With 3D shapes sometimes you need to use Pythagoras more than once to find a particular length.**Cuboids**

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

*ac*2 = a*b*2 + b*c*2

*ac*2 = 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:

*av*2 = ay2 + x2

so:

*x*2 = av2 – ay2 = 400 – (8.48528…)2 = 400 – 72 = 328

So x is the square root of 328

So x = 18.11 (2 d.p.)