Quantitative Aptitude Ques 2049

Question: If two medians BE and CF of a $ \Delta ABC, $ intersect each other at G and if $ BG=CG, $ $ \angle BGC=60{}^\circ , $ $ BC=8cm, $ then area of the $ \Delta ABC $ is

Options:

A) $ 96\sqrt{3}cm^{2} $

B) $ 48\sqrt{3}cm^{2} $

C) $ 48cm^{2} $

D) $ 64\sqrt{3}cm^{2} $

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Answer:

Correct Answer: B

Solution:

  • Given, $ BG=GC, $ $ BC=8cm $ Let $ \angle GBC=\angle GCB=x $ $ \angle BGC+\angle GCB+\angle CBG=180{}^\circ $

$ \Rightarrow $ $ 60{}^\circ +x+x=180{}^\circ $

$ \Rightarrow $ $ 2x=120{}^\circ $
$ \Rightarrow $ $ x=60{}^\circ $

$ \therefore $ $ \Delta BCG $ Is an equilateral triangle as all the angles are $ 60{}^\circ . $ Area of $ \Delta BCG=\frac{\sqrt{3}}{4}\times 8^{2}=\frac{\sqrt{3}}{4}\times 8\times 8=16\sqrt{3}cm^{2} $

$ \therefore $ Area of $ \Delta ABC=3\times \Delta BCG=3\times 16\sqrt{3}=48\sqrt{3}cm^{2} $