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 using 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 is 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.  The hypotenuse is the longest side of a right-angled triangle.  Sides a and b are known as the adjacent sides.  They are adjacent, or next to, the right angle.

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 side of length 3 = 3 \times 3=3^2=9

The area of the side of length 4 = 4\times 4=4^2=16

The area of the side of length 5 = 5\times5=5^2=25

We can see that when we add together the areas of 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 sides 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 .

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

### What is Pythagoras’ theorem? ## 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.

### Explain how to use Pythagoras’ theorem ## 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 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{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 &= 129\\\\ x &=\sqrt{129} \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?

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 &= 169 \end{aligned}

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

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?

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.

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 .

• Right-angled triangles may be in different orientations

The triangles can be drawn in different orientations.  These are right-angled triangles in different orientations

• Make sure you write what your calculator shows, before rounding

It is very easy to make a mistake at the end with rounding.  So make sure you have shown your working out.

• 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} .

• Pythagoras’ theorem only works on right-angled triangles

Pythagoras’ theorem only works on right-angle triangles, but with a bit of thought it can be used on other triangles.  For example 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 8.60 cm 8.6 cm 4.90 cm 4.9 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 17.2 cm 17.20 cm 9.79 cm 9.80 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 3 decimal places 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 6.75 cm 6.76 cm 7.62 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. Say if this a right-angled triangle and give your reason: Yes because 12^2+5^2=13^2 No because 12^2+5^2 13^2 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 \\\\ 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. Say if this a right-angled triangle and give your reason: No because 6^2+13^2 14^2 Yes 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

## The next lessons are

• Trigonometry – using sine, cosine and tangent ratios
• Coordinate geometry – the distance between two points
• 3D Pythagoras

## Still stuck?

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