Solutions of Exercises in Godfrey and Siddon's Solid Geometry (Paperback)

This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1912 edition. Excerpt: ...of four sides by planing off an edge of a tetrahedron. N 0 larger number of sides is possible for a section, as a tetrahedron has only four faces. Let A, A; B, B; C, C be the three pairs of opposite vertices. Then BCB C' is a rhombus with equal diagonals, i.e. a 281. 282. 283. 284. 285. Let A, A; B, B; C, C be the mid-points of opposite faces of the cube. Then the solid figure formed would be the figure whose edges are got by joining A and A to B, B, C, C: 11.12. would be a regular octahedron, from the next example. If A, A, &c. are the mid-points of the faces, then AA, BB, CC' bisect each other at O, and are mutually perpendicular. Hence all the lines joining A and A to B, B, C, C are equaL That is the figure ABCA B'C is a solid figure whose faces are equilateral triangles, and all of whose solid angles are bounded by four plane angles. It is therefore a regular octahedron. Let A, A; B, B; C, C be the mid-points of opposite edges of a regular tetrahedron. Then the figure BC B C for example is a rhombus, for each of its sides is equal to half the edge of the tetrahedron that is parallel to it. BB bisects CC at right angles, and similarly for each pair of lines AA, BB, CC . by Ex. 283 the points A, A, &c. are the vertices of a regular octahedron. If we plane off a vertex, or take a section through an edge, we get a four-sided figure. If we take a section through a vertex and cutting two adjacent faces of the four faces which form that vertex, we get a four-sided figure. If we take a section through a vertex and cutting two non-adjacent...