# Areas of Rhombuses and Kites

Although a rhombus is a type of

parallelogram, whereas a kite is not, they are similar in that their sides

have important properties. Recall that all four sides of a rhombus are congruent.

Kites, on the other hand, have exactly two pairs of consecutive sides that are congruent.

This characteristic of kites does not allow for both pairs of opposite sides to

be parallel. Let’s look at the image below to examine the properties that make these

figures distinguishable.

*The key characteristics of rhombus ABCD and kite EFGH are shown in the figure above.*

Despite how different they are, when it comes to

area, we will see that rhombuses and parallelograms are quite similar. The

areas of rhombuses and kites are equal to one half the product of their diagonals.

Mathematically, we express this as

where ** A** is the area of the of the

quadrilaterals (in square units),

**is the length**

*d*_{1}of one diagonal, and

**is the length of the other diagonal.**

*d*_{2}

Recall that every quadrilateral has exactly two diagonals. This is because diagonals

are line segments which connect vertices to each other. Since there already exist

line segments which connect one vertex to two other vertices in a quadrilateral,

the only other line segment to draw is to the vertex diagonal from the chosen vertex.

Let’s work on the following exercises, to help us apply the area formula for rhombuses

and kites.

## Exercise 1

**Find the area of rhombus PQRS.**

**Answer:**

The only two parts of the rhombus we need to figure out are the diagonals because

that is all that is required when we find the areas of rhombuses. We are given the

length of one diagonal of rhombus ** PQRS**, which will be our

**.**

*d*_{1}The length of diagonal

**is**

*PR***.**

*12*centimeters
Before we can find the area of rhombus ** PQRS**, we need to find the length

of

**. We see that the length of**

*d*_{2}**, is just**

*SQ*the sum of two smaller segments. So, we need to take the sum of segments

*ST*and

**to find the length of**

*TQ***. However, we do not**

*SQ*know what the length of

**is, so we must rely on our knowledge of**

*TQ*rhombuses to help us out at this point.

We know that the diagonals of a rhombus bisect each other. This means that the point

** T** is the midpoint of segment

**. Thus, we know that**

*SQ*the length of segment

**is equal to the length of segment**

*TQ***;**

*ST*they are both

**long. Now, we can find the length of**

*4*centimeters**, which is our**

SQSQ

**:**

*d*_{2}

So, we know that ** d_{2}** has a length of

**.**

*8*centimetersNow, we have all the requirements we need in order to solve for the area of rhombus

**. Let’s plug our values of**

*PQRS***and**

*d*_{1}

*d*_{2}in:

We see that rhombus ** PQRS** has an area of

**.**

*48*square centimeters## Exercise 2

**Find the value of y given that the area of kite YDOC is 552 square feet.**

**Answer:**

In this exercise, we will not be solving for area, since it has already been given

to us. Rather, we will have to use properties of kites and the area formula to deduce

the value of variable ** x**.

We have been given the length of segment ** CD**, which we will use as

**in our area formula. Therefore,**

*d*_{1}

*d*_{1}is equal to

**.**

*24*centimeters
Now, let’s look at the other diagonal of kite ** YDOC**. It has been given

a length of

**. However, we are not sure what the**

*12x+5y*centimetersvalues of

**or**

*x***are, so we must deduce key information**

*y*from the facts that have been given to us. Let’s take a look at the area formula

of our kite as we have it right now.

At this point in the exercise, we are stuck because we do not know the values of

** x** or

**. If we look at the sides of the kite, however,**

*y*we will notice that we can determine the value of

**. We know that**

*x*kites have two pairs of consecutive sides that are congruent. In this case, we see

that

**and**

*CY***are congruent, so we can set those values**

*DY*equal to each other to solve for

**. We have**

*x*

Therefore, we have determined that the value of ** x** is

**.**

*3*We can use this value to plug into our area formula in order to solve for

**.**

*y*We have

After simplifying, we get

Then, we divide both sides of the equation by ** 12**, which yields

So, we have determined that the value of ** y** is

**.**

*2*## Summary

As we progress through the lessons on area, it will be important to recognize the

properties of different polygons in order to help us find valuable information about

certain problems. While the different classifications of polygons have features

which set them apart from each other, it can also help to group polygons up by their

relationship with one another. In the same way that parallelograms and triangles

have completely different properties but share a distinct relationship in terms

of their areas, we see that rhombuses and kites are also two different polygons

with the same formula to determine their areas.