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Math Help - Heron's Formula for equilateral triangle

  1. #1
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    Heron's Formula for equilateral triangle

    Hello everyone and thanks in advance for any help!

    So I want to reduce Heron's formula to simplify equilateral triangles. But I'm getting caught up near the end:

    Formula:

    S=(a+b+c)/2, In this case S=3a/2 since a=b=c

    A=(S(S-a)(S-a)(S-a))^1/2

    So, ((3a/2)(3a/2-a)^3)^1/2

    This is where I seem to lose it.

    My calculator and wolfram agree that (3a/2)(3a/2-a)^3=(3a^4)/16
    However, I am getting (57a^4)/16. I am cubing everything ((3^3a^3)/2^3-a^3) and multiplying that by 3a/2. I'm positive that my problem lays in cubing.

    ((3^3a^3)/2^3-a^3)=(27a^3)/8-a^3? wolfram claims a^3/8. I don't understand that at all! Where did that 3^3 go?? It has to be something simple that I'm missing or not doing correctly.

    This emote: is definitely the most accurate representation of my current state.

    Thanks for the help!
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  2. #2
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    Notice that \displaystyle \frac{3a}{2} - a = \frac{3a}{2} - \frac{2a}{2} = \frac{a}{2}.

    So \displaystyle \sqrt{\frac{3a}{2}\left(\frac{3a}{2} - a\right)^3} = \sqrt{\frac{3a}{2}\left(\frac{a}{2}\right)^3}

    \displaystyle = \sqrt{\frac{3a}{2}\left(\frac{a^3}{8}\right)}

    \displaystyle = \sqrt{\frac{3a^4}{16}}

    \displaystyle = \frac{\sqrt{3}\,a^2}{4}.
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  3. #3
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    Quote Originally Posted by Bowlbase View Post
    So I want to reduce Heron's formula to simplify equilateral triangles.
    Why get equilateral triangle area from Heron's? Too much unecessary work...
    Divide equilateral triangle in 2 right triangles, each with hypotenuse = a and one leg = a/2 : get it?
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  4. #4
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    Thank you! I knew it would be something simple that I wasn't doing correctly.

    Wilmer, the point of the exercise wasn't to actually use the formula but to practice such problems. I appreciate the input though!
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