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Thread: Deriving a generic formula:

  1. #1
    Jun 2010

    Deriving a generic formula:

    Hello MHF community,

    For my first post to this website, leave it to me to ask a very silly question. I am going to write what I know (which is incorrect). I am just missing some basic rules on how to solve equations to make it work, and I am hoping you can fill it in.

    I am trying to derive a generic formula for finding the altitude for an equilateral triangle. I know this:

    I know this is incorrect because when I graph the equilateral triangle, the height of the point does not fall where my formula suggests it will. So I went looking on the internet to find what I was doing wrong. However, the internet has just the answer to my question. Not the reason why my formula is wrong. Can you please help?
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  2. #2
    MHF Contributor undefined's Avatar
    Mar 2010
    You forgot to square (x/2), and I'm not sure how you did the last step. Also the line on the radical sign doesn't extend over the x in the second-to-last-line, don't know if that was intentional.

    Should be

    $\displaystyle a = \sqrt{x^2-\left(\frac{x}{2}\right)^2}$

    $\displaystyle a = \sqrt{x^2\left(1-\frac{1}{4}\right)}$

    $\displaystyle a = x\sqrt{\frac{3}{4}}$

    $\displaystyle a = \left( \frac{\sqrt{3}}{2}\right)x$

    Edit: By the way, hello and welcome to Math Help Forum!
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