Question

In 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 $△ABC$ is

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

Answer 1

The correct option is C $\frac{32}{3\sqrt{3}}$

$In.△ABG$

$\cos \frac{\pi }{6}=\frac{AG}{AB}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{8}{3AB}$

$\Rightarrow AB=\frac{16}{3\sqrt{3}}$

$\therefore ar\left(\Delta ABD\right)=\frac{1}{2}\times AB\times AD\sin \frac{\pi }{6}$

$=\frac{1}{2}\times \frac{16}{3\sqrt{3}}\times 4\times \frac{1}{2}=\frac{16}{3\sqrt{3}}$

$\therefore$ Area of $\Delta ABC=2$(Area of$\Delta ABD$)

$=2\times \frac{16}{3\sqrt{3}}$ $=\frac{32}{3\sqrt{3}}$ sq units

Option $\left(C\right)$ is correct answer.