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:
https://magoosh.com/gmat/2012/polygons-a ... -the-gmat/Does this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)