2012 AMC 10B Problems/Problem 16

Revision as of 20:24, 8 February 2014 by Flamedragon (talk | contribs)

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]

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


Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them?

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



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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS