Difference between revisions of "Heron's Formula"

Latest revision as of 19:11, 24 February 2025

Heron's Formula (sometimes referred to as Hero's formula) is a mathematical formula for finding the area of a triangle given the three side lengths.

History

The formula is credited to Hero of Alexandria, and a proof can be found in his book Metrica. Mathematical historian Thomas Heath suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.

A formula equivalent to Heron's was discovered by Chinese mathematician Qin Jiushao, published in Mathematical Treatise in Nine Sections in 1247: $A = \frac{1}{2} \sqrt{a^2 c^2 - (\frac{a^2 + c^2 - b^2}{2}^2)}$

Statement

In any triangle with side lengths $a$ , $b$ , $c$ , and semiperimeter $s$ the area $A$ is equal to

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

Proof

A common formula for area states that $[ABC]=\frac{ab}{2}\sin C$ 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: \begin{align*} [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)((a - b)^2 - c^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)} \end{align*}

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

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.

Note

In general, it is a good advice not to use Heron's formula in computer programs whenever we can avoid it. For example, whenever vertex coordinates are known, vector product is a much better alternative. Main reasons:

Computing the square root is much slower than multiplication.
For triangles with area close to zero Heron's formula computed using floating point variables suffers from precision problems.

@@ Line 66: / Line 66: @@
 * [[Brahmagupta's formula]]
-* [[Geometry]]
+* [[Triangle]]
+* [[Area]]
 [[Category:Geometry]]
 [[Category:Theorems]]
 {{stub}}

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Difference between revisions of "Heron's Formula"

Latest revision as of 19:11, 24 February 2025

Contents

History

Statement

Proof

Isosceles Triangle Simplification

Square root simplification/modification

Note

Note

Problems

Introductory

Intermediate

Olympiad

See Also