Area of a triangle: SSS (Part 5)

triangle

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: