2011 AMC 10B Problem 9

Problem 9 of 25EasierGeometry

The area of

\triangle EBD

is one third of the area of the

\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3},

-

4

-

5

triangle

ABC.

Segment

DE

is perpendicular to segment

AB.

What is

BD?

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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 .

More practice

Concepts: similarity · area ratio

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.