jmc

Each of sector $ABD$ and $BDC$ is one-sixth of a full circle of radius $12,$ so each has area one-sixth of the area of a circle of radius $12.$ Therefore, each sector has area $$\frac{1}{6}(\pi (12^2))=\frac{1}{6}(144\pi )=24\pi .$$ Thus, the area of figure $ABCD$ is $2(24\pi )=\boxed{48\pi }.$