Mythizle wrote:

So This is Part of the Question:

State a rule to determine the number of diagonals that can be drawn from a chosen vertex (d) given the number of sides(n).

For Example

3 Sides = 0 Diagonals from one vertex

4 = ?

5 = ?

6 = ?

7 = ?

15 ?

45 = ?

? = 82

I need the rule :S

I'm happy to help with this.

In a standard n-sided polygon, from any one vertex, we cannot draw a diagonal back to that same vertex, or two either of the two adjacent vertices. For example, in hexagon ABCDEF, from D we cannot draw diagonals to C & D & E, and we can draw diagonals to A & B & F. Three forbidden vertices.

Therefore, for each of the n vertices, there are (n - 3). That counts every diagonal twice (once at each end, so we need divide by two.

The number of diagonals in an n-sided polygon is

number of diagonals = [n*(n-3)]/2

n= 4 ---> 2 diagonals

n= 5 ---> 5 diagonals

n= 6 ---> 9 diagonals

n= 7 ---> 14 diagonals

n= 8 ---> 20 diagonals

n= 9 ---> 27 diagonals

n= 15 ---> 90 diagonals

n= 45 ---> 945 diagonals

BTW, you may find this post informative:

http://magoosh.com/gmat/2012/polygons-a ... -the-gmat/Does this make sense?

Mike

_________________

Mike McGarry

Magoosh Test Prep