If a polygon of n sides has (n/2)(n - 3) diagonals, how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals?
Must I equate (n/2)(n - 3) to 65 and then again to 80 and solve for n individually?
(n/2)(n - 3) = 65
2[(n/2)(n - 3)] = 65(2)
n(n - 3) = 130
n^2 - 3n = 130
n^2 - 3n - 130 = 0
(n - 13)(n + 10) = 0
n - 13 = 13
n + 10 = -10 is rejected.
A polygon with 65 diagonals has 13 sides.
What about a polygon with 80 diagonals?
(n/2)(n - 3) = 80
n(n - 3) = 160
n^2 - 3n - 160 = 0
After using the quadratic formula, I got the following values for n:
n = (3/2) - [(sqrt{649})/2]
n = (3/2) + [(sqrt{649})/2]
The question now is: which one is correct?
Irrelevant. The discriminant must equal a positive integer
in order to get a valid solution; sqrt(649) <> integer.
So answer here is: no polygon has 80 diagonals.
Let d = number of diagonals.
d = n(n-3) / 2
d = (n^2 - 3n) / 2
So (n^2 - 3n) / 2 must equal an integer.
Since minimum n = 3 (triangle, 3 sides), you get this series:
n,d
3,0 (triangle)
4,2 (rectangle)
5,5 (5 sided)
6,9 (6 sided
7,14 (7 sided)
....and so on.
That's what I meant by:
If a polygon has n sides, then n must be an positive integer (greater than or equal to 3 because the smallest polygon has three sides).When you've done the solving, check your answers back in the context of the question.
Both values you have found are NOT integers (the first one BTW is negative so that's out for two reasons).
So even though you have found the solutions to the equation, it does not answer the question. Because of this, you can conclude that there is no polygon with 80 diagonals.
Again, in word problems especially, always take your answer back to the context of the question.