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**Joy of Mathematics Class 8 Solutions Chapter 19 Recognition of Solids**

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 19 Recognition of Solids. Here students can easily find step by step solutions of all the problems for Recognition of Solids. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 19.1 and 19.2

**Recognition of Solids Exercise 19.1 Solution : **

**Question no – (1)**

**Solution :**

Here, length = 6 cm

Breadth = 4 cm

Height = 3 m

**Question no – (2)**

**Solution :**

Here, cube of side = 5 cm

**Question no – (3)**

**Solution :**

Here,

Length = 22 cm

Breadth = 10 cm = Height of the cylinder.

**∴** Then circumference = 22 cm

According to the problem,

2πr = 22

=> r = 22/2π

= 22 × 7/2 × 22

= 7/2

= 3.5

**∴** Thus height of the cylinder 10 cm and radius of the cylinder 3.5 cm

**Question no – (4)**

**Solution :**

**(b)** Net of Pyramid

**Question no – (5)**

**Solution :**

**(a) **

This is two dimensional shape. Name of the shape is triangle.

**(b)**

This is three dimensional shape. Name of the shape is pyramid.

**(c)**

This is three dimensional shape. Name of the shape is cylinder.

**(d)**

This is three dimensional shape. Name of the shape is sphere.

**(e)** This is three dimensional shape. Name of the shape is prism.

**(f)** This is three dimensional shape. Name of the shape is cube.

**(g)**

This is three dimensional shape. Name of the shape is cone.

**(h)**

This is two dimensional shape. Name of the shape is rectangle.

**(i)** This is three dimensional shape and it’s a cube.

**Question no – (6)**

**Solution :**

Solids |
Nets |

(a) | (iv) |

(b) | (iii) |

(c) | (v) |

(d) | (i) |

(e) | (ii) |

**Question no – (7)**

**Solution :**

The cube can be used to make cube are, a, b, c, d.

**Question no – (8)**

**Solution :**

**(a)**

**(b)**

**Recognition of Solids Exercise 19.1 Solution : **

**Question no – (1)**

**Solution :**

**(a)** A triangular prism has __5__ faces 6 vertices and __9__ edges.

**(b)** A book is in the shape of __cuboid__.

**(c)** A torch cell is in the shape of __cylinder__.

**(d)** A candle is in the shape of __cylinder__.

**(e)** A triangular pyramid whose all faces are equilateral triangles of the same size is called __tetrahedron__.

**(f)** A pentagonal prism has __7__ faces __10__ vertices and __15__ edges.

**Question no – (2)**

**Solution :**

**(a)** This is a square pyramid.

Here,

Faces = 5

Vertices = 5

Edges = 8

Now, Euler’s formula we get,

F + V – E

= 5 + 5 – 8

= 2 (Proved)

**(b)** This is pentagonal prism

Here, Faces = 7

Vertices = 10

Edges = 15

From Euler’s formula we get,

F + V – E

= 7 + 10 – 15

= 17 – 5

= 12 (Proved)

**(c)** This is a Triangular prism

Here,

Faces = 5

Vertices = 6

Edges = 9

Now, Euler’s formula we get,

F + V – E

= 5 + 6 – 9

= 11 – 9

= 2 (Proved)

**Question no – (3)**

**Solution :**

**(a)** Here,

Faces = 4

Vertices = 4

Edges = x

Now, from Euler’s formula,

F + V – E = 2

=> 4 + 4 – x = 2

=> 8 – x = 2

=> x = 6

**(b)** and Here,

Faces = y

Vertices = x = 6

Edges = 12

From Euler’s formula,

F + V – E = 2

=> Y + 6 – 12 = 2

=> Y – 6 = 2

=> Y = 2 + 6 = 8

**(c)** Here, Faces = x = 6

Vertices = y = 8

Edges = z

From Euler’s formula,

F + V – E = 2

=> 6 + 8 – z = 2

=> 14 – z = 2

=> – z = 2 – 14

=> – z = – 12

=> z = 12

**Previous Chapter Solution : **