Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.3

Question 1.

Can a polyhedron have four its faces

- 3 triangles?
- 4 triangles?
- a square and four triangles?

Solution:

A polyhedron is bounded by four or more than four polygonal faces.

- No, it is not possible that a polyhedron has 3 triangles for its faces.
- Yes, 4 triangles can be the faces of a polyhedron.
- Yes, a square and 4 triangles can be the faces of a polyhedron.

Question 2.

Is it possible to have a polyhedron with any given number of faces?

Hint:

Think of a pyramid.

Solution:

Yes, it can be possible only if the number of faces is four or more than four.

Question 3.

Which are prisms among the following?

Solution:

Since, a prism is a polyhedron having two of its faces congruent and parallel, whereas other faces are parallelogram.

- No, a nail is not a prism.
- Yes, unsharpened pencil is a prism.
- No, table weight is not a prism.
- Yes, box is a prism.

Question 4.

- How are prisms and cylinders alike?
- How are pyramids and cones alike?

Solution:

- Both of the prisms and cylinders have their base and top as congruent faces and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.
- The pyramid and cones are alike because their lateral faces meet at a vertex. Also a pyramid becomes a cone as the number of sides of its base becomes larger and larger.

Question 5.

Is a square prism same as a cube? Explain?

Solution:

No, not always, because it can be a cuboid also.

Question 6.

Verify Eulerās formula for these solids?

Solution:

(i) In figure (i), we have

F = 7, V= 10 and E = 15

ā“ F + F = 7 + 10 = 17

F + V – E = 17 – 15 = 2

i.e., F + V – E = 2

Thus, Eulerās formula is verified.

(ii) In figure (ii), we have

F = 9, V = 9 and E = 16

ā“ F + V = 9 + 9 = 18 and

F + V – E = 18 – 16 = 2

i ,e., F + V – E = 2

Thus, Eulerās formula is verified.

Question 7.

Using Eulerās formula find the unknown?

Solution:

1. Here, V = 6 and E = 12

Since F + V – E = 2

ā“ F + 6 – 12 = 2 or F – 6 = 2

or F = 2 + 6 = 8

2. Here, F = 5 and E = 9

Since F + V – E = 2 or

ā“ 5 + V – 9 = 2

or V – 4 = 2

or V = 2 + 4 = 6

3. Here F = 20 and V = 12

Since, F + V – E = 2

ā“ 20 + 12 – E = 2

32 – E = 2

or E = 32 – 2 = 30

Question 8.

Can polyhedron have 10 faces, 20 edges and 15 vertices?

Solution:

Here, F = 10, E = 20 and V = 15

We have: F + V – E = 2

ā“ 10 + 15 – 20 = 2 or

25 – 20 = 2

or 5 = 2 which is not true

i.e., F + V – E ā 2

Thus, such a polyhedron is not possible.