Area of "Scalene" rectangles My friend told me that from antiquity land revenue officials compute area of "reasonably" rectangular ( diagonal no matter) fields to assess tax taking the opposite sides average as .. 
$$ A = \dfrac{a+c}{2} \cdot \dfrac {b+d}{2} $$ 
What could be the (relative) error?
 A: I will approximate a "scalene" rectangle as a trapezium, that is a quadrilateral with two opposite sides which are parallel. I chose $a$ and $c$ as bases. Let's denote the angles adjacent to the greatest base $\alpha$ and $\beta$. We have that
$$\alpha = \frac\pi2 - \theta,\qquad\beta = \frac\pi2 - \phi,$$
where
$$\theta,\phi \in \left[0, \frac\pi{36}\right].$$
The area of the trapezium is therefore
$$A = \frac12(a + c)h = \frac12(a + c)b\sin\alpha = \frac12(a + c)d\sin\beta.$$
Let $A_0$ be the approximated area; then the relative error is
$$\delta A = \frac{A_0 - A}A = \frac{A_0}A - 1,$$
or
$$\begin{align}
\delta A &= \frac{(a + c)(b + d)}4\cdot\frac2{(a + c)b\sin\alpha} - 1 =\\[1.5ex]
&= \frac{b + d}{2b\sin\alpha} - 1 = \frac{b + b\frac{\sin\alpha}{\sin\beta}}{2b\sin\alpha} - 1 =\\[1.5ex]
&= \frac{\sin\alpha + \sin\beta}{2\sin\alpha\sin\beta} - 1 =\\[1.5ex]
&= \frac12(\sec\theta + \sec\phi - 2)
\end{align}$$
I don't know how to maximize this function only over the specified interval. However, with numerical methods, I computed the absolute maximum to be, approximately,
$$\delta A_{\text{max}} = 0.0038198375433473597,$$
which seems quite small. I would be glad if someone else verified my result.
