Any smart ideas on finding the area of this shaded region?

We observe that $\triangle PRT$ can be partitioned into five congruent sub-triangles. Therefore, the entire shaded region has area given by ... $$\begin{align} 3 u + |\text{region}\; PAT| &= 3u + |\square OAPT| - |\text{sector}\;OAT| \\[6pt] &= 3u + \frac{3}{5}\,|\triangle PRT| - |\text{sector}\;OAT| \\[6pt] &= 3\cdot\frac{1}{4} r^2 \left( 4 - \pi \right) \;+\; \frac{3}{5}\cdot r^2 \;-\; \frac{1}{2}r^2\cdot 2\theta \end{align}$$ Since $\theta = \operatorname{atan}\frac{1}{2}$, this becomes

$$r^2\left(\; \frac{18}{5} - \frac{3}{4}\pi - \operatorname{atan}\frac{1}{2} \;\right) \qquad\stackrel{r=5}{\to}\qquad 90 - \frac{75}{4}\pi - 25\;\operatorname{atan}\frac{1}{2}$$

[Note: My second answer is much better.]

I'll focus on the unshaded region at the bottom-left.

By an aspect of the Inscribed Angle Theorem, we know that $\angle AOB = 2\;\angle ABP$ (justifying marking these $\theta$ and $2\theta$). By a related result, we have that $$\phi = \frac{1}{2}\left(\angle BOC - \angle AOB\right) = 45^\circ - \theta$$ Moreover, we know that $$\phi = \operatorname{atan}\frac{1}{2} \approx 26.56^\circ \qquad\to\qquad \theta = 45^\circ - \operatorname{atan}\frac{1}{2} \approx 18.43^\circ$$

From here, knowing the circle's radius, one may calculate the lower-left area as ... $$\begin{align} &|\triangle PAB| + |\triangle OAB| - |\text{sector } OAB| \\ \end{align}$$ ... from which we readily derive the area in the original question. For now, I'll leave these details to the reader.

Put $\arctan{1\over2}=:\alpha$. Then $$\sin(2\alpha)={2\tan\alpha\over1+\tan^2\alpha}={4\over5}\ .$$ The area $A$ in question consists of three "arrow heads" plus the area shaded in the following figure. The latter is a right triangle minus a sector and a smaller triangle. We therefore obtain $$A={3\over4}(10^2- 25\pi)+{25\over2}\bigl(2-2\alpha-\sin(2\alpha)\bigr)=90-{75\over4}\pi-25\alpha\ .$$