# Heron’s Area Formula

Here is the definition of Heron’s Formula which is an another way of finding the area of a triangle:

Given any triangle with sides of lengths a, b, and c, the area of the triangle is Area = √[s(s – a)(s – b)(s – c)] where s = (a + b + c)/2

And here is the proof:

The area of an oblique triangle is 1/2bc sin A where the height of the triangle is b sin A. If we square each side of the equation for area we get (Area)*2 = 1/4b*2c*2sin*2A. Now if we take the square root of each side of the equation we get Area = √[1/4b*2c*2sin*2A]. If we replace  sin*2A with the corresponding pythagorean identity, (1 – cos*2A), we get the equation √[1/4b*2c*2(1 – cos*2 A)]. Now we can factor the right side of the equation to get √[(1/2bc(1 + cos A))(1/2bc(1 – cos A)].

Using the Law of Cosines, we can show that 1/2bc(1 + cos A) = [(a + b + c)/2] · [(-a + b + c)/2] and 1/2bc(1 – cos A) = [(a – b + c)/2] · [(a + b – c)/2]. The middle dot signifies a multiplicative operation and not the dot product. I have uploaded a proof  below showing how to arrive at the first equivalence

. Now letting s = (a + b + c)/2, these two equations can be written as 1/2bc(1 + cos A) = s(s – a) and 1/2bc(1 – cos A) = (s – b)(s – c). By substituting into the last formula for area, we can conclude that Area = √[s(s – a)(s – b)(s – c)]