imc

Using the property that opposite angles are equal in a rhombus, $\angle DAB=\angle DCB=60^{\circ }$ and $\angle ADC=\angle ABC=120^{\circ }$. It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$. Let the lengths of the sides of rhombus $ABCD$ be $s$. The longer diagonal of rhombus $BFDE$ is $BD$. Since $BD$ is a side of an equilateral triangle with a side length of $s$, $BD=s$. The longer diagonal of rhombus $ABCD$ is $AC$. Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$, $AC=2\cdot \frac{s\sqrt{3}}{2}=s\sqrt{3}$. The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $\frac{s}{s\sqrt{3}}=\frac{\sqrt{3}}{3}$. Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is ${\left(\frac{\sqrt{3}}{3}\right)}^2=\frac{1}{3}$. Let $x$ be the area of rhombus $BFDE$. Then $\frac{x}{24}=\frac{1}{3}$, so $x=\boxed{8}$.