Difference between revisions of "2006 AMC 8 Problems/Problem 19"
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== Solution == | == Solution == | ||
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>. | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/HzzzRJEppRA Soo, DRMS, NM | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/abSgjn4Qs34?t=2124 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:47, 2 January 2023
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 ?
Solution
Since triangle is congruent to triangle and , . Since , . Because point is the midpoint of , .
Video Solution
https://youtu.be/HzzzRJEppRA Soo, DRMS, NM
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=2124
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.