Correct. The straight line path is the hypotenuse of a right triangle with legs of length 150 and 200. Letting c be the length of the straight line path, $\displaystyle{{c}}^{{2}}={{150}}^{{2}}+{{200}}^{{2}}={22500}+{40000}={62500}\usepackage{cmbright}\Huge\$ , so $\displaystyle{c}=\sqrt{{{62500}}}={250}{f{{t}}}\usepackage{cmbright}\Huge\$ .