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 $△OAB$ 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 $BD=\frac{b-0}{0-\frac{a}{2}}=-\frac{2b}{a}=m_1$ (let say)
Also, slope of $AE=\frac{\frac{b}{2}-0}{0-a}=-\frac{b}{2a}=m_2$ (let say)
$tan\angle DGA=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$\Rightarrow tan30^{\circ }=\left|\frac{\left(-\frac{2b}{a}\right)-\left(-\frac{b}{2a}\right)}{1+\left(-\frac{2b}{a}\right)\left(-\frac{b}{2a}\right)}\right|$
$\Rightarrow \frac{1}{\sqrt{3}}==\frac{\frac{3b}{2a}}{1+\frac{b^2}{a^2}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3ba}{2\left(a^2+b^2\right)}$
$\Rightarrow \frac{1}{2}ab=\left(\frac{a^2+b^2}{3\sqrt{3}}\right)...\left(i\right)$
Since, by Pythagoras theorem in $△OAB$ ,
${\left(OA\right)}^2+{\left(OB\right)}^2={\left(AB\right)}^2\Rightarrow a^2+b^2=3^2=\mathrm{9...}\left(ii\right)$
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.