well, if we take the whole area of the circle, including the trapezoid in it, and then we get the area of the trapezoid and subtract it from that of the circle's, what's leftover is what we did not subtract, namely the shaded area. Let's notice the circle has a diameter of 20, thus a radius of half that, or 10.
[tex]\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ a=12\\ b=20\\ h=8 \end{cases}\implies A=\cfrac{8(12+20)}{2}\implies A=128 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\stackrel{circle}{\pi (10)^2}~~ - ~~\stackrel{trapezoid}{128}}\implies 100\pi ~~ - ~~128~~\approx~~186.16[/tex]