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Triangle Area Based on Segment Division

  1. Sep 25, 2004 #1
    Hi all,

    I'm working on this problem in my Geometry class, and am having a bit of trouble coming to a conclusion. The problem as given is (I've attached the figure to this thread):

    From this information, I have deduced that triangle DCB is right isosceles, with DC = CB. Also, since <DBC and <DBA are supplementary (by LPP), <DBA = 135, and thus, by Triangle Sum Theorem, <BDA = 15. Then, by AAP, this means that <CDA = 60, and thus triangle ADC is a 30-60-90 right triangle.

    I've gotten to this point, and I know the ratios of this special triangle, but I'm having trouble forming the right equations. I was wondering if anyone could give me a lead onto which I could follow.

    Thanks a lot! :)

    Attached Files:

  2. jcsd
  3. Sep 25, 2004 #2
    Nevermind, I figured it out! :)
    For anyone who wants to know how I solved it, this is how I did it:

    I assigned "x" as the length DC and CB. Since triangle ADC is a 30-60-90 right triangle, the hypotenuse (DA) must be 2x. Then, using Pythagorean theorem, I derived the following equation:

    [itex](2x)^2 - x^2 = (x + 3 - \sqrt{3})^2[/itex]
    [itex]4x^2 - x^2 = (x + 3 - \sqrt{3})^2[/itex]
    [itex]3x^2 = (x + 3 - \sqrt{3})^2[/itex]
    [itex]x\sqrt{3} = x + 3 - \sqrt{3}[/itex]
    [itex]x\sqrt{3} - x = 3 - \sqrt{3}[/itex]
    [itex]x(\sqrt{3} - 1) = 3 - \sqrt{3}[/itex]
    [itex]x = \frac{3 - \sqrt{3}}{\sqrt{3} - 1}[/itex]
    [itex]x = \sqrt{3}[/itex]

    From there, I used x^2/2 = area of triangle BCD, so I got 3/2.

    By the way, how do I do multi-line LaTeX equations? I tried using the "\\" symbol, both at the end of the previous line and at the start of the next line, and neither worked? (I ended up using multiple [ itex ] tags to get the effect I wanted.)
    Last edited: Sep 25, 2004
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