We are given a triangle $ABC$ with a line segment $DE$ parallel to $AC$. We are given that $BD = 6$, $DA = 4$, $BE = 3$, $EC = 4$, $AC = 5$, and the area of triangle $DBE$ is $3.6 \, \text{cm}^2$. We need to find the area of the quadrilateral $ADEC$.

1. Problem Description

We are given a triangle $ABC$ with a line segment $DE$ parallel to $AC$. We are given that $BD = 6$, $DA = 4$, $BE = 3$, $EC = 4$, $AC = 5$, and the area of triangle $DBE$ is $3.6 \, \text{cm}^2$. We need to find the area of the quadrilateral $ADEC$.

2. Solution Steps

Since $DE$ is parallel to $AC$, triangle $DBE$ is similar to triangle $ABC$. The ratio of corresponding sides is given by

$$\frac{BD}{BA} = \frac{BE}{BC} = \frac{DE}{AC}$$

We have $BD = 6$, $DA = 4$, so $BA = BD + DA = 6 + 4 = 10$.

Also, $BE = 3$, $EC = 4$, so $BC = BE + EC = 3 + 4 = 7$.

The ratio of similarity between the triangles is

$$\frac{BD}{BA} = \frac{6}{10} = \frac{3}{5}$$

$$\frac{BE}{BC} = \frac{3}{7}$$

Since $\frac{BD}{BA} \neq \frac{BE}{BC}$ the given values must contain some mistake. Assuming the given values for $BD, DA, BE, EC$ are ratios, we assume that $\frac{BD}{DA} = \frac{6}{4} = \frac{3}{2}$, therefore $\frac{BD}{BA} = \frac{3}{3+2} = \frac{3}{5}$, and $\frac{BE}{EC} = \frac{3}{4}$, therefore $\frac{BE}{BC} = \frac{3}{3+4} = \frac{3}{7}$.

We assume that the $DE$ line is not parallel to $AC$.

The ratio of the sides $\frac{BD}{BA}$ is $\frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$.

The ratio of the sides $\frac{BE}{BC}$ is $\frac{3}{3+4} = \frac{3}{7}$.

Assuming $DE || AC$

Then we can use the fact that $\triangle DBE \sim \triangle ABC$ to say

$\frac{DE}{AC} = \frac{BD}{BA} = \frac{BE}{BC}$

However, this would require $\frac{6}{10} = \frac{3}{7}$ which is false.

Let's assume $DBE \sim ABC$

The ratio of areas is $\left( \frac{BD}{BA} \right)^2 = \frac{Area(DBE)}{Area(ABC)}$

$\frac{Area(DBE)}{Area(ABC)} = \left( \frac{6}{10} \right)^2 = \left( \frac{3}{5} \right)^2 = \frac{9}{25}$

$Area(DBE) = 3.6$

$3.6 / Area(ABC) = 9/25$

$Area(ABC) = 3.6 \cdot \frac{25}{9} = 0.4 \cdot 25 = 10$

$Area(ADEC) = Area(ABC) - Area(DBE) = 10 - 3.6 = 6.4$

3. Final Answer

The area of quadrilateral $ADEC$ is $6.4 \, \text{cm}^2$.