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With ganand

Attachment:
regular-octagon-with-diagonals-from-a-vertex.jpg
regular-octagon-with-diagonals-from-a-vertex.jpg [ 9.35 KiB | Viewed 3728 times ]


There are 8 points A, B,C, D, E, F, G & H

From each point say point A , there can be 5 Diagonals as -

1. AC
2. AD
3. AE
4. AF
5. AG

Now, Imagine Diagonals from point C , one of them will be AC , which is counted twice, one from Point A and the other from Point C...

Total there will be 8*5 Diagonals = 40 Diagonals

Due to counting Diagonals twice, actual no of Diagonals will be 40/2 = 20 diagonals...

Thus, answer must be (A) 20
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SajjadAhmad
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
Octagon has 8 Vertices.

So, the easiest way to do this one is by Combination. To draw a line, we need to pick 2 points. Therefore,

Number of ways to select 2 points out of 8 = 8C2

Combination because order of picking points is not important.

8 lines out of these will be the sides of the octagon.

Therefore, Total Number of "interior segment" or Diagonals = 8C2 - 8 = \(\frac{8!}{(2! * 6!)} - 8\) = 20

P.S. This is also the derivation of general formula for finding the Number of Diagonals of a closed polygon.
In other words, Formula for finding the Number of Diagonals of a closed polygon is nC2 - n
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Here, we need to find the diagonals which connects 2 points of the octagan.
Formula is : n*(n-3)/2
8*(5)/2 = 20.

Hence, A.

Hope it helps.
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