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Here we will learn about finding the area of a rhombus, including compound area questions, questions with missing side lengths and questions involving unit conversion.

There are also area of a rhombus worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The **area of a rhombus **is the amount of space inside the rhombus. It is measured in units squared ( cm^2, \ m^2, \ mm^2 etc).

Area formula:

A=\frac{1}{2}\times{D}\times{d}Where D is the long diagonal of the rhombus (the base) and d is the short diagonal in the rhombus (the height of the rhombus).

A rhombus is a **polygon **and a **quadrilateral **( 4 sided shape) where opposite sides are parallel, all sides are equal length and opposite angles are equal. The interior angles of a rhombus add up to 360^{\circ} . A rhombus is a special type of a parallelogram.

The diagonals of a parallelogram **bisect** each other at a right angle as shown below:

To calculate the area of a rhombus, we need to know the **length of each diagonal** of the rhombus.

If we move the two triangles from half of the rhombus to the opposing side, we get a rectangle:

The width of the rectangle is half of the long diagonal of the rhombus ( D ), and the height of the rectangle is the length of the short diagonal in the rhombus ( d ).

This gives the area of the rhombus formula:

A=\frac{1}{2}\times{D}\times{d}In order to find the area of a rhombus:

**Identify the length of the diagonals.****Write down the formula for the area of a rhombus.****Substitute the given values of the diagonals and solve.****Write down your final answer, including the units.**

Get your free area of a rhombus worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free area of a rhombus worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Area of a rhombus** is part of our series of lessons to support revision on **area**. You may find it helpful to start with the main area lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

- Area
- Area of a circle
- Area of a quadrilateral
- Area of a trapezium
- Area of an equilateral triangle
- Area of compound shapes
- Pi r squared
- Area of a parallelogram
- Area of a right angled triangle
- Area of an isosceles triangle
- Area of a triangle
- How to work out area
- Area of a rectangle
- Area of a hexagon
- Area of a pentagon

Find the area of the rhombus below:

**Identify the length of the diagonals.**

These are the diagonals of a rhombus which correspond to the base and height of the corresponding rectangle.

2**Write down the formula for the area of a rhombus.**

3**Substitute the given values of the diagonals and solve.**

4**Write down your final answer, including the units.**

In this case we are working with metres so our final answer must be in square metres.

A=9m^2Calculate the area of the rhombus below:

**Identify the length of the diagonals.**

\begin{aligned} &MO = 50cm \\\\ &NP= 1m \end{aligned}

We have two different units for the length and width so we must change the measures to a common unit.

In this case, let us change both units to centimetres.

As 1m=100cm .

\begin{aligned}
&MO = 50cm \\\\
&NP = 100cm
\end{aligned}

**Write down the formula for the area of a rhombus.**

A=\frac{1}{2}\times{D}\times{d}

**Substitute the given values of the diagonals and solve.**

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{100}\times{50}\\\\
&=2500
\end{aligned}

**Write down your final answer, including the units.**

As the area has been calculated using the lengths in centimetres, the area unit is square centimetres.

A=2500\text{ cm}^2

A rhombus ABCD has the point O as the point of intersection of the two diagonals AC and BD . The length OB=3cm , and OC=5cm . Calculate the area of the rhombus.

**Identify the length of the diagonals.**

As the point O is the intersection of the two diagonals, the lengths OB and OC are half of the length of each diagonal. This means that as OB=3cm, \ AB=3 \times 2=6cm and as OC=5cm, \ CD=5 \times 2=10cm.

**Write down the formula for the area of a rhombus.**

A=\frac{1}{2}\times{D}\times{d}

**Substitute the given values of the diagonals and solve.**

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{10}\times{6}\\\\
&=30
\end{aligned}

**Write down your final answer, including the units.**

As the diagonal lengths are given in centimetres, the area unit is square centimetres.

A=30\text{ cm}^2

Calculate the length of AC given the length of BD=14m and the area of the rhombus is 35m^2 . Use the diagram below to help you.

**Identify the length of the diagonals.**

\begin{aligned} &BD = 14m \\\\ &AC = x \end{aligned}

In this case we are also given Area = 35m^2 .

**Write down the formula for the area of a rhombus.**

A=\frac{1}{2}\times{D}\times{d}

**Substitute the given values of the diagonals and solve.**

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d} \\\\
35&=\frac{1}{2}\times{14}\times{x} \\\\
35&=7x \\\\
x&=5
\end{aligned}

**Write down your final answer, including the units.**

In this case we are asked to find the length of a diagonal and not area so our answer should be written in metres:

AC=5\text{ m}

ABCD is a rhombus where the edges AB and CD are parallel. Given that AB=20cm and BD=32cm, calculate the area of the rhombus.

