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Difference between revisions of "2003 AMC 12A Problems/Problem 15"
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== Problem ==
 
== Problem ==
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A semicircle of diameter <math>1</math> sits at the top of a semicircle of diameter <math>2</math>, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a ''lune''. Determine the area of this lune.  
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A [[semicircle]] of [[diameter]] <math>1</math> sits at the top of a semicircle of diameter <math>2</math>, as shown. The shaded [[area]] inside the smaller semicircle and outside the larger semicircle is called a ''lune''. Determine the area of this lune.  
  
 
[[Image:2003amc10a19.gif]]
 
[[Image:2003amc10a19.gif]]
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[[Image:2003amc10a19solution.gif]]
 
[[Image:2003amc10a19solution.gif]]
  
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The shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle.  
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The shaded area is equal to the area of the smaller semicircle minus the area of a [[sector]] of the larger circle plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle.  
  
 
The area of the smaller semicircle is <math>\frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>.  
 
The area of the smaller semicircle is <math>\frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>.  
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== See Also ==
 
== See Also ==
 
*[[2003 AMC 12A Problems]]
 
*[[2003 AMC 12A Problems]]
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*[[2003 AMC 12A/Problem 14|Previous Problem]]
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*[[2003 AMC 12A Problems/Problem 14|Previous Problem]]
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*[[2003 AMC 12A/Problem 16|Next Problem]]
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*[[2003 AMC 12A Problems/Problem 16|Next Problem]]
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 10:16, 11 November 2006

Problem

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

2003amc10a19.gif

$\mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi$

Solution

2003amc10a19solution.gif

The shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle.

The area of the smaller semicircle is $\frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi$.

Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures $60^\circ$.

The area of the $60^\circ$ sector of the larger semicircle is $\frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi$.

The area of the triangle is $\frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$

So the shaded area is $\frac{1}{8}\pi-\frac{1}{6}\pi+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}-\frac{1}{24}\pi \Rightarrow C$

See Also

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