How do you derive Heron's Formula for scalene triangle?

Answer 1

We stat with the law of cosines:

• c2 = a2 + b2 - 2ab*cos(C)

Then rearrange it:

• cos(C) = a2 + b2 - c2/2ab

Use the identity sin(x)=SQRT(1-cos2(x))

• sin(C) = SQRT( 1 - (a2 + b2 - c2/2ab)2)

Use the operator A = 1/2ab*sin(C) where A is area. Also, set one equal to 4a2b2 and factor it out.

• 2A/ab = SQRT(4a2b2 - (a2 + b2 - c2)2)/2ab

ab's cancel, and the term inside the square root is the difference of two squares.

• A = 1/4*SQRT((2ab - (a2 + b2 - c2))(2ab + (a2 + b2 - c2)))

when the two groups are simplified, the can be factored in binomial squares.

• A = 1/4*SQRT((c2 - (a - b)2)((a + b)2 - c2)

Once again, we have differences of squares.

• A = 1/4*SQRT((c - (a - b))(c + (a - b))((a + b) - c)((a + b + c))

Simplify.

• A = 1/4*SQRT((c + b - a)(c + a - b)(a + b - c)(a + b +c))

Here comes the tricky part. We have four parts here. Three have two terms positive and one negative. Having a + b + c is like having the P. If we have a + b - c, that is like saying P - 2c, right? So to make it even easier, we can call s, the semi-perimeter, P/2. Then we can say a + b - c is 2s - 2c, or 2(s - c). We can apply that to all parts except the last one, which is just 2s.

• A = 1/4*SQRT(2(s - a)*2(s - b)*2(s - c)*2s)

The two's can multiply together to 16 and come out of the root, canceling with the 1/4. we are left with good old Heron's formula.

• A = SQRT(s(s - a)(s - b)(s - c))