One to one maths interventions built for GCSE success

Weekly online one to one GCSE maths revision lessons available in the spring term

In order to access this I need to be confident with:

Algebraic expressionsUnderstanding and using formulae

Area of a quadrilateral Rearranging equationsThis topic is relevant for:

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.

A **right angled triangle** is a triangle with one** **right angle (

There are two types of right triangles:

Isosceles right angled triangle

-two equal angles (

-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)*

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**

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 **congruent** (identical) right angled triangles.

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

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

This can be shortened to

\[A=\frac{1}{2} b h\]

where

Your final answer must be given in units^{2} (e.g. cm^{2}, m^{2}, mm^{2}).

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)**

2**Write the formula**

\[A=\frac{1}{2} b h\]

3**Substitute the values for base and height**

4**Calculate**

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

DOWNLOAD NOWGet your free area of a Right angle triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD NOWFind the area of the triangle below:

**Identify the height and base length of the triangle**

2**Write down the formula**

\[A=\frac{1}{2} b h\]

3**Substitute the values for the base and height**

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

4**Calculate**

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

*Remember: Your final answer must be in units squared.*

Find the area of the triangle below:

**Identify the height and base length of the triangle**

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

**Write down the formula**

\[A=\frac{1}{2} b h\]

**Substitute the values for the base and height**

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

**Calculate**

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

*Remember: Your final answer must be in units squared.*

Find the area of triangle ABC below:

**Identify the height and base length of the triangle**

In this question we are given the length of the hypotenuse which is

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

**Write down the formula**

\[A=\frac{1}{2} b h\]

**Substitute the values for the base and height**

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

**Calculate**

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

*Remember: Your final answer must be in units squared.*

Below is the floor plan for a new deck that needs to be painted. One can of paint costs £^{2}

**Identify the height and base length of the triangle**

Split the plan into

For the triangle:

**Write down the formula**

\[A=\frac{1}{2} b h\]

**Substitute the values for the base and height**

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

**Calculate**

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

**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 = **11.875 m ^{2}**

Now we need to divide ^{2}

This means you have to buy

Now to work out the cost:

The total cost to paint the deck is **£ 60**.

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

Triangle XYZ is a right triangle with an area of ^{2}

**Rearrange the formula**

\[A = \frac{1}{2}bh\]

to make

\[2A = bh\]

Next we need to divide both sides of the formula by

\[\frac{2(A)}{h}=b\]

**Now substitute the given values**

Area =

Height =

\[b=\frac{2(15)}{5}\]

**Calculate**

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

**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=

Height =

Area=

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

We can ignore the value of the hypotenuse (

**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.

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

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)**

**(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)**

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

- Area of compound shapes
- Pythagoras’ Theorem
- Trigonometry
- Area of a parallelogram
- Area of an isosceles triangle
- Area of an equilateral triangle

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths revision programme.