GCSE Maths Geometry and Measure

Pythagoras Theorem

Pythagoras Theorem

Here we will learn about Pythagoras theorem, including how to find sides of a right-angled triangle and how to use Pythagoras theorem to check if a triangle has a right angle or not.

There are also Pythagoras theorem 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 Pythagoras theorem?

Pythagoras theorem states that the square of the longest side of a right angled triangle (called the hypotenuse) is equal to the sum of the squares of the other two sides.

Pythagoras theorem is:

$a^2+b^2=c^2$

Side c is known as the hypotenuse, which is the longest side of a right-angled triangle and is opposite the right angle.
Side a and side b are known as the adjacent sides because they are adjacent (next to) the right angle.

If we know any two sides of a right angled triangle, we can use Pythagoras theorem to work out the length of the third side.

We can only use Pythagoras theorem with right-angled triangles.

E.g.
Let’s look at this right-angled triangle:

We can see that three squares have been drawn next to each of the sides of the triangle.
The area of the square with side length 3 = 3 \times 3=3^2=9
The area of the square with side length 4 = 4\times 4=4^2=16
The area of the square with side length 5 = 5\times5=5^2=25

We can see that when we add together the areas of the squares on the two shorter sides we get the area of the square on the longest side.

$9+16=25$

We can see that when we square the lengths of the two shorter sides of a right angled triangle and add them together, we get the square of the longest side.

$3^2+4^2=5^2$

3, 4, 5 is known as a Pythagorean triple
There are other Pythagorean triples such as 5, 12, 13 and 8, 15, 17 .

How to use Pythagoras theorem

In order to use Pythagoras theorem:

1. Label the sides of the triangle.
2. Write down the formula and apply the numbers.

Pythagoras theorem examples

Example 1: finding the length of the hypotenuse (finding the length of the longest side)

1. Label the sides of the triangle.

It is very important to label the hypotenuse (the longest side) correctly with c .  The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.

2Write down the formula and apply the numbers.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 3^2 + 8^2 &= x^2\\\\ x^2 &= 3^2+8^2\\\\ x^2 &= 9 + 64\\\\ x^2 &=73\\\\ x &= \sqrt{73} \end{aligned}

Make sure you give your final answer in the correct form including units where appropriate.

$x=\sqrt{73}=8.5440037…$

x = 8.54 cm to 2 decimal places

An alternative method is to rearrange the formula and put the entire calculation into a calculator.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ c^2 &=a^2 + b^2 \\\\ c &= \sqrt{a^2 + b^2} \\\\ x &= \sqrt{3^2 + 8^2}\\\\ x &= 8.5440037… \end{aligned}

Example 2: finding the length of the hypotenuse (finding the length of the longest side)

It is very important to label the hypotenuse (the longest side) correctly with c .  The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 7^2 + 9^2 &= x^2\\\\ x^2 &= 7^2 + 9^2\\\\ x^2 &= 49+81\\\\ x^2 &= 130\\\\ x &=\sqrt{130} \end{aligned}

Make sure you give your final answer in the correct form including units where appropriate.

$x=\sqrt{130}=11.40175.…$

x = 11.4 cm to 3 significant figures

An alternative method is to rearrange the formula and put one calculation into a calculator.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ c^2 &=a^2 + b^2 \\\\ c &= \sqrt{a^2 + b^2} \\\\ x &= \sqrt{7^2 + 9^2}\\\\ x &= 11.40175… \end{aligned}

Example 3: finding an adjacent side (a short side)

It is very important to label the hypotenuse (the longest side) correctly with c .  The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.

We will label the short side we are trying to find as side a .

\begin{aligned} a^2 + b^2 &= c^2 \\\\ x^2 + 5^2 &= 8^2\\\\ x^2+25 &= 64 \\\\ x^2 &= 64 – 25 \\\\ x^2 &= 39\\\\ x &=\sqrt{39} \end{aligned}

Make sure you give your final answer in the correct form; including units where appropriate.

