2006 AMC 10B Problems/Problem 6

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]

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

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

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
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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