# Difference between revisions of "2012 AMC 10B Problems/Problem 21"

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 $b:a= \sqrt{3}=(A)$