Difference between revisions of "2011 AMC 10B Problems/Problem 16"
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<cmath> \frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath> | <cmath> \frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath> | ||
+ | |||
+ | == Solution 2== | ||
+ | area of a regular octagon = 2(1+sqrt{2})a^2 where a is the side hence the answer is obvious now. | ||
== See Also== | == See Also== |
Revision as of 02:50, 16 August 2020
Contents
[hide]Problem
A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
Solution
If the side lengths of the dart board and the side lengths of the center square are all then the side length of the legs of the triangles are .
Use Geometric probability by putting the area of the desired region over the area of the entire region.
Solution 2
area of a regular octagon = 2(1+sqrt{2})a^2 where a is the side hence the answer is obvious now.
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
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