153 diagonals.
A polygon with 14 diagonals will have 7 sides
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 normal convex polygon cannot have 15 diagonals. If it has n sides, it has n*(n-3)/2 diagonals and this is equal to 15 if n = 7.18. However, it is not possible for a polygon to have a fractional side.
Apothem!
153 diagonals.
A polygon with 14 diagonals will have 7 sides
There are 560 diagonals by using the diagonal formula
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.
9 sides because a nonagon has 27 diagonals
An 11 sided polygon has 44 diagonals
There can be 14 lines in a seven side shape * * * * * That is the total number of diagonals from ALL vertices. Not what the question asked, though. From one vertex, there can be 4. One to every other vertex except for itself and one each on either side.
A normal convex polygon cannot have 15 diagonals. If it has n sides, it has n*(n-3)/2 diagonals and this is equal to 15 if n = 7.18. However, it is not possible for a polygon to have a fractional side.
n side polygon has n(n-3)/2 diagonals therefore a hexagon has 9 diagonals
Apothem!
16 diagonals* * * * *5. To all vertices except itself and one each on either side.
The altitude is usually drawn from the base.