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# How many diagonals does a polygon with 18 sides have if three of its

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How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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Updated on: 16 Oct 2016, 06:32
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95% (hard)

Question Stats:

35% (01:54) correct 65% (01:29) wrong based on 195 sessions

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How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?

A. 10
B. 25
C. 58
D. 90
E. 91

Originally posted by Manonamission on 16 Oct 2016, 05:53.
Last edited by Bunuel on 16 Oct 2016, 06:32, edited 1 time in total.
Renamed the topic and edited the question.
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How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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16 Oct 2016, 06:10
1
1
For any n sided polygon, there are $$\frac{n*(n-3)}{2}$$ diagonals.

There are 18 sides or 18 vertices of which 3 vertices don't send out any diagonals.

Therefore, the number of diagonals is $$\frac{15*12}{2} = 90$$ (Option D)
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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17 Oct 2016, 05:54
Can someone provide the OA?
I think its D and not E.
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18 Oct 2016, 09:27
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pushpitkc wrote:
For any n sided polygon, there are (n*n-3)/2 diagonals.

There are 18 sides or 18 vertices of which 3 vertices don't send out any diagonals. Hence, number of diagonals is (15*12)/2 = 90(Option D)

Polygon has 18 vertices and 3 vertices do not send a diagonal. This means the number of vertices which will send diagonals is 15.

Total number of lines possible with 15 vertices = C(15,2) = 105 ; using combination formula

[Total number of lines = Number of sides+ Number of possible diagonals ]

Polygon Number of vertices Polygon Number of sides

18 vertices 18 sides

15 vertices 14 sides {since 3 vertices do not send a diagonal they cannot form a side.
Hence the count will be 14 and not 15}

Total number of diagonals if 3 vertices do not send any diagonal = 105 – 14 = 91
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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30 Oct 2016, 19:37
Can you please elaborate how 15 vertices are leading to 14 sides. I got it diagrammatically but it will be great if you can elaborate on some more details.
Further why 90 is a wrong answer?
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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30 Oct 2016, 22:03
Manonamission wrote:
pushpitkc wrote:
For any n sided polygon, there are (n*n-3)/2 diagonals.

There are 18 sides or 18 vertices of which 3 vertices don't send out any diagonals. Hence, number of diagonals is (15*12)/2 = 90(Option D)

Polygon has 18 vertices and 3 vertices do not send a diagonal. This means the number of vertices which will send diagonals is 15.

Total number of lines possible with 15 vertices = C(15,2) = 105 ; using combination formula

[Total number of lines = Number of sides+ Number of possible diagonals ]

Polygon Number of vertices Polygon Number of sides

18 vertices 18 sides

15 vertices 14 sides {since 3 vertices do not send a diagonal they cannot form a side.
Hence the count will be 14 and not 15}

Total number of diagonals if 3 vertices do not send any diagonal = 105 – 14 = 91

please explain this part
15 vertices 14 sides {since 3 vertices do not send a diagonal they cannot form a side.
Hence the count will be 14 and not 15}

Total number of diagonals if 3 vertices do not send any diagonal = 105 – 14 = 91
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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22 Apr 2017, 03:31
1
1
Total number of diagonals with no restriction:

(18*(18-3))/2 = 135

Restriction: three of its vertices, which are adjacent to each other, do not send any diagonals:

Each versicle sends 15 diagonals (15 = 18 - 1(the vertice) - 2(adjacent vertices)). Thus 3 vertices DO NOT send 15*3 = 45 diagonals.

Then, out of these 3 vertices, 2 are not adjacent to each other, thus they share 1 diagonal (double counted in 45).

Total number of diagonals: 135 - (45-1) = 91 -> E.
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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27 Apr 2017, 13:35
1
Manonamission wrote:
How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?

A. 10
B. 25
C. 58
D. 90
E. 91

Here is my simple approach. There are 18 points. Let us pick up any 1 point. Lets call it A. Now there are 4 points to which A can't be connected (3 adjacent as per the question + 1 adjacent on the other side of A). Therefore we can have 18 - 5 (3+1+1{A itself})=13 diagonals from A.
Similarly The one on the other side of A which we left out can't be joined to 3,A,itself and 1 more adjacent to it. Therefore it will have 18-6=12 diagonals.

Therefore it's nothing but a simple AP 13+12+11+...+1=13*14/2=91
Ans = E
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How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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17 Jun 2017, 00:24
1
Manonamission wrote:
How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?

A. 10
B. 25
C. 58
D. 90
E. 91

OFFICIAL EXPLANATION FROM VERITAS PREP

We will use two different methods to solve this question:

Method 1:

Number of diagonals in a polygon of 18 sides = 18*(18 – 3)/2 = 135 diagonals

Each vertex makes a diagonal with n-3 other vertices.

So each vertex will make 15 diagonals.

Total number of diagonals if 3 vertices do not send any diagonals = 135 – 15*3 = 90 diagonals.

Method 2:

The polygon has a total of 18 vertices. 3 vertices do not participate so we need to make all diagonals that we can with 15 vertices.

Number of lines you can make with 15 vertices = 15C2 = 15*14/2 = 105

But this 105 includes the sides as well. A polygon with 18 vertices has 18 sides. Since 3 adjacent vertices do not participate, 4 sides will not be formed. 15 vertices will have 14 sides which will be a part of the 105 we calculated before.

Total number of diagonals if 3 vertices do not send any diagonals = 105 – 14 = 91

Note that the two answers do not match. Method 1 gives us 90 and method 2 gives us 91. Both methods look correct but only one is actually correct. Your job is to tell us which method is correct and why the other method is incorrect.
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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17 Jun 2017, 04:43
1
Total number of diagonals possible= 18*(18-3)/2= 135

Now we have to subtract the diagonals which the three adjacent vertices would have formed.

Please note that these are adjacent vertices. Let's name them a, b and c. Such that the vertex b is connected to a and c.

Let's start counting
vertex a has 15 diagonals (out of which one diagonal is with vertex c)
vertex b has 15 diagonals
vertex c has 15 diagonals out of which one we have already counted
Therefore the answer is 135-15-15-14= 91

Kudos if this helps!!!!!!

What if the vertex were not adjacent? What would be the answer?
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Re: How many diagonals does a polygon with 18 sides have if three of its  [#permalink]

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21 Jun 2017, 10:34
1
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Easy but tricky question.
We can easily find out the total diagonals of an 18 sided polygon using the formula n(n-3)/2. Which will give us the no. of diagonals = 135.
Now if you draw any polygon and try to assess the total diagonals one point makes is number of sides - 3. For e.g., take a hexagon so no. diagonals a point makes in a hexagon = 6-3 =3.
So, for 18 sided figure, one point will make 15 diagonals and 3 points will make 45 total which we are supposed to deduct from total no. of diagonals.
But the catch is two points out of 3 will make one common diagonal which we are subtracting twice. So we need to add 1 to the final answer.
Therefore answer will be 135-45+1 =91.
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22 Jul 2018, 00:51
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