$ABCD$ forms a rhombus. $E$ is the intersection of $AC$ and $BD$. $F$ lie on $AD$ such that $EF$ is perpendicular to $FD$. Given
$EF=2$ and $FD=1$. Find the area of the rhombus $ABCD$
1. Given that $ABCD$ forms a rhombus, we know that the diagonals $AC$ and $BD$ intersect at right angles at point $E$ and bisect each other. Therefore, $\triangle AED$ is a right triangle with $E$ as the right angle.
2. Since $EF$ is perpendicular to $FD$, $\triangle EFD$ is a right triangle with $EF$ as one leg and $FD$ as the other leg. Given $EF = 2$ and $FD = 1$, we can use the Pythagorean theorem to find $ED$:
   \[
   ED = \sqrt{EF^2 + FD^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}
   \]
3. Since $F$ is the foot of the altitude from $E$ to $AD$, we use the property of right triangles that states the product of the segments of the hypotenuse equals the square of the altitude. Thus, $AF \cdot FD = EF^2$:
   \[
   AF \cdot 1 = 2^2 \implies AF = 4
   \]
4. The area of $\triangle AED$ can be calculated using the base $AD$ and height $EF$. Since $AD = AF + FD = 4 + 1 = 5$:
   \[
   \text{Area of } \triangle AED = \frac{1}{2} \times AD \times EF = \frac{1}{2} \times 5 \times 2 = 5
   \]
5. Since the diagonals of the rhombus divide it into four congruent right triangles, the total area of the rhombus is four times the area of one of these triangles:
   \[
   \text{Total area of } ABCD = 4 \times \text{Area of } \triangle AED = 4 \times 5 = 20
   \]

The final answer is $\boxed{20}$