Area Of Quadrilateral - GCSE Maths - Steps, Examples & Worksheet

Area Of A Quadrilateral

Here we will learn about about the area of a quadrilateral including how to find the area of a rectangle, square, parallelogram and trapezium. You will also learn how to find the area of compound shapes made from more than one quadrilateral and find missing lengths given an area.

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

What is area of a quadrilateral?

Area of a quadrilateral is a measure of how much space there is inside of a 2 dimensional shape four sided shape. To find the area of a shape we can either count the number of unit squares within a shape or use the appropriate area formula for that shape.

Area is measured in square units e.g. cm2, m2, mm2.

E.g.

This rectangle contains 15 squares.

\[\begin{aligned} \text{Area of a rectangle} &= \text{base} \times \text{height}\\\\ &= 5 \times 3\\\\ &= 15cm^2 \end{aligned}\]

What is area?

What are quadrilaterals?

A quadrilateral is a closed 2 dimensional shape with 4 straight sides and 4 vertices.

Regular quadrilaterals have sides that are all the same length and interior angles that are the same size.

Irregular quadrilaterals have sides that have different lengths and interior angles that have different sizes.

There are several different types of quadrilaterals:

Square:

A square is a closed 2 dimensional shape with four straight sides of equal length and four right angles (90°).

Rectangle:

A rectangle is aclosed 2 dimensional shape with four straight sides and four right angles (90°). It has two pairs of parallel sides that are equal.

Parallelogram:

A parallelogram is aclosed 2 dimensional shape with four straight sides. The opposite sides of a parallelogram have the same lengths and are parallel.

Rhombus:

A rhombus is aclosed 2 dimensional shape with four equal straight sides.

Trapezium:

A trapezium is aclosed 2 dimensional shape with four straight sides and only one pair of parallel sides.

Kite:

A Kite is aclosed 2 dimensional shape with four straight sides. It has two pairs of adjacent sides (sides that are next to each other )that are equal to each other.

What are quadrilaterals?

How to name quadrilaterals

We usually refer to quadrilaterals by using assigning letters to each of their vertices.

E.g.This is quadrilateral ABCD:

How to calculate the area of a rectangle, square, parallelogram and trapezium

We can use formulae to calculate the area of the following shapes:

Area of a rectangle/square:

\[\text {Area of rectangle/square = base × height}\]

Area of a parallelogram:

\[\text {Area of parallelogram = base × height} \]

In order to calculate the area of a rectangle, square or parallelogram:

1. Substitute the values into the formula. (Make sure the units are the same for all measurements e.g. all cm).
2. Work out the calculation.
3. Add the correct units.

Area of a trapezium:

A Trapezium is a trapezoid shape, meaning that it only has one pair of parallel sides. In order to calculate the area of a trapezium we need to use the following formula.

\[\text { Area of a trapezium }=\frac{1}{2}(a+b) h \]

In order to calculate the area of a trapezium:

1. Substitute the values into the formula.
2. Do the calculation.
3. Add the correct units.

We can use the following formula to work out the area of a rhombus and the area of a kite:

\[\text { Area of a rhombus or kite }=\frac{1}{2} \times d_{1} \times d_{2} \]

Where d1 and d2 are the diagonal lengths.

Area of quadrilateral worksheet

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

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Area of quadrilateral worksheet

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

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Related lessons on area

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

Area of quadrilateral examples

Example 1: area of a square

Find the area of the quadrilateral:

1. Here the height is 8. Since it is a square, the base is also 8.

\begin{aligned} \text { Area }&=\text { base } \times \text { height } \\\\ \text { Area }&=8 \times 8 \end{aligned}

2 Work out the calculation.

\[Area = 64\]

3 Add the correct units.

The measurements on this square are in cm so the area will be measured in cm2:

\[Area = 64cm^2\]

Example 2: area of a rectangle, different units

Find the area of the rectangle:

Here the base is 11m and the height is 400cm.

Notice that the base measurement is in m whereas the height measurement is in cm.

