four * * * * * The correct answer is 3. You cannot have a diagonal from a vertex to itself, nor to either of the two adjacent vertices (these would form sides of the polygon). So 3 out of the other vertices cannot be used. In a hexagon, that leaves 3 that can be used. Hence the answer.
Just one diagonal will divide a hexagon into two halves
110.Improved Answer:-It is 4
six
An n-gon has n(n-3)/2 total diagonals. You can draw n-3 diagonals from each vertex ( n>3) ( A triangle doesn't really have a diagonal) An alternative way of seeing this: from any vertex, you can draw a diagonal to any other vertex except itself and the immediate neighbour on either side (the latter would be sides of the n-gon). This gives n-3 diagonals.
3
four * * * * * The correct answer is 3. You cannot have a diagonal from a vertex to itself, nor to either of the two adjacent vertices (these would form sides of the polygon). So 3 out of the other vertices cannot be used. In a hexagon, that leaves 3 that can be used. Hence the answer.
Just one diagonal will divide a hexagon into two halves
110.Improved Answer:-It is 4
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.
six
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.
13 The correct answer is 12. From any one vertex, you can draw a diagonal to all but 3 vertices: the vertex itself and the next vertex on either side of your vertex (these would be sides of your shape, not diagonals).
false
In a regular pentagon, the lines of symmetry are drawn from each vertex to the midpoint of the edge directly opposite the vertex, so there are five in all.
An n-gon has n(n-3)/2 total diagonals. You can draw n-3 diagonals from each vertex ( n>3) ( A triangle doesn't really have a diagonal) An alternative way of seeing this: from any vertex, you can draw a diagonal to any other vertex except itself and the immediate neighbour on either side (the latter would be sides of the n-gon). This gives n-3 diagonals.
That would depend on which hexagon and what triangles. A small hexagon might not have room for any large triangles. A large hexagon will have room fro many small triangles.If you have a regular hexagon and connect the vertices you will have drawn six equilateral triangles