Question

Two medians drawn from the acute angles of a right angled triangle intersect at an angle $\frac{\pi }{6}$. If the length of the hypotenuse of the triangle is $3$ units, and the area of the triangle(in square units) is $\sqrt{K}$, then $K$ is:

• A. $3$
• B. $\frac{3\sqrt{5}}{2}$
• C. $\sqrt{3}$
• D. None of these

Answer 1

The correct option is A $3$


Let $OAB$ be the triangle right angled at $O$ aligned along the coordinate axes, such that $OB$ is along the $y$ axis and $OA$ is along the $x$ axis.
Let coordianres of $A$ and $B$ be $(a,0)$ and $(0,b)$ respectively. Also, let $G$ be the point of intersection of the two medians, drawn through $A$ and $B$.
Coordinate of $G(\frac{a}{3},\frac{b}{3})$
Now, from figure we have:
slope of $GB=\frac{-2b}{a}$ and slope of $AG=\frac{-b}{2a}$
$\tan {30}^{\circ }=\frac{\frac{3b}{2a}}{1+\frac{b^2}{a^2}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3ab}{2(a^2+b^2)}=\frac{3ab}{2.9}=\frac{ab}{6}$
$\Rightarrow \frac{1}{2}ab=\frac{3}{\sqrt{3}}=\sqrt{3}$
Hence on comparing, we have $K=3$