Difference between revisions of "2003 AMC 12A Problems/Problem 15"

Revision as of 10:12, 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.

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

Solution

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 $\frac{1}{8}\pi-\frac{1}{6}\pi+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}-\frac{1}{24}\pi \Rightarrow C$

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Difference between revisions of "2003 AMC 12A Problems/Problem 15"

Revision as of 10:12, 11 November 2006

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Revision as of 22:50, 9 November 2006 (view source) Xantos C. Guin (talk \| contribs) (added problem and solution)	Revision as of 10:12, 11 November 2006 (view source) JBL (talk \| contribs) m (2003 AMC 12A/Problem 15 moved to 2003 AMC 12A Problems/Problem 15) Newer edit →
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