Pythagoras Questions, Worksheets and Revision

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 AQA Edexcel OCR WJEC
Pythagoras’ Theorem

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.

Pythagoras’ Theorem
Level 4-5 GCSE AQA Edexcel OCR WJEC
Pythagoras Finding a Length

Example 1: Finding a Length

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

[2 marks]

Pythagoras Finding a Length

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 AQA Edexcel OCR WJEC
Calculating the Length of a Line

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]

Calculating the Length of a Line

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 AQA Edexcel OCR WJEC

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:Pythagoras Graph

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=8.9cm (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.)

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