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Substitution Squares and square roots Solving equations Rounding3D shapes

Pythagoras’ TheoremThis topic is relevant for:

Here we will learn about 3D Pythagoras including how to apply Pythagoras’ theorem in three dimensional situations.

There are also 3D Pythagoras worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**3D Pythagoras** is applying Pythagoras’ theorem to problems with

We can do this by breaking down the problem into

Remember Pythagoras’ theorem is:

\[a^2+b^2=c^2\]

Where

**See also:** Pythagoras theorem

Let’s take a look at an example:

We are going to find the length AG in the cuboid ABCDEFGH.

We can see that we can make a right angled triangle ACG which we can use to work out AG.

In order to use Pythagoras’ theorem we need to know two sides of the triangle. So in order to work out the longest side AG we first need to work out one of the shorter sides AC.

Let’s call this side

We can see that the side we have labeled

We can see that the triangle ABC is a right angled triangle so we can use Pythagoras’ theorem to calculate

\[AC=\sqrt{10^2 + 4^2} = 2\sqrt{29} = 10.7703…\]

So,

\[AG=\sqrt{10.7703…^2 + 6^2} = \sqrt{152}=2\sqrt{38} = 12.328…\]

So the required length is

We can also use a special 3D Pythagoras’ Theorem for cuboids where

\[x^2+y^2+z^2=s^2\]

E.g

\[\begin{aligned}
&s^{2}=10^{2}+4^{2}+6^{2} \\
&s=\sqrt{10^{2}+4^{2}+6^{2}} \\
&s=12.3 \mathrm{~cm} \quad(1 . \mathrm{d} \cdot \mathrm{p})
\end{aligned}\]

If angles and lengths are involved in the problem, we may also need to use trigonometry.

**See also: **Trigonometry

In order to solve problems in three-dimensions using Pythagoras’ theorem:

**Identify the useful right angled triangle(s).****Apply Pythagoras’ Theorem**.**Finalise your answer**.

Get your free 3D Pythagoras worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free 3D Pythagoras worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEFind the slant length of the cone with height

**Identify the useful right angled triangle(s)**.

We can draw a diagram of the cone and look for the right-angled triangle. The slant length,

2**Apply Pythagoras’ Theorem**.

\[\begin{align*}
a^2 + b^2 &= c^2 \\
8^2 + 10^2 &= l^2\\
l^2 &= 8^2+10^2\\
l^2 &= 64 + 100\\
l^2 &=164\\
l &= \sqrt{164} \\
\end{align*}\]

3**Finalise your answer**.

\[l = \sqrt{164}=12.8062…\]

The slant length of the cone is

This is useful as we need the slant length in order to find the curved surface area of a cone.

Find the height of the cone with radius

**Identify the useful right angled triangle(s)**.

We can draw a diagram of the cone and look for the right-angled triangle. The slant length,

**Apply Pythagoras’ Theorem**.

\[\begin{align*}
a^2 + b^2 &= c^2 \\
h^2 + 12^2 &= 17^2\\
h^2 &= 17^2-12^2\\
h^2 &= 289 -144\\
h^2 &=145\\
h &= \sqrt{145} \\
\end{align*}\]

**Finalise your answer**.

\[h = \sqrt{145}=12.04159…\]

The height of the cone is

This is useful as we need the height in order to find the volume of a cone.

Here is a cuboid ABCDEFGH. Find the length AG. Give your answer correct to

**Identify the useful right angled triangle(s)**.

**Apply Pythagoras’ Theorem**.

\[AC=\sqrt{20^2 + 7^2} =\sqrt{449} = 21.1896…\]

So,

\[AG=\sqrt{21.1896…^2 + 11^2} = \sqrt{570} = 23.87467…\]

**Finalise your answer**.

\[AG = \sqrt{570}=23.87467…\]

The length AG is

Alternatively we could have done this in one step:

\[AG=\sqrt{20^2 + 7^2 + 11^2}=\sqrt{570}=23.87467…\]

Here is a cuboid ABCDEFGH. Find the length BH. Give your answer correct to

**Identify the useful right angled triangle(s)**.

**Apply Pythagoras’ Theorem**.

\[BD=\sqrt{5.6^2 + 8.1^2} = 9.84733…\]

So,

\[AG=\sqrt{9.84733…^2 + 12.3^2} = 15.75626…\]

**Finalise your answer**.

\[AG=15.75626…\]

The length AG is

Alternatively we could have done this in one step:

\[AG=\sqrt{8.1^2 + 5.6^2 + 12.3^2}=15.75626…\]

The diagram shows a square-based pyramid ABCDX.

\[AB=12 cm\]

\[AX=BX=CX=DX=17 cm\]

The point M is the midpoint of AB.

Calculate the length MX.

Give your answer correct to

**Identify the useful right angled triangle(s)**.

Triangle ABX is an isosceles triangle. MX and AB are perpendicular.

**Apply Pythagoras’ Theorem**.

\[MX=\sqrt{17^2 – 6^2} = 15.90597…\]

**Finalise your answer**.

\[MX=15.90597…\]

The length MX is

This is useful as we need the height of the triangles in order to find the surface area of a pyramid.

The diagram shows a square-based pyramid ABCDX.

\[AB=20 cm\]

\[AX=BX=CX=DX=30 cm\]

M is the centre of the square base ABCD.

Calculate the length MX.

