Question

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

• A. $\frac{64}{3\sqrt{3}}$
• B. $\frac{8}{3\sqrt{3}}$
• C. $\frac{16}{3\sqrt{3}}$
• D. $\frac{32}{3\sqrt{3}}$

Answer 1

The correct option is D $\frac{32}{3\sqrt{3}}$

Given, $AD=4\Rightarrow AG=\frac{2}{3}AD=\frac{8}{3}$

In $△AGD,$ applying sine rule,

$\frac{AG}{\sin {60}^{\circ }}=\frac{BG}{\sin {30}^{\circ }}\Rightarrow BG=\frac{8}{3\sqrt{3}}$

Area of $△AGD=\frac{1}{2}\times AG\times BG(\because \angle AGB={90}^{\circ })$

$=\frac{1}{2}\times \frac{8}{3}\times \frac{8}{3\sqrt{3}}=\frac{32}{9\sqrt{3}}$

Area of $△ABC=3\times$ Area of $△AGB(\because$ Centroid divides the traingle in $3$ equal parts$)$

$=3\times \frac{32}{9\sqrt{3}}=\frac{32}{3\sqrt{3}}$ sq. units.