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Difference between revisions of "2012 AMC 10B Problems/Problem 16"

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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 <math>4</math>. 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: <math>1: \sqrt{3}: 2.</math> The height is <math>2\sqrt{3}</math> and the base is <math>2</math>. Multiplying the height and base together with <math>\dfrac{1}{2}</math>, we get <math>2\sqrt{3}</math>. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by <math>2</math>: <cmath>2\cdot 2\sqrt{3} = 4\sqrt{3}.</cmath>
 
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 <math>4</math>. 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: <math>1: \sqrt{3}: 2.</math> The height is <math>2\sqrt{3}</math> and the base is <math>2</math>. Multiplying the height and base together with <math>\dfrac{1}{2}</math>, we get <math>2\sqrt{3}</math>. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the 30-60-90 triangle by <math>2</math>: <cmath>2\cdot 2\sqrt{3} = 4\sqrt{3}.</cmath>
  
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To find the area of the remaining sectors, which are <math>\dfrac{5}{6}</math> of the original circles once we remove the triangle, we know that the sectors have a central angle of <math>300^\circ</math> since the equilateral triangle already covered that area. Since there are <math>3</math> <math>\dfrac{1}{6}</math> pieces gone from the equilateral triangle, we have, in total, <math>\dfrac{1}{2}</math> of a circle (with radius <math>2</math>) gone. Each circle has an area of <math>\pi r^2 = 4pi</math>, so three circles gives a total area of <math>12\pi</math>. Subtracting the half circle, we have: <cmath>12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.</cmath>  
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To find the area of the remaining sectors, which are <math>\dfrac{5}{6}</math> of the original circles once we remove the triangle, we know that the sectors have a central angle of <math>300^\circ</math> since the equilateral triangle already covered that area. Since there are <math>3</math> <math>\dfrac{1}{6}</math> pieces gone from the equilateral triangle, we have, in total, <math>\dfrac{1}{2}</math> of a circle (with radius <math>2</math>) gone. Each circle has an area of <math>\pi r^2 = 4\pi</math>, so three circles gives a total area of <math>12\pi</math>. Subtracting the half circle, we have: <cmath>12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.</cmath>  
  
 
Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>.
 
Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>.

Revision as of 10:31, 10 February 2017

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

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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