# Pythagoras Questions, Worksheets and Revision

GCSE 4 - 5KS3AQAEdexcelOCRWJECFoundationAQA 2022Edexcel 2022OCR 2022WJEC 2022

## 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.

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Level 4-5 GCSE    ## 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. Level 4-5 GCSE    ## 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 our 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.

Level 4-5 GCSE    ## 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.

Level 4-5 GCSE    ## Example Questions

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).

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).

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:

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

$QR=8.9$cm ($1$ dp).

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).

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.)

Level 6-7GCSEKS3

Level 4-5 GCSE

Level 4-5 GCSE

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