Now, notice further how the top left and bottom right sides of the kite shown below are parallel to each other. Similarly, the top right and bottom left sides of this kite are parallel to each other.

Any guesses as to what kind of quadrilateral might this be? That's correct! It is a parallelogram.

Say you are told to find the area of this kite. Since this is a type of parallelogram, we could use a particular formula to calculate the area of this kite.

Throughout this article, we will be introduced to **the area formula of a parallelogram** and look at some worked examples where it is applied.

## Recap on parallelograms

Before we get into our main subject at hand, let us conduct a quick review on parallelograms to ease ourselves into this topic.

As the name implies, a parallelogram has parallel sides. Thus, we can define a parallelogram as below.

A **parallelogram** is a quadrilateral with two pairs of parallel opposite sides. A parallelogram is a special case of a quadrilateral.

A four-sided plane figure is known as a quadrilateral.

The following figure describes a parallelogram with sides, AB, BD, CD and AC.

### Properties of parallelograms

We shall return to our parallelogram ABCD above. Let us look at some properties that distinguish this shape.

The opposite sides of ABCD are parallel. In this case, AB is parallel to CD and AC is parallel to BD. We write this as AB // CD and AC // BD,

The opposite angles of ABCD are equal. Here, ∠CAB = ∠CDB and ∠ACD = ∠ABD,

The diagonals of a parallelogram bisect each other at a point, say M. Then, AM = MD and BM = MC. This is shown below,

Property of a parallelogram , StudySmarter Originals

Each diagonal of a parallelogram divides the parallelogram into two congruent triangles. Triangle CAB is congruent to triangle CDB and triangle ACD is congruent to triangle ABD.

### Types of parallelograms

There are three types of parallelograms we must consider throughout this syllabus, namely

Square

Rhombus

Each of these parallelograms has its distinct features that differentiate them from one another. A more detailed explanation of parallelograms can be found here, Parallelograms.

## Area of parallelogram definition

The **area of a parallelogram** is defined as the region enclosed by a parallelogram in a two-dimensional space.

In the diagram above, the total area enclosed by ABCD is the area of the parallelogram ABCD.

## Area of Parallelogram Formula

Referring to our initial parallelogram ABCD, we shall add two new components to this figure called b and h. This is displayed in the diagram below.

The variable b is called the base of the parallelogram. Either of the long sides of ABCD can be used as the base. For the diagram above, b can be either AB or CD. Here, here we have taken b = AB.

Note that this notion is a convention and not a hard and fast rule.

The variable h is called the height of the parallelogram. This may also be referred to as the altitude. The altitude is the line segment perpendicular to a pair of adjacent sides of the parallelogram with one endpoint on one side and the other endpoint on the other side.

Now that we have defined our variables b and h, we can thus present the area of a parallelogram as follows.

The area of any parallelogram is given by the formula,

$A=b\times h$

where b = base and h = height.

## Area of parallelogram examples

With that in mind, let us now observe the following worked examples that make use of this formula.

Find the area of the following parallelogram,

**Solution**

Here, the base is b = 24 units and the height is h = 10 units. Using the area of a parallelogram formula, we obtain,

$A=b\times h=24\times 10=240unit{s}^{2}$Thus, the area of this parallelogram is 240 units^{2}.

A parallelogram with an altitude of 5 units of length has an area of 20 units^{2}. What is the length of the base?

**Solution**

Here, we are given the area of the parallelogram and the altitude (or height), that is,

A = 20 and h = 5.

To find the base, we simply have to substitute these values into our area of a parallelogram formula and rearrange the equation as below.

$A=b\times h\phantom{\rule{0ex}{0ex}}20=b\times 5\phantom{\rule{0ex}{0ex}}5b=20$Making b the subject, we obtain

$b=\frac{20}{5}=4units$

Thus, the base of this parallelogram is 4 units.

## Finding the Area of a Parallelogram from a Rectangle

Suppose we want to find the area of a parallelogram where the height (or altitude) is unknown. Instead, we are given the lengths of two sides of the parallelogram, namely the lengths of AB and AC.

