Difference between revisions of "2016 AMC 12B Problems/Problem 10"
(→Solution: Added a solution) |
(→Solution) |
||
Line 6: | Line 6: | ||
==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> | + | 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>\boxed{\textbf{(A)}\ 4}</math> |
+ | |||
Solution by I_Dont_Do_Math | Solution by I_Dont_Do_Math | ||
Revision as of 17:23, 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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.