Polygons

A polygon is a closed, flat shape made of straight line segments. Triangles, squares, hexagons, and trapezoids are all polygons. If a problem says a polygon is regular, that means:

• All side lengths are equal.
• All interior angles are equal.

So, if you can find one side length or one angle in a regular polygon, you know them all.

Many questions ask for the area of an odd-looking shape. Most irregular shapes don’t have a single “area formula” you can plug into. The usual strategy is to split the shape into familiar pieces (like triangles and rectangles), find each area, and then add or subtract.

The house-like shape above can be split into a triangle and a square. If you find the area of each piece and add them, you’ll get the area of the whole polygon.

The total height of the house is $5$. The square portion has height $3$, so the remaining height is $5-3=2$. That remaining height is the height of the triangle. The base of the triangle is $3$, the same as the base of the square.

• Triangle area: $bh/2=(3\times 2)/2=3$
• Square area: side${}^2=3\times 3=9$

Add them to get the total area:

$$3+9=12.$$

Example of a hard polygon GRE question

Here’s a more challenging one.

What is the area of the shape below?

Take your time solving this one. If you’re not sure what to do at first, spend a minute or two looking for familiar shapes or special triangles before you scroll down.

Ready to check your answer?

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

We aren’t given many lengths, which is a clue that geometry relationships (especially special triangles) will do most of the work.

Notice that the height of the figure is $3\sqrt{3}$. That height is also a leg of the $30$-$60$-$90$ triangle on the left side of the polygon. The $30^{\circ }$ angle isn’t labeled, but the other angles are $60^{\circ }$ and $90^{\circ }$, so the third angle must be $30^{\circ }$.

That means we can use the $x:x\sqrt{3}:2x$ ratio to find the hypotenuse of that triangle, which is also the top-left side of the polygon.

Next, look at the angles inside the figure. The $60^{\circ }$ angle in the bottom left and the $90^{\circ }$ angle at the top center suggest another $30$-$60$-$90$ triangle. This triangle isn’t fully drawn because it extends past the right side of the polygon, but you can draw it in and use the $x:x\sqrt{3}:2x$ ratio again.

Now the large triangle is fully labeled, so you can compute its area:

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

From the diagram, the area of this big triangle equals:

• the area of the original polygon, plus
• the area of the small triangle sticking out on the right.

That small triangle has a $30^{\circ }$ angle and a $90^{\circ }$ angle, so it’s also a $30$-$60$-$90$ triangle. Use the $x:x\sqrt{3}:2x$ ratio to find its side lengths.

Now compute the area of the small triangle:

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

Finally, subtract to get the polygon’s area:

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

This problem has several steps, but each step uses the same core ideas:

• Break a complicated shape into simpler shapes.
• Look for special triangles (like $30$-$60$-$90$ triangles).
• Add or subtract areas to get the final result.

No matter how complicated a polygon looks at first, you can usually make it manageable by splitting it into pieces you know how to handle.