The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
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.
Two triangles can be formed by the diagonals of one vertex of a four-sided 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.
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.
An hexagon
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.
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.
Two triangles can be formed by the diagonals of one vertex of a four-sided 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.
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.
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.
Heptagon
An hexagon
An eight-sided polygon has 8 vertices. Consider1st vertex = 7 diagonals to other vertices2nd vertex = 6 diagonals to other vertices (since one has now been used)3rd vertex = 54th vertex = 45th vertex = 36th vertex = 27th vertex = 18th vertex = 0, so7+6+5+4+3+2+1 = 28Improved Answer:-Formula for finding diagonals of a polygon 0.5*(n2-3n) when n is the number of sides0.5*(82-24) = 20 diagonals
It is 18 diagonals
Number of sides minus 2 equals the number of triangles within the polygon.