A1. The sides of the squares are long $|A B|=\frac{8}{\sqrt{2}} \mathrm{~cm}$ and $|B E|=\frac{11}{\sqrt{2}} \mathrm{~cm}$. Since point $P$ is the midpoint of diagonal $B F$, the area of triangle $A P F$ is equal to the area of triangle $A B P$, which is equal to

$$
\frac{|A B| \cdot \frac{|B G|}{2}}{2}=\frac{\frac{8}{\sqrt{2}} \cdot \frac{11}{2 \sqrt{2}}}{2}=\frac{88}{8}=11 \mathrm{~cm}^{2}
$$
A1. The diagonals of squares $A B C D$ and $B E F G$ are 8 cm and 11 cm long, respectively. Point $P$ is the intersection of the diagonals of square $B E F G$ (see figure). How many square centimeters is the area of triangle $A P F$?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-10.jpg?height=317&width=486&top_left_y=955&top_left_x=1413)