Suppose a polygon has n vertices (and sides). From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex. Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals. number of diagonals = n(n-3)/2
Suppose a polygon has n vertices (and sides).From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex.Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals.number of diagonals = n(n-3)/2
Number of diagonals = 1/2*(n2-3n) where n = the number of sides of the polygon.
You can dertimine a number of vertices a polygon has by counting all the dots around the shape
You can find the number of diagonals in a polygon using the formula n(n-3)/2, where n is the number of sides. Therefore an 11 sided polygon has 44 diagonals.
No, as long as the polygon is convex.
None.A polygon is made up of straight line edges between its vertices. There are the same number of edges as vertices in a polygon.In the case of a polygon, it is convex if all interior angles are less than 180o.
Suppose a polygon has n vertices (and sides). From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex. Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals. number of diagonals = n(n-3)/2
10 ... any polygon it is 2 less than the number of sides or vertices wince they are the same.
The five sided Polygon has 5 diagonals
Using the formula 0.5(n^2 -3n) whereas n is number of sides, altogether there are 104 diagonals in a 16 sided polygon
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.
Suppose a polygon has n vertices (and sides).From each vertex, a diagonal can be drawn to all vertices, excluding itself and the two adjacent vertices. So n-3 diagonals can be drawn from each vertex.Multiplying by the full complement of n vertices gives n(n-3). However, as things stand we have counted each diagonal twice: once at both ends. Dividing by two gives the actual number of diagonals.number of diagonals = n(n-3)/2
A 7 sided polygon has 14 diagonals
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.
A regular pentagon is convex. By taking a regular pentagon and shortening or lengthening one or more sides, an infinite number of possible convex pentagons can be created. A convex polygon is defined as a polygon such that all internal angles are less than or equal to 180 degrees, and a line segment drawn between any two vertices remains inside the polygon. It is possible to have non-convex (concave) pentagons; there are infinite number possible ways to do this, too.
It is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon