Solution

By rotating the square $90^{\circ}$ counterclockwise, we can conclude that triangle $A B Q$ is equal to triangle $B P C$, and consequently, $\angle B A Q=\angle P B C$. Furthermore, by rotating the square $90^{\circ}$ clockwise, we can also conclude that triangle $A D P$ is equal to triangle $D Q C$ and that $\angle Q D C=\angle D A P$. Therefore,

$$
\begin{aligned}
& \angle P B C+\angle Q A P+\angle Q D C= \\
& \angle B A Q+\angle Q A P+\angle P A D=90^{\circ} .
\end{aligned}
$$
In the drawing below, each square in the figure has a side length of $1 \mathrm{~m}$. Determine the sum of the angles $\angle P B C+\angle Q A P+\angle Q D C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_ffb48c2069f209c1311cg-05.jpg?height=552&width=534&top_left_y=1921&top_left_x=841)

#