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JAMB 1987 · UME · Q43✱✱

The base of a pyramid is a square of side 8 cm. If its vertex is directly above the centre, find the height, given that each slant edge is 434\sqrt3 cm.

Worked solution (try it first)
  1. The diagonal of the square base is 828\sqrt2 cm, so each corner is 424\sqrt2 cm from the centre.
  2. The height, that distance and a slant edge form a right-angled triangle, so h2=(43)2−(42)2h^2 = (4\sqrt3)^2 - (4\sqrt2)^2, which is 48−32=1648 - 32 = 16.
  3. So h=4h = 4 cm, option C.

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