**Identify the length of the diagonals.**

\begin{aligned} &BD = 32cm \\\\ &AC = x \end{aligned}

We need to calculate the length of AC. In this question we are given one side of the rhombus which is AB=20cm .

Note: Since we are given the length of the side AB we now know the length of any side of the rhombus since all sides of a rhombus are the same length.

We can split the rhombus into 4 right-angled triangles. If we take triangle OAB , the hypotenuse AB=20cm and OB=16cm since it is half the length of BD . We now need to calculate the length of OA .

As OAB is a right-angled triangle, we can use Pythagoras’ theorem to calculate OA .

\begin{aligned} c^{2}&=a^{2}+b^{2}\\\\ 20^{2}&=16^{2}+x^{2}\\\\ 400&=256+x^{2}\\\\ x^{2}&=144\\\\ x&=\sqrt{144}\\\\ x&=12 \end{aligned}

As x=OA=12cm, \ AC=12 \times 2=24cm

Now that we have the length of two of the diagonals of the rhombus, we can calculate the area of the rhombus ABCD .

**Write down the formula for the area of a rhombus.**

A=\frac{1}{2}\times{D}\times{d}

**Substitute the given values of the diagonals and solve.**

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{32}\times{24}\\\\
&=384
\end{aligned}

**Write down your final answer, including the units.**

The area of the rhombus ABCD is 384cm^2 .

A rhombus has a side length of 13cm . One pair of interior angles is double the other pair of interior angles. Calculate the area of the rhombus to 2 decimal places.

**Identify the length of the diagonals.**

Sketching a diagram first, we have:

As the sum of angles in a rhombus is 360^{\circ} , we can quickly calculate the value of x by forming and solving an equation:

\begin{aligned} 2x+x+2x+x&=360 \\\\ 6x&=360 \\\\ x&=60^{\circ} \end{aligned}

So the angles in the rhombus are 60^{\circ} and 120^{\circ} . If we filled these in on the diagram, we have:

We can calculate the length AC by using the cosine rule c^{2}=a^{2}+b^{2}-2ab\cos(\theta) , looking at the top half of the rhombus, we have:

Labelling the sides and angles according to the formula ( \theta must be the included angle between the two sides a and b , we have:

Substituting the values of a, b, and \theta into the cosine rule, we have:

\begin{aligned} c^{2}&=a^{2}+b^{2}-2ab\cos(\theta) \\\\
c^{2}&=13^{2}+13^{2}-2\times{13}\times{13}\times\cos(120)\\\\
c^{2}&=507\\\\
c&=13\sqrt{3}
\end{aligned}

So the long diagonal AC=13\sqrt{3}\text{ cm}

Repeating the same process for the diagonal BD, we have:

Substituting the values of a, b, and \theta into the cosine rule, we have:

\begin{aligned} c^{2}&=a^{2}+b^{2}-2ab\cos(\theta) \\\\ c^{2}&=13^{2}+13^{2}-2\times{13}\times{13}\times\cos(60) \\\\ c^{2}&=169 \\\\ c&=13 \end{aligned}

So the short diagonal BD=13\text{ cm} .

(For this example, BCD is an equilateral triangle!)

**Write down the formula for the area of a rhombus.**

A=\frac{1}{2}\times{D}\times{d}

**Substitute the given values of the diagonals and solve.**

As the long diagonal AC=13\sqrt{3}\text{ cm} and the short diagonal BD=13\text{ cm}, we have:

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{13\sqrt{3}}\times{13}\\\\
&=146.3582932…\\\\
&=146.36\text{ cm}^2\text{ (2dp)}
\end{aligned}

**Write down your final answer, including the units.**

The area of the rhombus ABCD is 146.36cm^2 \ (2dp). .

**Using incorrect units for the answer**

A common error is to forget to include squared units when asked to calculate area.

**Forgetting to convert measures to a common unit**

Before using the formula for calculating the area of a rectangle we need to ensure that units are the same. If different units are given (E.g. length = 4m and width = 3cm ) then you must convert them either both to cm or both to m.

**Using half of the diagonal**

As the area of a rhombus is half the length of one diagonal multiplied by the length of the other diagonal, the other diagonal is also halved within the calculation, leaving a solution that is out by a factor of 2.

**Calculating the area of a square (using side lengths)**

The area of a rhombus is calculated by using the side lengths, instead of the lengths of the diagonals.

1. Find the area of the rhombus below:

24m^2

48m^2

16m^2

48m

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{12}\times{4}\\\\
&=24
\end{aligned}

A=24\text{ m}^2

2. The rhombus MNOP has the following information:

- MP=250cm

- NO=4.5m

Calculate the area of the rhombus in square centimetres.

