for each of the five Platonic solids count the number of vertices, the number of faces and the edges check that in each case vertices-Edges+Faces=2i know its somsething simple but i fail to...

for each of the five Platonic solids count the number of vertices, the number of faces and the edges check that in each case vertices-Edges+Faces=2

i know its somsething simple but i fail to understand what to do

Asked on by dotta12345

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merefletch | Elementary School Teacher | (Level 1) Adjunct Educator

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Faces are the flat surfaces of three dimensional figures. Edges are the lines where two faces meet. Vertices are the points where multiple edges meet. The five Platonic solids are solids in which the regular (congruent or equal sides) polyons (multisided closed figure) of the figure are all the same and the number of edges at each vertex is the same. They are the tetrahedron, cube, octahedron, dodcahedron, and icosahedron. The number of edges is always greater than either the number of faces or the number if vertices; therefore, the formula in your question will not work. However, (faces + vertices) - edges does, in fact, equal 2. So the number of vertices, edges, and faces for each and proof of (faces + vertices) - edges = 2 are as follows: Tetrahedron Vertices = 4 Edges = 6 Faces= 4 (4+4)-6=2 Cube Vertices = 8 Edges = 12 Faces= 6 (6+8)-12=2 Octahedron Vertices = 6 Edges = 12 Faces= 8 (8+6)-12=2 Dodcahedron Vertices = 20 Edges = 30 Faces= 12 (12+20)-30=2 Icosahedron Vertices = 12 Edges = 30 Faces= 20 (20+12)-30=2
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