Class 8 Exam  >  Class 8 Notes  >  RD Sharma Solutions for Class 8 Mathematics  >  RD Sharma Solutions - Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths

Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 19.9:

Question 1:

What is the least number of planes that can enclose a solid? What is the name of the solid?

ANSWER:

The least number of planes that can enclose a solid is 4.Tetrahedron is a solid with four planes (faces).


Question 2:

Can a polyhedron have for its faces:
 (i) 3 triangles?
 (ii) 4 triangles?
 (iii) a square and four triangles?

ANSWER:

(i)No, because in order to complete a polyhedron, we need at least four triangular faces. 

(ii)Yes, a polyhedron with 4 triangular faces is a tetrahedron.

(iii)Yes, with the help of a square bottom and four triangle faces, we can form a pyramid.


Question 3:

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

ANSWER:

Yes, it is possible to have a polyhedron with any number of faces. 

The only condition is that there should be at least four faces. 

This is because there is no possible polyhedron with 3 or less faces.


Question 4:

Is a square prism same as a cube?

ANSWER:

Yes, a square prism and a cube are the same.Both of them have 6 faces, 8 vertices and 12 edges.

The only difference is that a cube has 6 equal faces, while a square prism has a shape like a cuboid with two squarefaces, one at the top and the other at the bottom and with, possibly, 4 rectangular faces in between. 


Question 5:

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

ANSWER:

No, because every polyhedron satisfies Euler's formula, given below: F + V = E + 2

Here, number of faces F  =  10

Number of edges E  =  20

Number of vertices V  =  15

So, by Euler's formula:

LHS: 10 + 15  =  25

RHS:  20 + 2  =  22,

which is not true because 25≠22

Hence, Eulers formula is not satisfied and no polyhedron may be formed.


Question 6:

Verify Euler's formula for each of the following polyhedrons:
Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 MathematicsChapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 MathematicsChapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 MathematicsChapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 MathematicsChapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

ANSWER:

(i)In the given polyhedron:Edges E = 15

Faces F = 7

Vertices V = 10

(i)In the given polyhedron:Edges E = 15

Faces F = 7

Vertices V = 10
Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics
Now, putting these values in Euler's formula:

LHS: F + V =  7 + 10 =  17

LHS: E + 2 = 15 + 2 = 17

LHS  =  RHS 

Hence, the Euler's formula is satisfied.Now, putting these values in Euler's formula:

LHS: F + V =  7 + 10 =  17

LHS: E + 2 = 15 + 2 = 17

LHS  =  RHS

 Hence, the Euler's formula is satisfied.

(ii)In the given polyhedron:Edges E = 16

Faces F = 9

Vertices V = 9

(ii)In the given polyhedron:Edges E = 16

Faces F = 9

Vertices V = 9

Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics
Now, putting these values in Euler's formula:

RHS: F + V =  9 + 9 =  18

LHS: E + 2 = 16 + 2 = 18

LHS  =  RHS

Hence, Euler's formula is satisfied.

Now, putting these values in Euler's formula:

RHS: F + V =  9 + 9 =  18

LHS: E + 2 = 16 + 2 = 18

LHS  =  RHS

Hence, Euler's formula is satisfied.

(iii)In the following polyhedron:Edges E = 21

Faces F = 9

Vertices V = 14

(iii)In the following polyhedron:Edges E = 21

Faces F = 9

Vertices V = 14
Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics
Now, putting these values in Euler's formula:

LHS: F + V =  9 + 14 =  23

RHS: E + 2 = 21 + 2 = 23

This is true. 

Hence, Euler's formula is satisfied.Now, putting these values in Euler's formula:

LHS: F + V =  9 + 14 =  23

RHS: E + 2 = 21 + 2 = 23

This is true. 

Hence, 

Euler's formula is satisfied.

(iv)In the following polyhedron:Edges E = 8

Faces F = 5

Vertices V = 5

(iv)In the following polyhedron:Edges E = 8

Faces F = 5

Vertices V = 5
Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics
Now, putting these values in Euler's formula:

LHS: F + V =  5 + 5 =  10

RHS: E + 2 = 8 + 2 = 10

LHS  =  RHS

Hence, Euler's formula is satisfied.Now, putting these values in Euler's formula:

LHS: F + V =  5 + 5 =  10

RHS: E + 2 = 8 + 2 = 10

LHS  =  RHS

Hence, Euler's formula is satisfied.

