In this chapter, we provide RD Sharma Solutions for Class 8 Chapter 19 Visualising Shapes Exercise 19.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Chapter 19 Visualising Shapes Exercise 19.1 pdf, free RD Sharma Solutions for Class 8 Chapter 19 Visualising Shapes Exercise 19.1 book pdf download. Now you will get step by step solution to each question.

### Access answers to RD Sharma Maths Solutions For Class 8 Exercise 19.1 Chapter 19 Visualising Shapes

**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:**

**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

**All Chapter RD Sharma Solutions For Class 8 Maths**

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