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.
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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