Difference between revisions of "2002 AMC 12A Problems/Problem 7"
(New page: {{duplicate|2002 AMC 12A #7 and 2002 AMC 10A #7}} ==Problem== A <math>45^\circ</math> arc of circle A is equal in length to a <math>3...) |
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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? | 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>\ | + | <math>\textbf{(A)}\ 4/9 \qquad \textbf{(B)}\ 2/3 \qquad \textbf{(C)}\ 5/6 \qquad \textbf{(D)}\ 3/2 \qquad \textbf{(E)}\ 9/4</math> |
− | ==Solution== | + | ==Solutions== |
− | Let <math>r_1</math> and <math>r_2</math> be the radii of circles A and B, respectively. | + | ===Solution 1=== |
+ | Let <math>r_1</math> and <math>r_2</math> be the radii of circles <math>A</math> and<math> B</math>, 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 | + | It is well known that in a circle with radius <math> r </math>, 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 | + | 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{\textbf{(A) } 4/9}</math>. |
+ | |||
+ | ===Solution 2=== | ||
+ | The arc of circle <math>A</math> is <math>\frac{45}{30}=\frac{3}{2}</math> that of circle <math>B</math>. | ||
+ | |||
+ | The circumference of circle <math>A</math> is <math>\frac{2}{3}</math> that of circle <math>B</math> (<math>B</math> is the larger circle). | ||
+ | |||
+ | The radius of circle <math>A</math> is <math>\frac{2}{3}</math> that of circle <math>B</math>. | ||
+ | |||
+ | The area of circle <math>A</math> is <math>{\left(\frac{2}{3}\right)}^2=\boxed{\textbf{(A) } 4/9}</math> that of circle <math>B</math>. | ||
==See Also== | ==See Also== | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:48, 8 November 2021
- 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?
Solutions
Solution 1
Let and be the radii of circles and, respectively.
It is well known that in a circle with radius , a subtended arc opposite an angle of degrees has length .
Using that here, the arc of circle A has length . The arc of circle B has length . 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 .
Solution 2
The arc of circle is that of circle .
The circumference of circle is that of circle ( is the larger circle).
The radius of circle is that of circle .
The area of circle is that of circle .
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.