65. Two squares $A B C D$ and $B E F G$ with side lengths of $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively are placed side by side as shown in the figure. Then the area of the shaded part is $\qquad$
Answer: $32 \mathrm{~cm}^{2}$.
Solution: Connect $C F$.
Since
$E F / / B C$,

Therefore,
the area of $\triangle C P F$ = the area of $\triangle C P E$.

Thus, the area of the shaded part in the figure $=$ the area of $\triangle A C F$
$$
\begin{array}{l}
=8^{2}+6^{2}-\frac{1}{2} \times 8^{2}-\frac{1}{2}(8+6) \times 6+\frac{1}{2}(8-6) \times 6 \\
=32
\end{array}
$$