GCSE Maths Geometry and Measure Area

Area Of A Triangle

# Area Of A Triangle

Here we will learn about the area of a triangle, including how to find the area of a triangle with given dimensions.

There are also area of a triangle 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 the area of a triangle?

The area of a triangle is the amount of enclosed two-dimensional space that three connected straight lines take up.

In order to derive the area of a triangle, we start with the area of a rectangle.

The area of the rectangle below would be calculated by multiplying the base by the height ( b \times h ). This is usually pronounced as  “base times height”.

We can split a rectangle into two smaller rectangles by drawing a perpendicular line through any two parallel sides of the rectangle.

Splitting these two rectangles in half by joining the diagonals together, we can create two pairs of congruent (identical) right-angled triangles.

The area of the rectangle is now halved to get the area of the triangle, so the area of the triangle is exactly half the area of the rectangle, when we know the base and the perpendicular height of the triangle.

### Area of a right angled triangle formula

The formula for the area of a triangle is:

\text{Area of a triangle}=\frac{\text{base }\times \text{ height}}{2}

This can be shortened to

A=\frac{1}{2}bh

where b is the base length and h is the perpendicular height of the triangle.

(Perpendicular means that the base and the height are at right angles to each other.)

Your final answer must be given in square units, e.g. cm^2, \ m^2, or mm^2 .

Higher – There is an alternative method to find the area of a triangle which involves sine, two sides and the included angle. This is part of trigonometry.

\text{Area of a triangle}=\frac{1}{2}ab\sin{C}

Step-by-step guide: Area of a triangle ( \frac{1}{2}ab \sin C )

On this page we’ll look at the area of scalene triangles. A scalene triangle is a triangle where all of the sides and all of the angles are different.

This means that a scalene triangle can contain three acute angles, a right-angle, or one obtuse angle (and two acute angles).

Remember: All of the angles within any triangle must add up to 180^{\circ} .

### Labelling triangles

The triangle below has the three sides labelled a, b, and c, and the three vertices labelled A, B, and C. The side a opposes the vertex A and so on.

You should also be familiar with the notation that this triangle is called triangle ABC .

Each side could also be labelled using the capital letters that denote the start and end point of the line. For example, the side c is also known as the side AB , the side a can be referred to as side BC , and side b can be expressed as side AC .

## How to calculate the area of a triangle

In order to calculate the area of a triangle:

1. Identify the base and perpendicular height of the triangle.
2. Write the area formula.
3. Substitute known values into the area formula.
4. Solve the equation.
5. Write the answer, including the units.

## Related lessons on area

Area of a triangle 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 of a triangle examples

### Example 1: right-angled scalene triangle

Calculate the area of the right-angled triangle ABC . Write your answer in square centimetres.

1. Identify the base and perpendicular height of the triangle.

The sides AC and BC are perpendicular to each other (the angle at C is a right-angle) and so we can state the values for the base and the perpendicular height of the triangle to be:

The base: b=3.2 \ cm

The perpendicular height: h=5.3 \ cm

2Write the area formula.

A=\frac{1}{2}bh

3Substitute known values into the area formula.

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(3.2)(5.3) \end{aligned}

4Solve the equation.

Work out the value for A by solving the equation:

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(3.2)(5.3) \\\\ &=\frac{1}{2}\times 3.2\times 5.3\\\\ &=8.48 \end{aligned}

5Write the answer, including the units.

A=8.48\text{ cm}^2

### Example 2: obtuse scalene triangle

For triangle ABC , the angle at C is an obtuse angle, the length AC=12cm , and the height of the triangle is 9cm . Calculate the area of triangle ABC.

As AC is perpendicular to the height stated for the triangle, we can state:

\begin{aligned} &b=12 \ cm \\\\ &h=9 \ cm. \end{aligned}

A=\frac{1}{2}bh

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(12)(9) \end{aligned}

Solving for A , we have

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(12)(9) \\\\ &=\frac{1}{2}\times 12\times 9\\\\ &=54 \end{aligned}

A=54\text{ cm}^2

### Example 3: acute scalene triangle – mixed units

Calculate the area of triangle ABC below. Write your answer in square metres.

Currently we have 2 different units here. We must convert them to a common unit (here we will use metres as the question requires the answer to be in square metres):

240 \ cm = 2.4 \ m.

So the base and the perpendicular height of the triangle are:

\begin{aligned} &b=2.4 \ m \\\\ &h= 2\ m \end{aligned}

A=\frac{1}{2}bh

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(2)(2.4) \end{aligned}

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(2)(2.4) \\\\ &=\frac{1}{2}\times 2\times 2.4\\\\ &=2.4 \end{aligned}

A=2.4\text{ m}^2

### Example 4: obtuse scalene triangle – too much information

PQR is an obtuse scalene triangle. The point S lies on the line PQ such that angle PSR=90^{\circ} .

Calculate the area of the triangle PQR below, correct to 2 decimal places.

Here, we need to distinguish which two lengths are perpendicular to one another as we can use these as the base and the height of the triangle PQR .

