Question

What is Herons formula ?

Answer 1

You can calculate the area of a triangle if you know the lengths of all three sides, using a formula that has been known for nearly $2000$ years.
It is called "Heron's Formula" after Hero of Alexandria (see below)
Just use this two step process:
Step 1: Calculate '$s$' (half of the triangles perimeter): $s=\frac{a+b+c}{2}$
Step 2: Then calculate the Area : $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

Derivation of Herons formula:

Let in a $\Delta ABC$,

The sides are $AB=c,BC=a$ and $AC=b$

Draw a $AD\perp BC$

Let, $BC=x$

$\Rightarrow CD=a-x$

Now, In $\Delta ADC,\angle ADC={90}^{\circ }$

$\Rightarrow b^2=(a-x{)}^2+h^2$

$\Rightarrow b^2-(a-x{)}^2=h^2$----(1)

Similarly,

In $\Delta ADB,\angle ADB={90}^{\circ }$

$\Rightarrow c^2=x^2+h^2$

$\Rightarrow c^2-x^2=h^2$---(2)

From (1) and (2),

$b^2-(a-x{)}^2=c^2-x^2$

$\Rightarrow b^2-a^2-x^2+2ax=c^2-x^2$

$\Rightarrow b^2-a^2+2ax=c^2$

$\Rightarrow 2ax=c^2+a^2-b^2$

$\Rightarrow x=\frac{c^2+a^2-b^2}{2a}$ ---(3)

Now, from (2),

$c^2-x^2=h^2$

$\Rightarrow c^2-(\frac{c^2+a^2-b^2}{2a}{)}^2=h^2$

$\Rightarrow \frac{(2ac{)}^2-(c^2+a^2-b^2{)}^2}{(2a{)}^2}=h^2$

$\Rightarrow h=\sqrt{\frac{(2ac{)}^2-(c^2+a^2-b^2{)}^2}{(2a{)}^2}}$

$\Rightarrow h=\frac{\sqrt{(2ac{)}^2-(c^2+a^2-b^2{)}^2}}{2a}$ ---(4)

Now,

Area of $\Delta ABC=\frac{1}{2}ah$

$=\frac{1}{2}a\frac{\sqrt{(2ac{)}^2-(c^2+a^2-b^2{)}^2}}{2a}$

$=\frac{1}{4}\sqrt{(2ac{)}^2-(c^2+a^2-b^2{)}^2}$

$=\frac{1}{4}\sqrt{(2ac+c^2+a^2-b^2)(2ac-(c^2+a^2-b^2)}$

$=\frac{1}{4}\sqrt{(2ac+c^2+a^2-b^2)(b^2-(c^2+a^2-2ac\left)\right)}$

$=\frac{1}{4}\sqrt{\left(\right(a+c{)}^2-b^2\left)\right(b^2-(a-c{)}^2)}$

$=\frac{1}{4}\sqrt{(a+b+c)(a-b+c)(-a+b+c)(a+b-c)}$----(5)

Let, Perimeter of triangle $ABC$ is,

$2s=a+b+c$ ---(i)

$\Rightarrow s=\frac{a+b+c}{2}$ ---(ii)

$2s=a+b+c$

$\Rightarrow 2s-2a=a+b+c-2a=-a+b+c$---(iii)

$2s-2b=a+b+c-2b=a-b+c$ ----(iv)

$2s-2c=a+b+c-2c=a+b-c$ -----(v)

Now, from (5)

Area of $\Delta ABC=\frac{1}{4}\sqrt{(a+b+c)(a-b+c)(-a+b+c)(a+b-c)}$

$=\frac{1}{4}\sqrt{\left(2s\right)(2s-2b)(2s-2a)(2s-2c)}$

$=\frac{1}{4}\sqrt{\mathrm{2.2.2.2.}\left(s\right)(s-b)(s-a)(s-c)}$

$=\frac{1}{4}\times 4\sqrt{s(s-b)(s-a)(s-c)}$

$=\sqrt{s(s-b)(s-a)(s-c)}$

Therefore, Area of $\Delta ABC=\sqrt{s(s-b)(s-a)(s-c)}$

i.e., $\Delta =\sqrt{s(s-b)(s-a)(s-c)}$

Hence, proved