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

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 ：\(\boxed{h = 6}\)