Let us try looking at this scenario graphically. Referring back to our initial parallelogram ABCD, let us draw two altitudes for each pair of adjacent sides, AC and AB as well as CD and BD.

We thus obtain two new points on this parallelogram, namely S and T. Now observe the shape formed by BTCS. Does this look familiar to you? That's right! It's a rectangle, which is also a type of parallelogram. We now need to find a way to obtain the lengths of either CS or BT in order for us to deduce the height of this parallelogram.

Notice that from the construction of these two line segments, we have obtained a pair of right-angle triangles, CAS and BDT. Since CS = BT, it suffices for us to only calculate one of them. Let us take a look at triangle CAS.

For simplicity, we shall denote the following sides as: x = AS, y = CS and z = AC. Since this is a right-angle triangle, we can use Pythagoras' theorem to obtain the length of CS, which is the height of the parallelogram ABCD. Given the lengths of AS and AC, we have

x^{2} + y^{2} = z^{2}

Rearranging this and applying the square root, we obtain

$y=\sqrt{{z}^{2}-{x}^{2}}$

As we have now found the length of CS, we can carry on finding the area of parallelogram ABCD by the formula given. We shall take the base as the length of AB. Thus, the area of ABCD is

$Are{a}_{ABCD}=AB\times CS$

Let us show this with an example.

Given parallelogram PQRS below, find its area.

The line OQ is the altitude of the adjacent sides PQ and PS. The lengths of QR, PQ and PO are given by 12 units, 13 units and 5 units, respectively.

**Solution **

Since QR = PS, we can take base as QR = 12 units. We now need to find the height of this parallelogram in order to find its area. This is given by the line segment OQ.

The diagram shows that triangle QPO is a right-angle triangle. Since we have the length of PO = 5 units, we can use Pythagoras' theorem to find OQ.

$P{O}^{2}+O{Q}^{2}=P{Q}^{2}\phantom{\rule{0ex}{0ex}}{5}^{2}+O{Q}^{2}={13}^{2}$

Rearranging this and applying the square root, we obtain the following value for OQ,

$O{Q}^{2}={13}^{2}-{5}^{2}\phantom{\rule{0ex}{0ex}}OQ=\sqrt{{13}^{2}-{5}^{2}}=\sqrt{169-25}=\sqrt{144}=12units$

Thus, the height of this parallelogram is 12 units. We can now find the area of PQRS as shown below,

$Are{a}_{PQRS}=QR\times OQ=12\times 12=144unit{s}^{2}$

Therefore, the area of this parallelogram is 144 units^{2}.

### Parallelogram Inscribed in a Rectangle Example

In this example, we shall look at a case where a parallelogram is inscribed inside a rectangle. We want to identify the area inside the rectangle that is not occupied by the parallelogram.

The figure below shows a parallelogram, PXRY inside a rectangle PQRS. Find the area of the region shaded in blue.

The line segment XZ is the altitude of the adjacent sides XP and PY. Here, QP = RS = XZ, PX = RY and QR = PS. The lengths of QP, PY and SY are given by 19 units, 21 units and 7 units, respectively.

**Solution**

Here, the height of the rectangle PQRS is h = QP = 19 units. The base is PS which is the sum of lengths PY and SY. Thus, the base is equal to

$PS=PY+YS=21+7=28units$

Thus, b = 28 units. The formula for the area of a rectangle is the product of its base and height. Thus, the area of the rectangle PQRS is

${A}_{PQRS}=b\times h=PS\times QP=28\times 19=532unit{s}^{2}$

Let us now find the area of the parallelogram PXRY. The height of the parallelogram is given by XZ. Since XZ = QP, then h = XZ = 19 units . The base is given by the length of PY. Thus, b = PY = 21 units. Using the area of a parallelogram formula, we obtain

${A}_{PXRY}=b\times h=PY\times XZ=21\times 19=399unit{s}^{2}$Thus, the areas of the rectangle PQRS and parallelogram PXRY are 532 units^{2} and 399 units^{2}, respectively.

We now need to find the area shaded in blue that is not occupied by the parallelogram inside the rectangle. This can be found by calculating the difference between the area of the rectangle PQRS and parallelogram PXRY. In doing so, we obtain

${A}_{blueregion}={A}_{PQRS}-{A}_{PXRY}=532-399=133unit{s}^{2}$

Hence the area of the remaining region shaded in blue is 133 units^{2}.

