2012 AMC 10B Problems/Problem 16


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]

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


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 $\pi r^2 = 4\pi$, 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: $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$.

Alternate Solution

First, the area of the 3 circles is simply $3*\pi*2^2 = 12 \pi$. Notice that the middle area is a little more than a rectangle formed by completely filling the rectangle formed by connecting two 90 degrees partial circles and then subtracting the two 90 degrees partial circles. The area of the rectangle is $4*2=8$ and the area of the 90 degrees partial circles are $2*(1/4)*\pi*2^2 = 2\pi$. Therefore, the area of the shape in between the three circles is a little less than $8 - 2\pi$. Summing up the 3 circles we got and the approximate area of the middle shape, we get $10\pi+8$, which is a little more than what we want. We see that all answer choices except $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$ is greater than $10\pi+8$, therefore it's the only answer. -dchang0524

Solution 3 (answer choices, not recommended)

Notice as in solution 1 that the figure is simply an equilateral triangle and $3$ sectors of $\frac{5}{6}$ of a circle which makes a total of $\frac{5}{2}$ circles. Calculating this area with $\pi r^2$, we get $10\pi$. We also know that the equilateral triangle will give us some constant multiplied by $\sqrt3$. The only answer choice with $10\pi$ and $\sqrt3$ is $\boxed{\textbf{(A)}}$. ~chrisdiamond10

Solution 4

Notice that we can connect the centers of the 3 circles to make an equilateral triangle with length $4$. The equilateral triangle with length $4$ has area $4$$\sqrt{3}$. Because the triangle took $1/6$ th of all the 3 circles away, we have the area of the circles as $(5/6)*(4\pi)*3 = 10\pi$. We add the area of the triangle to get $10 \pi$ + $4$$\sqrt{3}$ = $\boxed {A}$

Video Solution



Video Solution by OmegaLearn


~ pi_is_3.14

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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