How do you solve the following geometry problem? The problem is asking for the area of the shaded regions, the only ones given are the measures of parts of the triangle.
I tried viewing the unshaded parts as sectors but I would need the radius of the shaded circle to solve it this way and I do not know how to get the radius.

This question was found on a sample CET (College Entrance Exam) Quiz Booklets me and my friends had.
 A: This type of question can be solved very easily by eliminating options. The answer will be the result of area(circle) + area(triangle) - area(sector).
Notice that the ${\sqrt 3}$ term of the answer can only come from the area of the triangle. So you can find the area of the triangle and then compare the coefficient with the answers. As all the coefficients of ${\sqrt 3}$ are different, you will easily get the answer.
A: The presence of $\sqrt 3$ in the answers suggests that the triangle equilateral and so the radius of the circle is $8$. Given this, the answer is (a). The drawing is not very good and does not reflect this.
A: The triangle indicated seems equilateral, then we have


*

*area of the circle: $\pi R^2$

*area of the triangle: $\frac12 \cdot (2R)\cdot (2R)\frac {\sqrt 3} 2$

*white sectors: $2\cdot R^2\cdot \frac \pi 6$

*shaded area shared by triangle and circle: $R^2\cdot \frac \pi 6$
and then
$$S=\pi R^2+\frac12 \cdot (2R)\cdot (2R)\frac {\sqrt 3} 2-3\cdot R^2\cdot \frac \pi 6$$
$$=\pi\frac{R^2}2+\sqrt 3 R^2$$
