Just one diagonal will divide a hexagon into two halves
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47 sides. Take a vertex of an n-sided polygon. There are n-1 other vertices. It is already joined to its 2 neighbours, leaving n-3 other vertices not connected to it. Thus n-3 diagonals can be drawn in from each vertex. For n=50, n-3 = 50-3 = 47 diagonals can be drawn from each vertex. The total number of diagonals in an n-sided polygon would imply n-3 diagonals from each of the n vertices giving n(n-3). However, the diagonal from vertex A to C would be counted twice, once for vertex A and again for vertex C, thus there are half this number of diagonals, namely: number of diagonals in an n-sided polygon = n(n-3)/2.
hexagon
The length of on side of an equilateral hexagon is half the diameter of the circumscribing circle.
A regular hexagon folded in half would look like an isosceles trapezoid.
The diagonals bisect each other at right angles. So you could use Pythagoras on half the diagonals. So, if the diagonals are a and b units long, then half the diagonals are a/2 and b/2 units long. Then, by Pythagoras, the sides of the rhombus are sqrt[(a/2)2 + (b/2)2]