The easiest thing to do here is to change 400cm to 4m so that both measurements are in m. (You could also change 11m to 1100cm.

\begin{aligned} \text { Area }&=\text { base } \times \text { height } \\\\ \text { Area }&=11 \times 4 \end{aligned} \[Area = 44\]

The measurements on this rectangle are in m so the area will be in m2:

\[Area = 44m^2\]

Example 3: area of a parallelogram

Calculate the area of the parallelogram:

\begin{aligned} \text { Area }&=\text { base } \times \text { height } \\\\ \text { Area }&=12 \times 8 \end{aligned} \[Area = 96\]

The measurements on this parallelogram are in mm so the area will be in mm2.

\[Area = 96mm^2\]

Example 4: area of a trapezium

Find the area of the following trapezium:

Here a = 9, b = 13 and h = 5

\begin{aligned} \text { Area }&=\frac{1}{2}(a+b) h \\\\ \text { Area }&=\frac{1}{2} \times(9+13) \times 5 \end{aligned}

Remember to apply BIDMAS when working this out.

\begin{aligned} \text { Area }&=\frac{1}{2} \times(9+13) \times 5 \\\\ \text { Area }&=\frac{1}{2} \times 22 \times 5 \\\\ \text { Area }&=55 \end{aligned}

The measurements on this trapezium are in cm so the area will be in cm2 .

\[Area = 55cm^2\]

Example 5: area of a trapezium

Find the area of the following trapezium:

Here a = 5, b = 6 and h = 4

\begin{aligned} \text { Area }&=\frac{1}{2}(a+b) h \\\\ \text { Area }&=\frac{1}{2} \times(5+6) \times 4 \end{aligned} \begin{aligned} \text { Area }&=\frac{1}{2} \times(5+6) \times 4 \\\\ \text { Area }&=\frac{1}{2} \times 11 \times 4 \\\\ \text { Area }&=22 \end{aligned}

The measurements on this trapezium are in km so the area will be in km2.

\[Area = 22km^2\]

How to calculate the area of compound shapes with quadrilaterals

Sometimes a shape is made from two or more quadrilaterals put together. We can calculate the area of these shapes by breaking them down into individual quadrilaterals.

1. Draw lines to split the shape into quadrilaterals (this step is not always necessary). Label the quadrilaterals A, B, C,…
2. Consider each quadrilateral individually a) work out any measurements that you need. b) calculate the area using the methods above.
3. Add or subtract the relevant areas to find the total area.
4. Add the correct units.

Example 6: compound shape made from 2 quadrilaterals

Find the total area of the shape shown below:

This shape is already split into two quadrilaterals. Label them A and B.

Quadrilateral A:

a) For the rectangle we need to know the base and the height, which are shown.

b)

\begin{aligned} \text { Area }&=\text { base } \times \text { height } \\\\ \text { Area }&=6 \times 3 \\\\ \text { Area }&=18 \mathrm{~cm}^{2} \end{aligned}

Quadrilateral B:

a) For the trapezium, we need to know a, b and h.

We can see a = 4. b is the top of the rectangle and so it will be 6cm.

To get h we must subtract the height of the rectangle from the total height of the shape:

\begin{aligned} h&=5-3 \\\\ h&=2 \mathrm{~cm} \end{aligned}

b)

\begin{aligned} \text { Area } &=\frac{1}{2}(a+b) h \\\\ \text { Area } &=\frac{1}{2} \times(4+6) \times 2 \\\\ \text { Area } &=\frac{1}{2} \times 10 \times 2 \\\\ \text { Area } &=10 \mathrm{~cm}^{2} \end{aligned}

Add the areas together:

\[18 + 10 = 28\]

Total area = 28cm2

Example 7: compound shape made from three quadrilaterals

Find the total area of the shape shown below:

Split the shape into quadrilaterals and label them A, B and C. This can be done in two different ways – whichever way you choose will work.

