2011 AMC 10B Problems/Problem 9: Difference between revisions

Revision as of 15:45, 4 June 2011

Problem 9

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 $\triangle ABC$ . $\begin{align*} \frac{3}{8} BD^2 &= 6 \times \frac{1}{3}\\ 9BD^2 &= 48\\ BD^2 &= \frac{16}{3}\\ BD &= \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}} \end{align*}$

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 \end{align*}
 </cmath>
+== See Also==
+{{AMC10 box|year=2011|ab=B|num-b=8|num-a=10}}

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2011 AMC 10B Problems/Problem 9: Difference between revisions

Revision as of 15:45, 4 June 2011

Problem 9

Solution

See Also