Diagonals of an n-Sided Polygon | Geometry

In today’s post, let us discuss $n$ sided polygons and the number of diagonals they have.

We will discuss the following:

1. How do we find the number of diagonals an $n$ sided polygon has?

2. How many diagonals are subtended by each vertex?

3. What happens when one or more vertices do not make diagonals?

4. How to handle cases when ignored vertices are adjacent/not adjacent?

First of all, a very basic question:

Given an ${n}$ sided polygon, how many diagonals will it have?

An $n$ sided polygon has $n$ vertices. If we join every distinct pair of vertices we will get $^nC_2$ lines. These $^nC_2$ lines include the $n$ sides of the polygon as well as its diagonals.

So the number of diagonals is given by $^nC_2 - n$ .

$^nC_2 - n = \frac{n(n-1)}{2} - n = \frac{n(n - 3)}{2}$

Alternatively, think of it this way – every vertex makes a diagonal with $(n - 3)$ vertices. It does not make a diagonal with itself, and the two vertices next to it on either side (since it forms sides with these two). So we get $n\times(n - 3)$ diagonals. But here, each diagonal is double counted, once for each of its two vertices. Hence we divide $n\times(n - 3)$ by 2 to get the actual number of diagonals. That is another way of arriving at $\frac{n(n - 3)}{2}$

Let’s take some examples to solidify this concept:

Example 1: How many diagonals does a polygon with 25 sides have?

No. of diagonals = $^nC_2 - n = \frac{n(n - 3)}{2} = 25 \times \frac{(25 - 3)}{2} = 275$

Example 2: How many diagonals does a polygon with 20 sides have, if one of its vertices does not send any diagonal?

The number of diagonals of a 20 sided figure $= 20\times \frac{(20 - 3)}{2} = 170$

But one vertex does not send any diagonals. Each vertex makes a diagonal with $(n-3)$ other vertices – it makes no diagonal with 3 vertices: itself, the vertex immediately to its left, and the vertex immediately to its right. With all other vertices, it makes a diagonal.

So, we need to remove $20 - 3 = 17$ diagonals from the total.

Total number of diagonals if one vertex does not make any diagonals $= 170 - 17 = 153$ diagonals.

We hope everything done till now makes sense. Now let’s we will give we a question with two solutions and two different answers. We have to find out the correct answer and explain why the other is wrong.

Question: 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?

Solution: We will use two different methods to solve this question:

Method 1: Using the formula discussed above

Number of diagonals in a polygon of 18 sides $= 18\times \frac{(18 - 3)}{2} = 135$ diagonals

Each vertex makes a diagonal with $n-3$ other vertices – as discussed before.

So, each vertex will make 15 diagonals.

Total number of diagonals if 3 vertices do not send any diagonals $= 135 - 15\times3 = 135 - 45 = 90$ 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 we can make with 15 vertices $=\ ^{15}C_2 = 15 \times \frac{14}{2} = 105$

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 and the correct one is method 2. So then, what is the problem with method 1?

When we subtract 45 from 135 (for each of the 15 diagonals made by the 3 vertices which we need to ignore), we are double counting 1 diagonal in this figure of 45. We actually need to subtract only 44 diagonals. Which diagonal are we double counting? The one which connects 2 of the three ignored vertices. Try to make a polygon with a few vertices. Make a few diagonals. Remove 3 vertices next to each other. 2 of the three vertices which have a vertex between them will be joined by a diagonal. When we remove 15 diagonals for each vertex, we are removing that diagonal twice.

Hence, what we need to do is $135 - 44 = 91$ .

The correct answer is 91 diagonals.

Mind you, we assumed that the vertices which were removed were next to each other. If they are not, the answer would be different since there would be more double counting. Let’s take an example of that.

Question: How many diagonals does a polygon with 18 sides have if three of its non-adjacent vertices, do not send any diagonals? (No two of the three vertices are next to each other)

Solution: Number of diagonals in a polygon of 18 sides $= 18 \times \frac{(18 - 3)}{2} = 135$ diagonals

Each vertex makes a diagonal with $n-3$ other vertices – as discussed before.

So, each vertex will make 15 diagonals.

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

Vertex 1 makes diagonals with 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 total of 15 diagonals)

Vertex 16 makes diagonals with vertices 18, 1, 2, 3, 4, … 15, 16 (a total of 15 diagonals)

In our total of 45, we have counted the diagonal of vertices 1 and 4 twice. We have also counted the diagonal of vertices 1 and 16 twice and diagonal of vertices 4 and 16 twice. So, we have double counted 3 diagonals in our figure of 45. We need to subtract only 42 diagonals from 135.

Total number of diagonals if 3 vertices do not send any diagonals $= 135 - 42 = 93$ diagonals.

Answer here is 93 diagonals.

Note that in the previous example, only 1 diagonal was double counted while here, 3 diagonals were double counted. Understand why – in the previous example, since the vertices were adjacent, they made sides of the polygon. So, assuming vertices 1, 2 and 3 were ignored, 1 and 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 hence this diagonal was double counted. In our formula $\frac{n(n - 3)}{2}$ , we have already gotten rid of the sides so they don’t come in the picture at all. The formula only gives us the number of diagonals. So, in the previous example, we had already ignored the two sides formed by adjacent vertices (when we used the formula $\frac{n(n-3)}{2}$ ) and had to take care of double counting of only one diagonal. In this example, each ignored vertex made a diagonal with the other ignored vertex and hence we had to handle the double counting of 3 diagonals.

We hope all this is clear to you and you will be able to effortlessly handle any question regarding diagonals of polygons now.

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