Continuing our Geometry journey, let’s discuss polygons today. Some years back, I used to often get confused in the polygon sum-of-the-interior-angles formula if I had to recall it after a gap of some months because I had seen two variations of it:

Sum of interior angles of a polygon = (n – 2)*180

Sum of interior angles of a polygon = (2n – 4)*90

Now, I don’t want you to judge me. Of course, in the second formula, 2 has been removed from 180 and multiplied to the first factor. It is quite simple so why would anyone get tricked here, you wonder? The problem was that after a few months, I would somehow remember (2n – 4) and 180. So I was mixing up the two and I wasn’t sure of the logic behind this formula. That is until I came across the simple explanation of this formula in our Veritas Prep Geometry book (the one which explains how you can divide every polygon with n sides into (n – 2) triangles and hence get the sum of (n – 2)*180). Now it made perfect sense! I couldn’t believe that I had not come across that explanation before and had just learned up (well, tried to!) the formula blindly. So now I ensure that all my students understand every formula that I teach them.

Usually, we are given a regular polygon and we need to find the measure of interior angles or the number of sides. But what if we are given a polygon instead, not a regular polygon. Does this formula still apply? We wouldn’t know if we didn’t understand how the formula came into being. But since we know that we obtain the formula by dividing the polygon into (n-2) triangles, we know that the sum of all interior angles of a triangle is 180 irrespective of the kind of triangle. So it doesn’t matter whether the polygon is regular or not. The sum of all interior angles will still be (n-2)*180.

Let’s look at a question to see the application of this formula in *irregular* polygon scenario.

**Question**: The measures of the interior angles in a polygon are consecutive odd integers. The largest angle measures 153 degrees. How many sides does this polygon have?

A) 8

B) 9

C) 10

D) 11

E) 12

**Solution**:

The interior angles are: 153, 151, 149, 147 … and so on.

Now there are two ways to approach this question – one which is straight forward but uses algebra so is time consuming, another which makes you think but doesn’t take much time. You can guess which one we are going to focus on! But before we do that let’s take a quick look at the algebraic solution too.

**Method 1**: Algebra

Sum of interior angles of this polygon = 153 + 151 + 149 + … (153 – 2(n-1)) = (n – 2)*180

If there are n sides, there are n interior angles. The second largest angle will be 153 – 2*1. The third largest will be 153 – 2*2. The smallest will be 153 – 2*(n-1). This is an arithmetic progression.

Sum of all terms = [(First term + Last term)/2] * n = [(153 + 153 – 2(n-1))/2] * n

Equating, we get [(153 + 153 – 2(n-1))/2] * n = (n – 2)*180

Solving this you get, n = 10

But let’s figure out a solution without going through this painful calculation.

**Method 2**: Capitalize on what you know

Angles of the polygon: 153, 151, 149, 147, 145, 143, 141, … , (153 – 2(n-1))

The average of these angles must be equal to the measure of each interior angle of a regular polygon with n sides since the sum of all angles is the same in both the cases.

Measure of each interior angle of n sided regular polygon = Sum of all angles / n = (n-2)*180/n

Using the options:

Measure of each interior angle of 8 sided regular polygon = 180*6/8 = 135 degrees

Measure of each interior angle of 9 sided regular polygon = 180*7/9 = 140 degrees

Measure of each interior angle of 10 sided regular polygon = 180*8/10 = 144 degrees

Measure of each interior angle of 11 sided regular polygon = 180*9/11 = 147 degrees apprx

and so on…

Notice that the average of the given angles can be 144 if there are 10 angles.

The average cannot be higher than 144 i.e. 147 since that will give us only 7 sides (153, 151, 149, 147, 145, 143, 141 – the average is 147 is this case). But the regular polygon with interior angle measure of 147 has 11 sides. Similarly, the average cannot be less than 144 i.e. 140 either because that will give us many more sides than the required 9.

Hence, the polygon must have 10 sides.

**Answer (C).**

Interesting, eh? Well, it will be when you understand method 2 well and can do it intuitively!

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*