Show that the area of a triangle with side lengths $a$, $b$, and $c$ is:
\[A = \sqrt{s(s-a)(s-b)(s-c)}\]
where $s$ is the semiperimeter (half the perimeter):
\[s = \frac{a + b + c}{2}.\]
This is called Heron’s formula, after the ancient mathematician Heron (or Hero) of Alexandria.
Show hint
The area $A$ of the triangle is $bh/2$ where $h$ is the height of the triangle:
with $b_1 + b_2 = b$. From Pythagoras’ theorem,
\[c^2 = h^2 + b_1^2\] and \[a^2 = h^2 + b_2^2.\]
We need to find an expression for $h$ in terms of $a$, $b$, and $c$, so let’s try to eliminate $b_1$ and $b_2$:
\begin{align*}
a^2 &= h^2 + b_2^2\\
&= h^2 + (b - b_1)^2\\
&= h^2 + b^2 - 2bb_1 + b_1^2\\
&= h^2 + b^2 - 2bb_1 + (c^2 - h^2)\\
&= b^2 - 2bb_1 + c^2\\
b_1 &= \frac{b^2 + c^2 - a^2}{2b}
\end{align*}
Now we can solve for $h$:
\begin{align*}
h^2 &= c^2 - b_1^2\\
&= c^2 - \frac{(b^2 + c^2 - a^2)^2}{4b^2}\\
&= \frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{4b^2}
\end{align*}
And find $A$:
\begin{align*}
A &= \frac{bh}{2}\\
A^2 &= \frac{b^2h^2}{4}\\
&= \frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{16}
\end{align*}
Now we could expand this, and expand $s(s-a)(s-b)(s-c)$ and show that they’re the same, but that wouldn’t be very satisfying. Instead we can factorise it using the difference of two squares ($x^2 - y^2 = (x-y)(x+y)$, see Section 7.4.3: Expanding and Factorising):
\begin{align*}
A^2 &= \tfrac{1}{16}\left(4b^2c^2 - (b^2 + c^2 - a^2)^2\right)\\
&= \tfrac{1}{16}\left(2bc - (b^2 + c^2 - a^2)\right)\left(2bc + (b^2 + c^2 - a^2)\right)\\
&= \tfrac{1}{16}(a^2 - b^2 + 2bc - c^2)(b^2 + 2bc + c^2 - a^2)\\
&= \tfrac{1}{16}\left(a^2 - (b - c)^2\right)\left((b + c)^2 - a^2\right)\\
&= \tfrac{1}{16}\left(a - (b - c)\right)\left(a + (b-c)\right)\left((b + c) - a\right)\left((b + c) + a\right)\\
&= \frac{a - b + c}{2}\cdot\frac{a + b - c}{2}\cdot\frac{b + c - a}{2}\cdot\frac{a + b + c}{2}\\
&= (s - b)(s - c)(s - a)s\\
A &= \sqrt{s(s-a)(s-b)(s-c)}
\end{align*}