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

- All four sides are equal.
- Opposite sides are parallel.
- Opposite angles are equal.

In this tutorial, we’ll focus on the area of a rhombus. We’ll look at the different formulas and use them to solve example problems.

Let’s begin with a quick summary of what a rhombus is.

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

- All four sides are equal.
- Opposite sides are parallel.
- Opposite angles are equal.

If you are familiar with parallelograms, you might have already realized that a rhombus is just a parallelogram whose sides are all equal.

So all the formulas for the area of a parallelogram work for rhombuses too. The converse isn’t true though.

The area of a rhombus refers to a measure of the region in the plane enclosed within the rhombus.

So the area of the rhombus in the figure above is the number that accurately represents the yellow region.

1. Area of a rhombus with diagonals $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$,

$A = \frac{d_1d_2}{2}$

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

$A = bh$

3. Area using side length, $\hspace{0.2em} a \hspace{0.2em}$, and one internal angle $\hspace{0.2em} \theta \hspace{0.2em}$,

$A = a^2 \sin \theta$

Let's see how we can derive each of the three formulas mentioned above.

If the length of the diagonals of a rhombus are $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$, its area is given by the formula –

$A = \frac{d_1d_2}{2}$

Here's how we derive the Formula

Diagonals of a rhombus bisect (cut in half) each other at right angles. So they divide the rhombus into 4 congruent (equal) right triangles.

That means if the diagonals measure $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_1 \hspace{0.2em}$ in length, each of the four triangles would like this.

So the area of the rhombus would be 4 times the area of the triangle.

$\begin{align*} Area(◇) &= 4 \cdot Area(◺) \\[1.3em] &= 4 \cdot \frac{1}{2} \cdot \text{base} \cdot \text{height} \\[1.3em] &= 4 \cdot \frac{1}{2} \cdot \frac{d_1}{2} \cdot \frac{d_2}{2} \\[1.3em] &= \frac{d_1 d_2}{2} \end{align*}$

That’s it. We have derived the formula.

Like the area of any parallelogram, the area of a rhombus is given by –

$A = bh$

For a rhombus, its base is the same as its side length (all sides are equal in length) and its height is the distance between any pair of opposite sides (same for both pairs).

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

The area of a rhombus with a side length $\hspace{0.2em} a \hspace{0.2em}$ and one internal angle $\hspace{0.2em} \theta \hspace{0.2em}$ is –

$A = a^2 \sin \theta$

For the derivation, consider the rhombus below. It has a side length $\hspace{0.2em} a \hspace{0.2em}$ and one internal angle $\hspace{0.2em} \theta \hspace{0.2em}$.

From the figure, we have

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

And we know the area of a rhombus is –

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

Substituting the values of base and height,

$\begin{align*} A &= a \cdot a \sin \theta \\[1em] &= a^2 \sin \theta \end{align*}$

And voila! Formula derived.

Alright, let’s solve some example problems using the formulas learned above.

Example

Find the area of a rhombus with a side length of $\hspace{0.2em} 18$ inches and a height of $\hspace{0.2em} 10$ inches.

Solution

We have,

$b = 18 \text{ ft}, \hspace{0.2cm} h = 10 \text{ ft}$

For a rhombus with a base, $\hspace{0.2em} b \hspace{0.2em}$, and corresponding height, $\hspace{0.2em} h \hspace{0.2em}$, the area is given by

$A = bh$

Plugging in the values of $\hspace{0.2em} b \hspace{0.2em}$ and $\hspace{0.2em} h \hspace{0.2em}$ –

$\begin{align*} A &= 18 \cdot 10 \\[1em] &= 180 \end{align*}$

The lengths were in inches, so the area of the rhombus is $\hspace{0.2em} 180 \text{ in}^2 \hspace{0.2em}$.

Example

Solution

Here,

$d_1 = 12, \hspace{0.2cm} d_2=9$

When the lengths of the diagonals ($\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$) are known, we can get the area of the rhombus using the formula –

$A = \frac{d_1d_2}{2}$

Substituting the values of $\hspace{0.2em} d_1 \hspace{0.2em}$ and $\hspace{0.2em} d_2 \hspace{0.2em}$ into the formula, we have

$\begin{align*} A &= \frac{12 \cdot 9}{2} \\[1.3em] &= 54 \end{align*}$

Example

The diagonals of a rhombus measure $\hspace{0.2em} 8 \hspace{0.2em}$ and $\hspace{0.2em} 15 \hspace{0.2em}$ and its side length is $\hspace{0.2em} 10 \hspace{0.2em}$. Find the area and height of the rhombus.

Solution

Here again, we know the diagonal lengths and so we’ll use the following formula.

$A = \frac{d_1d_2}{2}$

That gives us

$\begin{align*} A &= \frac{8 \cdot 15}{2} \\[1.3em] &= 60 \end{align*}$

Now that we know the relationship between the area, base $\hspace{0.2em} (b)$, and height $\hspace{0.2em} (h)$ of a rhombus.

$A = bh$

Substituting the values of $\hspace{0.2em} A \hspace{0.2em}$ and $\hspace{0.2em} b \hspace{0.2em}$ and solving for $\hspace{0.2em} h \hspace{0.2em}$, we have –

$\begin{align*} 60 &= 10 \cdot h \\[1em] h &= 6 \end{align*}$

Example

The side length of a rhombus is $\hspace{0.2em} 20 \hspace{0.2em}$ and the angle between two of its adjacent sides is $\hspace{0.2em} 45 \degree \hspace{0.2em}$. Find the area of the rhombus.

Solution

If the side-length of a rhombus is $\hspace{0.2em} a \hspace{0.2em}$ and the angle between two of its sides is $\hspace{0.2em} θ \hspace{0.2em}$, its area is given by

$A = a^2 \sin \theta$

So here

$\begin{align*} A &= 20^2 \cdot \sin 45 \degree \\[1em] &= 282.84 \end{align*}$

And with that, we come to the end of this tutorial on how to find the area of a rhombus. Until next time.

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