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Difficulty:
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Question Stats:
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69%
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wrong
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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. 171
D. 357
E. 420
A. 21
B. 170
C. 171
D. 357
E. 420
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?
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?
10 Nov 2010, 08:17
Kudos
Bookmarks
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
Ques.jpg [ 18.05 KiB | Viewed 43095 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
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 43095 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
23 Sep 2010, 03:49
Kudos
Bookmarks
rraggio
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.
