19) The radius of the circle on the side is $5 \mathrm{~cm}$; furthermore, points $A, B$ and $C$ divide the circle into three arcs of equal length. Calculate the area delimited by the chords $A C$ and $B C$ and the arc with endpoints $A$ and $B$ containing $D$.
(A) $25\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right) \mathrm{cm}^{2}$
$\mathrm{cm}^{2}$, (D) $\frac{25 \sqrt{3}}{2} \mathrm{~cm}^{2}$,
(B) $25\left(\frac{\pi}{6}+\sqrt{3}\right) \mathrm{cm}^{2}$, (C)
(E) $\frac{25}{2}\left(\frac{\pi}{3}+\sqrt{3} / 2\right) \mathrm{cm}^{2}$.
19. The answer is $(\mathbf{A})$.

Let $O$ be the center of the circle. The angle $\widehat{A O B}$ is $120^{\circ}$, and therefore the angle $\widehat{A C B}$ is $60^{\circ}$. Moreover, since $A C$ and $A B$ are chords subtending arcs of equal length, they have equal length. We deduce that triangle $A B C$ is equilateral. Consequently,

$$
\operatorname{area}(A O C)=\operatorname{area}(B O C)=25 \frac{\sqrt{3}}{4} \mathrm{~cm}^{2} .
$$

The area of the circular sector bounded by the arc $A D B$ is one-third of the area of the circle, i.e., it is

$$
25 \frac{\pi}{3} \mathrm{~cm}^{2} .
$$

The area sought is the sum of the areas of triangles $A O C$ and $B O C$ and the area of the circular sector we just calculated:

$$
25 \frac{\sqrt{3}}{4} \mathrm{~cm}^{2}+25 \frac{\sqrt{3}}{4} \mathrm{~cm}^{2}+25 \frac{\pi}{3} \mathrm{~cm}^{2}=25\left(\frac{\sqrt{3}}{2}+\frac{\pi}{3}\right) \mathrm{cm}^{2} .
$$