Calculate the percentage error of the area of the floor given its length and width measurements.

The dimensions are given to the nearest 10 cm, which is 0.1 m.
The absolute error for each measurement is half of the precision.
Absolute error for length ($\Delta L$) = $\frac{0.1}{2} = 0.05$ m.
Absolute error for width ($\Delta W$) = $\frac{0.1}{2} = 0.05$ m.

Stated length ($L$) = 12.5 m.
Stated width ($W$) = 9