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You are watching: How many diagonals does a 17-gon have

In today’s post, permit us comment on n face polygons and also the variety of diagonals lock have.

We will discuss the following:

1. How do we discover the number of diagonals one n sided polygon has?2. How plenty of diagonals are subtended by every vertex?3. What happens as soon as one or an ext vertices execute not make diagonals?4. How to handle instances when ignored vertices space adjacent/not adjacent?

First that all, let’s start with a relatively basic question:

*Given an n sided polygon, how plenty of diagonals will it have?*

An n sided polygon has actually n vertices. If we join every unique pair that vertices us will gain nC2 lines. These nC2 lines incorporate the n political parties of the polygon and also its diagonals.So the variety of diagonals is given by the expression nC2 – n.nC2 – n = n(n-1)/2 – n = n(n – 3)/2

Alternatively, you can think of the this way – every vertex makes a diagonal through (n – 3) vertices. It does no make a diagonal v itself, and the two vertices next to it on either side (since it creates sides with these two). For this reason we acquire n*(n – 3) diagonals. But here, each diagonal is twin counted, as soon as for each of its two vertices. For this reason we divide n*(n – 3) through 2 to acquire the actual variety of diagonals. That is another means of getting here at n(n – 3)/2

Let’s take some instances to solidify this concept:

**Example 1:** How many diagonals go a polygon v 25 sides have?No. Of diagonals = nC2 – n = n(n – 3)/2 = 25*(25 – 3)/2 = 275

**Example 2:** How numerous diagonals go a polygon through 20 sides have, if one of its vertices does no send any diagonal?

The variety of diagonals of a 20 sided figure = 20*(20 – 3)/2 = 170

But one vertex does no send any type of diagonals. Each vertex makes a diagonal through (n-3) various other vertices – it provides no diagonal through 3 vertices: itself, the vertex automatically to that is left, and also the vertex immediately to its right. Through all other vertices, it renders a diagonal. So, we should remove 20 – 3 = 17 diagonals from the total.

Total variety of diagonals if one peak does no make any diagonals = 170 – 17 = 153 diagonals.

We hope everything done till currently makes sense so that we can build on the ide a bit. Now let’s us will provide we a inquiry with 2 solutions and two various answers. We have actually to find out the exactly answer and also explain why the other is wrong.

**Question:** How many diagonals walk a polygon through 18 sides have if 3 of the vertices, i beg your pardon are surrounding to every other, carry out not send any kind of diagonals?

Answer: we will use two different methods to settle this question:

**Method 1:** utilizing the formula discussed above

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

Each vertex makes a diagonal v n-3 various other vertices – as discussed before.

So, each vertex will make 15 diagonals.

Total number of diagonals if 3 vertices do not send any type of diagonals = 135 – 15*3 = 135 – 45 = 90 diagonals.

**Method 2:**The polygon has a total of 18 vertices. 3 vertices execute not take part so we need to make every diagonals that we can with 15 vertices.

Number of lines we can make v 15 vertices = 15C2 = 15*14/2 = 105

But this 105 includes the sides together well. A polygon with 18 vertices has 18 sides. Since 3 adjacent vertices carry out not participate, 4 sides will not be formed. 15 vertices will have actually 14 sides which will certainly be a component of the 105 us 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. An approach 1 gives us 90 and method 2 gives us 91. Both methods look correct but only one is actually correct and the exactly one is method 2. Therefore then, what is the problem with method 1?

When us subtract 45 native 135 (for every of the 15 diagonals made by the 3 vertices which we should ignore), we are twin counting 1 diagonal in this figure of 45. Us actually should subtract only 44 diagonals. I m sorry diagonal room we twin counting? The one which connects 2 the the 3 ignored vertices. Shot to do a polygon with a few vertices. Make a couple of diagonals. Remove 3 vertices alongside each other. 2 the the 3 vertices which have actually a vertex between them will certainly be join by a diagonal. Once we eliminate 15 diagonals because that each vertex, we room removing the diagonal twice.

Hence, what we must do is 135 – 44 = 91.

The exactly answer is 91 diagonals.

Mind you, us assumed the the vertices i beg your pardon were gotten rid of were beside each other. If they room not, the answer would certainly be different since there would be more twin counting. Let’s take an instance of that.

**Question:** How many diagonals walk a polygon through 18 sides have if three of the non-adjacent vertices, carry out not send any kind of diagonals? (No 2 of the three vertices are beside each other)

Number the diagonals in a polygon that 18 political parties = 18*(18 – 3)/2 = 135 diagonals

Each vertex provides a diagonal through n-3 various other vertices – as discussed before.

So, every vertex will certainly make 15 diagonals.

Since 3 vertices space not counted, they will certainly not send the end each of these 15 diagonals i.e. 45 diagonals. However in this figure of 45, us have twin counted 3 diagonals. To know this, say we number the vertices native 1 come 18. Say, us leave out vertices 1, 4 and 16.

Vertex 1 provides diagonals v vertices 3, 4, 5 … 16, 17 (a complete of 15 diagonals)

Vertex 4 renders diagonals through vertices 6, 7, 8, … 16, 17, 18, 1, 2 (a full of 15 diagonals)

Vertex 16 renders diagonals through vertices 18, 1, 2, 3, 4, … 15, 16 (a full of 15 diagonals)

In our total of 45, we have counted the diagonal line of vertices 1 and also 4 twice. We have also counted the diagonal line of vertices 1 and also 16 twice and also diagonal of vertices 4 and 16 twice. So, us have twin counted 3 diagonals in our number of 45. We have to subtract just 42 diagonals native 135.Total variety of diagonals if 3 vertices do not send any kind of diagonals = 135 – 42 = 93 diagonals.

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The answer right here is 93 diagonals.

Note that in the ahead example, just 1 diagonal line was twin counted while here, 3 diagonals were twin counted. Understand why – in the ahead example, due to the fact that the vertices were adjacent, lock made sides of the polygon. So, presume vertices 1, 2 and also 3 to be ignored, 1 and also 2 joined to do a next of the polygon and also 2 and 3 joined to do a next of the polygon. Just vertices 1 and also 3 joined to make a diagonal and hence this diagonal was twin counted. In our formula n(n – 3)/2, we have already gotten rid that the political parties so they nothing come in the picture at all. The formula only offers us the number of diagonals. So, in the ahead example, us had already ignored the two sides developed by nearby vertices (when we supplied the formula n(n-3)/2) and had to take care of twin counting of only one diagonal. In this example, each ignored vertex do a diagonal v the various other ignored vertex and hence we had to take care of the double counting of 3 diagonals.

We hope all this is clean to you and also you will be able to effortlessly handle any question about diagonals of polygon now.

*Karishma, a computer Engineer v a keen interest in alternative Mathematical approaches, has actually mentored college student in the continent of Asia, Europe and also North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content advancement projects such as this blog!*

*Want more practice with diagonals-of-polygons and other geometric concepts? examine out the Veritas prepare Question financial institution for hundreds of totally free practice problems.*