B3. A cylindrical cheese wheel with a height of $10 \mathrm{~cm}$ and a diameter of $30 \mathrm{~cm}$ is cut into 8 equal pieces, as we would cut a cake. Each piece is then wrapped in foil. For each piece, we use $20 \%$ more foil than the surface area of the piece of cheese. How many $\mathrm{dm}^{2}$ of foil will we use to wrap the cheese? Round the result to the nearest hundredth. Write down the answer.
B3. The surface area of one piece of cheese is $P_{1}=\frac{1}{8} \cdot 2 \pi r^{2}+\frac{1}{8} \cdot 2 \pi r+2 \cdot r \cdot v=594.375 \mathrm{~cm}^{2}$. Eight pieces of cheese have eight times the surface area $\stackrel{8}{P}=\left(\frac{1}{8} \cdot 2 \pi r^{2}+\frac{1}{8} \cdot 2 \pi r+2 \cdot r \cdot v\right) \cdot 8=47.55 \mathrm{dm}^{2}$. We also consider an additional $20 \%$ for the wrapping foil: $8 \cdot 1.2 \cdot P_{1}=57.06 \mathrm{dm}^{2}$. For wrapping the cheese, we will need $57.06 \mathrm{dm}^{2}$ of foil.

The surface area of one piece of cheese is: $P=\frac{1}{8} \cdot 2 \pi r^{2}+\frac{1}{8} \cdot 2 \pi r+2 \cdot r \cdot v \ldots \ldots \ldots \ldots \ldots 1+1+1$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-14.jpg?height=54&width=1636&top_left_y=1092&top_left_x=276)

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-14.jpg?height=63&width=1639&top_left_y=1139&top_left_x=274)
Answer: For 8 pieces, we need $57.06 \mathrm{dm}^{2}$ of foil. .................................................. OR

![](https://cdn.mathpix.com/cropped/2024_06_07_4401d59a37898c9fd2b4g-14.jpg?height=115&width=1641&top_left_y=1276&top_left_x=276)
$P=594.38 \mathrm{~cm}^{2}$ 1 point

Considering $20 \%$ 1 point

Answer: For 8 pieces, we need $57.06 \mathrm{dm}^{2}$ of foil.

1 point