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PROBLEM:
The length and width of a rectangular room are measured to be 3.955 ±0.005 m and 3.050 ± 0.005 m . Calculate the area of the room and its uncertainty in square meters.
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SOLUTION:
The average area of the room is
\begin{align*}
A & =l\times w \\
& =3.955\:\text{m}\times 3.050\:\text{m} \\
& =12.06\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}Compute for the percent uncertainties of each dimension.
\begin{align*}
\text{\%\:unc}_{width} & =\frac{0.005\:\text{m}}{3.050\:\text{m}}\times 100\%=0.1639\% \\
\text{\%\:unc}_{length} & =\frac{0.005\:\text{m}}{3.955\:\text{m}}\times 100\%=0.1264\:\%
\end{align*}The percent uncertainty in the area is the combined effect of the uncertainties of the length and width.
\text{\%\:unc}_{area}=0.1639\%+0.1264\%=0.2903\%The uncertainty in the area is
\delta _{area}=\frac{0.2903\:\%}{100\:\%}\times 12.06\:\text{m}^2=0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)Therefore, the area is
A=12.06\pm 0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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