Answer
$$r=5\sqrt{3}$$
Work Step by Step
We know that for an isosceles triangle, the radius of the inscribed circle is:
$$r=\frac{b}{2} \sqrt{\frac{2a-b}{2a+b}}$$
Where b is the length of the base and a is the length of the two congruent sides. Using the two 30-60-90 triangles within the isosceles triangle, we find that b is equal to:
$$6\sqrt3$$
Thus, we find:
$$r=\frac{6\sqrt3}{2} \sqrt{\frac{2(6)-6\sqrt3}{2(6)+6\sqrt3}}$$
$$r=5\sqrt{3}$$
