4. As shown in the right figure, in the square $A B C D$ with a side length of 12 cm, an isosceles triangle $P A B$ is constructed with $A B$ as the base and a leg length of 10 cm. Then the area of triangle $P A C$ is equal to  square cm.
Draw $\mathrm{PE} \perp \mathrm{AB}$ through point $\mathrm{P}$. Since triangle $\mathrm{PAB}$ is an isosceles triangle, $\mathrm{AE}=\mathrm{EB}=6 \mathrm{~cm}$. According to the Pythagorean theorem: $\mathrm{PE}^{2}=10^{2}-6^{2}=64=8^{2}$, so $\mathrm{PE}=8 \mathrm{~cm}$.
$$
\begin{array}{l}
S \triangle \mathrm{PAB}=12 \times 8 \div 2=48 \mathrm{~cm}^{2}, S \triangle \mathrm{PCB}=12 \times 6 \div 2=36 \mathrm{~cm}^{2} \\
\mathrm{~S} \triangle \mathrm{PAC}=48+36-12 \times 12 \div 2=12 \mathrm{~cm}^{2} 。
\end{array}
$$