Problem
2011 AMC 10B Problem 9
The area of \triangleEBD is one third of the area of 3-4-5 \triangleABC. Segment DE is perpendicular to segment AB. What is BD?
\textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2}
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