19) Let's draw a rectangle with sides $5 \mathrm{~cm}$ and $12 \mathrm{~cm}$, the circle in which it is inscribed, and the semicircles that have the sides of the rectangle as diameters and are external to it, as indicated in the adjacent figure. What is the area of the shaded part?
(A) $45 \mathrm{~cm}^{2}$,
(B) $13 \pi \mathrm{cm}^{2}$,
(D) $60 \mathrm{~cm}^{2}$,
(E) $20 \pi \mathrm{cm}^{2}$.
(C) $19 \pi \mathrm{cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_17_cd047c2caa32abeb5404g-2.jpg?height=358&width=347&top_left_y=1559&top_left_x=998)
19. The answer is $(\mathbf{D})$.

We can obtain the required area by difference, considering the area of the figure obtained by joining to the rectangle the four semicircles constructed on its sides and subtracting from the area of this figure the area of the circle in which the rectangle is inscribed.
![](https://cdn.mathpix.com/cropped/2024_04_17_e415f128e14398de33bcg-6.jpg?height=434&width=858&top_left_y=1708&top_left_x=658)

The area of the rectangle is $60 \mathrm{~cm}^{2}$. Referring to the figure on the left, the area of each of the two semicircles $a$ and $c$ is $\frac{25}{8} \pi \mathrm{cm}^{2}$, while the area of each of the two semicircles $b$ and $d$ is $18 \pi \mathrm{cm}^{2}$. Therefore, the total area of the first figure is: $\left(60+\frac{25}{4} \pi+36 \pi\right) \mathrm{cm}^{2}$. The radius $R$ of the circle in which the rectangle is inscribed can be found using the Pythagorean Theorem applied to the rectangle $A B C$ in the right figure. This gives:

$$
r=\left(\sqrt{\frac{25}{4}+36}\right) \mathrm{cm}
$$

Thus, the area of the circle in which the rectangle is inscribed is

$$
\left(\frac{25}{4} \pi+36 \pi\right) \mathrm{cm}^{2}
$$

By taking the difference of the areas of the two figures, we find that the area required by the problem is $60 \mathrm{~cm}^{2}$.

[Problem proposed by D. Lombardo.]