Also, SO BC § 587 (in two similar polyhcılrons any two homologous lines are in the same ratio as uny two homologous edges). 589. COROLLARY 1. Two similar polyhedrons are to each other as the cubes of any two homologous edges. For, two similar polyhedrons may be decomposed into tetrahedrons similar, each to each, and similarly placed, of which any two homologous edges have the same ratio as any two homologous edges of the polyhedrons. And, since any pair of the similar tetrahedrons are to each other as the cubes of any two homologous edges, the entire polyhedrons are to each other as the cubes of any two homologous edges. § 266 590. Cor. 2. Similar prisms or pyramids are to each other as the cubes of their altitudes ; and similar polyhedrons are to each other as the cubes of any two homologous lines. Ex. 1. The portion of a tetrahedron cut off by a plane parallel to any face is a tetrahedron similar to the given tetrahedron. Ex. 2. Two tetrahedrons, having a dihedral angle of one equal to a dihedral angle of the other, and the faces including these angles respectively similar, and similarly placed, are similar. Ex. 3. Given two similar polyhedrons, whose volumes are 125 feet and 12.5 feet respectively; find the ratio of two homologous edges. - On Regular POLYHEDRONS. 591. DEF. A Regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhedral angles are equal. The regular polyhedrons are the tetrahedron, octahedron and icosahedron, all of whose faces are equal equilateral triangles ; the hexahedron, or cube, whose faces are squares; the dodecahedron, whose faces are regular pentagons. Only these five regular polyhedrons are possible, for a polyhedral angle must have at least three face angles, and must have the sum of its face angles less than four right angles, (§ 488). Hence : I. If the faces be equilateral triangles, polyhedral angles may be formed of them in groups of 3, 4, or 5. only, as in the tetrahedron, octahedron and icosahedron. Since each angle of an equilateral triangle is two-thirds of a right angle, the sum of six such angles is four right angles, and therefore greater than a convex polyhedral angle. II. If the faces be squares, polyhedral angles may be formed of them in groups of three only, as in the regular hexahedron, or cube ; since four such angles would be four right angles. III. If the faces be regular pentagons, polyhedral angles may be formed of them in groups of three only, as in the regular dodecahedron ; since four such angles would be greater than four right angles. IV. We can proceed no farther ; for a group of three angles of regular hexagons would equal four right angles, and of regular heptagons, etc., would be greater than four right angles. $ 232 PROPOSITION XXIV. PROBLEM. 592. Given an edge, to construct the five regular polyhedrons. Let A B be the given edge. Upon A B construct the equilateral A Find the centre 0 of this A, § 238 $ 450 $ 492 G II. To construct a regular hexahedron. Upon the given edge AB construct the square A BCD, and upon the sides of this square conC struct the squares E B, FC, G D, HAI to the plane A B C D. Then AG is the regular hexahedron required. III. To construct a regular octahedron. Upon the given edge A B construct the square A B C D. Through its centre o pass a I to its plane A B C D. ck-------- --In this I take two points E and F, one above and the other below the plane, so that A E and A Fare each equal to A B. Join E and F to each of the vertices of the square. $ 450 and hence the faces are equal equilateral A. • And, since the A D E F and D A C are equal, $ 108 DEBF is a square and the pyramid A-D EBF is equal in all its parts to the pyramid E-A B C D. Hence, the polyhedral As A and E are equal. B IV. To construct a regular dodecahedron. Upon A B construct the regular pentagon A B C D E. § 395 On each side of this pentagon construct an equal pentagon, so inclined that trihedral & shall be formed at A, B, C, D, E. The convex surface thus formed is composed of six regular pentagons. In like manner, upon an equal pentagon A' B'C' D' E' construct an equal convex surface. Apply one of these surfaces to the other, with their convexities turned in opposite directions, so that PO and P' Q' shall fall upon PO and P Q. Then every face z of the one will, with two consecutive face of the other, form a trihedral 2. The solid thus formed is the regular dodecahedron required. and the polyhedral As are all equal. § 492 DI V. To construct a regular icosahedron. At its centre 0 erect a I to its plane. Join P with each of the vertices of the pentagon ; thus forming a regular pentagonal pyramid whose vertex is P, and whose dihedral és formed on the edges PA, PB, PC, etc. are all equal. $ 571 Taking A and B as vertices, construct two pyramids each equal to the first, and having for bases B P E F G and A G HCP respectively. There will thus be formed a convex surface consisting of ten equal equilateral A. In like manner upon an equal pentagon A' B'C'D' E' construct an equal convex surface. Apply one of these surfaces to the other with their convexities turned in opposite directions, so that every combination of two face ts of the one, as P' D'C', P D' E', shall with a combination of three face s of the other, as BCH, BC P, PC D, form a pentahedral Z. The solid thus formed is the regular icosahedron required. the faces are all equal; Cons. and the polyhedral & are all equal, $ 571 Q. E. D. For, TETRAHEDRON. OCTAHEDRON. HEXAHEDRON. ICOSAHEDRON. DODECAHEDRON. 593. SCHOLIUM. The regular polyhedrons can be formed thus : Draw the above diagrams upon card-board. Cut through the exterior lines and half through the interior lines. The figures will then readily bend into the regular forms required. |