Answer
$8 \%$
Work Step by Step
Let the area of the rectangle with width $B$, length $L$ and area A:
\[
B \times L=A
\]
Total differential $d A$ of $A$ is given by
\[
B d L+L d B=d A
\]
$\frac{L d B}{A}+\frac{B d L}{A} =\frac{d A}{A}\quad$
\[
\frac{d L}{L}+\frac{d B}{B}=\frac{d A}{A}
\]
Thus, we are given the errors in length and width as:
\[
\begin{array}{l}
0.05=\frac{\Delta B}{B} | \\
0.03=\left|\frac{\Delta L}{L}\right|
\end{array}
\]
Thus:
\[
\begin{aligned}
\frac{\Delta A}{A}|\approx| \frac{d A}{A} | &=\left|\frac{\Delta B}{B}+\frac{\Delta L}{L}\right| \\
& 0.05+0.03 =\frac{\Delta A}{A} |\\
&\left|\frac{\Delta A}{A}\right|=0.08
\end{aligned}
\]
