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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]
Harcourt's theorem is a formula in geometry for the area of a triangle, as a function of its side lengths and the perpendicular distances of its vertices from an arbitrary line tangent to its incircle. [1] The theorem is named after J. Harcourt, an Irish professor. [2]
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.
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 approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
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 formula of the area of an equilateral triangle can be obtained by substituting the altitude formula. [7] Another way to prove the area of an equilateral triangle is by using the trigonometric function. The area of a triangle is formulated as the half product of base and height and the sine of an angle. Because all of the angles of an ...
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]