Difference between revisions of "2020 AMC 12B Problems/Problem 11"

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==Problem==
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#REDIRECT [[2020 AMC 10B Problems/Problem 14]]
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?
 
 
 
[asy]
 
size(140);
 
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
 
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
 
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
 
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
 
fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white);
 
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
 
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
 
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
 
draw(arc((2,0),1,180,0)--(2,0)--cycle);
 
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
 
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
 
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);
 
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
 
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
 
label("<math>2</math>",(3.5,3sqrt(3)/2),NE);
 
[/asy]
 
 
 
<math>\textbf{(A)}\ 6\sqrt3-3\pi \qquad\textbf{(B)}\ \frac{9\sqrt3}{2}-2\pi \qquad\textbf{(C)}\ \frac{3\sqrt3}{2}-\frac{\pi}{3} \qquad\textbf{(D)}\ 3\sqrt3-\pi \\ \qquad\textbf{(E)}\ \frac{9\sqrt3}{2}-\pi</math>
 
 
 
==Solution==
 
 
 
==See Also==
 
 
 
{{AMC12 box|year=2020|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 

Latest revision as of 16:55, 9 February 2020