SOLUTION: if polygons of n sides has 1/2n(n-3) diagonals, how many sides will a polygon with 65 diagonals have? is there a ploygon 80 diagonal

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Question 750743: if polygons of n sides has 1/2n(n-3) diagonals, how many sides will a polygon with 65 diagonals have? is there a ploygon 80 diagonal
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Let n be the number of sides = number of vertices.
Then each vertex can be connected to n-3 other vertices, giving n%28n+-3%29 connections. However, this counts each vertex pair connection twice, so there are:
n%28n+-+3%29+%2F+2 diagonals.
Thus if there are 65 diagonals:
n%28n+-+3%29+%2F+2+=+65
n%5E2+-+3n+=+65%2A2
n%5E2+-+3n+-+130+=+0...write -3n as 10n-13n
n%5E2%2B10n-13n-130=0...group
%28n%5E2%2B10n%29-%2813n%2B130%29=0
n%28n%2B10%29-13%28n%2B10%29=0
%28n-13%29%28n%2B10%29+=+0
so n+=+13 or n=-10
As a polygon cannot have a negative number of sides, n+=+13.



if there are 80 diagonals:
n%28n+-+3%29+%2F+2+=+80
n%5E2+-+3n+=+80%2A2
n%5E2+-+3n+-+160+=+0...use quadratic formula

n+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

n+=+%28-%28-3%29+%2B-+sqrt%28%28-3%29%5E2-4%2A1%2A%28-160%29+%29%29%2F%282%2A1%29+
n+=+%283+%2B-+sqrt%289%2B640%29+%29%29%2F2+
n+=+%283+%2B-+sqrt%28649%29+%29%29%2F2+
n+=+%283+%2B-+25.48+%29%29%2F2+
solutions:
n+=+%283+%2B+25.48+%29%29%2F2+
n+=+28.48%2F2+
n+=+28.48%2F2+
n+=+14.24+
so, since n+=+14.24 and n is equal to the number of sides, means that a polygon cannot have a number of sides as decimal number
there is no a polygon with 80 diagonals