Difference between revisions of "2006 AMC 10B Problems/Problem 6"

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== Solution ==
 
== Solution ==
Since the side of the square is the diameter of the semicircle, the radius of the semicircle is <math> \frac{1}{2}\cdot\frac{2}{\pi}=\frac{1}{\pi} </math>
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Since the side of the square is the diameter of the semicircle, the radius of the semicircle is <math> \frac{1}{2}\cdot\frac{2}{\pi}=\frac{1}{\pi} </math>.
  
Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is <math> \frac{1}{2}\cdot2\cdot\pi\cdot\frac{1}{\pi}=1</math>  
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Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is <math> \frac{1}{2}\cdot2\cdot\pi\cdot\frac{1}{\pi}=1</math>.
  
 
Since the desired perimeter is made up of four of these arcs, the perimeter is <math>4\cdot1=4\Rightarrow D</math>
 
Since the desired perimeter is made up of four of these arcs, the perimeter is <math>4\cdot1=4\Rightarrow D</math>

Revision as of 13:40, 18 July 2006

Problem

A region is bounded by semicircular arcs constructed on the side of a square whose sides measure $\frac{2}{\pi}$, as shown. What is the perimeter of this region?

2006amc10b06.gif

$\mathrm{(A) \ } \frac{4}{\pi}\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } \frac{8}{\pi}\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } \frac{16}{\pi}$

Solution

Since the side of the square is the diameter of the semicircle, the radius of the semicircle is $\frac{1}{2}\cdot\frac{2}{\pi}=\frac{1}{\pi}$.

Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is $\frac{1}{2}\cdot2\cdot\pi\cdot\frac{1}{\pi}=1$.

Since the desired perimeter is made up of four of these arcs, the perimeter is $4\cdot1=4\Rightarrow D$

See Also