100 sided polygon has = n(n-3)/2 diagonals
100(100-3)/2
4850 diagonals
A drawing of diagonals for four different polygons is listed below. What is the number of diagonals for a 100-sided polygon?
2 answers
The 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.
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.