# What is the Herons formula

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### Heron's formula for the area of the triangle

Heron of Alexandria lived in the first century CE and was mainly concerned with geometry and applied mathematics in surveying. In his work Metrika he proved a formula for calculating the triangular area, which differs from the formula A = ½ · g · h commonly used today. According to Arabic sources, his formula goes back to Archimedes. The formula makes it possible to calculate the area of the triangle based solely on the three side lengths of the triangle, without knowing a height. Strangely enough, despite this advantage, it is largely unknown and usually does not appear in school lessons. Probably the reason is that the square root has to be taken, and that "may" not appear in school before the 9th grade. There is also no clear derivation - the proof presented here is probably too demanding for normal middle school levels, although it gets by with the geometric and algebraic means known in 9th grade.

The formula is as follows:

A._{triangle} = | (1) |

where s is half of the circumference:

s: = (a + b + c) / 2

##### Calculation example

Consider a triangle with sides a = 4cm, b = 6cm and c = 7.2cm.

First calculate s = (a + b + c) / 2 = (4 + 6 + 7.2) / 2 = 17.2 / 2 = 8.6.

Then compute A = √ (s * (sa) * (sb) * (sc)) = √ (8.6 * (8.6-4) * (8.6-6) * (8.6-7, 2)) = √ (8.6 x 4.6 x 2.6 x 1.4) = √143.9984 = 11.9999333331481 ... ~ 12.

The triangle has an area of 12cm² (rounded to the nearest thousandth).

#### proof

I shall now show that Heron's formula is equivalent to the well-known area formula. Drawing the exclusively positive square root while neglecting the negative root is permissible for geometric subjects, since negative lines are not relevant.

The Pythagorean theorem is used. (In right triangles, the sides that form the right angle are called the cathetus, and the third side opposite the right angle is called the hypotenuse. If you square the catheters, the sum of their squares is equal to the square of the hypotenuse.)

(Heron himself proved the formula differently. I may give his proof later.)

We consider a triangle with sides a, b, c and height h on a:

The base of the height divides the distance a into the sections p and q. In addition, the height divides the triangle into two right-angled sub-triangles.

Compare equations (18) and (23). They are identical. (18) is based on the division of the triangle into two right-angled triangles, whereby h could be calculated and inserted into the formula A = ½ · g · h.

(23) is based on Heron's formula. If both approaches lead to identical equations, they are equivalent, i.e. Heron's formula is equivalent to the known formula!

It remains to be clarified whether the formula also applies to triangles where the height lies outside the triangle or coincides with one side.

Both cases can be dealt with quickly:

If the height is outside then p = a + q, i.e. q = p - a = - (a - p)

In equation (5) the difference (a-p) is squared. But the square of - (a-p) or (p-a) gives the same result, namely also a

^{2}- 2ap + p^{2}.From (6) onwards, the path is identical again.

- If h = b, then the conventional area formula (see (13)) is: A = ½ * a * b (24).

Since the triangle in this case is right-angled with c as the hypotenuse, c holds^{2} = a^{2} + b^{2}, and the numerator in equation (23) simplifies as follows:

-a^{4} - b^{4} - c^{4} + 2a^{2}b^{2} + 2a^{2}c^{2} + 2b^{2}c^{2} (Counter from 23)

= -a^{4} - b^{4} - (a^{2} + b^{2})^{2} + 2a^{2}b^{2} + 2a^{2}(a^{2} + b^{2}) + 2b^{2}(a^{2} + b^{2})

= -a^{4} - b^{4} - a^{4} - 2a^{2}b^{2} - b^{4} + 2a^{2}b^{2} + 2a^{4} + 2a^{2}b^{2} + 2a^{2}b^{2} + 2b^{4}

= 4a^{2}b^{2}

This simplifies (23) overall to √ (4a^{2}b^{2}/ 16) = √ (a^{2}b^{2}/ 4) = a · b / 2, and this is the same as (24).

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© Arndt Brünner

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Version: March 20, 2003

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