If the length of each edge of regular octahedron ABCDEF is 3, then find the area of triangle ACF.
Because of symmetry, the area of triangle ACF will be the same as the area of triangle EAC
EC is the base = the diagonal of a square with a side of 3 = 3sqrt 2
The height of this triangle = sqrt [3^2 - [(3/2)sqrt2]^2 ] = sqrt [ 9 - 9/2 ] = sqrt [ 9/2] = 3 /sqrt 2
So.....the area of triangle ACF = (1/2) [3sqrt 2] [3 /sqrt 2 ] = 9/2 = 4.5 units^2