KAREN BUURMAN

Posted on Tuesday, 25 May, 2004 - 07:55 am:

To make a rhombic dodecahedron the same size as gold mined - about 2500 tonnes, given the density of gold is 19.3g/cm^3. What is the length of an edge of our dodecahedron, correct to the nearest cm?

Matthew Smith

Posted on Tuesday, 25 May, 2004 - 09:45 pm:

There's a formula for the volume of a rhombic dodecahedron towards the bottom of this page. The result is that

V = (16 / 9) \sqrt{3} a^{3}

, where

a

is the edge length. If you need to prove this formula, then you can consider the rhombic dodecahedron as being made up of a cube of side

s

, with a square-based pyramid of height

s / 2

stuck on each side (see if you can show that this figure really is a rhombic dodecahedron), and hence work out a relation between the volume and side length.

You can work out the volume of gold (because density is just mass per unit volume, and you know the mass). Hence, using the formula above, you can find the edge length.

Matthew.