Consider a regular polygon with n sides (and n vertices).
Select any vertex. This can be done in n ways.
There is no line from that vertex to itself.
The lines from the vertex to the immediate neighbour on either side is a side of the polygon and so a diagonal.
The lines from that vertex to any one of the remaining n-3 vertices is a diagonal.
So, the nuber of ways of selecting the two vertices that deefine a diaginal seem to be n*(n-3). However, this process counts each diagonal twice - once from each end.
Therefore a regular polygon with n sides has n*(n-3)/2 diagonals.
Now n*(n-3)/2 = 4752
So n*(n-3) = 9504
that is n2 - 3n - 9504 = 0
using the quadratic equation, n = [3 + sqrt(9 + 4*9504)]/2 = 99 sides/vertices.
The negative square root in the quadratic formula gives a negative number of sides and that answer can be ignored.
It is: 0.5*(144-36) = 54 diagonals
For a polygon with n sides, there would be n*(n-3)/2
how many diagonals are there iin a polygon of 11 sides
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
That polygon is called a "triangle". It has no diagonals.
It has 9 sides and it can be an regular or a irregular polygon which is called a nonagon Check: 0.5*(81-27) = 27 diagonals
It is: 0.5*(144-36) = 54 diagonals
pentagon
For a polygon with n sides, there would be n*(n-3)/2
how many diagonals are there iin a polygon of 11 sides
A polygon that has 104 diagonals will have 16 sides
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
That polygon is called a "triangle". It has no diagonals.
A 7 sided polygon has 14 diagonals
38 diagonals
Let the number of sides be x and by solving the equation for diagonals 0.5(x*x-3x) = 135 the solution is -15 or 18 and so therefore it has 18 sides irrespective if it is an irregular or a regular polygon
There are 77 diagonals