If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ∠BGC = 60°, BC = 8 cm, then area of the triangle ABC is:

Answer: Option B

According to question,
⇒ ∵ ∠BGC = 60° (given)
⇒ ∠GBC = ∠GCB = x°
⇒ x° + x° + 60° = 180°
⇒ x = 60°

⇒ So ΔBGC is an equilateral triangle with side 8 cm each
Then,
Area pf triangle ΔBGC
= $$\frac{\sqrt{3}}{4}$$ a²
= $$\frac{\sqrt{3}}{4}$$ 8²
= 16$$\sqrt{3}$$ cm²
⇒ Area of ΔABC
= Area (ΔBGC + ΔAGC + ΔAGB)
⇒ Area of ΔABC = 3 × 16$$\sqrt{3}$$
⇒ Area of ΔABC = 48$$\sqrt{3}$$ cm² {∵ ΔBGC = ΔAGC = ΔAGB}