Difference between revisions of "2006 AMC 10B Problems/Problem 6"
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== Problem == | == Problem == | ||
| + | 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? | ||
| + | |||
| + | <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> \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>. | ||
| + | |||
| + | 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=\boxed{\textbf{(D) }4}</math>. | ||
| + | |||
== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2006|ab=B|num-b=5|num-a=7}} | |
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | [[Category:Circle Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:45, 26 January 2022
Problem
A region is bounded by semicircular arcs constructed on the side of a square whose sides measure 
, 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]](http://latex.artofproblemsolving.com/5/e/2/5e2f7d7427c74567de47a93edfad961d23cd8633.png)
Solution
Since the side of the square is the diameter of the semicircle, the radius of the semicircle is 
.
Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is 
.
Since the desired perimeter is made up of four of these arcs, the perimeter is 
.
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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