two triangles and three circles

As described in Alak Bhattacharya's solution, we find the radius of the circle is r = 6. Draw three line segments connecting each vertex of the small triangle with the corresponding vertex of the larger triangle. This partitions the big triangle into three congruent trapezoids plus the small triangle. Therefore, the area of the larger triangle is A = 3 (area of trapezoid) + (area of smaller triangle). But, the (area of the trapezoid) = (1/2) (height) (sum of the bases) = (1/2) r (2r+2 r sqrt(3)+r) (the last equality follows from 30-60-90 triangles). Substitute r = 6, and recall that the area of the smaller triangle is given to be 36 sqrt(3) and simplify. This yields A = 144*sqrt(3)+216, which is equivalent to the third choice.

Let the length of each side of the smaller triangle be $a$ . Then $\dfrac{√3a^2}{4}=36√3$ or $a=12$ . Therefore the radius of each circle is $6$ , and the length of each side of the larger triangle is $12(1+\cot 30\degree)=12(√3+1)$ . Therefore the area of this triangle is $\dfrac{√3}{4}\times 144\times (√3+1)^2=\boxed {\dfrac{√3}{4}(576+288√3)}$