Revision as of 12:59, 28 January 2009 by Mgiordan (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


In this problem, I considered the fact that in a convex polygon, diagonals do not connect to three of the other sides. For example in an octagon, at one point there are five connecting diagonals. Therefore the amount of diagonals for the entire octagon would be 8(5), but to account for overcount you have to divide by two. I believe this holds true for all convex polygons, therefore my equation was (n(n-3))/2.

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn