In this tutorial, you’ll learn how to find the area of a parallelogram. We’ll look at the different formulas and use them to solve example problems.

Let’s begin with a quick summary of what parallelograms are.

A parallelogram is a quadrilateral (4-sided polygon) with certain specific properties. Here are three of those properties –

- Opposite sides are parallel and equal in length.
- Opposite angles are equal.
- Each pair of adjacent angles are supplementary (their sum is $\hspace{0.2em} 180 \degree \hspace{0.2em}$).

## Area of a Parallelogram

The area of a parallelogram refers to the area or region – in the 2-dimensional plane – enclosed within the parallelogram.

So the area of the parallelogram above is a quantity that accurately represents the yellow region.

1. Area of a parallelogram with a base $\hspace{0.2em} b \hspace{0.2em}$ and height $\hspace{0.2em} h \hspace{0.2em}$,

$A = bh$

2. Area of a parallelogram with diagonals $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$ and included angle $\hspace{0.2em} \theta \hspace{0.2em}$,

$A = \frac{1}{2} \cdot d_1 \cdot d_2 \cdot \sin \theta$

3. Area using adjacent sides $\hspace{0.2em} (a \hspace{0.2em}$ & $\hspace{0.2em} b) \hspace{0.2em}$ and the included angle $\hspace{0.2em} ( \theta ) \hspace{0.2em}$,

$A = ab \sin \theta$

Skip to examples Let's take a closer look and derive each of the three formulas mentioned above.

As we saw above, a simple formula for the area of a prallelogram is –

$\begin{align*} \text{Area} &= \text{base} \cdot \text{height} \\[1em] A &= bh \end{align*}$

Any of the four sides of the parallelogram can act as the base. The corresponding height would be the length of the perpendicular drawn from the base to the opposite side.

Now, here's how we get this formula.

A parallelogram consists of two equal triangular sections and a rectangular section sandwiched between them. If we cut one triangular section and move it to the other side as shown below, we end up with a rectangle.

Now, the base and height of the rectangle are the same as those of the parallelogram.

Also, it’s clear from the figure that the area of the rectangle is exactly the same as that of the parallelogram. So –

Note – If you don’t have a basic understanding of trigonometric ratios, feel free to skip the rest of this section.

##### Using Adjacent Sides and the Included Angle

For a parallelogram with adjacent sides $\hspace{0.2em} a \hspace{0.2em}$ and $\hspace{0.2em} b \hspace{0.2em}$ and the included angle $\hspace{0.2em} \theta \hspace{0.2em}$, the area is given by –

$A = ab \sin \theta$

Here's how.

Consider the parallelogram below. Its adjacent sides measure $\hspace{0.2em} a \hspace{0.2em}$ and $\hspace{0.2em} b \hspace{0.2em}$ in length and the angle between them is $\hspace{0.2em} \theta \hspace{0.2em}$. So how can we get its area?

Well, if we can find its height $\hspace{0.2em} h \hspace{0.2em}$, we can use the standard formula $\hspace{0.2em} (A = bh) \hspace{0.2em}$. So let’s find $\hspace{0.2em} h \hspace{0.2em}$ in terms of the known quantities.

In $\hspace{0.2em} \triangle ABM \hspace{0.2em}$,

$\begin{align*} \sin \theta &= \frac{h}{b} \\[1.3em] h &= b \sin \theta \end{align*}$

Great! Now that we know the height –

$\begin{align*} A &= \text{base} \cdot \text{height} \\[1em] &= a \cdot b \sin \theta \\[1em] &= ab \sin \theta \end{align*}$

And there we have our formula.

##### Diagonals and the Included Angle

If a parallelogram has diagonals of length $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$ and the angle between the diagonals is $\hspace{0.2em} \theta \hspace{0.2em}$, its area is given by –

$A = \frac{1}{2} \cdot d_1 \cdot d_2 \cdot \sin \theta$

And we derive it as follows.

Diagonals of a parallelogram bisect each other (cut in half). So they divide the parallelogram into 2 pairs of congruent triangles.

In the figure above triangles with the same color are congruent and hence have the same area.

Also in each of the four triangles, the two dotted sides have lengths $\hspace{0.2em} d_1 / 2 \hspace{0.2em}$ and $\hspace{0.2em} d_2 / 2 \hspace{0.2em}$. The figure below shows one triangle from each pair.

Here’s how it helps us find the area of the parallelogram.

First, the area of the pink triangle would be –

Similarly,

Now, the area of the parallelogram would be the sum of the areas of the four triangles. Or two times the sum of the two triangles shown above.

Formula derived!

## How to Find the Area of a Parallelogram | Examples

So far, so good. Now let’s use the formulas we just learned to solve some examples.

I have divided the examples into three categories based on what information the question provides.

