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03 Feb 2017, 04:05
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
58% (01:27) correct
42%
(01:45)
wrong
based on 106
sessions
History
Date
Time
Result
Not Attempted Yet
An "interior segment" of a polygon is defined as any line segment that can be drawn between two points of a polygon except for the sides of the polygon itself. If polygon P is a regular octagon, how many interior segments does P have?
(A) 20
(B) 28
(C) 36
(D) 48
(E) 56
(A) 20
(B) 28
(C) 36
(D) 48
(E) 56
04 Feb 2017, 06:12
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rishit1080
Hi rishit1080,
In this question, we have to find the number of diagonals of an octagon.
The number of diagonals of any \(n\)-sided polygon is = \(\frac{n(n-3)}{2}\)
In this case, n = 8. Number of diagonals = 8*5/2 = 20.
Alternate Method:
Please refer the attached diagram. From each vertex, we can draw five lines. For example, from vertex A we can draw AC, AD, AE, AF, and AG.
Total number of lines = 8*5 = 40 lines. Please note that each line we are counting twice, so we have to divide it by 2.
Hence, total number of unique lines = 40/2 = 20.
Hope this helps.
Attachments
Octagon.jpg [ 6.2 KiB | Viewed 3830 times ]
General Discussion
04 Feb 2017, 01:29
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SajjadAhmad
Tough for sub 600 as this question makes you draw an octagon and show the pattern so you don't get it wrong. Got A, but took time to show the different routes. 5 for first point, 5 for second point, 4 for third point, 3 for fourth point, 2 for fifth point, 1 for sixth point, and 0 for seventh and eighth points. Add them up to get 20.
