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

Let $c_1$ and $c_2$ be the circumference of circles $A$ and $B$ , respectively.

The length of a $45^{\circ}$ arc of circle $A$ is $\frac{c_1}{\frac{360}{45}}=\frac{c_1}{8}$ , and the length of a $30^{\circ}$ arc of circle $B$ is $\frac{c_2}{\frac{360}{30}}=\frac{c_2}{12}$ . We know that the length of a $45^{\circ}$ arc on circle $A$ is equal to the length of a $30^{\circ}$ arc of circle $B$ , so $\frac{c_1}{8}=\frac{c_2}{12}$ . Manipulating the equation, we get $\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$ . Because the ratio of the areas is equal to the ratio of the circumferences squared, our answer is $\frac{2^2}{3^2}=\boxed{\text{(A)}\ 4/9}$

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

2002 AMC 10A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 12A Problems/Problem 7

Contents

Problem

Solutions

Solution 1

Solution 2

See Also