3. As shown in the figure, the side lengths of the two squares $\mathrm{ABCD}$ and $\mathrm{DEFG}$ are both integer centimeters, point $\mathrm{E}$ lies on line segment $\mathrm{CD}$, and $\mathrm{CE}<\mathrm{DE}$, line segment $\mathrm{CF}=5$ centimeters, then the area of pentagon $\mathrm{ABCFG}$ is $\qquad$ square centimeters.
3. Solution: $\mathrm{CF}=5$, and both $\mathrm{CD}$ and $\mathrm{DF}$ are integers. According to the Pythagorean theorem, $\mathrm{CE}=3, \mathrm{DF}=4, \mathrm{CD}=7$,

Therefore, the area of pentagon $\mathrm{ABCFG}$ is:
$$
4^{2}+7^{2}+\frac{1}{2} \times 4 \times 3=16+49+6=71 \text { (square centimeters) }
$$