B4. A sheet of paper has the shape of a rectangle with sides measuring $12.6 \mathrm{~cm}$ and $10 \mathrm{~cm}$.

a) This sheet is rolled into the mantle of a cylinder such that the shorter side of the rectangle is the height of the cylinder. Calculate the volume of the cylinder formed by this rolled paper, accurate to the nearest cubic centimeter.

b) If we cut out squares with a side length of $3 \mathrm{~cm}$ from the corners of the rectangle, we get a net of a box without a lid. Draw a sketch of this net, determine the lengths of the edges of the box, and calculate the volume of this box.

c) Calculate the area of the smallest lid that will cover the box.

## Solutions to Problems and Scoring

## A contestant who arrives at the correct solution by any valid method (even if the scoring guide does not anticipate it) receives all possible points.

A valid method is one that

- sensibly takes into account the wording of the problem,
- leads to the solution of the problem,
- is mathematically correct and complete.

A contestant who has only partially solved the problem, from otherwise correct steps in solving but does not show the path to the final solution of the problem, cannot receive more than half of the possible points.

## PART I

The correct answers are recorded in the table. A correct answer by the contestant is scored with 2 points, an incorrect answer with -1 point, and a blank field in the table with 0 points.

| Problem | A1 | A2 | A3 | A4 | A5 | A6 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | C | A | B | C | D | E |
B4. The height of the cylinder is $10 \mathrm{~cm}$, and the circumference of its base is $12.6 \mathrm{~cm}$. From the circumference, we express the radius $r=\frac{12.6}{2 \pi}=\frac{6.3}{\pi} \mathrm{cm}$. The volume of the cylinder is $V=\pi r^{2} \cdot h=\pi \frac{(6.3)^{2}}{\pi^{2}} \cdot 10=\frac{(6.3)^{2}}{\pi} \cdot 10 \doteq 126 \mathrm{~cm}^{3}$. The lengths of the edges of the box are $6.6 \mathrm{~cm}, 4 \mathrm{~cm}$, and $3 \mathrm{~cm}$ (see the figure). The volume of the box is $6.6 \cdot 4 \cdot 3=79.2 \mathrm{~cm}^{3}$.

The smallest cover that can cover the box is a rectangle with side lengths of $6.6 \mathrm{~cm}$ and $4 \mathrm{~cm}$. Its area is $26.4 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_e4711ece0d44ad9bc5c3g-8.jpg?height=372&width=491&top_left_y=565&top_left_x=1405)

Scoring: Total: 7 points

$\longmapsto 12.6 \longrightarrow$

a) The height and circumference of the base of the cylinder ($h=10 \mathrm{~cm}, o=12.6 \mathrm{~cm}$) and the radius $r=\frac{12.6}{2 \pi} \mathrm{cm}$ ..... $1 \mathrm{p}$
Calculated volume of the cylinder: $V \doteq 126 \mathrm{~cm}^{3}$ ..... 1 p
b) Drawn sketch: ..... 1 p
The lengths of the edges of the box are: $6.6 \mathrm{~cm}, 4 \mathrm{~cm}$, and $3 \mathrm{~cm}$ ..... $2 \mathrm{p}$
Volume of the box: $V=79.2 \mathrm{~cm}^{3}$ ..... $1 \mathrm{p}$
c) The area of the cover is $26.4 \mathrm{~cm}^{2}$. ..... $1 \mathrm{p}$