Spinning Dodecahedron It has 20 Vertices (corner points) and at each vertex 3 edges meet It is one of the Platonic Solids Volume and Surface Area Volume = (15+7×√5)/4 × (Edge Length)3 Surface Area = 3×√(25+10×√5) × (Edge Length)2 It is called a dodecahedron because it is a polyhedron that has 12 faces (from Greek dodeca- meaning 12). When we have more than one dodecahedron they are called dodecahedra When we say "dodecahedron" we often mean "regular dodecahedron" (in other words all faces are the same size and shape), but it doesn't have to be - this is also a dodecahedron, even though all faces are not the same. 12-Sided Dice? Yes! A dodecahedron which has 12 equal faces has an equal chance of landing on any face. In fact, you can make fair dice out of all of the Platonic Solids. cut out the shape and glue it together.

How many edges does a dodecahedron have? | Reference.com How many edges does a dodecahedron have? A: Quick Answer A dodecahedron has 30 edges that are shared by 12 faces. A regular dodecahedron is a three-dimensional shape where all the faces are pentagons. Full Answer There are many different types of dodecahedron, such as the rhombic or pyritohedral dodecahedron. Both of these variations lack regular pentagonal faces, which means that all of their edge lengths are not the same. Different types of dodecahedrons have different numbers of edges. A regular dodecahedron is a highly symmetrical shape, having multiple lines of symmetry. There are other 12-sided shapes that have limited or no symmetry, usually consisting of a mixture of pentagonal, square or triangular faces. The Dodecahedron The dodecahedron has 30 edges, 20 vertices and 12 faces. Dodeca is a prefix meaning “twelve.” The dodecahedron is the only polyhedron I know of which is composed entirely of pentagons.

Buckminster Fuller describes what he calls a '6 pentagonal tensegrity sphere' in Synergetics I, 726.01, and which is shown at Marvin Solit's website at but I don't believe that structure can be built without tensegrity struts and tension wires. Figure 1 The dodecahedron is pentagonal both inside and out, as can be seen from Figure 1. Like the icosahedron, it has many golden section relationships, which we shall see. The dodecahedron is even more versatile then the icosahedron. The icosahedron contains and geometry, but the dodecahedron contains , Figure 2 Cube and tetrahedron in dodecahedron. Cube in gray, tetrahedron in green This view of the dodecahedron is significant in that it shows the 2 dimensional shadow of the decagon. The decagon itself is based upon the pentagon, the building block of the dodecahedron. See Pentagon and Decagon for more information.

Figure 2A -- Another view of tetrahedron (green) inside cube (blue) inside dodecahedron (orange) Figures 2 and 2A show how a cube and a tetrahedron can be placed inside a dodecahedron. These placements are 'nice' meaning that the vertices of the placed-in solids are all vertices of the dodecahedron. The cube, octahedron and tetrahedron are all based on root 2 and root 3 geometry: The relationship of the side of the cube to the radius of its enclosing sphere is r = sqrt(3) / 2. For the tetrahedron, The dodecahedron is capable of elegantly sustaining these and relationships, along with its own many relationships. Figure 3 -- octahedron inside tetrahedron Notice that the octahedron fits precisely on the bisected sides of the tetrahedron. The icosahedron cannot contain any of the other 5 solids 'nicely' on its vertices. The icosehedron and the dodecahedron are 'duals' (as are the cube and the octahedron).

By 'dual' is meant that if you put a vertex in the middle of all of the faces and connect the lines, you get the dual. By placing a vertex at the middle of all the faces of the dodecahedron you get an icosahedron, and vice-versa. Figure 4 shows the dual nature of the icosahedron and dodecahedron. Figure 4 Duals ---- dodecahedron inside icosahedron Notice that to create the dodecahedron, all we did was draw lines from each vertex of the icosahedron to every other vertex. The vertices of the dodecahedron are at the intersection points. We could just as easily have found the vertices of the dodecahedron by drawing lines on every triangular face of the icosahedron. Where those lines intersect is the center of the face, and a vertex of the dodecahedron. That occurs because the dodecahedron has 12 faces and the icosahedron has 12 vertices. Now for the standard analysis: What is the volume of the dodecahedron? We will use the pyramid method.

There are 12 pentagonal pyramids, 1 for each face, each pyramid beginning at O, the centroid. See Figure 1 and Figure 5. The volume of any n-sided pyramid is 1/3 * area of base * pyramid height. First we need to get the area of the base, which is the area of each pentagonal face: Figure 5 One pyramid on face BCHLG The area of the pentagon is the area of the 5 triangles which compose it. From Area of Pentagon we know This is approximately 1.720477401 . Now we need to find the height of the pyramid, OU. To do that, we need to find the distance from O to a vertex, lets say, OH. This distance will be the hypotenuse of the right triangle OUH. Since we already know UH, we can then get OU from the good ol’ Pythagorean Theorem. Imagine a sphere surrounding the dodecahedron and touching all of its vertices. OH is just the radius of the enclosing sphere. If you look at Figure 6, HOZ = GON = diameter. There is a line through HOZ to show the diameter.

