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2012 AMC 8 Problem 24 Mensuration


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Understand the problem

A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle? [asy] size(0,50); draw((-1,1)..(-2,2)..(-3,1)..(-2,0)..cycle); dot((-1,1)); dot((-2,2)); dot((-3,1)); dot((-2,0)); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy] $\textbf{(A)}\hspace{.05in}\frac{4-\pi}{\pi}\qquad\textbf{(B)}\hspace{.05in}\frac{1}\pi\qquad\textbf{(C)}\hspace{.05in}\frac{\sqrt2}{\pi}\qquad\textbf{(D)}\hspace{.05in}\frac{\pi-1}{\pi}\qquad\textbf{(E)}\hspace{.05in}\frac{3}\pi$
Source of the problem

2012 AMC(American Mathematical Contest) 8 Problem 24

Topic
Mensuration
Difficulty Level
Easy

Start with hints

Do you really need a hint? Try it first!

Hint figure :   [asy] dot((0,0),red); dot((0,2),red); dot((2,0),red); dot((2,2),red); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red); size(0,50); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]
Assume that the radius of the circle is \( r \) unit and proceed to compare .
    Note that side length of the square is \( 2r \) unit .      
Clearly , $$ area(red \ square ) – area(circle ) = area(star) \\ \Rightarrow \frac{area(red \ square )}{area(circle )} – 1 =\frac{area(star)}{area(circle )} \\ \Rightarrow \frac{area(star)}{area(circle )} = \frac {(2r)^2}{\pi r^2} – 1 = \frac {4 – \pi}{\pi} $$

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