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Difference between revisions of "2002 AMC 12A Problems/Problem 7"

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

Revision as of 20:45, 12 May 2018

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 $\frac{2^2}{3^2}=\boxed{\text{(A)}\ 4/9}$

Solution 3

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 (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 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

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