What is the measure of the diameter?

From the tangent points we draw the radii to the center of the circle $O$ which lies on segment AB, then we divide $\triangle ABC$ into two smaller triangles $\triangle AOC$ and $\triangle BOC$ . The sum of the small triangles areas is equal to the area of $\triangle ABC$ . Let the length of the radius be $r$ , then

$[AOC] = \dfrac{11}{2} r$ , and

$[BOC] = \dfrac{13}{2} r$

Further, from Heron's formula, we can compute the area of $\triangle ABC$ . The semi-perimeter $s = \frac{1}{2} (20+11+13) = 22$ . Hence

$[ABC] = \sqrt{ s(s-a)(s-b)(s-c) } = \sqrt{ (22)(2)(11)(9) } = (11)(2)(3) = 66$

Thus, $\dfrac{11}{2} r + \dfrac{13}{2} r = 66$ . Solving for $r$ , we get $r = \dfrac{66}{12} = \dfrac{11}{2}$ .

Therefore, the diameter $d = 2 r = 11$