jmc

We begin by drawing a diagram: Since $\angle CDA=60^{\circ }$ and $AD=DC$, $△ACD$ is an equilateral triangle, so $AC=10$ and $$[△ACD]=\frac{10^2\sqrt{3}}{4}=25\sqrt{3}.$$Now we want to find $[△ABC]$. To find the height of this triangle, we drop a perpendicular from $B$ to $AC$ and label the intersection point $E$: Let $BE=h$, $CE=x$, and $EA=10-x$. Using the Pythagorean Theorem on $△BCE$ yields $$x^2+h^2=16$$and on $△ABE$ yields $$(10-x{)}^2+h^2=64.$$Expanding the second equation yields $x^2-20x+100+h^2=64$; substituting $16$ for $x^2+h^2$ yields $16+100-20x=64$. Solving yields $x=\frac{13}{5}$ and $h=\sqrt{16-x^2}=\frac{\sqrt{231}}{5}$. It follows that $$[△ABC]=\frac{1}{2}(BE)(AC)=\frac{1}{2}\cdot \frac{\sqrt{231}}{5}\cdot 10=\sqrt{231}.$$Finally, $$[ABCD]=[△ADC]+[△ABC]=25\sqrt{3}+\sqrt{231}=\sqrt{a}+b\sqrt{c}.$$Thus we see $a=231$, $b=25$, and $c=3$, so $a+b+c=\boxed{259}$.