Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle $\frac{\mathrm{\pi }}{6}$. If the length of the hypotenuse of the triangle is $3$ units, then area of the triangle (in sq. units) is

• Correct answer
  A

  $\sqrt{3}$
• B

  $3$
• C

  $\sqrt{2}$
• D

  $9$

Solution

Slope of $AG=-\frac{b}{2a},\tan {30}^o=\frac{\frac{3b}{2a}}{1+\frac{b^2}{a^2}}$



${\mathrm{a}}^2+{\mathrm{b}}^2=9\Rightarrow \mathrm{}\frac{1}{\sqrt{3}}=\frac{3\mathrm{b}.\mathrm{a}}{2\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}$

$\Rightarrow \mathrm{ }\frac{1}{2}\mathrm{ }\mathrm{a}\mathrm{b}=\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{3\sqrt{3}}\right)=\frac{9}{3\sqrt{3}}=\sqrt{3}$

Hence, the correct answer is option: 'A'.