Difference between revisions of "2003 AMC 12A Problems/Problem 15"
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+ | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #15]] and [[2003 AMC 10A Problems|2003 AMC 10A #19]]}} | ||
== Problem == | == Problem == | ||
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. | 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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The area of the triangle is <math>\frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</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 | + | So the shaded area is <math>\frac{1}{8}\pi-\frac{1}{6}\pi+\frac{\sqrt{3}}{4}=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math> |
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2003|ab=A|num-b=18|num-a=20}} | |
{{AMC12 box|year=2003|ab=A|num-b=14|num-a=16}} | {{AMC12 box|year=2003|ab=A|num-b=14|num-a=16}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 17:31, 31 July 2011
- The following problem is from both the 2003 AMC 12A #15 and 2003 AMC 10A #19, so both problems redirect to this page.
Problem
A semicircle of diameter sits at the top of a semicircle of diameter , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
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 .
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 .
The area of the sector of the larger semicircle is .
The area of the triangle is
So the shaded area is
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |