2007 AMC 12B Problems/Problem 11

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Solution

The angles in a quadrilateral add up to 360, regardless of what shape or size the figure takes. Let $x$ be $\angle A$, so then:

$\angle B=.5x$

$\angle C=(1/3)x$

$\angle D=(1/4)x$

$\angle A+\angle B+\angle C+\angle D=360$

$x+.5x+(1/3)x+(1/4)x=360$

$(25/12)x=360$

$x=172.8$

or x is around 173, choice D.

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 12 Problems and Solutions