NCERT solution class 8 chapter 10 Visualising Solid Shapes

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.

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