2006-11-22 03:31:50 · 4 answers · asked by paul b 1 in Science & Mathematics ➔ Mathematics
This sounds like plane geometry where you have an n sided polygon and want to know how many diagonals you can draw. From each vertex, you can draw a diagonal to every other vertex except to yourself and the two next to you (those two line make sides of the polygon not diagonals). So for n sides (which means n vertices), you can draw n-3 diagonals at each vertex. This gives (n-3)*n. But this just counts the starting end of each line. Whenever you draw one of these lines, it goes to another vertex. If we drew (n-3)*n lines, we would traces over each diagonal twice, once from each end so there are only half as many diagonals as that.
The answer then is: (n-3)*n/2
2006-11-22 04:28:02 · answer #1 · answered by Pretzels 5 · 0⤊ 0⤋
eulers formula:-
V-E+F=2
where E=number of edges
V=number of vertices
F=number of faces
i hope that this helps
2006-11-22 04:03:09 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Go on Sam Learning, www.samlearning(.co.uk or .com) if that doesn't work go on google and type in Sam Learning.
2006-11-22 03:43:39 · answer #3 · answered by Daree 2 · 0⤊ 0⤋
Do you mean Euler's formula?
2006-11-22 03:38:59 · answer #4 · answered by raz 5 · 0⤊ 1⤋