Difference between revisions of "2010 AMC 10B Problems/Problem 16"
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− | The radius of the circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>. Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}</math>. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle Therefore the picture will look something like this: | + | The radius of the circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>. Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}</math>. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this: |
<center><asy> | <center><asy> |
Revision as of 21:26, 7 December 2018
Problem
A square of side length and a circle of radius share the same center. What is the area inside the circle, but outside the square?
Solution
The radius of the circle is . Half the diagonal of the square is . We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:
Then we proceed to find: 4 * (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits in half. Let this half-length be . Also note that because it is half the sidelength of the square. Because this is a right triangle, we can use the Pythagorean Theorem to solve for
Solving, and . Since , is an equilateral triangle and the central angle is . Therefore the sector has an area .
Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is
Putting it together, we get the answer to be
See Also
2010 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.