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PROPOSITION XX. THEOREM

605. The volumes of two triangular pyramids, that have a triedral angle of one equal to a triedral angle

e to each other as the product of the three edges of these angles.

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SIMILAR POLYEDRONS 606. DEF. Similar polyedrons are polyedrons that have the same number of faces similar each to each and similarly placed, and have the homologous polyedral angles equal.

PROPOSITION XXI. THEOREM

607. Two similar polyedrons can be decomposed into the same number of tetraedrons similar each to each and similarly placed.

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Hyp. R and s are two similar polyedrons; A and A' homologous vertices.

To prove R and S can be decomposed into the same number of tetraedrons, similar each to each, and similarly placed.

Proof. Divide all the faces of R and S except those having A and A' as vertices, i.e. the faces FGHIK, GC, etc., and F'G' H'I'K', G'C", etc., into A.

Pass planes through A and the vertices of the A in the faces of R, and through A' and the vertices of A in the faces of S.

Let AKFG and A'K'F'G' be two homologous tetraedrons formed by these planes.

A-KFG is similar to A-K'F'G'.

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. A AKG ~ A'K'G'.

(Why ?) The tetraedrons A-KFG and A-K'F"G' have all homologous triedral angles equal. .. tetraedron A-KFG is similar to tetraedron A-KFG'

(606) When A-KFG and A-K'F'G' are removed, the polyedrons are similar, for their faces remain similar, and the polyedral angles are equal.

In like manner any other two homologous tetraedrons can be proved similar.

Since the homologous faces of R and S will be divided into the same number of similar A, there will be the same number of planes passed through A and A', and consequently the polyedrons will be decomposed into the same number of tetraedrons similar each to each and similarly placed. Q.E.D.

608. Cor. 1. The homologous edges of similar polyedrons are proportional.

609. Cor. 2. Any two homologous lines in two similar polyedrons have the same ratio as any two homologous edges.

610. CoR. 3. Two homologous faces of similar polyedrons are to each other as the squares of any two homologous edges.

611. Cor. 4. The total areas of two similar polyedrons are to each other as the squares of any two homologous edges. PROPOSITION XXII. THEOREM 612. The volumes of two similar tetraedrons are to each other as the cubes of their homologous edges.

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Ex. 1125. In the diagram for Prop. XXII, if SA = 3 and TE = 2, find the ratio of V to V'.

Ex. 1126. In the same diagram, find TF, if SB = 2, and V:V = 1:2.

613. Cor. The volumes of two similar polyedrons are to each other as the cubes of any two homologous edges.

Similar polyedrons can be decomposed into the same num.

ber of similar tetraedrons similarly placed which are to each other as the cubes of homologous edges.

And in a series of equal ratios, the sum of antecedents is to sum of consequents as any antecedent is to its consequent.

REGULAR POLYEDRONS

614. DEF. A regular polyedron is a polyedron whose faces are equal regular polygons, and whose polyedral angles are equal.

PROPOSITION XXIII. PROBLEM 615. To determine the number of regular convex polyedrons possible :

(a) A convex polyedral angle must have at least three faces.

(6) The sum of its face angles must be less than 360°

(523) 1. A convex polyedral angle may be formed by combining three, four, or five equilateral triangles. Since each angle of an equilateral A is 60°, the sum of six such angles is 360°, a sum greater than that of the face angles •of a convex polyedral angle (b).

Hence three regular convex polyedrons are possible with equilateral A as faces.

2. A convex polyedral angle may be formed by combining three squares.

(Why cannot four squares be used ?) Hence one regular convex polyedron is possible with squares as faces.

3. A convex polyedral angle may be formed by combining three regular pentagons.

(Why cannot four be used ?) Hence one regular convex polyedron can be formed having regular pentagons as faces.

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