# Regular Polygons and the Irregular Ones Continuing our GMAT Geometry journey, let’s discuss polygons today. While many of the classic shapes – triangles, rectangles, and squares, most notably – count as polygons and their formulas have been burned in our minds since elementary school, some of the less-common (we won’t say irregular because that’s an important definition as you’ll see) polygons require a bit more knowledge and understanding. And I can attest to that personally: 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 – that extra level of understanding makes remembering the rule for interior angles in a polygon exponentially easier!

But while remembering the rule can be made much easier (and I am thankful for that!), knowing how and when to apply the formulas for sum of interior angles of a polygon requires some additional understanding. Let’s investigate how we can put that formula to good use.

On the GMAT, we are usually 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 an irregular polygon instead, not a regular polygon? Does this formula for the interior angles of a polygon 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 we’re dealing with a regular or irregular polygon: 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 for the sum of the interior angles in an 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 a good bit of algebra so is time consuming, and 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: The Algebra

The sum of the interior angles of this irregular 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, because as you know on the GMAT time matters, and there’s usually a shortcut for those who deeply understand these concepts!

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.

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

Some important takeaways here that you should remember from this post:

1) While an irregular polygon doesn’t allow you to determine the measures of each individual interior angle via the (n – 2)*180 formula the way a regular polygon does, you can certainly apply that formula to determine the sum of the interior angles of an irregular polygon. It’s just that the angles aren’t all congruent, so you can’t divide by n to get the individual measures.

2) You can use the (n – 2)*180 formula to find both the sum of the interior angles of an irregular polygon and the average measure of an angle in an irregular polygon (just divide the sum by n).

3) It is helpful to understand why the formula for the sum of the interior angles of a polygon works – you’re essentially dividing every n-sided polygon into (n-2) triangles, and each of those triangles has an interior angle sum of 180. (It’s also just good standard “whenever you’re stuck” practice on any GMAT geometry problem to look for ways to divide regular or irregular shapes into triangles.)

4) Recognize that while you may memorize formulas – like the formula for the sum of the interior angles of a polygon – in terms of “here’s the input (e.g. n) and the formula will give you the output (e.g. the sum of the interior angles),” the GMAT loves to ask you to do these problems in reverse. As you saw in the problem above, your job was to solve for n when given some information about the interior angles. This is why understanding these formulas can be so important – the better you understand them, the easier it is to apply them in different directions based on what the question is asking you to do.

In summary, know that the (n – 2)*180 formula for the sum of the interior angles of a polygon applies regardless of whether you’re dealing with a regular or irregular polygon, and that while it’s a formula that you should have memorized it’s also a formula worth understanding. Polygon questions can get tricky because the number of sides and angles can increase dramatically over your standard triangles and quadrilaterals, but the good news is that there’s a one-size-fits-all formula that can get you through just about any polygon question – regular or irregular, no matter how large n is – and hopefully now you understand it thoroughly.

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! If she were a polygon, she’d be an irregular polygon (if only because “irregular” means “unique”).

Interested in applying your knowledge of regular and irregular polygons toward an irregularly-high score on the GMAT? Consider joining us for a GMAT class, where there are plenty more lessons like this!