Hello, Chris11!

Another approach . . . Code:

D * - - - - - * - * C
| *:::::::* |
| *:::::::* |
| *::::::* |
| ::::::* | 6
|*::::* |
|:::* |
*:* |
A * - - - - - - - * B
6

The shaded region is:

. . $\displaystyle \begin{array}{cccccc}

\text{a quarter-circle:} & \frac{1}{4}\pi(6^2) &=& 9\pi \\

\text{minus} \\

\text{half the square:} & \frac{1}{2}(6^2) &=& 18 \end{array}$

Hence: .$\displaystyle 9\pi - 18 \;=\;9(\pi-2)$

Therefore, the lens-shaped region has area:

. . . . $\displaystyle 2\cdot9(\pi - 2) \:=\:18(\pi - 2) $