Ancient Egyptian Method for getting an Area of Quadrilateral

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.

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.