Difference between revisions of "2010 AMC 10B Problems/Problem 16"

(Undo revision 37207 by Sanchitb (Talk))
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Radius of circle = <math>\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} = 0.577</math>
  
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Half the diagonal of the square = <math>\sqrt{.5^2 + .5^2} = \frac{\sqrt{2}}{2} = 0.707</math>
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Therefore the picture will look something like this:
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[[Image:squarecircle.png]]
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Then you proceed to find the 4 * (area of the sector - are of the triangle) to get answer (B)

Revision as of 03:12, 1 March 2011

Radius of circle = $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} = 0.577$

Half the diagonal of the square = $\sqrt{.5^2 + .5^2} = \frac{\sqrt{2}}{2} = 0.707$

Therefore the picture will look something like this:

Squarecircle.png

Then you proceed to find the 4 * (area of the sector - are of the triangle) to get answer (B)

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