5. The area of triangle $ABC$ is $18 \, \text{cm}^2$. On side $\overline{AC}$, there is a point $D$ such that $|DC| = 2 \cdot |AD|$. Calculate the areas of triangles $ABD$ and $DBC$.

## Each task is scored out of 10 points.

The use of a pocket calculator or any reference materials is not allowed.
5. 

![](https://cdn.mathpix.com/cropped/2024_05_30_f844036739139e74d8f5g-2.jpg?height=303&width=482&top_left_y=565&top_left_x=844)

Let $b = |AC|$ and $v_b$ be the length of the altitude to that side. For the area of triangle $ABC$, we have $P(ABC) = \frac{b \cdot v_b}{2} = 18$, hence $b \cdot v_b = 36$.

Since $|DC| = 2|AD|$, it follows that $b = |AC| = |AD| + |DC| = |AD| + 2|AD| = 3|AD|$, so $|AD| = \frac{1}{3} b$.

Furthermore, since triangles $ABD$ and $ABC$ share the same altitude of length $v_b$, the area of triangle $ABD$ is

$$
P(ABD) = \frac{|AD| \cdot v_b}{2} = \frac{\frac{b}{3} \cdot v_b}{2} = \frac{b \cdot v_b}{6} = \frac{36}{6} = 6 \text{ cm}^2
$$

Finally, the area of triangle $DBC$ is $P(DBC) = P(ABC) - P(ABD) = 18 - 6 = 12 \text{ cm}^2$.