DEFINITION. A parallelepiped is a solid figure bounded by three pairs of parallel plane faces. THEOREM 3. (i) The faces of a parallelepiped are parallelograms, of which those which are opposite are identically equal. (ii) The four diagonals of a parallelepiped are concurrent and bisect one another. B' B A Let ABA'B' be a parped, of which ABCD, C'D'A'B' are opposite faces. (i) Then all the faces shall be parms, and the opposite faces shall be identically equal. For since the planes DA', AD' are par), XI. Def. 15. and the plane DB meets them, .. the common sections AB and DC are par! XI. 16. Similarly AD and BC are par'. the fig. ABCD is a parm, and AB=BC; also AD=BĆ. I. 34. Similarly each of the faces of the parped is a parm; so that the edges AB, C'D', B'A', DC are equal and parl: so also are the edges AD, C'B', D'A', BC; and likewise AC', BD', CA', DB'. Then in the opp. faces ABCD, C'D'A'B', we have AB=C'D' and BC=D'A'; Proved. and since AB, BC are respectively parł to C'D', D'A', the L ABC=the L C'D'A'; XI. 10. .: the parm ABCD=the parm C'D'A'B' identically. Ex. 11, p. 70. (ii) The diagonals AA', BB', CC', DD' shall be concurrent and bisect one another. Join AC and A'C'. Then since AC' is equal and par? to A'C, :: the fig. ACA'C' is a parm; :. its diagonals AA’, CC' bisect one another. Ex. 5, p. 70. That is, AA' passes through O, the middle point of CC'. Similarly if BC' and BẮC were joined, the fig. BCB'C' would be a parm; .. the diagonals BB', CC' bisect one another. Similarly it may be shewn that DD' passes through, and is bisected at, O. Q. E. D. THEOREM 4. The straight lines which join the vertices of a tetrahedron to the centroids of the opposite faces are concurrent. B Х Let ABCD be a tetrahedron, and let 91, 92, 93, 94 be the centroids of the faces opposite respectively to A, B, C, D. Then shall Agi, B92, Cg3, Dgy be concurrent. Take X the middle point of the edge CD; 91 Ex. 4, p. 113. and AX=3. Xg; :: 9192 is par to AB. And A91, Bg, must intersect one another, since they are both in the plane of the A AXB : let them intersect at the point G. Then by similar A$, AG : Gg1=AB : 9192 =AX : Xg2 3 : 1. :. Bg, cuts Ag, at a point G whose distance from g=1. Agi. Similarly it may be shewn that C9z and D94 cut Ag, at the same point; ... these lines are concurrent. Q.E.D. THEOREM 5. (i) If a pyramid is cut by planes drawn parallel to its base, the sections are similar to the base. (ii) The areas of such sections are in the duplicate ratio of their perpendicular distances from the vertex. .. Let SABCD be a pyramid, and abcd the section formed by a plane drawn parl to the base ABCD. (i) Then the figs. ABCD, abcd shall be similar. Because the planes abcd, ABCD are par?, and the plane ABba meets them, :. the common sections ab, AB are par?. Similarly is parl to BC; cd to CD; and da to DA. And since ab, bc are respectively parl to AB, BC, the L abc=the L ABC. XI. 10. Similarly the remaining angles of the fig. abcd are equal to the corresponding angles of the fig. ABCD. And since the As Sab, SAB are similar, = bc : BC, for the As Sbc, SBC are similar. Or, ab : bc=AB : BC. In like manner, bc : cd=BC: CD; and so on. .. the figs. abcd, ABCD are equiangular to one another, and have their sides about the equal angles proportional; :. they are similar. (ii) From S draw SxX perp. to the parl planes abcd, ABCD and meeting them at x and X. Then shall fig. abcd : fig. ABCD=Sx? : SX?. Join ax, AX. VI. 20. Q.E.D. DEFINITION. A polyhedron is regular when its faces are similar and equal regular polygons. THEOREM 6. There cannot be more than five regular polyhedra. This is proved by examining the number of ways in which it is possible to form a solid angle out of the plane angles of various regular polygons ; bearing in mind that three plane angles at least are required to form a solid angle, and the sum the plane angles forming a solid angle is less than four right angles. XI. 21. Suppose the faces of the regular polyhedron to be equilateral triangles. Then since each angle of an equilateral triangle is į of a right angle, it follows that a solid angle may be formed (i) by three, (ii) by four, or (iii) by five such faces; for the sums of the plane angles would be respectively (i) two right angles, (ii) % of a right angle, (iii) 130 of a right angle ; that is, in all three cases the sum of the plane angles would be less than four right angles. But it is impossible to form a solid angle of six or more equilateral triangles, for then the sum of the plane angles would be equal to, or greater than four right angles. Again, suppose that the faces of the polyhedron are squares. (iv) Then it is clear that a solid angle could be formed of three, but not more than three, of such faces. Lastly, suppose the faces are regular pentagons. (v) Then, since each angle of a regular pentagon is of a right angle, it follows that a solid angle may be formed of three such faces; but the sum of more than three angles of a regular pentagon is greater than four right angles. Further, since each angle of a regular hexagon is equal to $ of a right angle, it follows that no solid angle could be formed of such faces; for the sum of three angles of a hexagon is equal to four right angles. Similarly, no solid angle can be formed of the angles of a polygon of more sides than six. Thus there can be no more than five regular polyhedra. 6 NOTE ON THE REGULAR POLYHEDRA. (i) The polyhedron of which each solid angle is formed by three equilateral triangles is called a regular tetrahedron. (ii) The polyhedron of which each solid angle is formed by four equilateral triangles is called a regular octahedron. (iii) The polyhedron of which each solid angle is formed by five equilateral triangles is called a regular icosahedron. It has twenty faces, twelve vertices, thirty edges. |