JAMB - Mathematics (1987 - No. 34)
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
Giải thích
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