How many diagonals does a regular 11 -sided polygon contain?

Question

How many diagonals does a regular 11-sided polygon contain?
 
A. 35
 
B. 38
 
C. 40
 
D. 44
 
E. 55

souvonik2k · Accepted Answer

11-sided polygon contains 11 vertices which can be joined with each other in 11C2 ways=55 ways.
But this includes the sides of the polygon also, hence subtracting 11, gives 44.
Answer D.

BrentGMATPrepNow · Answer

A nice/fast approach that doesn't involve any counting techniques is to recognize that, for each of the 11 points (vertices), there are 8 possible points that we can connect to to create a diagonal.

ASIDE : there are 8 possible points because we cannot create a diagonal by connecting 2 adjacent points. 
So, there are 8 possible diagonals for each of the 11 points
These means that there are 88 possible diagonals in total (8x11=88). 
 However , we need to recognize that this method counts each diagonal twice. For example, it counts the diagonal AB and the diagonal BA as 2 separate diagonals.

So, to account for this duplication, we'll divide 88 by 2 to get 44
Answer:  D

Cheers, 
Brent

Abhishek009 · Answer

No of Diagonals of a Polygon =  \frac{n(n - 3)}{2}

No of Diagonals of a Polygon =  \frac{11(11 - 3)}{2}

No of Diagonals of a Polygon =  44 
  
Hence, answer will be (D) 44

suramya26 · Answer

Straight 15 second answer!!!!!!!!!!!!!!!!!
Its D i.e. 44

The formula for the number of diagonals for an n sided polygon is represented as  nC2-n 
where n represents the number of sides of the polygon

Here, n=11
so by placing the value of n we get the number of diagonals as 44

JeffTargetTestPrep · Answer

Given a vertex, in order to form a diagonal, we need to choose a vertex other than the vertex itself and the two adjacent vertices; thus, we have 8 possibilities for a diagonal from any given vertex. This idea is true for all 11 vertices; however, we must divide out the overlap since each diagonal is counted twice.

Thus, the number of diagonals created from an 11-sided polygon is (11 x 8) / 2 = 44 diagonals.

Answer: D

Indraksh · Answer

No. of diagonals of a regular n sided polygon is nc2-n

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vineet6316 · Answer

Approach 1: Excluding the side lengths from the total line segments =  nc2 - n

Approach 2: Every vertex can yield n-3 diagonals (excluding itself and two adjacent points). So, n vertices would yield n(n-3) diagonals. However, it has double counting since the diagonal from A to B is essentially the same as that from B to A. Hence,  no. of unique diagonals n(n-3)/2

himanshu1105 · Answer

Simply use this formula: n(n-3)/2 = 11*(11-3)/2 = 44

Seryozha · Answer

Here we see that in the polygon containing 11 angles, the angle B have only 8 diagonals in  RED   ( 2 closest sides plus 1 itself are ignored). 
We see that the angle C also has 8 diagonals because there is no overlap. However, we also see that overlap begins when we draw diagonals from the angle A (in the blue color). 
Therefore, visually and logically we can conclude that to count the number of diagonals without counting any overlap, we will have the following:
8+8+7+6+5+4+3+2+1=44

Salsanousi · Answer

11c2 - 11 or 11(11-3)/2

11*10/2 = 55 - 11 = 44

11 * 8/2 = 11*4 = 44

Posted from my mobile device

AnishTrivedi99 · Answer

There are 11 sides.

A diagonal connects any 2 points in the interior.

11 c 2 = 55

But these also contain the 11 edges of the polygon.

Hence, 55 - 11 = 44

How many diagonals does a regular 11 -sided polygon contain?

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How many diagonals does a regular 11 -sided polygon contain?

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