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### Access answers to Maths RS Aggarwal Solutions for Class 6 Chapter 21 Concept of Perimeter and Area Exercise 21D

**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**

**Solutions**

**(i) **Given

Length = 46 cm

Breadth = 25 cm

Area of the rectangle = (length Ã— breadth) sq units

= (46 Ã— 25) cm^{2}

= 1150 cm^{2}

(ii) Given

Length = 9 m

Breadth = 6 m

Area of the rectangle = (length Ã— breadth) sq units

= (9 Ã— 6) m^{2}

= 54 m^{2}

(iii) Given

Length = 14.5 m

Breadth = 6.8 m

Area of the rectangle = (length Ã— breadth) sq units

= (14.5 Ã— 6.8) m^{2}

= (145 / 10 Ã— 68 / 10) m^{2}

= (9860 / 100) m^{2}

= 98.6 m^{2}

(iv) Given

Length = 2 m 5 cm

= 200 cm + 5 cm {1 m = 100 cm}

= 205 cm

Breadth = 60 cm

Area of the rectangle = (length Ã— breadth) sq units

= (205 Ã— 60) cm^{2}

= 12300 m^{2}

(v) Given

Length = 3.5 km

Breadth = 2 km

Area of the rectangle = (length Ã— breadth) sq units

= (3.5 Ã— 2) km^{2}

= (35 / 10 Ã— 2) km^{2}

= 7 km^{2}

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

**Solution**

** **Given

Side of square plot = 14 m

Area of a square = (side)^{2} sq units

= (14)^{2} m^{2}

= (14 Ã— 14) m^{2}

= 196 m^{2}

Hence, area of the square plot = 196 m^{2}

**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 the table = 2 m 25 cm

= (2 + 0.25) m [1 m = 100 cm]

= 2.25 m

Breadth of the table = 1 m 20 cm

= (1 + 0.20) m [1 m = 100 cm]

= 1.20 m

Area of the rectangle = (length Ã— breadth) sq units

= (2.25 Ã— 1.20) m^{2}

= [(225 / 100) Ã— (120 / 100)] m^{2}

= 2.7 m^{2}

Area of the table is 2.7 m^{2}

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

**Solution**

** **Given

Length of the carpet = 30 m 75 cm

= (30 + 0.75) cm [1 m = 100 cm]

= 30.75 m

Breadth of the carpet is given 80 cm

= 0.80 m [1 m = 100 cm]

Area of the carpet = (Length Ã— breadth) sq units

= (30.74 Ã— 0.80) m^{2}

= (3074 / 100 Ã— 80 / 100) m^{2}

= 24.6 m^{2}

Carpet cost of 1 m^{2} = Rupees 150

Carpet cost of 24.6 m^{2} = (24.6 Ã— 150)

= Rupees 3690

Hence, cost of carpet at rupees 150 per square metre is rupees 3690

**5. How many envelopes can be made out of a sheet of 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**

Given

Length of sheet of paper = 3 m 24 cm

= 324 cm [1 m = 100 cm]

Breadth of sheet of paper = 1 m 72 cm

= 172 cm [1 m = 100 cm]

Area of the sheet of paper = (length Ã— breadth)

= (324 Ã— 172) cm^{2}

= 55728 cm^{2}

Length of the piece of paper to make 1 envelope = 18 cm

Breadth of the piece of paper to make 1 envelope = 12 cm

Area of the piece of paper to make 1 envelope = (length Ã— breadth)

= (18 Ã— 12) cm^{2}

= 216 cm^{2}

Number of envelope = (Area of the sheet of paper) / (Area of the piece of paper to make 1 envelope)

Number of envelopes = 55728 / 216

Number of envelopes = 258 envelopes

Number of envelopes that can be made is 258

**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**

** **Given

Length of the room = 12.5 m

Breadth of the room = 8 m

Area of the room = (length Ã— breadth)

= (12.5 Ã— 8) m^{2}

= 100 m^{2}

Square carpet side = 8 m

Area of the square carpet = (side)^{2}

= (8)^{2}

= 8 Ã— 8

= 64 m^{2}

Area of the floor which is not carpeted = (Area of the room) â€“ ( Area of the carpet)

= (100 â€“ 64) m^{2}

= 36 m^{2}

Hence, area of the room which is not carpeted is 36 m^{2}

**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**

** **Given

Length of the road = 150 m

= 15000 cm [1 m = 100 cm]

Breadth of the road = 9 m

= 900 cm [1 m = 100 cm]

Area of the road = (length Ã— breadth)

= (150000 Ã— 900)

= 13500000 cm^{2}

Given, length of the brick = 22.5 cm

Breadth of the brick = 7.5 cm

Area of the brick = (length Ã— breadth)

= (22.5 Ã— 7.5)

= 168.75 cm^{2}

Number of bricks = Area of the road / Area of one brick

= 13500000 / 168.75

Number of bricks = 80000 bricks

**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 the rate of Rupees 65 per metre.**

**Solution**

** **Given,

Length of the room = 13 m

Breadth of the room = 9 m

Area of the room = (length Ã— breadth)

= (13 Ã— 9) m^{2}

= 117 m^{2}

Let the required carpet length be x m

Breadth of the carpet = 75 cm = 0.75 m [1 m = 100 cm]

Area of the carpet = (0.75 Ã— x) m^{2}

= 0.75x m^{2}

For carpeting the room

Area covered by the carpet = Area of the room

0.75x = 117

x = 117 / 0.75

x = 156 m

Hence, length of the carpet = 156 m

1 m carpet cost = rupees 65

156 m carpet cost will be = (156 Ã— 65)

** **= Rupees 10140

**9. The length and the 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 be 5x

Breadth of the rectangular park be 3x

Perimeter of the rectangular park = 2 (length + breadth)

= 2 (5x + 3x)

= 2 (8x)

= 16x

Given perimeter of the rectangular park = 128 m

128 = 16x

x = 128 / 16

x = 8

Hence, length of the park = 5x

= 5 Ã— 8

= 40 m

Breadth of the park = 3x

= 3 Ã— 8

= 24 m

Area of the park = (length Ã— breadth)

= (40 Ã— 24)

= 960 m^{2}