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.
It works out that a polygon with 1175 diagonals has 50 sides
1/2*(142-42) = 77 diagonals
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
38 diagonals
A polygon with n sides has n*(n - 3)/2 diagonals. So n = 23 gives 23*20/10 = 230 diagonals
9 sides because a nonagon has 27 diagonals
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.
It works out that a polygon with 1175 diagonals has 50 sides
1/2*(142-42) = 77 diagonals
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
In a 54-sided polygon, 53 possible diagonals can be drawn from one vertex to another. These diagonals will not intersect. Therefore, the interior will be divided into 54 regions by the 53 diagonals plus the two sides of the original polygon that adjoin the vertex from which the diagonals are drawn.
In a polygon with n sides, the number of diagonals that can be drawn from one vertex is given by the formula (n-3). Therefore, in a 35-sided polygon, you can draw (35-3) = 32 diagonals from one vertex.
Number of sides minus two equals number of diagonals drawn from one vertex.
In a polygon with n sides, we have n(n-3)/2 diagonals. In a convex polygon with n sides, you can draw n-3 diagonals from each vertex, but you are counting each one twice you so you need to divide by do. That is why we have n(n-3) divided by 2
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon