2002 AMC 10A Problem 7

Let the radii be $r_A$ and $r_B$. An arc of $\theta$ degrees has length $\frac{\theta}{360}\cdot 2\pi r$, so the two arcs have lengths $\frac{45}{360}\cdot 2\pi r_A$ and $\frac{30}{360}\cdot 2\pi r_B$. Setting them equal and cancelling $\frac{2\pi}{360}$:
$$45\, r_A = 30\, r_B \quad\Longrightarrow\quad \frac{r_A}{r_B} = \frac{30}{45} = \frac{2}{3}.$$

Circle $A$ needs a bigger angle to produce the same length, so it is the smaller circle, which matches $r_A < r_B$. Areas scale with the square of the radius:
$$\frac{\text{area}_A}{\text{area}_B} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}.$$

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

Arc length is proportional to both the central angle and the radius, so equal arcs force the radii to be inversely proportional to the angles. All circles are similar, so any ratio of lengths (radius, circumference) becomes its square for areas. A quick sanity check on which circle is bigger catches inverted ratios.

Pick numbers: let the common arc length be $\pi$. Eight $45^\circ$ arcs make circle $A$'s circumference $8\pi$, so $r_A = 4$; twelve $30^\circ$ arcs make circle $B$'s circumference $12\pi$, so $r_B = 6$. Areas $16\pi$ and $36\pi$ give $\frac{16}{36} = \frac{4}{9}$.

• Reporting the radius ratio 2/3 (choice B) instead of squaring it, or inverting the ratio to get 3/2 or 9/4.
• Writing $\frac{r_A}{r_B} = \frac{45}{30}$ because "$A$ has the bigger angle," when the bigger angle actually means the smaller circle.