GCSE Maths Geometry and Measure

Area

Area of a Right Angled Triangle

# Area of a Right Angled Triangle

Here we will learn about the area of a right angled triangle including how to find the area of a right angled triangle with given lengths and how to calculate those lengths if they are not given.

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 a right angled triangle?

A right angled triangle is a triangle with one right angle (90° angle). The little square in the corner tells us it is a right angled triangle.

There are two types of right triangles:

Isosceles right angled triangle

-two equal angles (45° each)

-two equal side lengths

Scalene right angled triangle

-all angles different sizes

-all sides different lengths

• The relationship between the angles and side lengths of the triangle form the basis of the trigonometric functions sine, cosine and tangent.

Step by step guide: Trigonometry

• The side of the triangle opposite the right angle is the longest side of the triangle and is referred to as the hypotenuse.

Step by step guide: Hypotenuse (coming soon)

• Right triangles are also used in the Pythagorean Theorem:
$a^{2}+b^{2}=c^{2}$

Step by step guide: Pythagoras’ Theorem (coming soon)

### How to name a triangle

We can identify a triangle by putting a capital letter on each vertex (corner).

We can then refer to each of the sides of the triangle by using two letters to describe where the line starts and ends.

We can refer to the entire triangle by using all three letters.

E.g.

Name of sides:
side AB, side AC, side BC

Name of triangle:
triangle ABC

### How do we find the area of a triangle?

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

To find the area of a rectangle you must multiply adjacent sides together.

The area of the rectangle below would be calculated by multiplying the base x height
(b x h).

We can split a rectangle into 2 congruent (identical) right angled triangles.

So the area of each of the right triangles is exactly half the area of the rectangle.

### Area of a right angled triangle formula

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

This can be shortened to

$A=\frac{1}{2} b h$

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

Your final answer must be given in units2 (e.g. cm2, m2, mm2).

## How to find the area of a right angled triangle

In order to find the area of a right angled triangle:

1Identify the height and base length of your triangle (you might need to calculate these values)

2Write the formula

$A=\frac{1}{2} b h$

3Substitute the values for base and height

4Calculate

## Area of right angled triangle examples

### Example 1: given base length and height

Find the area of the triangle below:

1. Identify the height and base length of the triangle

h = 7cm

b = 8cm

2Write down the formula

$A=\frac{1}{2} b h$

3Substitute the values for the base and height

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(8)(7) \end{aligned}

4Calculate

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(8)(7)\\ &=28cm^2 \end{aligned}

Remember: Your final answer must be in units squared.

### Example 2: given base length and height

Find the area of the triangle below:

h = 200cm

b = 6m

Note: You have 2 different units here. You must convert them to a common unit 200cm = 2m.

h = 2m

b = 6m

$A=\frac{1}{2} b h$

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(6)(2) \end{aligned}

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(6)(2)\\ &=6m^2 \end{aligned}

Remember: Your final answer must be in units squared.

### Example 3: missing height

Find the area of triangle ABC below:

b = 5cm

In this question we are given the length of the hypotenuse which is 13cm. We need to calculate the height using Pythagoras’ Theorem.

$\begin{array}{l} h^{2}=13^{2}-5^{2} \\ h^{2}=169-25 \\ h^{2}=144 \\ h=\sqrt{144} \\ h=12 \mathrm{~cm} \end{array}$

h = 12cm

$A=\frac{1}{2} b h$

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(5)(12) \end{aligned}

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(5)(12)\\ &=30cm^2 \end{aligned}

Remember: Your final answer must be in units squared.

### Example 4: compound shape

Below is the floor plan for a new deck that needs to be painted. One can of paint costs £12 and covers an area of 2.5m2. How much would it cost to paint the entire deck?

Split the plan into 2 shapes. You have a rectangle and a right angled triangle.

For the triangle:

b = 2.5m     h = 3.5m

$A=\frac{1}{2} b h$

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(2.5)(3.5) \end{aligned}

\begin{aligned} A &=\frac{1}{2} b h \\ &=\frac{1}{2}(2.5)(3.5)\\ &=4.375m^2 \end{aligned}

Now you must find the area of the rectangle.

\begin{aligned} \text { Area of a rectangle }&=l \times w \\ &=2.5 \times 3 \\ &=7.5 m^{2} \\ \end{aligned}

Total Area = 4.375 + 7.5 = 11.875 m2

Now we need to divide 11.875 by 2.5 since one can of paint only covers an area of 2.5m2:

11.875 ÷ 2.5 = 4.75

This means you have to buy 5 cans of paint as we can only buy whole cans of paint.

Now to work out the cost: 5 cans of paint x £12 = £60

The total cost to paint the deck is £60.