As the question states that angle PSR=90^{\circ} , the side of the triangle PQ is perpendicular to the length RS which is the height of the triangle. We therefore can state:

\begin{aligned} &b= 11.02 \ cm \\\\ &h= 8.97\ cm. \end{aligned}

A=\frac{1}{2}bh

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(11.02)(8.97) \end{aligned}

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(11.02)(8.97)\\\\ &=\frac{1}{2}\times 11.02\times 8.97\\\\ &=49.4247 \end{aligned}

A=49.42 \text{ cm}^2 \text{ (2dp)}

### Example 5: compound shape

A rectangle and a triangle are placed together to make a compound shape ABCDE .

Calculate the area of the compound shape to 1 decimal places.

If we split the compound shape into two polygons, we have a rectangle and a triangle.

As AB and DE are two parallel sides of a rectangle, AB=DE and so we can calculate the vertical height of the triangle by subtracting 4m from 6m. This gives us the following dimensions on the original diagram:

If we sketch another copy of the triangle and label the dimensions of it, we get:

We can now state the values for the base and perpendicular height:

\begin{aligned} &b= 5 \ m \\\\ &h= 2 \ m. \end{aligned}

A=\frac{1}{2}bh

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(5)(2) \end{aligned}

Work out the calculation

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(5)(2)\\\\ &=\frac{1}{2}\times 5\times 2\\\\ &=5 \end{aligned}

Now we must find the area of the rectangle.

\begin{aligned} A &=l\times w \\\\ &= 5\times 4\\\\ &=20 \end{aligned}

\text{Total area}=5+20=25

A=25\text{ cm}^2.

### Example 6: compound shape – including Pythagoras’ theorem

An isosceles triangle ACD and a right-angled triangle BCD share a congruent side such that when they are attached, they produce the right-angled scalene triangle ABC . Using the diagram below, determine the area of the triangle ABC. Write your answer to 2 decimal places.

The two sides that we can use for the base and the perpendicular height are AB and BC because they are perpendicular to each other. We therefore need to calculate the lengths of AB and BC separately.

To do this, we must use Pythagoras’ theorem as we do not know any other angle in the two triangles except for the angle at B.

To calculate the length of AB, we need to know the length of AD .

Let us calculate the length of the line AD

ADE is a right-angled triangle with DE=4m and AE=AC\div{2}=4\text{ cm} .

We can calculate the length of the line AD using Pythagoras’ theorem:

As AD=\sqrt{32}\text{ m} , we can say that AB=\sqrt{32}+2\text{ m}

The base b=\sqrt{32}+2\text{ m} .

Now we need to calculate the perpendicular height of the triangle BC . We can use the value for AD as AD=CD and so we have the following information for the triangle BCD :

Using Pythagoras’ theorem again, we can calculate the length of the side BC:

\begin{aligned} BC^{2}&=CD^{2}-BD^{2}\\\\ &=(\sqrt{32})^{2}-2^{2}\\\\ &=32-4\\\\ &=28\\\\ AD&=\sqrt{28} \end{aligned}

So the length AD or the perpendicular height of the triangle is:

h=\sqrt{28}\text{ m} .

So we now know:

The base b=\sqrt{32}+2\text{ m} .

The height h=\sqrt{28}\text{ m} .

A=\frac{1}{2}bh

\begin{aligned} A&=\frac{1}{2}bh\\\\ &=\frac{1}{2}(\sqrt{32}+2)(\sqrt{28}) \end{aligned}

\begin{aligned} A&=\frac{1}{2}(\sqrt{32}+2)(\sqrt{28})\\ &=\frac{1}{2}(8\sqrt{14}+4\sqrt{7})\\\\ &=4\sqrt{14}+2\sqrt{7}\\\\ &=20.25813217… \end{aligned}

A=20.26\text{ m}^2\text{ (2dp)}.

## How to find a missing side length given the area

Sometimes a question might give you the area and ask you to work out the missing length. In order to do this you must rearrange the formula to make the missing value the subject, and then solve for this letter.

In order to find a missing length given the area:

1. Write the area formula.
2. Substitute the known values into the area formula.
3. Rearrange the equation and solve.
4. Write the answer, including the units.

## Missing side length example

### Example 7: calculating the base length given the area

The area of a triangle is 45 \ cm^2 . The height of the triangle is 10 \ cm. Find the length of the base of the triangle.

A=\frac{1}{2}bh

\begin{aligned} A&=\frac{1}{2}bh\\\\ 45&=\frac{1}{2}\times b\times 10 \end{aligned}

We can solve the equation by rearranging to find the unknown b.

First we can tidy up the right-hand-side

\begin{aligned} 45&=\frac{1}{2}\times b\times 10\\\\ 45&=b\times 5 \end{aligned}

Then we divide both sides by 5.

\begin{aligned} 45&=b\times 5\\\\ \frac{45}{5}&=b\\\\ b&=9 \end{aligned}

b=9\text{ cm}

Remember: Your final answer must be in linear units only as it is a length.

