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Difference between revisions of "2012 AMC 10B Problems/Problem 16"
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Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them?
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==Solution==
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<math> \text{(A)}\ 10\pi+4\sqrt{3}\qquad\text{(B)}\ 13\pi-\sqrt{3}\qquad\text{(C)}\ 12\pi+\sqrt{3}\qquad\text{(D)}\ 10\pi+9\qquad\text{(E)}\ 13\pi </math>
 
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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 <math>4</math>. Find the area of this triangle to include the figure formed in between the circles. This area is <math>4\sqrt{3}</math>.
 
To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle of length <math>4</math>. Find the area of this triangle to include the figure formed in between the circles. This area is <math>4\sqrt{3}</math>.
  
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This means <math>(A)</math> is the right answer.
 
This means <math>(A)</math> is the right answer.
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==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2012|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2012|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:24, 8 February 2014

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]

$\textbf{(A)}\ 10\pi+4\sqrt{3}\qquad\textbf{(B)}\ 13\pi-\sqrt{3}\qquad\textbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi$


Solution

To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle of length $4$. Find the area of this triangle to include the figure formed in between the circles. This area is $4\sqrt{3}$.


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 $2^2 \pi * 5/6 = 10\pi/3$. Then this area is multiplied by three to find the total area of the sectors $(10 \pi)$. This result is added to area of the equilateral triangle to get a final answer of $10\pi + 4\sqrt3$.

This means $(A)$ 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

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