Area of a Triangle

10.6 Area of a Triangle — Four Formulas

(1) SAS form: $\Delta=\frac12bc\sin A=\frac12ca\sin B=\frac12ab\sin C$.

Proof. Drop a perpendicular from $C$ to $AB$, meeting it at $D$, so $CD=b\sin A$ (in right triangle $ACD$). Taking $AB=c$ as the base, $\Delta=\frac12\times\text{base}\times\text{height}=\frac12c(b\sin A)=\frac12bc\sin A$. The other two forms follow by choosing a different base.

(2) Heron's formula: $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$.

Proof. From §10.5, $\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}$, so $\sin A=2\sin\frac A2\cos\frac A2=\dfrac{2\sqrt{s(s-a)(s-b)(s-c)}}{bc}$. Substituting into $\Delta=\frac12bc\sin A$ gives $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$.

(3) $\Delta=\dfrac{a^2\sin B\sin C}{2\sin A}$.

Proof. By the sine rule, $c=\dfrac{a\sin C}{\sin A}$. Substituting into $\Delta=\frac12ac\sin B$ gives $\Delta=\frac12a\cdot\dfrac{a\sin C}{\sin A}\cdot\sin B=\dfrac{a^2\sin B\sin C}{2\sin A}$.

(4) $\Delta=2R^2\sin A\sin B\sin C$.

Proof. By the sine rule, $b=2R\sin B$, $c=2R\sin C$. Substituting into $\Delta=\frac12bc\sin A$ gives $\Delta=\frac12(2R\sin B)(2R\sin C)\sin A=2R^2\sin A\sin B\sin C$.

Each formula is the natural choice for a different combination of given data: (1) for SAS, (2) for SSS, (3) for ASA/AAS with one side, (4) when the circumradius is known.

Worked example. For $a=13,b=14,c=15$: $s=21$, so $\Delta=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7056}=84$. This value threads through §§10.7–10.8: $r=\Delta/s=84/21=4$ and $r_1=\Delta/(s-a)=84/8=10.5$.

A second worked example (using the circumradius). Suppose $A=45^\circ,B=60^\circ,C=75^\circ$ and $R=10$. By formula (4),

This situation — three angles and the circumradius known, but no side lengths at all — is exactly where formulas (1)–(3) cannot even get started, making formula (4) indispensable rather than merely an algebraic curiosity. …