Heron's Formula

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Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths.

Theorem

For any triangle with side lengths ${a}, {b}, {c}$, the area ${A}$ can be found using the following formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

where the semi-perimeter $s=\frac{a+b+c}{2}$.

Proof

Using basic Trigonometry, we have $$[ABC]=\frac{ab}{2}\sin C,$$ which simplifies to $$[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.$$

The Law of Cosines states that in triangle $ABC$, $c^2 = a^2 + b^2 - 2ab\cos C$, which can be written as $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Thus, $[ABC]=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}.$

Now, we can simplify: $$[ABC]=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$$

$$=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$$

$$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$$

$$=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$$

$$=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$$

$$=\sqrt{s(s-a)(s-b)(s-c)}$$

Isosceles Triangle Simplification

$A=\sqrt{s(s-a)(s-b)(s-c)}$ for all triangles

$b=c$ for all isosceles triangles

$A=\sqrt{s(s-a)(s-b)(s-b)}$ simplifies to $A=(s-b)\sqrt{s(s-a)}$ $\blacksquare$

Square root simplification/modification

From $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ We can "take out" the $1/2$ in each $s$, then we have $$A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$Using the difference of squares on the first two and last two factors, we get $$A=\frac{1}{4}\sqrt{(b^2+2bc+c^2-a^2)(a^2-b^2+2bc-c^2)}$$and using the difference of squares again, we get $$A=\frac{1}{4}\sqrt{(2bc)^2-(-a^2+b^2+c^2)^2}$$ From this equation (although seemingly not symmetrical), it is much easier to calculate the area of a certain triangle with side lengths with quantities with square roots. One can remember this formula by noticing that when finding the cosine of an angle in a triangle, the formula is $$\cos{A}=\frac{-a^2+b^2+c^2}{2bc}$$ and the two terms in the formula are just the denominator and numerator of the fraction for $\cos{A}$, only they're squared. This can also serve as a reason for why the area $A$ is never imaginary. This is equivalent of ending at step $4$ in the proof and distributing.

Note

Replacing $-a^2+b^2+c^2$ as $2bc\cos{A}$, the area simplifies down to $\frac{1}{4}\sqrt{(2bc\sin{A})^2}$, or $\frac{1}{4}\cdot2bc\sin{A}$, or $\frac{1}{2}bc\sin{A}$, another common area formula for the triangle.

Example

Let's say that you have a right triangle with the sides $3$ ,$4$ , and $5$. Your semi- perimeter would be $6$ since $(3+4+5)$ ÷ $2$ is $6$. Then you have $6-3=3$, $6-4=2$, $6-5=1$. $1\cdot 2\cdot 3=6.$ $6\cdot 6 = 36$ The square root of $36$ is $6$. The area of your triangle is $6$.