Hello, Phil!

Find the area of the middle region.

Answer: .$\displaystyle \dfrac{4\pi + 6\sqrt{3}}{3}$ Code:

* * *
* *
A ♥ - - - + - - - ♥ B
* * | * *
* |1 *
* * | * *
* ♥O *
* * | * *
* |1 *
* * | * *
* - - - + - - - *
* *
* * *

Note all the $\displaystyle 60^o$ angles.

We find that the radius is 2.

The area of the circle is: .$\displaystyle \pi r^2 \:=\:\pi(2^2) \:=\:4\pi$

The area of sector $\displaystyle OAB$ is: .$\displaystyle \frac{1}{3}(4\pi)$

The area of $\displaystyle \Delta OAB\:=\:\frac{1}{2}(2)(2)\sin120^o \;=\;\sqrt{3}$

The area of the segment is: .$\displaystyle \dfrac{4\pi}{3} - \sqrt{3}$

The area of the middle region is: .$\displaystyle 4\pi - 2\left(\dfrac{4\pi}{3} - \sqrt{3}\right)$

. . $\displaystyle =\;\;4\pi - \dfrac{8\pi}{3} + 2\sqrt{3} \;\;=\;\;\dfrac{4\pi}{3} + 2\sqrt{3} \;\;=\;\;\dfrac{4\pi + 6\sqrt{3}}{3} $