Art of Problem Solving

Euler's Polyhedral Formula: Difference between revisions

Temperal (talk | contribs)
Arb95 (talk | contribs)
No edit summary
Line 1: Line 1:
Let <math>P</math> be any [[convex]] [[polyhedron]], and let <math>V</math>, <math>E</math> and <math>F</math> denote the number of [[vertex|vertices]], [[edge]]s, and [[face]]s, respectively. Then <math>V-E+F=2</math>.
Let <math>P</math> be any [[convex]] [[polyhedron]], and let <math>V</math>, <math>E</math> and <math>F</math> denote the number of [[vertex|vertices]], [[edge]]s, and [[face]]s, respectively. Then <math>V-E+F=2</math>.
==Observe!==
Apply Euler's Polyhedral Formula on the following polyhedra:
<math> \begin{tabular}{|c|c|c|c|}\hline Shape & Vertices & Edges & Faces\\ \hline Tetrahedron &4  &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline \end{tabular} </math>


== See Also ==
== See Also ==


* [[Euler characteristic]]
* [[Euler characteristic]]


{{stub}}
{{stub}}

Revision as of 07:59, 23 June 2009

Let $P$ be any convex polyhedron, and let $V$, $E$ and $F$ denote the number of vertices, edges, and faces, respectively. Then $V-E+F=2$.

Observe!

Apply Euler's Polyhedral Formula on the following polyhedra:

$\begin{tabular}{|c|c|c|c|}\hline Shape & Vertices & Edges & Faces\\ \hline Tetrahedron &4  &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline \end{tabular}$

See Also

This article is a stub. Help us out by expanding it.