11250cm^2

28125cm^2

56.25cm^2

56250cm^2

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{250}\times{450}\\\\
&=56250
\end{aligned}

A=56250\text{ cm}^2

3. Mr. Perry owns a farm and is looking to make the chicken enclosure shown below. Each chicken needs 4 square metres to move around freely. What is the maximum number of chickens that can fit into the enclosure?

24

48

96

192

Calculating the area of the rhombus, we have

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{12}\times{16}\\\\ &=96 \end{aligned}

A=96\text{ m}^2

As the field is 96m^2 and each chicken requires 4m^2, \ 96\div{4}=24 chickens

4. The area of rhombus ABCD is 54m^2. The centre of the rhombus is located at point P . Given that AP=3m , calculate the length BD.

9m

18m

54m

324m

As A=54m^2 and AP=3m, \ AC=32=6m.

Using the formula for the area of a rhombus, we have

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ 54&=\frac{1}{2}\times{6}\times{x}\\\\ 54&=3x\\\\ x&=18 \end{aligned}

x=18\text{ m}

5. In the rhombus below AB = 5m, \ BD = 8m. Calculate the area of the rhombus.

24m^2

48m^2

6m^2

20m^2

We can calculate half the length of AC by using Pythagoras’ theorem. Let M be the centre of the rhombus. The distance BM is half the distance BD and so BM=4m.

Using Pythagoras’ theorem to calculate the length of AM , we have:

\begin{aligned} AM^{2}&=5^{2}-4^{2}\\\\ &=25-16\\\\ &=9\\\\ AM&=3 \end{aligned}

As AM=3m, \ AC=3 \times 2=6m.

The area of the rhombus is therefore:

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{6}\times{8}\\\\ &=24 \end{aligned}

A=24\text{ m}^2

6. (Higher only) A rhombus PQRS has side length 5cm . The rhombus has two pairs of interior angles, equal to x and 3x-20 .

Use the cosine rule c^2=a^2+b^2-2ab \cos(\theta) to calculate the area of the rhombus to 1 decimal place.

19.2cm^2

12.5cm^2

25.0cm^2

20.0cm^2

First we need to calculate the value of the two pairs of angles in the rhombus:

\begin{aligned} x+3x-20+x+3x-20&=360\\\\ 8x-40&=360\\\\ 8x&=400\\\\ x&=50 \end{aligned}

x=50^{\circ}

3x-20=130^{\circ}

So we have a rhombus with a side length of 5cm and the interior angles are 50^{\circ} and 130^{\circ}.

Calculating the distance PR using the cosine rule, we have:

PR=\sqrt{5^{2}+5^{2}-2\times{5}\times{5}\times\cos(130)}=9.06307787

Calculating the distance QS using the cosine rule, we have:

QS=\sqrt{5^{2}+5^{2}-2\times{5}\times{5}\times\cos(50)}=4.226182617

The area of the rhombus is therefore:

\begin{aligned}
A&=\frac{1}{2}\times{D}\times{d}\\\\
&=\frac{1}{2}\times{9.06307787}\times{4.226182617}\\\\
&=19.15111108
\end{aligned}

A=19.2\text{ cm}^2\text{ (1dp)}

1. Shown below is a rhombus where AM=6m , and BM=10m . Calculate the area of the rhombus.

**(3 marks)**

Show answer

AC=12m, \ BD=20m

**(1)**

\frac{12 \times 20}{2}

**(1)**

120m^2

**(1)**

2. (a) A rectangular logo is composed of 8 congruent rhombuses. Calculate the area of one rhombus.

(b) Calculate the perimeter of the logo. Write your answer as a surd in its simplest form.

**(6 marks)**

Show answer

(a)

12\div{3}=4\text{m and }4\div{2}=2\text{m}

**(1)**

\frac{4 \times 2}{2}

**(1)**

4\text{ m}^{2}

**(1)**

(b)

2^{2}+1^{2}=5

**(1)**

Side length of a rhombus =\sqrt{5}

**(1)**

16\sqrt{5}

**(1)**

3. Ms. Polly is looking to cover her garden with new grass. One roll of grass covers 6 square metres and costs £9. How much would it cost to cover the entire ground with grass?

**(5 marks)**

Show answer

\sqrt{25^{2}-7^{2}}

**(1)**

24m

**(1)**

\frac{48 \times 14}{2}

**(1)**

(336 \div 6) \times 9

**(1)**

£504

**(1)**

You have now learned how to:

- Calculate the area of a rhombus and related composite shapes
- Calculate the area of parallelograms
- Recognise when it is possible to use formulae for area
- Derive and apply formulae to calculate and solve problems involving: perimeter and area of triangles and parallelograms

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