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Shoelace scheme for determining the area of a polygon with point coordinates (,),..., (,). The shoelace formula, also known as Gauss's area formula and the surveyor's formula, [1] is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by their Cartesian coordinates in the plane. [2]
A different proof that these triangles have area is based on the use of Minkowski's theorem on lattice points in symmetric convex sets. [10] Subdivision of a grid polygon into special triangles. This already proves Pick's formula for a polygon that is one of these special triangles.
Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero. Brahmagupta's formula gives the area K {\displaystyle K} of a cyclic quadrilateral whose sides have lengths a , {\displaystyle a,} b , {\displaystyle b,} c , {\displaystyle c ...
The second moment of area about the origin for any simple polygon on the XY-plane can be computed in general by summing contributions from each segment of the polygon after dividing the area into a set of triangles. This formula is related to the shoelace formula and can be considered a special case of Green's theorem. A polygon is assumed to ...
A short elementary proof of Pascal's theorem in the case of a circle was found by van Yzeren (1993), based on the proof in (Guggenheimer 1967). This proof proves the theorem for circle and then generalizes it to conics. A short elementary computational proof in the case of the real projective plane was found by Stefanovic (2010).
The honeycomb conjecture states that hexagonal tiling is the best way to divide a surface into regions of equal area with the least total perimeter. The optimal three-dimensional structure for making honeycomb (or rather, soap bubbles) was investigated by Lord Kelvin , who believed that the Kelvin structure (or body-centered cubic lattice) is ...
It is also related to the densest circle packing of the plane, in which every circle is tangent to six other circles, which fill just over 90% of the area of the plane. The case when the problem is restricted to a square grid was solved in 1989 by Jaigyoung Choe who proved that the optimal figure is an irregular hexagon. [4] [5]
The area within a circle is equal to the radius multiplied by half the circumference, or A = r x C /2 = r x r x π.. Liu Hui argued: "Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the ...