The altitude of $M$ to $DC$ is also $h$ and has base point $I$ on $DC$. As $F$ is the midpoint to $DM$ using scaling the altitude of $F$ to $DC$ is $h/2$. Thus $EF=3/2\cdot h$. $EF$ is a radius and $DM$ a diameter, thus $DM=2EF=3h$.

Because $AD=BC$ your trapezoid is symmetric. Thus $CI=((DC-AB)/2)/2=1$ and $DI=DC-CI=6$.

Apply pythagoras to $DIM$: $6^2+h^2=(3h)^2 \Rightarrow h=\frac{3}{\sqrt{2}}$.

The altitude is $2h=3\sqrt{2}$.

I am using $h$ as defined by the OP.

The altitude of $M$ to $DC$ is also $h$ and has base point $I$ on $DC$. As $F$ is the midpoint to $DM$ using scaling the altitude of $F$ to $DC$ is $h/2$. Thus $EF=3/2\cdot h$. $EF$ is a radius and $DM$ a diameter, thus $DM=2EF=3h$.

Because $AD=BC$ your trapezoid is symmetric. Thus $CI=((DC-AB)/2)/2=1$ and $DI=DC-CI=6$.

Apply pythagoras to $DIM$: $6^2+h^2=(3h)^2 \Rightarrow h=\frac{3}{\sqrt{2}}$.

The altitude is $2h=3\sqrt{2}$.