Let's solve this step-by-step.

Step 1: Determine the Coordinates of Points

1. Triangle ABC is a right-angled triangle at $B$, with $AB = 8$ units and $BC = 6$ units.
2. Place $B$ at the origin $(0, 0)$.
3. Let $A$ be at $(8, 0)$ (since $AB = 8$).
4. Let $C$ be at $(0, 6)$ (since $BC = 6$).

Step 2: Locate Point $D$ on $AC$

1. Since $D$ divides $AC$ in the ratio $AD : DC = 2 : 3$, use the section formula.
2. Coordinates of $D$ (using $(x_1, y_1) = (8, 0)$ for $A$ and $(x_2, y_2) = (0, 6)$ for $C$): $D = \left( \frac{2 \cdot 0 + 3 \cdot 8}{2 + 3}, \frac{2 \cdot 6 + 3 \cdot 0}{2 + 3} \right) = \left( \frac{24}{5}, \frac{12}{5} \right)$

Step 3: Locate Point $E$ on $BC$

1. $E$ divides $BC$ in the ratio $BE : EC = 3 : 4$.
2. Coordinates of $E$ (using $(x_1, y_1) = (0, 0)$ for $B$ and $(x_2, y_2) = (0, 6)$ for $C$): $E = \left( \frac{3 \cdot 0 + 4 \cdot 0}{3 + 4}, \frac{3 \cdot 6 + 4 \cdot 0}{3 + 4} \right) = \left( 0, \frac{18}{7} \right)$

Step 4: Calculate the Area of $\triangle ABC$

Since $\triangle ABC$ is a right triangle with base $AB = 8$ and height $BC = 6$, $\text{Area of } \triangle ABC = \frac{1}{2} \times 8 \times 6 = 24$

Step 5: Calculate the Area of $\triangle BDE$

Using vertices $B = (0, 0)$, $D = \left(\frac{24}{5}, \frac{12}{5}\right)$, and $E = \left(0, \frac{18}{7}\right)$, apply the area formula for a triangle given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ Substitute the coordinates: $\text{Area of } \triangle BDE = \frac{1}{2} \left| 0 \cdot \left(\frac{12}{5} - \frac{18}{7}\right) + \frac{24}{5} \cdot \left(\frac{18}{7} - 0\right) + 0 \cdot \left(0 - \frac{12}{5}\right) \right|$ This simplifies to: $= \frac{1}{2} \left| \frac{24}{5} \cdot \frac{18}{7} \right| = \frac{1}{2} \cdot \frac{432}{35} = \frac{216}{35}$

Step 6: Find the Ratio of the Areas of $\triangle BDE$ and $\triangle ABC$

$\text{Ratio} = \frac{\text{Area of } \triangle BDE}{\text{Area of } \triangle ABC} = \frac{\frac{216}{35}}{24} = \frac{216}{35 \times 24} = \frac{216}{840} = \frac{9}{35}$

Final Answer

The ratio of the areas of triangles $BDE$ and $ABC$ is: $\boxed{\frac{9}{35}}$

Would you like a further breakdown or additional questions related to this problem?

Related Questions

1. How do you apply the section formula for internal division?
2. What is the significance of maximizing the area in such geometric problems?
3. How do you calculate the area of a triangle given vertices?
4. What are some other methods to find the area of a triangle besides base and height?
5. How does changing the ratio $AD : DC$ affect the area of $\triangle BDE$?

Tip

In coordinate geometry, using section and area formulas often simplifies problems involving division points and area calculations of triangles.