Heron's Formula!

If in \(\triangle ABC\) \(a=\dfrac{\overline{BC}}{2},\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{c}{b}\) then the maximum area of the triangle \(= \dfrac{2a^2bc}{|b^2-c^2|}\)

Proof:

BC=2a\overline{BC}=2a ABAC=cb\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{c}{b} let AB=2cx,AC=2bx\overline{AB}=2cx,\overline{AC}=2bx s=a+x(b+c)\Rightarrow s=a+x(b+c) sBC=x(b+c)a\Rightarrow s-\overline{BC}=x(b+c)-a sAB=a+x(bc)\Rightarrow s-\overline{AB}=a+x(b-c) sAC=ax(bc)\Rightarrow s-\overline{AC}=a-x(b-c) If the area is \triangle then 2=(x2(b+c)2a2)(a2x2(bc)2)\Rightarrow \triangle ^{2} = (x^2(b+c)^2-a^2)(a^2-x^2(b-c)^2) =x4(b2c2)2+2a2x2(b2+c2)a4= -x^4(b^2-c^2)^2+2a^2x^2(b^2+c^2)-a^4 let γ=x2\gamma = x^2 γ2(b2c2)2+2a2γ(b2+c2)a4=2\Rightarrow -\gamma ^2(b^2-c^2)^2+2a^2\gamma (b^2+c^2)-a^4=\triangle ^2 Now, maxima of =\triangle = maxima of 2=\triangle ^2 = maxima of γ2(b2c2)2+2a2γ(b2+c2)a4-\gamma ^2(b^2-c^2)^2+2a^2\gamma (b^2+c^2)-a^4

\therefore maximum value of \triangle is at γ=a2(b2+c2)(b+c)2(bc)2=x2\gamma=\dfrac{a^2(b^2+c^2)}{(b+c)^2(b-c)^2}=x^2

\therefore maximum area =2a2bcb2c2= \dfrac{2a^2bc}{|b^2-c^2|}


Inspiration

Note :

  • I have skipped the last part for the readers.

Note by Zakir Husain
1 month, 2 weeks ago

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