Martin cut out a rectangle from paper, whose side lengths were in the ratio $3: 4$ and the circle circumscribed around this rectangle had a radius of $8 \mathrm{~cm}$. Then he folded the rectangle along the diagonal and a pentagon was formed. Calculate the perimeter of Martin's pentagon.

(Dillingerová)
In the picture, the situation is depicted - the description is based on the fact that the given figure is axially symmetric. For the perimeter of this pentagon, we have: $o=16+a+x+x+a$, where $a$ is the shorter side of the rectangle and $x$ is the segment on the longer side of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_04_17_384553f2501bf62a691ag-3.jpg?height=477&width=868&top_left_y=1886&top_left_x=617)

First, we calculate $a$:

We denote the sides of Martin's rectangle as $a=3 y$ and $b=4 y$. From the Pythagorean theorem, we then get that the diagonal of the rectangle measures $5 y$.

Since the center of the circle circumscribed around the rectangle lies at the intersection of its diagonals and its diameter is equal to the length of this diagonal, we have: $5 y=2 \cdot 8$, from which $y=\frac{16}{5} \mathrm{~cm}$. Substituting into the expressions for the lengths of the sides of Martin's rectangle, we get $a=9.6 \mathrm{~cm}$ and $b=12.8 \mathrm{~cm}$.

Now we calculate $x$:

For this, we will use the Pythagorean theorem again and start from the strongly marked triangle:

$$
\begin{aligned}
(12.8-x)^{2} & =x^{2}+9.6^{2}, \\
163.84-25.6 x+x^{2} & =x^{2}+92.16, \\
-25.6 x & =-71.68, \\
x & =2.8 \mathrm{~cm} .
\end{aligned}
$$

The perimeter of Martin's pentagon is therefore equal to

$$
o=16+9.6+2.8+2.8+9.6=40.8 \mathrm{~cm} .
$$