Difference between revisions of "Diagonal"

(diagonal formula, someone please fix the proof.)
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[[Triangle]]s have no diagonals while [[convex polygon|convex]] [[quadrilateral]]s have two interior diagonals and [[concave]] quadrilaterals have on interior and one exterior diagonal.
 
[[Triangle]]s have no diagonals while [[convex polygon|convex]] [[quadrilateral]]s have two interior diagonals and [[concave]] quadrilaterals have on interior and one exterior diagonal.
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The number of diagonals in a polygon can be determined by the formula <math>\frac{n(n-3)}{2}</math>.
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To prove this is quite simple. We have an n-gon and we choose one point. From this point, there are <math>(n-3)</math> diagonals we can make form this point; the ones we don't count are itself and the two vertices that form sides with it. There are n vertices, so for each of these there are <math>(n-3)</math> diagonals, thus, there are <math>n(n-3)</math> diagonals. However, we are not done. When counting our diagonals, we counted some twice; we used the same vertices, just backwards. (i.e. In quadrilateral ABCD, diagonals AC and CA are the same.) So we simply divide by 2, to get our final formula, <math>\frac{n(n-3)}{2}</math>
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The number of edges plus the number of diagonals of a polygon with n vertices is equal to <math>n \choose 2</math>
 
The number of edges plus the number of diagonals of a polygon with n vertices is equal to <math>n \choose 2</math>
 
==Polyhedra==
 
==Polyhedra==

Revision as of 10:58, 13 July 2006

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

Polygons

A diagonal of a polygon is any segment joining two vertices other than an edge.

Triangles have no diagonals while convex quadrilaterals have two interior diagonals and concave quadrilaterals have on interior and one exterior diagonal. The number of diagonals in a polygon can be determined by the formula $\frac{n(n-3)}{2}$.

To prove this is quite simple. We have an n-gon and we choose one point. From this point, there are $(n-3)$ diagonals we can make form this point; the ones we don't count are itself and the two vertices that form sides with it. There are n vertices, so for each of these there are $(n-3)$ diagonals, thus, there are $n(n-3)$ diagonals. However, we are not done. When counting our diagonals, we counted some twice; we used the same vertices, just backwards. (i.e. In quadrilateral ABCD, diagonals AC and CA are the same.) So we simply divide by 2, to get our final formula, $\frac{n(n-3)}{2}$

The number of edges plus the number of diagonals of a polygon with n vertices is equal to $n \choose 2$

Polyhedra

Polyhedra have two different kinds of diagonals, face diagonals and space diagonals. A face diagonal of a polyhedron is a diagonal of one of the faces of the polyhedron, while a space diagonal is any segment joining two vertices which is neither an edge nor a face diagonal.

Tetrahedra have no space or face diagonals. Octahedra have no face diagonals but have 3 space diagonals. Cubes have 12 face diagonals (2 on each face) and 4 space diagonals. The number of edges plus the number of face diagonals plus the number of space diagonals of a polyhedron with n vertices is equal to $n \choose 2$.

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