Difference between revisions of "2006 AMC 8 Problems/Problem 19"
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Since triangle <math> ABD </math> is congruent to triangle <math> ECD </math> and <math> \overline{CE} =11 </math>, <math> \overline{AB}=11 </math>. Since <math> \overline{AB}=\overline{BC}</math>, <math> \overline{BC}=11 </math>. Because point <math> D </math> is the midpoint of <math> \overline{BC} </math>, <math> \overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5} </math>. | Since triangle <math> ABD </math> is congruent to triangle <math> ECD </math> and <math> \overline{CE} =11 </math>, <math> \overline{AB}=11 </math>. Since <math> \overline{AB}=\overline{BC}</math>, <math> \overline{BC}=11 </math>. Because point <math> D </math> is the midpoint of <math> \overline{BC} </math>, <math> \overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5} </math>. | ||
| + | ==See Also== | ||
{{AMC8 box|year=2006|n=II|num-b=18|num-a=20}} | {{AMC8 box|year=2006|n=II|num-b=18|num-a=20}} | ||
Revision as of 19:14, 24 December 2012
Problem
Triangle 
is an isosceles triangle with 
. Point 
is the midpoint of both 
and 
, and 
is 11 units long. Triangle 
is congruent to triangle 
. What is the length of 
?
![[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]](http://latex.artofproblemsolving.com/4/b/4/4b49c3097ed0eacc7cdb3ab3850b7fdf1f67d407.png)
Solution
Since triangle is congruent to triangle and 
, 
. Since , 
. Because point is the midpoint of , 
.
See Also
| 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 | ||
