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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 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.
The three splitters concur at the Nagel point of the triangle. A cleaver of a triangle is a line segment that bisects the perimeter of the triangle and has one endpoint at the midpoint of one of the three sides. So any cleaver, like any splitter, divides the triangle into two paths each of whose length equals the semiperimeter.
giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths. A related formula, which was proved by Coolidge, also gives the area of a general convex quadrilateral. It is [2]
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. [3]
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
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]
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