If an arc of $60^{\circ}$ in circle I has the same length as an arc of $45^{\circ}$ in circle II, find the ratio between the area of circle I and the area of circle II.
(a) $\frac{16}{9}$
(b) $\frac{9}{16}$
(c) $\frac{4}{3}$
(d) $\frac{3}{4}$
(e) $\frac{6}{9}$
The correct option is (b).

Since the $60^{\circ}$ arc of circle I has the same length as the $45^{\circ}$ arc in circle II, we conclude that the radius of circle I is smaller than that of circle II. Let's denote the radii of circles I and II by $r$ and $R$, respectively.

In circle I, the length of the $60^{\circ}$ arc is equal to $1 / 6$ of its circumference, i.e., $2 \pi r / 6$. Similarly, in circle II, the length of the $45^{\circ}$ arc is equal to $1 / 8$ of its circumference, i.e., $2 \pi R / 8$. Therefore, $2 \pi r / 6 = 2 \pi R / 8$, or $r / R = 6 / 8 = 3 / 4$. Finally, we have

$$
\frac{\text { area of circle I }}{\text { area of circle II }}=\frac{\pi r^{2}}{\pi R^{2}}=\left(\frac{r}{R}\right)^{2}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}
$$