## 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 height or missing length. In order to do this you must rearrange the formula.

To find a missing length given the area:

1Rearrange the formula

2Substitute in the values you know

3Calculate

### Example 5: calculating base length

Triangle XYZ is a right triangle with an area of 15cm2. The height of the triangle is 5cm. Find the base length of the triangle.

$A = \frac{1}{2}bh$

to make b the subject, start by multiplying both sides of the formula by 2

$2A = bh$

Next we need to divide both sides of the formula by h

$\frac{2(A)}{h}=b$

Area = 15

Height = 5

$b=\frac{2(15)}{5}$
\begin{aligned} b&=\frac{2(15)}{5} \\ b&=\frac{30}{5} \\ b&=6 \mathrm{~cm} \end{aligned}

### Common misconceptions

• Identifying the correct information to use

A question may give extra information that is not needed to answer it. Carefully identify the relevant pieces of information.
E.g.
To calculate the area here we only need the base and height.

Base= 4cm,

Height = 3cm

Area=

$\frac{3 \times 4}{2} = 6cm^{2}$

We can ignore the value of the hypotenuse (5cm)

• Units

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

### Practice area of right angled triangles questions

1. Find the area of the triangle below:

108cm^{2}

54cm^{2}

27cm^{2}

36cm^{2}

The lengths needed for the base and the height are 9cm and 12cm , so the calculation we need to do is \frac{1}{2} \times 9 \times 12 = 54cm^{2}

2. Find the area of the triangle below:

22.5cm^{2}

45cm^{2}

2250cm^{2}

4500cm^{2}

We need to convert the units so that they are the same.

0.9m=90cm

Then the calculation needed is \frac{1}{2} \times 90 \times 50= 2250cm^{2}

3. Shown below is a outline of a meerkat enclosure. Each meerkat needs a minimum of 9m^{2} to roam around. What is the maximum number of meerkats that can fit into this enclosure?

2

3

24

25

We can treat the shape as a rectangle with area 3.2 \times 4.5 = 14.4m^{2} and a triangle with area \frac{1}{2} \times 4.5 \times 4.5 = 10.125m^{2}.

This gives a total area of 14.4 + 10.125 = 24.525m^{2}.

By considering multiples of 9 , we conclude that 2 meerkats can fit into the cage.

4. Find the area of the triangle below:

54cm^{2}

101.82cm^{2}

72cm^{2}

50.91cm^{2}

Using Pythagoras’ Theorem to find length AB

AB=\sqrt{18^{2}-6^{2}}

AB=16.97…

The area is then given by \frac{1}{2} \times 6 \times 16.97 = 50.91cm^{2}

5. Triangle PQR is a right angled triangle with an area of 40cm^{2} . The base length of the triangle is 0.1m. Find the height of the triangle.

8cm

800cm

2cm

200cm

Starting with the formula for the area of a triangle: Area=\frac{1}{2}  \times base \times height

We substitute in the known information (convert lengths to the same units), so

40=\frac{1}{2} \times 10  \times height

40=5  \times height

Therefore:

height = 8cm

### Area of a right angled triangleGCSE questions

1. A logo is in the shape of a right angled triangle. It has a base length of 10cm and a height of 4.5cm . Calculate the area of the logo.

(2 marks)

Show answer
A = \frac{1}{2}\times 10 \times 4.5

(1)

A=22.5cm^{2}

(1)

2. The diagram below shows the plan of a rectangular garden:

Calculate the area of the lawn.

(3 marks)

Show answer

Area of rectangle:

5 \times 12 = 60m^{2}

(1)

Area of triangle:

\frac{1}{2} \times 3 \times 5 = 7.5m^{2}

(1)

Area of lawn:

60-7.5=52.5m^{2}

(1)

3. The area of the square and the area of the right angled triangle below are equal.

Work out the height of the triangle.

(4 marks)

Show answer

Area of square:

6 \times 6 = 36cm^{2}

(1)

Rearrange area of triangle:

\begin{aligned} A &= \frac{1}{2} bh \\ 2A &= bh\\ h &= \frac{2A}{b} \end{aligned}

(1)

Substitute in values:

h= \frac{2 \times 36}{9} \begin{array}{l} h= \frac{72}{9}\end{array}\begin{array}{l} h=8cm \end{array}

(1)

## Learning checklist

You have now learned how to:

•  Apply formula to calculate and solve problems involving the area of triangles
•  Use Pythagoras’ Theorem to solve problems involving triangles

## Still stuck?

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#### FREE GCSE Maths Practice Papers - 2022 Topics

Practice paper packs based on the advanced information for the Summer 2022 exam series from Edexcel, AQA and OCR.

Designed to help your GCSE students revise some of the topics that will come up in the Summer exams.