RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Ex 14.1 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Exercise 14.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Perimeter and Area |

Exercise |
Ex 14.1 |

Number of Questions |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Ex 14.1

Question 1.

Find the perimeter of each (RBSESolutions.com) of the following figures

Solution:

(i) Perimeter of given figure = Sum of all sides

= 2 + 5 + 1 + 3.5 + 3.5 + 1 + 5

= 21 cm

(ii) Perimeter (RBSESolutions.com) of given figure = Sum of all sides

= 1 + 8 + 4 + 5 + 1 + 4 + 3 + 8 + 1 + 8 + 4 + 5 + 1 + 4 + 3 + 8

= 68 cm

(iii) Perimeter of given sides = Sum of all sides

= 6 + 6 + 2 + 2 + 2 + 2 + 2 + 2

= 24 cm

Question 2.

Find the perimeter of a regular pentagon with each side equal to 4 cm.

Solution:

Total sides in (RBSESolutions.com) regular pentagon = 5

Length of one side = 4 cm

Perimeter of regular pentagon = Total sides × Length of one side

= 5 × 4 = 20 cm

Question 3.

A piece of string is 36 cm long. What will be the length of each (RBSESolutions.com) side if the string is used to form

(i) a square?

(ii) an equilateral?

(iii) a regular hexagon?

Solution:

Length of piece of string = 36 cm

(i) Total sides in square = 4

∴ Length of each side of square

[latex s]=\frac { 36 }{ 6 } [/latex] = 9 cm

(ii) Total side in (RBSESolutions.com) equilateral triangle = 3

∴ Length of each side of regular hexagon

(iii)Total sides in regular hexagon = 6

∴ Length of each side of regular hexagon

Question 4.

Geeta runs around a square (RBSESolutions.com) field of side 50 m. Pooja runs around a rectangular field with length 65 m and breadth 25 m. Who covers less distance?

Solution:

If we find perimeter of both field then, person runs around the field of least perimeter will covers the less distance.

Perimeter of squared field with side 50 m

= 4 × side = 4 × 50 = 200 m

Perimeter of rectangular field with length 65 m and breadth 25 m

= 2 (l + b) = 2 (65 + 25)

= 2 (90) = 180 m

∵ 200 > 180

∴ Pooja will cover less distance (200 – 180 = 20 m)

Question 5.

Find the side of the regular (RBSESolutions.com) pentagon whose perimeter is 30 cm.

Solution:

Perimeter of regular pentagon = 30 cm

Number of sides in it = 5

Length of each side of regular pentagon

= 6 cm

Thus, length of one side is 6 cm.

Question 6.

Madhu has a rectangular field of length and (RBSESolutions.com) breadth 23.5 m and 15.5 m respectively. He want to fence his field with steel wire. What is the total length of steel wire he must use?

Solution:

Length of wire required for fencing the field = Perimeter of field

= 2 (l + b) = 2 (23.5 + 15.5)

= 2 (39) = 78 m

Thus, 78 m long steel wire needed.

Question 7.

Perimeter of a football ground is 270 m. Find (RBSESolutions.com) the breadth of the ground if the length of the ground is 90 m.

Solution:

Perimeter of ground = 270 m

⇒ 2 (l + b) = 270

⇒ 2 (90 + breadth) = 270

⇒ 90 + breadth = [latex s]\frac { 270 }{ 2 } [/latex] = 135

⇒ breadth = 135 – 90 = 45

⇒ Thus, breadth of ground is 45 m.

We hope the RBSE Solutions for Class 6 Maths Chapter 14 Perimeter and Area Ex 14.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Exercise 14.1, drop a comment below and we will get back to you at the earliest.

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