The area $A$ of the figure $K$ is in fact $\leq{\pi\over4}$, and this bound is attained only if $K$ is a circular disk of diameter $1$.
Let $\bigl(u(\phi),v(\phi)\bigr)$ be the midpoint of the median with slope $\phi$, and let $\mu$ be the locus of these midpoints. Then the boundary curve $\gamma$ of $K$ has a parametric representation of the following form: $\left.\eqalign{x(\phi)&=u(\phi)+r(\phi)\cos\phi\cr y(\phi)&=v(\phi)+r(\phi)\sin\phi\cr}\right\}\qquad(0\leq\phi\leq 2\pi)\ ,$ where $u(\cdot)$, $v(\cdot)$ and $r(\cdot)$ are periodic with period $\pi$ (and not $2\pi$ !).
Denoting by $A_\phi$ the part of $A$ to the right of the median with slope $\phi$, $\ 0<\phi<\pi$, and by $A(\phi)$ its area, we have 2A(\phi)=\int_{\partial A_\phi}(x\ dy-y\ dx)=\int_{\phi-\pi}^\phi(x y'-yx')\ dt +\int_\sigma (x\ dy -y\ dx)\ , where $\sigma$ denotes the directed segment from $\bigl(x(\phi),y(\phi)\bigr)$ to $\bigl(x(\phi-\pi),y(\phi-\pi)\bigr)$. Using that $u$, $v$ and $r$ have period $\pi$ one computes A'(\phi)=2 r(v'\cos\phi-u'\sin\phi)\ . As this should vanish identically we necessarily have v'(\phi)\cos\phi-u'(\phi)\sin\phi\equiv0\ .\qquad\qquad(1) Geometrically this means that where (u',v')\ne(0,0) the midpoint locus $\mu$ is the envelope of the medians. Maybe there is a simpler way to prove that.
Let $R$ be the rectangle $[0,1]\times[0,2\pi]$ in the $(t,\phi)$-plane and consider the map $g:\ R\to K\ ,\quad (t,\phi)\mapsto\cases{x:=u(\phi)+tr(\phi)\cos\phi \cr y:=v(\phi)+t r(\phi)\sin\phi \cr}\quad.$ This map is surjective, since through each point $(x,y)\in K$ there is at least one median, so that the forward half of a median turning around $360^\circ$ will pass over this point. Therefore the function $\nu(x,y)$ counting the inverse images of the point $(x,y)$ is $\geq1$ on $K$. The Jacobian of $g$ computes to J_g(t,\phi)=r(v'\cos\phi-u'\sin\phi) + tr^2=t\> r^2(\phi)\geq 0\ , where we have used $(1)$. Using the (intuitively evident) formula $\int\nolimits_K\nu(x,y)\ {\rm d}(x,y)=\int\nolimits _R J_g(t,\phi)\ {\rm d}(t,\phi)$ from geometric measure theory it now follows that $A=\int\nolimits_K\ {\rm d}(x,y)\leq\int\nolimits_K\nu(x,y)\ {\rm d}(x,y)={1\over2}\int_0^{2\pi}r^2(\phi)\ d\phi\leq{\pi\over4}\ .$