imc

The shaded area is equivalent to the area of sector $DOE$ minus the area of triangle $DOE$ plus the area of triangle $DBE$. Using the Pythagorean Theorem, $(DA{)}^2=(CE{)}^2=2^2-1^2=3$ so $DA=CE=\sqrt{3}$. Clearly, $DOA$ and $EOC$ are $30-60-90$ triangles with $\angle EOC=\angle DOA=60^{\circ }$. Since $OABC$ is a square, $\angle COA=90^{\circ }$. $\angle DOE$ can be found by doing some subtraction of angles. $\angle COA-\angle DOA=\angle DOC=\angle EOA$ $90^{\circ }-60^{\circ }=\angle EOA=30^{\circ }$ $\angle DOA-\angle EOA=\angle DOE$ $60^{\circ }-30^{\circ }=\angle DOE=30^{\circ }$ So, the area of sector $DOE$ is $\frac{30}{360}\cdot \pi \cdot 2^2=\frac{\pi }{3}$. The area of triangle $DOE$ is $\frac{1}{2}\cdot 2\cdot 2\cdot \sin 30^{\circ }=1$. Since $AB=CB=1$ , $DB=EB=(\sqrt{3}-1)$. So, the area of triangle $DBE$ is $\frac{1}{2}\cdot (\sqrt{3}-1{)}^2=2-\sqrt{3}$. Therefore, the shaded area is $(\frac{\pi }{3})-(1)+(2-\sqrt{3})=\boxed{\text{(A) }\frac{\pi }{3}+1-\sqrt{3}}$ OR $△ODA$ has the same height as $△OBD$ which is $1.$ We already know that $BD=\sqrt{3}-1.$ Therefore the area of $△OBD$ is $(\sqrt{3}-1)\cdot 1\cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2}.$ Since $△OBD=△OBE=\frac{\sqrt{3}-1}{2}.$ Therefore the sum of the areas is $2\cdot \frac{\sqrt{3}-1}{2}=\sqrt{3}-1.$ Then the area of the shaded area becomes $\frac{\pi }{3}-(\sqrt{3}-1)=\boxed{\frac{\pi }{3}+1-\sqrt{3}}.$