A paper rectangle with dimensions $48 \mathrm{~mm}$ and $64 \mathrm{~mm}$ was folded along the diagonal, resulting in a pentagon. $\mathrm{O}$ By how many $\mathrm{mm}^{2}$ is the area of the resulting pentagon smaller than the original rectangle?

(Dillingerová)
The pentagon thus formed has a smaller area than the original rectangle by the area of the triangle, which is formed by the overlapping parts of the original rectangle (this triangle is shaded in the following image).

![](https://cdn.mathpix.com/cropped/2024_04_17_7f9cac6bd796c2fe4728g-4.jpg?height=457&width=851&top_left_y=754&top_left_x=631)

To find the area of this triangle (which is isosceles - the angles at side $c$ are equal), we need to know the length of the base $c$ and the corresponding height. Let $a = 48 \, \text{mm}$ and $b = 64 \, \text{mm}$ be the sides of the original rectangle. For side $c$, the following holds:

$$
c^{2} = a^{2} + b^{2}
$$

Thus, $c = 80 \, \text{mm}$.

To calculate the height, we first need to determine the length of the legs. We will use the right triangle, one of whose legs is $a$, i.e., $48 \, \text{mm}$, and the other leg we will denote as $x$. The hypotenuse can be expressed as $64 - x$ (if we add the length of the leg denoted as $x$ and the length of the hypotenuse, we get the length of the longer side of the original rectangle $b$, see the image). Using the Pythagorean theorem, we get:

$$
\begin{aligned}
\sqrt{48^{2} + x^{2}} & = 64 - x, \\
48^{2} + x^{2} & = 64^{2} - 128 x + x^{2}, \\
128 x & = 1792, \\
x & = 14 \, \text{mm}.
\end{aligned}
$$

The legs of the triangle are therefore $64 - 14 = 50 \, \text{mm}$.

The height in the triangle can again be calculated using the Pythagorean theorem.

![](https://cdn.mathpix.com/cropped/2024_04_17_7f9cac6bd796c2fe4728g-5.jpg?height=468&width=853&top_left_y=363&top_left_x=630)

It holds that $v^{2} + \left(\frac{80}{2}\right)^{2} = 50^{2}$, hence $v = 30 \, \text{mm}$. The area of this triangle is $S = \frac{80 \cdot 30}{2} = 1200 \, \text{mm}^{2}$.

Conclusion: The pentagon thus formed has an area $1200 \, \text{mm}^{2}$ smaller than the original rectangle.