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Magoosh GMAT Instructor
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Re: Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4) [#permalink]
abhinav11 wrote:
mikemcgarry wrote:
Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4), K = (–2, 1), L = (6, 7), and M = (6, –4). What is its perimeter?

(A) 34
(B) 36
(C) 38
(D) 40
(F) 42


For a discussion of how to handle with elegance both this question and distance in the x-y plane in general see this post. Other experts, feel free to share any other insights about elegance in x-y plane problems.



Is there any shortcut to these type of questions than just calculating distance between two points and then adding all of them, that seems bit of time killer for me. But if we dont have any alternative we have to follow the approach.

Regards,
Abhinav


I think all sides are pretty straight forward. The only side that needs calculation is the one that is not parallel to the axis (y and x)
and it can be found using the triangle formula (draw a triangle that originates in the vertices to the base.)
you get 10+8+5+11=34
Magoosh GMAT Instructor
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Re: Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4) [#permalink]
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abhinav11 wrote:
mikemcgarry wrote:

For a discussion of how to handle with elegance both this question and distance in the x-y plane in general see this post. Other experts, feel free to share any other insights about elegance in x-y plane problems.


Is there any shortcut to these type of questions than just calculating distance between two points and then adding all of them, that seems bit of time killer for me. But if we dont have any alternative we have to follow the approach.

Regards,
Abhinav

Dear Abhinav,
The reason I posted this problem was precisely to demonstrate that an elegant solution was possible. Please follow the link given in the problem to a page that demonstrates this elegant time-saving solution.
Mike :-)
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Re: Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4) [#permalink]
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Re: Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4) [#permalink]
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