Polygons

A polygon is a fancy word for a shape. Triangles, squares, hexagons, and trapezoids are all polygons. If a question prompt states that a shape is a regular polygon, it means that all the side lengths and angles have the same measurement. So, if you can find one side or angle of a regular polygon, you know them all.

Many questions will ask for the area of an odd-looking shape. Most odd-looking shapes do not have a simple equation to solve for the area. However, if you cut up the shape into familiar-looking parts, you may be able to find the total area with ease.

The house-like shape above can be split up into a triangle and a square. If we can find the sum of the area of both those shapes, it will be much easier to find the area of the whole polygon.

The height of the house is $5$, and the square has a height of just $3$, so we’re left with an extra height of $2$ for the height of the triangle. The base of the triangle is $3$, the same as the base of the square.

The area of the triangle is $bh/2=(3\times 2)/2=3$, and the area of the square is the side length squared, $3\times 3=9$. Adding these together gives us the area of the whole polygon: $3+9=12$.

Example of a hard polygon GRE question

Now that you’ve done an easy polygon question, are you ready for a hard one?

What is the area of the shape below?

Take your time solving this one. Even if you don’t know what to do at first, spend a minute or two thinking about it before you scroll down - trying to figure out what to do when you come across tough questions is important. So give it a try - we’ll wait!

Ready to check your answer?

Let’s look at the figure again and walk through it step by step.

We don’t have that much information to start, and that’s a good thing because it gives us a hint about the right steps to take. Did you notice that the height of the figure ($3\sqrt{3}$) is also one of the sides of the 30-60-90 triangle that makes up the left side of the polygon? The $30^{\circ }$ isn’t written in, but with the other two angles of this triangle being $60^{\circ }$ and the $90^{\circ }$ right angle, we can be sure the last one is $30^{\circ }$. That’s great news, because we can apply the $x:x\sqrt{3}:2x$ ratio to calculate the hypotenuse of the triangle, which is also the top left side of the polygon.

Alright, we’re getting somewhere. But what do we do next? It seems like we hit a dead end… but we can take another look at the interior angles we’re given. That same $60^{\circ }$ in the bottom left and the $90^{\circ }$ right angle in the top center seems a lot like… another 30-60-90 triangle! This triangle isn’t fully drawn and extends past the right of our polygon, but we can draw it in ourselves, and then use the $x:x\sqrt{3}:2x$ ratio again.

So now we’ve got this big triangle fully labeled, and we have enough information to calculate its area:

$$\begin{array}{rl}{A}_{\text{big}△}& =bh/2\\ & =12\ast 3\sqrt{3}/2\\ & =18\sqrt{3}\end{array}$$

Looking at the shape again, we can see that the area of this big triangle is equal to the area of our original polygon plus the area of the little triangle off to the right. That little triangle to the right has a $30^{\circ }$ and a $90^{\circ }$ right angle, which means… you guessed it… it’s another 30-60-90 triangle. So we can again use the $x:x\sqrt{3}:2x$ ratio to find the sides.

Now we can calculate the area of this small triangle:

$$\begin{array}{rl}{A}_{\text{small}△}& =bh/2\\ & =3\sqrt{3}\ast 3/2\\ & =\frac{9\sqrt{3}}{2}\end{array}$$

And now we have enough information to calculate the size of our polygon!

$$\begin{array}{rl}{A}_{\text{polygon}}& =A_{\text{big}△}-A_{\text{small}△}\\ & =18\sqrt{3}-\frac{9\sqrt{3}}{2}\\ & =\frac{36\sqrt{3}}{2}-\frac{9\sqrt{3}}{2}\\ & =\frac{27\sqrt{3}}{2}\end{array}$$

That was a lot of steps, but not so hard after all, right? Even if a question feels unapproachable at first, remember that you can break it down into multiple smaller steps, and every hard question will become easy!

We’ve said it many times throughout this course, but once more for good measure: no matter how hard the question, you can solve it using a combination of the same fundamental techniques you’ve learned.