As we’ve progressed through the quadrilaterals section, we have become more and more specific about the

type of figures we are dealing with. Initially, we considered all sorts of polygons, and then

we narrowed it down to four-sided polygons called quadrilaterals. From there, we

learned about a special type of quadrilateral whose opposite sides were parallel,

called a parallelogram. In this section, we will get even more specific by studying

the properties of different parallelograms. Let’s learn about what makes rectangles,

rhombuses, and squares special figures.

## Rectangles

**Definition:** A rectangle is a quadrilateral with four right angles.

Notice that we use “quadrilateral” in our definition of rectangles. We could have

also said that a rectangle is a parallelogram with four right angles, since

and quadrilateral with four right angles is also a parallelogram (because their

opposite sides would be parallel).

Rectangles have a couple of properties that help distinguish them from other parallelograms.

By studying these properties, we will be able to differentiate between various types

of parallelograms and classify them more specifically. Keep in mind that all of

the figures in this section share properties of parallelograms. That is, they all

have

**(1)** opposite sides that are parallel,

**(2)** opposite angles that are congruent,

**(3)** opposite sides that are congruent,

**(4)** consecutive angles that are supplementary, and

**(5)** diagonals that bisect each other.

Now, let’s look at the properties that make rectangles a special type of parallelogram.

**(1)** All four angles of a rectangle are right angles.

**(2)** The diagonals of a rectangle are congruent.

## Rhombuses

**Definition:** A rhombus is a quadrilateral with four congruent sides.

Similar to the definition of a rectangle, we could have used the word “parallelogram”

instead of “quadrilateral” in our definition of rhombus. Thus, rhombuses have all

of the properties of parallelograms (stated above), along with a few others. Let’s

look at these properties.

**(1)** Consecutive sides of a rhombus are congruent.

**(2)** The diagonals of a rhombus bisect pairs of opposite angles.

**(3)** The diagonals of a rhombus are perpendicular.

## Squares

**Definition:** A square is a parallelogram with four congruent sides and four

congruent angles.

Notice that the definition of a square is a combination of the definitions of a

rectangle and a rhombus. Therefore, a square is both a rectangle and a rhombus,

which means that the properties of parallelograms, rectangles, and rhombuses all

apply to squares. Because squares have a combination of all of these different properties,

it is a very specific type of quadrilateral.

Look at the hierarchy of quadrilaterals below. This figure shows the progression

of our knowledge of polygons, beginning with quadrilaterals, and ending with squares.

*Notice that there are two arrows pointing to the square. That is because a square has all the properties of a rectangle and rhombus.*

Now that we are aware of the properties of rectangles, rhombuses, and squares, let’s

work on a few exercises that will gauge our understanding of this material.

### Exercise 1

**Identify each parallelogram as a rectangle, rhombus, or a square.**

**Answer:**

First, let’s take a look at Parallelogram A. The figure shows that it has four congruent

sides and that its diagonals intersect perpendicularly. Because its sides are congruent,

we know that the parallelogram is not a rectangle. The fact that Parallelogram A’s

diagonals intersect perpendicularly does not help us because both rhombuses and

squares share this characteristic. The angle at the top of Parallelogram A is not

a right angle, however. Therefore, we know that it is not a square. Parallelogram

A is a rhombus.

In Parallelogram B, we see that there are four right angles and that the pairs of

opposite sides are congruent. However, consecutive sides are not congruent, so we

can eliminate rhombuses and squares from our options. Thus, Parallelogram B is a

rectangle.

Let’s take a look at Parallelogram C now. We note that it has a pair of right angles

and four congruent sides. Our inclination leads us to think that this parallelogram

is a square, but let’s make sure just in case. We know that the two right angles

given to us have a sum of ** 180°**. Because the interior angles of a quadrilateral

is

**, we know that the remaining two angles must have a sum of**

*360°***(because**

180°180°

**). Opposite angles of parallelograms**

*360-180=180*are congruent, which means that each other the missing angles must have a measure

of

**(since**

*90°***). This tells us that there are**

*180÷2=90*actually four right angles in Parallelogram C, so we know that it is a square (and

a rhombus).

### Exercise 2

**Find the value of x given rectangle ABCD below.**

**Answer:**

We know that ** ABCD** is a rectangle, so let’s use some rectangle properties

to help us figure out what

**is. It appears as though the focus of**

*x*this exercise is on the diagonals of the figure. From above, we know that the diagonals

of a rectangle are congruent, so let’s set segments

**and**

*AC*

*BD*equal to each other:

So, we get ** x=12**.

### Exercise 3

**Answer:**

Let’s examine the information we have been given from the exercise, in order to

deduce more information from it. We know that ** EKIN** is a parallelogram,

and that

**. Since**

*?1??2***is a parallelogram, we know**

*EKIN*that its opposite sides are parallel. Therefore, segments

**and**

*EK***are parallel.**

ININ

Next, we can use the **Alternate Interior Angles Theorem** to claim that *?1??4*

and ** ?2??3**. Recall, that the alternate interior angles are congruent

if and only if a transversal intersects a pair of parallel lines. In this case,

our pair of parallel lines is

**and**

*EK***, and our transversal**

*IN*is segment

**.**

*NK* By transitivity, we can say that ** ?1??3** and

**. Let’s**

*?2??4*look at our chain of congruences to be assured that the previous statements are

true.

The diagonal splits our parallelogram into two triangles. In fact, because two angles

of each triangle are congruent, we can say that ** ?EKN** and

*?INK*are isosceles triangles. The converse of the

**Isosceles Triangle Theorem**states

that the sides opposite of congruent angles of isosceles triangles are congruent,

so we know that segment

**is congruent to segment**

*EK***,**

*EN*and that segments

**and**

*IK***are congruent.**

*IN* Now, can say that segments ** EK** and

**are congruent,**

*IN*as are

**and**

*EN***because opposite sides of a parallelogram**

*IK*are congruent. By transitivity, we know that

**. Let’s look**

*EN?EK?IN?IK*at our new illustration.

Thus, parallelogram ** EKIN** is a rhombus because it has four congruent

sides. Our two-column geometric proof for this exercise is shown below.

### Exercise 4

**What must the value of y be in order for rhombus PQRS to be a square?**

**Answer:**

Before we can figure our ** y**, we must determine what the value of

**is. Ultimately, we want rhombus**

xx

**to be a square,**

*PQRS*which means that

**should have four right angles.**

*PQRS* Let’s begin by figuring out what ** x** is. This is relatively simple because

we can just set segment

**equal to**

*PQ***:**

*PS*

Now that we know what ** x** is, we can plug it into the measure of the angle given to

us. But, first, we need to figure out what the total measure of

**is. We know**

*?QSR*that we want

**to be**

*?PSR***. Also, we know that the diagonals of a square bisect**

*90°*pairs of opposite angles. Therefore,

**should be bisected by segment**

*?PSR***, splitting**

*QS*the angle up into two congruent angles of

**(because**

*45°***). Now, we can set**

*90÷2=45***equal to**

*?QSR***. We get:**

*45°*

Now, we substitute ** 7** in for

**:**

*x*

So, the value of ** y** must be

**in order for rhombus**

*4***to also be a square.**

*PQRS*