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

Using Pythagoras' theorem in $3$â€‹ dimensions

It is best to show this with a sketch. Here is a cuboid $ABCDEFGH$â€‹:

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

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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 $ABH$:

$a^2+b^2=(AH)^2$â€‹â€‹â€‹â€‹

â€‹$3^2+4^2=(AH)^2$â€‹â€‹

â€‹$(AH)^2=25$â€‹â€‹

â€‹$AH=\sqrt{25}=5cm$â€‹â€‹

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

â€‹$(AH)^2+c^2=d^2$â€‹â€‹

â€‹$5^2+c^2=13^2$â€‹â€‹

â€‹â€‹â€‹$25+c^2=169$â€‹â€‹

â€‹$c^2=144$â€‹â€‹

$c=\sqrt{144}=12cm$â€‹â€‹â€‹â€‹â€‹â€‹

â€‹$\underline{c=12cm}$â€‹â€‹

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Other shapes

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

Example 2

A cube $ABCDEFGH$â€‹ has side lengths $2cm$â€‹. What is the exact length of $BE$â€‹ 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 $BAG$:

$(BA)^2+(AG)^2=(BG)^2$â€‹â€‹

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

â€‹$2^2+2^2=(BG)^2$â€‹â€‹

â€‹$(BG)^2=8$â€‹â€‹

â€‹$BG=\sqrt{8} cm$â€‹â€‹

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

How do you know when to use Pythagoras' theorem in 3D?

When you have a 3d shape and have sides that travel across 3 different dimensions (width, height, and depth), you know you have to use Pythagoras' theorem twice.

How do you use Pythagoras' theorem in 3 dimensions?

Apply Pythagoras' theorem twice on two different triangles to find the 3 dimensional length.

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