2011 AMC 10B Problem 9

Question

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}

Use similarity between

\triangle BDE

and

\triangle BCA

Area ratio

1:3

gives side-length ratio

1:\sqrt3

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 .

Problem 9 in other years