### Common misconceptions

• Identifying the correct information to use

The base and the height must be at a right angle to each other. They must be perpendicular. You may be given additional sides of a triangle that are not needed. You may also need to rotate the triangle around so that you can identify the base and height more easily.

• Units

It is common to forget the units for area in the final answer. When calculating area, your answer must always have units squared (or square units).

### Practice area of a triangle questions

1. Calculate the area of the triangle:

8\text{ cm}^2

7.5\text{ cm}^2

15\text{ cm}^2

16\text{ cm}^2

The base and the perpendicular height are 5 \ cm and 3 \ cm respectively, so the calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(5)(3) \\\\ &=\frac{1}{2}\times 5\times 3 \\\\ &=7.5 \end{aligned}

The units will be cm^2.

2. Find the area of the triangle. Give your answer in cm^2:

126\text{ cm}^2

252\text{ cm}^2

12.6\text{ cm}^2

25.2\text{ cm}^2

The base is 63 \ mm which needs to be converted to 6.3 \ cm. The height is 4 \ cm. The calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(6.3)(4) \\\\ &=\frac{1}{2}\times 6.3\times 4 \\\\ &=12.6 \end{aligned}

The units will be cm^2.

3. Find the area of the triangle:

24\text{ cm}^2

21\text{ cm}^2

12\text{ cm}^2

10.5\text{ cm}^2

It may be easier to see the base and height if you turn the triangle. The base and the height are 3 \ cm and 7 \ cm. The 8 \ cm is not needed. So the calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(3)(7) \\\\ &=\frac{1}{2}\times 3\times 7 \\\\ &=10.5 \end{aligned}

The units will be cm^2.

4. Triangle ABC is an obtuse scalene triangle. The angle at A=120^{\circ} . Calculate the area of the triangle. Give your answer in m^2 :

13.8 \text{ m}^2

13 \ 800 \text{ cm}^2

6.9\text{ m}^2

69 \ 000\text{ cm}^2

The base is 460 \ cm which needs to be converted to 4.6 \ m. The height is 3 \ m. The calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(4.6)(3) \\\\ &=\frac{1}{2}\times 4.6 \times 3 \\\\ &=6.9 \end{aligned}

The units will be m^2.

5. Find the area of the compound shape. Give your answer in m^2 :

1250\text{ m}^2

500\text{ m}^2

750\text{ m}^2

2000\text{ m}^2

The compound shape is made up of a rectangle and a triangle.

The area of the rectangle is 10\times{50}=500\text{ m}^2

The area of the triangle is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(50)(30) \\\\ &=\frac{1}{2}\times 50 \times 30 \\\\ &=750 \end{aligned}

The total area will be 500+750=1250

The units will be m^2.

6. The area of a triangle is 72 \ cm^2 . It has a height of 12 \ cm . Find the base length of the triangle.

24 \ cm

12 \ cm

6 \ cm

18 \ cm

The area is 84 \ cm^2 and the height is 14 \ cm. The calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ 84&=\frac{1}{2}(b)(14) \\\\ 84&= b\times 7 \\\\ b&=84\div 7 \\\\ b&=12 \end{aligned}

The units will be cm.

### Area of a triangleGCSE questions

1. A triangle has a base of 6 \ cm and perpendicular height 7.2 \ cm.

Work out the area of the triangle.

(2 marks)

\frac{1}{2}\times 6\times 7.2

(1)

21.6 \ cm^2

(1)

2. The diagram shows a triangular lawn.

A gardener wants to buy grass seed for this triangular lawn.

A packet of grass seed covers 15 \ m^2 of ground.

A packets of grass seed costs £6.99.

Calculate the cost of buying grass seed to cover the lawn.

(4 marks)

\frac{1}{2}\times 9.5\times 8.7=41.325

(1)

41.325\div 15=2.755\text{ or }3

(1)

3\times 6.99

(1)

£20.97

(1)

3. Here is a triangle and a rectangle.

The area of the rectangle is 5 times the area of the triangle.

Work out the width x of the rectangle.

(4 marks)

\frac{1}{2}\times 6\times 4=12

(1)

12\times 5=60

(1)

60\div 8 = 7.5

(1)

7.5 \ cm

(1)

## Learning checklist

You have now learned how to:

• Calculate the area of triangles
• Derive and apply the formula to calculate and solve problems involving area of triangles

## The next lessons are

### Beyond GCSE maths – Heron’s formula

The ancient Greeks used a different method for finding the area of a triangle. This is a useful method as we can work out the area of a triangle if we know the side lengths of the triangle. It is based on the semi-perimeter of the triangle (half of the perimeter).

The semi-perimeter is s and the side lengths of the triangle are a, b, and c.

s=\frac{a+b+c}{2}

\text{Area of triangle}=\sqrt{s(s-a)(s-b)(s-c)}

This formula is known as Heron’s formula.

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#### GCSE Maths Papers - November 2022 Topics

Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

Designed to help your GCSE students revise some of the topics that are likely to come up in November exams.