Let AB = b, and AC = c. Moreover, let [ABX]=A1 and [ACX]=A2, where [⋅] denote "area of".
By Stewart's theorem, we have
(13+5)(132+13(5))=5b2+13c25b2+13c2=4212
By basic geometry, we have A1A2=135.
Recall that the circumradius of a triangle with sides a, b, c and area A is abc4A. Then, since the circumradius of triangles ABX and ACX are the same, we have
132b4A1=5(13)c4A2A1A2=13b5c135=13b5cb=c.
Hence,
18b2=18c2=4212b=c=3√26
We calculate the half perimeter of triangle ABC to be 18+b+c2=3√26+9. Then, by Heron's formula, the area required is:
√(3√26+9)(3√26−9)(92)=27√17.