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 $ABCDEFGH$ :

Maths; Trigonometry; KS4 Year 10; 3D Pythagoras - Higher

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 $AF$ .

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 $AFH$ :

$(AH)^2+c^2=d^2$

$5^2+c^2=13^2$

$25+c^2=169$

$c^2=144$

$c=\sqrt{144}=12cm$

$\underline{c=12cm}$

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?

Maths; Trigonometry; KS4 Year 10; 3D Pythagoras - Higher

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 $BG$ :

$(BG)^2+(EG)^2=(BE)^2$ 

$(\sqrt{8})^2+2^2=(BE)^2$ 

$8+4=(BE)^2$ 

$(BE)^2=12$

$BE=\sqrt{12}=2\sqrt{3}cm$

$\underline{BE=2\sqrt{3}cm}$