n(n-3)/2
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A polygon with n sides had n*(n-3)/2 diagonals. So a 20 sided polygon would have 20*17/3 = 170
Such a polygon does not exist. A polygon with n sides has 0.5*n*(n-3) diagonals If there are 10 diagonals then 0.5*n*(n-3) = 10 which requires n2 - 3n - 20 = 0 which has no integer roots.
For a polygon with n sides, there would be n*(n-3)/2
A normal convex polygon cannot have 15 diagonals. If it has n sides, it has n*(n-3)/2 diagonals and this is equal to 15 if n = 7.18. However, it is not possible for a polygon to have a fractional side.
A polygon with n sides has n*(n-3)/2 diagonals. Simple as that. So a 9 sided polygon has 9*6/2 = 27 diagonals.