3D Pythagoras

In a nutshell

So far, you have used Pythagoras' theorem to find lengths of $$2$$​ dimensional shapes. However, Pythagoras' theorem can also applied to find distances in $$3$$​ dimensions.

Using Pythagoras' theorem in $$3$$​ dimensions

It is best to show this with a sketch. Here is a cuboid $$ABCDEFGH$$​:

To find the distance $$AF$$​ ($$=d$$), first use Pythagoras' theorem on triangle $$ABH$$​ to find the distance $$AH$$​, and then use it again on triangle $$AFH$$​ to find the distance $$AF$$​.

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Example 1

In the cuboid shown above, $$a=3cm,b=4cm,d=13cm$$. Find the value of $$c$$.

Find $$AH$$ by focusing on the right-angled triangle $$ABH$$:

$$a^2+b^2=(AH)^2$$​​​​

​$$3^2+4^2=(AH)^2$$​​

​$$(AH)^2=25$$​​

​$$AH=\sqrt{25}=5cm$$​​

Then, find $$c$$ by focusing on the triangle $$AFH$$:

​$$(AH)^2+c^2=d^2$$​​

​$$5^2+c^2=13^2$$​​

​​​$$25+c^2=169$$​​

​$$c^2=144$$​​

$$c=\sqrt{144}=12cm$$​​​​​​

​$$\underline{c=12cm}$$​​

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Other shapes

This formula can be applied to other 3D shapes, such as cubes and pyramids.

Example 2

A cube $$ABCDEFGH$$​ has side lengths $$2cm$$​. What is the exact length of $$BE$$​ shown in the diagram?

The length $$BE$$ can be found by using Pythagoras on the triangle $$BEG$$. However, to find the length $$BG$$, first use Pythagoras on the triangle $$BAG$$:

$$(BA)^2+(AG)^2=(BG)^2$$​​

Since it's a cube, all sides are of length $$2cm$$, so substitute this for the lengths $$BA$$ and $$AG$$​:

​$$2^2+2^2=(BG)^2$$​​

​$$(BG)^2=8$$​​

​$$BG=\sqrt{8} cm$$​​

Now, use Pythagoras on the triangle $$BEG$$ to find $$BG$$:

$$(BG)^2+(EG)^2=(BE)^2$$​​

$$(\sqrt{8})^2+2^2=(BE)^2$$​​

$$8+4=(BE)^2$$​​

​$$(BE)^2=12$$​​

​​$$BE=\sqrt{12}=2\sqrt{3}cm$$​​

​$$\underline{BE=2\sqrt{3}cm}$$​​