Any smart ideas on finding the area of this shaded region? 
Don't let the simplicity of this diagram fool you. I have been wondering about this for quite some time, but I can't think of an easy/smart way of finding it.
Any ideas?

For reference, the Area is:
$$\bbox[10pt, border:2pt solid grey]{90−18.75\pi−25\cdot \arctan\left(\frac 12\right)}$$


Considering the right triangle with legs $1-s\sin(\theta)$ and $1-s\cos(\theta)$, we have
$$
1=(1-s\cos(\theta))^2+(1-s\sin(\theta))^2\tag{1}
$$
Solving for $s$, we get
$$
s=\sin(\theta)+\cos(\theta)-\sqrt{2\sin(\theta)\cos(\theta)}\tag{2}
$$
Setting $t=\tan(\theta)$ yields
$$
\begin{align}
s\sin(\theta)&=\frac{t}{1+t^2}\left(t-\sqrt{2t}+1\right)\\
s\cos(\theta)&=\frac{1}{1+t^2}\left(t-\sqrt{2t}+1\right)\\
\tan(\phi/2)&=\frac{1-s\cos(\theta)}{2-s\sin(\theta)}
\end{align}\tag{3}
$$
Setting $\tan(\theta)=\frac12$ gives
$$
\begin{align}
s\sin(\theta)&=\frac15\\
s\cos(\theta)&=\frac25\\
\tan(\phi/2)&=\frac13
\end{align}\tag{4}
$$
The area of the green piece is
$$
\frac\phi2-\frac12(1-s\cos(\theta))=\tan^{-1}\left(\frac13\right)-\frac3{10}\tag{5}
$$
The sum of the areas of the purple and green pieces is
$$
\frac12s\sin(\theta)=\frac1{10}\tag{6}
$$
Therefore, the area of the purple piece is
$$
\frac25-\tan^{-1}\left(\frac13\right)\tag{7}
$$

$(7)$ used a circle with radius $1$. For a circle with radius $5$, we get an area of
$$
10-25\tan^{-1}\left(\frac13\right)\tag{8}
$$
The area of half the $10\times20$ rectangle minus the two circles of radius $5$ is
$$
100-25\pi\tag{9}
$$
Therefore, the area in the given image is the difference of $(9)$ and $(8)$:
$$
\bbox[5px,border:2px solid #C0A000]{90+25\tan^{-1}\left(\frac13\right)-25\pi}\tag{10}
$$

$(10)$ is a different form of the same area:
$$
\begin{align}
90+25\tan^{-1}\left(\frac13\right)-25\pi
&=90+25\tan^{-1}\left(\frac13\right)-25\tan^{-1}\left(1\right)-\frac{75}4\pi\\
&=90+25\tan^{-1}\left(\frac{\frac13-1}{1+\frac13\cdot1}\right)-\frac{75}4\pi\\
&=\bbox[5px,border:2px solid #C0A000]{90-25\tan^{-1}\left(\frac12\right)-\frac{75}4\pi}\tag{11}
\end{align}
$$
