Difference between revisions of "2007 AMC 12B Problems/Problem 11"

Latest revision as of 16:22, 5 June 2011

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-==Solution==
+#REDIRECT [[2007 AMC 10B Problems/Problem 15]]
-The angles in a quadrilateral add up to 360, regardless of what shape or size the figure takes. Let <math>x</math> be <math>\angle A</math>, so then:
-<math>\angle B=.5x</math>
-<math>\angle C=(1/3)x</math>
-<math>\angle D=(1/4)x</math>
-<math>\angle A+\angle B+\angle C+\angle D=360</math>
-<math>x+.5x+(1/3)x+(1/4)x=360</math>
-<math>(25/12)x=360</math>
-<math>x=172.8</math>
-or x is around 173, choice D.
-==See Also==
-{{AMC12 box|year=2007|ab=B|num-b=10|num-a=12}}