(v)In the following polyhedron:

Edges E = 16

Faces F = 9

Vertices V = 9(v)

In the following polyhedron:Edges E = 16

Faces F = 9

Vertices V = 9
Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics
Now, putting these values in Euler's formula:

LHS: F + V =  9 + 9 =  18

RHS: E + 2 = 16 + 2 = 18

LHS  =  RHS

Hence, Euler's formula is satisfied.

 

PAGE NO 19.10:

Question 7:

Using Euler's formula find the unknown:

Faces?520
Vertices6?12
Edges129?

ANSWER:

We know that the Euler's formula is: F + V  =  E + 2

(i) The number of vertices V is 6 and the number of edges E is 12.

Using Euler's formula:

F + 6  =  12 + 2

F + 6 =  14

F  =  14 - 6

F  =  8

So, the number of faces in this polyhedron is 8.

(ii)Faces, F  =  5

Edges, E  =  9.

We have to find the number of vertices.

Putting these values in Euler's formula:

5 + V  =  9 + 25 + V  =  11

V  =  11 - 5

V  =  6

So, the number of vertices in this polyhedron is 6.

(iii)Number of faces F  =  20

Number of vertices V  =  12

Using Euler's formula:

20 + 12  =   E + 2

32  =  E + 2

E + 2  =  32

E  =  32 - 2

E  =  30.

So, the number of edges in this polyhedron is 30.

The document Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
88 docs

Top Courses for Class 8

FAQs on Chapter 19 - Visualising Shapes (Part - 1), Class 8, Maths RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. What are the different types of shapes that can be visualized in mathematics?
Ans. In mathematics, there are various types of shapes that can be visualized. Some common examples include points, lines, line segments, rays, angles, triangles, quadrilaterals, circles, and polygons.
2. How can visualizing shapes help in solving mathematical problems?
Ans. Visualizing shapes can greatly assist in solving mathematical problems. It allows us to understand the properties and relationships between different shapes, making it easier to identify patterns and solve geometric problems. Additionally, visualizing shapes helps in creating mental images, aiding in problem-solving and mathematical reasoning.
3. What is the importance of visualizing shapes in real-life applications?
Ans. Visualizing shapes is essential in real-life applications such as architecture, engineering, design, and construction. It helps professionals in these fields to visualize and plan structures, create accurate blueprints, and determine the dimensions and proportions of various components. Visualizing shapes also plays a crucial role in fields like computer graphics, animation, and computer-aided design.
4. How can one improve their ability to visualize shapes in mathematics?
Ans. Improving the ability to visualize shapes in mathematics can be achieved through practice and exposure to various geometric concepts. Engaging in activities like drawing and sketching different shapes, solving geometry problems, and using interactive visual aids can enhance spatial reasoning skills and strengthen the ability to mentally manipulate shapes.
5. Are there any strategies or techniques to enhance visualization skills in mathematics?
Ans. Yes, there are several strategies and techniques to enhance visualization skills in mathematics. Some effective approaches include breaking complex shapes into simpler components, using concrete objects or models to represent abstract shapes, employing spatial reasoning puzzles and games, and practicing mental rotations or transformations of shapes. Regular practice and perseverance are key in developing strong visualization skills.
88 docs
Download as PDF
Explore Courses for Class 8 exam

Top Courses for Class 8

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

practice quizzes

,

MCQs

,

Class 8

,

ppt

,

Chapter 19 - Visualising Shapes (Part - 1)

,

Free

,

Objective type Questions

,

Class 8

,

pdf

,

Sample Paper

,

past year papers

,

mock tests for examination

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

study material

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

,

Extra Questions

,

Chapter 19 - Visualising Shapes (Part - 1)

,

video lectures

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

,

Semester Notes

,

Exam

,

Chapter 19 - Visualising Shapes (Part - 1)

,

Viva Questions

,

Summary

,

Important questions

,

Class 8

;