There cannot be a polygon with 6 diagonals.
A polygon with n vertices has n*(n-3)/2 diagonals.
If this is 6, then n*(n-3)/2 = 6
therefore n*(n-3) = 12
and so n^2 - 3n - 12 = 0
The only solutions to this equation are n = 1.5-sqrt(57)/2 or 1.5+sqrt(57)/2.
However, n, being the NUMBER of vertices cannot be an irrational number and so there is no solution.
9
There is no such polygon that has 95 diagonals because a 15 sided polygon has 90 diagonals and a 16 sided polygon has 104 diagonals.
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.
There are three diagonal lines in a six sided Polygon(Hexagon)
A polygon that has 104 diagonals will have 16 sides
That polygon is called a "triangle". It has no diagonals.
In a polygon with n sides there are n*(n-3)/2 diagonals. For a hexagon, n = 6 giving 9 diagonals in all.
If it is a regular polygon with each interior angle of 120 degrees then it is an hexagon which will have 0.5((6^2)-3*6) equals 9 diagonals
An 11 sided polygon has 44 diagonals.
There are 90 diagonals in a 15-sided polygon.
A four-sided polygon has two diagonals.
It depends. It doesn't have to always be the same shape. So it can be 0, it can be 1, as long as it has straight lines and is a closed figure, it's a polygon. * * * * * What? 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.