Figure 6 Now look at the rectangle MIFK. The diagonal of it, MF, is also a diameter (MF = HZ). Notice that the long sides of the rectangle, MK, and IF, are diagonals of the two large pentagons ADMQK and FEIRP. We know from Composition of the Pentagon that the diagonal of a pentagon is * side of pentagon. We also can see from Figure 6 and more clearly in Figure 1 that the sides of the large pentagons themselves are diagonals of the pentagonal faces of the dodecahedron! (For instance, DA is a diagonal of the face ABCDE). That means each side of the large pentagons is * s and that MK (or any of the diagonals of a large pentagon) is * So MK = s. In fact, like the icosahedron, the dodecahedron is composed of rectangles divided in Extreme and Mean Ratio. In the icosahedron, we found these rectangles to be rectangles. In the dodecahedron, they are rectangles. In rectangle MKIF, MK = IF = , MI = KF = , as shown in figure 7.

Figure 7 -- showing vertices of the rectangle MKIF. There are 30 sides to the dodec, and therefore 15 different rectangles. All of this as explanation of finding the distance OH from Figure 5! Because we are not using trigonometry, we need OH in order to get the pyramid height, OU in Figure 5. Notice that MHFZ is also a rectangle and that HZ is the diagonal of it. If we can find HZ, then OH is just 1/2 of that. d² = HZ² = MZ² + HM² = diameter = HZ = Now we can find OU, the height of the pyramid. From Area of Pentagon we know the distance mid-face to any vertex of a pentagon = HOC = central angle = 41.81031488° Since each face of the dodecahedron is a pentagon, the surface angle = 108° While we're at it, lets get OX, the distance from the centroid to any mid- edge. What is the dihedral angle of the dodecahedron? Figure 9 AXH.

AH is one of the long sides of any of the 15 rectangles which compose the dodec. AX and HX are the height h of the pentagon. We know from Construction of the Pentagon Part 2 that the height h of the pentagon is: We know from Figure 7 that AH is , so IH = sin( IXH) = IH / XH = We recognize this ratio as our old friend the Phi Triangle with sides in ratio of So dihedral angle AXH = 116.5650512°. Distance from centroid to mid-face (h) = = 1.113516365s. Distance from centroid to mid-edge = = 1.309016995. Distance from centroid to vertex = = 1.401258539. Go back to Figure 6. We have colored the 4 internal pentagonal planes of the dodecahedron. U,X, W and V are the centers of these 4 planes which line up with the centroid O. What is the distance UX = WV? What is XW?

If we can find these out we can figure out more deeply how the dodecahedron is constructed. Figure 6, repeated In Figure 6 we can see that UH on the top plane is the distance from the pentagon center to a vertex. On plane ADMQK, XM is parallel to UH and is also the distance from that pentagon center to a vertex. UH is connected to XM by HM, a side of the pentagon face CDIMH. So we have a quadrilateral UHMX, with UH parallel to XM. From here we can derive UX, the distance between the two planes. Figure 10 -- dodecahedron planar distance. We know from Construction of the Pentagon Part 2 that the distance from the center of pentagon to a vertex = Therefore UH = The side of the large pentagon ADMQK in Figure 6 is, as we have seen, a diagonal of a dodecahedron face and so the side of the large pentagon = Therefore, XM = *UH, and XM is divided in Mean and Extreme Ratio at N (See Figure 10). = HM = s. Triangle MNH is right by construction.

So = HN = Notice: UH = UX. So the dodecahedron is designed such that the distance to the 2 large pentagonal planes from the top or bottom faces is exactly equal to the distance between the center and a vertex of any of the faces of the dodecahedron. This relationship is precisely what we saw in the icosahedron! That makes sense because the two are duals of each other. The difference is that the dodecahedron is entirely pentagonal, both internally, and externally, on its faces. What is the distance XW between the 2 large pentagonal planes ADMQK and FEIRP? Figure 6 is misleading, it looks like the distance must be MI or KF, the dodecahedron side, but it isn't. We already have enough information to establish this distance. UX = VW. UV = 2*height of any pyramid = So XW = UV - 2*UX = XW = UW is divided in Mean and Extreme Ratio at X. XV / WV = . XV is divided in Mean and Extreme Ratio at W.

