## What you need to know

The **area** of a 2D shape is the amount of space it takes up in 2 dimensions, and its units are always squared, e.g \text{cm}^2,\hspace{1mm}\text{m}^2, and so on. You should know how to calculate the areas of the following four shapes: rectangles, triangles, parallelograms, and trapezia. The formulae you need for these four shapes are shown below.

It’s important to know that wherever the height of a shape is used in a formula, it refers to the perpendicular height. In other words, the lines marked ‘h’ must be at right-angles to the base, marked ‘b’.

**Example: **The shape below is a trapezium with a perpendicular height of 4mm. Work out its area.

Our formula is \frac{1}{2}(a+b)h where in this case:

a = 8,\hspace{1mm}b = 12.5, and h = 4 . So, we get:

\begin{aligned}\text{Area } &= \dfrac{1}{2}(8+12.5) \times 4 \\ &=\dfrac{1}{2} \times 20.5 \times 4 \\ &= 41\text{mm}^2 \end{aligned}

**Example: **The triangle below has a base of 6cm and an area equal to 24\text{cm}^2. Calculate its perpendicular height.

Our formula is \dfrac{1}{2}(a+b)h where in this case:

a = 8,\hspace{1mm}b = 12.5, and h = 4 . So, we get:

\begin{aligned}\text{Area } &= \dfrac{1}{2}(8+12.5) \times 4 \\ &=\dfrac{1}{2} \times 20.5 \times 4 \\ &= 41\text{mm}^2 \end{aligned}

## GCSE Maths Revision Cards

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On the higher course, there is another formula for calculating the area of a triangle that you’ll need to remember.

You’ll notice that the triangle is labelled so the angles are capital-letter angles correspond to the lower-case letter sides, i.e. they’re always opposite one another. This is a useful way of labelling a triangle, and it’s important to remember because it comes up again for the sine rule and cosine rule. In this case, what you need to understand is that the three things needed to calculate the area using this formula are two side-lengths and the angle which is formed by those sides (the ones that you know the length of).

**Example: **Find the area of the triangle below to 2 significant figures.

We have the lengths of two sides as well as the angle between them, so this triangle is primed for our latest formula. In this case, our a\text{ and }b are 6.7 and 7.2 (and it’s not important which way round they go), whilst our C is 68\degree. So:

\text{Area }= \dfrac{1}{2}\times 7.2 \times 6.7 \times \sin(68) = 23 \text{cm}

(to 2s.f.).

If you are looking for areas of shapes revision resources, look no further. From trapezium area questions to finding the area of a non-right angled triangle, the resources on this page cover the full spectrum of foundation and higher GCSE Maths levels so that students can practise the entire shapes and area topic. For more great GCSE Maths revision materials visit our homepage.

### Example Questions

1) The triangle below has a base of 11.5cm and a perpendicular height of 12cm. Calculate its area.

The formula for the area is \frac{1}{2}bh, where b = 11.5 \text{ and } h = 12. So, we get:

\text{Area } = \dfrac{1}{2} \times 11.5 \times 12 = 69\text{cm}^2

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2) Below is a trapezium with sides of length 8cm, 5cm, and 5cm as shown below. Calculate the perpendicular height and use it to find the total area. **HINT: **Pythagoras might be useful here.

This is a bit of a weird-looking trapezium, but it is a trapezium nonetheless. To work out the area, we will, as the question states, need to find the perpendicular height. The word ‘perpendicular’ is key here, because it means that we can draw one line and form a right-angled triangle, as seen below.

Because the top of a trapezium is always parallel to the base, we can find the base of the small right-angled triangle

by subtracting the length of the top from the length of the base to get 3cm.

Then, this is where Pythagoras comes in. The hypotenuse of the right-angled triangle is 5cm and the base is 3cm, so we get the other side (the perpendicular height of the trapezium) as such:

\text{Perpendicular height } = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{cm}

Now we know the perpendicular height, we can calculate the area.

\text{Area } = \dfrac{1}{2}(a + b)h = \dfrac{1}{2}(5 + 8) \times 4 = 26\text{cm}^2

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3) **HIGHER ONLY. **The triangle below has area 1.47\text{m}^2. Work out the missing side-length to 2d.p.

As we need to find a missing side-length rather than the area, we’re going to have to set up an equation and rearrange it to find x. The formula for the area we’ll need here is

\dfrac{1}{2}ab \sin(C),

so, our equation is

\dfrac{1}{2} \times 2.15 \times x \times \sin(26) = 1.47

Simplifying the left-hand-side, we get

1.075 \sin (26) \times x = 1.47

Finally, dividing through by 1.075\sin(26), and putting it into a calculator, we get

x = \dfrac{1.47}{1.075\sin(26)} = 3.12\text{m (2 d.p.)}

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### Worksheets and Exam Questions

### Videos

#### Areas of Shapes Q1

GCSE MATHS#### Areas of Shapes Q2

GCSE MATHS#### Areas of Shapes Q3

GCSE MATHS### Other worksheets

## Areas of Shapes Worksheets and Revision

## Areas of Shapes Teaching Resources

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#### Surface Area Worksheets, Questions and Revision

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