### Using the Base and Height

Example

Find the area of the parallelogram shown below.

Solution

We know the area of a parallelogram with a base $\hspace{0.2em} b \hspace{0.2em}$ and height $\hspace{0.2em} h \hspace{0.2em}$ is given by –

$A = bh$

For the parallelogram above, we can take either $\hspace{0.2em} AB \hspace{0.2em}$ or $\hspace{0.2em} BC \hspace{0.2em}$ as the base. However, the corresponding height is known only for $\hspace{0.2em} AB \hspace{0.2em}$. So, we take $\hspace{0.2em} AB \hspace{0.2em}$ as the base and $\hspace{0.2em} AE \hspace{0.2em}$ as the height. That means

$b = 15, \hspace{0.15cm} h = 20$

Substituting the values in our formula for area, we get

$\begin{align*} A &= 15 \cdot 20 \\[1em] &= 300 \end{align*}$

### When the Sides and the Angle Are Known

Example

$\hspace{0.2em} ABCD \hspace{0.2em}$ is a parallelogram in which adjacent sides $\hspace{0.2em} AB \hspace{0.2em}$ and $\hspace{0.2em} BC \hspace{0.2em}$ are $\hspace{0.2em} 10$ cm and $\hspace{0.2em} 12$ cm long respectively. The height corresponding to the side $\hspace{0.2em} AB \hspace{0.2em}$ is $\hspace{0.2em} 6$ cm. Find the area of the parallelogram and its height corresponding to $\hspace{0.2em} BC \hspace{0.2em}$.

Solution

Again, we’ll be using the same formula.

$\text{Area} = \text{base} \cdot \text{height}$

For the first part of the question (finding the area), we take the side $\hspace{0.2em} AB \hspace{0.2em}$ as the base. That’s because the height corresponding to $\hspace{0.2em} AB \hspace{0.2em}$ is known. So, the area would be –

$\begin{align*} \text{Area} &= AB \cdot h_{AB} \\[1em] &= 10 \cdot 6 = 60 \end{align*}$

So the area of the parallelogram is $\hspace{0.2em} 60 \text{ cm}^2 \hspace{0.2em}$.

Great. For the second part, we take $\hspace{0.2em} BC \hspace{0.2em}$ as the base. That gives us,

$\text{Area} = BC \cdot h_{BC}$

Substituting the values for $\hspace{0.2em} A \hspace{0.2em}$ and $\hspace{0.2em} BC \hspace{0.2em}$, and solving for $\hspace{0.2em} h_{BC} \hspace{0.2em}$ –

$\begin{align*} 60 &= 12 \cdot h_{BC} \\[1em] h_{BC} &= 5 \end{align*}$

So the height corresponding to the side BC is $\hspace{0.2em} 5$ cm.

### When the Diagonals and the Angle Between Them Are Known

Example

Two adjacent sides of a parallelogram are $\hspace{0.2em} 8$ inches and $\hspace{0.2em} 11$ inches long. If the angle between them is $\hspace{0.2em} 60 \degree \hspace{0.2em}$, what is the area of the parallelogram?

Solution

This time the formula we need is

$A = ab \sin \theta$

Here $\hspace{0.2em} a \hspace{0.2em}$ and $\hspace{0.2em} b \hspace{0.2em}$ are adjacent sides of the parallelogram and $\hspace{0.2em} \theta \hspace{0.2em}$ is the angle between them.

Substituting the values from the question, we get

$\begin{align*} A &= 8 \cdot 11 \cdot \sin 60 \degree \\[1em] &\approx 76.21 \end{align*}$

So the area of the parallelogram is $\hspace{0.2em} 76.21 \text{ in}^2 \hspace{0.2em}$.

Example

$\hspace{0.2em} PQRS \hspace{0.2em}$ is a parallelogram such that the diagonals $\hspace{0.2em} PR \hspace{0.2em}$ and $\hspace{0.2em} QS \hspace{0.2em}$ measure $\hspace{0.2em} 8 \hspace{0.2em}$ and $\hspace{0.2em} 7 \hspace{0.2em}$ respectively. If the angle between them is $\hspace{0.2em} 45 \degree \hspace{0.2em}$, find the area of $\hspace{0.2em} PQRS \hspace{0.2em}$.

Solution

If the lengths of the diagonals are $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$ and the angle between them is $\hspace{0.2em} \theta \hspace{0.2em}$, the area of the parallelogram would be

$A = \frac{1}{2} \cdot d_1 \cdot d_2 \cdot \sin \theta$

Plugging in the values from the question, we have

$\begin{align*} A &= \frac{1}{2} \cdot 8 \cdot 7 \cdot \sin 45 \degree \\[1.3em] &\approx 19.8 \end{align*}$

And with that, we come to the end of this tutorial. Until next time.