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Math Help - Acute Angled Triangle...difference of two sides

  1. #1
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    Acute Angled Triangle...difference of two sides

    I am totally clueless on this. I have tried Pythagoras theorem and tried taking a lot of ratios but don't seem to get anywhere:

    The length of the sides of the acute angled triangle ABC are x-1, x and x+1. BD is perpendicular to AC. Then CD-DA equals

    (a) x/8
    (b) x/9
    (c) 2
    (d) 4

    ?
    Attached Thumbnails Attached Thumbnails Acute Angled Triangle...difference of two sides-34.png  
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  2. #2
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    Re: Acute Angled Triangle...difference of two sides

    OK guys I got it!

    The answer is four.

    First we equate

    cos C = CD/(x+1) and cos C = [ (x+1)^2 + x^2 - (x-1)^2 ] / [2(x+1)(x)]

    to find an expression for CD in terms of x then

    cos A = AD/(x-1) and cos A = [ (x-1)^2 + x^2 - (x+1)^2 ] / [2(x-1)(x)]

    to find an expression for DA in terms of x. Then subtracting CD-DA gives 4.
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  3. #3
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    Re: Acute Angled Triangle...difference of two sides

    Quote Originally Posted by cosmonavt View Post
    OK guys I got it!

    The answer is four.

    First we equate

    cos C = CD/(x+1) and cos C = [ (x+1)^2 + x^2 - (x-1)^2 ] / [2(x+1)(x)]

    to find an expression for CD in terms of x then

    cos A = AD/(x-1) and cos A = [ (x-1)^2 + x^2 - (x+1)^2 ] / [2(x-1)(x)]

    to find an expression for DA in terms of x. Then subtracting CD-DA gives 4. <--- correct!
    Here is a slightly different approach:

    1. I've modified your sketch a little bit (see attachment)

    2. Using Pythagorean theorem you'll get:

    (x-1)^2 - (x-k)^2 = (x+1)^2-k^2

    which simplifies to

    2kx-2x=x^2+2x~\implies~k=\frac12 x +2

    3. Now determine using the result from #2

    k - (x-k)

    which yields 4.
    Attached Thumbnails Attached Thumbnails Acute Angled Triangle...difference of two sides-hypotendifferenz.png  
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  4. #4
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    Re: Acute Angled Triangle...difference of two sides

    Yeah! That's way more simpler. Wonder why I didn't get that in the first place.
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