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Homework Help: Stuipd Simple Geometry Question

  1. Dec 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Ok, I need to find the dimensions of a triangle. I have the length of the hypotenuse(sp?) and a ratio of the other sides. And yes, in case you are wondering, I am figuring out TV dimensions. :-/

    2. Relevant equations
    If using a+b+c, I know c, and the values of A and B are a 16:9 ratio, whats the equation to solve?

    3. The attempt at a solution
    Dunno, that's why I am here. Forgive the simple question, but if you dont ask you wont know. TIA.
  2. jcsd
  3. Dec 29, 2009 #2


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    I'm assuming you're looking for the values of a and b.

    If you have the the lengths a:b in the ratio 16:9 then you have:


    and you're also using a right-angled triangle, so use pythagoras' theorem:


    So all you need to do is solve these two equations simultaneously since there are two variables to find (c is a constant since you know what it is).
  4. Dec 29, 2009 #3
    Forgive me but my algebra days are long gone. I remember some things but I couldnt figure out how to solve those two equations together. I just kept going round in circles.
    So there has to be an easier way than what I figured out.
    9^2 + 16^2 = c^2
    81 + 256 = 337
    then I took those number and made my own ratio
    337 / 256 = 1.31640625
    337 / 81 = 4.16049382
    take those answers and use a different value for c, lets say 32
    32 * 32 = 1024
    1024 / 1.31640625 = 777.8753709
    777.8753709^1/2(sqrt) = 27.890417 which equals b

    1024 / 4.16049382 = 246.1246290
    246.1246290^1/2(sqrt) = 15.6883596 which equals a

    Then I just played around and found that a 15 by 28 measurement gives 31.7 which is actually what a 32" TV is measured at.

    Again I say, there has to be an easier way.
  5. Dec 29, 2009 #4
    I couldn't quite follow everything you did, and there is a more straightforward way, so:

    The ratio of the length and width of the TV is 16:9 , or let a/b = 16/9. We don't know what a or b are, so let's not assume or work with them as 16 or 9.
    [tex]\frac{a}{b} = \frac{16}{9} \Longrightarrow a = \frac{16b}{9}[/tex]
    Using the Pythagorean theorem for a TV with length and width a and b, and a diagonal length c of 32, we have a2 + b2 = 322 = 1,024.

    Substitute a = 16b/9 into the previous equation to get
    [tex]\left(\frac{16b}{9}\right)^2 + b^2 = 1024[/tex]

    Solve for b, then substitute that value into a = 16b/9 to get a, and you'll have the width and length of a 32" TV screen.
  6. Dec 30, 2009 #5


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    I am quite astounded at your approach to the problem. Especially since it simply worked :smile:
    Such methods were probably the logic used before algebraic equations had been "invented".
  7. Dec 30, 2009 #6
    I was trying to do that, but I couldnt remeber how to square a fraction with a variable in it. I think I did figure it out, finally.

    Welcome to most of my mathmatics career. I can get the right answer, just not the conventional way. That caused a few issues with certain teachers.
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