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

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 21D

**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.**

**Find the area of a rectangle whose**

**(i) length = 46 cm and breadth = 25 cm**

**(ii) length = 9 m and breadth = 6 m**

**(iii) length = 14.5 m and breadth = 6.8 m**

**(iv) length = 2 m 5 cm and breadth = 60 cm**

**(v) length = 3.5 km and breadth = 2 km**

**Solution:**

(i) Length (l) = 46 cm

Breadth (b) = 25 cm

Area of rectangle = l x b

= 46 x 25 sq. cm

= 1150 sq.cm.

(ii) Length (l) = 9 m

Breadth (b) = 6 m

Area = l x b = 9 x 6

= 54 sq. metre Ans.

(iii) Length (l) = 14.5 m

Breadth (b) = 6.8 m

Area = l x b

= 14.5 x 6.8 sq. m

= 98.6 sq. m Ans.

(iv) Length (l) = 2m 5cm

= 2x 100 cm + 5 cm

= 200 cm + 5 cm

= 205 cm

Breadth = 60 cm

Area = l x b

= 205 cm x 60 cm

= 12300 cm²

(v) Length (l) = 3.5 km

Breadth (b) = 2 km

Area = l x b

= 3.5 x 2

= 7 sq. km. Ans.

**Question 2.**

**Find the area of a square plot of side 14m**

**Solution:**

Side of a square plot = 14 m

Area = (Side)²

= (14)²

= 196 m²

**Question 3.**

**The top of a table measures 2 m 25 cm by 1 m 20 cm. Find its area in square metres.**

**Solution:**

Length of top of table (l) = 2 m, 25 cm = 2.25 m

Breadth (l) = 1 m 20 cm = 1.20 m

Area of the top of the table = l x b

= (2.25 x 1.20) sq. m

= 2.7 sq. m. Ans.

**Question 4.**

**A carpet is 30 m 75 cm long and 80 cm wide. Find its cost at Rs. 150 per square metre.**

**Solution:**

Length of carpet (l) = 30 m 75 cm

= 30.75 m

Breadth (b) = 80 cm = 0.80 m

Area of the carpet = l x b

= (30.75 x 0.80) sq. m

= 24.6 sq. m

Cost of one square metre = Rs. 20

Total cost = 24.6 x 150

= Rs. 3690. Ans

**Question 5.**

**How many envelopes can be made out of a sheet of a paper 3 m 24 cm by 1 m 72 cm, if each envelope requires a piece of paper of size 18 cm by 12 cm ?**

**Solution:**

Length of the sheet of paper 3 m 24 cm

= 300 cm + 24 cm

= 324 cm

Breadth of the sheet of the paper 1 m 72 cm

= 100 cm + 72 cm

= 172 cm

Area of the sheet of paper = (324 x 172) cm²

Also, area of the piece of paper required for an envelope = (18 x 12) cm².

Number of envelopes that can be made

=

= 258

**Question 6.**

**A room is 12.5 m long and 8 m wide. A square carpet of side 8 m is laid on its floor. Find the area of the floor which is not carpeted**

**Solution:**

Length of room (l) = 12.5 m

Breadth (b) = 8 m

Area = l x b

= (12.5 x 8) sq. m

= 100 sq. m

Side of square carpet (a) = 8 m

Area of carpet = a² = (8 x 8) sq. m.

= 64 square metre

Area left without carpet = 100 sq. m – 64 sq. m

= 36 sq. m Ans.

**Question 7.**

**A lane 150 m long and 9 m wide is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. Find the number of bricks required**

**Solution:**

Length of the lane = 150 m

Breadth of the lane = 9 m

Area of the lane = (150 x 9) m²

= 1350 m²

Area of the brick = 22.5 cm x 7.5 cm

= 168.75 cm²

**Question 8.**

**A room is 13 m long and 9 m broad. Find the cost of carpeting the room with a carpet 75 cm broad at titrate of Rs. 65 per metre.**

**Solution:**

Length of room (l) = 13 m

Breadth (b) = 9 m

Area of floor or carpet = l x b

= 13 x 9

= 117 sq. m

**Question 9.**

**The length and breadth of a rectangular park are in the ratio 5:3 and its perimeter is 128 m. Find the area of the park.**

**Solution:**

Let the length of the rectangular park = 5x metres

and the breadth of the rectangular park = 3x metres

**Question 10.**

**Two plots of land have the same perimeter. One is a square of side 64 m and the other is a rectangle of length 70 m. Find the breadth of the rectangular plot. Which plot has the greater area and by how much.**

**Solution:**

Side of the square plot = 64 m

Perimeter of the square plot = 4 x Side

= 4 x 64

= 256 m

Perimeter of the rectangular plot = Perimeter of the square plot = 256 m

Length of the rectangular plot = 70 m

Perimeter = 2 x (Length + Breadth)

256 = 2 (70 + b)

256 = 140 + 2b

=> 2b = 256 – 140

=> 2b = 116

b = = 58 cm

Area of the rectangular plot = (length x breadth)]

= (70 x 58) m²

= 4060 m²

Area of the square plot = Side x Side

= (64 x 64) m²

= 4096 m².

Square plot has the greater area than that of the rectangular plot by

(4096 – 4060)

= 36 m².

**Question 11.**

**The cost of cultivating a rectangular field at Rs. 35 per square metre, is Rs. 71400. If the width of the field is 40 m, find its length. Also, find the cost of fencing the field at Rs. 50 per metre.**

**Solution:**

Total cost of cultivating the rectangular field = Rs. 71400

Rate of cultivating = Rs. 35 per sq. m

**Question 12.**

**The area of a rectangle is 540 cm² and its length is 36 cm. Find its width and perimeter.**

**Solution:**

Area of rectangle = 540 sq. cm

Length (l) = 36 cm

**Question 13.**

**A marble tile measures 12 cm x 10 cm. How many tiles will be required to cover a wall of size 4 m by 3 m ? Also, find the total cost of the tiles at Rs. 22.50 per tile.**

**Solution:**

Measure of a marble tile = 12cm x 10cm

Area of wall = 4 m x 3 m

= 12 m²

Area of one marble tile

= 12 x 10

= 120 cm²

**Question 14.**

**Find the perimeter of a rectangle whose area is 600 cm² and breadth is 25 cm**

**Solution:**

Area of a rectangle = 600 cm²

Breadth = 25 cm

**Question 15.**

**Find the area of a square whose diagonal is 5√2 diagonal**

**Solution:**

diagonal of square = 5 √2

**Question 16.**

**Calculate the area of each one of the shaded regions given below :**

**Solution:**

(i) We name the given region as shown in the figure

**Question 17.**

**Calculate the area of each one of the shaded regions given below (all measures are given in cm):**

**Solution:**

Measures are in cm

(i) In the figure, there are three rectangles and one square

= 5(6 x 6) cm²

= 180 cm²

Hope given RS Aggarwal Solutions Class 6 Chapter 21 Concept of Perimeter and Area Ex 21D 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.

Deepak says

It is a good idea

Ansh Mishra king says

Bhy maray shath videos banao gay

Bhagirath says

Ossum😍😍