Let's make a chart of these planar distances along the diameter of the dodecahedron as we did with the icosahedron: (see Figure 6): Relative Chart of Distances –– Pentagonal Planes of Dodecahedron Relative to the side of the dodecahedron (Available in the book) Here is a table of these relationships, letting UX = 1: (Available in the book) Note that the diameter of the enclosing sphere is HZ, not UV. We already know that the diameter is, from page 92, So what is the distance from U to the top of the sphere, and from V to the bottom of the sphere? Let T’ be the top of the sphere and B’ be the bottom of the sphere. Refer to Figure 6. If the radius is and the distance OU is , then Finally, let us demonstrate how the dodecahedron may be constructed from the interlocking vertices of 5 tetrahedron. We have already seen how the cube fits inside the dodecahedron, and how 2 interlocking tetrahedron may be formed from the diagonals of the cube.

As Buckminster Fuller has pointed out, however, the cube and the dodecahedron are structurally unsound unless bolstered by the additional struts supplied by the tetrahedron. Fuller concludes logically that the tetrahedron is the basic building block of Universe; yet it is the dodecahedron that provides the blueprint and forms the structure for the interlocking tetrahedrons. The dodecahedron unites the geometry of crystals and lattices (root 2 and root 3) with the geometry of Phi (root 5), found in the biology of organic life. The dodecahedron is entirely pentagonal, consisting of the geometry of Phi. Yet it contains the and geometry of the cube, tetrahedron, and octahedron. Remarkably, the sides of the cube are * side of the dodecahedron, because the cube side is the diagonal of a pentagonal face. Here is the key to the relationship of the first three Regular Solids and the much more complex icosahedron and dodecahedron.

Later on in this book we will discover a remarkable polyhedron that defines the relationship and provides the proper nesting for all 5 Platonic Solids, including the icosahedron, directly on its vertices. If a polyhedron could be called exciting, this one is IT! If you can’t wait, go to the last chapter of the book. Dodecahedron Reference Tables How many faces does an icosahedron have? | Reference.com How many faces does an icosahedron have? A: Quick Answer An icosahedron is a regular polyhedron with 20 faces, all of which are equilateral triangles. An icosahedron has 30 edges and 12 vertices. Five equilateral triangles meet at each vertex. Full Answer An icosahedron is an example of a Platonic solid, which is a polyhedron that is regular, convex, made of congruent faces of regular polygons, and has the same number of faces meeting at each vertex. Only five known shapes meet this criteria.

The other Platonic solids include the tetrahedron with four triangular faces, the cube, or hexahedron, with six square faces, the octahedron with eight triangular faces and the dodecahedron with twelve pentagonal faces. Dodecahedral Facts Dodecahedral Facts U 23, C 26, W 5 Properties Net Dodecahedron Dodecahedral describes all things related to geometic dodecahedrons. The dodecahedron (Greek δωδεκάεδρον, from δώδεκα 'twelve' + εδρον 'base', 'seat' or 'face') is any polyhedron with twelve faces, but usually a regular dodecahedron is meant: a Platonic solid . It is composed of 12 regular pentagonal faces, with three meeting at each vertex, and is represented by the Schläfli symbol {5,3}. It has 20 vertices and 30 edges. Its dual polyhedron is the icosahedron , with Schläfli symbol {3,5}.

A large number of other (nonregular) polyhedra also have 12 sides, but are given other names. The most frequently named other dodecahedron is the rhombic dodecahedron . Area and volume The area A and the volume V of a regular Dodecahedral of edge length a are: Cartesian Coordinates The following Cartesian coordinates define the vertices of a Dodecahedral centered at the origin: (±1, ±1, ±1) where φ = (1+√5)/2 is the golden ratio (also written τ). The dihedral angle of a Dodecahedral is approximately 116.565 degrees. Trivia The dihedral angle of a dodecahedron is 2 arctan(φ) or approximately 116.565 degrees. If the original dodecahedron has edge length 1, its dual icosahedron has edge length If one were to make every one of the Platonic solids with edges of the same length, the dodecahedron would be the largest. It has 43,380 nets .

If one were to color the dodecahedron faces such that no two adjacent faces had the same color, one would need 4 colors. The distance between the vertices on the same face not connected by an edge is 2 cos(36°). Geometric Relations The regular dodecahedron is the third in an infinite set of truncated trapezohedra which can be constructed by truncating the two axial vertices of a pentagonal trapezohedron . The stellations of the dodecahedron make up three of the four Kepler-Poinsot polyhedra . A rectified dodecahedron forms an icosidodecahedron . The regular dodecahedron has 120 symmetries, forming the group . Icosahedron vs dodecahedron When a dodecahedron is inscribed in a sphere , it occupies more of the sphere's volume (66.49%) than an icosahedron inscribed in the same sphere (60.54%).

A regular dodecahedron with edge length 1 has more than three and a half times the volume of an icosahedron with the same length edges (7.663... compared with 2.181...). Related Polyhedra The dodecahedron can be transformed by a truncation sequence into its dual , the icosahedron: Picture Vertex arrangement The dodecahedron shares its vertex arrangement with four nonconvex uniform polyhedra and three uniform polyhedron compounds . Five cubes fit within, with their edges as diagonals of the dodecahedron's faces, and together these make up the regular polyhedral compound of five cubes. Since two tetrahedra can fit on alternate cube vertices, five and ten tetrahedra can also fit in a dodecahedron.