$x=\sqrt{39}=6.244997.…$

x = 6.24 cm to 2 significant figures

An alternative method is to rearrange the formula and put one calculation into a calculator.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ a^2 &=c^2 – b^2 \\\\ a &= \sqrt{c^2 – b^2} \\\\ x &= \sqrt{8^2 – 5^2}\\\\ x &= 6.244997… \end{aligned}

Example 4: finding an adjacent side (a short side)

It is very important to label the hypotenuse (the longest side) correctly with c .  The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.

We will label the short side we are trying to find as side a .

\begin{aligned} a^2 + b^2 &= c^2 \\\\ x^2 + 11^2 &= 20^2\\\\ x^2+121 &= 400 \\\\ x^2 &= 400 – 121 \\\\ x^2 &= 279\\\\ x &=\sqrt{279} \end{aligned}

Make sure you give your final answer in the correct form; including units where appropriate.

$x=\sqrt{279}=16.70329.…$

x = 16.7 cm to 3 significant figures

An alternative method is to rearrange the formula and put one calculation into a calculator.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ a^2 &=c^2 – b^2 \\\\ a &= \sqrt{c^2 – b^2} \\\\ x &= \sqrt{20^2 – 11^2}\\\\ x &= 16.70329… \end{aligned}

Example 5: checking if a triangle has a right angle

Is the triangle below a right-angled triangle?

If the triangle is a right-angled triangle, then Pythagoras theorem should work:

It is very important to label the hypotenuse (the longest side) correctly with c .  The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 8^2 + 10^2 &= 13^2\\\\ 64+100 &= 169 \\\\ 164 &\neq 169 \end{aligned}

Pythagoras theorem only works with right-angled triangles.

So because,

$8^2 + 10^2 \neq 13^2$

The sides of the triangles do not fit with Pythagoras theorem.  Therefore the triangle is NOT a right-angled triangle.

Example 6: checking if a triangle has a right angle

Is the triangle below a right-angled triangle?

If the triangle is a right-angled triangle, then Pythagoras theorem should work:

It is very important to label the hypotenuse (the longest side) correctly with c .  The adjacent sides, next to the right angle can be labelled a and b either way around as they are interchangeable.

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 6^2 + 8^2 &= 10^2\\\\ 36+64 &= 100 \\\\ 100 &= 100 \end{aligned}

This is correct

Pythagoras theorem only works with right-angled triangles.
So because,

$6^2 + 8^2 = 10^2$

The sides of the triangles fit with Pythagoras theorem.  Therefore the triangle is a right-angled triangle.

Common misconceptions

• Make sure you identify the hypotenuse

It is very important to make sure that the hypotenuse is correctly identified and labelled c . You can find the hypotenuse by identifying the side which is opposite the right angle.

• Right-angled triangles may be in different orientations

The triangles can be drawn in different orientations.
E.g.

• Lengths of sides do not have to be whole numbers

Lengths can be decimals, fractions or even irrational numbers such as surds,

E.g.
\sqrt{2} .

• Right-angled triangles

Pythagoras theorem only works on right-angle triangles, however we can sometimes calculate the lengths of other triangles by splitting them into right angled triangles.
E.g.
An isosceles triangle can be made into 2 right-angled triangles by putting in the line of symmetry.

• Pythagoras theorem can also be applied to other shapes

An application of Pythagoras theorem is to extend it to work on other shapes such as a trapezium.

• Do not round too early

If you need to use Pythagoras theorem in a question with multiple steps, do not round until the very end of the question or you will lose accuracy.  For example, you may need to find the height of a triangle, and then use that height to find its area.

Practice Pythagoras theorem questions

4.9 cm

8.61 cm

4.90 cm

8.60 cm

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 7^2 + 5^2 &= x^2 \\\\ x^2 &= 7^2 + 5^2\\\\ x^2 &= 49+25 \\\\ x^2 &= 74\\\\ x &= \sqrt{74}\\\\ x &= 8.602325… \end{aligned}

x=8.60 cm to 2 decimal places

17.3 cm

17.2 cm

9.7 cm

9.8 cm

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 14^2 + 10^2 &= x^2\\\\ x^2 &= 14^2 + 10^2\\\\ x^2 &= 196+10\\\\ x^2 &= 296\\\\ x &= \sqrt{296}\\\\ x &= 17.20465… \end{aligned}

x=17.2 cm to 1 decimal place.

