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Difference between revisions of "2005 AMC 10A Problems/Problem 12"

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The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?
 
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?
  
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[[Image:2005amc10a12.gif]]
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<asy>
 +
unitsize(1.5cm);
 +
defaultpen(linewidth(.8pt)+fontsize(12pt));
  
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<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math>
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pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180);
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pair E=B+C;
 +
 
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draw(D--E--B--O--C--B--A,linetype("4 4"));
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draw(Arc(O,1,0,60),linewidth(1.2pt));
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draw(Arc(O,1,120,180),linewidth(1.2pt));
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draw(Arc(C,1,0,60),linewidth(1.2pt));
 +
draw(Arc(B,1,120,180),linewidth(1.2pt));
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draw(A--D,linewidth(1.2pt));
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draw(O--dir(40),EndArrow(HookHead,4));
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draw(O--dir(140),EndArrow(HookHead,4));
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draw(C--C+dir(40),EndArrow(HookHead,4));
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draw(B--B+dir(140),EndArrow(HookHead,4));
 +
 
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label("2",O,S);
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draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar);
 +
draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar);
 +
</asy>
 +
 
 +
<math> \textbf{(A) }\frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \textbf{(B) } \frac{2}{3}\pi\qquad \textbf{(C) } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \textbf{(D) } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \textbf{(E) } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math>
  
 
==Solution==
 
==Solution==
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The area of the ''trefoil'' is equal to the area of a small equilateral triangle plus the area of four <math>60^\circ</math> sectors with a radius of <math>\frac{2}{2}=1</math> minus the area of  a small equilateral triangle.  
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The area of the ''trefoil'' is equal to the area of the big equilateral triangle plus the area of four <math>60^\circ</math> sectors with a radius of <math>\frac{2}{2}=1</math> minus the area of  a small equilateral triangle.  
  
 
This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>.  
 
This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>.  
  
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So the answer is:
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So the answer is <math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \boxed{\textbf{(B) }\frac{2}{3}\pi} </math>
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<math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math>
 
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==Video Solution==
 
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CHECK OUT Video Solution: https://youtu.be/77AjOy7zDwI
 
  
 
==See also==
 
==See also==

Latest revision as of 15:40, 2 June 2024

Problem

The figure shown is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $2$?

[asy] unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(12pt)); pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180); pair E=B+C; draw(D--E--B--O--C--B--A,linetype("4 4")); draw(Arc(O,1,0,60),linewidth(1.2pt)); draw(Arc(O,1,120,180),linewidth(1.2pt)); draw(Arc(C,1,0,60),linewidth(1.2pt)); draw(Arc(B,1,120,180),linewidth(1.2pt)); draw(A--D,linewidth(1.2pt)); draw(O--dir(40),EndArrow(HookHead,4)); draw(O--dir(140),EndArrow(HookHead,4)); draw(C--C+dir(40),EndArrow(HookHead,4)); draw(B--B+dir(140),EndArrow(HookHead,4)); label("2",O,S); draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar); draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar); [/asy]

$\textbf{(A) }\frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \textbf{(B) } \frac{2}{3}\pi\qquad \textbf{(C) } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \textbf{(D) } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \textbf{(E) } \frac{2}{3}\pi+\frac{\sqrt{3}}{2}$

Solution

The area of the trefoil is equal to the area of the big equilateral triangle plus the area of four $60^\circ$ sectors with a radius of $\frac{2}{2}=1$ minus the area of a small equilateral triangle.

This is equivalent to the area of four $60^\circ$ sectors with a radius of $1$.

So the answer is $4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \boxed{\textbf{(B) }\frac{2}{3}\pi}$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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