Difference between revisions of "2011 AMC 10B Problems/Problem 9"

(Problem)
m (Solution)
Line 32: Line 32:
  
 
<math>\triangle ABC \sim \triangle EBD</math> by AA Similarity. Therefore <math>DE = \frac{3}{4} BD</math>. Find the areas of the triangles.
 
<math>\triangle ABC \sim \triangle EBD</math> by AA Similarity. Therefore <math>DE = \frac{3}{4} BD</math>. Find the areas of the triangles.
<cmath>\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6\\
+
<cmath>\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6</cmath>
\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2</cmath>
+
<cmath>\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2</cmath>
 
The area of <math>\triangle EBD</math> is one third of the area of <math>\triangle ABC</math>.
 
The area of <math>\triangle EBD</math> is one third of the area of <math>\triangle ABC</math>.
 
<cmath>
 
<cmath>

Revision as of 12:27, 27 July 2018

Problem

The area of $\triangle$$EBD$ is one third of the area of $3-4-5$ $\triangle$$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

$\triangle ABC \sim \triangle EBD$ 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*}

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png