How many diagonals does a polygon with 21 sides have, if one

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How many diagonals does a polygon with 21 sides have, if one [#permalink]

23 Sep 2010, 02:59

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How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

A. 21
B. 170
C. 340
D. 357
E. 420

[Reveal] Spoiler:

I was considering a different approach of 20 vertices (21 vertices -1 vertex not linking) linking to 17 (21 vertices - 1 vertix I'm considering -2 vertices next to the one I'm considering - 1 vertex not linking) others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170.

Consider the parallel case in which we have 21 people who shake hands to each other.
In that case we'll have 21*20/2 handshakes.
This because, if we consider all the 21 people in a line, the first one will shake hands to 20, the second to 19, the third to 18 people and so on. So we have:
20+19+18+...+1
This sum equals to 21*20/2=210.

Handshaking and diagonals are different in counting since diagonals don't include sides.
To calculate diagonals we have to subtract the number of sides (that are 21) from 21*20/2.
So:
N° of diagonals = (21*20/2)-21=210-21=189

Since one vertex does not connect to any diagonals, we have to subtract the diagonals of this vertex.
Diagonals of one vertex equals to: 21 (number of vertex) - 1 (itself) - 2 (sides with vertices next to it).

So the answer should be: 189-18=171
That's different from 170 obtained above.

What's wrong with my second approach?

[Reveal] Spoiler: OA

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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

23 Sep 2010, 03:49

rraggio wrote:

How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

a) 21
b) 170
c) 340
d) 357
e) 420

I don't think B (170) is a correct answer.

Generally the # of diagonals in n sided polygon equals to C^2_n-n=\frac{n(n-3)}{2}: C^2_n choose any two vertices out of n to connect minus n sides, which won't be diagonals.

So, # of diagonals in 21 sided polygon is C^2_n-n=210-21=189. Since the diagonals from 1 particulat vertex shouldn't be counted then 189-(21-3)=171 (one vertex makes n-3 diagonals).

Answer: 171.
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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

23 Sep 2010, 04:04

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Take any vertex in the polygon. To form a diagnol, it can be connected to any one of 18 vertices (counting out itself, and the 2 vertices adjacent to it).

If we count diagnols like this, we will double count everything (once for every end point). So the total number of diagnols = (18*21)/2 = 189

Each vertex has 18 diagnols, so if we take all the diagnols from this one vertex out .. we are left with 171 diagnols

Alternate Solution

Consider a polygon with 20 sides. By logic similar to above, the number of diagnols is (17*20)/2 = 170

Now take a new point X and any side of the 20-sided figure AB. Join XA and XB to form a 21 sided figure. In this new figure all the old diagnols are still diagnols + the side AB also becomes a diagnol. Also note that there are no diagnols originating from X

Hence the number of diagnols = 170 + 1 (AB) = 171 diagnols
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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

23 Sep 2010, 04:05

rraggio wrote:

Actually first approach is wrong: 17*20 is not right. As 2 vertices which are next to excluded vertex can be connected each to 21-3=18 vertices, so 2*18. Other 18 vertices can be connected to 21 - 3 - 1 (excluded vertex) = 17 vertices. So we would have 2*18+18*17=342 --> divided by 2 to exclude double counting --> 342/2=171.
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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

23 Sep 2010, 06:47

shrouded1 wrote:

Consider a polygon with 20 sides. By logic similar to above, the number of diagnols is (17*20)/2 = 170

Now take a new point X and any side of the 20-sided figure AB. Join XA and XB to form a 21 sided figure. In this new figure all the old diagnols are still diagnols + the side AB also becomes a diagnol. Also note that there are no diagnols originating from X

Hence the number of diagnols = 170 + 1 (AB) = 171 diagnols

I did same way and got 170 :-D

forgot to add +1. Thanks !!
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Polygon problems a sucking combination---HELP! [#permalink]

10 Nov 2010, 06:59

Hey Guys, I need some help out here...I picked up this question from a list of questions posted by a fellow gmat geek.
I am not able to figure out out how to solve this question in terms of whether I should I directly work out a 20c2 for this question.

To sum it up please let me know how to solve problems like these in terms of what should be the basic approach.

How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

21
170
340
357
420

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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

10 Nov 2010, 07:32

Thanks guys! I appreciate your quick response!

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Re: How many diagonals does a polygon with 21 sides have... [#permalink]

10 Nov 2010, 08:17

It is good to remember that a polygon with n sides has nC2 - n diagonals and follow up from there as Bunuel did or follow shrouded1's approach, which I thought was pretty cool too.

Though, remember that if you are lost with the 21 sides, cannot think of the formula and are hard pressed for time, try and work it on a smaller scale.
Lets see what happens in a 6 sided figure which is extremely easy to draw:
6 sided figure. 1 vertex left alone. When you start from the first point, you cant join it to 3 of the 6 points - itself, point left alone and point next to it.
Total diagonals drawn: 3 + 2 + 1

Attachment:

Ques.jpg [ 18.05 KiB | Viewed 2398 times ]

21 sided figure. When you will start with the first point, you will not join it to 3 of the 21 points - itself, point left alone and point next to it.
Total diagonals drawn will be: 18 + 17 + 16 + ... +1 = 18*19/2 = 171
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Re: How many diagonals does a polygon with 21 sides have... [#permalink] 10 Nov 2010, 08:17

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