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 $\Delta ABC$ is

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

Answer 1

Answer: (A)

$\frac{8}{3}$

Given:$AD=4,$ and $BD=DC$

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

$\therefore AO=\frac{8}{3}$ and $DO=\frac{4}{3}$

In $△ABO,\tan \frac{\pi }{3}=\frac{AO}{BO}$

$\Rightarrow BO=AO\cot \frac{\pi }{3}$

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

Area of $△ADB=\frac{1}{2}\times AD\times BO$

$=\frac{1}{2}\times 4\times \frac{8}{3\sqrt{3}}=\frac{16}{3\sqrt{3}}$

Since, median divides a triangle in to two triangles of equal area.

$\therefore$Area of $△ABC=2\times$Area of $△ADB$

$=2\times \frac{16}{3\sqrt{3}}=\frac{32}{3\sqrt{3}}$sq.units