5.
5
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.
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.
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
An octagon has 8 vertices which is the plural of vertex
5
40
No.
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:ACADAEAFAG
If all of the diagonals are drawn from a vertex of an octagon, how many triangles are formed
6 triangles because there are 180 degrees in a triangle and 6*180 = 1080 degrees which is the sum of the angles in octagon when added up.
it really isn't hard 5
The number of Diagonals in one vertex of a Triangle is 0 (zero)..
5 triangles are formed.
You Can Get 6 triangles
Number of sides minus two equals number of diagonals drawn from one vertex.
16 diagonals* * * * *5. To all vertices except itself and one each on either side.