I'm having real troubles solving this one geometry problem. I've attempted to draw it as best as I could.
Shape $[ABCD]$ is partly inside the Unit Circle. The only other information states that tan$\left(-\theta -\pi \right)=-\frac{2}{5}$. With this I managed to find sin$\left(\theta \right)=\frac{2\sqrt{29}}{29}$ and cos$\left(\theta \right)=\frac{5\sqrt{29}}{29}$.
Having done that, I concluded that side $AD$ must be $2\cdot$ sin$\left(\theta \right)$ and side $BC$ is $2\cdot$ tan$\left(\theta \right)$. The height is $1-$cos$\left(\theta \right)$. Therefore, the area of $[ABCD]$ is given by: $$\frac{1}{2}\left(\frac{4\sqrt{29}}{29}+\frac{4}{5}\right)\left(1-\frac{5\sqrt{29}}{29}\right)$$ But I must be making a mistake somewhere, since I'm unable to reach the textbook solution by solving that expression, which is: $$\frac{33}{145}-\frac{\sqrt{29}}{29}$$
