A Circle Problem .

$\angle OBD=90^\circ$ and $|OC|=|CD|$ Hence $|CB|=r$ . That makes $△OCB$ equilateral triangle with all of its angles are $60^\circ$

$\angle ABC=90^\circ$ and $\angle BCA=60^\circ$ Hence $|AB|=r\sqrt{3}$

Let $K$ be the midpoint of $[AB]$

$|KB|=|BE|=\frac{r\sqrt{3}}{2}$ , $\angle KBE=60^\circ$ makes $△KBE$ equilateral with $|KE|=\frac{r\sqrt{3}}{2}$

$|KE|=|AK|=|KB|$ Hence $\angle AEB=90^\circ$ . So $|AE|=\frac{3r}{2}$

$A(ABE)=\frac{|BE| \cdot |AE|}{2}$

$=\frac{\frac{r\sqrt{3}}{2} \cdot \frac{3r}{2}}{2}$

$=\frac{3\sqrt{3}}{8} \cdot r^{2}$

So $A=\frac{3\sqrt{3}}{8} \approx \boxed{0.65}$