Propagation of uncertainties leading to lower error in calculated product than measured quantities

Question

I was doing a propagation of uncertainty on the following numbers, and I got a lower error from the calculated quantity than the measured quantities. So why is it that my compounded error smaller than the individual errors it is made up of?
The two measured quantities are $A = 1.0 pm 0.2 hspace{0.1cm} m$ and $ B = 2.0 pm 0.2  hspace{0.1cm}  m$. I am supposed to find $ sqrt{AB}$. I calculated it to be $1.41 pm 0.158  hspace{0.1cm} m$. I somehow end up getting a lower error than my beginning errors!
I'm pretty sure my work is right (I verified it here). Am I getting a lower error because I'm using too many digits to report the calculation (i.e. it's a sig fig mistake)? Even if it is, why is it theoretically possible to get a lower error from a function having the same dimensions than the errors I began with?

ProfRob · Accepted Answer

You get a smaller error because $sqrt{AB}$ is the geometric mean of the two measurements, i.e. a form of averaging.
When you take the average of a set of numbers, the uncertainty on the average is smaller than the uncertainty in the individual measurements (if the measurements have similar uncertainties), and so it is here.
$sqrt{AB}$ will always be in between $A$ and $B$. It is this centralising tendency that results in a smaller standard deviation.

Propagation of uncertainties leading to lower error in calculated product than measured quantities

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