Difference between revisions of "2002 AMC 10A Problems/Problem 7"
(New page: ==Problem== A <math>45^\circ</math> arc of circle A is equal in length to a <math>30^\circ</math> arc of circle B. What is the ratio of circle A's area and circle B's area? <math>\text{(...) |
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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>. | ||
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2002|ab=A|num-b=6|num-a=8}} | ||
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+ | [[Category:Introductory Geometry Problems]] |
Revision as of 17:33, 26 December 2008
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 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 |