a heptagon has 7 sides. you cannot draw a diagonal to the 2 adjacent vertices, so 7-2 = 5. there would be 5 diagonals.
Three fewer than the total number of vertices.
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
Three
Depends on the shape.
Number of sides minus two equals number of diagonals drawn from one vertex.
The number of Diagonals in one vertex of a Triangle is 0 (zero)..
There are 10 possible diagonals drawn from one vertex of the 13-gon which divide it into 11 nonoverlapping triangles.
If you are merely drawing from one vertex to all the others, the number of triangles in any n-sided figure is equal to n-2. In this case, a heptagon has seven sides, and thus (7 - 2) = 5 triangles can be drawn.
8
If all of the diagonals are drawn from a vertex of an octagon, how many triangles are formed
A diagonal of a polygon is a segment drawn from one vertex to another non-adjacent vertex in a polygon. This leaves 32 diagonals that can be drawn from one vertex in a 35 sided polygon.
Three fewer than the total number of vertices.
If you mean "How many diagonals can be drawn from one vertex of a figure with 16 sides", the formula is n-3, where "n" being the number of sides of the figure. So 16-3 = 13 diagonals that can be drawn from one vertex.
No.
It is 18 diagonals
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.