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Difference between revisions of "2011 AMC 10B Problems/Problem 9"
 
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<math> \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} </math>
 
<math> \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} </math>
−
[[Category: Introductory Geometry Problems]]
 
  
 
== Solution 1==
 
== Solution 1==
Line 48: Line 47:
 
<cmath>\frac{1}{3}[\triangle ABC] = [\triangle EBD]</cmath>
 
<cmath>\frac{1}{3}[\triangle ABC] = [\triangle EBD]</cmath>
 
<cmath>\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB</cmath>
 
<cmath>\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB</cmath>
−
In order to scale the sides of ED and DB to make <math>\frac{1}{3}</math> (since the ratios of sides are the same), we take the square root of <math>\frac{1}{3} = \frac{\sqrt(3)}{3}</math> to scale each side by the same amount.  
+
In order to scale the sides of ED and DB to make <math>\frac{1}{3}</math> (since the ratios of sides are the same), we take the square root of <math>\frac{1}{3} = \frac{\sqrt{3}}{3}</math> to scale each side by the same amount.  
  
−
Thus <math>BD = 4 \times \frac{\sqrt(3)}{3}</math> and the answer is <math>BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}</math>
+
Thus <math>BD = 4 \times \frac{\sqrt{3}}{3}</math> and the answer is <math>BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}</math>
  
 
==Solution 3 (Shortcut)==
 
==Solution 3 (Shortcut)==
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==Remark (slightly more vigorous than Solution 3)==
+
==Solution 4 (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 <math>\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\sqrt{3}.</math> By similarity, <math>\frac{ED}{DB}=\frac{\sqrt{3}}{DB}=\frac{3}{4}</math>, so solving for DB, we get <math>\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}</math>.
 
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 <math>\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\sqrt{3}.</math> By similarity, <math>\frac{ED}{DB}=\frac{\sqrt{3}}{DB}=\frac{3}{4}</math>, so solving for DB, we get <math>\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}</math>.
  
 
~JH. L
 
~JH. L
  
−
==Solution 4==
+
==Solution 5==
−
It is apparent that <math>\Delta ABC~\Delta EBD</math> by <math>AA</math> similarity (<math>\angle B=\angle B</math> and <math>\angle EDB=\angle ACB</math>). Thus, let the side length of <math>ED</math> equal <math>3x</math> and <math>DB=4x.</math> We can then see that <math>[EDB]=\dfrac{3x\cdot4x}2=6x^2</math>, and we are given that <math>[ABC]=3\cdot[EDB]</math>. Thus, <math>\dfrac{3\cdot4}2=3\cdot6x^2\implies6=18x^2\implies x=\dfrac{\sqrt{3}}3</math>. Since we let <math>BD=4x</math>, we know that <math>BD=\boxed{\textbf{D}~\dfrac{4\sqrt3}3}</math>.  
+
It is apparent that <math>\Delta ABC \sim \Delta EBD</math> by AA similarity (<math>\angle B=\angle B</math> and <math>\angle EDB=\angle ACB</math>). Thus, let the side length of <math>ED</math> equal <math>3x</math> and <math>DB=4x.</math> We can then see that <math>[EDB]=\dfrac{3x\cdot4x}2=6x^2</math>, and we are given that <math>[ABC]=3\cdot[EDB]</math>. Thus, <math>\dfrac{3\cdot4}2=3\cdot6x^2\implies6=18x^2\implies x=\dfrac{\sqrt{3}}3</math>. Since we let <math>BD=4x</math>, we know that <math>BD=\boxed{\textbf{(D) }\dfrac{4\sqrt{3}}3}</math>.  
 
~Technodoggo
 
~Technodoggo
  
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{{AMC10 box|year=2011|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2011|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 17:56, 7 September 2025

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


Solution 4 (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 5

It is apparent that $\Delta ABC \sim \Delta EBD$ 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 $BD=\boxed{\textbf{(D) }\dfrac{4\sqrt{3}}3}$. ~Technodoggo

See Also

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

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