Difference between revisions of "2005 AMC 8 Problems/Problem 25"
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</math>r=\frac{2}{\sqrt{\pi}}<math> | </math>r=\frac{2}{\sqrt{\pi}}<math> | ||
− | So the answer is </math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi | + | So the answer is </math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}$. |
==Solution 2== | ==Solution 2== |
Revision as of 17:50, 14 November 2016
Problem
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
Solution
Let the region within the circle and square be . In other words, it is the intersection of the area of circle and square. Let be the radius. We know that the area of the circle minus is equal to the area of the square, minus .
We get:
So the answer is .
Solution 2
We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.
So the answer is .
Solution 2
We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.
r^2=\frac{4}{\pi}$$ (Error compiling LaTeX. ! Missing $ inserted.)r=\frac{2}{\sqrt{\pi}}\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}$.
Solution 2
We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.
So the answer is .
See Also
2005 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.