2015 AMC 10B Problems/Problem 9

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 $\dfrac{9\pi}{4}-\dfrac{9\pi}{8}=\boxed{\mathbf{(B)}\ \dfrac{9\pi}{8}}$

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

Template:MAA notice

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2015 AMC 10B Problems/Problem 9

Problem

Solution

See Also