Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade None
Question:
If points $D$, $E$ and $F$ divide sides $BC$, $CA$ and $AB$ respectively in ratio $\lambda : 1$ (in order) and $ar(\triangle DEF) = 0.4 \cdot ar(\triangle ABC)$, then $\lambda$ is equal to:
\frac{2 - \sqrt{3}}{2}
\frac{3 - \sqrt{5}}{2}
\frac{2 + \sqrt{3}}{2}
\frac{3 + \sqrt{5}}{2}
Step-by-Step Solution
Key Concept: Area ratios of similar or proportionally-divided triangles depend on the square of linear scaling factors.
For triangle $AEF$ inscribed in triangle $ABC$ with $AE:EB = AF:FC = \lambda$, the area is $ar(\triangle AEF) = \frac{\lambda}{2(\lambda+1)^2} \cdot \frac{\lambda c}{\lambda+1} \cdot b \sin A = \frac{\lambda\Delta}{(\lambda+1)^2}$. Similarly, $ar(\triangle BDF) = ar(\triangle CDE) = \frac{\lambda\Delta}{(\lambda+1)^2}$, where the geometry shows how the triangles partition the original figure.
Correct Answer: 2,4