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Geometry problem (kinda)

  1. Sep 30, 2004 #1
    First off, my attached image will make things make sense. I have a triangle (with a 90 degrees corner), I know that the height is 20' and that the length is more than 58'. I need to find out what the length is to the left of the 6' height mark (trying to find the length of the red line). Can anyone help me?

    Attached Files:

  2. jcsd
  3. Sep 30, 2004 #2


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    I'll call the distance to be found x
    Use similar triangles the 20 is to the 6 as x+58 is to x. From here you can solve for x.
  4. Sep 30, 2004 #3
    Excuse me for being dumb, but I still can't figure it out.

    c^2 = 20^2 + (58+x)^2 gives me really hard to work with numbers, I don't think i'm doing it right.
  5. Oct 1, 2004 #4


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    Don't use the Pythagorean theorem, us "similar triangles". Since the two triangles, the small one on the left and the large triangle, have the same angles, they are similar and their sides have the same ratios: x/6= (58+x)/20.

    That's much easier to solve!
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