**Answer**

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}

**Attribution***Source : Link , Question Author : The Artist , Answer Author : Blue*