polygon

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pmFollowed by: 6 members

by BTGmoderatorRO » Sat Sep 30, 2017 7:14 pm

A polygon has 12 edges. How many different diagonals does it have?
(A) 54
(B) 66
(C) 108
(D) 132
(E) 144
oa is a

What is the best approach to solve this question. Why is the answer C and not E?

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Source: Beat The GMAT — Problem Solving | Sat Sep 30, 2017 7:14 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Oct 01, 2017 4:07 am

Roland2rule wrote:A polygon has 12 edges. How many different diagonals does it have?
(A) 54
(B) 66
(C) 108
(D) 132
(E) 144

A 12-sided polygon has 12 vertices.
Any combination of 2 NON-ADJACENT vertices can serve to form a diagonal.
From the 12 vertices, the number of combinations of 2 that can be formed = 12C2 = (12*11)/(2*1) = 66.
These 66 combinations include the 12 sides of the polygon.
Subtracting the 12 sides of the polygon from the 66 combinations that can be formed, we get:
66-12 = 54 diagonals.

The correct answer is A.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Oct 01, 2017 6:14 am

Roland2rule wrote:A polygon has 12 edges. How many different diagonals does it have?
(A) 54
(B) 66
(C) 108
(D) 132
(E) 144

Here's another approach.

The polygon in question has 12 vertices.

Let's focus on ONE VERTEX, which we'll call vertex A
How many diagonal can be drawn from vertex A?
We'll there are 11 other vertices to connect to, HOWEVER the vertices on either side of vertex A are out of contention, since joining two adjacent points will not create a diagonal.
So, we can create a diagonal with vertex A by connecting to any of the 9 eligible vertices.

Using the same logic, we can conclude that we can create a diagonal with vertex B by connecting to any of the 9 eligible vertices.

And we can create a diagonal with vertex C by connecting to any of the 9 eligible vertices.

There are 12 vertices, so the TOTAL number of diagonals = (12)(9) = 108
HOWEVER, this is not the final answer, since we have counted each diagonal TWICE.
For example, in the first step, we counted the diagonal created by joining vertex A with vertex B, and in the second step, we counted the diagonal created by joining vertex B with vertex A

To account for this duplication, we must take 108 and divide it by 2 to get: 54

Answer: A

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8149; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: polygon

by Scott@TargetTestPrep » Wed Jan 29, 2020 5:15 am

BTGmoderatorRO wrote: ↑
Sat Sep 30, 2017 7:14 pm
A polygon has 12 edges. How many different diagonals does it have?
(A) 54
(B) 66
(C) 108
(D) 132
(E) 144
oa is a

What is the best approach to solve this question. Why is the answer C and not E?

Recall that the number of diagonals of an n-sided polygon n(n - 3)/2. So the number of diagonals of a 12-sided polygon is 12(12 - 3)/2 = 6(9) = 54.

Answer: A

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