# College Physics Problem 1.27

## Solution:

The average area of the room is

$\displaystyle A=l\times w$

$\displaystyle A=3.955\:m\times 3.050\:m$

$\displaystyle A=12.06\:m^2$

Compute for the percent uncertainties of each dimension.

$\displaystyle \%\:unc_{width}=\frac{0.005\:m}{3.050\:m}\times 100\%=0.1639\%$

$\displaystyle \%\:unc_{length}=\frac{0.005\:m}{3.955\:m}\times 100\%=0.1264\:\%$

The percent uncertainty in the area is the combined effect of the uncertainties of the length and width.

$\displaystyle \%\:unc_{area}=0.1639\%+0.1264\%=0.2903\%$

The uncertainty in the area is

$\displaystyle \delta _{area}=\frac{0.2903\:\%}{100\:\%}\times 12.06\:m^2=0.035\:m^2$

Therefore, the area is

$\displaystyle A=12.06\pm 0.035\:m^2$