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Gre Math prep geometry problem 8C

  1. Jul 12, 2011 #1
    1. The problem statement, all variables and given/known data

    You have a parallelogram ABCD. You know one side has a length of 12. They have given a right triangle extended from one side of the parallelogram height 4 and base 2 hypotenuse is the side of the parallelogram. There is a diagonal that goes from A to C that they want you to find the length of. (the right triangle is off of the DC side of the parallelogram so I can't see a way to use the Pythagorean theorem here) Direct link to the pdf with the problems
    http://www.ets.org/Media/Tests/GRE/pdf/GREmathPractice.pdf" [Broken]

    2. Relevant equations
    The Pythagorean theorem and the area of a triangle (1/2 base times height) are all I can see as being useful here. May be others though that I don't see.

    3. The attempt at a solution
    The 4 2 right triangle can be used to find the length of the other side of the parallelogram which turns out to be 2[itex]\sqrt{5}[/itex] If this was a rectangle then since the diagonals would be equal this would be easy. Tried also using the area of the parallelogram and the fact that the diagonal makes two similar triangles but that only gives you the length of the side of the parallelogram you all ready know. At a loss now...any help very much appreciated!

    Thanks very very much! Any location for geometry problems for more practice (this type especially) would be great. Have a geometry book that I have worked through but it didn't have anything like this.
    Last edited by a moderator: May 5, 2017
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  3. Jul 12, 2011 #2


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    I might be misunderstanding the problem (can't find it in your link) since this seems awfully easy to me. The length of the long diagonal is sqrt(160) and the length of the short diagonal is sqrt(116)

    OOPS ... make that long is sqrt(212)
  4. Jul 12, 2011 #3


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    Info: OP refers to Problem 8 on p. 48 of the document at the link.
  5. Jul 12, 2011 #4


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    Hm ... how did you get from "8c" to "Prob 8 on p 48" ??? You're right of course so I guess I'm missing something.

    Anyway, my answer is correct for that figure.
  6. Jul 14, 2011 #5
    Sorry it seems easy for you. It is the 8th problem in the geometry section sorry I didn't make that clear enough. Can you explain HOW you got that answer? Got the one for the long diagonal, the short diagonal is what is giving me an issue.
  7. Jul 14, 2011 #6


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    I tossed my figure but as I recall, the long one is sqrt (4**2 + (10+2+2)**2) and the short one is sqrt(4**2 + 10**2). It's all just simple right triangles and you know the base and side of each so the hypotenuse is just that trivial.
  8. Jul 17, 2011 #7
    hmm still missing something. Unless you drew something in I don't see how you have the base and the side for the right triangle formed by the long diagonal. The short diagonal is the only one you have the right triangle drawn in outside of the parallelogram.

    Thanks for all your help, sorry it is taking me so long to see this.
  9. Jul 17, 2011 #8


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    Look for similar triangles (draw a couple that are obvious) and if you still can't get it, post again and I'll draw a figure.
  10. Jul 18, 2011 #9
    I think that hint did it, and I feel like an idiot now. To make sure I am on the right track, you can draw a straight line down from a to b in the INSIDE of the parallelogram this gives you two right triangles one that you use to find the length of the new side (lets call it b to e). You can find this length using the Pythagorean theorem. Then subtract this from twelve, to get the length of the triangle e to c. Then you can use the Pythagorean theorem again to get the diagonal length.

    Doable but not in a minute thirty...at least for me. Thanks for the help guys going to work this now see if I get the right answer then find some more geometry problems where you have to draw stuff in.
  11. Jul 18, 2011 #10


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    Yes, you've got it. I think whether or not you can do this kind of thing in 10 seconds or 5 minutes is a function of (1) how conversant are you with math basics and (2) how much practice do you have?

    For basic stuff like this, I have lots of both so it jumped right out at me, but I see no reason why you should feel bad that it didn't for you. I've been doing this sort of thing for 50+ years.
  12. Jul 18, 2011 #11
    Ah I've been teaching chemistry so ratios and algebra and linear equations are more my thing. Geometry (and statistics) seem to be the weak point. Got any good locations for other problems like this? The high school geometry text book I had didn't have that many this complicated. (though I am going to look again).

    What made me especially mad about this one was I could get 11 which was more difficult I thought. (there it is a rectangle so everything is 90 degrees which actually made it easier).

    Anyway thanks for all the help guys. Anyone have knowledge about the pHD in chemical education degree from purdue that is what I am going for here.
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