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
- 6cm
- 5cm
-
4cm
- 3cm
The correct answer is: C
Explanation
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