Difference between revisions of "2016 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
+ | By distance formula we have <math>(a-b)^2+(b-a)^2*2*(a+b)^2 = 256</math>. SImplifying we get <math>(a-b)(a+b) = 8</math>. Thus <math>a+b</math> has to be a factor of 8 and the only answer that's a factor of 8 is <math>\textbf{(A)}\ 4</math> | ||
+ | Solution by I_Dont_Do_Math | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:21, 21 February 2016
Problem
A quadrilateral has vertices , , , and , where and are integers with . The area of is . What is ?
Solution
By distance formula we have . SImplifying we get . Thus has to be a factor of 8 and the only answer that's a factor of 8 is Solution by I_Dont_Do_Math
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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All AMC 12 Problems and Solutions |
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