Question

In a right angled triangle, medians drawn from the acute angles make an angle of $\theta$ which each other and L is the length of the hypotenuse. Then the area of the triangle is equal to:

• A. $\Delta =\frac{L^{2}tan\theta }{6}$
• B. $\Delta =\frac{L^{2}tan\theta }{3}$
• C. $\Delta =\frac{L^{2}cot\theta }{3}$
• D. $\Delta =\frac{L^{2}cot\theta }{6}$

Answer 1

The correct option is B $\Delta =\frac{L^{2}tan\theta }{3}$





Given: angle ABC =${90}^{\circ }$ as $\Delta$ is right rt. angle
triangle and G is centroid of $\Delta$ as DC & AE are medians .
As, AE median AE;
Length of median AE ;
$AE=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$
Length $DC=\frac{1}{2}\sqrt{2b^2+2a^2-c^2}$
Centroid G divides median in ratio of 1: 2
$\therefore GC=\frac{2}{3}DC=\frac{1}{3}\sqrt{2b^2+2a^2-c^2}$
$\&GE=\frac{1}{3}AE=\frac{1}{6}\sqrt{2b^2+2c^2-a^2}$
$\&EC=\frac{a}{2}$
$\therefore$ In $\Delta GEC$ ; applying cosine formula
$cos\theta =\frac{G{E}^2+G{C}^2-E{C}^2}{2GE.GC}$
$=\frac{\frac{2b^2+2c^2-a^2}{36}+\frac{2b^2+2a^2-c^2}{9}-\frac{a^2}{4}}{2\frac{1}{6}\sqrt{2b^2+2c^2-a^2-\frac{1}{3}\sqrt{2b^2+2a^2-c^2}}}$
$=\frac{2b^2+2c^2-a^2+8b^2+8a^2-4c^2-9a^2}{4\sqrt{(4c^2+a^2)(4a^2+c^2)}}$
[using $a^2+c^2=b^2$ as angle B=90]
$\Rightarrow cos\theta =\frac{2b^2}{\sqrt{(4c^2+a^2)(4a^2+c^2)}}$




$\therefore$ perpendicular = ${\left[\right(4c^2+a^2\left)\right(4a^2+c^2)-4b^4]}^{\frac{1}{2}}$
$={[17a^2{c}^2+4a^4+4c^4-4a^4-4c^4-8a^{2}c]}^{\frac{1}{2}}$
$={\left[9a^2{c}^2\right]}^{\frac{1}{2}}$
$=3ac$
$\therefore tan\theta =\frac{3ac}{2b^2}$
But $tan\Delta =\frac{1}{2}ac$
$\therefore tan\theta =\frac{3\Delta }{b^2}$
$\therefore \Delta =\frac{b^{2}tan\theta }{3}$
$\therefore \Delta =\frac{L^{2}tan\theta }{3}$