Question

In a triange $ABC$, Median $AD$ and $BE$ are drawn. If $AD=4$units, $\angle DAB={30}^0$ and $\angle ABE={60}^0$, then area of triangle $ABC$ is

• A. $\frac{32\sqrt{3}}{3}$
• B. $\frac{32\sqrt{3}}{9}$
• C. $\frac{8}{3}$
• D. $\frac{61}{3}$

Answer 1

Answer: (B)

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

Given, In a triangle ABC,Median AD,BE are drawn.if AD=4 units,$\angle DAB=30,\angle ABE=60$,AG=$\frac{2}{3}\times AD=\frac{8}{3}$

In $△AGB,$

$\frac{AG}{sin60}=\frac{BG}{sin30}$

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

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

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

$=\frac{16}{3\sqrt{3}}$

Area of $△ABC$=2\times Area of $△ADB$

$=2\times \frac{16}{3\sqrt{3}}=\frac{32\sqrt{3}}{9}$

Option B is correct.