This video shows how to find the faces, edges, and vertices of a polyhedron. To find how many vertices a rhombic triacontahedron with 30 faces has, you will need to use Euler’s formula: V+F=E+2. Because the number of faces is 30, you can replace the F with 30. To find the number of edges, multiply the number of edges a rhombus has by the number of faces the triacontahedron has and divide it by two: 30×4/2 = 120/2 = 60. This means there are 60 edges, making your formula V+30=60+2. To solve for the number of vertices, subtract F from both sides. This gives you an answer of 32. This triacontrahedron has 32 vertices, 30 faces, and 60 edges.

Faces, Edges, and Vertices of a Polyhedron

Faces, Edges, and Vertices of a Polyhedron

A rhombic triacontahedron has 30 faces. How many vertices does it have? To solve this problem, we’re going to use Euler’s formula which is vertices plus faces is equal to the number of edges, plus 2.

This works on all polyhedrons. We are told that this figure has 30 faces, so we can substitute 30 for faces; we’re trying to find the number of vertices, so I know that’s going to be our V; but I can’t solve unless I know how many edges our figure has.

To find the number of edges we’re going to take the total number of edges, which each rhombus has 4 edges and there are 30 faces, so 30 times 4, and then we’re going to divide it by 2 because each edge is shared by 2 faces. 30 times 4 is 120, divided by 2 is 60, which means there are a total of 60 edges, so we substitute 60 for edges.

Now we can solve for vertices. The number of vertices plus 30 is equal to 60 plus 2 is 62. To solve for vertices, we need to subtract 30 from both sides. 30 minus 30 is 0, so vertices is 62 minus 30, 32. This polyhedron has 30 faces, 60 edges, and 32 vertices.



by Mometrix Test Preparation | Last Updated: August 15, 2019