Difference between revisions of "2003 AMC 12A Problems/Problem 15"
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The shaded area <math>[A]</math> is equal to the area of the smaller semicircle <math>[A+B]</math> minus the area of a [[sector]] of the larger circle <math>[B+C]</math> plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle <math>[C]</math>. | The shaded area <math>[A]</math> is equal to the area of the smaller semicircle <math>[A+B]</math> minus the area of a [[sector]] of the larger circle <math>[B+C]</math> plus the area of a [[triangle]] formed by two [[radius | radii]] of the larger semicircle and the diameter of the smaller semicircle <math>[C]</math>. | ||
− | The area of the smaller semicircle is <math>[A+B] = \frac{1}{2}\pi\cdot(\frac{1}{2})^{2}=\frac{1}{8}\pi</math>. | + | The area of the smaller semicircle is <math>[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi</math>. |
− | Since the radius of the larger semicircle is equal to the diameter of the smaller | + | Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, 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>[B+C] = \frac{60}{360}\pi\cdot(\frac{2}{2})^{2}=\frac{1}{6}\pi</math>. | + | The area of the <math>60^\circ</math> sector of the larger semicircle is <math>[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi</math>. |
The area of the triangle is <math>[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>. | The area of the triangle is <math>[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>. | ||
− | So the shaded area is <math>[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>. | + | So the shaded area is <math>[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>. We have thus solved the problem. |
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+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=-4qnOdE0Dyk | ||
== See Also == | == See Also == | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
[[Category:Area Problems]] | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:06, 3 June 2024
- 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 half circle, 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 . We have thus solved the problem.
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=-4qnOdE0Dyk
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.