SOLUTION: An n-sided polygon has (n(n - 3))/2 diagonals. a How many sides has a polygon with 665 diagonals? b Why can’t a polygon have 406 diagonals?

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Question 1086341: An n-sided polygon has
(n(n - 3))/2 diagonals. a How many sides has a polygon with 665 diagonals? b Why can’t a polygon have 406 diagonals?
Found 2 solutions by MathLover1, htmentor:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

An n-sided polygon has
n%28n+-+3%29%2F+2 diagonals
if a polygon has 665 diagonals, we have
n%28n+-+3%29%2F+2=665
n%28n+-+3%29=665%2A2
n%5E2+-+3n=1330
n%5E2+-+3n-1330=0...factor
%28n+-+38%29+%28n+%2B+35%29+=+0
solutions:
%28n+-+38%29+=+0=>n=38
%28n+%2B+35%29+=+0=>n=-35=> disregard negative solution if looking for number of the sides

b. Why can’t a polygon have 406 diagonals?
n%28n+-+3%29%2F+2=406}
n%28n+-+3%29=406%2A2}
n%5E2+-+3n=812}
n%5E2+-+3n-812=0}...can't be factored, so 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%28-812%29+%29%29%2F%282%2A1%29+
n+=+%283+%2B-+sqrt%28+9%2B3248+%29%29%2F2+
n+=+%283+%2B-+sqrt%28+3257+%29%29%2F2+
n+=+%283+%2B-+57.07%29%2F2+
solutions: use positive only
n+=+%283+%2B+57.07%29%2F2+
n+=+60.07%2F2+
n+=+30.035+=> decimal number cannot be solution to number of sides, it have to be an integer because a polygon has integer number of sides





Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Number of diagonals = d = (n/2)(n-3) -> n^2 - 3n - 2d = 0
a) Solve for n, when d = 665
n^2 - 3n - 1330 = 0
This can be factored as (n-38)(n+35) = 0
We take the positive solution, n = 38
For d = 406, we have the equation n^2 - 3n - 812 = 0
This does not give an integer solution for n. Thus no polygon can have 406 diagonals.