# 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:

$\overline{BC}=2a$ $\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{c}{b}$ let $\overline{AB}=2cx,\overline{AC}=2bx$ $\Rightarrow s=a+x(b+c)$ $\Rightarrow s-\overline{BC}=x(b+c)-a$ $\Rightarrow s-\overline{AB}=a+x(b-c)$ $\Rightarrow s-\overline{AC}=a-x(b-c)$ If the area is $\triangle$ then $\Rightarrow \triangle ^{2} = (x^2(b+c)^2-a^2)(a^2-x^2(b-c)^2)$ $= -x^4(b^2-c^2)^2+2a^2x^2(b^2+c^2)-a^4$ let $\gamma = x^2$ $\Rightarrow -\gamma ^2(b^2-c^2)^2+2a^2\gamma (b^2+c^2)-a^4=\triangle ^2$ Now, maxima of $\triangle =$ maxima of $\triangle ^2 =$ maxima of $-\gamma ^2(b^2-c^2)^2+2a^2\gamma (b^2+c^2)-a^4$

$\therefore$ maximum value of $\triangle$ is at $\gamma=\dfrac{a^2(b^2+c^2)}{(b+c)^2(b-c)^2}=x^2$

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

Inspiration

Note :

• I have skipped the last part for the readers. Note by Zakir Husain
5 months ago

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