14.97 cm

14.96 cm

20.59 cm

2.58 cm

\begin{aligned} a^2 + b^2 &= c^2 \\\\ x^2 + 10^2 &= 18^2\\\\ x^2+100 &= 324 \\\\ x^2 &= 324 – 100 \\\\ x^2 &= 224\\\\ x &=\sqrt{224}\\\\ x &= 14.96662… \end{aligned}

x=14.97 cm to 2 decimal places

7.62 cm

6.76 cm

6.75 cm

7.63 cm

\begin{aligned} a^2 + b^2 &= c^2 \\\\ x^2 + 2.5^2 &= 7.2^2\\\\ x^2+6.25 &= 51.84 \\\\ x^2 &= 51.84 – 6.25 \\\\ x^2 &= 45.59\\\\ x &=\sqrt{45.59}\\\\ x &= 6.75203… \end{aligned}

x=6.75 cm to 2 decimal places

5. Is this triangle a right-angled triangle? Give a reason for you answer:

No because I measured the angle and it was not 90^{\circ}

No because 12^2+5^2 13^2

Yes because I measured the angle and it was 90^{\circ}

Yes because 12^2+5^2=13^2

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 12^2 + 5^2 &= 13^2\\\\ 144+25 &= 169 \\\\ 169 &= 169 \end{aligned}

This is correct. Pythagoras’ theorem only works with right-angled triangles.

Therefore the triangle is a right-angled triangle.

6. Is this triangle a right-angled triangle? Give a reason for you answer:

Yes because 6^2+13^2=14^2

No because 6^2+13^2 14^2

Yes because I measured the angle and it was 90^{\circ}

No because I measured the angle and it was not 90^{\circ}

\begin{aligned} a^2 + b^2 &= c^2 \\\\ 6^2 + 13^2 &= 14^2\\\\ 36+169 &= 196 \\\\ 205 &= 169 \end{aligned}

This is NOT correct. Pythagoras’ theorem only works with right-angled triangles.

Therefore the triangle is NOT a right-angled triangle.

Pythagoras theorem GCSE questions

1. ABC is a right-angled triangle.

Calculate the length of AC.

(3 marks)

7.3^2 + 12.7 ^2 = 214.58

(1)

\sqrt{214.58}

(1)

AC = 14.6485… = 14.6 cm

(1)

2. Triangle ABC has a perimeter of 19 cm.

AB = 5 cm \\ BC = 6 cm

By calculation, deduce whether triangle ABC is a right-angled triangle.

(4 marks)

19-(5+6)=8

Third side is 8 cm and is the longest so it is the hypotenuse

(1)

5^2 + 6 ^2 = 61

(1)

\sqrt{61}=7.81…

OR

8^2=64

(1)

Triangle ABC is NOT a right-angled triangle

(1)

3. A frame is made from wire

The frame is in the shape of a rectangle 10 cm by 15 cm.

The diagonal of the rectangle is also made from wire.

Calculate the total length of wire needed to make the frame and the diagonals.

(4 marks)

10^2 + 15 ^2 = 325

(1)

\sqrt{325}=18.02775

(1)

18.02775.. + (2\times 15) + (2\times 10)

(1)

68.027756…. = 68.0 cm

(1)

Did you know?

Pythagoras theorem is named after a Greek mathematician who lived about 2500 years ago, however the ancient Babylonians used this rule about 4 thousand years ago!  At the same time the Egyptians were using the theorem to help them with right angles when building structures.

Learning checklist

You have now learned how to:

• Use Pythagoras theorem to find the length of the longest side of a right-angled triangle – its hypotenuse
• Use Pythagoras theorem to find the length of one of the shorter sides of a right-angled triangle

Still stuck?

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GCSE Maths Papers - November 2022 Topics

Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

Designed to help your GCSE students revise some of the topics that are likely to come up in November exams.