Ancient Egyptian Method for getting an Area of Quadrilateral

Question

I am currently learning the history of calculus, and I am really curious about the Ancient Egyptian way to get an Area of Quadrilateral. So, I hope I could get some enlightenment.

Here is the Egyptian Way:

Consider Quadrilateral with 4 different lengths a,b,c,d. (a,c are facing each other, and b,d are facing each other.) (Each set of sides may not be parallel.)

They calculated the area as follows, $$A=\frac{a+c}{2}\times \frac{b+d}{2}$$

However, we know this is wrong, but I want to show if this method is overestimated or underestimated, and why.

P.S. I think they tried this way because getting an Area of rectangular is the easiest way, and they decided to make rectangular with lengths of averages of two opposite sides. And this new area somehow looks similar (close) to the original area.

Answer 1

Connect the vertices $A$ and $C$ to observe the following: $$ab+cd \geq ab\sin(x)+cd\sin(y)=2\bigtriangleup ABC+2\bigtriangleup ACD=2\cdot Area(ABCD)$$ Likewise: $$bc+ad \geq2\cdot Area(ABCD)$$ So, this method always overestimates. Also, note that the equality occurs only when quadrilateral $ABCD$ is a rectangle.

Answer 2

Let's see on a rhombus how much the area can be off.

Let call $c$ the side of the rhombus. Its area is half the product of the diagonals, so, with Pythagoras' theorem, if the half-diagonals are $a$ and $b$,

$$c^2=a^2+b^2$$

And the area is

$$A=2ab=2a\sqrt{c^2-a^2}$$

While the approximate area is

$$A'=c^2$$

Hence

$$\frac{A}{A'}=2\frac{a}{c}\sqrt{1-\frac{a^2}{c^2}}=2\sqrt{\frac{a^2}{c^2}\left(1-\frac{a^2}{c^2}\right)}=2\sqrt{x(1-x)}$$

with $x=\frac{a^2}{c^2}$.

The last expression is maximum for $x=\frac12$, and equals then $1$.

This ratio tends to $0$ as $\frac{a}{c}$ tends to $0$ or $1$, that is, when the rhombus is nearly flat. One can expect this: the approximate area is constant, while the exact area varies from $0$ to a maximum of $c^2$ obtained for $a=\frac{c}{\sqrt2}$. For the rhombus, the approximate area is then always larger or equal to the true one, and is correct in only one case, i.e. when it's a square.

Answer 3

Wikipedia article Quadrilateral, section Area says that the area of a convex figure is always less than or equal to the Egyptian estimate: $$A\le\tfrac 14(a+c)(b+d)$$ Of course, a concave one has even less area.