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

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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>
 
<math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math>
  
==See Also==
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==See also==
*[[2005 AMC 10A Problems]]
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{{AMC10 box|year=2005|ab=A|num-b=11|num-a=13}}
  
*[[2005 AMC 10A Problems/Problem 11|Previous Problem]]
 
 
*[[2005 AMC 10A Problems/Problem 13|Next Problem]]
 
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Area Problems]]
 
 
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Latest revision as of 12:57, 25 November 2020

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

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions

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