Difference between revisions of "2007 AMC 12B Problems/Problem 11"
(New page: The angles in a quadrilateral add up to 360, regardless of what shape or size the figure takes. Let x be <A, so then: <B=.5x <C=(1/3)x <D=(1/4)x <A+<B+<C+<D=360 x+.5x+(1/3)x+(1/4)x=360 (2...) |
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| + | ==Solution== | ||
| + | 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> | |
| − | <B=.5x | + | |
| − | <C=(1/3)x | + | <math>\angle C=(1/3)x</math> |
| − | <D=(1/4)x | + | |
| − | <A+ | + | <math>\angle D=(1/4)x</math> |
| − | x+.5x+(1/3)x+(1/4)x=360 | + | |
| − | (25/12)x=360 | + | <math>\angle A+\angle B+\angle C+\angle D=360</math> |
| − | x=172.8 | + | |
| − | or | + | <math>x+.5x+(1/3)x+(1/4)x=360</math> |
| − | x is around 173, choice D. | + | |
| + | <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}} | ||
Revision as of 02:23, 10 February 2009
Solution
The angles in a quadrilateral add up to 360, regardless of what shape or size the figure takes. Let 
be 
, so then:
or x is around 173, choice D.
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 | |
