2011 AMC 10B Problem 9

Below is the professionally curated solution for Problem 9 of the 2011 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

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Concepts:similarity area ratio

Difficulty rating: 1280

The area of $\triangle EBD$ is one third of the area of the $3$ - $4$ - $5$ triangle $ABC.$ Segment $DE$ is perpendicular to segment $AB.$ What is $BD?$

$\dfrac{4}{3}$

$\sqrt{5}$

$\dfrac{9}{4}$

$\dfrac{4\sqrt{3}}{3}$

$\dfrac{5}{2}$

Solution:

By angle angle similarity, we have $BDE \sim BCA .$

Then, since the ratio of the areas is $\frac 13,$ the ratio of the sidelengths is $\frac{1}{\sqrt 3}.$

As such, $\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3},$ making $BD = \dfrac 4{ \sqrt 3} = \dfrac{4\sqrt{3}}{3} .$

Thus, the correct answer is D .

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