Question 705592: if the sum of the measure of the angles of a apolygon is 2340 how many diagonals does it have?
how do you do this problem...
Answer by josgarithmetic(39630)   (Show Source):
You can put this solution on YOUR website! This is a (regular) polygon of 15 sides. If you wanted to, you could literally draw this on paper, and individually draw and count all of the diagonals. The harder way is to do this for just the first simpler regular polygons and try to develop a formula, and then use it to compute the number of diagonals for a polygon of 15 sides.
You can figure the number of sides using 180*(n-2) being the sum of the interior angles of a regular polygon (which you can find by using any polygon cut into separate triangles cutting at exactly one vertex.
MORE:
This was difficult to analyze but it seems to give me something that may work.
___n_________diagonals
___3_________0
___4_________2
___5_________5
___6_________9
___7_________14
There seems to be a (n-3) happenig.
Adjusting the n=4, the n-3 give a 1, and so 4/2=2.
At n=5, the n-3 gives 2, so to get 5 diags, I tried 5/2 times this n-3.
I believe but not yet sure, that n(n-3)/2 may give the number of diagonals in a polygon of n sides.
Just to keep checking, will this formula give me 14 when n=7?
7(7-3)/2=7(4)/2=7(2)=14, okay here.
What about n=6?
6(6-3)/2=6(3)/2=3*3=9, okay here too.
To be more certain, maybe a proof my mathematical induction?
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