2006 AMC 10B Problem 15

Because $AB=AD$ and $\angle BAD=60^\circ,$ triangle $ABD$ is equilateral, and so is triangle $CBD.$ Points $E$ and $F$ split the rhombus into six congruent triangles, each of area $\tfrac{24}{6}=4.$ Rhombus $BFDE$ is the union of triangles $BED$ and $BFD,$ so its area is $2\cdot4=8.$ Thus, the correct answer is C.