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The excentral triangle of a reference triangle has vertices at the centers of the reference triangle's excircles. Its sides are on the external angle bisectors of the reference triangle (see figure at top of page ).
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 :
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.
Its surface area is four times the area of an equilateral triangle: = =. [7] Its volume can be ascertained similarly as the other pyramids, one-third of the base times height. Because the base is an equilateral, it is: [ 7 ] V = 1 3 ⋅ ( 3 4 a 2 ) ⋅ 6 3 a = a 3 6 2 ≈ 0.118 a 3 . {\displaystyle V={\frac {1}{3}}\cdot \left({\frac {\sqrt {3 ...
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.
More formulas of this nature can be given, as explained by Ramanujan's theory of elliptic functions to alternative bases. Perhaps the most notable hypergeometric inversions are the following two examples, involving the Ramanujan tau function τ {\displaystyle \tau } and the Fourier coefficients j {\displaystyle \mathrm {j} } of the J-invariant ...
Using the usual notations for a triangle (see the figure at the upper right), where a, b, c are the lengths of the three sides, A, B, C are the vertices opposite those three respective sides, α, β, γ are the corresponding angles at those vertices, s is the semiperimeter, that is, s = a + b + c / 2 , and r is the radius of the inscribed circle, the law of cotangents states that
The fraction of the triangle's area that is filled by the square is no more than 1/2. Squaring the circle In 1882, the task was proven to be impossible as a consequence of the Lindemann–Weierstrass theorem , which proves that pi ( π ) is a transcendental number rather than an algebraic irrational number ; that is, it is not the root of any ...