First slide

Equation of line in Straight lines

Question

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

Difficult

A

$\sqrt{3}$
B

3
C

$\sqrt{2}$
D

9

Solution

Let A(a,0) and B(0,b) then G= $(\frac{a}{3}, \frac{b}{3})$
Slope of is AN = $\frac{- b}{2 a}$ and slope of BM = $\frac{- 2 b}{a}$
Now $i f θ i s t h e a n g l e b e t w e e n t h e m e d i a n s t h e n \tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}|$
$\tan 30^{0} = |\frac{\frac{- b}{2 a} - (\frac{- 2 b}{a})}{1 + \frac{- b}{2 a} (\frac{- 2 b}{a})}| = |\frac{\frac{3 b}{2 a}}{1 + \frac{b^{2}}{a^{2}}}|$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{3 b a}{2 (a^{2} + b^{2})}$
$\Rightarrow \frac{1}{2} a b = (\frac{a^{2} + b^{2}}{3 \sqrt{3}})$ taking $a^{2} + b^{2} = 9$
ab= $2 \sqrt{3}$
Area of the triangle = $\frac{1}{2} a b = \frac{1}{2} \times 2 \sqrt{3}$
$= \sqrt{3}$

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