You are watching: How many edges does a dodecahedron have

Its is clear that their space $20$ vertices, $30$ edges, and $12$ faces. Ns am not sure exactly how to prove this though.

For vertices, there room $12$ encounters times $5$ vertices per face but since each face is associated to $3$ vertices that is counted 3 times. Therefore, $V = 12 imes 5 div 3 = 20$.

For edges, there space $12$ encounters times $5$ edges per face but since each edge joins $2$ deals with it is counted twice. Therefore, $E = 12 imes 5 div 2 = 30$.

Do you view the pattern? shot the same exercise with a tetrahedron and an octahedron to check out if you obtain it.

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What space the formulas because that the variety of vertices, edges, faces, cells, 4-faces, ..., $n$-faces, of convex continual polytopes in $n geq 5$ dimensions?

Probability the connecting 2 vertices in a dodecahedron will result in a segment inside the dodecahedron

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