Continuing this topic, look at the following diagram, with two quarter circles in a unit square, the question is: what are the area of of each of the 4 regions inside the square?
To answer the question, we need to draw line $\overline{BE}$ and $\overline{CE}$, as shown in the following diagram:
Obviously, $\triangle{BCE}$ is an equalateral triangle, with its area as $\sqrt{3}{4}$. Additionally, the size of the fan-shaped area $[CBE]$ and $[BCE]$ is $\dfrac{\pi}{6}$. The area of the wedge-shaped area $[CED]$ and $[BEA]$ is $\dfrac{\pi}{12}$.
Therefore, the area of the region bounded by line $\overline{CE}$ and arc $\overparen{EC}$ is $$[CEC]=\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}$$
We have the area of the region bounded by $\overline{BC}$, arc $\overparen{BE}$ and $\overparen{CE}$ is $$\dfrac{\pi}{6}+(\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4})=\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}$$
And the area of the region bounded by $\overline{CD}$, arc $\overparen{DE}$ and $\overparen{EC}$ (and congruent region on the left side) is $$\dfrac{\pi}{12}-(\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4})=\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}$$
Finally, the area of the region bounded by line $\overline{AD}$, arc $\overparen{AE}$ and $\overparen{DE}$ is $$1-(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4})-2\cdot(\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12})=1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}$$