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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.
See: Area of a triangle § Using coordinates. The expression is equivalent to h = 2 A b {\textstyle h={\frac {2A}{b}}} , which can be obtained by rearranging the standard formula for the area of a triangle: A = 1 2 b h {\textstyle A={\frac {1}{2}}bh} , where b is the length of a side, and h is the perpendicular height from the opposite vertex.
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:
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
Altitudes can be used in the computation of the area of a triangle: one-half of the product of an altitude's length and its base's length (symbol b) equals the triangle's area: A = h b /2. Thus, the longest altitude is perpendicular to the shortest side of the triangle.
This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points (x 1,y 1), (x 2,y 2), and (x 3,y 3). The shoelace formula can also be used to find the areas of other polygons when their vertices are known.
The equation in trilinear coordinates x, y, z of any circumconic of a triangle is [1]: p. 192 l y z + m z x + n x y = 0. {\displaystyle lyz+mzx+nxy=0.} If the parameters l, m, n respectively equal the side lengths a, b, c (or the sines of the angles opposite them) then the equation gives the circumcircle .
If the point is not inside the triangle, then we can still use the formulas above to compute the barycentric coordinates. However, since the point is outside the triangle, at least one of the coordinates will violate our original assumption that . In fact, given any point in cartesian coordinates, we can use this fact to determine where this ...