Q:

Why is the Pythagorean Theorem used for error calculation?

They say that if $A = X \times Y$, with $X$ statistically independent of $Y$, then
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
I can't understand why that is so geometrically.
If $X$ and $Y$ are interpreted as lengths and $A$ as area, it is pretty easy to understand, geometrically, that
$$\Delta{A} = X\times\Delta{Y} + Y\times\Delta{X} + \Delta{X}\times\Delta{Y}$$
Ignoring the term $\Delta{X}\times\Delta{Y}$ and dividing the both sides by $A$ ($= X \times Y$), that expression becomes
$$\frac{\Delta{A}}{A} = \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$
which is different from
$$\frac{\Delta{A}}{A}=\sqrt{ \left(\frac{\Delta{X}}{X}\right)^2 + \left(\frac{\Delta{Y}}{Y}\right)^2 }$$
which looks like a distance calculation. I just can't see how a distance is related to $\Delta{A}$.
Interpreting $A$ as the area of a rectangle in a $XY$ plane, I do see that $\Delta{X}^2+\Delta{Y}^2$ is the how much the distance between two opposite corners of that rectangle varies with changes $\Delta{X}$ in $X$ and $\Delta{Y}$ in $Y$. But $\Delta{A}$ is how much the area, not that distance, would vary.

A:

The general formula for error propagation is:
$$\Delta f(x_1,x_2,\ldots)=\sqrt{(\frac{\partial f}{\partial x_1}\Delta x_1)^2  + (\frac{\partial f}{\partial x_2}\Delta x_2)^2 + \cdots}$$
where $\Delta m$ means "standard deviation of lots of repeated measurements of m".
Where does this come from? By calculus, when all the $x_i$s vary, it causes the following variation of $f$:
$$\delta f = \sum_i (\partial f / \partial x_i) \delta x_i$$
where $\delta x_i$ is the difference between this particular measurement of $x_i$ and its true value, and $\delta f$ is ditto for $f$. We are assuming that the errors are relatively small (ignore $\delta x_i \delta x_j$ terms etc.)
I think you had all this so far. The part that you're missing is:
For independent random processes, the variance of the sum is the sum of the variances.
The analogous statement is not true for standard deviations. It is only true for variance, i.e. standard deviation squared.
Since we want the standard deviation of $\delta f$, we need to add up the variances of $(\partial f / \partial x_i) \delta x_i$ and then take the square root. So we wind up with the formula that I wrote at the beginning.

