Question

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

• A. $\sqrt{3}$
• B. $3$
• C. $\sqrt{2}$
• D. $9$

Answer 1

Answer: (A)

Let the sides of the right-angle triangle be coinciding with the coordinate axes and the vertices of $\triangle OAB$ be $O\left(0 , 0\right),A\left(a , 0\right)\&B\left(0 , b\right)$ .
Also, let $BD\&AE$ are the medians cutting the sides at $D\&E$ .
So, coordinates of $D=\left(\frac{a}{2} , 0\right)\&E=\left(0 , \frac{b}{2}\right)$ .
Now, Slope of $BD=\frac{b - 0}{0 - \frac{a}{2}}=-\frac{2 b}{a}=m_{1}$ (let say)
Also, slope of $AE=\frac{\frac{b}{2} - 0}{0 - a}=-\frac{b}{2 a}=m_{2}$ (let say)
$tan\angle DGA=\left|\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}\right|$
$\Rightarrow tan30^\circ =\left|\frac{\left(- \frac{2 b}{a}\right) - \left(- \frac{b}{2 a}\right)}{1 + \left(- \frac{2 b}{a}\right) \left(- \frac{b}{2 a}\right)}\right|$
$\Rightarrow \frac{1}{\sqrt{3}}==\frac{\frac{3 b}{2 a}}{1 + \frac{b^{2}}{a^{2}}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3 b a}{2 \left(a^{2} + b^{2}\right)}$
$\Rightarrow \frac{1}{2}ab=\left(\frac{a^{2} + b^{2}}{3 \sqrt{3}}\right)...\left(i\right)$
Since, by Pythagoras theorem in $\triangle OAB$ ,
$\left(O A\right)^{2}+\left(O B\right)^{2}=\left(A B\right)^{2}\Rightarrow a^{2}+b^{2}=3^{2}=9...\left(ii\right)$
Putting equations $\left(ii\right)$ in equation $\left(i\right)$ , we get
$\frac{1}{2}ab=\left(\frac{9}{3 \sqrt{3}}\right)=\sqrt{3}$
Hence, area of triangle is $\sqrt{3}$ Sq. units.