Pythagoras' theorem in 3 dimensions - Higher
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
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 is the middle of a line segment. It divides the line segment in half. of the square base ABCD.
Lengths AD, DC, BC and AB are all 4 cm.
The If the angle between two lines is a right angle, the lines are said to be 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.
\(a^2 + b^2 = c^2\)
\(\text{CD}^2 + \text{AD}^2 = \text{AC}^2\)
\(4^2 + 4^2 = \text{AC}^2\)
\(32 = \text{AC}^2\)
\(\text{AC} = \sqrt{32}\)
\(\sqrt{32}\) 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)