1. ## 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?

2. Originally Posted by matgrl
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

3. 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%.

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