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Types of quadrilateralsMultiplication and division

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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.

**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. ^{2}, m^{2}, mm^{2}

E.g.

This rectangle contains 15 squares.

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

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

**Rhombus**:

A rhombus is a** **closed 2 dimensional shape with four equal straight sides.

**Trapezium**:

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

**Kite**:

A Kite is a** **closed 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.

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

E.g.

This is quadrilateral

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**:

**Substitute the values into the formula. (Make sure the units are the same for all measurements e.g. all**.cm )**Work out the calculation.****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**:

**Substitute the values into the formula.****Do the calculation.****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 _{1}_{2}

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

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

DOWNLOAD FREE**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:

Find the area of the quadrilateral:

**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 ^{2}

\[Area = 64cm^2\]

Find the area of the rectangle:

**Substitute the values into the formula.**

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}

**Do the calculation.**

\[Area = 44\]

**Add the correct units.**

The measurements on this rectangle are in m so the area will be in m^{2}:

\[Area = 44m^2\]

Calculate the area of the parallelogram:

**Substitute the values into the formula.**

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

**Do the calculation.**

\[Area = 96\]

**Add the correct units.**

The measurements on this parallelogram are in mm so the area will be in mm^{2}.

\[Area = 96mm^2\]

Find the area of the following trapezium:

**Substitute the values into the formula.**

Here

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

**Do the calculation.**

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}

**Add the correct units.**

The measurements on this trapezium are in cm so the area will be in cm^{2} .

\[Area = 55cm^2\]

Find the area of the following trapezium:

**Substitute the values into the formula.**

Here

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

**Do the calculation.**

\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}

**Add the correct units.**

The measurements on this trapezium are in km so the area will be in km^{2}.

\[Area = 22km^2\]

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.

**Draw lines to split the shape into quadrilaterals (this step is not always necessary). Label the quadrilaterals A, B, C,…****Consider each quadrilateral individually**a) work out any measurements that you need.

b) calculate the area using the methods above.**Add or subtract the relevant areas to find the total area.****Add the correct units.**

Find the total area of the shape shown below:

**Draw lines to split the shape into quadrilaterals if necessary. Label the quadrilaterals A, B, C,…**

This shape is already split into two quadrilaterals. Label them

**Consider each quadrilateral individually and:****a) work out any measurements that you need.b) calculate the area using the methods above**.

**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

We can see

To get

\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 or subtract the areas**.

Add the areas together:

\[18 + 10 = 28\]

**Add the units.**

Total area ^{2}

Find the total area of the shape shown below:

**Draw lines to split the shape into quadrilaterals if necessary. Label the quadrilaterals A, B, C,… **

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.

**Consider each quadrilateral individually and:****a) work out any measurements that you need.b) calculate the area using the methods above**.

Using the first way of splitting:

**Quadrilateral A**:

**a) **The base of quadrilateral

**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

**b**)

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

**Add or subtract the areas.**

Add the areas together:

\[140 + 90 + 160 = 390\]

**Add the units.**

Total area ^{2}

Find the shaded area:

**Draw lines to split the shape into quadrilaterals if necessary. Label the quadrilaterals A, B, C,…**

The shape is already split into two quadrilaterals. Label them

**Consider each quadrilateral individually and:****a) work out any measurements that you need.b) calculate the area using the methods above**.

**Quadrilateral A:**

**a**) Here

**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) **

**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}

**Add or subtract the areas.**

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

\[300 – 60 = 240\]

**Add the units.**

Shaded area ^{2}

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

**Put the values you know into the formula.****Solve the equation**.

Work out the height of the rectangle:

**Put the values you know into the formula.**

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}

**Solve the equation.**

To solve this equation we need to divide by 10:

\[4 = h\]

The height of the rectangle is

Find the height of the following trapezium given that it has an area of 56cm^{2}:

**Put the values you know into the formula.**

Here

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

**Solve the equation.**

\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.

**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).

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}

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)**

You have now learned how to:

- Calculate the area of a square, rectangle, parallelogram and trapezium
- Calculate the area of a compound shape
- Calculate missing lengths given an area

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