Difference between revisions of "2003 AMC 10A Problems/Problem 19"

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== Problem ==
#REDIRECT[[2003 AMC 12A Problems/Problem 15]]
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.
<math> \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 </math>
== Solution ==
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 <math>\frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>.
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 <math>60^\circ</math>.
The area of the <math>60^\circ</math> sector of the larger semicircle is <math>\frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi</math>.
The area of the triangle is <math>\frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>
So the shaded area is <math>\frac{1}{8}\pi-\frac{1}{6}\pi+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}-\frac{1}{24}\pi \Rightarrow C</math>
== See Also ==
*[[2003 AMC 10A Problems]]
*[[2003 AMC 10A Problems/Problem 18|Previous Problem]]
*[[2003 AMC 10A Problems/Problem 20|Next Problem]]
[[Category:Introductory Geometry Problems]]

Latest revision as of 17:31, 31 July 2011

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