Question

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

Hint:

Since AD:DB and AE:EC are different ratios here, DE is not parallel to BC -- do not treat triangle ADE and ABC as similar. Instead compare their areas through a shared base or shared height, using a helper triangle like ADC to bridge the two ratios.

Answer 1

 

Given:

Area of triangle \( \triangle ADE \) is calculated as:

\[ \text{Area}_{\triangle ADE} = \frac{1}{2} \times AD \times AE \times \sin A \]

Let \( AD = 2x \) and \( AE = 2y \) 
So: \[ \text{Area}_{\triangle ADE} = \frac{1}{2} \times 2x \times 2y \times \sin A = 8 \]

Simplifying: \[ \Rightarrow 2x \cdot 2y = 4xy \Rightarrow 4xy \cdot \sin A = 8 \Rightarrow xy \cdot \sin A = 2 \] (Corrected as per proper simplification; your original said 4, but let’s assume area is **8**, so: \( 4xy \sin A = 8 \Rightarrow xy \sin A = 2 \))

Now, calculating Area of Triangle \( \triangle ABC \):

Given: \[ AB = 3x, \quad AC = 5y \] So: \[ \text{Area}_{\triangle ABC} = \frac{1}{2} \times AB \times AC \times \sin A \] \[ = \frac{1}{2} \times 3x \times 5y \times \sin A = \frac{15}{2} \cdot xy \cdot \sin A \]

We already have: \[ xy \cdot \sin A = 2 \] So, \[ \text{Area}_{\triangle ABC} = \frac{15}{2} \cdot 2 = \boxed{15} \]

✅ Final Answer: Area of triangle ABC is 30 cm²

Answer 2

 

Area of △ADE using Similar Triangles

By the theorem for similar triangles:
The ratio of their areas is the product of the ratios of corresponding sides: 
\[ \text{Area of } \triangle ADE = \frac{AD}{AB} \times \frac{AE}{AC} \times \text{Area of } \triangle ABC \]

Plug in the values given:
\[ 8 = \frac{2}{3} \times \frac{2}{5} \times \text{Area of } \triangle ABC \]

Now solve for the area of \( \triangle ABC \): 
\[ \text{Area of } \triangle ABC = \frac{8 \times 3 \times 5}{2 \times 2} \] \[ = \frac{120}{4} \] \[ = 30 \]

<h3>Therefore, the area of \( \triangle ABC \) is 30 cm².</h3>

Answer 3

Concept:

• Since AD:DB and AE:EC are different ratios, DE is not parallel to BC, so triangle ADE is not similar to triangle ABC. Compare areas using a shared base or shared height through an intermediate triangle instead of trigonometry.

Step 1: Set up the ratios and draw the helper line.
Given $AD:BD = 2:1$, so $AD:AB = 2:3$. Given $AE:CE = 2:3$, so $AE:AC = 2:5$. Draw the diagonal $DC$ to form the helper triangle $ADC$.

Step 2: Compare triangle ADE with triangle ADC.
Triangles $ADE$ and $ADC$ share the same vertex $D$, and their third vertices $E$ and $C$ both lie on line $AC$, so both triangles have the same height measured from $D$ to line $AC$. Triangles with equal height have areas in the ratio of their bases:
$\frac{\text{Area}(ADE)}{\text{Area}(ADC)} = \frac{AE}{AC} = \frac{2}{5}$, so $\text{Area}(ADE) = \frac{2}{5} \times \text{Area}(ADC)$

Step 3: Compare triangle ADC with triangle ABC.
Triangles $ADC$ and $ABC$ share the same vertex $C$, and their third vertices $D$ and $B$ both lie on line $AB$, so both triangles have the same height measured from $C$ to line $AB$:
$\frac{\text{Area}(ADC)}{\text{Area}(ABC)} = \frac{AD}{AB} = \frac{2}{3}$, so $\text{Area}(ADC) = \frac{2}{3} \times \text{Area}(ABC)$

Step 4: Combine both ratios and solve.
$\text{Area}(ADE) = \frac{2}{5} \times \frac{2}{3} \times \text{Area}(ABC) = \frac{4}{15}\text{Area}(ABC)$
Given $\text{Area}(ADE) = 8$ sq cm: $8 = \frac{4}{15}\text{Area}(ABC) \Rightarrow \text{Area}(ABC) = \frac{8 \times 15}{4} = 30$ sq cm

Final Answer: The area of triangle ABC is 30 sq cm.