jmc

Let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=24$. Let $K$ denote the area of $△ABC$.

Heron's formula tells us that $$\begin{array}{rl}K& =\sqrt{s(s-AB)(s-AC)(s-BC)}\\ & =\sqrt{24\cdot 14\cdot 7\cdot 3}\\ & =\sqrt{2^4\cdot 3^2\cdot 7^2}\\ & =84.\end{array}$$The area of a triangle is equal to its semiperimeter multiplied by the radius of its inscribed circle ($K=rs$), so we have $$84=r\cdot 24,$$which yields the radius $r=\boxed{3.5}$.