The 3 stellations of the dodecahedron are all regular ( nonconvex ) polyhedra: ( Kepler-Poinsot polyhedra ) 0 Facet diagram Other Dodecahedra The term dodecahedron is also used for other polyhedra with twelve faces, most notably the rhombic dodecahedron which is dual to the cuboctahedron (an Archimedean solid ) and occurs in nature as a crystal form. The Platonic solid dodecahedron can be called a pentagonal dodecahedron or a regular dodecahedron to distinguish it. The pyritohedron is an irregular pentagonal dodecahedron.

Other dodecahedra include: Elongated square dipyramid - 8 triangles and 4 squares Metabidiminished icosahedron - 10 triangles and 2 pentagons Congruent nonregular faced: ( face-transitive ) Rhombic dodecahedron (mentioned above) - 12 rhombi , dual of cuboctahedron Other nonregular faced: Trapezo-rhombic dodecahedron - 6 rhombi, 6 trapezoids - dual of Triangular orthobicupola Rhombo-hexagonal dodecahedron or Elongated Dodecahedron - 8 rhombi and 4 equilateral hexagons . In all there are 6,384,634 topologically distinct dodecahedra. [1] History and Uses Dodecahedral objects have found some practical applications, and have also played a role in the visual arts and in philosophy. Plato 's dialogue Timaeus (c.

360 B.C.) associates the other four platonic solids with the four classical elements , adding that "there is a fifth figure (which is made out of twelve pentagons), the dodecahedron—-this God used as a model for the twelvefold division of the Zodiac." [2] Aristotle postulated that the heavens were made of a fifth element, aithêr (aether in Latin, ether in American English), but he had no interest in matching it with Plato's fifth solid. A few centuries later, small, hollow bronze Roman dodecahedra were made and have been found in various Roman ruins in Europe. Their purpose is not certain. In twentieth century art, dodecahedra appear in the work of M. C. Escher , such as his lithograph Reptiles (1943), and in his Gravitation . In Salvador Dalí 's painting The Sacrament of the Last Supper (1955), the room is a hollow dodecahedron. In modern role-playing games , the dodecahedron is often used as a twelve-sided die , one of the more common polyhedral dice .

As a Graph The skeleton of the dodecahedron – the vertices and edges – form a graph . The high degree of symmetry of the polygon is replicated in the properties of this graph, which is distance-transitive , distance-regular , and symmetric . The automorphism group has order 120. The vertices can be colored with 3 colors, as can the edges, and the diameter is 5. [3] The dodecahedral graph is Hamiltonian – there is a cycle containing all the vertices. Indeed, this name derives from a mathematical game invented in 1857 by William Rowan Hamilton , the icosian game . The game's object was to find a Hamiltonian cycle along the edges of a dodecahedron. See Also The Rhombic Triacontahedron is an extremely fascinating polyhedron. It is built around the dodecahedron, and like the dodecahedron, it has many phi relationships within it. This polyhedron contains all five of the Platonic Solids directly on its vertices, and shows the proper relationship between them.

The Rhombic Triacontahedron (hereinafter referred to as r.t.) has 30 faces, 60 edges, and 32 vertices. Figure 1 -- showing the front portion of the Rhombic Triacontahedron (red) with the sides of the dodecahedron (gray) Notice pentagon CDNJO, then notice Z. Z is raised off the pentagon (a distance we shall find later on) and Z is connected to all 5 vertices of the pentagonal face of the dodecahedron. This can be seen even more clearly at U. The pentagon GPIJO has all of its vertices connected to U. Of course, we could have drawn this figure without the dodecahedron faces, which is obviously not part of the r.t., but it helps for clarity. Notice that the faces of the r.t. are diamond-shaped, somewhat like the rhombic dodecahedron, but these faces are longer and skinnier. Look at the r.t. face UOZJ at the upper left. U and Z are the long-axis vertices.

Notice that the line between O and J, the short-axis vertices, form one of the edges (sides) of the dodecahedron. The dual of the r.t. is the icosa-dodecahedron. The r.t. is more clearly spherical than any of the polyhedra we have studied so far. Notice also: In Figure 2 below, we see that the rhombic triacontahedron has internal pentagonal planes just like the icosahedron and the dodecahedron! Figure 2 -- showing the internal pentagonal planes of the r.t. Figure 2 is rotated 180 degrees with respect to Figure 1. Note the large highlighted pentagons. The lengths of all of the large internal pentagon sides are precisely the long axis of every r.t. face! For example, look at VZ and ZFo in the top pentagonal plane and EoBo in the lower pentagonal plane. The r.t. has a circumsphere and an inner sphere. The circumsphere goes around all 20 of the vertices that rise off the faces of the dodecahedron. The diameter of the circumsphere is UW.

