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

• A

  $\frac{64}{3}$
• B

  $\frac{8}{3}$
• C

  $\frac{16}{3}$
• Correct answer
  D

  None of these

Solution

$\tan 60^{\circ}=\frac{8 / 3}{x}$ or $x=\frac{8}{3 \sqrt{3}}$
Area of $\triangle \mathrm{ABD}=\frac{1}{2} \times 4 \times \frac{8}{3 \sqrt{3}}=\frac{16}{3 \sqrt{3}}$

Hence, the correct answer is option: 'D'.