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

Correct Answer: 30

Given:

• \( AD = 2x \)
• \( AE = 2y \)
• \( AB = 3x \)
• \( AC = 5y \)
• \( \angle A \) is common to both triangles \( \triangle ADE \) and \( \triangle ABC \)
• Area of \( \triangle ADE = 8 \)

Step 1: Area of \( \triangle ADE \)

Using the area formula for a triangle: \[ \text{Area} = \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{included angle}) \] Therefore, \[ \text{Area of } \triangle ADE = \frac{1}{2} \times AD \times AE \times \sin A \] Substituting the given values: \[ = \frac{1}{2} \times 2x \times 2y \times \sin A = 8 \] This simplifies to \( 2xy \cdot \sin A = 8 \), which further simplifies to \( xy \cdot \sin A = 4 \) (Equation 1).

Step 2: Area of \( \triangle ABC \)

Applying the area formula again: \[ \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times AC \times \sin A \] Substituting the values: \[ = \frac{1}{2} \times 3x \times 5y \times \sin A = \frac{15}{2} \cdot xy \cdot \sin A \] Using the result from Equation 1, where \( xy \cdot \sin A = 4 \): \[ \text{Area} = \frac{15}{2} \cdot 4 = 30 \]

Final Answer:

∴ The area of triangle \( ABC \) is 30.