Textbook

There are four main *quadrilaterals* you will see in the GRE: *squares*, *rectangles*, *rhombuses*, and *trapezoids*.

The definitions of these different types of quadrilaterals can be a bit confusing because some categories encompass others. For instance, all squares are rectangles, but not all rectangles are squares. But fear not! We’ll cover each of them with examples.

You probably already know what a rectangle is, but it’s good to review the basics briefly from a technical perspective.

The opposite sides of a rectangle are parallel and congruent, and the two diagonals are also congruent.

The **perimeter of a rectangle** is the sum of all four sides, but since the opposite pairs are equal, it can be simplified to $P=2b+2h$.

The **area of a rectangle** is simply the base times the height: $A=bh$.

We’ll keep this one really short 😄

Since all sides are the same, the **perimeter of a square** is $P=4s$.

Since the base and the height of a square are the same side length $s$, the **area of a square** is $A=s_{2}$.

You might *not* have heard of a *rhombus* before.

Basically, a rhombus is a square leaning over to the side to make a diamond. The lengths of the sides are all still the same, but the angles get squished, so instead of having four right angles, you end up with two identical acute angles and two identical obtuse angles.

Since the sides are all equal in length, the **perimeter of a rhombus** is the same as for a square: $P=4s$.

There are two main ways to get the **area of a rhombus**. The first is very similar to that of a rectangle: $A=bh$. However, since our shape is tilted, the height $h$ is not the full height of a side. Instead, it’s the inside height that goes up at a right angle from the base.

The second way to get the **area of a rhombus** is using the inside diagonals that reach from one corner across the shape to the other opposite corner. Using these $p$ and $q$ diagonals, the area is $A=pq/2$.

Note that the square form of a rhombus maximizes the area! The more the rhombus leans over and deviates from the 90-degree interior angles, the smaller the area will become.

*Parallelograms* are quadrilaterals that have parallel opposite sides of equal lengths. The opposite angles inside a parallelogram are also of equal measure.

Since the definition of a *parallelogram* is broad, it encompasses many of the other types of quadrilaterals: *squares*, *rectangles*, *rhombuses*, and even *trapezoids* might also be *parallelograms* depending on their shape.

There aren’t any special formulas to memorize for parallelograms since it’s just a general category of shape.

The key thing to remember is that if the shape is described as a *parallelogram*, you only need to worry about two adjacent sides and two adjacent angles, since the opposite ones will be identical.

*Trapezoids* can be a bit ambiguous since there are various definitions. Since this is a GRE course, we’ll use the ETS definition: *a quadrilateral in which at least one pair of opposite sides is parallel is a trapezoid*. Just be aware that other definitions vary, sometimes saying that a trapezoid must have *only* one pair of parallel opposite sides, or that every quadrilateral is a trapezoid.

The opposite parallel sides are the *bases*, although they might not necessarily be the top and the bottom of the shape. The formulas will still be accurate even if the shape is rotated.

The **perimeter of a trapezoid** is the sum of all four sides. Since the sides could all be different lengths, there is no shortcut version of this formula.

The **area of a trapezoid** is equal to the average of the two parallel sides (“*bases*”) times the height, i.e., $A=21 (b_{1}+b_{2})(h)$. Note that the height is the inside distance from one base to another, and most likely won’t be the length of one of the sides!

Now that you know the fundamentals, let’s try a question!

The area of the trapezoid is greater than the area of the rectangle. Both shapes have the same height.

Quantity A: $a$

Quantity B: $7$

Give it a try, and then check your answer!

(spoiler)

Answer: *D. The relationship cannot be determined*

First, recall the equation for the area of a trapezoid: $A_{t}=21 (b_{1}+b_{2})(h)$.

And the equation for area of a rectangle: $A_{r}=bh$.

The question tells us that the area of the trapezoid is greater than the area of the rectangle, so we can set up an inequality. Both shapes have the same height, so we can use the same $h$ for both.

$A_{t}21 (b_{1}+b_{2})(h)21 (14+a)(h)21 (14+a)14+aa >A_{r}>b_{r}h>10h>10>20>6 $

We’ve figured out that $a>6$, and we’re comparing it against $7$. The question doesn’t say anything about the numbers needing to be integers, so it’s completely possible that $a$ could be something like $6.1$, $7$, $8$, $13$… any number where $a>6$ and $a<14$ (since it’s shown to be smaller than the base). We don’t have enough information to determine a more specific relationship, so the answer must be *D*.

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