Three.
5
The diagonals of a square bisect each corner or vertex of the square.
Three
n-3 from each vertex.
how many diagonals from a vertex a heptagon have
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
There are no 'diagonals' in a triangle. Each vertex is connected to both of the other vertices, by the sides.
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.
A diagonal is normally defined as a straight line joining a vertex of a polygon with any vertex other than an adjoining vertex (lines joining a vertex to an adjoining vertex would simply be a side of the polygon). Since a triangle has only got adjoining vertices, it has no diagonals. Since there are no diagonals, they cannot bisect one another.
it really isn't hard 5
None. By definition a diagonal goes from one vertex to another vertex and so each diagonal MUST have two vertices.