A decagon.
Proof:
In an n sided polygon, each vertex would be have (n-3) diagonals attached to it, as it would be connected to every vertex other than itself and the two next to it by a diagonal. There are n sides, so there are n(n-3) ends of diagonals. Therefore there are (n(n-3))/2 diagonals in the polygon.
Taking the number of diagonals to be 35, we have:
(n(n-3))/2 = 35
n(n-3) = 70
which gives the quadratic
n2-3n-70 = 0
Solving this gives n = 10 and -7. -7 can be ignored, so the answer is 10.
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.
There are 560 diagonals by using the diagonal formula
A polygon that has 104 diagonals will have 16 sides
A regular octadecagon (an eighteen sided polygon) would have nine diagonals.
38 diagonals
It has 35 diagonals
35 diagonals
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.
(35*(35-3))/2 (35*32)/2 1120/2 560 diagonals
There are 560 diagonals by using the diagonal formula
It will have ten sides
It has 10 sides because using the formula 0.5*(102-30) = 35 diagonals
The number of diagonals is n(n-1)/2 - n substitute n=35 35(34)/2 - 35 = 560
100 diagonals * * * * * No, it is 0.5*10*(10-3) = 35
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
There is no such polygon that has 95 diagonals because a 15 sided polygon has 90 diagonals and a 16 sided polygon has 104 diagonals.
A polygon that has 104 diagonals will have 16 sides