imc

Denote $D=(0,0)$. Then $A=(0,2),F=\left(\frac{1}{2},0\right),H=\left(\frac{1}{2},1\right)$. Let the intersection of $AF$ and $DH$ be $X$, and the intersection of $BF$ and $CH$ be $Y$. Then we want to find the coordinates of $X$ so we can find $XY$. From our points, the slope of $AF$ is $\left(\frac{-2}{\frac{1}{2}}\right)=-4$, and its $y$-intercept is just $2$. Thus the equation for $AF$ is $y=-4x+2$. We can also quickly find that the equation of $DH$ is $y=2x$. Setting the equations equal, we have $2x=-4x+2⟹x=\frac{1}{3}$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\frac{1}{3}$, so $XY=1-2\cdot \frac{1}{3}=\frac{1}{3}$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF=1$, our answer is $\frac{\frac{1}{3}\cdot 1}{2}=\boxed{\frac{1}{6}}$.