2012 AMC 10B Problems/Problem 21

Revision as of 18:38, 12 March 2012 by Dongha (talk | contribs)

When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that $b=\sqrt{3a}$. Drawing the points out, it is possible to have a diagram where b=$\sqrt{3a}$ So, $b=\sqrt{3a}$, so $b:a= \sqrt{3}=(A)$

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