Exercise 10.1 Exercise 10.2 Exercise 10.3
Exercise 10.3
Question 1
Can a polyhedron have for its faces
(i) 3 triangles? (ii) 4 triangles?
(iii) a square and four triangles?
Sol :
(i) No, such a polyhedron is not possible. A polyhedron has minimum 4 faces.
(ii) Yes, a triangular pyramid has 4 triangular faces.
(iii) Yes, a square pyramid has a square face and 4 triangular faces.
Question 2
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).
Sol :
A polyhedron has a minimum of 4 faces.
Question 3
Which are prisms among the following?
(i) |
(ii) |
(iii) |
(iv) |
Sol :
(i) It is not a polyhedron as it has a curved surface. Therefore, it will not be a prism also.
(ii) It is a prism.
(iii) It is not a prism. It is a pyramid.
(iv) It is a prism.
Question 4
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Sol :
(i) A cylinder can be thought of as a circular prism i.e., a prism that has a circle as its base.
(ii) A cone can be thought of as a circular pyramid i.e., a pyramid that has a circle as its base.
Question 5
Is a square prism same as a cube? Explain.
Sol :
A square prism has a square as its base. However, its height is not necessarily same as the side of the square. Thus, a square prism can also be a cuboid.
Question 6
Verify Euler’s formula for these solids.
(i) |
(ii) |
Sol :
(i) Number of faces = F = 7
Number of vertices = V = 10
Number of edges = E = 15
We have, F + V − E = 7 + 10 − 15 = 17 − 15 = 2
Hence, Euler’s formula is verified.
(ii) Number of faces = F = 9
Number of vertices = V = 9
Number of edges = E = 16
F + V − E = 9 + 9 − 16 = 18 − 16 = 2
Hence, Euler’s formula is verified.
Question 7
Using Euler’s formula, find the unknown.
Faces |
? |
5 |
20 |
Vertices |
6 |
? |
12 |
Edges |
12 |
9 |
? |
Sol :
By Euler’s formula, we have
F + V − E = 2
(i) F + 6 − 12 = 2
F − 6 = 2
F = 8
(ii) 5 + V − 9 = 2
V − 4 = 2
V = 6
(iii) 20 + 12 − E = 2
32 − E = 2
E = 30
Thus, the table can be completed as
Faces |
8 |
5 |
20 |
Vertices |
6 |
6 |
12 |
Edges |
12 |
9 |
30 |
Question 8
Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Sol :
Number of faces = F = 10
Number of edges = E = 20
Number of vertices = V = 15
Any polyhedron satisfies Euler’s Formula, according to which, F + V − E = 2
For the given polygon,
F + V − E = 10 + 15 − 20 = 25 − 20 = 5 ≠ 2
Since Euler’s formula is not satisfied, such a polyhedron is not possible.