Question

Two medians drawn from 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, then area of the triangle (in sq. units) is

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

Answer 1

Answer: (A)

$\sqrt{3}$

$G\equiv (\frac{b}{3},\frac{a}{3})$

Slope of line AG $=\frac{a-\frac{a}{3}}{0-\frac{b}{3}}=\frac{-2a}{b}$

Slope of MG $=\frac{\frac{a}{2}-\frac{a}{3}}{0-\frac{b}{3}}=\frac{-a}{2b}$

$\tan {30}^o=\frac{\frac{-a}{2b}-\left(\frac{-2a}{b}\right)}{1+\frac{a^2}{b^2}}$

$⟹\frac{1}{\sqrt{3}}=\frac{3ab}{2(a^2+b^2)}$

$⟹\frac{1}{2}ab=\frac{(a^2+b^2)}{3\sqrt{3}}$

In a right angled triangle $A{C}^2+B{C}^2=A{B}^2$

$⟹a^2+b^2=9$

$\therefore Area=\frac{1}{2}ab=\frac{9}{3\sqrt{3}}=\sqrt{3}$ sq.units