Correct. Given $\displaystyle{4}\pi{{r}}^{{2}}={A}\usepackage{cmbright}\Huge\$ , you know $\displaystyle{{r}}^{{2}}=\frac{{A}}{{{4}\pi}}\usepackage{cmbright}\Huge\$ , so $\displaystyle\sqrt{{{{r}}^{{2}}}}=\sqrt{{\frac{{A}}{{{4}\pi}}}}\usepackage{cmbright}\Huge\$ . Therefore $\displaystyle{r}=\frac{{1}}{{2}}\sqrt{{\frac{{A}}{\pi}}}\usepackage{cmbright}\Huge\$ .