Search results
Results from the WOW.Com Content Network
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]
In any triangle, the distance along the boundary of the triangle from a vertex to the point on the opposite edge touched by an excircle equals the semiperimeter. The semiperimeter is used most often for triangles; the formula for the semiperimeter of a triangle with side lengths a, b, c = + +.
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 = that avoids the usual procedure of doubling the area of the triangle and then halving it.
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. If the semiperimeter is not used, Brahmagupta's formula is
The three splitters of a triangle all intersect each other at the Nagel point of the triangle. A cleaver of a triangle is a segment from the midpoint of a side of a triangle to the opposite side such that the perimeter is divided into two equal lengths. The three cleavers of a triangle all intersect each other at the triangle's Spieker center.
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.
For any shape, there is a similar equable shape: if a shape S has perimeter p and area A, then scaling S by a factor of p/A leads to an equable shape. Alternatively, one may find equable shapes by setting up and solving an equation in which the area equals the perimeter. In the case of the square, for instance, this equation is
This formula can be derived by partitioning the n-sided polygon into n congruent isosceles triangles, and then noting that the apothem is the height of each triangle, and that the area of a triangle equals half the base times the height. The following formulations are all equivalent: