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Geometry of Methane Question

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data

    In a methane molecule, determine the length of the distance between a hydrogen atom at A and the carbon atom at O (see diagram) in terms of the length of the edge (e) of the cube at four of whose corners the hydrogen atoms rest.

    Screen Shot 2016-04-06 at 10.22.09 AM.png

    2. Relevant equations

    pythagorean theorem?

    3. The attempt at a solution


    The first part of the problem asks to find AB (see diagram). AB= e(sqrt-2).

    The book gives the answer as AO=e(sqrt-3)/2, but I don't know how they got it.

    I'm sure this is simple! I'm just not having any luck.

    Thanks!
     
    Last edited: Apr 6, 2016
  2. jcsd
  3. Apr 6, 2016 #2

    gneill

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    Staff: Mentor

    Hi crastinus, you can UPLOAD your file and attach it to your post. Then you can insert it in the text where you want it. Use the UPLOAD button in the lower right corner of the edit pane.
     
  4. Apr 6, 2016 #3
    Ah. I see it now. Thanks! Edits made.
     
  5. Apr 6, 2016 #4

    gneill

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    Staff: Mentor

    From your image:
    upload_2016-4-6_10-22-57.png

    Note the plane image of the triangle in question. APO is a right triangle with AO as its hypotenuse. Can you find values for AP and OP?
     
  6. Apr 6, 2016 #5
    If I'm right that AB= e(sqrt2), then AP=e(sqrt2)/2.

    How to find AO? If I knew that this were a certain type of triangle, 45-45 or 30-60, I could determine AO. So, my question is: How can I determine AO knowing only one side and the right angle of triangle AOP. I know the various angles involved, but the point is to get AO in terms of e. That's what I'm not sure how to do.

    Thanks!
     
  7. Apr 6, 2016 #6

    gneill

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    Staff: Mentor

    You should be able to determine OP. Where is O located in the cube? Where is P located?
     
  8. Apr 6, 2016 #7
    Wow. OK. Yes. OP=e/2.

    And so, A0=(e/2)^2 + (e(sqrt2)/2)^2, which gives us AO=(e(sqrt3)/2)!

    Thanks. Sometimes I just don't see what I'm looking at.
     
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