Difference between revisions of "2006 AMC 8 Problems/Problem 19"

Revision as of 00:22, 5 July 2013

Problem

Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$ . Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$ , and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$ . What is the length of $\overline{BD}$ ?

$[asy] size(100); draw((0,0)--(2,4)--(4,0)--(6,4)--cycle--(4,0),linewidth(1)); label("$A$", (0,0), SW); label("$B$", (2,4), N); label("$C$", (4,0), SE); label("$D$", shift(0.2,0.1)*intersectionpoint((0,0)--(6,4),(2,4)--(4,0)), N); label("$E$", (6,4), NE);[/asy]$

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6$

Solution

Since triangle $ABD$ is congruent to triangle $ECD$ and $\overline{CE} =11$ , $\overline{AB}=11$ . Since $\overline{AB}=\overline{BC}$ , $\overline{BC}=11$ . Because point $D$ is the midpoint of $\overline{BC}$ , $\overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5}$ .

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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 AJHSME/AMC 8 Problems and Solutions

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

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 ==See Also==
 {{AMC8 box|year=2006|n=II|num-b=18|num-a=20}}
+{{MAA Notice}}

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Difference between revisions of "2006 AMC 8 Problems/Problem 19"

Revision as of 00:22, 5 July 2013

Problem

Solution

See Also