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Difference between revisions of "2011 AMC 12B Problems/Problem 10"

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== Solution ==
 
== Solution ==
 
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Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CDM</math> hence <math>CM=CD=6</math>.  Therefore <math>\angle BMC = 30^\circ</math>.  Therefore <math>\angle AMD=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}}

Revision as of 01:46, 3 July 2011

Problem

Rectangle $ABCD$ has $AB=6$ and $BC=3$. Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD$. What is the degree measure of $\angle AMD$?

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 75$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Since $AB \parallel CD$, $\angle AMD = \angle CDM$ hence $CM=CD=6$. Therefore $\angle BMC = 30^\circ$. Therefore $\angle AMD=75^\circ$

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
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