2002 AMC 12A Problems/Problem 7

The following problem is from both the 2002 AMC 12A #7 and 2002 AMC 10A #7, so both problems redirect to this page.

Problem

A $45^\circ$ arc of circle A is equal in length to a $30^\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area?

Let $r_1$ and $r_2$ be the radii of circles A and B, respectively.

It is well known that in a circle with radius r, a subtended arc opposite an angle of $\theta$ degrees has length $\frac{\theta}{360}\cdot{2\pi{r}$ .

Using that here, the arc of circle A has length $\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}$ . The arc of circle B has length $\frac{30}{360}\cdot{2\pi{r_2}=\frac{r_2\pi}{6}$ . We know that they are equal, so $\frac{r_1\pi}{4}=\frac{r_2\pi}{6}$ , so we multiply through and simplify to get $\frac{r_1}{r_2}=\frac{2}{3}$ . As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is $\boxed{\text{(A)}\ 4/9}$ .