In a hexagon, each vertex can connect to all other non-adjacent vertices to form diagonals. Since a hexagon has 6 vertices, each vertex can form diagonals with the other 4 non-adjacent vertices. Therefore, each vertex of a hexagon can create 4 diagonals. This results in a total of 6 vertices x 4 diagonals = 24 diagonals in a hexagon.
None. By definition a diagonal goes from one vertex to another vertex and so each diagonal MUST have two vertices.
A. The hexagon is circumscribed about the circle . D. Each vertex of the hexagon lies outside the circle . E. The circle is tangent to each side of the hexagon .
A hexagon (six-sided polygon) can be divided into 4 triangles by drawing all of the diagonals from one vertex (only three lines can be drawn in this case, since each vertex already connects to two others on the edges of the form). If you instead drew lines from the center to each vertex, you would get 6 triangles.
it really isn't hard 5
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
Hexagon (6 sides)
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
None. By definition a diagonal goes from one vertex to another vertex and so each diagonal MUST have two vertices.
A. The hexagon is circumscribed about the circle . D. Each vertex of the hexagon lies outside the circle . E. The circle is tangent to each side of the hexagon .
A hexagon (six-sided polygon) can be divided into 4 triangles by drawing all of the diagonals from one vertex (only three lines can be drawn in this case, since each vertex already connects to two others on the edges of the form). If you instead drew lines from the center to each vertex, you would get 6 triangles.
In a regular hexagon, you can draw a total of 9 distinct lines connecting the vertices. This is achieved by connecting each vertex to every other vertex, resulting in 6 edges and 3 diagonals. Specifically, each vertex connects to 3 other vertices not adjacent to it, leading to a total of ( \binom{6}{2} = 15 ) lines, which includes the 6 edges. However, if you only consider the diagonals, there are 9 lines.
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The diagonals of a square bisect each corner or vertex of the square.
From each vertex to its opposite vertex. These will be centered on a shared point at the center of the hexagon. Each complete line will be a line of symmetry for the hexagon.
There are 9 diagonals in every hexagon.To see why, select any one of the 6 vertices. There are 5 other vertices to which it can be joined. But the lines joining it to the two adjacent vertices (one on each side) are not diagonals but sides of the hexagon. So, from each vertex, you get 3 diagonals.The first vertex can be selected in 6 ways so it would seem that you have 6*3 = 18 diagonals. However, you will then be counting each diagonal twice: once from each end. So the correct answer is 18/2 = 9.In general, a polygon with n sides has n*(n-3)/2 diagonals. Also, whether or not it is regular is totally irrelevant.
n-3 from each vertex.
There are no 'diagonals' in a triangle. Each vertex is connected to both of the other vertices, by the sides.