A prism with 10 faces would have 16 vertices. A pyramid with 10 faces would have 10 vertices. Since your figure is neither of these, you need to tell us what it is. In case your numbers were incorrect, here are the formulas: A prism with an n-sided base will have 2n vertices, n + 2 faces, and 3n edges. A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edges.
A cube or a cuboid both would fit the given description both of which have 8 vertices, 12 edges and 6 faces.
Well, it has 24 edges, 10 faces, and finally,16 vertices's
A square-based pyramid would fit the given description. (A triangular prism would have 9 edges and 6 vertices)
According to Euler none; for all 3d shapes: Vertices + Faces = Edges + 2 ⇒ 12 + 8 = 19 + 2 ⇒ 20 = 21 So unless 20 does equal 21, no 3d shape has 8 faces, 19 edges and 12 vertices. Any 3d shape with 8 faces would be an octahedron.
There are 4 visible faces, 5 vertices, and 8 edges. But considering you can see the bottom of the pyramid, there would then be 5 visible faces, 5 vertices, and 8 edges. * * * * * That is an incomplete answer. A pyramid need not have a quadrilateral base. "Pyramid" is a generic term for a 3-dimensional shape with a base which is an n-sided polygon and n triangular faces that meet at an apex. A pyramid with a n-polygon base has n + 1 faces, n + 1 vertices and 2n edges.
A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edges. 6 faces, 10 edges, 6 vertices
The shape would be impossible. The faces and vertices have to add up to two more than the edges.
Oh, dude, it's like a math riddle! So, if a polyhedron has 10 more edges than vertices, we can use Euler's formula: Faces + Vertices - Edges = 2. Since we know the relationship between edges and vertices, we can substitute that in and solve for faces. So, it would have 22 faces. Math can be fun... sometimes.
A prism with 10 faces would have 16 vertices. A pyramid with 10 faces would have 10 vertices. Since your figure is neither of these, you need to tell us what it is. In case your numbers were incorrect, here are the formulas: A prism with an n-sided base will have 2n vertices, n + 2 faces, and 3n edges. A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edges.
A prism has 2n vertices, where n is the number of sides in its base polygon. This is because each side of the base polygon is connected to another side by a vertical edge, and each vertex is formed at the intersection of two edges. So, a prism with a pentagonal base would have 10 vertices (5 vertices from the base polygon and 5 vertices from the top polygon).
A cube or a cuboid both would fit the given description both of which have 8 vertices, 12 edges and 6 faces.
If it's a pyramid, it has 10 edges.
Well, it has 24 edges, 10 faces, and finally,16 vertices's
A square-based pyramid would fit the given description. (A triangular prism would have 9 edges and 6 vertices)
The shape you're looking for that has 2 faces, 0 edges, and 0 vertices would be a cylinder.
Such a polyhedron cannot exist. According to the Euler characteristics, V + F - E = 2, where V = vertices, F = faces, E = edges. This would require that the polyhedron had only two faces.