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

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$\sqrt{3}$

$\frac{3 \sqrt{5}}{2}$

None of these

answer is C.

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Detailed Solution

Slope $G C = \frac{- 2 b}{a}, slope of A G = \frac{- b}{2 a}$
$\tan 30^{\circ} = \frac{\frac{3 b}{2 a}}{1 + \frac{b^{2}}{a^{2}}} and a^{2} + b^{2} = 9$

$\begin{array}{ll} \Rightarrow \frac{1}{\sqrt{3}} = \frac{3 a b}{2 (a^{2} + b^{2})} \\ \Rightarrow \frac{1}{2} a b = (\frac{a^{2} + b^{2}}{3 \sqrt{3}}) \\ \Rightarrow \sqrt{k} = \frac{9}{3 \sqrt{3}} = \sqrt{3} \\ \Rightarrow k = 3 \end{array}$