In geometry, Heron's formula provides an alternative method for finding the area of a triangle. The formula is very useful when only the lengths of the sides are known.

This formula is credited to mathematician Heron of Alexandria. The formula is a special case of Brahmagupta's formula.

The proof and demonstration of Heron's formula can be done from algebra and elementary consideration of geometry. Also applications of the law of cosine help to derive the Heron's formula for the area of a triangle in which all sides of the triangle are given. One of the numbers found in this formula is the semiperimeter of a triangle, which defined as one half the perimeter.

In this article, we are going to study about Heron's formula and its derivation in detail.

The area of a triangle is often calculated using the formula A = $\frac{1}{2}$ bh ("one half the base times the altitude"), where b is the length of the side of given triangle taken as base and h is the length of the corresponding altitude on it.

Heron’s formula provides a method to compute the area when the lengths of all the three sides of the triangles are known. In other words we can say that, Heron's formula gives the area of a triangular region in terms of the lengths of its sides.
Heron’s formula was given by the great mathematician from Egypt, Heron of Alexandria. His name is also referred to as Hero and hence the formula is also known as Hero’s formula.  

This formula for the area of triangle and its proof are found in his work “Metrica” which deals with measurements.
The heron's formula for the area of a triangle is stated as follows:

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

where a, b c are the lengths of 

the sides of the triangle and S is

the semi perimeter i.e. s= $\frac{a+b+c}{2}$

Area of Heron's Formula

The algebraic proof for Heron's formula can be given, using the general formula for area of a triangle.
Proof of Heron's Formula

The general formula for area $\Delta$ =$\frac{1}{2}$ x Base x Height

From the expression for P = a + b + c = 2s

we can also derive,

2s -2a = b + c-a,     2s -2b = c + a -b  and  2s - 2c = a + b - c

Taking side BC as base,   base = a   and height = h.

Using Pythagorean theorem the height h is related to sides of the triangles in two ways.

In right triangle ABD,   $h^2$= $c^2$-$x^2$   ..............(1)

In right triangle ACD,   $h^2$= $b^2$ - $(a-x)^2$  ..............(2)

Eliminating $h^2$ between (1) and (2), we get,

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

$c^2 -x^2= b^2-(a^2-2ax+x^2)$

$c^2-x^2= b^2 -a^2+2ax-x^2$  and x can be solved as

x= $\frac{c^2+a^2-b^2}{2a}$

Substituting the value of x in equation 1, we get

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

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

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

The numerator is a difference of squares and hence can be factored.

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

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

Substituting the factors suitably with the expressions given in terms of s given in the beginning of proof we get,

$h^2$ = $\frac{2s(2s-2b)(2s-2a)(2s-2c)}{4a^2}$=$\frac{16s(s-a)(s-b)(s-c)}{4a^2}$= $\frac{4s(s-a)(s-b)(s-c)}{a^2}$

Thus we get  $h$= $\frac{2\ \sqrt{s(s-a)(s-b)(s-c)}}{a}$

Substituting base = a and height $h$= $\frac{2\ \sqrt{s(s-a)(s-b)(s-c)}}{a}$

Area $\Delta$ =$\frac{1}{2}$ x $a$ x $\frac{2\ \sqrt{s(s-a)(s-b)(s-c)}}{a}$

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

Heron’s formula can also be derived from some trigonometric formulas known:

The trig formula for the area of the triangle $\Delta$ =$\frac{1}{2}$ $bc sinA$

According to cosine formula, $Cos A$ =$\frac{b^2+c^2-a^2}{2bc}$

And 

$Sin^2 A$= 1- $Cos^2 A$ = 1 - ($\frac{b^2+c^2-a^2}{2bc}$)$^2$

$\frac{4b^2c^2-(b^2+c^2-a^2)^2}{4b^2c^2}$ which simplifies to

$Sin^2 A$ =$\frac{4s(s-a)(s-b)(s-c)}{b^2 c^2}$
 or 
$Sin A$ = $\frac{2\ \sqrt{s(s-a)(s-b)(s-c)}}{bc}$

Substituting in area formula,

Area of triangle ABC $\Delta$ = $\frac{1}{2}$ x bc x $\frac{2\ \sqrt{s(s-a)(s-b)(s-c)}}{bc}$

which is indeed the Heron’s Formula.

Example 1: Solve for the area of the triangle with side lengths 13 in, 14 in and 15 in?

Solution: S= $\frac{a+b+c}{2}$ = $\frac{13+14+15}{2}$ = 21

Area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

$\sqrt{21(21-13)(21-14)(21-15)}$

$\sqrt{21 \times 8 \times 7 \times 6}$

$\sqrt{7056}$  = 84 sq.inches.

Example 2: Find the area of the triangle whose sides measure 10 cm, 17 cm and 21 cm.  Also determine the length of the altitude on the side which measures 17 cm? 

Solution: 
S=$\frac{a+b+c}{2}$= $\frac{10+17+21}{2}$ = 24

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

$\sqrt{24 \times 14 \times 7 \times 3}$

$\sqrt{7056}$ = 84 sq.cm

Taking 17 cm as the base length we need to find the height

Area A = $\frac{1}{2}$ x base x height

$\frac{1}{2}$  x 17 x h = 84   or   h = $\frac{168}{17}$  = 9.88 cm   (Rounded to the nearest hundredth).