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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 T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
A triangle with sides a, b, and c. In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths , , . Letting be the semiperimeter of the triangle, = (+ +), the area is [1]
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.
Shoelace scheme for determining the area of a polygon with point coordinates (,),..., (,). The shoelace formula, also known as Gauss's area formula and the surveyor's formula, [1] is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. [2]
The area A of any triangle is the product of its inradius (the radius of its inscribed circle) and its semiperimeter: A = r s . {\displaystyle A=rs.} The area of a triangle can also be calculated from its semiperimeter and side lengths a, b, c using Heron's formula :
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d (or any one side) approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
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.
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]