## Pythagoras

Pythagoras’ theorem is an equation that describes a relationship between the 3 sides of a **right-angled triangle**. We can use it to determine a missing length when given the two other lengths.

Being able to rearrange equations will help with this topic.

## Pythagoras’ Theorem

The equation is:

\textcolor{red}{a}^2+\textcolor{limegreen}{b}^2=\textcolor{blue}{c}^2

where \textcolor{blue}{c} is the **hypotenuse** and \textcolor{red}{a} and \textcolor{limegreen}{b} are the two other sides. The **hypotenuse** is always the longest side of the triangle and can be found opposite the right angle.

**Example 1: Finding a Length**

Find the length of the side marked \textcolor{red}{x} on the right angled triangle shown.

**[2 marks]**

In order to find this using Pythagoras’ theorem, we need to work out which side corresponds to each of the letters \textcolor{red}{a}, \textcolor{limegreen}{b}, and \textcolor{blue}{c} in the equation.

\textcolor{blue}{c} = \textcolor{blue}{10} cm

\textcolor{limegreen}{b} = \textcolor{limegreen}{6} cm

\textcolor{red}{a} = \textcolor{red}{x}

Then we can substitute are values in so the equation a^2+b^2=c^2 becomes:

\textcolor{red}{x}^2+\textcolor{limegreen}{6}^2=\textcolor{blue}{10}^2

\textcolor{red}{x}^2+\textcolor{limegreen}{36}=\textcolor{blue}{100}

Now we can solve for \textcolor{red}{x}

\textcolor{red}{x}^2=\textcolor{blue}{100}-\textcolor{limegreen}{36}=64.

\textcolor{red}{x}=\sqrt{64}=8 cm.

**Example 2: Calculating the Length of a Line**

Calculate the distance between (-4, -1) and (3, 4).

Give your answer to 1 dp.

**[2 marks]**

First we plot the two points on a pair of axes, draw a line connecting them and then draw a triangle underneath making a note of the lengths of the sides:

Side 1 (height) = 4 -(-1) = \textcolor{red}{5}

Side 2 (width) = 3 -(-4) = \textcolor{limegreen}{7}

Next we substitute these lengths into Pythagoras’ theorem:

\textcolor{red}{a}^2+\textcolor{limegreen}{b}^2 = \textcolor{blue}{c}^2

\begin{aligned} \textcolor{blue}{c}^2 &= \textcolor{red}{5}^2+\textcolor{limegreen}{7}^2 \\ \textcolor{blue}{c}^2 &= \textcolor{red}{25}+\textcolor{limegreen}{49}=74 \\ \textcolor{blue}{c} &= \sqrt{74}=8.602325267... \\ \textcolor{blue}{c} &= 8.6 \, \, (1 \text{ dp}) \end{aligned}

**Note:** The length you are trying to calculate, when finding the length of a line between two points, will always be the hypotenuse.

## GCSE Maths Revision Guide

£14.99### Example Questions

**Question 1:**

Find length BC.

Give your answer to 1 decimal place.

**[2 marks]**

The missing side is the hypotenuse, by substituting the 2 sides we know into the equation a^2+b^2=c^2 we get:

8^2+14^2=c^2

Evaluating the two squares gives us:

c^2=64+196

c^2=260

Then square rooting both sides of the equation gives us:

BC=\sqrt{260}=16.1245155...

= 16.1 cm (1 dp).

**Question 2:** Work out the distance between the two points (-6, 2) and (4, 5).

Give your answer to 3 significant figures.

**[2 marks]**

To do this, we plot the points on a graph and draw a line connecting them. We then construct a right-angled triangle. The resulting picture looks like:

We can see that the distance between the two points is the longest side of a right-angle triangle – the hypotenuse, so by substituting the known sides into the equation a^2+b^2=c^2 we get:

10^2+3^2=c^2

Evaluating the squares gives us:

c^2=100+9

c^2=109

Finally, by square rooting both sides, we get:

c=\sqrt{109}=10.44030651...

= 10.4 cm (3 sf).

**Question 3: ** PQR is a right-angled triangle. Work out the length of the missing side, QR, to 1 decimal place.

**[2 marks]**

The missing side is the hypotenuse, so by substituting the known sides into the equation a^2+b^2=c^2 we get:

5.9^2+6.7^2=c^2

Evaluating the two squares gives us:

c^2=34.81+44.89

c^2=79.7

Then by square rooting both sides of this equation, we get:

QR=\sqrt{79.7}=8.927485648...

QR=8.9cm (1 dp).

**Question 4:** LMN is a right-angled triangle. Work out the length of the missing side, LN, to 1 decimal place.

**[2 marks]**

By substituting the known sides into Pythagoras’ Theorem c^2=a^2+b^2 becomes:

5.1^2=LN^2+3.1^2

Next we evaluate the two squares to give us:

26.01=LN^2+9.61

Subtracting 9.61 from both sides:

16.4=LN^2

Then we square root both sides of this equation:

LN=\sqrt{16.4}

LN=4.049691346...

LN=4.0 cm (1 dp).

**Question 5: **A builder places a 2.9 m ladder on horizontal ground, resting against a vertical wall. To be safe to use, the base of this ladder must be 1.3 m away from the wall. How far up the wall does the ladder reach?

Give your answer to 1 decimal place.

**[2 marks]**

By substituting the known sides into Pythagoras’ Theorem c^2=a^2+b^2 (and letting the height of the wall be a) we get:

2.9^2=a^2+1.3^2

Calculating the two square roots and rearranging gives us:

a^2=8.41-1.69=6.72

Then by square rooting both sides of this equation, we get:

a=\sqrt{6.72}=2.592296279

a=2.6 cm (1 d.p.)

### Worksheets and Exam Questions

#### (NEW) Pythagoras Exam Style Questions - MME

Level 4-5 New Official MME### Drill Questions

#### Pythagoras 1 - Drill Questions

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