UW is in green in Figure 2. O' is the centroid. The circumsphere touches all of the vertices that are raised off the pentagonal planes of the dodecahedron. There are 12 of these, because there are 12 faces on the dodecahedron. The inner sphere touches all of the 20 vertices of the dodecahedron. The diameter of the inner sphere would be, for example, IL. UW > IL. This leads us to think that possibly, the relationship between U, Go,O',Ho and W will be similar to those of the dodecahedron and the icosahedron. Later on, we will see that this is indeed the case. Notice also that the rhombic triacontahedron inherently contains within it not only a dodecahedron, but an icosahedron as well! Notice that the 12 vertices which rise off the dodecahedron faces provide the 12 vertices of the icosahedron. Remember that the dual of the dodecahedron is the icosahedron and that the dual is formed by taking points at the center of the faces.

Figure 3 -- showing the 12 vertices of the icosahedron (blue) within the rhombic triacontahedron The sides of the icosahedron are the long axes of the r.t. faces. The sides of the dodecahedron are the short axes of the r.t. faces. Therefore the rhombic triacontahedron is just the combination of the dodecahedron with its dual, the icosahedron. The rhombic triacontahedron is the model nature uses to demonstrate the true relationship between the side of the icosahedron and the side of the dodecahedron. We will see later on that these relationships are based on . Note also that when a cube is inscribed within 8 of the 12 vertices belonging to the dodecahedron within the rhombic triacontahedron, the lengths of the cube sides are precisely equal to the long axis of any of the rhombic triacontahedron rhombi.

Furthermore, the sides of the cube are precisely equal to any of the diagonals of the pentagonal faces of the dodecahedron: Figure 3A The sides of the cube, in blue, are also the diagonals of the pentagonal faces of the dodecahedron within the rhombic triacontahedron. Since the tetrahedron (one tetrahedron in green, the other in purple) and the octahedron (in orange) can be inscribed within the cube, the rhombic triacontahedron shows the precise relationship between the Platonic Solids! Figure 3B -- The cube, 2 interlocking tetrahedrons, and the octahedron inside a cube The Rhombic Triacontahedron therefore elegantly describes the nesting of the five Platonic Solids: icosahedron, dodecahedron, cube, tetrahedron, octahedron. When the sides of the octahedron are divided in Mean and Extereme (Phi) Ratio, another icosahedron is formed.

This begins the process all over again, and shows that the 5 nested Platonic Solids may not only grow and contract to infinity, but do so in a perfectly harmonious way. Figure 3B -- Showing how the icosahedron nests within the octahedron. Interestingly, the icosahedron is formed within the octahedron by dividing each edge of the octahedron in Phi Ratio. What is the volume of the rhombic triacontahedron? As before we use the pyramid method. There are 30 faces, so there are 30 pyramids. Imagine a point at the very center of the r.t. If you connect that point up with one of the faces, you will have a pyramid that looks like this: Figure 4 -- one of the 30 pyramids of the rhombic triacontahedron. O'Z is the height of the pyramid. O'U = O'V = radius of outer sphere which touches the 12 r.t. vertices which are raised off the center of the dodecahedron face.

O'I = O'J = radius of inner sphere which touches all 20 of the short-axis r.t. vertices, which are also the vertices of the dodecahedron We want to find the volume of the r.t. in terms of the r.t. side. However, we don't know the length of the r.t side! But we do know the length of the side of the dodecahedron, in terms of a unit sphere which touches all the vertices of the dodecahedron. So let's try to get the r.t. side (hereinafter referred to as rts) in terms of the dodecahedron side. Figure 5 Figure 5A Figure 5 and 5A show the following important data: V is one of the 12 raised vertices of the r.t. off the face of the dodecahedron. C is the center of the dodecahedron face. C lies in the plane of IJOGP, and is directly below V. Z is the center of the r.t. face IUJV (see Figure 4), and also the mid-edge of the side of the dodec.

IVJ (in red) is one-half of the r.t. diamond face IUJV (see Figure 4) O'Z is the distance from centroid to mid-edge of the dodec face. This is the height of the r.t. pyramid. ZV is the distance from mid-edge of dodecahedron to the raised vertex V off the dodecahedron face. ZC is the distance from mid-edge of dodecahedron to center of dodecahedron face. CV is the distance of V off the center of the dodecahedron face. Figure 5A shows that O'CV is a straight line, so that ZO'C = O'ZV is right. O'CZ is right. From this data we can show that the triangles O'ZV, O'CZ, and CZV are similar by angle-side-angle. First we show the triangle O'ZV and O'CZ are similar: O'ZV is right, so is O'CZ. O'Z is common to both triangles. ZO'V = ZO'C. Therefore both triangles are similar by ASA.

