Question

In a triangle $ABC,$ median $AD$ and $BE$ are drawn. If $AD=4,DAB=\frac{\pi }{4}$ and $\angle ABE=\pi /3,$

then the area of the triangle $ABC$ is?

• A. $\frac{16}{3}$
• B. $\frac{32\sqrt{3}}{9}$
• C. $\frac{64}{3}$
• D. $\frac{8}{3}$

Answer 1

Answer: (B)

$\frac{32\sqrt{3}}{9}$

Given $AD=4,\angle BAD=\frac{\pi }{4},\angle ABE=\frac{\pi }{3}$

$\angle AOB=\pi -\left(\frac{\pi }{6}\frac{\pi }{3}\right)=\left(\frac{\pi }{2}\right)$

O is centroid $AO=\frac{2}{3}\left(AD\right)=\left(\frac{8}{3}\right)\sin \left(60\right)=\left(\frac{AO}{AB}\right),AB=\left(\frac{16}{3\sqrt{3}}\right)$

Area of $\left(△ABD\right)=\frac{1}{2}\left(AB\right)\left(AD\right)\sin \left(30\right)=\frac{1}{4}\left(\frac{16}{3\sqrt{3}}\right)\left(4\right)=\frac{16}{3\sqrt{3}}$

Area of $\left(△ABC\right)$ = 2 (Area of $△$ABD)

$2\times \left(\frac{16}{3\sqrt{3}}\right)=\frac{32\sqrt{3}}{9}$