Base of pyramid of a square of side 8cm vertex directly above the centre edge = \(4\sqrt{3}\)cm From the diagram, the diagonal of one base is AC\(^2\) = 8\(^2\) + 8\(^2\) AC\(^2\) = 64 + 64 = 128 AC = \(8\sqrt{2}\) but OC = \(\frac{1}{2}\)AC = \(\frac{1}{2}\) x 8\(\sqrt{2}\) = \(4\sqrt{2}\)cm OE = h = height h\(^2\) = EC\(^2\) - OC\(^2\) h\(^2\) = (\(4\sqrt{3}\))\(^2\) - (\(4\sqrt{2}\))\(^2\) h\(^2\) = 16 x 3 - 16 x 2 48 - 32 = 16 h = \(\sqrt{16}\) = 4cm