2011 AMC 10B Problems/Problem 9


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

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$\textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2}$

Solution 1

$\triangle ABC \sim \triangle EBD$ by AA Similarity. Therefore $DE = \frac{3}{4} BD$. Find the areas of the triangles. \[\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6\] \[\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2\] The area of $\triangle EBD$ is one third of the area of $\triangle ABC$. \begin{align*} \frac{3}{8} BD^2 &= 6 \times \frac{1}{3}\\ 9BD^2 &= 48\\ BD^2 &= \frac{16}{3}\\ BD &= \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}} \end{align*}

Solution 2

$\triangle ABC \sim \triangle EBD$ by AA Similarity. Since the area of $\triangle EBD$ is $\frac{1}{3}$ of $\triangle ABC$ and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus, \[\frac{1}{3}[\triangle ABC] = [\triangle EBD]\] \[\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB\] In order to scale the sides of ED and DB to make $\frac{1}{3}$ (since the ratios of sides are the same), we take the square root of $\frac{1}{3} = \frac{\sqrt(3)}{3}$ to scale each side by the same amount.

Thus $BD = 4 \times \frac{\sqrt(3)}{3}$ and the answer is $BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}$

Solution 3 (Shortcut)

The ratio of the areas of $\triangle$$EBD$ and $\triangle$$ABC$ is $1 : 3$, meaning the ratio of the sides is $1 : \sqrt{3}$. The only answer choice involving $\sqrt{3}$ is $\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}$.

-Solution by Joeya

Remark (slightly more vigorous than Solution 3)

The ratio of the areas is equal to twice the ratio of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is $\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\sqrt{3}.$ By similarity, $\frac{ED}{DB}=\frac{\sqrt{3}}{DB}=\frac{3}{4}$, so solving for DB, we get $\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}$.

~JH. L

Solution 4

It is apparent that $\Delta ABC~\Delta EBD$ by $AA$ similarity ($\angle B=\angle B$ and $\angle EDB=\angle ACB$). Thus, let the side length of $ED$ equal $3x$ and $DB=4x.$ We can then see that $[EDB]=\dfrac{3x\cdot4x}2=6x^2$, and we are given that $[ABC]=3\cdot[EDB]$. Thus, $\dfrac{3\cdot4}2=3\cdot6x^2\implies6=18x^2\implies x=\dfrac{\sqrt{3}}3$. Since we let $BD=4x$, we know that $BD=\boxed{\textbf{D}~\dfrac{4\sqrt3}3}$. ~Technodoggo

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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