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August 1, 2014

August 1, 2014

Posted by **dan** on Tuesday, March 30, 2010 at 11:46pm.

- math -
**miley**, Wednesday, March 31, 2010 at 12:17am100 sided polygon has = n(n-3)/2 diagonals

100(100-3)/2

4850 diagonals

- math -
**tchrwill**, Wednesday, March 31, 2010 at 3:22pmThe number of diagonals in the first series of polygons are

Number of sides...........n = 3....4....5....6....7....8

Number of diagonals.....N = 0....2....5....9...14..20

1st Difference.......................2....3....4....5....6

2nd Difference.........................1....1....1....1

We therefore, have a finite difference sequence with the 2nd differences constant at 1. This means that the general expression for the number of diagonals in any n-gon is of the form N = an^2 + bn + c.

Using the data, we can write

a(3^2) + b(3) + c = 0 or 9a + 3b + c = 0

a(4^2) + b(4) + c = 2 or 16a + 4b + c = 2

a(5^2) + b(5) + c = 5 or 25a + 5b + c = 5

Solving this set of equations leads us to a = 1/2, b = -3/2, and c = 0 resulting in N = n^2/2 - 3n/2 = n(n - 3)/2.

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