Using the first way of splitting:

Quadrilateral A:

a) The base of quadrilateral A is 7m and the height is 20.

b)

\begin{aligned} \text { Area }&=7 \times 20 \\\\ \text { Area }&=140 \mathrm{~m}^{2} \end{aligned}

Quadrilateral B:

a) The base is 10m. The height is 20 – 11 = 9m.

b)

\begin{aligned} \text { Area }&=10 \times 9 \\\\ \text { Area }&=90 \mathrm{~m}^{2} \end{aligned}

Quadrilateral C:

a) The base is 8m and the height is 20m.

b)

\begin{aligned} \text { Area }&=8 \times 20 \\\\ \text { Area }&=160 \mathrm{~m}^{2} \end{aligned}

Add the areas together:

\[140 + 90 + 160 = 390\]

Total area = 390m2

Example 8: subtracting areas

Find the shaded area:

The shape is already split into two quadrilaterals. Label them A and B (A is the big trapezium, B is the small trapezium).

Quadrilateral A:

a) Here a = 12, b = 18 and h = 20.

b)

\begin{aligned} \text { Area } &=\frac{1}{2} \times(12+18) \times 20 \\\\ \text { Area } &=\frac{1}{2} \times 30 \times 20 \\\\ \text { Area } &=300 \mathrm{~cm}^{2} \end{aligned}

Quadrilateral B:

a) a = 4, b = 8 and h = 10

b)

\begin{aligned} \text { Area } &=\frac{1}{2} \times(4+8) \times 10 \\\\ \text { Area } &=\frac{1}{2} \times 12 \times 10 \\\\ \text { Area } &=60 \mathrm{~cm}^{2} \end{aligned}

This time we need to subtract the areas as the non-shaded area is being removed from the shaded area.

\[300 – 60 = 240\]

Shaded area = 240cm2

How to find a missing length

We will sometimes be given the area of a quadrilateral and then need to calculate an unknown length.

1. Put the values you know into the formula.
2. Solve the equation.

Example 9: missing length in a rectangle

Work out the height of the rectangle:

Here the area is 40 and the base is 10.

\begin{aligned} \text { Area }&=\text { base } \times \text { height } \\\\ 40&=10 \times h \end{aligned}

To solve this equation we need to divide by 10:

\[4 = h\]

The height of the rectangle is 4cm.

Example 10: missing length in a trapezium

Find the height of the following trapezium given that it has an area of 56cm2:

Here a = 6, b = 10 and the area is 56.

\begin{aligned} \text { Area }&=\frac{1}{2}(a+b) h \\\\ 56&=\frac{1}{2} \times(6+10) \times h \end{aligned} \begin{aligned} 56&=\frac{1}{2} \times(6+10) \times h \\\\ 56&=\frac{1}{2} \times 16 \times h \\\\ 56&=8 \times h \\\\ 7&=h \end{aligned}

The height of the trapezium is 7cm.

Common misconceptions

• Calculating perimeter instead of area

• Height of parallelogram

Using the wrong measurement for the height of a parallelogram.

• Wrong formula for area of a trapezium

A common error is to use the wrong formula for area of a trapezium, for example using the formula for area of a triangle instead

• Units

Using measurements with different units. Remember to make sure the units are the same for each length (e.g. all cm).

Practice area of quadrilateral questions

1. Find the area of the following square:

9 \mathrm{~m}^{2} 12 \mathrm{~m}^{2} 6 \mathrm{~m}^{2} 27 \mathrm{~m}^{2} \begin{aligned} \text{Area }&=\text{ base }\times \text{ height}\\\\ &= 3 \times 3\\\\ &=9 \mathrm{m}^{2} \end{aligned}

2. Find the area of the following parallelogram. Give your answer in cm^2.

420 \mathrm{~cm}^{2} 42 \mathrm{~cm}^{2} 36 \mathrm{~cm}^{2} 4200 \mathrm{~cm}^{2} First we need to make the units the same. Here 70mm=7cm. \begin{aligned} \text{Area }&=\text{ base }\times \text{ height}\\\\ &= 7 \times 6\\\\ &=42 \mathrm{~cm}^{2} \end{aligned}

