Difference between revisions of "2015 AMC 10B Problems/Problem 9"
(One intermediate revision by one other user not shown) | |||
Line 21: | Line 21: | ||
==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> | ||
+ | |||
+ | ==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}} | ||
− | {{MAA | + | {{MAA Notice}} |
[[Category: Prealgebra Problems]] | [[Category: Prealgebra Problems]] |
Latest revision as of 13:16, 24 July 2020
Contents
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?
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
~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.