Every Sunday, Pedro prepares cannelloni for lunch. First, he cuts rectangles of pasta measuring $16 \mathrm{~cm}$ by $12 \mathrm{~cm}$ and then glues the two longer sides, overlapping a strip of $2 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-07.jpg?height=300&width=505&top_left_y=739&top_left_x=1184)

This way, he obtains cylinders that he fills with ricotta, and he already knows that he always uses $500 \mathrm{~g}$ of ricotta. One fine Sunday, with the same number of pasta rectangles measuring $16 \mathrm{~cm}$ by $12 \mathrm{~cm}$, he decides to produce the cylinders by gluing the shorter sides, always overlapping a strip of $2 \mathrm{~cm}$. In this situation, will he use more or less ricotta than before? How much?
By gluing the rectangles of dough along the longer side, Pedro obtains a cylinder with a circular base that is $10 \mathrm{~cm}$ in length and $16 \mathrm{~cm}$ in height. The volume that he fills with ricotta is the volume of this cylinder:

$$
V=\text { area of the base } \times \text { height }
$$

The area of the base is given by $\pi \times r^{2}$, where $r$ is the radius of the base. Let's calculate the radius knowing that the circumference of the base is $10 \mathrm{~cm}$; we have:

$$
2 \pi r=10 \Rightarrow r=\frac{5}{\pi}
$$

Therefore, the volume of ricotta for each cannelloni is

$$
V=\pi \times \frac{5^{2}}{\pi^{2}} \times 16=\frac{16 \times 25}{\pi}=\frac{400}{\pi} \mathrm{cm}^{3}
$$

Now, by gluing the rectangles of dough along the shorter side, Pedro obtains a cylinder with a circular base that is $14 \mathrm{~cm}$ in length and $12 \mathrm{~cm}$ in height. The radius of the base is $r^{\prime}=\frac{14}{2 \pi}=\frac{7}{\pi}$, so the volume of ricotta for each cannelloni will be:

$$
V^{\prime}=\pi \times \frac{7^{2}}{\pi^{2}} \times 12=\frac{588}{\pi} \mathrm{cm}^{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-28.jpg?height=333&width=545&top_left_y=278&top_left_x=1186)

Finally, to calculate the new cost of ricotta, we use the following direct Rule of Three:

$$
\begin{array}{ccc}
\text { Volume }\left(\mathrm{cm}^{3}\right) & & \text { Ricotta }(\mathrm{g}) \\
\frac{400}{\pi} & \longrightarrow & 500 \\
\frac{588}{\pi} & \longrightarrow & x
\end{array}
$$

It follows that

$$
x=\frac{500 \times 588}{400}=735 \mathrm{~g} .
$$