Answer
$3\pi\approx9.42\ in$,
$5\pi\approx15.71\ in$
Work Step by Step
Step 1. In 15 minutes, the hand will move $\theta_1=\frac{15}{60}(2\pi)=\frac{\pi}{2}$. With $r=6\ in$, we have the distance $s_1=r\theta_1=6(\frac{\pi}{2})=3\pi\approx9.42\ in$
Step 1. In 25 minutes, the hand will move $\theta_2=\frac{25}{60}(2\pi)=\frac{5\pi}{6}$. With $r=6\ in$, we have the distance $s_2=r\theta_2=6(\frac{5\pi}{6})=5\pi\approx15.71\ in$
