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Math Help - Real Hard Geo Question, need help

  1. #1
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    Real Hard Geo Question, need help

    Let ABC be a triangle with right angle C. The mid-point of the hypotenuse AB is twice as far from the side BC as it is from side AC. Find the length of the side AC, if AB = 10cm.
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  2. #2
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    Wording

    The wording of your question is a bit confusing. By the distance from midpoint of AB to BC and AC, do you mean the distance DE and DF as I have in my diagram?

    This is how I did the calculation, if this is wrong, just let me know.

    Answer
    Let DE=x, DF=2x
    As point, by letting \angle{B}=\theta and \angle{A}=90-\theta, we see that the triangles AED and DFB are the same. Thus
    AE=EC=2x, CF=FB=x

    Calculate for x with Pythagoras' theorem, and you'll find AC.

    If my assumption is wrong, just let me know.
    Attached Thumbnails Attached Thumbnails Real Hard Geo Question, need help-assumption.jpg  
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  3. #3
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    wait give me a minute.
    can you give your final answer as well.
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  4. #4
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    Quote Originally Posted by Vinsanity View Post
    Let ABC be a triangle with right angle C. The mid-point of the hypotenuse AB is twice as far from the side BC as it is from side AC. Find the length of the side AC, if AB = 10cm.

    The 'Pythagoras' reference:
    The distance BC is 2x and the disance CA is 4x

     (2x)^2 + (4x)^2 = 10^2

     4x^2 + 16x^2 = 10^2

     x^2 (4+16) = 10^2

     x^2 = \frac {10^2}{(4+16)}

     x = \sqrt { \frac {10^2}{(4+16)} }


    NOTE:
    This is being done ( not specificly or exactly for your benefit ) so that I can practice using Latex.
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  5. #5
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    Why did you result using 2x in the first place
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  6. #6
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    Quote Originally Posted by Vinsanity View Post
    Why did you result using 2x in the first place
    I'm sorry that wasn't clear.
    My apologies.

    Obviously my attempt at clarity wasn't.
    You probably used this to answer the question anyway:



     \bar{AC} = 10 \times \sin \left ( \tan^{-1} \left ( \frac{2x}{x} \right ) \right )

    Again, sorry for the obfuscation.
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  7. #7
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    Quote Originally Posted by aidan View Post
    I'm sorry that wasn't clear.
    My apologies.

    Obviously my attempt at clarity wasn't.
    You probably used this to answer the question anyway:



     \bar{AC} = 10 \times \sin \left ( \tan^{-1} \left ( \frac{2x}{x} \right ) \right )

    Again, sorry for the obfuscation.
    You are very humble, though how can you use trig?
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  8. #8
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    Quote Originally Posted by Vinsanity View Post
    ..., though how can you use trig?
    I can use trig because I had a very good 6th grade teacher. The teacher was adamant that we know what a trig function did -- not merely able to write that tangent = sine/cosine or that sine = opposite / hypotenuse.

    I can use trig because I practiced solving and creating trig problems.

    There are many reasons that I can use trig, but those are the primaries.
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