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Difference between revisions of "2002 AMC 12A Problems/Problem 7"
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Let <math>r_1</math> and <math>r_2</math> be the radii of circles A and B, respectively.  
 
Let <math>r_1</math> and <math>r_2</math> 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 <math>\theta</math> degrees has length <math>\frac{\theta}{360}\cdot{2\pi{r}</math>.  
+
It is well known that in a circle with radius r, a subtended arc opposite an angle of <math>\theta</math> degrees has length <math>\frac{\theta}{360} \cdot 2\pi r</math>.  
  
−
Using that here, the arc of circle A has length <math>\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}</math>. The arc of circle B has length <math>\frac{30}{360}\cdot{2\pi{r_2}=\frac{r_2\pi}{6}</math>. We know that they are equal, so <math>\frac{r_1\pi}{4}=\frac{r_2\pi}{6}</math>, so we multiply through and simplify to get <math>\frac{r_1}{r_2}=\frac{2}{3}</math>. 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 <math>\boxed{\text{(A)}\ 4/9}</math>.
+
Using that here, the arc of circle A has length <math>\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}</math>. The arc of circle B has length <math>\frac{30}{360} \cdot 2\pi{r_2}=\frac{r_2\pi}{6}</math>. We know that they are equal, so <math>\frac{r_1\pi}{4}=\frac{r_2\pi}{6}</math>, so we multiply through and simplify to get <math>\frac{r_1}{r_2}=\frac{2}{3}</math>. 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 <math>\boxed{\text{(A)}\ 4/9}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:33, 14 May 2015

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?

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

Solution

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

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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