Search results
Results from the WOW.Com Content Network
The surface area A and the volume V of the rhombic dodecahedron with edge length a are: [4] =, =. The rhombic dodecahedron can be viewed as the convex hull of the union of the vertices of a cube and an octahedron where the edges intersect perpendicularly.
There exist tetrahedra having integer-valued edge lengths, face areas and volume. These are called Heronian tetrahedra. One example has one edge of 896, the opposite edge of 990 and the other four edges of 1073; two faces are isosceles triangles with areas of 436 800 and the other two are isosceles with areas of 47 120, while the volume is 124 ...
The surface area and the volume of the truncated icosahedron of edge length are: [2] = (+ +) = +. The sphericity of a polyhedron describes how closely a polyhedron resembles a sphere. It can be defined as the ratio of the surface area of a sphere with the same volume to the polyhedron's surface area, from which the value is between 0 and 1.
The volume of a rhombicuboctahedron can be determined by slicing it into two square cupolas and one octagonal prism. Given that the edge length a {\displaystyle a} , its surface area and volume is: [ 7 ] A = ( 18 + 2 3 ) a 2 ≈ 21.464 a 2 , V = 12 + 10 2 3 a 3 ≈ 8.714 a 3 . {\displaystyle {\begin{aligned}A&=\left(18+2{\sqrt {3}}\right)a^{2 ...
Therefore, the circumradius of this rhombicosidodecahedron is the common distance of these points from the origin, namely √ φ 6 +2 = √ 8φ+7 for edge length 2. For unit edge length, R must be halved, giving R = √ 8φ+7 / 2 = √ 11+4 √ 5 / 2 ≈ 2.233.
The lengths of the five simple closed geodesics on a snub disphenoid with unit-length edges are 2 3 ≈ 3.464 {\displaystyle 2{\sqrt {3}}\approx 3.464} (for the equatorial geodesic), 13 ≈ 3.606 {\displaystyle {\sqrt {13}}\approx 3.606} , 4 {\displaystyle 4} (for the geodesic through the midpoints of opposite edges), 2 7 ≈ 5.292 ...
AOL Mail welcomes Verizon customers to our safe and delightful email experience!
In algebraic terms, doubling a unit cube requires the construction of a line segment of length x, where x 3 = 2; in other words, x = , the cube root of two. This is because a cube of side length 1 has a volume of 1 3 = 1, and a cube of twice that volume (a volume of 2) has a side length of the cube root of 2.