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How many diagonals does a polygon with 21 sides have?

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by MBA2010HereWeGo » Tue Sep 22, 2009 2:47 pm
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

I am confused...pls help!
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by sanjib » Tue Sep 22, 2009 3:04 pm
Clue- Its a combination question.
How about solving this question:
how many triangle could be drawn from a polygon with n amount of sides.?
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by mjsobo » Tue Sep 22, 2009 5:10 pm
sanjib wrote:Clue- Its a combination question.
How about solving this question:
how many triangle could be drawn from a polygon with n amount of sides.?
multiply 20*17 to get 340. (17 bc it is the number of sides -3). Next divide by 2 because each line connects to two vertices. so 340/2 = 170

ignore the 21st side that does not connect to.

hope this helps and happy testing!
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by aa2kash » Wed Sep 23, 2009 2:21 pm
Remember the general formulas

Number of diagonals can be drawn from 1 vertex in n sided polygon is (n - 3)

Total number of diagonals that can be drawn is n(n-3)/2

as per ur question n is 20. as one of its vertices doesn't connect.
Hope it helps.
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by sanjib » Sun Sep 27, 2009 11:13 am
How many diagonals does a polygon with 20 sides have, if one of its vertices does not connect to any diagonal
Is it 20.17/2
or 19.16/2
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by rajiishere » Tue Sep 29, 2009 8:01 am
Since your question says that the polygon has 20 sides and one of its vertices does not connect to a diagonal, it should be 19x16/2...
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by sanju09 » Thu Apr 01, 2010 4:24 am
MBA2010HereWeGo wrote: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

I am confused...pls help!
An n-sided convex polygon has nC2 different line segments that could possibly be drawn, out of which n are the sides of the polygon, and hence the total number of diagonals can be given by

nC2 - n = [n (n - 1)/2] - n = n (n - 3)/2

AB and BA are two different representations of the same line segment, call it AB or BA; the division by 2 is hence there in the resulting formula. If it's known that each of the n vertices has exactly (n - 3) one-way-read diagonals to its name, then the n vertices would have a total of n (n - 3) two-way-read diagonals or just n (n - 3)/2 diagonals to name. When one vertex does not participate in the diagonal formation, its share of exactly (n - 3) one-way-read diagonals is out from the total, and the remaining number of diagonals can be given by

[n (n - 3)/2] - (n - 3) = (n - 2) (n - 3)/2.

We have, n = 21, so our answer must be (21 - 2) (21 - 3)/2 = [spoiler]171, oopsy!![/spoiler]
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by eaakbari » Thu Apr 01, 2010 5:42 am
Number of diagonals in any polygon can be arrived it

A vertex can form diagonals with all other vertices in the polygon but its neighbouring vertices and itself
that makes n-3 for each vertex

Number of vertices = n
hence n (n-3)

but each diagonal will be counted twice hence the formula for a n sided polygon will be

n(n-3)/2

Now for 21 vertices 21(18)/2 = 189
Since one vertex is not participating we subtract the number of diagonals it forms
It forms n-3 that is 18 diagonals
Hence 189-18 = 171
Answer is 171
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by eaakbari » Thu Apr 01, 2010 5:46 am
Remember the general formulas

Number of diagonals can be drawn from 1 vertex in n sided polygon is (n - 3)

Total number of diagonals that can be drawn is n(n-3)/2

as per ur question n is 20. as one of its vertices doesn't connect.
Hope it helps.
You cannot ignore one of the vertices
Take an example of a pentagon
It has 5 sides
Apply formula and you obtain 5 which is the number of diagonals
If you say one vertex is not participating and ignore it you get diagonals as two which is like a quad
but actually there are 3 left (draw it and try it out)

Hence answer is 171
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