Pythagoras' theorem in 3 dimensions - Higher
Explore updated revision resources for GCSE Maths: Higher - Solving 3D problems using Pythagoras's, with step-by-step slideshows, quizzes, practice exam questions, and more!
An introduction to turning 3D problems into 2D problems and using Pythagoras' theorem
Pythagoras's theorem applies to right-angled triangles. The area of the square drawn on the hypotenuse is equal to the sum of the squares drawn on the other two sides. can be used to solve An object with width, height and depth, eg a cube. 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.
Example
ABCDEFGH is a cuboid.
AB is 6 cm, BG is 3 cm and FG is 2 cm.
Calculate AF.
Draw the right-angled triangle ACF and label the sides.
This is the right-angled triangle that contains the unknown AF.
To calculate AF, AC is needed.
To calculate 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.
AC is \(\sqrt{40}\) cm.
In the right-angled triangle AFC, 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. The square base has sides of length 4cm. O is the Midpoint is the middle of a line segment. It divides the line segment in half. of the base.
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 AV, OA is needed.
Draw the right-angled triangle ACD and label the sides.
\(a^2 + b^2 = c^2\)
\(\text{CD}^2 + \text{AD}^2 = \text{AC}^2\)
\(4^2 + 4^2 = \text{AC}^2\)
\(32 = \text{AC}^2\)
\(\text{AC} = \sqrt{32}\)
\(\sqrt{32}\) is a surd. Do not round this answer yet.
AC is \(\sqrt{32}\) cm.
The point O is in the centre of AC, so OA is half of the length AC.
OA is \(\frac{\sqrt{32}}{2}\) cm.
In the right-angled triangle AOV, OA is now known.
\(a^2 + b^2 = c^2\)
\(\text{OV}^2 + \text{OA}^2 = \text{AV}^2\)
\(3^2 + (\frac{\sqrt{32}}{2})^2 = \text{AV}^2\)
\(17 = \text{AV}^2\)
\(\text{AV} = \sqrt{17}\)
\(AV = 4.1~\text{cm}\) (to one decimal place)
3D Pythagoras – quick method
2D Pythagoras can be extended to 3D Pythagoras when the three lengths are all at right-angles to each other.
Example one – cuboid
\(\text{AF} = \sqrt{(2^2+6^2+3^2)} = \sqrt{49} = 7~\text{cm}\).
Example 2 – squared based pyramid
Previously, we used Pythagoras with 4 cm and 4 cm to find AC and then halved it to find AO. Instead, we could have used Pythagoras with 2 cm and 2 cm to find AO directly.
Now using 3D Pythagoras with 2 cm, 2 cm and 3 cm, we can find AV.
\(\text{AV} = \sqrt{(2^2+2^2+3^2)} = \sqrt{17}~\text{cm}\)
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- Angles, lines and polygons - OCR
- Loci and constructions - OCR
- 2-dimensional shapes - OCR
- 3-dimensional solids - OCR
- Circles, sectors and arcs - OCR
- Circle theorems - Higher - OCR
- Transformations - OCR
- Units of measure - OCR
- Trigonometry - OCR
- Vectors - OCR
