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 AB‾AC‾=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) ⇒s−BC‾=x(b+c)−a\Rightarrow s-\overline{BC}=x(b+c)-a ⇒s−AB‾=a+x(b−c)\Rightarrow s-\overline{AB}=a+x(b-c) ⇒s−AC‾=a−x(b−c)\Rightarrow s-\overline{AC}=a-x(b-c) If the area is △\triangle then ⇒△2=(x2(b+c)2−a2)(a2−x2(b−c)2)\Rightarrow \triangle ^{2} = (x^2(b+c)^2-a^2)(a^2-x^2(b-c)^2) =−x4(b2−c2)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(b2−c2)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(b2−c2)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(b−c)2=x2\gamma=\dfrac{a^2(b^2+c^2)}{(b+c)^2(b-c)^2}=x^2

∴\therefore maximum area =2a2bc∣b2−c2∣= \dfrac{2a^2bc}{|b^2-c^2|}

Inspiration

Note :

  • I have skipped the last part for the readers.
#Geometry

Note by Zakir Husain
3 months, 2 weeks ago

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