Video solution:

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Written solution:

Define $I$ to be the midpoint of $\overline{AD}$ and $G$ to be the midpoint of $\overline{EF}.$ Also define $H$ to be the intersection of $\overline{CF}$ and $\overline{BE}.$

The area of $\triangle BCE$ equals $$\dfrac{1}{2} \cdot 1 \cdot 4 = 2.$$ By symmetry, we can see that $\triangle BCH$ and $\triangle EFH$ are similar. Since their bases are in a $1 : 3$ ratio, so are their altitudes. This means that $3IH = HG,$ which implies that $IH = 1.$

Therefore, the area of $$\begin{align*}\triangle BCH &= \dfrac{1}{2} \cdot 1 \cdot 1\\ &= \dfrac{1}{2}.\end{align*}$$ This implies that the area of $$\begin{align*}\triangle ECH &= 2 - \dfrac{1}{2} \\&= \dfrac{3}{2}.\end{align*}$$ Since the figure is symmetric, the total area of the bat wings is $2 \cdot \dfrac{3}{2} = 3.$

Thus, C is the correct answer.