Check the picture below.
[tex]\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}[/tex]
notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.