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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 A of any triangle is the product of its inradius (the radius of its inscribed circle) and its semiperimeter: =. The area of a triangle can also be calculated from its semiperimeter and side lengths a, b, c using Heron's formula:
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.
A version of the isoperimetric inequality for triangles states that the triangle of greatest area among all those with a given perimeter is equilateral. [35] The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. [36]
In particular, to find the quadrilateral, or the triangle, or another particular figure, with the largest area amongst those with the same shape having a given perimeter. The solution to the quadrilateral isoperimetric problem is the square , and the solution to the triangle problem is the equilateral triangle .
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 version of the isoperimetric inequality for triangles states that the triangle of greatest area among all those with a given perimeter is equilateral. That is, for perimeter p {\displaystyle p} and area T {\displaystyle T} , the equality holds for the equilateral triangle: [ 8 ] p 2 = 12 3 T . {\displaystyle p^{2}=12{\sqrt {3}}T.}