jmc

Let $E$, $F$ and $G$ be the points of tangency of the incircle with $BC$, $AC$ and $AB$, respectively. Without loss of generality, let $AC<AB$, so that $E$ is between $D$ and $C$. Let the length of the median be $3m$. Then by two applications of the Power of a Point Theorem, $D{E}^2=2m\cdot m=A{F}^2$, so $DE=AF$. Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE=CF=c$ and thus $AC=AF+CF=DE+CE=CD=10$. Then $DE=AF=AG=10-c$ so $BG=BE=BD+DE=20-c$ and thus $AB=AG+BG=30-2c$. Now, by Stewart's Theorem in triangle $△ABC$ with cevian $\overline{AD}$, we have $$(3m{)}^2\cdot 20+20\cdot 10\cdot 10=10^2\cdot 10+(30-2c{)}^2\cdot 10.$$ Our earlier result from Power of a Point was that $2m^2=(10-c{)}^2$, so we combine these two results to solve for $c$ and we get $$9(10-c{)}^2+200=100+(30-2c{)}^2⟹c^2-12c+20=0.$$ Thus $c=2$ or $=10$. We discard the value $c=10$ as extraneous (it gives us a line) and are left with $c=2$, so our triangle has area $\sqrt{28\cdot 18\cdot 8\cdot 2}=24\sqrt{14}$ and so the answer is $24+14=\boxed{38}$.