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

In today’s write-up, let us discuss n sided polygons and the number of diagonals they have.

We will talk about the following:

1. How execute we discover the variety of diagonals an n sided polygon has?2. How many type of diagonals are subtfinished by each vertex?3. What happens when one or even more vertices perform not make diagonals?4. How to take care of instances when ignored vertices are adjacent/not adjacent?

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

*Given an n sided polygon, exactly how many type of diagonals will certainly it have?*

An n sided polygon has n vertices. If we join every distinct pair of vertices we will certainly gain nC2 lines. These nC2 lines incorporate the n sides of the polygon and also its diagonals.So the variety of diagonals is offered by the expression nC2 – n.nC2 – n = n(n-1)/2 – n = n(n – 3)/2

Additionally, you can think of it this means – every vertex makes a diagonal via (n – 3) vertices. It does not make a diagonal with itself, and the 2 vertices beside it on either side (because it forms sides via these two). So we get n*(n – 3) diagonals. But here, each diagonal is double counted, once for each of its 2 vertices. Hence we divide n*(n – 3) by 2 to get the actual number of diagonals. That is an additional means of getting here at n(n – 3)/2

Let’s take some examples to solidify this concept:

**Example 1:** How many kind of diagonals does a polygon with 25 sides have?No. of diagonals = nC2 – n = n(n – 3)/2 = 25*(25 – 3)/2 = 275

**Example 2:** How many diagonals does a polygon via 20 sides have actually, if among its vertices does not sfinish any diagonal?

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

But one vertex does not sfinish any type of diagonals. Each vertex renders a diagonal with (n-3) other vertices – it provides no diagonal through 3 vertices: itself, the vertex instantly to its left, and the vertex automatically to its ideal. With all other vertices, it provides a diagonal. So, we must remove 20 – 3 = 17 diagonals from the complete.

Total variety of diagonals if one vertex does not make any type of diagonals = 170 – 17 = 153 diagonals.

We hope whatever done till now provides feeling so that we deserve to develop on the principle a little. Now let’s we will certainly give we a question with 2 options and two different answers. We have to find out the correct answer and define why the various other is wrong.

**Question:** How many kind of diagonals does a polygon with 18 sides have if three of its vertices, which are nearby to each various other, carry out not send any type of diagonals?

Answer: We will certainly usage 2 various techniques to settle this question:

**Method 1:** Using the formula disputed above

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

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

So, each vertex will certainly make 15 diagonals.

Total variety of diagonals if 3 vertices do not sfinish any diagonals = 135 – 15*3 = 135 – 45 = 90 diagonals.

**Method 2:**The polygon has a full of 18 vertices. 3 vertices do not take part so we must make all diagonals that we deserve to via 15 vertices.

Number of lines we deserve to make through 15 vertices = 15C2 = 15*14/2 = 105

But this 105 has the sides as well. A polygon through 18 vertices has 18 sides. Because 3 surrounding vertices do not take part, 4 sides will certainly not be created. 15 vertices will certainly have 14 sides which will certainly be a component of the 105 we calculated prior to.

Total variety of diagonals if 3 vertices perform not send any type of diagonals = 105 – 14 = 91

Keep in mind that the two answers perform not complement. Method 1 gives us 90 and also strategy 2 offers us 91. Both approaches look correct but just one is actually correct and the correct one is strategy 2. So then, what is the problem with strategy 1?

When we subtract 45 from 135 (for each of the 15 diagonals made by the 3 vertices which we have to ignore), we are double counting 1 diagonal in this number of 45. We actually must subtract only 44 diagonals. Which diagonal are we double counting? The one which connects 2 of the 3 ignored vertices. Try to make a polygon through a couple of vertices. Make a couple of diagonals. Rerelocate 3 vertices beside each various other. 2 of the three vertices which have actually a vertex between them will be joined by a diagonal. When we rerelocate 15 diagonals for each vertex, we are rerelocating that diagonal twice.

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

The correct answer is 91 diagonals.

Mind you, we assumed that the vertices which were removed were beside each various other. If they are not, the answer would certainly be various since there would be even more double counting. Let’s take an instance of that.

**Question:** How many diagonals does a polygon via 18 sides have actually if 3 of its non-nearby vertices, perform not send any kind of diagonals? (No 2 of the 3 vertices are next to each other)

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

Each vertex renders a diagonal through n-3 various other vertices – as questioned prior to.

So, each vertex will certainly make 15 diagonals.

Due to the fact that 3 vertices are not counted, they will certainly not sfinish out each of these 15 diagonals i.e. 45 diagonals. But in this number of 45, we have double counted 3 diagonals. To understand also this, say we number the vertices from 1 to 18. Say, we leave out vertices 1, 4 and also 16.

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

Vertex 4 makes diagonals with 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 complete of 15 diagonals)

In our complete of 45, we have counted the diagonal of vertices 1 and 4 twice. We have actually likewise counted the diagonal of vertices 1 and also 16 twice and also diagonal of vertices 4 and also 16 twice. So, we have double counted 3 diagonals in our figure of 45. We must subtract only 42 diagonals from 135.Total variety of diagonals if 3 vertices perform not sfinish any diagonals = 135 – 42 = 93 diagonals.

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The answer here is 93 diagonals.

Note that in the previous instance, just 1 diagonal was double counted while below, 3 diagonals were double counted. Understand why – in the previous instance, given that the vertices were nearby, they made sides of the polygon. So, assuming vertices 1, 2 and also 3 were ignored, 1 and also 2 joined to make a side of the polygon and 2 and 3 joined to make a side of the polygon. Only vertices 1 and 3 joined to make a diagonal and also for this reason this diagonal was double counted. In our formula n(n – 3)/2, we have currently acquired rid of the sides so they don’t come in the photo at all. The formula just gives us the variety of diagonals. So, in the previous example, we had already ignored the two sides developed by surrounding vertices (when we offered the formula n(n-3)/2) and had to take care of double counting of just one diagonal. In this instance, each ignored vertex made a diagonal through the other ignored vertex and thus we had actually to take care of the double counting of 3 diagonals.

We hope all this is clear to you and you will certainly be able to effortlessly manage any kind of question regarding diagonals of polygons currently.

*Karishma, a Computer Engineer through a keen interemainder in different Mathematical approaches, has mentored students in the continents of Asia, Europe and The United States and Canada. She teaches the **GMAT** for Veritas Prep and also frequently participates in content advance projects such as this blog!*

*Want even more practice via diagonals-of-polygons and various other geometric concepts? Check out the Veritas Prep Concern Bank for thousands of cost-free practice troubles.*