There are 10 possible diagonals drawn from one vertex of the 13-gon which divide it into 11 nonoverlapping triangles.
The number of Diagonals in one vertex of a Triangle is 0 (zero)..
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.
12-3 = 9
Oh, dude, a decagon has 10 sides, right? So, if you pick one vertex and want to draw diagonals, you can draw diagonals to all the other vertices except the adjacent ones, which are already sides. So, you can draw 7 diagonals from that one vertex. Math can be fun, like a puzzle... or a headache, depending on how you look at it.
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)..
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.
8
If all of the diagonals are drawn from a vertex of an octagon, how many triangles are formed
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.
The formula for the number of diagonals is: 0.5*(n^2-3n) whereas 'n' is the number of sides of the polygon
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.