2002 AMC 10A Problem 7

The arc length is not given, so pick a convenient one and see what it tells us about each circle.

1. Pick a number. Say each arc has length $12$.

1. Circle A. A $45^\circ$ arc is $\frac{45}{360} = \frac{1}{8}$ of the whole circle. So the whole circumference of A is $8 \times 12 = 96$.

1. Circle B. A $30^\circ$ arc is $\frac{30}{360} = \frac{1}{12}$ of the circle. So the circumference of B is $12 \times 12 = 144$.

4. Compare radii. The radius is the circumference divided by $2\pi$, so the radii compare the same way the circumferences do:
$$\frac{r_A}{r_B} = \frac{96}{144} = \frac{2}{3}.$$
This makes sense: A needs a bigger angle to make the same length, so A is the smaller circle.

5. Compare areas. Area is $\pi r^2$, so the ratio gets squared. For example, radii $2$ and $3$ give areas $4\pi$ and $9\pi$:
$$\frac{\text{area of A}}{\text{area of B}} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}.$$

The answer is $\boxed{\textbf{(A)}\ 4/9}$.

Arc length depends on both the angle and the radius, so equal arcs with different angles tell you how the radii compare. Every circle is a scaled copy of every other, and when lengths scale by some factor, areas scale by that factor squared. A quick "which circle is bigger?" check catches an upside-down ratio.

Use the arc-length formula. An arc of $\theta$ degrees on a circle of radius $r$ has length $\frac{\theta}{360} \cdot 2\pi r$. Setting the two arcs equal and cancelling the common $\frac{2\pi}{360}$ gives
$$45\, r_A = 30\, r_B, \quad\text{so}\quad \frac{r_A}{r_B} = \frac{30}{45} = \frac{2}{3},$$
and the area ratio is $\left(\frac23\right)^2 = \frac49$.