No, I think 5 diagonals would be the maximum from a particular vertex. Let's label the vertices [A through H].
For this example, they go clockwise, starting with A and going around to H, so looking at A: B & H are adjacent and both of the segments AB and AH are actually sides of the octagon, not diagonals. The remaining diagonals, from A are:
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.
3
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.
8
No.
16 diagonals* * * * *5. To all vertices except itself and one each on either side.
The number of Diagonals in one vertex of a Triangle is 0 (zero)..
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.
There are 10 possible diagonals drawn from one vertex of the 13-gon which divide it into 11 nonoverlapping triangles.
3
Three
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.
36
six
5.