SOLUTION: If a polygon of n sides has (1/2)(n)(n - 3) diagonals, how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals?
Let me see.
I understan
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Let me see.
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Question 1207480: If a polygon of n sides has (1/2)(n)(n - 3) diagonals, how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals?
Let me see.
I understand this problem leading to this equation:
For first equation, you have
= 65
n*(n-3) = 2*65
n*(n-3) = 130
and the solution / (the answer) is, OBVIOUSLY, n= 13.
It also can be done solving quadratic equation formally.
For second equation, you have
= 80
n*(n-3) = 2*80
n*(n-3) = 160.
Try to solve it formally as a quadratic equation,
and make sure that it does not have solution in integer positive numbers.
It means that such a polygon with precisely 80 diagonals does not exist.
(n/2)(n - 3) = 65 solves to n = -10 and n = 13
Ignore the negative answer. A negative number of sides doesn't make sense.
n = 13 is the only solution
Therefore a 13 sided polygon has 65 diagonals.
(n/2)(n - 3) = 80 produces two non-integer solutions (roughly n = -11.2377 and n = 14.2377)
This will mean there aren't any polygons that have 80 diagonals.
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