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 (10 pi). | + | 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>10\pi + 4\sqrt3</math>. | ||
This means (A) is the right answer. | This means (A) is the right answer. | ||
Revision as of 21:15, 16 February 2013
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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 (A) is the right answer.
