Textbook

**Regular polygons** have special characteristics that make them easier to work with.

You already know some of these. For example, a *regular triangle* is an *equilateral triangle*, and a *regular quadrilateral* is a *square*. When reading geometry questions, always be on the lookout for the word *regular*, since this piece of information will likely make the question much easier to solve.

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, i.e., they can have a perfect circle drawn around them with all corners precisely on the circumference. Some irregular polygons can also be circumscribed, but not always.

*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 these shapes might show up 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?

(spoiler)

Answer: Quantity A is greater

First, note that the question stated these are *regular polygons*. This means each shape has sides with equal lengths.

$a=b=cw=x=y=z $

Remember the equations for the area of a polygon?

- $A_{△}=bh/2$
- $A_{□}=s_{2}$

The question states that the area of our square is twice as large as the area of the triangle, and we can write this as an equation.

$A_{□}s_{2}s_{2} =2∗A_{△}=2∗bh/2=bh $

The question asks us to compare $b$ with $z$. This gets a little easier when we understand what we’re really being asked:

- Quantity A: $b$ (any side of the triangle)
- Quantity B: $z$ (and side of the square)

Because this is a regular triangle, i.e., an equilateral triangle, the base of the triangle is greater than the height. Since the side lengths are all equal, it follows that the length of side $b$ is also greater than the height.

We’re almost done. Let’s revisit that equation we set up earlier: $s_{2}=bh$

With our knowledge that $b>h$, we can infer that $b>s$ and $h<s$, and therefore, Quantity A is greater.

GRE figures are not drawn to scale, but now we have enough information to make a sketch of how this would look if drawn to scale:

For the triangle to have half the area of the square, the base needs to be a little big bigger than the height. It checks out!

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