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.
A square or other rhombus (= parallelogram with 4 equal sides, whether or not they're at right angles).Also, any other regular 4n-gon -- i.e., regular polygon whose number of sides is a multiple of 4 -- will have pairs of perpendicular diagonals. So a regular octagon has two such pairs.
It works out that a polygon with 1175 diagonals has 50 sides
The formula is: 0.5*(n2-3n) where n is the number of sides of the polygon
If the sides and angles are equal, yes. Otherwise, no.
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
For a polygon with n sides, there would be n*(n-3)/2
A polygon that has 104 diagonals will have 16 sides
how many diagonals are there iin a polygon of 11 sides
That polygon is called a "triangle". It has no diagonals.
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
A 7 sided polygon has 14 diagonals
There are 77 diagonals
An heptagon has 7 sides and 14 diagonals