Areas of Triangles

Posted by Beetle B. on Tue 23 November 2021

When You Know 2 Sides and 1 Angle

\begin{equation*} A=\frac{1}{2} b c \sin \alpha \end{equation*}

When You Know 1 Side and 2 Angles

\begin{equation*} \frac{a^{2} \sin \beta \sin \gamma}{2 \sin \alpha} \end{equation*}

Proof:

Utilize the law of sines and plug into the previous formula.

Hero’s Formula

\begin{equation*} A=\sqrt{s(s-a)(s-b)(s-c)} \end{equation*}

Proof:

Start with:

\begin{equation*} A=\frac{1}{2} b c \sin \alpha \end{equation*}

Then:

\begin{equation*} \begin{array}{l}=\frac{1}{2} b c \cdot 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \quad\left(\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right) \\ =b c \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{s(s-a)}{b c}} \text { (by half angle formulas) } \\ =b c \frac{\sqrt{s(s-a)(s-b)(s-c)}}{b c}\end{array} \end{equation*}