3D Pythagoras & Trigonometry (Edexcel IGCSE Maths)

Spec point — 6F

3D Pythagoras & trigonometry

How do I use Pythagoras' theorem in a 3D shape?

• You can often find right-angled triangles within 3D shapes

  • If two sides of the triangle are known, you can use Pythagoras’ theorem

Is there a 3D version of the Pythagoras' theorem formula?

• There is a 3D version of Pythagoras’ theorem: $d^{2}=x^{2}+y^{2}+z^{2}$

  • $d$ is the distance between two points
  • $x,y$ and $z$ are the distances in the three different perpendicular directions between the two points

• However, all 3D situations can be broken into two 2D problems

  • Form two right-angle triangles

Exam Hint You are not given the 3D Pythagoras formula in the exam.

You can always split 3D problems into two 2D problems (which don't need this formula).

How do I use SOHCAHTOA in 3D?

• Again, look for right-angled triangles to use with SOHCAHTOA

  • You may need combinations of triangles that lead to the missing side or angle

Exam Hint If you are stuck in the exam with a complicated 3D diagram, it is always better to just start finding any lengths and angles in the shape, as:

• these may end up being useful
• you may score more marks than if you had left the question blank

Worked Example A pencil is being put into a cuboid shaped box.

The box has dimensions 3 cm by 4 cm by 6 cm.

(a) Find the length of the longest pencil that can fit inside the box.

Answer:

The longest possible pencil will fit between diagonally opposite vertices, e.g. AF

Form a 2D right-angled triangle, such as triangle ABF

Method 1

To find the length AF, there are a few different options One option is to find length BF (from triangle BEF) then AF (from triangle ABF) Draw triangle BEF flat and use Pythagoras' theorem to find BF

$a^{2}=4^{2}+6^{2}\,a^{2}=16+36\,a^{2}=52$

Draw triangle ABF flat and use Pythagoras' theorem to calculate AF

$\begin{gathered} b^{2} & = 3^{2} + a^{2} \\ b^{2} & = 9 + 52 \\ b^{2} & = 61 \\ b = \sqrt{61} = 7.81024... \end{gathered}$

Final answer: The longest pencil that can fit inside the box is 7.81 cm (3 s.f.)

Method 2

*Apply the 3D version of Pythagoras’ theorem: *$d^{2}=x^{2}+y^{2}+z^{2}$

The distance in the x direction is 4 cm The distance in the y direction is 6 cm The distance in the z direction is 3 cm

$\begin{gathered} d^{2} & = 4^{2} + 6^{2} + 3^{2} \\ d = \sqrt{4^{2} + 6^{2} + 3^{2}} \\ d = \sqrt{61} = 7.81024... \end{gathered}$

Final answer: The longest pencil that can fit inside the box is 7.81 cm (3 s.f.)