Now we show that triangle CZV is similar to triangle O'ZV by angle-side-angle. ZCV and ZV is common to both triangles. O'VZ = CVZ. Therefore both triangles are similar by angle-side-angle. With this information we can determine CV, the distance of the vertex V off of the dodec face, and ZV, which will enable us to get the side of the r.t. in terms of the side of the dodecahedron. Since all 3 triangles are similar, we can write the following relationship: CZ / O'C = CV / CZ. The distances O'Z, O'C, and CZ are known. From Dodecahedron it is known that (ds means dodecahedron side) Now we have CV, the distance from the plane of the dodecahedron to the rhombic triacontahedron “cap” over the dodecahedron face, in terms of the dodecahedron side. Now we need to find ZV, so that we can get IV, the length of the side or edge of the rhombic triacontahedron.

We can write the following relationship: Refer back to Figures 4 and 5. Now that we have ZV, we can find the side of the r.t., IV. The triangle IVJ in Figure 4 and 5 is one-half of an r.t face. We have already calculated ZV, and we know that IZ is just one–half the side of the side of the dodecahedron, ds. We also know that the angle IZV is right by construction. Therefore, by the Pythagorean Theorem, Here we have established an important fact: we have related the side of the rhombic triacontahedron, or rts, to the side of the dodecahedron and, therefore, to the radius of the unit sphere which encloses all 20 vertices of the dodecahedron and, in turn, the short axis vertices of the rhombic triacontahedron. We can now describe the distance of any vertex of the r.t. off the plane of the dodecahedron, CV, in terms of the side of the r.t. We may now write Let's now find the radius of the outer sphere of the r.t.

in terms of the side of the r.t. itself. Remember that the outer sphere touches all 12 long-axis vertices of the r.t., and that these 12 vertices are the vertices of an icosahedron We need to convert these distances so that they are related to rts, not ds. In the Rhombic Triacontahedron, the relationship between the side and the radius of the enclosing sphere is Phi. What is ? This is just the unit sphere which touches all 20 vertices of the dodecahedron. We know from Dodecahedron that Converting this to the side of the r.t. we have: Now we have enough information to calculate the volume of the rhombic triacontahedron in terms of it's own side. Refer back to Figures 4 and 5 for diagrams. We can calculate the area of the r.t. face, because we know IJ and ZV, which we can use to give us the area of one-half the face of the r.t. We see from Figure 4 that O’Z, the height of the r.t.

pyramid, is just the distance from the centroid to the mid-edge of any of the sides of the dodecahedron. We know from Dodecahedron , that this distance is . We have to do some conversions of these values first, to get them all in terms of the rts. To get the area of the r.t. face, divide it into 2 identical triangles at the short axis IJ, find the area of one triangle, and multiply by 2.. The area of any triangle is 1/2 * base * height, so the area of the r.t. face is twice this value. IJ is just the side of the dodecahedron, and it is the base of our triangle. There are 30 pyramids for 30 faces so total volume is: The volume of the r.t. can be calculated another way. Since the r.t. is built upon the dodecahedron, the r.t. volume is just the volume of the dodecahedron + the extra volume of all of the little 12 pentagonal pyramids formed from the raised vertices off the 12 pentagonal faces of the dodecahedron.

To see this, check Figure 1 again and look at U-IJOGP or V- IJNFM. The volume of the dodecahedron is, from Dodecahedron , Converting this to the side of the r.t. we get: The volume of each of the 12 "extra" pyramids = 1/3 * area of pentagon * CV (height of each raised vertex off of the face of the dodecahedron) = We only have one problem: one of our values is in terms of the dodecahedron side. We need to convert that to the side of the r.t. What is the surface area of the rhombic triacontahedron? Before we calculate the central and surface angles of the rhombic triacontahedron, let us complete our research into the distances from the centroid to various points of interest on this polyhedron. We have already calculated the distances to the small and large axis vertices, and to the mid-face. Now we need to find the distance from the centroid to any mid-edge. Figure 4a Showing the right triangle O’ZA’ We are looking for OA’. We may write .

We know O’Z, but what is ZA’? IZ = ½ IJ by construction. JA’ = ½ JV by construction. The triangles VIJ and ZA’J are congruent by angle-angle-angle. Triangle VIJ is isosceles by construction, therefore triangle ZA’J is isosceles and ZA’ = ½rts. What are the central angles of the rhombic triacontahedron? Figure 4, repeated There are 3 central angles of the r.t. that are of interest. Refer to Figure 4. The first is UO'V, central angle of the long-axis The second is IO'J, central angle of the short-axis. The third and primary central angle is IO'V, the central angle of each adjacent side. O'Z is perpendicular to IJ, and to the plane of the r.t. face, IUJV. O'Z bisects IJ at Z. Triangles UO'Z and VO'Z are right. So we can write We recognize this angle (see Phi Ratio Triangle ) as being part of a ratio triangle.

