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Intern
Joined: 05 Nov 2012
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

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05 Nov 2012, 16:01
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
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3301
Followers: 1121

Kudos [?]: 4914 [0], given: 54

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05 Nov 2012, 16:31
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

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