# Area of a triangle: SSS (Part 5) In yesterday’s post, we discussed how the area $K$ of a triangle can be found using SAS: two sides and the angle between the two sides. We found that $K = \displaystyle \frac{1}{2} a b \sin C$

This can be used as the starting point for the derivation of Heron’s formula, which determines the area of a triangle using SSS (i.e., only the three sides). I won’t give the full derivation in this post — there’s no point in me retyping the details — but will refer to the Wikipedia page and the MathWorld page for the details. However, I will give the big ideas behind the derivation.

1. We begin by recalling that $\sin^2 C + \cos^2 C = 1$. Since $0 < C < 180^o$, we know that $\sin C$ must be positive, so that $\sin C = \sqrt{1 - \cos^2 C}$

2. From the Law of Cosines, we know that $c^2 = a^2 + b^2 - 2 a b \cos C$,

or $\cos C = \displaystyle \frac{a^2 + b^2 - c^2}{2ab}$

3. Substituting, we see that $K = \displaystyle \frac{1}{2} ab \sqrt{1 - \cos^2 C}$ $K = \displaystyle \frac{1}{2} ab \sqrt{1 - \frac{(a^2+b^2-c^2)^2}{4a^2b^2}}$

4. This last expression only contains the side lengths $a$, $b$, and $c$. So the “only” work that’s left is simplifying this right-hand side and seeing what happens. After considerable work — requiring factoring the difference of two squares on two different steps — we end up with Heron’s formula: $K = \sqrt{s (s-a) (s-b) (s-c)}$

where $s = \displaystyle \frac{a+b+c}{2}$ is the semiperimeter, or half the perimeter of the triangle.

A final note: If you actually are able to start with Step 3 and end with Heron’s formula on your own — without consulting a textbook or the Internet if you get stuck — feel free to cry out “More power!” and grunt like Tim “The Toolman” Taylor: