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

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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.  
 
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.  
  
This is equivilant to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>.  
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This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>.  
  
 
So the answer is:  
 
So the answer is:  

Revision as of 12:12, 4 February 2010

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 $2$?

2005amc10a12.gif

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

Solution

The area of the trefoil is equal to the area of a small 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:

$4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B$

See Also

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