Question 1202866
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Answer:  <font color=red size=4>170</font>



Reason:


Let n be positive integer such that {{{n >= 3}}}. 
The value of n can be selected from this set {3, 4, 5, 6, ...}


Consider a convex polygon with n sides.
Select a vertex at random. That particular vertex connects to n-3 other vertices to form n-3 diagonals.
We subtract off 3 because we exclude the selected vertex itself and its two adjacent neighbors. 
Do this for all n vertices to have n(n-3) diagonals. 
But wait, we've double counted things so we must divide by 2. 
The true count of diagonals will be n(n-3)/2


I recommend testing that formula with shapes like squares, pentagons and hexagons.
I also recommend to draw the accompanying diagram to help cement your understanding.
Take careful note that the expression n(n-3) will always be even. The proof is left for the reader.


For a polygon with n = 20 sides there are 
n(n-3)/2 = 20*(20-3)/2 = <font color=red size=4>170</font> different diagonals
The diagram for this would be very messy to draw out, so I don't recommend it.
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