Difference between revisions of "2011 AMC 12B Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CMD</math> hence <math>CM=CD=6</math>. Therefore <math>\angle BMC = 30^\circ</math>. Therefore <math>\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}</math> | + | Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM=CD=6</math>. Therefore, <math>\angle BMC = 30^\circ</math>. Therefore <math>\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:11, 5 August 2019
Problem
Rectangle 
has 
and 
. Point 
is chosen on side 
so that 
. What is the degree measure of 
?
Solution
Since 
, 
, so 
, so 
is isosceles, and hence 
. Therefore, 
. Therefore 
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
