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

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)

Can someone provide the OA?
I think its D and not E.

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?

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

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

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.

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?

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.

Posted from my mobile device

Approach:

Total Number of diagonals:\(\frac{n(n-3)}{2}\) => \(\frac{(18 * 15)}{2}\)= 135

Total number of diagonals which could have sent from 3 adjacent vertices : 15+15+14 = 44

Please check the attached image, for representation, point A & C will have one common diagonal counted each in numbers for both A & C. So we'll reduce the count by 1 from either of A or C.

Hence, our final result would be : 135 - 44 = 91

Option E

Hope it helps!

Asked: 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?

number of sides = Number of vertices - 1 = n-4
The number of such diagonals = \(^{n-3}C_2 - (n-4) = \frac{(n-3)(n-2)}{2} +4-n = \frac{15*14}{2} + 4 - 18 = 105 -14 = 91\)

IMO E

I solved it like this:

Since 3 vertices cannot make diagonals, total vertices = 18-3 = 15

Consider this to be a 15-side polygon.
Therefore, total number of diagonals in a 15-side polygon = [n(n-3)]/2 = 15*12/2 = 90

However, in this case, 1 side of this 15-sided polygon will actually be a diagonal since its originally a 18-sided polygon. This is the side that is made by the 2 vertices adjacent to the 3 vertices that send no diagonals and which has not been counted in the 90 diagonals.

Thus, total number of diagonals = 90+1 = 91.

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