 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.