Answer
(a) The area is $ A = 14.44 \pm 0.095 \,\textrm{cm}^2 $.
(b) The fractional uncertainty in area is $0.66 \,\%$, while the sum of the fractional uncertainties in lengths is $0.13 \,\% + 0.53 \,\% = 0.66 \,\%$.
Work Step by Step
(a) The area is computed as
$$\begin{align} A &= \left( 7.60 \,\pm 0.01 \,\textrm{cm}\right) \times \left( 1.90 \,\pm 0.01 \,\textrm{cm} \right) \\
&= \left( 7.60 \,\textrm{cm} \right) \left( 1.90 \,\textrm{cm} \right) \pm \left( 7.60 \,\textrm{cm} \right) \left( 0.01 \,\textrm{cm} \right) \\
&\qquad \pm \left( 0.01 \,\textrm{cm} \right) \left( 1.90 \,\textrm{cm} \right) \pm \left( 0.01 \,\textrm{cm} \right) \left( 0.01 \,\textrm{cm} \right)
\end{align}
$$ The last term is negligible in essence and can be dropped from the calculations. Simplifying, we have $$ A = 14.44 \pm 0.095 \,\textrm{cm}^2 $$
(b) From this result, we can calculate the fractional uncertainty in area as $$
\dfrac{0.095 \,\textrm{cm}^2}{14.44 \,\textrm{cm}^2} = 0.66 \,\% $$ On the other hand, the fractional uncertainly in length is computed as $$ \dfrac{0.01 \,\textrm{cm}}{7.60 \,\textrm{cm}} = 0.13 \,\% $$ and the fractional uncertainty in width as $$ \dfrac{0.01 \,\textrm{cm}}{1.90 \,\textrm{cm}} = 0.53 \,\% $$ The sum of these fractional uncertainties is $$
0.13 \,\% + 0.53 \,\% = 0.66 \,\% $$ showing the agreement with the fractional uncertainty in area.
