RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A

Other Exercises

Question 1.
Solution:
(i) Length (l) = 16.8 cm
Breadth (b) = 6.2 cm
Perimeter = 2 (l + b)
= 2 (16.8 + 6.2) cm
= 2 x 23
= 46 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q1.1
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q1.2
= 30m 6 dm

Question 2.
Solution:
Length of rectangular field (l) = 62 m
and breadth (b) = 33 m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q2.1

Question 3.
Solution:
Perimeter of field = 128 m
Length + Breadth = \(\\ \frac { 128 }{ 2 } \) = 64 m
Ratio in length and breadth = 5:3
Let length (l) = 5x
Then breadth = 3x
5x + 3x = 64
=> 5x = 64
=> x = \(\\ \frac { 64 }{ 8 } \) = 8
Length of the field = 5x = 5 x 8 = 40m
and breadth = 3x = 3 x 8 = 24m

Question 4.
Solution:
Cost of fencing a rectangular field
= Rs. 18 per m
Total cost = Rs. 1980
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q4.1

Question 5.
Solution:
Total cost of fencing a rectangular field
= Rs 3300
Rate of fencing = Rs 25 per m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q5.1

Question 6.
Solution:
(i) Side of square = 3.8 cm
Perimeter = 4 x side
= 4 x 3.8 cm
= 15.2 cm
(ii) Side of a square = 4.6 m
Perimeter = 4 x side
= 4 x 4.6 m
= 18.4 m
(iii) Side of a square = 2 m 5 dm
= 2.5 m
Perimeter = 4 x side
= 4 x 2.5 m
= 10 m

Question 7.
Solution:
Total cost of fencing a square field = Rs. 4480
Rate of fencing = Rs. 35 per m
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q7.1

Question 8.
Solution:
Side of a square field (a) = 21 m
Perimeter = 4a = 4 x 21 = 84m
Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let length (l) = 4x
and breadth (b) = 3x
Perimeter = 2 (l + b)
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q8.1

Question 9.
Solution:
(i) Sides of a triangle are 7.8 cm, 6.5 cm and 5.9 cm
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q9.1
RS Aggarwal Class 6 Solutions Chapter 21 Concept of Perimeter and Area Ex 21A Q9.2

Question 10.
Solution:
(i) Each side of a regular pentagon
= 8 cm
Perimeter = 5 x Side
= 5 x 8
= 40 cm
(ii) Each side of an octagon = 4.5 cm
Perimeter = 8 x Side
= 8 x 4.5
= 36 cm
(iii) Each side of a regular decagon = 3.6 cm
Perimeter = 10 x Side
= 10 x 3.6
= 36 cm

Question 11.
Solution:
We know that perimeter of a closed figure or a polygon = Sum of its sides
(i) In the figure, sides of a quadrilateral are 45 cm, 35 cm, 27 cm, 35 cm
Its perimeter = Sum of its sides
= (45 + 35 + 27 + 35) cm
= 142 cm
(ii) Sides of a quadrilateral (rhombus) are 18 cm, 18 cm, 18 cm, 18 cm
i.e., each side = 18 cm Perimeter
= 4 x Side
= 4 x 18
= 72 cm
(iii) Sides of the polygon given are 16 cm, 4 cm, 12 cm, 12 cm, 4 cm, 16 cm and 8 cm
Its perimeter = Sum of its sides
= (16 + 4 + 12 + 12 + 4 + 16 + 8) cm
= 72 cm

 

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A are helpful to complete your math homework.

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