Exercise 21.3, mainly deals with the surface area of a cuboid and a cube. We shall also discuss the surface area of the walls of a room. Solutions here are formulated by our expert faculty team with utmost care. Students who are willing to clear their doubts and improve their problem-solving skills can make use of RD Sharma Class 8 Solutions. This is the best resource students can access and prepare for their exams. Students can download the PDF easily that is provided below.

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### Access answers to RD Sharma Maths Solutions For Class 8 Exercise 21.3 Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)

**1. Find the surface area of a cuboid whose**

**(i) length = 10 cm, breadth = 12 cm, height = 14 cm(ii) length = 6 dm, breadth = 8 dm, height = 10 dm(iii) length = 2m, breadth = 4 m, height = 5 m(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.**

**Solution:**

**(i)** Given details are,

Length of a cuboid = 10 cm

Breadth of a cuboid = 12 cm

Height of a cuboid = 14 cm

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm^{2}

= 2 (10Ã—12 + 12Ã—14 + 14Ã—10)

= 2 (120 + 168 + 140)

= 2 (428)

= 856 cm^{2}

**(ii)** Given details are,

Length of a cuboid = 6 dm

Breadth of a cuboid = 8 dm

Height of a cuboid = 10 dm

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm^{2}

= 2 (6Ã—8 + 8Ã—10 + 10Ã—6)

= 2 (48 + 80 + 60)

= 2 (188)

= 376 dm^{2}

**(iii)** Given details are,

Length of a cuboid = 2m

Breadth of a cuboid = 4m

Height of a cuboid = 5m

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm^{2}

= 2 (2Ã—4 + 4Ã—5 + 5Ã—2)

= 2 (8 + 20 + 10)

= 2 (38)

= 76 m^{2}

**(iv)** Given details are,

Length of a cuboid = 3.2 m= 32 dm

Breadth of a cuboid = 30 dm

Height of a cuboid = 250 cm= 25 dm

We know that,

surface area of cuboid = 2 (lb + bh + hl) cm^{2}

= 2 (32Ã—30 + 30Ã—25 + 25Ã—32)

= 2 (960 + 750 + 800)

= 2 (2510)

= 5020 dm^{2}

**2. Find the surface area of a cube whose edge is(i) 1.2 m(ii) 27 cm(iii) 3 cm(iv) 6 m(v) 2.1 m**

**Solution:**

**(i)** Given,

Edge of cube = 1.2 m

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 1.2^{2}

= 6 Ã— 1.44

= 8.64 m^{2}

**(ii)** Given,

Edge of cube = 27 cm

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 27^{2}

= 6 Ã— 729

= 4374 cm^{2}

**(iii)** Given,

Edge of cube = 3 cm

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 3^{2}

= 6 Ã— 9

= 54 cm^{2}

**(iv)** Given,

Edge of cube = 6 m

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 6^{2}

= 6 Ã— 36

= 216 m^{2}

**(v)** Given,

Edge of cube = 2.1 m

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 2.1^{2}

= 6 Ã— 4.41

= 26.46 m^{2}

**3. A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.**

**Solution:**

Given details are,

Dimensions of cuboidal box = 5cm Ã— 5cm Ã— 4cm

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm^{2}

= 2 (5Ã—5 + 5Ã—4 + 4Ã—5)

= 2 (25 + 20 + 20)

= 2 (65)

= 130 cm^{2}

**4. Find the surface area of a cube whose volume is (i) 343 m ^{3 }(ii) 216 dm^{3}**

**Solution:**

**(i)** Given details are,

Volume of cube = 343 m^{3}

Side of cube, a = ^{3}âˆš(343) = 7m

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 7^{2}

= 6 Ã— 49

= 294 m^{2}

**(ii)** Given details are,

Volume of cube = 216 dm^{3}

Side of cube a = ^{3}âˆš(216) = 6dm

We know that,

Surface area of cube = 6 Ã— side^{2}

= 6 Ã— 6^{2}

= 6 Ã— 36

= 216 dm^{2}

**5. Find the volume of a cube whose surface area is**

**(i) 96 cm ^{2 }(ii) 150 m^{2}**

**Solution:**

**(i)** Given details are,

Surface area of cube = 96 cm^{2}

6 Ã— side^{2} = 96cm^{2}

Side^{2} = 96/6

= 16

Side = âˆš16 = 4cm

âˆ´ Volume of a cube = 4^{3} = 64cm^{3}

**(ii)** Given details are,

Surface area of cube = 150 m^{2}

6 Ã— side^{2} = 150cm^{2}

Side^{2} = 150/6

= 25

Side = âˆš25 = 5cm

âˆ´ Volume of a cube = 5^{3} = 125m^{3}

**6. The dimensions of a cuboid are in the ratio 5: 3: 1 and its total surface area is 414 m ^{2}. Find the dimensions.**

**Solution:**

Given details are,

Ratio of dimensions of a cuboid = 5:3:1

Total surface area of cuboid = 414 m^{2}

The dimensions are = 5x Ã— 3x Ã— x

Surface area of cuboid = 414 m^{2}

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm^{2}

2 (lb + bh + hl) cm^{2} = 414

2 (15x^{2} + 3x^{2} + 5x^{2}) = 414

2 (23x^{2}) = 414

46x^{2} = 414

x^{2} = 414/46

= 9

x = âˆš9

= 3

âˆ´ Dimensions are,

5x = 5 (3) = 15m

3x = 3 (3) = 9m

x = 3m

**7. Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm.**

**Solution:**

Given details are,

Dimensions of closed box = 25cm Ã— 0.5m Ã— 15cm = 25cm Ã— 50cm Ã— 15cm

We know that,

Area of cardboard required = 2 (lb + bh + hl) cm^{2}

= 2 (25Ã—50 + 50Ã—15 + 15Ã—25)

