This seems to be a problem of arc length. Let me show you ...
In this case, the minute hand goes from 0 to 15min, so theta would be 90°
We are told that the length of the minute hand is 3.6 cm = r
Thus, S which is the distance that the tip of the minute hand travels, is...
[tex]\begin{gathered} S=2\pi r\cdot\frac{\theta}{360} \\ S=2\cdot\pi\cdot(3.6)\cdot\frac{90}{360} \\ S=\frac{9}{5}\pi \end{gathered}[/tex]
S is approximately 5.65 cm
