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.
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:
a2+b2=(AH)2
32+42=(AH)2
(AH)2=25
AH=25=5cm
Then, find c by focusing on the triangle AFH:
(AH)2+c2=d2
52+c2=132
25+c2=169
c2=144
c=144=12cm
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?
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:
22+22=(BG)2
(BG)2=8
BG=8cm
Now, use Pythagoras on the triangle BEG to find BG:
(BG)2+(EG)2=(BE)2
(8)2+22=(BE)2
8+4=(BE)2
(BE)2=12
BE=12=23cm
BE=23cm
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FAQs - Frequently Asked Questions
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.