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. |
− | + | <asy> | |
+ | import graph; | ||
+ | size(150); | ||
+ | defaultpen(fontsize(8)); | ||
+ | pair A=(-2,0), B=(2,0); | ||
+ | filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); | ||
+ | filldraw(Arc((0,0),2,0,180)--cycle,white); | ||
+ | draw(2*expi(2*pi/6)--2*expi(4*pi/6)); | ||
+ | |||
+ | label("1",(0,sqrt(3)),(0,-1)); | ||
+ | label("2",(0,0),(0,-1)); | ||
+ | </asy> | ||
<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> | <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 == | == Solution == | ||
− | [[ | + | <asy> |
+ | import graph; | ||
+ | size(150); | ||
+ | defaultpen(fontsize(8)); | ||
+ | pair A=(-2,0), B=(2,0); | ||
+ | filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); | ||
+ | fill(Arc((0,0),2,0,180)--cycle,white); | ||
+ | draw(Arc((0,0),2,0,180)--cycle); | ||
+ | draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0)); | ||
+ | |||
+ | label("A",(0,2),(0,4)); | ||
+ | label("B",(0,2),(0,-1)); | ||
+ | label("C",(0,sqrt(3)/2),(0,2)); | ||
+ | label("1",(-0.5,sqrt(3)/2),(-1,0)); | ||
+ | label("1",(0.5,sqrt(3)/2),(1,0)); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | 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 | + | 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 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\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi</math>. | |
− | The area of the | + | 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>. We have thus solved the problem. | |
− | + | ==Video Solution by SpreadTheMathLove== | |
+ | https://www.youtube.com/watch?v=-4qnOdE0Dyk | ||
== 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}} | |
− | |||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry 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.