[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360} ~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=27.93\\ \theta =50 \end{cases}\implies 27.93=\cfrac{(50)\pi r^2}{360}\implies 27.93=\cfrac{5\pi r^2}{36} \\\\\\ 27.93(36)=5\pi r^2\implies \cfrac{27.93(36)}{5\pi }=r^2\implies \sqrt{\cfrac{27.93(36)}{5\pi }}=r[/tex]
since now we know what the radius of the circle is, le's get the area of the whole circle and subtract the orange sector from it, what's leftover is the part we didn't subtract, namely the blue shaded area.
[tex]\textit{area of a circle}\\\\ A=\pi r^2\hspace{5em}A=\pi \left( \sqrt{\cfrac{27.93(36)}{5\pi }} \right)^2\implies A=\cfrac{27.93(36)}{5} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\LARGE Areas}}{\stackrel{whole~circle}{\cfrac{27.93(36)}{5}}~~ - ~~\stackrel{orange~sector}{27.93}}\implies 201.096~~ - ~~27.93 ~~ \stackrel{\textit{blue shaded area}}{\approx ~~ 173.17}[/tex]
