You can Download KSEEB Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.3 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.3

Question 1.

Can a polyhedron have for its faces

(i) 3 triangles?

(ii) 4 triangles?

(iii) a square and four triangles?

Solution:

(i) A polyhedron cannot have three triangles for its faces.

(ii) A polyhedron can have four triangles for its faces.

(iii) A polyhedron can have a square and four triangles for its faces.

Question 2.

Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).

Solution:

Yes, it is possible only if the number of faces is greater than or equal to four.

Question 3.

Which are prisms among the following?

Solution:

(i) A nail is not a prism.

(ii) An unsharpened pencil is a prism.

(iii) A table weight is not a prism.

(iv) Yes a box is a prism.

Question 4.

(i) How are prisms and cylinders alike?

(ii) How are pyramids and cones alike?

Solution:

(i) A prism becomes a cylinder as the number of sides of its base becomes larger and larger.

(ii) A pyramid becomes a cone as the number of sides of its base becomes larger and larger.

Question 5.

Is a square prism the same as a cube? Explain.

Solution:

A square prism is made up of two parallel end faces, each one of which is a triangle and three lateral faces, each one of which is a square. In a cube, its faces are congruent, regular polygon. Vertices are formed by the same number of faces. Thus, a square prism and cube have a square of its base. Clearly, their bases are square. Hence prism and cube are the same.

Question 6.

Verify Euler’s formula for these solids.

Solution:

(i) In this figure,

F = 7, V = 10, E = 15

∴ F + V = 7 + 10 = 17

and E + 2 = 15 + 2 = 17

⇒ F + V = E + 2

(ii) In this figure,

F = 9, V = 9, E = 16

∴ F + V = 9 + 9 = 18

and E + 2 = 16 + 2 = 18

F + V = E + 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 | ? |

Solution:

In I part,

Faces = ?, Vertices = 6, Edges = 12

By Euler’s formula, we know that

F + V = E + 2

F = E + 2 – V

= 12 + 2 – 6

= 8

Hence, Faces = 8

II part,

Faces = 5, Vertices = ?, Edges = 9

By Euler’s formula, we know that

F + V = E + 2

V = E + 2 – F

= 9 + 2 – 5

= 6

Hence, Vertices = 6

In III part,

Faces = 20, Vertices = 12, Edges = ?

By Euler’s formula, we know that

F + V = E + 2

E = F + V – 2

= 20 + 12 – 2

= 30

Hence, Edges = 30

Question 8.

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

Solution:

Since, F + V ≠ E + 2 (10 + 15 ≠ 20 + 2)

∴ A polyhedron cannot have 10 faces, 20 edges, and 15 vertices.