Difference between revisions of "2016 AMC 12B Problems/Problem 18"
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Because of symmetry, the area is the same in all four quadrants. | Because of symmetry, the area is the same in all four quadrants. | ||
The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math> | The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:57, 9 January 2024
Problem
What is the area of the region enclosed by the graph of the equation
Solution
Consider the case when , . Notice the circle intersect the axes at points and . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of and a triangle: Because of symmetry, the area is the same in all four quadrants. The answer is
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
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