2011 AMC 10B Problems/Problem 9

Problem

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

$[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3)); pair[] ps={A,B,C,D,E}; draw(A--B--C--cycle); draw(E--D); draw(rightanglemark(E,D,B)); dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); label("$3$",midpoint(A--C),NW); label("$4$",midpoint(C--B),NE); label("$5$",midpoint(A--B),SW); [/asy]$

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

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

2011 AMC 10B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search