17
5 diagonals * * * * * That is not correct since two of these would be lines joining the vertex to adjacent vertices (one on either side). These are sides of the polygon, not diagonals. The number of diagonals from any vertex of a polygon with n sides is n-3.
47 sides. Take a vertex of an n-sided polygon. There are n-1 other vertices. It is already joined to its 2 neighbours, leaving n-3 other vertices not connected to it. Thus n-3 diagonals can be drawn in from each vertex. For n=50, n-3 = 50-3 = 47 diagonals can be drawn from each vertex. The total number of diagonals in an n-sided polygon would imply n-3 diagonals from each of the n vertices giving n(n-3). However, the diagonal from vertex A to C would be counted twice, once for vertex A and again for vertex C, thus there are half this number of diagonals, namely: number of diagonals in an n-sided polygon = n(n-3)/2.
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
You can use the formula D=S-2 where D is the number of possible diagonals and S is the number of sides the polygon has.
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.
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.
It is 18 diagonals
Number of sides minus two equals number of diagonals drawn from one vertex.
5
17
98
5 diagonals * * * * * That is not correct since two of these would be lines joining the vertex to adjacent vertices (one on either side). These are sides of the polygon, not diagonals. The number of diagonals from any vertex of a polygon with n sides is n-3.
47 sides. Take a vertex of an n-sided polygon. There are n-1 other vertices. It is already joined to its 2 neighbours, leaving n-3 other vertices not connected to it. Thus n-3 diagonals can be drawn in from each vertex. For n=50, n-3 = 50-3 = 47 diagonals can be drawn from each vertex. The total number of diagonals in an n-sided polygon would imply n-3 diagonals from each of the n vertices giving n(n-3). However, the diagonal from vertex A to C would be counted twice, once for vertex A and again for vertex C, thus there are half this number of diagonals, namely: number of diagonals in an n-sided polygon = n(n-3)/2.
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
n-3 diagonals. Of the n vertices of the polygon, you cannot draw diagonals to the two adjacent vertices since these are sides of the polygon and so not diagonals. And you cannot draw a diagonal from a vertex to itself. So those are three vertices that are ruled out, leaving n-3.
N-2 according to yahoo answers