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

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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)]**