RD Sharma Solutions for Class 8 Chapter 19 Visualising Shapes

In Chapter 19, we shall discuss the visualisation of 3-D shapes from their plane figures. We will also learn about the representation of three-dimensional shapes on the plane of the paper. For a better understanding of the concepts, students can solve the exercise wise problems using the solutions, which are formulated by our expert faculty team at BYJU’S. Students aspiring to secure high marks in their examination are advised to practice the solutions on a regular basis. RD Sharma Class 8 pdf can be downloaded from the links provided below.

Chapter 19- Visualising Shapes contains two exercises and the RD Sharma Class 8 Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Polyhedra.
  • Prisms and pyramids.
  • Platonic solids.
  • Visualisation of 3-D shapes through nets.

Download the Pdf of RD Sharma for Class 8 Maths Chapter 19 Visualising Shapes

 

rd sharma class 8 maths chapter 19 ex 1
rd sharma class 8 maths chapter 19 ex 1
rd sharma class 8 maths chapter 19 ex 1
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rd sharma class 8 maths chapter 19 ex 2
rd sharma class 8 maths chapter 19 ex 2
rd sharma class 8 maths chapter 19 ex 2
rd sharma class 8 maths chapter 19 ex 2

 

Access answers to Maths RD Sharma Solutions For Class 8 Chapter 19 Visualising Shapes

EXERCISE 19.1 PAGE NO: 19.9

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

Solution:

The least number of planes that are required to enclose a solid is 4.

The name of solid is tetrahedron.

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

Solution:

(i) 3 triangles?

No, because a polyhedron is a solid shape bounded by polygons.


(ii) 4 triangles?

Yes, because a tetrahedron as 4 triangles as its faces.


(iii) a square and four triangles?

Yes, because a square pyramid has a square and four triangles as its faces.

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

Solution:

Yes, if number of faces is four or more.

4. Is a square prism same as a cube?

Solution:

Yes. We know that a square is a three dimensional shape with six rectangular shaped sides, out of which two are squares. Cubes are of rectangular prism length, width and height of same measurement.

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

Solution:

No.

Let us use Euler’s formula

V + F = E + 2

15 + 10 = 20 + 2

25 ≠ 22

Since the given polyhedron is not following Euler’s formula, therefore it is not possible to have 10 faces, 20 edges and 15 vertices.

6. Verify Euler’s formula for each of the following polyhedrons:

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 1

Solution:

(i) Vertices = 10

Faces = 7

Edges = 15

By using Euler’s formula

V + F = E + 2

10 + 7 = 15 + 2

17 = 17

Hence verified.


(ii) Vertices = 9

Faces = 9

Edges = 16

By using Euler’s formula

V + F = E + 2

9 + 9 = 16 + 2

18 = 18

Hence verified.


(iii) Vertices = 14

Faces = 8

Edges = 20

By using Euler’s formula

V + F = E + 2

14 + 8 = 20 + 2

22 = 22

Hence verified.


(iv) Vertices = 6

Faces = 8

Edges = 12

By using Euler’s formula

V + F = E + 2

6 + 8 = 12 + 2

14 = 14

Hence verified.

(v) Vertices = 9

Faces = 9

Edges = 16

By using Euler’s formula

V + F = E + 2

9 + 9 = 16 + 2

18 = 18

Hence verified.

7. Using Euler’s formula find the unknown:

Faces

?

5

20

Vertices

6

?

12

Edges

12

9

?

Solution:

(i)

By using Euler’s formula

V + F = E + 2

6 + F = 12 + 2

F = 14 – 6

F = 8

∴ Number of faces is 8


(ii)

By using Euler’s formula

V + F = E + 2

V + 5 = 9 + 2

V = 11 – 5

V = 6

∴ Number of vertices is 6


(iii)

By using Euler’s formula

V + F = E + 2

12 + 20 = E + 2

E = 32 – 2

E = 30

∴ Number of edges is 30


EXERCISE 19.2 PAGE NO: 19.12

1. Which among of the following are nets for a cube?

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 2

Solution:

Figure (iv), (v), (vi) are the nets for a cube.

2. Name the polyhedron that can be made by folding each net:

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 3

Solution:

(i) From figure (i), a Square pyramid can be made by folding each net.


(ii) From figure (ii), a Triangular prism can be made by folding each net.


(iii) From figure (iii), a Triangular prism can be made by folding each net.


(iv) From figure (iv), a Hexagonal prism can be made by folding each net.


(iv) From figure (v), a Hexagonal pyramid can be made by folding each net.


(v) From figure (vi), a Cube can be made by folding each net.

3. Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice?

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 4

Solution:

Figure (i), is a dice. Since the sum of numbers on opposite faces is 7 (3 + 4 = 7 and 6 + 1 = 7).

4. Draw nets for each of the following polyhedrons:

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 5

Solution:

(i) The net pattern for cube is

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 6

(ii) The pattern for triangular prism is

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 7

(iii) The net pattern for hexagonal prism is

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 8

(iv) The net pattern for pentagonal pyramid is

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 9

5. Match the following figures:

RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes image - 10

Solution:

(a)-(iv) Because multiplication of numbers on adjacent faces are equal, where 6×4 = 24 and 4×4 = 16

(b)-(i) Because multiplication of numbers on adjacent faces are equal, where 3×3 = 9 and 8×3 = 24

(c)-(ii) Because multiplication of numbers on adjacent faces are equal, where 6×4 = 24 and 6×3 = 18

(d)-(iii) Because multiplication of numbers on adjacent faces are equal, where 3×3 = 9 and 3×9 = 27


Also, access exercises of RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes

Exercise 19.1 Solutions

Exercise 19.2 Solutions

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