Difference between revisions of "2020 AMC 10B Problems/Problem 14"

(Solution: Explain why two equilateral triangles exist in each sixth of the hexagon)
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{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #14]] and [[2020 AMC 12B Problems|2020 AMC 12B #11]]}}
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==Problem==
 
==Problem==
  
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<asy>
 
<asy>
real x=sqrt(3);
+
size(140);
real y=2sqrt(3);
+
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
real z=3.5;
+
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
real a=x/2;
+
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
real b=0.5;
+
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
real c=3a;
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fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white);
pair A, B, C, D, E, F;
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fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
A = (1,0);
+
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
B = (3,0);
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draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
C = (4,x);
+
draw(arc((2,0),1,180,0)--(2,0)--cycle);
D = (3,y);
+
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
E = (1,y);
+
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
F = (0,x);
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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>
  
fill(A--B--C--D--E--F--A--cycle,grey);
 
fill(arc((2,0),1,0,180)--cycle,white);
 
fill(arc((2,y),1,180,360)--cycle,white);
 
fill(arc((z,a),1,60,240)--cycle,white);
 
fill(arc((b,a),1,300,480)--cycle,white);
 
fill(arc((b,c),1,240,420)--cycle,white);
 
fill(arc((z,c),1,120,300)--cycle,white);
 
draw(A--B--C--D--E--F--A);
 
draw(arc((z,c),1,120,300));
 
draw(arc((b,c),1,240,420));
 
draw(arc((b,a),1,300,480));
 
draw(arc((z,a),1,60,240));
 
draw(arc((2,y),1,180,360));
 
draw(arc((2,0),1,0,180));
 
label("2",(z,c),NE);
 
</asy>
 
 
<math> \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi </math>
 
<math> \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi </math>
  
==Solution==
+
==Solution 1==
 
<asy>
 
<asy>
 
real x=sqrt(3);
 
real x=sqrt(3);
Line 149: Line 139:
  
 
Since the area of a regular hexagon can be found with the formula <math>\frac{3\sqrt{3}s^2}{2}</math>, where <math>s</math> is the side length of the hexagon, the area of this hexagon is <math>\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}</math>. Since the area of an equilateral triangle can be found with the formula <math>\frac{\sqrt{3}}{4}s^2</math>, where <math>s</math> is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is <math>\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}</math>. Since the area of a circle can be found with the formula <math>\pi r^2</math>, the area of a sixth of a circle with radius 1 is <math>\frac{\pi(1^2)}{6} = \frac{\pi}{6}</math>. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of <math>\frac{\sqrt{3}}{4}</math>, and one sixth of a circle with radius 1 colored white, with an area of <math>\frac{\pi}{6}</math>. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is <math>2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}</math>, which equals <math>\frac{\sqrt{3}}{2} + \frac{\pi}{6}</math>, and the total area colored white is <math>6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})</math>, which equals <math>3\sqrt{3} + \pi</math>. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is <math>6\sqrt{3} - (3\sqrt{3} + \pi)</math>, which equals <math>\boxed{\textbf{(D) }3\sqrt{3} - \pi}</math>.
 
Since the area of a regular hexagon can be found with the formula <math>\frac{3\sqrt{3}s^2}{2}</math>, where <math>s</math> is the side length of the hexagon, the area of this hexagon is <math>\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}</math>. Since the area of an equilateral triangle can be found with the formula <math>\frac{\sqrt{3}}{4}s^2</math>, where <math>s</math> is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is <math>\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}</math>. Since the area of a circle can be found with the formula <math>\pi r^2</math>, the area of a sixth of a circle with radius 1 is <math>\frac{\pi(1^2)}{6} = \frac{\pi}{6}</math>. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of <math>\frac{\sqrt{3}}{4}</math>, and one sixth of a circle with radius 1 colored white, with an area of <math>\frac{\pi}{6}</math>. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is <math>2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}</math>, which equals <math>\frac{\sqrt{3}}{2} + \frac{\pi}{6}</math>, and the total area colored white is <math>6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})</math>, which equals <math>3\sqrt{3} + \pi</math>. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is <math>6\sqrt{3} - (3\sqrt{3} + \pi)</math>, which equals <math>\boxed{\textbf{(D) }3\sqrt{3} - \pi}</math>.
 +
 +
==Solution 2==
 +
First, subdivide the hexagon into 24 equilateral triangles with side length 1:
 +
<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);
 +
draw((1,0)--(3,2sqrt(3)));
 +
draw((3,0)--(1,2sqrt(3)));
 +
draw((4,sqrt(3))--(0,sqrt(3)));
 +
draw((2,0)--(3.5,3sqrt(3)/2));
 +
draw((3.5,sqrt(3)/2)--(2,2sqrt(3)));
 +
draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2));
 +
draw((2,2sqrt(3))--(0.5,sqrt(3)/2));
 +
draw((2,0)--(0.5,3sqrt(3)/2));
 +
draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2));
 +
</asy>
 +
Now note that the entire shaded region is just 6 times this part:
 +
<asy> size(100);
 +
fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4));
 +
fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white);
 +
draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle);
 +
label("$1$",(2.25,7sqrt(3)/4),NE);
 +
draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle);
 +
draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2));
 +
</asy>
 +
The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of:
 +
<cmath> 2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}</cmath>
 +
The arc that is not included has an area of:
 +
<cmath> \frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}</cmath>
 +
Hence, the area of the shaded region in that section is <cmath>\frac{\sqrt{3}}{2}-\frac{\pi}{6}</cmath>
 +
For a final area of:
 +
<cmath>6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}</cmath>
 +
~N828335
  
