# NCERT Solutions Class 8 Maths Chapter 10 Exercise 10.3

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## NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes (EX 10.3) Exercise 10.3

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### Access NCERT Solution for Class 8 Maths Chapter 10 – Visualising Solid Shapes

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### NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Exercise 10.3

The NCERT Solutions For Class 8 Maths Chapter 10 Exercise 10.3 covers the notion of Polyhedrons is explained in Visualizing Solid Shapes. Polyhedrons are solids made up of Polygonal areas called “faces” that converge at “edges.” These “vertices,” or points, are where these line segments, or “edges,” converge. Convex and regular Polyhedron types as well as others, have been discussed. Students will learn about the link between the vertices, edges, and faces of a Polyhedron with the use of Euler’s formula in NCERT Solutions For Class 8 Maths Chapter 10 Exercise 10.3. The main lesson to be learned from this section is that the Polyhedron family is made up mostly of Prisms and Pyramids. All these ideas are covered in the NCERT Solutions For Class 8 Maths Chapter 10 Exercise 10.3.

### NCERT Solutions for Class 8

The NCERT Solutions for Class 8 Mathematics provide answers to every question from the textbook, which the CBSE Board mandates. The offered syllabuses for all courses are solely based on the NCERT curriculum. Thus, students will surely have a chance to succeed if they use NCERT Solutions to study for exams. To make it simpler for students to approach the problems, Extramarks developed the NCERT Class 8 Solutions for Mathematics. The CBSE Class 8 Solutions for Mathematics offered here comprise well-planned exercises and comprehensive explanations organised by qualified teachers, substantially streamlining the process of learning and understanding subjects. The NCERT Solutions For Class 8 Maths Chapter 10 Exercise 10.3 are also available to students. They can find notes, important topics, and sample questions for NCERT Solutions For Class 8 Maths Chapter 10 Exercise 10.3. Students can access past years’ papers and get help with their solutions with the NCERT Solutions For Class 8 Maths Chapter 10 Exercise 10.3.

All the other chapters’ NCERT solutions for Class 8 are also available on Extramarks’ website and mobile application.

Chapter 1 Rational Numbers

Chapter 2 Linear Equation

Chapter 4 Applied Practical Geometry

Chapter 6 Square Square Roots

Chapter 7 Cube And Cube Roots

Chapter 8 Comparing Quantities

Chapter 9 Algebraic Expressions And Identities

Chapter 10 Visualizing Solid Shapes

Chapter 11 Mensuration

Chapter 12 Exponents And Powers

Chapter 13 Direct Inverse Proportions

Chapter 14 Factorisation

Chapter 15 Introduction To Graphs

Chapter 16 Playing With Number

Q.1 Can a polyhedron have for its faces

(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?

Ans

(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.

Q.2 Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).

Ans

Yes, it is possible, if a polyhedron has minimum of 4 faces

Q.3 Which are prisms among the following?

Ans

(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)Since the faces are triangular, so it is a pyramid. Hence, it is not a prism

(iv) It is a prism.

Q.4 (i) How are prisms and cylinders alike?

(ii) How are pyramids and cones alike?

Ans

(i) A prism that has a circle as its base can be thought of as a circular prism.

(ii) A pyramid that has a circle as its base can be thought of as a circular pyramid.

Q.5 Is a square prism same as a cube? Explain.

Ans

A square prism has a square as its base. Its height is not necessarily same as the side of the square. Thus, it is not necessary that always a square prism is same as a cube. It can be cuboid also.

Q.6 Verify Euler’s formula for these solids.

Ans

The Euler’s formula F+V–E = 2

(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, 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, verified.

Q.7 Using Euler’s formula find the unknown.

 Faces ? 5 20 Vertices 6 ? 12 Edges 12 9 ?

Ans

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

Q.8 Can a polyhedron have 10 faces, 20 edges and 15 vertices?

Ans

Number of faces = F = 10

Number of edges = E = 20

Number of vertices = V = 15

By using Euler’s Formula, we can find whether a polyhedron satisfies the given conditions.

Therefore, we have, F + V − E = 2

But, F + V − E = 10 + 15 – 20

= 25 – 20

= 5 ≠ 2

Since Euler’s formula is not satisfied, such a polyhedron is not possible.