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
jwilson.coe.uga.edu/emt725/Heron/Heron.html 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.
mathforum.org/library/drmath/view/54957.html 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.
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
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