Does anyone know how to find the sides of a polygon?

I need help!
A polygon has 135 diagonals.
How many sides does it have?

say the polygon has n sides. then it also has n vertices. the number of line segments we can draw between any two of those n points is "n choose 2" = n!/[(n-2)!*2!] = n*(n-1)/2. in general, "n choose k" is the number of ways to choose k things from a set of n things; in this case we have n points, and every time we draw a line segment we are choosing 2 of the points to connect.

now, n of those line segments are the sides of the polygon itself; all the others are what we call diagonals. so the total number of diagonals in an n-gon is

[n*(n-1)/2] - n.

we have to solve the equation

[n*(n-1)/2] - n = 135.
n^2 - 3*n - 270 = 0

(n-18)*(n+15)=0
the negative root is not valid, of course (a polygon can't have -15 sides!), so n=18 is the solution.

135 = n(n - 3) / 2
n^2 - 3n - 270 = 0
(n - 18)(n + 15) = 0
n > 0 yori, n = 18

answer = 18 sides

#If you have any other info about this subject , Please add it free.#

16 March 2010 | cameltoepants.com | edit