The ratio's of these distances are shown below in Figure 6: Figure 6 --- showing the relationship between VZ, O'Z, O'V Now Note that IZ / O’Z = , so that IZ and O’Z have a relationship based on the square of . Figure 7 -- showing the relationship between O'Z, IZ and O'I This division is in ratio . To find IO'V, recognize that the point V lies in a straight line directly above the center of the dodecahedron face. So the angle from IO'V is the same as the angle from I to O' to a point (G) in the middle of the dodecahedron face. Figure 8 -- showing that the central angle IO'V = the angle IO'G The triangle O'GI is right by construction, so we need only to know GI and O'G. We know from Area of the Pentagon that , and from Dodecahedron we know the distance . What are the surface angles of the rhombic triacontahedron?

Observe from Figure 4 that the angles we are looking for are IVJ and Z is a bisector of IZ so the triangle IVZ is right. IV = the side of the r.t. = rts. , which we already found above. So we can write . We recognize immediately, from The Phi Triangle , that this angle indicates a right triangle whose long and short sides are divided in Extreme and Mean Ratio (Phi ratio). Therefore And so the ratio between the long axis VU and the short axis IJ (see Figure 4) must also be in Mean and Extreme Ratio. Therefore VU / IJ = So the face of the rhombic triacontahedron is a rhombus. The face angle we want, IVJ, is then twice IVJ = 63.43494882° = short axis face angle. One-half of the other face angle, ZIV, is just 90 - So UIV = 116.5650512°. What are the lengths of the long axis VU and the short axis IJ of the r.t. face? To see this calculation, click on R.T. Dihedral Angle .

The dihedral angle = 144°. The distance from the centroid to any of the 12 long-axis vertices above the dodecahedron faces = Distance from centroid to any of the 20 short-axis vertices = . We need to go back to Figure 2 now and look at the diameter of the sphere which encloses the 12 long-axis vertices of the icosahedron. This line is shown in Figure 2 as UO'W. The diameter passes through the centroid at O' and also through the middle of the 2 large pentagonal planes marked in Figure 2. It also passes through the top and bottom faces of the dodecahedron. These faces are also marked in Figure 2. There are 4 highlighted pentagonal planes along the diameter UW, as well as the midpoint O'. Figure 2, repeated We have the extra distance off the top plane of the dodecahedron to the vertex U, marked as IoU, and the extra distance off the bottom plane of the dodecahedron to the vertex W, marked as JoW. What are the relationships between U, Io, Go, O', Ho, Jo, and W?

In Figures 5 and 5A, we found the distance off the plane of the dodecahedron face (IoU , JoW ) to be CV. CV we found to be So IoU = JoW = Refer back to Figure 2. Let's find the distance UGo, or the distance from U to the first large pentagonal plane VYCoFoZ. This will be easy, because we know that the sides of this pentagonal plane are just the long axes of the r.t. We can form a right triangle from U to any one of the vertices of VYCoFoZ, to the center Go. Let's take the right triangle UGoZ. From Construction of the Pentagon we can get GoZ, it is just the distance from pentagon center to one of the vertices. Inspection of Figure 2 shows that UZ is just the long axis of the r.t. face UJZO. Figure 9 -- showing the large internal pentagon VYCoFoZ and the right triangle UGoZ. Go is the center of the large internal pentagon. U is directly above Go. The sides of the large pentagon (in green) are all long axes of the r.t.

faces, indicated in red. Triangle UGoZ is right by construction. or long-axis of r.t. face. Now we can find UGo: So UGo = 2 * UIo and the plane VYCoFoZ of the dodecahedron is twice the distance from U as is the large internal pentagonal plane IPGOJ. That means UIo = IoGo = 1/2 UGo = What is the distance between the centroid O' and the plane IPGOJ, or O'Go? It is O'U - UGo. We know O'U, it is the radius of the outer sphere, or Therefore, Now we have enough information to make our distance chart of internal planar distances of the rhombic triacontahedron, just as we did for the icosahedron and the dodecahedron. Table 1. From the table of relationships we see that: O'Io is divided in Extreme and Mean Ratio at Go. O'Jo is divided in Extreme and Mean Ratio at Ho. UHo is divided in Mean and Extreme Ratio squared at Go. WGo is divided in Mean and Extreme Ratio squared at Ho.

