We measure two quantities as $\mathrm{A}=1.0 \mathrm{m} \pm 0.2 \mathrm{m}, \mathrm{B}=2.0 \mathrm{m}\pm 0.2 \mathrm{m}$. We should report correct value for $\sqrt{\mathrm{AB}}$ as:

Here, $\mathrm{A}=1.0 \mathrm{m}\pm 0.2 \mathrm{m}, \mathrm{B}=2.0 \mathrm{m} \pm 0.2 \mathrm{m}$

$$\begin{gathered} \mathrm{AB}=\left(1.0 \mathrm{m}\right)\left(2.0 \mathrm{m}\right)=2.0 {\mathrm{m}}^2 \\ \sqrt{\mathrm{AB}}=\sqrt{2.0}\mathrm{m}=1.414 \mathrm{m} \end{gathered}$$

Rounding off to two significant figures, we get:

$$\begin{gathered} \sqrt{\mathrm{AB}}=1.4 \mathrm{m} \\ \frac{∆\sqrt{\mathrm{AB}}}{\sqrt{\mathrm{AB}}}=\frac{1}{2}\left(\frac{∆\mathrm{A}}{\mathrm{A}}+\frac{∆\mathrm{B}}{\mathrm{B}}\right)=\frac{1}{2}\left(\frac{0.2}{1.0}+\frac{0.2}{2.0}\right)=\frac{0.3}{2} \\ ∆\sqrt{\mathrm{AB}}=\frac{0.3}{2}\times \sqrt{\mathrm{AB}}=\frac{0.3}{2}\times 1.414=0.212 \mathrm{m} \end{gathered}$$

Rounding off to one significant figure, we get:

$∆\sqrt{\mathrm{AB}}=0.2 \mathrm{m}$

The correct value for $\sqrt{\mathrm{AB}}$ is $1.4 \mathrm{m} \pm 0.2 \mathrm{m}$.