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Proposition 24. Theorem. 647. Two similar tetraedrons are to each other as the cubes of their homologous edges.

S Hyp. Let SABC, S'A'B'C' be two similar tetraedrons; and let

S' V and V' denote their volumes.

V

AB
V
A'B's
AE

CA
Proof. Let the vertices S and
S' be homologous.

B'
Then, since the triedral Zs S and S' are equal, (641)
V SA X SB X SC SA

SB

SC
V

Х Х
S'A' X S'B' X S'C' S'A' S'B' S'C

(637)
SA SB SC AB
But
S'A' S'B' S'C A'B':

(641) V AB AB AB AB

Х X
A'B' A'B' A'B'

Q.E.D.

A'B' 648. Cor. 1. Two similar polyedrons are to each other as the cubes of their homologous edges.

For, two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed (645); and any two homologous tetraedrons are to each other as the cubes of their homologous edges (647), or as the cubes of any two homologous edges of the polyedrons (646). Therefore the polyedrons themselves are to each other as the cubes of their homologous edges.

(296) 649. COR. 2. Similar prisms, or pyramids, are to each other as the cubes of their altitudes.

650. COR. 3. Similar polyedrons are to each other as the cubes of any two homologous lines.

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REGULAR POLYEDRONS. 651. A regular polyedron is one whose faces are all equal regular polygons, and whose polyedral angles are all equal.

Proposition 25. Theorem.* 652. There can be only five regular convex polyedrons.

Proof. At least three faces are necessary to form a polyedral angle, and the sum of its face-angles must be less than 360°.

(566) 1. Because the angle of an equilateral triangle is 60°, each convex polyedral angle may have 3, 4, or 5 equilateral triangles. It cannot have 6 faces, because the sum of 6 such angles is 360°, reaching the limit. Therefore no more than three regular convex polyedrons can be formed with equilateral triangles ; the tetraedron, octaedron, and icosaedron.

2. Because the angle of a square is 90°, each convex polyedral angle may have 3 squares. It cannot have 4 squares because the sum of 4 such angles is 360°. Therefore only one regular convex polyedron can be formed with squares ; the hexaedron, or cube.

3. Because the angle of a regular pentagon is 108°, each convex polyedral angle may have 3 regular pentagons. It cannot have 4 faces because the sum of 4 such angles is 432°. Therefore only one regular convex polyedron can be formed of regular pentagons; the dodecaedron.

Because the angle of a regular hexagon is 120°, and the angle of every regular polygon of more than 6 sides is yet greater than 120°, therefore there can be no regular convex polyedron formed of regular hexagons or of any regular polygons of more than 6 sides.

Therefore there can be only five regular convex poly edrons.

Proposition 26. Problem.* 653. To construct the regular polyedrons, having given one of the edges.

Given, the edge AB.

Required, to construct the regular polyedrons. 1. The regular tetraedron.

Af Cons. Upon AB construct the equilateral A ABC.

At its centre 0 erect OD I to the plane ABC so that AD = AB.

Join DA, DB, DC.
Then ABCD is a regular tetraedron.
Proof. Since the faces are each = the ABC, (496)

.. the triedral 8 A, B, C, D are all equal. (567)

... ABCD is a regular tetraedron. 2. The regular hexaedron, or cube.

HO
Cons. Upon AB construct the square
ABCD.

E

F Upon the sides of this square construct

D the four equal squares AF, BG, CH, DE, I to the plane ABCD.

Then AG is a regular hexaedron. Proof. Since the faces are equal squares, (Cons.) ... the triedral Z8 A, B, C, D, E, F, G, H are all

(567) .. ABCD-E is a regular hexaedron, or cube.

А

equal.

3. The regular octaedron.
Cons. Upon AB construct the

square ABCD.

А4

At its centre 0 erect EOF I to the plane ABCD, so that OE = OF = 0A.

Join the pts. E and F to all the vertices of the square.

Then E-ABCD-F is a regular octaedron.

B

Proof. The lines from E and F to A, B, C, D are all equal.

(496) And since the rt. As AOE, AOB are equal, (104)

.:. AE = AB.

... the twelve edges of the octaedron are all equal, and the faces are eight equal equilateral As.

Since the diagonals BD and EF are equal and bisect each other at rt. ZS,

(Cons.)

.:. BEDF is a square = ABCD.

And since AO is I to BD and EF,

.. AO is 1 to the plane of BEDF.

(500)

.: pyramid A-BEDF = pyramid E-ABCD. (633)

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Similarly, it can be shown that any other two polyedral Ls are equal.

.:. E-ABCD-F is a regular octaedron.

4. The regular dodecaedron.

Cons. Upon AB construct the regular pentagon ABCDE, and join to its sides the sides of five other equal pentagons, so inclined to the plane of ABCDE as to form five triedral

s at A, B, C, D, E.

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in fk,

There is then formed a convex surface FGHIK, etc., composed of six equal regular pentagons.

Construct a second convex surface fyhik, etc., equal to the first.

Then the two equal convex surfaces may be combined so as to form a single convex surface, which is the regular dodecaedron.

Proof. Because the face Zs of the triedral Zs in FK are equal respectively to the face s of the triedral Zs

(Cons.) ... the triedral Zs of FK and fk are equal, each to each.

(567) Now suppose the convexity of FK to be up, and the convexity of fk to be down.

Put the two surfaces together so that the pt. O and the side OP shall coincide with the pt. n and the side no, respectively. Then two consecutive face _s of one surface will unite with a single face of the other. Thus the three faces 1,5,7, will enclose a triedral , since the diedral L contained by 1 and 5 is the diedral Z of the equal triedral / formed at E.

The vertex N will coincide with m, and a like triedral / will be formed at that pt., and so of all the others.

.. the triedral _s are all equal, and the solid is a regular dodecaedron.

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