Difference between revisions of "2015 AMC 10B Problems/Problem 9"
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==Solution== | ==Solution== | ||
The area of the shark's fin falcata is just the area of the quarter-circle minus the area of the semicircle. The quarter-circle has radius <math>3</math> so it has area <math>\dfrac{9\pi}{4}</math>. The semicircle has radius <math>\dfrac{3}{2}</math> so it has area <math>\dfrac{9\pi}{8}</math>. Thus, the shaded area is <math>\dfrac{9\pi}{4}-\dfrac{9\pi}{8}=\boxed{\mathbf{(B)}\ \dfrac{9\pi}{8}}</math> | The area of the shark's fin falcata is just the area of the quarter-circle minus the area of the semicircle. The quarter-circle has radius <math>3</math> so it has area <math>\dfrac{9\pi}{4}</math>. The semicircle has radius <math>\dfrac{3}{2}</math> so it has area <math>\dfrac{9\pi}{8}</math>. Thus, the shaded area is <math>\dfrac{9\pi}{4}-\dfrac{9\pi}{8}=\boxed{\mathbf{(B)}\ \dfrac{9\pi}{8}}</math> | ||
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| + | ==Video Solution 1== | ||
| + | https://youtu.be/4fTjISCoRlU | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 16:12, 2 August 2022
Problem
The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius 
and center 
that lies in the first quadrant, the portion of the circle with radius 
and center 
that lies in the first quadrant, and the line segment from to 
. What is the area of the shark's fin falcata?
![[asy] import cse5;pathpen=black;pointpen=black; size(1.5inch); D(MP("x",(3.5,0),S)--(0,0)--MP("\frac{3}{2}",(0,3/2),W)--MP("y",(0,3.5),W)); path P=(0,0)--MP("3",(3,0),S)..(3*dir(45))..MP("3",(0,3),W)--(0,3)..(3/2,3/2)..cycle; draw(P,linewidth(2)); fill(P,gray); [/asy]](http://latex.artofproblemsolving.com/d/c/e/dce4e98a2133efe32043ca2e707377730ddad63c.png)
Solution
The area of the shark's fin falcata is just the area of the quarter-circle minus the area of the semicircle. The quarter-circle has radius so it has area 
. The semicircle has radius 
so it has area 
. Thus, the shaded area is 
Video Solution 1
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
