Difference between revisions of "2012 AMC 10A Problems/Problem 18"
(→Solution) |
(→Solution) |
||
Line 36: | Line 36: | ||
== Solution == | == Solution == | ||
− | Draw the hexagon between the centers of the circles, and compute its area <math>(6)(0.5)(2\sqrt{3})=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 (<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>. |
== See Also == | == See Also == |
Revision as of 21:11, 13 February 2012
- The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.
Problem 18
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?
Solution
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 (
) to get the area enclosed in the curved figure
, which 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 |