Give your answer correct to

**Identify the useful right angled triangle(s)**.

**Apply Pythagoras’ Theorem**.

\[AC=\sqrt{20^2 + 20^2} =\sqrt{800} = 20\sqrt{2}=28.28427…\]

So,

\[AM=\frac{1}{2}AC=10\sqrt{2}=14.14213…\]

\[MX=\sqrt{30^2 – 14.14213…^2} = 10\sqrt{7} = 26.45751…\]

**Finalise your answer**.

\[MX = 10\sqrt{7}=26.45751…\]

The length MX is

This is useful as we need the height of the pyramid in order to find the volume of a pyramid.

**Three dimensions can be difficult to visualise**

It can be tricky to identify where right angles are on a 2D diagram of a 3D shape. You may find it useful to use a box to help you to visualise it, such as an old tissue box or a shoe box.

**Use all of the digits on the calculator screen**

It is very easy to make a mistake at the end with rounding. Make sure you use all of the digits on the calculator screen and only round your final answer at the end of the question. Rounding too soon will lose accuracy.

**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} .

1. Find the slant length of the cone with height 15 cm and radius 10 cm . Give your answer correct to 1 decimal place.

18.0 cm

18.1 cm

11.2 cm

11.1 cm

l=\sqrt{15^2 + 10^2}=18.02775…

2. Find the radius of the cone with height 19 cm and slant length 23 cm . Give your answer correct to 1 decimal place.

12.9 cm

13.0 cm

29.8 cm

29.9 cm

l=\sqrt{23^2 – 19^2}=12.96148…

3. Here is a cuboid ABCDEFGH. Find the length AG. Give your answer correct to 1 decimal place.

13.1 cm

12.4 cm

12.5 cm

13.2 cm

AG=\sqrt{10^2 + 3^2 + 8^2}=\sqrt{173}=13.1529…

4. Here is a cuboid ABCDEFGH. Find the length BH. Give your answer correct to 1 decimal place.

20.2 cm

20.3 cm

12.4 cm

12.5 cm

BH=\sqrt{7^2 + 6^2 + 18^2}=\sqrt{409}=20.22374…

5. The diagram shows a square-based pyramid ABCDX.

AB=10 cm

AX=BX=CX=DX=18 cm

The point M is the midpoint of AB.

Calculate the length MX.

Give your answer correct to 1 decimal place.

17.2 cm

14.9 cm

17.3 cm

15.0 cm

MX=\sqrt{18^2 – 5^2}=\sqrt{299}=17.29161…

6. The diagram shows a square-based pyramid ABCDX.

AB=40 cm

AX=BX=CX=DX=50cm

M is the centre of the square base ABCD.

Calculate the length MX.

Give your answer correct to 1 decimal place.

41.3 cm

26.5 cm

26.4 cm

41.2 cm

AC =\sqrt{40^2 + 40^2} = \sqrt{3200}= 40\sqrt{2} = 56.5685…

So,

AM=\frac{1}{2}AC=20\sqrt{2}=28.28427…\\
MX=\sqrt{50^2 – 28.28427…^2} = 10\sqrt{17} = 41.2310…

1. Rectangle ABCD is the horizontal base of a triangular prism ABCDEF.

AE = BE\\
AB = 10 cm\\
AE = 17 cm\\
BC = 25 cm

Work out the volume of the prism. Give your answer correct to 3 significant figures.

**(5 marks)**

Show answer

h^2=17^2 – 5^2

for starting to use Pythagoras to find the height

**(1)**

h=\sqrt{17^2 – 5^2} = 2\sqrt{66} = 16.24807…

for finding the height of the triangle

**(1)**

Area = \frac{1}{2}\times 10 \times 2\sqrt{66} = 10\sqrt{66}=81.24038…

for finding the area of the triangle

**(1)**

Volume = 10\sqrt{66} \times 25 = 2031.0096…

for finding the volume of the triangular prism

**(1)**

Volume = 2030 cm^3

for the final volume

**(1)**

2. The diagram shows a box in the shape of a cuboid ABCDEFGH.

AB = 12.5 cm\\ AE = 6.7 cm\\ BC = 3.8 cmA string runs diagonally across the box from D to F.

Calculate the length of the string DF.

Give your answer correct to 3 significant figures.

**(3 marks)**

Show answer

DB=\sqrt{12.5^2 + 3.8^2} = 13.06483…

for using Pythagoras to find DB

**(1)**

DF=\sqrt{13.06483…^2 + 6.7^2} = 14.68264…

for using Pythagoras to find DF

**(1)**

DF = 14.7 cm

for the final length

**(1)**

3. A cone has a radius of 10 cm .

The curved surface area of the cone is 800 cm^2 .

Find the volume of the cone.

Give your answer correct to 3 significant figures.

**(5 marks)**

Show answer

\pi \times 10 \times l=800

for setting up the equation to find the slant length

**(1)**

l=\frac{800}{10\pi}=\frac{80}{\pi}

for finding the slant length

**(1)**

h=\sqrt{(\frac{80}{\pi})^2-10^2}=23.419128…

for finding the height

**(1)**

V=\frac{1}{3}\times \pi \times 10^2 \times 23.419128… = 2452.445…

for finding the volume

**(1)**

V=2452.445… = 2450 cm^3

insert for the final answer

**(1)**

You have now learned how to:

- Use and apply Pythagoras’ theorem to solve problems in three dimensions
- Solve multi-step problems

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#### GCSE Maths Papers - November 2022 Topics

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