2012 AMC 10B Problems/Problem 16

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 with a side of length $4$ . We must find the area of this triangle to include the figure formed in between the circles. Since the equilateral triangle has two 30-60-90 triangles inside, we can find the height and the base of each 30-60-90 triangle from the ratios: $1: \sqrt{3}: 2.$ The height is $2\sqrt{3}$ and the base is $2$ . Multiplying the height and base together with $\dfrac{1}{2}$ , we get $2\sqrt{3}$ . Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by $2$ : $2\cdot 2\sqrt{3} = 4\sqrt{3}.$

To find the area of the remaining sectors, which are $\dfrac{5}{6}$ of the original circles once we remove the triangle, we know that the sectors have a central angle of $300^\circ$ since the equilateral triangle already covered that area. Since there are $3$ $\dfrac{1}{6}$ pieces gone from the equilateral triangle, we have, in total, $\dfrac{1}{2}$ of a circle (with radius $2$ ) gone. Each circle has an area of $\pi r^2 = 4pi$ , so three circles gives a total area of $12\pi$ . Subtracting the half circle, we have: $12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.$

Summing the areas from the equilateral triangle and the remaining circle sections gives us: $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$ .

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2012 AMC 10B Problems/Problem 16

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