In a triangle ABC,AD and BE are medians. If A | vedclass

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(D) Let $G$ be the centroid of $	riangle ABC$. Since $AD$ and $BE$ are medians,$G$ divides $AD$ and $BE$ in the ratio $2:1$. Thus,$AG = \frac{2}{3} A

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$\frac{32}{3 \sqrt{3}}$

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(D) Let $G$ be the centroid of $	riangle ABC$. Since $AD$ and $BE$ are medians,$G$ divides $AD$ and $BE$ in the ratio $2:1$. Thus,$AG = \frac{2}{3} AD = \frac{2}{3} 	imes 4 = \frac{8}{3}$ and $BG = \frac{2}{3} BE$. In $	riangle ABG$,by the Sine Rule: $\frac{AG}{\sin(\angle ABG)} = \frac{BG}{\sin(\angle BAG)}$. Given $\angle BAG = \frac{\pi}{6}$ and $\angle ABG = \frac{\pi}{3}$,we have $\frac{8/3}{\sin(\pi/3)} = \frac{BG}{\sin(\pi/6)}$. $BG = \frac{8}{3} 	imes \frac{\sin(\pi/6)}{\sin(\pi/3)} = \frac{8}{3} 	imes \frac{1/2}{\sqrt{3}/2} = \frac{8}{3 \sqrt{3}}$. Area of $	riangle ABG = \frac{1}{2} 	imes AG 	imes BG 	imes \sin(\angle AGB)$. Since $\angle AGB = \pi - (\pi/6 + \pi/3) = \pi/2$,$\sin(\angle AGB) = 1$. Area of $	riangle ABG = \frac{1}{2} 	imes \frac{8}{3} 	imes \frac{8}{3 \sqrt{3}} = \frac{32}{9 \sqrt{3}}$. Since the centroid divides the triangle into three triangles of equal area,Area of $	riangle ABC = 3 	imes 	ext{Area of } 	riangle ABG = 3 	imes \frac{32}{9 \sqrt{3}} = \frac{32}{3 \sqrt{3}}$.

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

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