AD, BE and CF are medians of a triangle ABC, right angled at C. AD lies along the line (1−3)x+(3+1)y−6=0, BE lies along the line (1−23)x+(3+2)y−8=0. If the length of the hypotenuse is 60 and area of triangle ABC is ∆, then the value of Δ10.

mAD=2−3;mBE=53−8tan⁡θ=mAD−mBE1+mAD⋅mBE=13 Also, tan⁡α=b2a and tan⁡(θ+α)=2ba⇒b2a+131−b6a=2ba⇒2a2+b2=9(ab) Now, Area =12×ab=12×2a2+b29=400

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AD, BE and CF are medians of a triangle ABC, right angled at C. AD lies along the line $(1 - \sqrt{3}) x + (\sqrt{3} + 1) y - 6 = 0,$ BE lies along the line $(1 - 2 \sqrt{3}) x + (\sqrt{3} + 2) y - 8 = 0$ . If the length of the hypotenuse is 60 and area of triangle ABC is $∆$ , then the value of $\frac{Δ}{10} .$

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answer is 400.

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$m_{A D} = 2 - \sqrt{3}; m_{B E} = 5 \sqrt{3} - 8$

$\tan θ = |\frac{m_{A D} - m_{B E}}{1 + m_{A D} \cdot m_{B E}}| = \frac{1}{3}$

$Also, \tan α = \frac{b}{2 a} and$ $\tan (θ + α) = \frac{2 b}{a} \Rightarrow \frac{\frac{b}{2 a} + \frac{1}{3}}{1 - \frac{b}{6 a}} = 2 \frac{b}{a}$

$\Rightarrow 2 (a^{2} + b^{2}) = 9 (a b) Now, Area$

$= \frac{1}{2} \times a b = \frac{1}{2} \times \frac{2 (a^{2} + b^{2})}{9} = 400$

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