 
==Video Solution==
 
==Video Solution==
https://youtu.be/t6yjfKXpwDs
+
https://youtu.be/t6yjfKXpwDs?t=786 (for AMC 10)
 +
 
 +
https://youtu.be/0xgTR3UEqbQ (for AMC 12)
  
 
~IceMatrix
 
~IceMatrix
 +
 +
 +
https://youtu.be/oTqx8OqSMQI ~DSA_Catachu
  
 
==See Also==
 
==See Also==
  
 +
{{AMC12 box|year=2020|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2020|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2020|ab=B|num-b=13|num-a=15}}
 +
 +
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:41, 14 September 2020

The following problem is from both the 2020 AMC 10B #14 and 2020 AMC 12B #11, so both problems redirect to this page.

Problem

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("$2$",(3.5,3sqrt(3)/2),NE); [/asy]

$\textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi$

Solution 1

[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x);  fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label("2",(z,c),NE);  pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label("A",X,SW); label("B",Y,SE); label("C",Z,N); draw(X--Y--Z--A); label("1",(1.5,0),S); label("1",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label("$60^\circ$",anglemark(Y,X,Z),d); [/asy]

Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, $BC = 1$, since B is the center of the semicircle with radius 1 that C lies on, $AB = 1$, since B is the center of the semicircle with radius 1 that A lies on, and $\angle BAC = 60^\circ$, as a regular hexagon has angles of 120$^\circ$, and $\angle BAC$ is half of any angle in this hexagon. Now, using the sine law, $\frac{1}{\sin \angle ACB} = \frac{1}{\sin 60^\circ}$, so $\angle ACB = 60^\circ$. Since the angles in a triangle sum to 180$^\circ$, $\angle ABC$ is also 60$^\circ$. Therefore, $\triangle ABC$ is an equilateral triangle with side lengths of 1.

[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x);  fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label("2",(z,c),NE); label("1",(1.5,0),S); label("1",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label("$60^\circ$",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label("$60^\circ$",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label("$60^\circ$",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]

Since the area of a regular hexagon can be found with the formula $\frac{3\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\frac{\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}$. Since the area of a circle can be found with the formula $\pi r^2$, the area of a sixth of a circle with radius 1 is $\frac{\pi(1^2)}{6} = \frac{\pi}{6}$. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of $\frac{\sqrt{3}}{4}$, and one sixth of a circle with radius 1 colored white, with an area of $\frac{\pi}{6}$. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is $2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}$, which equals $\frac{\sqrt{3}}{2} + \frac{\pi}{6}$, and the total area colored white is $6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})$, which equals $3\sqrt{3} + \pi$. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is $6\sqrt{3} - (3\sqrt{3} + \pi)$, which equals $\boxed{\textbf{(D) }3\sqrt{3} - \pi}$.

Solution 2

First, subdivide the hexagon into 24 equilateral triangles with side length 1: [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);  draw((1,0)--(3,2sqrt(3))); draw((3,0)--(1,2sqrt(3))); draw((4,sqrt(3))--(0,sqrt(3))); draw((2,0)--(3.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(2,2sqrt(3))); draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2)); draw((2,2sqrt(3))--(0.5,sqrt(3)/2)); draw((2,0)--(0.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2)); [/asy] Now note that the entire shaded region is just 6 times this part: [asy] size(100); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white);  draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle); label("$1$",(2.25,7sqrt(3)/4),NE);  draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); [/asy] The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: \[2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\] The arc that is not included has an area of: \[\frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}\] Hence, the area of the shaded region in that section is \[\frac{\sqrt{3}}{2}-\frac{\pi}{6}\] For a final area of: \[6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}\] ~N828335

Video Solution

https://youtu.be/t6yjfKXpwDs?t=786 (for AMC 10)

https://youtu.be/0xgTR3UEqbQ (for AMC 12)

~IceMatrix


https://youtu.be/oTqx8OqSMQI ~DSA_Catachu

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 12 Problems and Solutions
2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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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