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A twelve-sided polygon consists of vertices A – L. How many
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29 Oct 2017, 01:46
29 Oct 2017, 01:46
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Question Stats:
50% (00:54) correct
50% (02:36) wrong based on 12 sessions
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A twelve-sided polygon consists of vertices A – L. How many lines can be drawn between any two vertices, such that a line is neither repeated, nor redundant, with any sides of the polygon?
(A) 45
(B) 54
(C) 55
(D) 66
(E) 110
Kudos for correct solution.
(A) 45
(B) 54
(C) 55
(D) 66
(E) 110
Kudos for correct solution.
Re: A twelve-sided polygon consists of vertices A – L. How many
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30 Oct 2017, 08:30
30 Oct 2017, 08:30
2
Let's start by computing the number of ways in which we can choose 2 points out of 12, i.e. \(12C2 = \frac{12!}{10!2!} = \frac{11*12}{2} = 66\). Then, in order to eliminate the lines that joint two adjacent vertices, i.e. sides, we have to subtract 12. Thus, we get 66 - 12 = 54. Answer B
Re: A twelve-sided polygon consists of vertices A L. How many
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01 Mar 2023, 12:44
01 Mar 2023, 12:44
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