Question

In a triangle $ABC,$ medians $AD$ and $BE$ are drawn. If $AD=4,\angle DAB=\frac{\pi }{6}$ and $\angle ABE=\frac{\pi }{3}$, then the area of the $\Delta ABC$ is :

• A. $8/3$
• B. $16/3$
• C. $\frac{32}{3\sqrt{3}}$
• D. $64/3$

Answer 1

Answer: (C)

Given $AD=4$ and $BD=DC$

In $△ABG,tan\frac{\pi }{3}=\frac{AG}{BG}$
$BG=AGcot\frac{\pi }{3}$
$BG=\frac{8}{3}\times \frac{1}{\sqrt{3}}=\frac{8}{3\sqrt{3}}$
Area of $△ADB=\frac{1}{2}\times BG\times AD$
$=\frac{1}{2}\times \frac{8}{3\sqrt{3}}\times 4=\frac{16}{3\sqrt{3}}$
Since, median divide a triangle into two triangles of equal area.
Therefore Area of $△ABC=2\times$ area of $△ADB$
$=2\times \frac{16}{3\sqrt{3}}=\frac{32}{3\sqrt{3}}$