With 3D shapes sometimes you need to use Pythagoras more than once to find a particular length.
There’s a formula in cuboids for finding the length of the longest diagonal.Other Shapes
Example: Find the height x for the below pyramid.First find the length AC
ac2 = ab2 + bc2
ac2 = 144 + 144 = 288
ac = 16.97056…
So ay is half this.
ay = 8.48528…
So now you have another right angled triangle you can work out:
av2 = ay2 + x2
x2 = av2 – ay2 = 400 – (8.48528…)2 = 400 – 72 = 328
So x is the square root of 328
So x = 18.11 (2 d.p.)