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How many vertices does a polygon have, if each vertex is the intersect

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How many vertices does a polygon have, if each vertex is the intersect  [#permalink]

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New post 16 Oct 2016, 03:44
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A
B
C
D
E

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63% (00:51) correct 37% (01:06) wrong based on 56 sessions

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How many vertices does a polygon have, if each vertex is the intersect  [#permalink]

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New post 16 Oct 2016, 04:27
Bunuel wrote:
How many vertices does a polygon have, if each vertex is the intersection of exactly three diagonals?

A. 3
B. 4
C. 5
D. 6
E. No such polygon exists.




if each vertex is the intersection of exactly three diagonals i.e. from one point 3 diagonals can be drawn

therefore total no. of diagonals = 3*x/2 for a polygon of x sides (we divide it by 2 because every diagonal has been counted twice

now 3x/2 = xC2-x (where xC2 represents total no. of diagonals)

now, 3x/2 = x(x-1)/2 - x
i.e. 5x/2 = x(x-1)/2
i.e. x = 6

Answer: Option D
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Re: How many vertices does a polygon have, if each vertex is the intersect  [#permalink]

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New post 16 Oct 2016, 04:46
I solved this Q with' pick and choose' method. Let me know if this correct?
Number of vertices mentioned here is given as 3,4,5 or 6.
Since we know that each vertex makes a diagonal with (n-3) other vertices – it makes no diagonal with 3 vertices: itself, the vertex immediately to its left, and the vertex immediately to its right therefore we can look out for options.

i) A polygon with three sides have zero diagonal
ii) A polygon with 4 sides will have two diagonals
iii) A polygon with 5 sides will have two diagonals
iv) A polygon with 6 sides will have three diagonals -> This exactly matches our options.
Hence D
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Re: How many vertices does a polygon have, if each vertex is the intersect   [#permalink] 16 Oct 2016, 04:46
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