The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4\(\sqrt3\)cm
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
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