3. Find the area of the following trapezium:

27 \mathrm{~m}^{2} 24 \mathrm{~m}^{2} 54 \mathrm{~m}^{2} 45 \mathrm{~m}^{2} \begin{aligned} \text{Area }&=\frac{1}{2}(a+b)h\\\\ &=\frac{1}{2}(6+9) \times 6\\\\ &=\frac{1}{2} \times 15 \times 6\\\\ &=45\mathrm{m}^{2} \end{aligned}

4. Find the area of the following shape:

96 \mathrm{~cm}^{2} 144 \mathrm{~cm}^{2} 132 \mathrm{~cm}^{2} 88 \mathrm{~cm}^{2}

Quadrilateral A: \begin{aligned} \text{Area }&=3 \times 4\\\\ &=12\mathrm{~cm}^{2} \end{aligned} Quadrilateral B: \begin{aligned} \text{Area }&=12 \times 7\\\\ &=84\mathrm{~cm}^{2} \end{aligned} \text{Total area: }12+84=96\mathrm{~cm}^{2}

5. Find the shaded area:

600 \mathrm{~m}^{2} 516 \mathrm{~m}^{2} 480 \mathrm{~m}^{2} 432 \mathrm{~m}^{2}

Quadrilateral A:

\begin{aligned} \text{Area }&=30\times20\\\\ &=600\mathrm~{~m}^{2} \end{aligned}

Quadrilateral B:

\begin{aligned} \text{Area }&=\frac{1}{2}(6+14)\times12\\\\ &=\frac{1}{2}\times 20 \times 12\\\\ &=120 \mathrm{~m}^{2} \end{aligned} \text{Shaded area: }600-120=480\mathrm{~mm}^{2}

6. Find the height of the following parallelogram:

 

144mm 4mm 18mm 8mm \begin{aligned} \text{Area }&= \text{ base } \times \text{ height}\\\\ 24&=6h\\\\ 4&=h \end{aligned}

Area of quadrilateral GCSE questions

1. A plan of Rosie’s garden is shown below.

Rosie wants to buy grass seed to grow a lawn in the spaces not covered by the patio and the vegetable patch.

Each box of grass seed covers 20m^2 and costs £5.50 .

How much will Rosie need to spend on grass seed?

(5 marks)

Show answer

Total area: \frac{1}{2} \times (8+12) \times 18

= 180 \mathrm{~m}^{2} 

(1)

Area of patio: 5 \times 8 = 40 \mathrm{~m}^{2}

Area of vegetable patch: 3 \times 9 = 27 \mathrm{~m}^{2} 

(1)

Area to be seeded: 180-40-27=113\mathrm{~m}^{2}

(1)

Boxes of seed: 113 \div 20 = 5.65

(1)

Need 6 boxes, so 6 \times \pounds 5.50 = \pounds 33 she will have to spend

(1)

2. Rita wants to tile a section of her kitchen wall, as shown below:

(a) Calculate the area of the wall that Rita wants to tile. Give your answer in cm^2.

(b) The tiles that Rita has chosen are square, with side length 20cm. How many tiles will Rita need?

(5 marks)

Show answer

(a)

1.2m=120cm, 1.6m=160cm, 1m=100cm

(1)

\begin{array}{l} 120 \times 160 = 19200 \mathrm{~cm}^{2}\\\\ 320 \times 40 = 12800 \mathrm{~cm}^{2} \end{array}

(1)

Total area: 19200 + 12800 = 32000 \mathrm{~cm}^{2}

(1)

(b)

20 \times 20 = 400

(1)

3200 \div 400 = 80

(1)

3. The shape below is made from two identical rectangles. The area of the shape is 120cm^2

Calculate the height of the shape.

(3 marks)

Show answer

Area of one rectangle: 60 \mathrm{~cm}^{2}

(1)

\begin{aligned} \text { Area }&=\text { base } \times \text { height } \\\\ 60&=5 \times h \end{aligned}

(1)

h = 12cm

(1)

Learning checklist

You have now learned how to:

The next lessons are

• How to work out perimeter
• Symmetry
• Circles, sectors and arcs

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