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

Latest revision as of 16:55, 9 February 2020

Redirect to:

2020 AMC 10B Problems/Problem 14

@@ Line 1: / Line 1: @@
-<!-- #REDIRECT [[2020 AMC 10B Problems/Problem 14]] -->
+#REDIRECT [[2020 AMC 10B Problems/Problem 14]]
-==Problem==
-As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length <math>2</math> 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("$2$",(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}}

Page

Toolbox

Search

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

Latest revision as of 16:55, 9 February 2020