## 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**

- RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21A
- RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21B
- RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21C
- RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D
- RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21E

**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

= 30m 6 dm

**Question 2.**

**Solution:**

Length of rectangular field (l) = 62 m

and breadth (b) = 33 m

**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

**Question 5.**

**Solution:**

Total cost of fencing a rectangular field

= Rs 3300

Rate of fencing = Rs 25 per m

**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

**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)

**Question 9.**

**Solution:**

(i) Sides of a triangle are 7.8 cm, 6.5 cm and 5.9 cm

**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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.