Difference between revisions of "2002 AMC 12A Problems/Problem 7"
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Revision as of 09:12, 4 July 2013
- 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 
arc of circle A is equal in length to a 
arc of circle B. What is the ratio of circle A's area and circle B's area?
Solution
Let 
and 
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 
degrees has length $\frac{\theta}{360}\cdot{2\pi{r}$ (Error compiling LaTeX. Unknown error_msg).
Using that here, the arc of circle A has length 
. The arc of circle B has length $\frac{30}{360}\cdot{2\pi{r_2}=\frac{r_2\pi}{6}$ (Error compiling LaTeX. Unknown error_msg). We know that they are equal, so 
, so we multiply through and simplify to get 
. 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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
