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Q. 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

NTA AbhyasNTA Abhyas 2020

Solution:

Slope of $A G = - \frac{b}{2 a} , \, tan 30^{o} = \frac{\frac{3 b}{2 a}}{1 + \frac{b^{2}}{a^{2}}}$
Solution
$a^{2}+b^{2}=9\Rightarrow \frac{1}{\sqrt{3}}=\frac{3 b . a}{2 \left(a^{2} + b^{2}\right)}$
$\Rightarrow \, \frac{1}{2} \, ab=\left(\frac{a^{2} + b^{2}}{3 \sqrt{3}}\right)=\frac{9}{3 \sqrt{3}}=\sqrt{3}$