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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- Sep 12th 2010, 06:28 PM #1

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- Sep 12th 2010, 07:10 PM #2

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- Sep 13th 2010, 12:09 PM #3

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

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

- Sep 13th 2010, 12:49 PM #4

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Hello, matgrl!

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

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

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

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

We have equilateral triangle $\displaystyle ABC$ with side $\displaystyle \,a.$

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

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

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

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

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

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

. . which is slightly smaller than the true area.

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

. . . . . about 1% as emakarov already stated.

- Sep 13th 2010, 05:10 PM #5

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- Sep 13th 2010, 05:15 PM #6

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- Sep 13th 2010, 07:22 PM #7

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