Difference between revisions of "2011 AMC 10B Problems/Problem 18"
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− | It is given that <math>\angle AMD \sim \angle CMD</math>. Since <math>\angle AMD</math> and <math>\angle CDM</math> are | + | It is given that <math>\angle AMD \sim \angle CMD</math>. Since <math>\angle AMD</math> and <math>\angle CDM</math> are alternate interior angles and <math>\overline{AB} \parallel \overline{DC}</math>, <math>\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM</math>. Use the [[Base Angle Theorem]] to show <math>\overline{DC} \cong \overline{MC}</math>. We know that <math>ABCD</math> is a [[rectangle]], so it follows that <math>\overline{MC} = 6</math>. We notice that <math>\triangle BMC</math> is a <math>30-60-90</math> triangle, and <math>\angle BMC = 30^{\circ}</math>. If we let <math>x</math> be the measure of <math>\angle AMD,</math> then |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2x + 30 &= 180\\ | 2x + 30 &= 180\\ |
Revision as of 12:40, 10 September 2017
Problem
Rectangle has and . Point is chosen on side so that . What is the degree measure of ?
Solution
It is given that . Since and are alternate interior angles and , . Use the Base Angle Theorem to show . We know that is a rectangle, so it follows that . We notice that is a triangle, and . If we let be the measure of then
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
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