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In a triangle $ABC$ , medians $AD$ and $BE$ are drawn. if $AD=4,\angle DAB=\pi /6\, and\, \angle ABE=\pi /3,$ then the area of the $\triangle ABC\, is$
Option: 1
$\frac{16}{3 \sqrt{3}}$

Option: 2
$32/3$

Option: 3
$64/3$

Option: 4
$\frac{32}{3 \sqrt{3}}$

Answers (1)

best_answer

$AG=\frac{2}{3} \: \ AD= \frac{8}{3}$

$In \bigtriangleup AGB, \angle GAB= \frac{\pi }{6}, \: \angle ABG=\frac{\pi }{3}$

Applying, Sine law to $\bigtriangleup AGB,$

$\frac{AG}{\sin \angle BAG}=\frac{BG}{\sin \angle BAG}\Rightarrow BG= \frac{\sin \angle BAG}{\sin \angle ABG}\times \frac{8}{3}$

$= \frac{1}{\sqrt{3}}\times \frac{8}{3}=\frac{8}{3\sqrt{3}}$

$BG=\frac{8}{3\sqrt{3}}$

Now, $\angle AGB= \pi -\left ( \frac{\pi }{3} +\frac{\pi }{6}\right )=\frac{\pi }{2}$

$\bigtriangleup AGB=\frac{1}{2}AG\cdot BG\sin \left ( \frac{\pi }{2} \right )= \frac{1}{2}\times \frac{8}{3}\times \frac{8}{3\sqrt{3}}=\frac{64}{18\sqrt{3}}$

Now, $\bigtriangleup ABC=3\times \bigtriangleup AGB= \frac{64}{6\sqrt{3}}=\frac{32}{9}\sqrt{3}$

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