Heron's formula proof

Question

I need to prove that the area of triangle ABC = √s(s-a)(s-b)(s-c) its Heron's formula

Answer 1

The classic formula is A=1/2 b*h, where b is the base and h is the triangle's height.

In the proof, we cannot assume we are given h so that we can transform h into b*sin C:

Area=ab/2 * sin C

Now we will use the law of cosines to prove Heron's formula since the law of cosines only has triangle side lengths.

We will use the Pythagorean identity to rewrite sin C.

sin²C + cos²C=1

sin²C =1 -cos²C

sin C = √1 - cos²C

Now using the law of cosines:

Area ABC = ab/2 * sqrt(1 - cos²C)

= ab/2 * sqrt(1- ((c²-a²-b²)² / (4a²b²))

= sqrt((4a²b² - (c²-a²-b²)) /16)

= sqrt(((2ab -c²+a²+b²)*(2ab -c²+a²+b²)) /16)

= sqrt( (((a+b)² - c²)*(c² - (a-b)²))/16 )

= sqrt(((a+b-c)*(a+b+c)*(a-b+c)*(-a+b+c))/16)

= sqrt(s*(s-a)*(s-b)*(s-c))

This completes the proof.

s is the semiperimeter of the triangle s = (a+b+c)/2