In a circle, a 90 degree sector has area of 36pi ftsqd. what is the radius of the circle?

[tex]\bf \textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}\quad \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ \theta =90\\ A=36\pi \end{cases}\implies 36\pi =\cfrac{90\pi r^2}{360} \\\\\\ \cfrac{36\pi }{90\pi }=\cfrac{r^2}{360}\implies \cfrac{2}{5}=\cfrac{r^2}{360}\implies \cfrac{360\cdot 2}{5}=r^2\implies 144=r^2 \\\\\\ \sqrt{144}=r\implies 12=r[/tex]