Pythagoras’ theorem can be used to calculate the length of any side in a right-angled triangle. Pythagoras’ theorem can be applied to solve 3-dimensional problems.

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Pythagoras' theorem can be used to solve 3-dimensional problems which involve calculating the length of a right-angled triangle.

It may be necessary to use Pythagoras' theorem more than once in a problem.

The shape ABCDEFGH is a cuboid.

Length AB is 6 cm, length BG is 3 cm and length FG is 2 cm.

Calculate the length AF.

Draw the right-angled triangle ACF and label the sides.

This is the right-angled triangle that contains the unknown length AF.

To calculate the length AF, the length AC is needed.

To calculate the length AC, draw the right-angled triangle ABC and label the sides.

\[a^2 + b^2 = c^2\]

\[\text{BC}^2 + \text{AB}^2 = \text{AC}^2\]

\[2^2 + 6^2 = \text{AC}^2\]

\[40 = \text{AC}^2\]

\[\text{AC} = \sqrt{40}\]

\(\sqrt{40}\) is a surd. Do not round this answer yet.

The length AC is \(\sqrt{40}\) cm.

In the right-angled triangle AFC the length AC is now known.

\[a^2 + b^2 = c^2\]

\[\text{FC}^2 + \text{AC}^2 = \text{AF}^2\]

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

\[49 = \text{AF}^2\]

\[\text{AF} = 7~\text{cm}\]

Length AF = 7 cm

- Question
ABCDV is a square based pyramid. O is the midpoint of the square base ABCD.

Lengths AD, DC, BC and AB are all 4 cm.

The perpendicular height of the pyramid (OV) is 3 cm.

Calculate the length AV. Give the answer to one decimal place.

Draw the right-angled triangle AOV and label the sides.

This is the right-angled triangle that contains the unknown length AV.

To calculate the length AV, the length OA is needed.

Draw the right-angled triangle ACD and label the sides.

\[a^2 + b^2 = c^2\]

\[\text{CD}^2 + \text{AD}^2 = \text{AC}^2\]

\[4^2 + 4^2 = \text{AC}^2\]

\[32^2 = \text{AC}^2\]

\[\text{AC} = \sqrt{32}\]

\(\sqrt{32}\) is a surd. Do not round this answer yet.

The length AC is \(\sqrt{32}\) cm.

The point O is in the centre of the length AC, so OA is half of the length AC.

The length OA is \(\frac{\sqrt{32}}{2}\) cm.

In the right-angled triangle AOV, the length OA is now known.

\[a^2 + b^2 = c^2\]

\[\text{OV}^2 + \text{OA}^2 = \text{AV}^2\]

\[3^2 + (\frac{\sqrt{32}}{2})^2 = \text{AV}^2\]

\[17 = \text{AV}^2\]

\[\text{AV} = \sqrt{17}\]

\(AV = 4.1 cm\) (to one decimal place)