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.
A polygon that has 104 diagonals will have 16 sides
There are three diagonal lines in a six sided Polygon(Hexagon)
That polygon is called a "triangle". It has no diagonals.
An 11 sided polygon has 44 diagonals.
There are 90 diagonals in a 15-sided polygon.
A four-sided polygon has two 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
No