In a triangle ABC,medians AD and BE are drawn | vedclass

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(C) Let $G$ be the centroid of $	riangle ABC$. The centroid $G$ divides the median $AD$ in the ratio $2:1$. Given $AD = 4$,we have $AG = \frac{2}{3}

Vedclass · Accepted Answer

$\frac{32}{3\sqrt{3}}$ sq. units

Vedclass · Answer

(C) Let $G$ be the centroid of $	riangle ABC$. The centroid $G$ divides the median $AD$ in the ratio $2:1$. Given $AD = 4$,we have $AG = \frac{2}{3} 	imes 4 = \frac{8}{3}$ and $GD = \frac{1}{3} 	imes 4 = \frac{4}{3}$. In $	riangle ABG$,we have $\angle GAB = \frac{\pi}{6}$ and $\angle GBA = \frac{\pi}{3}$. Therefore,$\angle AGB = \pi - (\frac{\pi}{6} + \frac{\pi}{3}) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$. Thus,$	riangle ABG$ is a right-angled triangle at $G$. In $	riangle ABG$,$	an(\angle GBA) = \frac{AG}{BG} \implies 	an(\frac{\pi}{3}) = \frac{8/3}{BG}$. $BG = \frac{8/3}{\sqrt{3}} = \frac{8}{3\sqrt{3}}$. The area of $	riangle ABD = \frac{1}{2} 	imes AD 	imes BG 	imes \sin(\angle AGB) = \frac{1}{2} 	imes 4 	imes \frac{8}{3\sqrt{3}} 	imes 1 = \frac{16}{3\sqrt{3}}$. Since the median $AD$ divides $	riangle ABC$ into two triangles of equal area,the area of $	riangle ABC = 2 	imes 	ext{Area}(	riangle ABD) = 2 	imes \frac{16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$ sq. units.

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 $\triangle ABC$ is

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