In a triangle ABC,if the medians AD and BE ar | vedclass

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(D) Let $G$ be the centroid of $	riangle ABC$. The medians $AD$ and $BE$ intersect at $G$. Given $AD=4$,$\angle GAB = \frac{\pi}{6}$,and $\angle GBA

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $G$ be the centroid of $	riangle ABC$. The medians $AD$ and $BE$ intersect at $G$. Given $AD=4$,$\angle GAB = \frac{\pi}{6}$,and $\angle GBA = \frac{\pi}{3}$. In $	riangle AGB$,the sum of angles is $\pi$,so $\angle AGB = \pi - (\frac{\pi}{6} + \frac{\pi}{3}) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$. Since $G$ is the centroid,$AG = \frac{2}{3} AD = \frac{2}{3} 	imes 4 = \frac{8}{3}$. In right-angled $	riangle AGB$,$\sin(\angle GBA) = \frac{AG}{AB} \implies \sin(\frac{\pi}{3}) = \frac{8/3}{AB} \implies \frac{\sqrt{3}}{2} = \frac{8}{3AB} \implies AB = \frac{16}{3\sqrt{3}}$. Also,$	an(\angle GBA) = \frac{AG}{BG} \implies 	an(\frac{\pi}{3}) = \frac{8/3}{BG} \implies \sqrt{3} = \frac{8}{3BG} \implies BG = \frac{8}{3\sqrt{3}}$. Area of $	riangle AGB = \frac{1}{2} 	imes AG 	imes BG = \frac{1}{2} 	imes \frac{8}{3} 	imes \frac{8}{3\sqrt{3}} = \frac{32}{9\sqrt{3}}$. Since the centroid divides the triangle into three triangles of equal area,Area$(	riangle ABC)$ = $3 	imes$ Area$(	riangle AGB)$ = $3 	imes \frac{32}{9\sqrt{3}} = \frac{32}{3\sqrt{3}}$ sq. units.

In a $\triangle ABC$,if the medians $AD$ and $BE$ are such that $AD=4$,$\angle DAB=\frac{\pi}{6}$ and $\angle ABE=\frac{\pi}{3}$,then the area of $\triangle ABC$ (in square units) is

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