You know the area of a triangle is $\frac{1}{2}ab\sin C$ (area with sine). But what if you know only the three sides? The law of cosines could find an angle first, but a formula found nearly two thousand years ago does it in one step.
Idea clave
For a triangle with sides $a$, $b$, $c$, let $s = \dfrac{a + b + c}{2}$ (the semiperimeter). Then $$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}.$$
Por qué funciona
Start with $$\text{Area}^2 = \frac{1}{4}a^2b^2\sin^2 C = \frac{1}{4}a^2b^2(1 - \cos^2 C),$$ and the law of cosines gives $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. After substituting, $$16\,\text{Area}^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2.$$ That is a difference of squares, and factoring twice more gives $$(a + b + c)(-a + b + c)(a - b + c)(a + b - c).$$ Each factor is $2s$, $2(s - a)$, $2(s - b)$, $2(s - c)$, and their product is $16\,s(s - a)(s - b)(s - c)$.
Worked examples
Ejemplo resuelto: A whole-number area
Find the area of a triangle with sides $13$, $14$ and $15$.
- $s = \dfrac{13 + 14 + 15}{2} = 21$.
- $s - a = 8$, $s - b = 7$, $s - c = 6$.
- Area $= \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$.
Respuesta: $84$ square units
Ejemplo resuelto: A radical area
Find the exact area of a triangle with sides $4$, $5$ and $6$.
- $s = 7.5 = \frac{15}{2}$; the other factors are $\frac{7}{2}$, $\frac{5}{2}$, $\frac{3}{2}$.
- Area $$= \sqrt{\dfrac{15 \cdot 7 \cdot 5 \cdot 3}{16}} = \dfrac{\sqrt{1575}}{4} = \dfrac{15\sqrt{7}}{4}.$$
Respuesta: $\frac{15\sqrt{7}}{4} \approx 9.92$
Ejemplo resuelto: Finding a height
A triangle has sides $5$, $5$ and $6$. How tall is it, measured to the side of length $6$?
- $s = 8$, so the area is $\sqrt{8 \cdot 3 \cdot 3 \cdot 2} = \sqrt{144} = 12$.
- Area is also $\frac{1}{2} \cdot 6 \cdot h$, so $3h = 12$.
Respuesta: $h = 4$
Error común
Using the whole perimeter where the formula needs half of it. With $P = 12$ for the $3$-$4$-$5$ right triangle you would get $\sqrt{12 \cdot 9 \cdot 8 \cdot 7}$, about $78$, but its area is $\frac{1}{2} \cdot 3 \cdot 4 = 6$. With $s = 6$: $\sqrt{6 \cdot 3 \cdot 2 \cdot 1} = 6$. ✓ Each factor $s - a$ must be positive; if one is zero or negative, the sides cannot form a triangle.