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# A polygon has 12 edges. How many different diagonals does it

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A polygon has 12 edges. How many different diagonals does it [#permalink]  30 Jul 2018, 11:06
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65% (01:03) correct 34% (00:42) wrong based on 23 sessions
A polygon has 12 edges. How many different diagonals does it have? (A diagonal is a line drawn from one vertex to any other vertex inside the given shape. This line cannot touch or cross any of the edges of the shape. For example, a triangle has zero diagonals and a rectangle has two.)

(A) 54
(B) 66
(C) 108
(D) 132
(E) 144
[Reveal] Spoiler: OA

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Re: A polygon has 12 edges. How many different diagonals does it [#permalink]  31 Jul 2018, 08:49
someone please explain how to calculate this?
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Re: A polygon has 12 edges. How many different diagonals does it [#permalink]  01 Aug 2018, 09:22
Total number of diagonals of n sided polygon is given by the formula n(n-3)/2

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Re: A polygon has 12 edges. How many different diagonals does it [#permalink]  21 Aug 2018, 18:37
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Explanation

A diagonal of a polygon is an internal line segment connecting any two unique vertices; this line segment does not lie along an edge of the given shape. Consider a polygon with 12 vertices.

Construct a diagonal by choosing any two vertices and connecting them with a line. Remember that this is order independent; the line is the same regardless of which is the starting vertex. Therefore, this is analogous to choosing any 2 elements from a set of 12, and can be written as $$\frac{12!}{10! \times 2!}= 6 \times 11 = 66$$.

However, this method includes the vertices connected to their adjacent vertices, which form edges instead of diagonals. In order to account for this, subtract the number of edges on the polygon from the above number: 66 – 12 = 54.

Alternatively, choose a random vertex of the 12-sided shape. There are 12 – 1 = 11 lines that can be drawn to other vertices since no line can be drawn from the vertex to itself. However, the lines from this vertex to the two adjacent vertices will lie along the edges of the polygon and therefore cannot be included as diagonals (see the figure of a pentagon below for an example):

Attachment:

Capture.PNG [ 116.3 KiB | Viewed 802 times ]

Thus, there are 12 – 1 – 2 = 9 diagonals for any given vertex. Since there are 12 vertices, it is tempting to think that the total number of diagonals is equal to (12)(9) = 108.

However, this scheme counts each diagonal twice, using each side of the diagonal once as the starting point. Therefore, there are half this many different diagonals: $$\frac{108}{2}= 54$$.
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Re: A polygon has 12 edges. How many different diagonals does it [#permalink]  13 Dec 2018, 03:21
n(n-1)/2 - n
12(12-1)/2 - 12
66 - 12 = 54
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Re: A polygon has 12 edges. How many different diagonals does it [#permalink]  01 Jan 2019, 23:28
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It's nice to know the diagonal formula, but in case you dont want to memorize the formula ( which will save you tons of time) then go about it this way..
a rectangle has 2 diagonals
a pentagon has 5 diagonals
a hexagon has 9 diagonals.
see the pattern?

the first increase of diagonals is from 0 to 2, then the next jump increases by 3 the next jump increases by 4, the next jump increases by 5.. and so on

diagonals 0 2 5 9 14 20 27 35 44 54
sides____ 3 4 5 6 7_ 8_ 9_ 10_ 11_ 12
_
Re: A polygon has 12 edges. How many different diagonals does it   [#permalink] 01 Jan 2019, 23:28
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