Subject: Geometrical solids
Name: Sarah Question:
I am a tutor and I unfortunately do not know the answer to either of these questions. Any help would be wonderful, thank you. Hi Sarah,There can be several answers to this type of question. It is basically testing your experience and memory (hardly testing your ability to do mathematics). However, there are a few pieces of math which are relevant. The first is Euler's Forumla for spherical polyhedra: #vertices - # edges + #faces = 2. For the first example, that means 5 - 8 + #faces = 2 so the polyhedron will have 5 faces. If all the faces were triangles, you would have 5x3 sides of edges, and each edge has two 'sides', so this does not make sense. However, four triangles and a quadrilateral give 4x3 + 4 = 16 = 2x8 sides of edges. This is, combinatorially, a possible shape. Can you figure that out now? (Try one Square face, with four of the five vertices, and then add one more vertex.) For the second example, 12 equal length edges, there are several examples. #vertices - 12 + #faces = 2. There are lots of combinatorial possibilities. Here are two: 6 vertices, 12 edges, 8 faces. 8 vertices, 12 edges, 6 faces. Both of these are possible, and occur among that well known group of polyhedra - the Platonic Solids. You could play with the number of edges at a vertex, and the number of edges around a face, and quickly identify at least one of these. If you happen to have access to the kinds of dice they use for role-playing games, you will find the Platonic Solids among them. Hope that helps. If not, web sites on Platonic Solids and polyhedra would show pictures you could check. Walter Whiteley
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