Proof of Hero’s Formula

Question: Proof of Hero’s formula for the area of triangle ABC with sides a, b and c and s being the semi-perimeter i.e. s=a+b+c/2, then

Area A = [pic]

Proof:

Let a triangle ABC with sides a, b and c whose area is equal to A = [pic]. Let the triangle be as follows:

Here, perimeter is the length of the sides, now as the sides are a, b & c, assume it is also the length of the sides, then the perimeter of this triangle is

P = a+b+c

And semi-perimeter i.e. half of the perimeter is

S = a+b+c/2

If we draw a perpendicular from C to base and call it ‘h’ which divides the base into two parts i.e. x and c-x, then the diagram looks as follows:

The perpendicular has divided the triangle into two right-angled triangles. Now for any right-angle triangle, according to Pythagorean Theorem,

[pic] = [pic] + [pic]

If Pythagorean is applied to the right-angled triangles in the above triangle, then in the case of left right-angle triangle in the above diagram, it would give us the equation

[pic] = [pic] + [pic] where ‘a’ = hypotenuse and ‘h’ = height/perpendicular and ‘x’ = base.

Re-writing it, the equation would become which we will call Eq. A

[pic] = [pic] - [pic] ---------------------( Eq. A

Similarly, for the right angle triangle on the right half to triangle ABC,

[pic] = [pic] + [pic] where ‘b’ = hypotenuse, ‘h’ = height/perpendicular and ‘c-x’ = base.

Re-writing this equation would result in

[pic] = [pic] – [pic]

Expanding [pic] would give us

[pic] = [pic] – ([pic] + [pic] - 2cx)

[pic] = [pic] – [pic] - [pic] + 2cx --------------------------( Eq. B

As, the left hand sides of Eq.A and Eq. B are equal, we can equate them. Equating Eq.A and Eq. B would give us

[pic] - [pic] = [pic] –