## A Special Case: Area of the Rhombus

The rhombus is a special type of quadrilateral that in fact has its own formula for calculating its area. It is sometimes referred to as equilateral quadrilateral. Let us recall the definition of a rhombus.

A **rhombus** is a parallelogram with all four sides of equal length.

We shall now consider the rhombus below. Two diagonals, AD (light blue line) and BC (dark blue line) are constructed on this parallelogram. The diagonals have lengths d_{1} and d_{2}, respectively.

Area of a rhombus, StudySmarterOriginals

**Area of a Rhombus**

The area of the rhombus is given by the formula,

$A=\mathit{}\frac{\mathit{1}}{\mathit{2}}{d}_{\mathit{1}}{d}_{\mathit{2}}$

where A = area, d_{1} = length of diagonal AD and d_{2} = length of diagonal BC.

### Example of the Area of a Rhombus

Here is an example involving the area of a rhombus formula.

A rhombus has diagonals of lengths 10 units and 15 units. What is the area of the rhombus?

**Solution **

Let us denote d_{1} = 10 units and d_{2} = 15 units. Applying the formula above, we obtain

$A\mathit{=}\mathit{}\frac{\mathit{1}}{\mathit{2}}{d}_{\mathit{1}}{d}_{\mathit{2}}=\frac{\mathit{1}}{\mathit{2}}\times 10\times 15=75unit{s}^{2}$

Thus, the area of this rhombus is 75 units^{2}.

**The formula for the area of a rhombus can also be used to find the area of a kite in a similar way.**

We shall end this article with a final example involving the area of a parallelogram, or more specifically a kite.

## Real-world Example of the Area of a Parallelogram

We shall now return to our example at the beginning of this article. As we now have a basic formula for calculating the area of a parallelogram, we can thus use it to find the area of our kite.

You decide to measure the two diagonal lengths of your kite with a tape measure. You find that the horizontal diagonal and vertical diagonal are equal to 18 inches and 31 inches, respectively. Using the formula for the area of a rhombus, find the area of this kite.

**Solution **

Let

d_{1} = horizontal diagonal = 18 inches

d_{2} = vertical diagonal = 31 inches

Applying the formula for the area of a rhombus, we obtain

$A\mathit{=}\mathit{}\frac{\mathit{1}}{\mathit{2}}{d}_{\mathit{1}}{d}_{\mathit{2}}=\frac{\mathit{1}}{\mathit{2}}\times 18\times 31=558inche{s}^{2}$

Thus, the area of this kite is 558 inches^{2}.

## Area of Parallelograms - Key takeaways

- A quadrilateral with two pairs of parallel opposite sides is called a parallelogram.
- There are three types of parallelograms: a rectangle, a square and a rhombus.
- Notable properties of a parallelogram:
The opposite sides are parallel

The opposite angles are equal

The diagonals bisect each other as a point

Each diagonal divides the parallelogram into two congruent triangles

- The area of a parallelogram is given by the formula:
**A = b × h**, where b = base, h = height. The area of the rhombus is given by the formula:$\mathit{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{d}}_{\mathbf{1}}{\mathit{d}}_{\mathbf{2}}$, where d

_{1}and d_{2 }are the lengths of the diagonals of the rhombus.

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##### Frequently Asked Questions about Area of Parallelograms

How to find the area of a parallelogram?

Area = b × h

where b=base, h=height.

What is the area of a parallelogram?

Area = b × h

where b=base, h=height.

What is the formula for the area of a parallelogram?

Area = b × h

where b=base, h=height.

How do you find the area of a parallelogram without the height or area?

Area=0.5×d1×d2×sin(α), where d1, d2 are the lengths of the respective diagonals and α is the angle between them.

What are the properties of a parallelogram?

- In a parallelogram, the opposite sides are equal.
- In a parallelogram, the opposite angles are equal.
- The diagonals of a parallelogram bisect each other.
- Each diagonal of a parallelogram divides the parallelogram into 2 congruent triangles.

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