## Task B-1.3.

Mia wants to wrap a Christmas gift, a box in the shape of a square prism. She has decorative paper in a square shape. Wrapping is only possible if the box is placed diagonally, as shown in the figure, with the center of the base of the box at the center of the square paper. The four corners of the paper will overlap at the center $T$ of the top base after wrapping, and the edges of the paper

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-02.jpg?height=269&width=585&top_left_y=728&top_left_x=1315)
will join along the diagonals without overlapping. Determine the area and dimensions of the wrapping paper if the dimensions of the box are $20 \mathrm{~cm} \times 20 \mathrm{~cm} \times 5 \mathrm{~cm}$.
## Solution.

Let's denote the dimensions of the box: $a=20 \mathrm{~cm}, h=5 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-02.jpg?height=1113&width=1196&top_left_y=1348&top_left_x=356)

(sketch with all labels or exact values)

When the triangle $A B C$ is wrapped, it is mapped to the triangle $A^{\prime} B^{\prime} C^{\prime}$.

Therefore, the desired surface area is

$$
\begin{aligned}
P & =a^{2}+4 \cdot \frac{1}{2} \cdot \frac{a}{2} \cdot a+4 a h+4 \cdot \frac{1}{2} h^{2} \\
& =2 a^{2}+4 a h+2 h^{2}=2(a+h)^{2} \\
& =2(20+5)^{2} \mathrm{~cm}^{2}=1250 \mathrm{~cm}^{2}
\end{aligned}
$$

Then, for the length of the side of the square paper, we have

$$
\begin{aligned}
x^{2} & =1250 \mathrm{~cm}^{2} \\
x & =25 \sqrt{2} \mathrm{~cm}
\end{aligned}
$$