Polygon -Geometry

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Soumita Ghosh: Master | Next Rank: 500 Posts; Posts: 111; Joined: Sat Apr 14, 2012 11:09 amFollowed by: 7 members; Thanked: 1 times

by Soumita Ghosh » Wed Jun 27, 2012 6:28 pm

A POLYGON HAS 20 DIAGONALS, How Many Sides Does It Have?

A)12

B) 11

C) 10

D) 9

E) 8

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Source: Beat The GMAT — Problem Solving | Wed Jun 27, 2012 6:28 pm

mdavidm_531: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Mon Jun 07, 2010 3:12 am; Thanked: 10 times

by mdavidm_531 » Wed Jun 27, 2012 9:42 pm

Number of diagonals in a polygon = [n(n-3)]/2 where n is the number of sides of the polgyon

Now, we have 20 = [n(n-3)]/2 = n^2 - 3n - 40 = 0 or (n-8)(n+5) = 0

So we have two values of n: n = 8 and n = -5

Since there's no such thing as -5 sides of a polygon, we have the other value, which is n = 8

The answer is (E)

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amit28it: Senior | Next Rank: 100 Posts; Posts: 82; Joined: Fri Nov 04, 2011 12:35 amFollowed by: 2 members

by amit28it » Thu Jun 28, 2012 12:21 am

Here is the solution
Formula to find diagonals in a polygon:
D = (n(n-3)/2)
D = number of diagonals that is 20
n = ?

20=(n^2-3n)/2

40=n^2-3n
n^2-3n-40= 0
Splitting the middle term,

n^2-8n+5n-40=0
n(n-8)+5(n-8)=0
(n-8)(n+5)= 0

So there are two values for n
n-8=0 or n+5=0
n = 8 or n = -5
n can't be negative
So n = 8

5th grade math

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tutorphd: Master | Next Rank: 500 Posts; Posts: 126; Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members; Location: Chicago, IL; Thanked: 36 times

by tutorphd » Thu Jun 28, 2012 5:28 am

Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals.

You can set n(n-3)/2 = 20, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 40. Ans: n=8.

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shekhar.kataria: Master | Next Rank: 500 Posts; Posts: 268; Joined: Mon Mar 15, 2010 12:10 am; Thanked: 13 times

by shekhar.kataria » Thu Jun 28, 2012 7:40 am

Thank You Tutorphd for the explanation but do you think that this question can be a part of GMAT. till date in case of polygons, i have seen questions only abt internal angles which has a basic formula.Moreover, the guides also teach only abt the angles formula.

@poster, what is the source of the question.??

tutorphd wrote:Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals.

You can set n(n-3)/2 = 20, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 40. Ans: n=8.

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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 » Thu Jun 28, 2012 8:41 am

shekhar.kataria wrote:Thank You Tutorphd for the explanation but do you think that this question can be a part of GMAT. till date in case of polygons, i have seen questions only abt internal angles which has a basic formula.Moreover, the guides also teach only abt the angles formula.

This is really a combinatorics question in disguise; it can be solved without knowing the formula for the number of diagonals.
Check my post here:

https://www.beatthegmat.com/polygon-geom ... tml#482906

That being said, questions about polygons are rare.

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quiquent: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Thu Mar 03, 2011 1:33 pm

by quiquent » Thu Jun 28, 2012 1:59 pm

By doing Excel with formula n*(n-3) / 2, you can have an idea of the different outcomes

sides diagonals
1 -1
2 -1
3 0
4 2
5 5
6 9
7 14
8 20
9 27
10 35
11 44
12 54
13 65
14 77
15 90
16 104
17 119
18 135
19 152
20 170
21 189
22 209
23 230
24 252
25 275
26 299

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tutorphd: Master | Next Rank: 500 Posts; Posts: 126; Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members; Location: Chicago, IL; Thanked: 36 times

by tutorphd » Thu Jun 28, 2012 2:07 pm

Yes as a previous poster said, this question has very little geometry in it, it's a combinatorics in disguise. Questions like that do happen on GMAT but not as frequent as more obvious combinatorics questions. I can think of about two problems like that in OG12.

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intercostalcom: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Thu Jun 28, 2012 2:01 pm; Thanked: 1 times

by intercostalcom » Thu Jun 28, 2012 2:21 pm

If question changes to 44 , then:

A polygon has 44 diagonals, How Many Sides Does It Have?

A) 12

B) 11

C) 10

D) 9

E) 8

Consider a polygon with n vertices. Each vertex can be connected to the n-1 other vertices but two of those connections (to the two adjacent vertices) form sides of the polygon, not diagonals. So there are n-3 diagonals emanating from each of the n vertices giving n(n-3) diagonals. This formula overcounts because it counts each diagonal twice: once for each vertex on the diagonal. Divide by two to compensate for the overcounting giving n(n-3)/2 diagonals.

You can set n(n-3)/2 = 44, expand and solve the quadratic equation by factorization but this is time consuming and inefficient. Way faster is to simply plug in the answer choices for n and see which one satisfies the equation: n(n-3) = 88. Ans: n=11.
The equation a*x^2 + b*x + c = 0 has as solutions: x = ( -b + sqrt(b^2-4*a*c) ) / (2*a) or x = ( -b - sqrt(b^2-4*a*c) )

44=(n^2-3n)/2

88=n^2-3n

So there are two values for n
n = 11 or n = -8
n can't be negative
So n = 11

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