Solution:

Consider the following diagram:

The shape is symmetric, so we can find the ratio of the areas of $EFCD$ to the area of $ZFCW.$ For this, we can extend $FE$ and $CD$ until they meet each other, and let this point be $P.$

Also, the distance from $ED$ to $ZW$ is the same as the distance from $ZW$ to $YX,$ and the distance from $ZW$ to $YX$ is twice the distance as the distance from $ZW$ to $FC.$

Thus, the distance from $ED$ to $ZW$ is twice the distance as the distance from $ZW$ to $FC.$ Let the altitude from $P$ to $ED$ be $s.$ Then, if the altitude from $P$ to $ZW$ is $s+d,$ the altitude from $P$ to $CF$ is $s+1.5d$ because of the ratio. Then, since $$\triangle FPC \sim \triangle PED,$$$$\dfrac{FP}{EP} = \dfrac{s+1.5d}{s}.$$ Also, since $FE= ED = PE$ as $PED$ is equilateral, we have $2s = s+1.5d,$ and so $s=1.5d.$

Thus, the ratio between side lengths of $\triangle PED$ and $\triangle PZW$ is $\frac 53$ which makes the ratio between the area $\frac{25}9.$ Also the ratio between side lengths of $\triangle PED$ and $\triangle PFC$ is $2,$ which makes the ratio between the area $4.$

This makes the area of $EDCF$ $3$ times the area of $\triangle PED$ and the area of $EDWZ$ $\frac{16}{9}$ times the area of $\triangle PED$ Therefore, the area of $ZWCF$ is $\frac{11}{9}$ times the area of $\triangle PED.$

Thus, the ratio between $ZWCF$ and $EDCF$ is $$\dfrac{\frac{11}9}{3} = \dfrac{11}{27} .$$

Thus, the correct answer is C .