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Math Help - [SOLVED] finding the aspect ratio of a rectangle

  1. #1
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    [SOLVED] finding the aspect ratio of a rectangle

    [I originally posted this as a HS level geometry problem but I'm now thinking it's more difficult than that.]

    It's been too long since I've had to figure something like this out. I'm hoping someone will be kind enough to help. I'm attaching a picture of what I'm working on as it will make it easier to describe. (I'll leave the image up until someone's given me an answer).



    I'm looking for the width (line dc) of this rectangle when the height (line de) is 5 inches (and for the formula needed to figure it out). Note that the blue line (da) and red line (ac) are equal in length and that angle abc is 90 degrees. Also note that line ec is a diagonal of the rectangle from the lower left corner to the upper right corner and line af runs from the middle of line dc to the lower right corner (f).

    Just to clarify:

    • Point a must bisect line dc.
    • Angle abc must be 90 degrees.
    • Points e, c and f must be in their respective corners.

    From experimenting a bit I know that line dc is about 7.1 inches. I'd like the formula for getting the exact length.

    Thank you in advance. I didn't think this would be that hard to figure out but I don't really know how to even approach solving it.
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  2. #2
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    Triangle abc is similar to fbe.
    => be/bc = ef/ac = dc/ac = 2

    Triangle abc is similar to ade.
    ac/ec = bc/dc. Let da = ac = x.
    x/ec = bc/2x
    2x^2 = ec * bc. Let bc=y => be=2y => ec=3y => 2x^2 = 3y^2.

    One last thing, from Pithagoras:
    (2x)^2 + 5^2 = (3y)^2. Both should give you the solution.

    Not very hard, but need to be careful.

    -O
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  3. #3
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    If the ratio ac:cf is the same as ed:dc (meaning triangle acf is just a smaller version of triangle edc) then it's pretty simple to show that the width of the rectangle is sqrt(50) and its aspect ratio (w/h) is sqrt(2).

    let x = the length of line dc
    ratio ac:cf is then = 5/(x/2) and ed:dc = x/5

    • 5/(x/2) = x/5 (<--- assumed)
    • 25/(x/2) = x
    • 25 = x(x/2)
    • 25 = x*x/2
    • 50 = x*x
    • x = sqrt(50)
    • sqrt(50)/5 = 1.41421356 = sqrt(2)

    So, is there a way of proving that ac:cf = ed:dc?
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  4. #4
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    "meaning triangle acf is just a smaller version of triangle edc"

    Ofcourse true. Try to see it from angles: a+c=90. Also a+f=90 => c=f (I am talking about angles) You can immediately say:
    acf is similar to edc.

    There are many ways of skinning this cat. Look at this one:
    bf^2=bc*be (why?)
    5^2 = y*(3y)
    Find y.
    Then find x. an so on...

    -O
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