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.
7-3 = 4
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 n-sided polygon wil have n*(n-3)/2 diagonals.Consider joining each vertex to every other vertex. That gives potentially n-1 vertices. However, two of these will be sides of the polygon and so not diagonals. So each vertex gives rise to (n-3) diagonals. There are n vertices in the polygon and so that gives n*(n-3) diagonals. But, this method counts each diagonal twice: once from each end and so the correct answer is n*(n-3)/2.
That polygon is called a "triangle". It has no diagonals.
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.
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.
7-3 = 4
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 polygon with n sides, we have n(n-3)/2 diagonals. In a convex polygon with n sides, you can draw n-3 diagonals from each vertex, but you are counting each one twice you so you need to divide by do. That is why we have n(n-3) divided by 2
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
n-3 diagonals. Of the n vertices of the polygon, you cannot draw diagonals to the two adjacent vertices since these are sides of the polygon and so not diagonals. And you cannot draw a diagonal from a vertex to itself. So those are three vertices that are ruled out, leaving n-3.
An n-sided polygon wil have n*(n-3)/2 diagonals.Consider joining each vertex to every other vertex. That gives potentially n-1 vertices. However, two of these will be sides of the polygon and so not diagonals. So each vertex gives rise to (n-3) diagonals. There are n vertices in the polygon and so that gives n*(n-3) diagonals. But, this method counts each diagonal twice: once from each end and so the correct answer is n*(n-3)/2.
A hexagon has 9 diagonals. Each vertex of a n-sided polygon can be connected to n - 3 others with diagonals. Thus n(n - 3) possible diagonals. However, when Vertex A is connected to vertex C, vertex C is also connected to vertex A, thus each diagonal is counted twice. Thus: number_of_diagonals = n(n - 3)/2 = 6(6-3)/2 = 6x3/2 = 9
3
By drawing all the diagonals from one vertex, the polygon is divided up into triangles. The sum of the interior angles of the polygon is equal to the sum of the internal angles in the triangles. With n vertices, each vertex is not directly connected to n-3 other vertices, thus n-3 diagonals can be drawn from a vertex which will create n-2 triangles (each with the sum of their interior angles as 180o); so: sum_of_interior_angles = 180 x (number_of_sides - 2)
Using the formula 0.5(n^2 -3n) whereas n is number of sides, altogether there are 104 diagonals in a 16 sided polygon