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2012 AMC 10A Problems/Problem 18: Difference between revisions

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The closed curve in the figure is made up of 9 congruent circular arcs each of length <math>\frac{2\pi}{3}</math>, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
The closed curve in the figure is made up of 9 congruent circular arcs each of length <math>\frac{2\pi}{3}</math>, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
<asy>
defaultpen(fontsize(6pt));
dotfactor=4;
label("$\circ$",(0,1));
label("$\circ$",(0.865,0.5));
label("$\circ$",(-0.865,0.5));
label("$\circ$",(0.865,-0.5));
label("$\circ$",(-0.865,-0.5));
label("$\circ$",(0,-1));
dot((0,1.5));
dot((-0.4325,0.75));
dot((0.4325,0.75));
dot((-0.4325,-0.75));
dot((0.4325,-0.75));
dot((-0.865,0));
dot((0.865,0));
dot((-1.2975,-0.75));
dot((1.2975,-0.75));
draw(Arc((0,1),0.5,210,-30));
draw(Arc((0.865,0.5),0.5,150,270));
draw(Arc((0.865,-0.5),0.5,90,-150));
draw(Arc((0.865,-0.5),0.5,90,-150));
draw(Arc((0,-1),0.5,30,150));
draw(Arc((-0.865,-0.5),0.5,330,90));
draw(Arc((-0.865,0.5),0.5,-90,30));
</asy>


<math>\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}</math>
<math>\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}</math>
Line 7: Line 34:
== Solution ==
== Solution ==


Draw the hexagon between the centers of the circles, and compute its area (6*.5*2*sqrt3 = 6sqrt3). Then add the areas of the three sectors outside the hexagon (2pi) and subtract the areas of the three sectors inside the hexagon (pi) to get the area enclosed in the curved figure (6sqrt3 + pi), which is E.
Draw the hexagon between the centers of the circles, and compute its area (<math>6(0.5)(2\sqrt{3})=6\sqrt{3}</math>). Then add the areas of the three sectors outside the hexagon (<math>2\pi</math>) and subtract the areas of the three sectors inside the hexagon (<math>\pi</math>) to get the area enclosed in the curved figure (<math>\pi+6\sqrt{3}</math>), which is <math>\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}</math>.
 
== See Also ==
 
{{AMC10 box|year=2012|ab=A|num-b=17|num-a=19}}

Revision as of 13:02, 12 February 2012

Problem 18

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?

[asy] defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]

$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$

Solution

Draw the hexagon between the centers of the circles, and compute its area ($6(0.5)(2\sqrt{3})=6\sqrt{3}$). Then add the areas of the three sectors outside the hexagon ($2\pi$) and subtract the areas of the three sectors inside the hexagon ($\pi$) to get the area enclosed in the curved figure ($\pi+6\sqrt{3}$), which is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}$.

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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