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| − | ==Problem==
| + | #redirect [[2002 AMC 12A Problems/Problem 7]] |
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| − | 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?
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| − | <math>\text{(A)}\ 4/9 \qquad \text{(B)}\ 2/3 \qquad \text{(C)}\ 5/6 \qquad \text{(D)}\ 3/2 \qquad \text{(E)}\ 9/4</math>
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| − | ==Solution==
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| − | Let <math>r_1</math> and <math>r_2</math> be the radii of circles A and B, respectively.
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| − | 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>.
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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>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=A|num-b=6|num-a=8}}
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| − | [[Category:Introductory Geometry Problems]] | |
