jmc

Let $d$ be the diameter of the inscribed circle, and let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=12$. Let $K$ denote the area of $△ABC$.

Heron's formula tells us that $$\begin{array}{rl}K& =\sqrt{s(s-AB)(s-AC)(s-BC)}\\ & =\sqrt{12\cdot 1\cdot 6\cdot 5}\\ & =\sqrt{6^2\cdot 10}\\ & =6\sqrt{10}.\end{array}$$The area of a triangle is equal to its semiperimeter multiplied by the radius of its inscribed circle ($K=rs$), so we have $$6\sqrt{10}=r\cdot 12,$$which yields the radius $r=\frac{\sqrt{10}}{2}$. This yields the diameter $d=\boxed{\sqrt{10}}$.