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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.

**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.

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 .

**See also: **3D Pythagoras’

In order to use Pythagoras’ theorem:

**Label the sides of the triangle.****Write down the formula and apply the numbers.****Work out the answer.**

Get your free pythagoras’ theorem worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free pythagoras’ theorem worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEFind x and give your answer to 2 decimal places.

**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.

2**Write down the formula and apply the numbers.**

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

3**Work out the answer.**

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

\[x=\sqrt{73}=8.5440037…\]

The final answer is:

x = 8.54 cm to 2 decimal places

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

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

Find x and give your answer to 3 significant figures:

**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.

**Write down the formula and apply the numbers.**

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

**Work out the answer.**

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

\[x=\sqrt{130}=11.40175.…\]

The final answer is:

x = 11.4 cm to 3 significant figures

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

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

Find x and give your answer to 2 significant figures:

**Label the sides of the triangle.**

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

**Write down the formula and apply the numbers.**

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

**Work out the answer.**

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

\[x=\sqrt{39}=6.244997.…\]

The final answer is:

x = 6.24 cm to 2 significant figures

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

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

Find x and give your answer to 3 significant figures:

**Label the sides of the triangle.**

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

**Write down the formula and apply the numbers.**

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

**Work out the answer.**

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

\[x=\sqrt{279}=16.70329.…\]

The final answer is:

x = 16.7 cm to 3 significant figures

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

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

Is the triangle below a right-angled triangle?

**Label the sides of the 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.

**Write down the formula and apply the numbers.**

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

**Work out the answer.**

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.

Is the triangle below a right-angled triangle?

**Label the sides of the triangle.**

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

**Write down the formula and apply the numbers.**

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

This is **correct**.

**Work out the answer.**

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**.

**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.

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 calculate the lengths of other tringles 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.

1. Find side x. Give your answer to 2 decimal places:

4.9 cm

8.6 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

2. Find side x. Give your answer to 3 decimal places:

17.20 cm

17.2 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

3. Find side x. Give your answer to 2 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

4. Find side x. Give your answer 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.

1. ABC is a right-angled triangle.

Calculate the length of AC.

Give your answer correct to 3 significant figures.

**(3 marks)**

Show answer

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)**

Show answer

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.

Give your answer correct to 1 decimal place.

**(4 marks)**

Show answer

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)**

Pythagoras’ theorem is named after a Greek mathematician who lived about

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

- Trigonometry – using sine, cosine and tangent ratios
- Alternate angles
- Congruent shapes

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