Conclusions: The rhombic triacontahedron is a combined icosahedron-dodecahedron dual, so it is not surprising to see so many relationships based on the division in Mean and Extreme Ratio. The rhombic triacontahedron contains all of the properties of the icosahedron and all of the properties of the dodecahedron and it tells us the proper nesting order of the 5 Platonic Solids. Dihedral Angle Of The Rhombic Triacontahedron Figure 1. We are looking for the dihedral angle, BIE . Figure 1 shows an r.t. “cap” (in red) over the face of the dodecahedron, and shows that J and G are on AH, the height of the pentagon. F is directly above the mid-face of the pentagon, at G. I is directly above J on BE. BJI is right, allowing us to calculate BIJ or BIE, the dihedral angle, is then just twice that. AH and BE lie on the pentagonal plane ABCDE. BI is a line along an r.t.

face, EI is a line along another of the faces. Here is our plan of attack: We know from Rhombic Triacontahedron that BFA is 63.43494882° This can be rewritten without loss of information as . Triangle BIF is right by construction. BF is the side of the r.t., or rts. Here we must use trigonometry to get BI, by taking the sine of BFA so we can find BI. BE is bisected by AH, the angle bisector of BAE and a diagonal of the pentagon. AJ is known from Pentagon Construction . To get BIJ, we can take the sine of BIJ = BJ / BI. The dihedral angle BIE is twice angle BIJ. First let’s find BJ. BJ is one-half the diagonal of the pentagon with sides equal to the side of the dodecahedron. We must change units in terms of the side of the r.t. Remember that How many sides does a dodecahedron have? How many sides does a dodecahedron have?

0 votes posted May 17, 2016 by Nikita Sehgal Share this puzzle solution May 17, 2016 by Neha Gupta Your comment on this solution: Email me at this address if a comment is added after mine:Email me if a comment is added after mine Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register . 0 votes solution May 17, 2016 by Salil Agrawal Your comment on this solution: Email me at this address if a comment is added after mine:Email me if a comment is added after mine Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register . 0 votes The dodecahedron has 12 sides solution May 26, 2016 by anonymous Your comment on this solution: Email me at this address if a comment is added after mine:Email me if a comment is added after mine Privacy: Your email address will only be used for sending these notifications.

Anti-spam verification: To avoid this verification in future, please log in or register . Your solution Thanks for contributing at QueryHome Puzzle Please be sure to solve the Puzzle, share your working and research. If you are not sure about solution use FB/G+ commenting option instead. Use ctrl+k to highlight, ctrl+b to make the text as bold. Preview Email me at this address if my solution is selected or commented on:Email me if my solution is selected or commented on Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register . Similar Puzzles +1 vote I have four sides i am the most common shape but i am skinnier than a rectangle what am I? 0 votes Two triangles have integral side lengths, with all sides being less than 50. They are similar but not congruent and smaller triangle has two side lengths identical with the larger triangle. What is the sum of the side lengths of the larger triangle? 0 votes A solid sphere of chocolate has a volume of 288Pi . The chocolate is then melted down and reshaped into a solid cube. Assume no chocolate was wasted.

Did the sphere have a greater surface area or does the cube? 0 votes The lengths of two adjacent sides of a parallelogram are 8 and 15, respectively. Find the sum of squares of the lengths of the two diagonals. 0 votes polygons and polyhedra : paul scott : dodecahedron 12 2 We notice that the conditions for this to be a regular polyhedron are all satisfied: congruent regular pentagons for faces, and vertices all alike. Each vertex figure is again an equilateral triangle. The face angles are clearly all 108°. At right is the Schlegel diagram for the dodecahedron. Does it have the right number of vertices, edges , and faces? What does it tell us about this polyhedron? All the usual measurements can be calculated by the enthusiastic: surface area, volume, inradius, midradius, circumradius, dihedral angle. There seems to be no great merit in doing this; why not just read how someone else did it?! Here ... For the record, here is a summary of some of the metric attributes of a regular dodecahedron of side length s. The values indicate some hard calculations!

Volume Now check your answers ... The projections of the dodecahedron are less interesting than those of some of the other Platonic solids but you should be able to obtain a (non-regular) hexagon, an octagon, and a regular decagon {10}.  You can try to find the sequence of cross-sections as a plane moves through the dodecahedron cutting it from vertex to vertex, edge to edge, and face to face. Alternatively, see if you can find a cross section which is a {3}, a {4}, a {5}, a {6} (there are two distinct types of these) and a {10}. This first diagram shows one of the projections, where we obtain {10} as a shadow figure. Then below are diagrams showing the regular cross-sections. It is perhaps surprising that we can obtain so many. The blue pentagonal section is of course no surprise, as each face of the dodecahedron is a regular pentagon. We can obtain a proper} section by taking a parallel plane which cuts the solid. The magenta decagonal section is obtained by cutting the solid suitably through its centre. Is the green section actually square?

The edge lengths are all equal (diagonals of congruent faces), so it must be at least a rhombus. Further, since the faces are congruent and regular, the two diagonals of the faces with a common edge are both centred with respect to that edge. So the green section is a square. The factor of 3 in the triangular and hexagonal sections seems a little out of place, except that we have three faces meeting at each vertex. So the equilateral triangle section is just a vertex figure of the solid. Notice that the two regular hexagonal sections are different in nature: the orange section has vertices at edge midpoints, but the purple section has to have its vertices at carefully chosen points on the edges of the dodecahedron. My calculation shows that such points have to be at a ratio of 5  1 :

Question: How many sides does a dodecahedron have?
Answer: 12