Difference between revisions of "2012 AMC 10B Problems/Problem 16"
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To find the area of the remaining sectors, notice that the sectors have a central angle of 300 because 60 degrees were "used up" for the triangle. The area of one sector is <math>2^2 \pi * 5/6 = 10\pi/3</math>. Then this area is multiplied by three to find the total area of the sectors <math>(10 \pi)</math>. | To find the area of the remaining sectors, notice that the sectors have a central angle of 300 because 60 degrees were "used up" for the triangle. The area of one sector is <math>2^2 \pi * 5/6 = 10\pi/3</math>. Then this area is multiplied by three to find the total area of the sectors <math>(10 \pi)</math>. | ||
| − | This result is added to area of the equilateral triangle to get a final answer of <math>10\pi + 4\sqrt3</math>. | + | This result is added to area of the equilateral triangle to get a final answer of <math>\boxed{10\pi + 4\sqrt3}</math>. |
This means <math>(A)</math> is the right answer. | This means <math>(A)</math> is the right answer. | ||
Revision as of 00:24, 18 January 2016
Problem
Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?
![[asy] filldraw((0,0)--(2,0)--(1,sqrt(3))--cycle,gray,gray); filldraw(circle((1,sqrt(3)),1),gray); filldraw(circle((0,0),1),gray); filldraw(circle((2,0),1),grey);[/asy]](http://latex.artofproblemsolving.com/f/d/3/fd36c0563f1447ec842cc791ac398c4cd8f20685.png)
Solution
To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle of length 
. Find the area of this triangle to include the figure formed in between the circles. This area is 
.
To find the area of the remaining sectors, notice that the sectors have a central angle of 300 because 60 degrees were "used up" for the triangle. The area of one sector is 
. Then this area is multiplied by three to find the total area of the sectors 
.
This result is added to area of the equilateral triangle to get a final answer of 
.
This means 
is the right answer.
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
