In the following equilateral triangle ABC, AF = 3FB,
and FH is the perpendicular to BC. If the area of the shaded triangle is 1 cm^{2},
then the area of the equilateral triangle is cm^{2}.

We form an equilateral triangle, using FB as a side length. We do this by flipping FBH over FH, such that B becomes D, on HC.

Since FB is a fourth of AB and BD is a fourth of BC, ABC represents 4 × 4 = 16 the area of FBD. Since FBD is double FBH, ABC is 16 × 2 = 32 times the size of FBH, so it has an area of 32 × 1 = 32.