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
\(=\frac { 36 }{ 6 } \) = 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 = \(\frac { 270 }{ 2 } \) = 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.