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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==
 +
https://youtu.be/22AUHebZNyk
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2015|ab=B|num-b=8|num-a=10}}
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{{MAA notice}}
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{{MAA Notice}}
  
 
[[Category: Prealgebra Problems]]
 
[[Category: Prealgebra Problems]]
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The problems on this page are copyrighted by the [http://www.maa.org Mathematical Association of America]'s [http://amc.maa.org American Mathematics Competitions]. [[File:AMC_logo.png|middle]]
 
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<div style="clear:both;"></div>
 

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 $3$ and center $(0,0)$ that lies in the first quadrant, the portion of the circle with radius $\tfrac{3}{2}$ and center $(0,\tfrac{3}{2})$ that lies in the first quadrant, and the line segment from $(0,0)$ to $(3,0)$. 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]

$\textbf{(A) } \dfrac{4\pi}{5} \qquad\textbf{(B) } \dfrac{9\pi}{8} \qquad\textbf{(C) } \dfrac{4\pi}{3} \qquad\textbf{(D) } \dfrac{7\pi}{5} \qquad\textbf{(E) } \dfrac{3\pi}{2}$

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 $3$ so it has area $\dfrac{9\pi}{4}$. The semicircle has radius $\dfrac{3}{2}$ so it has area $\dfrac{9\pi}{8}$. Thus, the shaded area is $\dfrac{9\pi}{4}-\dfrac{9\pi}{8}=\boxed{\mathbf{(B)}\ \dfrac{9\pi}{8}}$

Video Solution 1

https://youtu.be/4fTjISCoRlU

~Education, the Study of Everything

Video Solution

https://youtu.be/22AUHebZNyk

~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. AMC logo.png

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