The diagram below shows a square-based pyramid.
Point $\mathrm{O}$ is directly above the centre of the base.
a) Work out the length of $\mathrm{AC}$.
b) Work out the perpendicular height of the pyramid.
Give each of your answers to 3 s.f.
Not drawn accurately
\(\boxed{h = 6}\)
Step 1 :\(s^2 = 4^2 + 4^2 = 16 + 16 = 32\)
Step 2 :\(s = \sqrt{32} = 4\sqrt{2}\)
Step 3 :\(AC = \sqrt{AO^2 + OC^2} = \sqrt{6^2 + (4\sqrt{2})^2} = \sqrt{36 + 32} = \sqrt{68}\)
Step 4 :\(\boxed{AC \approx 8.25}\)
Step 5 :\(\text{Let } h \text{ be the perpendicular height of the pyramid}\)
Step 6 :\(\text{Applying Pythagorean theorem to triangle AOC, we get}\)
Step 7 :\(h^2 = AC^2 - OC^2 = 68 - 32 = 36\)
Step 8 :\(h = \sqrt{36} = 6\)
Step 9 :\(\boxed{h = 6}\)