number of diagonals for an n-gon
= C(n,2) - n
e.g. for a pentagon, n=5
number of diagonals = C(5,2) - 5 = 10-5 = 5
then n!/(2!(n-2)!) = 9 , where n must be a whole number
n(n-1)/2 = 9
n^2 - n -18 = 0
There is no integer solution for this quadratic ,
therefore no polygon has 9 diagonals
what polygon has 9 diagonals?
1 answer