Drop an altitude from C to AD to form a 30-60-90 triangle. Call the foot of this altitude E. CE is \(5\sqrt{3}\) and ED is 5. This means AE is \(10-ED\), which is 5. Thus, if we draw AC, we form another 30-60-90 triangle, and we find that AC is 10, and that triangle ACD is equilateral. An equilateral triangle with a side length of 10 has an area \(25\sqrt{3}\). Now, we just have to calculate the area of triangle BCA, then add the two areas. Triangle BCA has sides 4, 8, and 10. We drop another altitude, this one to AC, but unfortunately, this doesn't lead to any nice triangles. So, we resort to algebra. Call the foot of this altitude F. Let \(FC=x\), so \(AF=10-x\). Also, let \(BF=h\). Using the Pythagorean theorem on both triangle BCF and ABF, we get the two equations \(x^2+h^2=16\) and \((10-x)^2+h^2=64\). Subtracting the two equations and solving them gives \(x=\frac{13}{5}\). Substituting into either equations yields \(h=\frac{\sqrt{231}}{5}\) So, triangle BCA has a base of 10 and a height of \(\frac{\sqrt{231}}{5}\). This means that triangle BCA has an area of \(\sqrt{231}\), which can't be simplified. Adding, our total area is \(\sqrt{231}+25\sqrt{3}\). Thus, \(\mathrm{a+b+c \ is}\)231+25+3, which is \(\boxed{259}\).
+1164 