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

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== Problem ==
 
== Problem ==
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A region is bounded by semicircular arcs constructed on the side of a square whose sides measure <math> \frac{2}{\pi} </math>, as shown. What is the perimeter of this region?  
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A region is bounded by semicircular arcs constructed on the side of a square whose sides measure <math> \frac{2}{\pi} </math>, as shown. What is the perimeter of this region?
  
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[[Image:2006amc10b06.gif]]
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<asy>
 +
size(90); defaultpen(linewidth(0.7));
 +
filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle,gray(0.5));
 +
filldraw(arc((1,0),1,180,0, CCW)--cycle,gray(0.7));
 +
filldraw(arc((0,1),1,90,270)--cycle,gray(0.7));
 +
filldraw(arc((1,2),1,0,180)--cycle,gray(0.7));
 +
filldraw(arc((2,1),1,270,90, CCW)--cycle,gray(0.7));
 +
</asy>
  
−
<math> \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} </math>
+
<math> \textbf{(A) } \frac{4}{\pi}\qquad \textbf{(B) } 2\qquad \textbf{(C) } \frac{8}{\pi}\qquad \textbf{(D) } 4\qquad \textbf{(E) } \frac{16}{\pi} </math>
  
 
== 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 [[arc]]s 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 [[arc]]s 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=\boxed{\textbf{(D) }4}</math>.
  
 
== See Also ==
 
== See Also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=5|num-a=7}}
−
 
 
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*[[2006 AMC 10B Problems/Problem 5|Previous Problem]]
 
−
 
 
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*[[2006 AMC 10B Problems/Problem 7|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
[[Category:Circle Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:45, 26 January 2022

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?

[asy] size(90); defaultpen(linewidth(0.7)); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle,gray(0.5)); filldraw(arc((1,0),1,180,0, CCW)--cycle,gray(0.7)); filldraw(arc((0,1),1,90,270)--cycle,gray(0.7)); filldraw(arc((1,2),1,0,180)--cycle,gray(0.7)); filldraw(arc((2,1),1,270,90, CCW)--cycle,gray(0.7)); [/asy]

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

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 $4\cdot1=\boxed{\textbf{(D) }4}$.

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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