2011 AMC 10B Problem 9

In the $3$-$4$-$5$ triangle the right angle is at $C$, with $AC = 3$, $BC = 4$, $AB = 5$.

Triangle $EBD$ has a right angle at $D$ and shares $\angle B$ with triangle $ABC$, so by AA the two triangles are similar, with $D$ matching $C$, $E$ matching $A$, and $B$ matching $B$. Hence $BD$ corresponds to $BC = 4$.

Since the areas are in ratio $1 : 3$, corresponding lengths are in ratio $1 : \sqrt{3}$. Therefore
$$BD = \frac{BC}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{4\sqrt{3}}{3}}$.

A perpendicular dropped inside a right triangle creates a smaller triangle similar to the original, and areas of similar figures scale as the square of the length ratio. The one care point is the correspondence of vertices: $BD$ lies along the hypotenuse of $ABC$ but is a leg of $EBD$, matching leg $BC$, not $AB$.

Let $BD = x$. Since $\tan B = \frac{AC}{BC} = \frac{3}{4}$, $DE = \frac{3}{4}x$ and the area of $EBD$ is $\frac{3}{8}x^2$. Setting this equal to $\frac{1}{3} \cdot 6 = 2$ gives $x^2 = \frac{16}{3}$, so $x = \frac{4}{\sqrt{3}}$.

• Scaling sides by $1/3$ instead of $1/\sqrt{3}$, giving $4/3$, or matching $BD$ to the wrong side of the $3$-$4$-$5$ triangle.
• Placing the right angle of $ABC$ at $B$ instead of at $C$, which breaks the similarity.