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Math Help - Gerbert

  1. #1
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    Gerbert

    Gerbert (ca. 950-1003), who became Pope Sylvester 2nd, claimed that the area of an equilateral triangle of side a was (a/2)(a-a/7). Show he was wrong, but close.


    How can I prove that he is wrong, but close?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by matgrl View Post
    Gerbert (ca. 950-1003), who became Pope Sylvester 2nd, claimed that the area of an equilateral triangle of side a was (a/2)(a-a/7). Show he was wrong, but close.


    How can I prove that he is wrong, but close?
    What does "a" denote?

    Compare the actual area of an equilateral triangle with what Gerbert's formula predicts.

    CB
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  3. #3
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    The correct formula for the square of a triangle involves \sqrt{3}. If Gerbert were correct, then \sqrt{3} would be a rational number.

    According to my calculations, the real square is greater than Gerbert's estimate by only 1%.
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  4. #4
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    Hello, matgrl!

    I'll do this one "from scratch" . . .



    \text{Gerbert (ca. 950-1003), who became Pope Sylvester 2nd, claimed}

    \text{that the area of an equilateral triangle of side }a\text{ is: }\:\frac{a}{2}\left(a-\frac{a}{7}\right)

    \text{Show he was wrong, but close.}
    Code:
                  A
                  *
                 /|\
                / | \
               /  |  \
            a /   |   \ a
             /    |h   \
            /     |     \
           /      |      \
        B * - - - + - - - * C
                  D  a/2

    We have equilateral triangle ABC with side \,a.

    The altitude \,AD bisects the base, so DC = \frac{a}{2}

    Pythagorus says: . h^2 + \left(\dfrac{a}{2}\right)^2 \:=\:a^2 \quad\Rightarrow\quad h^2 + \dfrac{a^2}{4} \:=\:a^2

    . . . . . . . . . . . . . h^2 \:=\:\dfrac{3a^2}{4} \quad\Rightarrow\quad h \:=\:\dfrac{\sqrt{3}}{2}a


    The area of the triangle is: . A \;=\;\frac{1}{2}bh \;=\;\frac{1}{2}(a)\left(\dfrac{\sqrt{3}}{2}a\righ  t) \;=\;\dfrac{\sqrt{3}}{4}a^2


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Gerbert's formula has: . \dfrac{3}{7}a^2

    . . which is slightly smaller than the true area.


    The relative error is: . \dfrac{\frac{\sqrt{3}}{4} - \frac{3}{7}}{\frac{\sqrt{3}}{4}} \;=\;0.010256681

    . . . . . about 1% as emakarov already stated.
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  5. #5
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    Could you further explain this, I am not sure where you are getting the square root of 3 from. I am not sure where to even start off in this problem.
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  6. #6
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    This is wonderful Thank you! How did you know to substiute the square root of 3 over 2 for a. Where did this square root of 3 come from?
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by matgrl View Post
    This is wonderful Thank you! How did you know to substiute the square root of 3 over 2 for a. Where did this square root of 3 come from?
    Look up (or better yet, work out) the area of an equilateral triangle of side a

    CB
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