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 AC² = 8² + 8²
Ac² = 64 + 64 = 128
AC = \(8\sqrt{2}\)
but OC = \(\frac{1}{2}\)AC = 8\(\sqrt{\frac{2}{2}}\) = \(4\sqrt{2}\)cm
OE = h = height
h² = (\(4\sqrt{3}\))²
16 x 2 - 16 x 2
48 - 32 = 16
h = \(\sqrt{16}\)
= 4