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Heron's formula for the area of a triangle

Named for the Greek Mathematician, Heron, who lived circa AD 62.  He was also called Hero.

Consider triangle ABC with sides of length a, b, and c:

If you let b be the base, then the height of the triangle is a sin(C), so the area, K, of the triangle is given by:

K = (1/2)ab sin(C), where a and b are any two sides, and C is the included angle.

The law of cosines tells us that c² = a² + b² - 2ab cos(C)

Solving for cos(C), we get cos(C) = (a² + b² - c²)/(2ab)

Since sin²(C) + cos²(C) = 1, we get this expression for sin(C):

sin(C) = sqrt(1 - ((a² + b² - c²) / (2ab) )²)

Plugging this into the area formula, above,

K = (1/2)ab sqrt(1 - ((a² + b² - c²) / (2ab) )²)
K = (1/4)(2ab) sqrt(1 - ((a² + b² - c²) / (2ab) )²)
K = (1/4) sqrt((2ab)² - (a² + b² - c²)²)
K = (1/4) sqrt(((2ab) + (a² + b² - c²))((2ab) - (a² + b² - c²)))
K = (1/4) sqrt((a² + 2ab + b² - c²)(-a² + 2ab - b² + c²))
K = (1/4) sqrt(((a+b)² - c²)(-(a-b)² + c²))
K = (1/4) sqrt((a+b+c)(a+b-c)(a-b+c)(-a+b+c))
K = sqrt(s(s-c)(s-b)(s-a)), where s=(a+b+c)/2, the semiperimeter

Internet references: proves Heron's formula by expressing the height, h, of a triangle in terms of its sides a, b, and c, and then using the area formula K = (1/2) ch to derive the result. proves it using K = b c sin(A), and then using needlessly complicated trig substitutions to express sin(A) in terms of a, b, and c.  Sort of like my proof, except trickier.  It reveals some interesting factoids along the way, though.

Related pages in this website:

Summary of geometrical theorems

Law of Cosines

Point and Triangle -- a series of pages explaining how to determine whether a point is inside a triangle.  One of these pages gives an easy method using Heron's formula: Point D is inside triangle ABC if the sums of the areas of ABD, BCD and CAD equal the area of ABC.

Heron's Formula for the area of a triangle.

Brahmagupta's Formula for the area of a cyclic quadrilateral, which can be considered a generalization of Heron's Formula.

Bretschneider's Formula for the area of a quadrilateral, which can be considered a generalization of Brahmagupta's Formula.

Triangle Area using Determinant

Triangle Area using Vectors

Polygon Area using Determinant


The webmaster and author of this Math Help site is Graeme McRae.