EMAT 6690

Heron’s Formula

Ken Montgomery

Although the area formula for a triangle enables one to easily calculate the area of a triangular region, one must be able to measure the height (Equation 1).

**Equation
1: _{}**

In some cases, as in a farmer’s field for instance, a measure of the height is difficult to obtain, although the individual side lengths are easily measured (Figure 1).

**Figure
1:** _{}with
side lengths, *a*, *b* and *c*

A calculation of area,
given the
individual side lengths is readily made with use of Heron’s formula
(Equation
2), where *s*, is the semi-perimeter, and is equal to one-half of
the
Perimeter (Figure 1).

**Equation
2: _{}**

Equation 3 represents one of the three relations from the Law of Cosines.

**Equation
3:** _{}

Solving Equation 1 for *cos(
A)*,
we obtain Equation 4.

**Equation
4:** _{}

Equation 5 represents a fundamental trigonometric identity:

**Equation
5: _{}**

Solving Equation 3 for _{},
we obtain Equation 6.

**Equation
6: _{}**

Substituting from Equation 4, we have Equation 7.

**Equation
7: _{}**

Squaring the term in parenthesis, we obtain Equation 8.

**Equation
8: _{}**

Rewriting 1 with a common denominator gives Equation 9.

**Equation
9: _{}**

Distributing the minus sign yields Equation 10.

**Equation
10: _{}**

Combining like-terms and factoring yields Equation 11.

**Equation
11: _{}**

Taking the square root of both sides gives Equation 12.

**Equation
12: _{}**

The area formula for a triangle is given in Equation 13.

**Equation
13: _{}**

In figure 2, we have _{},
with height *h* and base, *b*, partitioned into
segments _{}and_{}.

**Figure
2:** _{}with
height, *h* and partitioned base, *b*

From Figure 2 and trigonometric ratio of sides, we obtain Equation 14.

**Equation
14: _{}**

Substitution from Equation 14 into Equation 13 yields Equation 15.

**Equation
15: _{}**

Substituting from Equation 12, we obtain Equation 16.

**Equation
16: _{}**

Simplifying yields Equation 17.

**Equation
17: _{}**

Distributing and rearranging terms yields Equation 18.

**Equation
18: _{}**

We then write the equivalent relation given in Equation 19.

**Equation
19: _{}**

Rearranging terms yields Equation 20.

**Equation
20: _{}**

We rewrite one of the terms in the discriminate to obtain Equation 21.

**Equation
21: _{}**

We add zero to the discriminate via canceling terms, obtaining Equation 22.

**Equation
22: _{}**

We then factor the discriminate in Equation 22, yielding the simpler Equation 23.

**Equation
23: _{}**

We add zero again, via canceling terms, to the two factors of the discriminate obtaining Equation 24.

**Equation
24: _{}**

These additions of zero allow for the factoring of the discriminate in Equation 24, yielding the more elegant Equation 25.

**Equation
25: _{}**

Distributing the fraction under the radical, we obtain Equation 26.

**Equation
26: _{}**

We factor out the one-half from each factor of the radical obtaining Equation 27.

**Equation
27: _{}**

We define semi-perimeter (s) to be half of the perimeter in Equation 28.

**Equation
28: _{}**

Substitution from Equation 28 allows for the familiar expression of Heron’s formula in Equation 29.

**Equation
29: _{}**

Click here to open HeronsFormula.gsp and explore the
relationship
between perimeter and area.

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