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Using Pythagoras in 3D

Pythagoras' theorem can also be used to solve 3D problems. You often need to use Pythagoras' theorem more than once in each problem as we create extra right-angled triangles in the 3D shapes.

Example

This cuboid has side lengths of 2cm, 3cm and 6cm.

Diagram of 3D cuboid with values 2cm, 3cm and 6cm

Work out the length of the diagonal AF.

Answer

First use Pythagoras' theorem in triangle ABC to find length AC.

\(A{C^2} = {6^2} + {2^2}\) (as BC must be 2cm)

\(AC = \sqrt {40}\)

You do not need to find the root as we will need to square it in the following step. Next we use Pythagoras' theorem in triangle ACF to find length AF.

\(A{F^2} = A{C^2} + C{F^2}\)

\(AF^{2}=(\sqrt{40})^{2}+3^{2}\)

\(AF^{2}= 40+9\)

\(AF^{2}=49\)

\(AF = \sqrt {49}\)

\(AF = 7cm\)

Now try the example question below.

Question

Work out the vertical height (VM) in this square-based pyramid and give your answer to one decimal place.

Diagram of a square based pyramid with values

The square base ABCD has side lengths of 5cm.

Lengths VA, VB, VC and VD are each 8cm.

M is the mid point of the square base.

  • Remember to use at least one more decimal place than you are asked for during your calculation.
  • When giving your final answer, always round according to what the question is asking for, in this case one decimal place.