2011 AMC 12B Problems/Problem 10

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 $\angle AMD$ ?

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

Solution

Since $AB \parallel CD$ , $\angle AMD = \angle CDM$ , so $\angle AMD = \angle CMD = \angle CDM$ , so $\bigtriangleup CMD$ is isosceles, and hence $CM=CD=6$ . Therefore, $\angle BMC = 30^\circ$ . Therefore $\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}$

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.

Page

Toolbox

Search

2011 AMC 12B Problems/Problem 10

Problem

Solution

See also