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

Let A(a,0) and B(0,b) then G=$\left(\frac{a}{3},\frac{b}{3}\right)$

Slope of is AN = $\frac{-b}{2a}$  and slope of BM = $\frac{-2b}{a}$ 

Now $if \theta is the angle between the medians then \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan {30}^0= \left|\frac{{\displaystyle \frac{-b}{2a}}-\left({\displaystyle \frac{-2b}{a}}\right)}{1+\frac{-b}{2a}\left(\frac{-2b}{a}\right)}\right| =\left|\frac{\frac{3b}{2a}}{1+\frac{b^2}{a^2}}\right|$

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

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

ab=$2\sqrt{3}$

Area of the triangle = $\frac{1}{2}ab=\frac{1}{2}\times 2\sqrt{3}$

$=\sqrt{3}$