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Practice each question, then open the answer or worked solution.
Can a polyhedron have for its faces:
(i) 3 Triangles?
(ii) 4 triangles?
(iii) A square and four triangles?
(i) No, such polyhedrons are not possible. A polyhedron should have a minimum of 4 faces.
(ii) Yes, a triangular pyramid has 4 triangular faces.
(iii) Yes, as a square pyramid has a square face and 4 triangular faces.
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid)
It is possible only if the number of faces is greater than or equal to 4.
Which are prisms among the following:
(i) A nail: It is not a prism.
(ii) Unsharpened pencil: It is a prism.
(iii) A table weight: It is not a prism.
(iv) A box: It is a prism.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
(i) A cylinder can look like a circular prism, a prism with a circular base.
(ii) A cone can be a circular pyramid, a pyramid with a circular base.
Is a square prism the same as a cube? Explain.
Yes, a square prism can also be a cube. A square prism has a square as its base. However, its height is not necessarily the same as the side of the square. Thus, a square prism can also be a cuboid.
Verify Euler’s formula for the given solids.
(i) Number of faces, F = 7
Number of edges, E = 15
Number of vertices, V = 10
As per formula, F + V – E = 2
Substitute the values, we have
F + V – E = 7 + 10 – 15
= 2
Hence, verified.
(ii) Here, F = 9, V = 9 and E = 16
Using formula, F+ V – E = 2
F + V – E = 9 + 9 – 16 = 2
Hence, Euler’s formula is verified.
Using Euler’s formula, find the unknown:
| Faces | ? | 5 | 20 |
| Vertices | 6 | ? | 12 |
| Edges | 12 | 9 | ? |
Euler’s formula: F + V – E = 2
Where, F = Faces, V = Vertices and E = Edges
(i) F + 6 – 12 = 2
F = 2 + 6
⇒ F = 8
(ii) 5 + V – 9 = 2
V – 4 = 2
⇒ V = 4 + 2
⇒ V = 6
(iii) 20 + 12 – E = 2
32 – E = 2
⇒ E = 32 – 2
⇒ E = 30
Can a polyhedron have 10 faces, 20 edges and 15 vertices?
From the given data, we have
F = 10
E = 20
V = 15
Every polyhedron satisfies Euler’s formula, which is stated as, F + V – E = 2
For the given polygon,
F + V – E = 10 + 15 – 20 = 25 – 20 = 5, which is not equal to 2.
Therefore, a polyhedron cannot have 10 faces, 20 edges and 15 vertices, as Euler’s formula is not satisfied.