A

slant height, and Fits middle point.

Now (last scholium),

S = +CD × 2π(AC+BD) = CD ×2+(AC+BD)

= CD × 2FE.

Proposition 23.

Theorem. The surfaces of similar polyedrons are proportional to the squares of their homologous edges or of any homologous lines.

Polyedrons are similar when they are composed of the same number of similar faces; hence the whole surfaces are proportional to any two homologous faces; but these faces are proportional to the squares of any two homologous sides; therefore the whole surfaces are proportional to the squares of any two homologous edges, or of any lines proportional to them.

Corollary 1.- The surfaces of cylinders and cones are proportional to the squares of the radii of their bases or of their altitudes.

Corollary 2.- The surfaces of all similar solids are proportional to the squares of their homologous lines, and their volumes to the cubes of their homologous lines.

The last statement is evident from (VIII. 18, Cor. 3).

Proposition 24.

Theorem. Only five regular polyedrons are possible.

A regular polyedron is one whose faces are equal regular polygons. The angles constituting any solid angle must be less (VII. 20) than four right angles, and there must be at least three plane angles at every solid angle.

The simplest regular polygon is the equilateral triangle. Three angles of an equilateral triangle will form an angle of a tetraedron; four of an octaedron; and five of an icosaedron; six would together be equal to four right angles, and would not form a solid angle.

The regular polygon of four sides is the square. Three angles of a square would form an angle of a cube or hexaedron; four would be equal to four right angles.