jmc

For $△ABC$ to be acute, all angles must be acute.

For $\angle A$ to be acute, point $C$ must lie above the line passing through $A$ and perpendicular to $\overline{AB}$. The segment of that line in the first quadrant lies between $P(4,0)$ and $Q(0,4)$.

For $\angle B$ to be acute, point $C$ must lie below the line through $B$ and perpendicular to $\overline{AB}$. The segment of that line in the first quadrant lies between $S(14,0)$ and $T(0,14)$.

For $\angle C$ to be acute, point $C$ must lie outside the circle $U$ that has $\overline{AB}$ as a diameter.

Let $O$ denote the origin. Region $R$, shaded below, has area equal to $$\begin{array}{rl}\text{Area}(△OST)-\text{Area}(△OPQ)-\text{Area(Circle }U)& =\frac{1}{2}\cdot 14^2-\frac{1}{2}\cdot 4^2-\pi {\left(\frac{\sqrt{50}}{2}\right)}^2\\ & =\boxed{90-\frac{25}{2}\pi }.\end{array}$$