jmc

Let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=10$. Let $K$ denote the area of $△ABC$.

Heron's formula tells us that $$\begin{array}{rl}K& =\sqrt{s(s-AB)(s-AC)(s-BC)}\\ & =\sqrt{10\cdot 3\cdot 3\cdot 4}\\ & =6\sqrt{10}.\end{array}$$

The area of a triangle is equal to its semiperimeter multiplied by the radius of its inscribed circle ($K=rs$), so we have $$6\sqrt{10}=r\cdot 10,$$ which yields the radius $r=\boxed{\frac{3\sqrt{10}}{5}}$.