2011 AMC 10B Problem 9 — Solution | LIVE by Po-Shen Loh

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2011 AMC 10B Problem 9

Attempt Problem 9 of the 2011 AMC 10B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

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 $BD?$

$\dfrac{4}{3}$

$\sqrt{5}$

$\dfrac{9}{4}$

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

$\dfrac{5}{2}$

Answer: D

Concepts:similarity area ratio

Difficulty rating: 1280

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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