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In this case, the semiperimeter will equal the longest side, causing Heron's formula to equal zero. If one of three given lengths is greater than the sum of the other two, then they violate the triangle inequality and do not describe the sides of a Euclidean triangle. In this case, Heron's formula gives an imaginary result.
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula = that avoids the usual procedure of doubling the area of the triangle and then halving it.
In geometry, a Heronian triangle (or Heron triangle) is a triangle whose side lengths a, b, and c and area A are all positive integers. [1] [2] Heronian triangles are named after Heron of Alexandria, based on their relation to Heron's formula which Heron demonstrated with the example triangle of sides 13, 14, 15 and area 84.
A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula. If the semiperimeter is not used, Brahmagupta's formula is
In geometry, the semiperimeter of a polygon is half its perimeter. Although it has such a simple derivation from the perimeter, the semiperimeter appears frequently enough in formulas for triangles and other figures that it is given a separate name. When the semiperimeter occurs as part of a formula, it is typically denoted by the letter s.
The area formula for a triangle can be proven by cutting two copies of the triangle into pieces and rearranging them into a rectangle. In the Euclidean plane, area is defined by comparison with a square of side length , which has area 1. There are several ways to calculate the area of an arbitrary triangle.
For any shape, there is a similar equable shape: if a shape S has perimeter p and area A, then scaling S by a factor of p/A leads to an equable shape. Alternatively, one may find equable shapes by setting up and solving an equation in which the area equals the perimeter. In the case of the square, for instance, this equation is
Any triangle subdivides its bounding box into the triangle itself and additional right triangles, and the areas of both the bounding box and the right triangles are easy to compute. Combining these area computations gives Pick's formula for triangles, and combining triangles gives Pick's formula for arbitrary polygons. [7] [8] [13]