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

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:

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

Revision as of 03:12, 1 March 2011 (view source) Sanchitb (talk \| contribs) ← Older edit		Revision as of 03:12, 1 March 2011 (view source) Sanchitb (talk \| contribs) Newer edit →
Line 7:		Line 7:
	[[Image:squarecircle.png]]		[[Image:squarecircle.png]]

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