Regular polygons

Regular polygons have special characteristics that often make geometry problems easier to solve.

You’ve already seen some regular polygons:

• A regular triangle is an equilateral triangle.
• A regular quadrilateral is a square.

When you see the word regular in a geometry question, treat it as an important clue. It usually lets you assume equal side lengths (and equal angles), which can simplify the work.

Here are some of the most common regular polygons:

Sides	Name	Interior angle	Figure
3	Triangle	60°
4	Square	90°
5	Pentagon	108°
6	Hexagon	120°
8	Octagon	135°
10	Decagon	144°

All regular polygons can be perfectly circumscribed. That means you can draw a circle around the polygon so that every vertex lies exactly on the circle. Some irregular polygons can also be circumscribed, but not all of them.

The figure directly above shows a circumscribed regular quadrilateral (a square).

The figure directly above shows an irregular quadrilateral that is not circumscribed.

Let’s see how regular polygons might appear in an example question.

The triangle and quadrilateral shown below are regular polygons. The area of the quadrilateral is twice as large as the area of the triangle.

Quantity A: $b$
Quantity B: $z$

Ready for the answer?

First, use the fact that the shapes are regular. That tells you all side lengths within each shape are equal.

$$\begin{array}{c}a=b=c\\ w=x=y=z\end{array}$$

Now recall the area formulas:

• $A_{△}=bh/2$
• $A_{\square }=s^2$

The question says the area of the square is twice the area of the triangle. Write that relationship as an equation.

$$\begin{array}{rl}{A}_{\square }& =2\ast A_{△}\\ s^2& =2\ast bh/2\\ s^2& =bh\end{array}$$

We’re asked to compare $b$ and $z$. Interpreting the labels helps:

• Quantity A: $b$ (a side of the triangle)
• Quantity B: $z$ (a side of the square)

Because the triangle is regular (equilateral), its height is shorter than its base. So the base is greater than the height, and since all sides are equal, $b$ is also greater than the height.

Now go back to the equation $s^2=bh$. Since $b>h$, the product $bh$ is less than $b^2$. That means:

$$s^2=bh<b^2$$

Taking positive square roots gives $s<b$. Also, since $w=x=y=z=s$, we have $z=s$. Therefore, $b>z$, so Quantity A is greater.

GRE figures are not drawn to scale, but you can sketch what this would look like if it were drawn to scale:

For the triangle to have half the area of the square, the base needs to be a little bigger than the height, which matches what you’d expect for an equilateral triangle.