Difference between revisions of "2012 AMC 10A Problems/Problem 18"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
| − | == Solution == | + | == Solution 1 == |
Draw the hexagon between the centers of the circles, and compute its area <math>(6)(0.5)(2\sqrt{3})=6\sqrt{3}</math>. Then add the areas of the three sectors outside the hexagon (<math>2\pi</math>) and subtract the areas of the three sectors inside the hexagon but outside the figure(<math>\pi</math>) to get the area enclosed in the curved figure <math>(\pi+6\sqrt{3})</math>, which is <math>\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}</math>. | Draw the hexagon between the centers of the circles, and compute its area <math>(6)(0.5)(2\sqrt{3})=6\sqrt{3}</math>. Then add the areas of the three sectors outside the hexagon (<math>2\pi</math>) and subtract the areas of the three sectors inside the hexagon but outside the figure(<math>\pi</math>) to get the area enclosed in the curved figure <math>(\pi+6\sqrt{3})</math>, which is <math>\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}</math>. | ||
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| + | == Solution 2 (Looking at the answer choices) == | ||
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| + | We see that after forming the hexagon using the sectors outside the hexagon, there will be three sectors left. Each sector has an area of <math>\frac{\pi}{3},</math> so the three combined make <math>\pi.</math> Since the side length of the hexagon is <math>2,</math> it's area doesn't have <math>\pi</math> in it, so we know that the final answer will be <math>\pi + \text{(area of hexagon)}.</math> Looking at the answer choices, the only answer with only one <math>\pi</math> is <math>\boxed{\textbf{(E)}}.</math> | ||
== See Also == | == See Also == | ||
Revision as of 11:01, 31 January 2021
- The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.
Problem 14
The closed curve in the figure is made up of 9 congruent circular arcs each of length 
, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
![[asy] defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]](http://latex.artofproblemsolving.com/c/2/5/c25ea7245d49ffc173e58d29fdca0f31dda4f437.png)
Solution 1
Draw the hexagon between the centers of the circles, and compute its area 
. Then add the areas of the three sectors outside the hexagon (
) and subtract the areas of the three sectors inside the hexagon but outside the figure(
) to get the area enclosed in the curved figure 
, which is 
.
Solution 2 (Looking at the answer choices)
We see that after forming the hexagon using the sectors outside the hexagon, there will be three sectors left. Each sector has an area of 
so the three combined make 
Since the side length of the hexagon is 
it's area doesn't have in it, so we know that the final answer will be 
Looking at the answer choices, the only answer with only one is 
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
| 2012 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
