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

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

Therefore, the area of \( \triangle ABC \) is 30 cm².