Question

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 $△ABC$ is

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

Answer 1

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

Given:-

$AD=4\text{units}BD=DC$

To Find:- $ar\left(△ABC\right)$

Sol:-

Since the centroid $G$ divides the line $AD$ in the ratio $2:1$

$\therefore AG=\frac{AD}{3}$ and $DG=\frac{4}{3}$

In $△ABG$

$\tan \frac{\pi }{3}=\frac{AG}{BG}$

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

Now,

$ar\left(△ADB\right)$ $=\frac{1}{2}\times AD\times BG=\frac{1}{2}\times 4\times \frac{8}{3\sqrt{3}}=\frac{16}{3\sqrt{3}}$

Since median divides a triangle into two triangles of equal area.

$\therefore ar\left(△ABC\right)$ $=2\times ar\left(△ADB\right)=2\times \frac{16}{3\sqrt{3}}=\frac{32}{3\sqrt{3}}$

Hence, this is the answer.