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> \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> | ||
+ | |||
== Solution == | == Solution == | ||
+ | Since the side of the [[square (geometry) | 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=4\Longrightarrow \boxed{\mathrm{(D)}}</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 23:22, 18 October 2020
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?
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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