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In a triangle $\mathrm{ABC}, \mathrm{AD}$ and $\mathrm{BE}$ are medians. If $\mathrm{AD}=4$, $\left\lfloor\mathrm{DAB}=\frac{\pi}{6}\right.$ and $\left\lfloor\mathrm{ABE}=\frac{\pi}{3}\right.$ then the area of $\triangle \mathrm{ABC}$ is
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The correct answer is:
$\frac{32}{3 \sqrt{3}}$
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