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?

$\text{(A)}\ 4/9 \qquad \text{(B)}\ 2/3 \qquad \text{(C)}\ 5/6 \qquad \text{(D)}\ 3/2 \qquad \text{(E)}\ 9/4$

Solutions

Solution 1

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 $\boxed{\text{(A)}\ 4/9}$.

Solution 2

The arc of circle $A$ is $\frac{45}{30}=\frac{3}{2}$ that of circle $B$.

The circumference of circle $A$ is $\frac{2}{3}$ that of circle $B$ ($B$ is the larger circle).

The radius of circle $A$ is $\frac{2}{3}$ that of circle $B$.

The area of circle $A$ is ${\left(\frac{2}{3}\right)}^2=\boxed{\text{(A)}\ 4/9}$ that of circle $B$.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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