Difference between revisions of "2011 AMC 10A Problems/Problem 18"

Latest revision as of 19:19, 27 June 2020

Redirect to:

Revision as of 01:39, 16 February 2011 (view source) Flyingpenguin (talk \| contribs) (Created page with '== Solution == Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can c…')		Latest revision as of 19:19, 27 June 2020 (view source) Jbala (talk \| contribs) (Redirected page to 2011 AMC 12A Problems/Problem 11) (Tag: New redirect)
(36 intermediate revisions by 16 users not shown)
Line 1:		Line 1:
−	~~== Solution ==~~	+	#redirect [[2011 AMC 12A Problems/Problem 11]]
−
−	~~Draw a rectangle with vertices at the centers of <math>A<~~/math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shades area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{2 \ \mathbf{(C)}}</math>.