4. (10 points) As shown in the 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 $\qquad$ square centimeters.
【Analysis】Draw $P E \perp A B, P F \perp B C$, the area of triangle $P A C$ = the area of triangle $P A B$ + the area of triangle $P B C$ - the area of triangle $A B C$.
【Solution】Solution: Since $10^{2}-(12 \div 2)^{2}=64$, then $P E=8, P F=6$,
Therefore, the area of triangle $P A B$ is: $12 \times 8 \div 2=48$ (square centimeters),
The area of triangle $P B C$ is: $12 \times 6 \div 2=36$ (square centimeters),
The area of the shaded part is: $48+36-12 \times 12 \div 2$,
$=84-72$,
$=12$ (square centimeters);
Answer: The area of triangle $P A C$ is 12 square centimeters.
Therefore, the answer is: 12.