= 2 (1250 + 750 + 375)

= 2 (2375)

= 4750 cm^{2}

**8. Find the surface area of a wooden box whose shape is of a cube, and if the edge of the box is 12 cm.**

**Solution:**

Given details are,

Edge of a cubic wooden box = 12 cm

We know that,

Surface area of cubic wooden box = 6 Ã— side^{2}

= 6 Ã— 12^{2}

= 6 Ã— 144

= 864 cm^{2} Â

**9. The dimensions of an oil tin are 26 cmÃ— 26 cmÃ— 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs Rs. 10, find the cost of tin sheet used for these 20 tins.**

**Solution:**

Given details are,

Dimensions of oil tin = 26cm Ã— 26cm Ã— 45cm

Then,

Area of tin sheet required for making one oil tin = total surface area of oil tin

= 2 (lb + bh + hl) cm^{2}

= 2 (26Ã—26 + 26Ã—45 + 45Ã—26)

= 2 (676 + 1170 + 1170)

= 2 (3016)

= 6032 cm^{2}

Area of tin sheet required for 20 oil tins = 20 Ã— 6032

= 120640 cm^{2}

= 12.064 m^{2}

Given, Cost of 1 m^{2}Â tin sheet = Rs 10

So, Cost of 12.064 m^{2}Â tin sheet = 10 Ã— 12.064

= Rs 120.60

**10. A classroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows etc.)**

**Solution:**

Given details are,

Dimensions of class room = 11m Ã— 8m Ã— 5m

Where, Length = 11m, Breadth = 8m, Height = 5m

We know,

Area of floor = length Ã— breadth

= 11 Ã— 8

= 88 m^{2}

Area of four walls (including doors & windows) = 2 (lh + bh) cm^{2}

= 2 (11Ã—5 + 8Ã—5)

= 2 (55 + 40)

= 2 (95)

= 190m^{2}

âˆ´Â Sum of areas of floor and four walls = area of floor + area of four walls

= 88 + 190

= 278 m^{2}

**11. A swimming pool is 20 m long 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs. 25 per square metre.**

**Solution:**

Given details are,

Dimensions of swimming pool are = 20m Ã— 15m Ã—3m

Where, Length = 20m , Breadth = 15m , Height = 3m

We know,

Area of floor = length Ã— breadth

= 20 Ã— 15

= 300 m^{2}

Area of walls of swimming pool = 2 (lh + bh) cm^{2}

= 2 (20Ã—3 + 15Ã—3)

= 2 (60 + 45)

= 2 (105)

= 210m^{2}

Sum of areas of floor and four walls = area of floor + area of walls

= 300 + 210

= 510 m^{2}

Given, Cost for repairing 1m^{2}Â area = Rs 25

âˆ´Â Cost for repairing 510 m^{2}Â = 510 Ã— 25

= Rs 12750

**12. The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of four walls of the room.**

**Solution:**

Given details are,

Height of floor = 3m

Perimeter of floor = 30m

So, perimeter = 30

2(l+b) = 30

l+b = 30/2

l+b = 15m

âˆ´ Area of four walls of room = 2 (lh + bh) m^{2}

= 2h (l+b)

= 2 (3) (15)

= 90m^{2}

**13. Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.**

**Solution:**

Let us consider length of cuboid as =Â l cm

Let us consider breadth of cuboid as = b cm

Let us consider height of cuboid as = h cm

We know,

Area of floor = l Ã— b = lb cm^{2}Â

Then,

Product of areas of two adjacent walls = (lÃ—h) Ã— (bÃ—h) = lbh^{2} cm^{4}Â

Product of areas of floor and two adjacent walls = lb Ã— lbh^{2} cm^{6}Â

= l^{2} Ã— b^{2} Ã— h^{2 }cm^{6}

= (lbh)^{2} cm^{6}

We know, volume of cuboid = lbh cm

Hence, areas of the floor and two adjacent walls of a cuboid is the square of its volume.

**14. The walls and ceiling of a room are to be plastered. The length, breadth nad height of the room are 4.5 m, 3 m and 350 cm, respectively. Find the cost of plastering at the rate of Rs. 8 per square metre.**

**Solution:**

Given details are,

Length of room = 4.5m

Breadth of wall = 3m

Height of wall = 350cm = 350/100 = 3.5mÂ

Area of ceiling = l Ã— b

= 4.5 Ã— 3

= 13.5 m^{2}

Area of walls = 2 (lh + bh) m^{2}

= 2 (4.5Ã—3.5 + 3Ã—3.5)

= 2 (15.75 + 10.5)

= 52.5 m^{2}

Sum of Area of ceiling + area of walls = 13.5m^{2} + 52.5m^{2}

= 66m^{2}

Â Given, Cost for plastering 1m^{2}Â area = Rs 8

âˆ´Â Cost for plastering 66 m^{2}Â area = 66 Ã— 8 = Rs 528

**15. A cuboid has total surface area of 50 m ^{2}Â and lateral surface area is 30 m^{2}. Find the area of its base.**

**Solution:**

Given details are,

Total surface area of cuboid = 50 m^{2}

Lateral surface area of cuboid = 30 m^{2}

Total Surface area = 2 (surface area of base) + (surface area of 4 walls)

50 = 2 (surface area of base) + (lateral surface area)

50 = 2 (surface area of base) + 30

50 – 30 = 2 (surface area of base)

20 = 2 (surface area of base)

Surface area of base = 20/2

= 10 m^{2}

âˆ´ Area of base is 10m^{2}

**16. A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m ^{2}. What is the cost of white washing the walls at the rate of Rs 1.50 per m^{2}?**

**Solution:**

Given details are,

Dimensions of class room = 7m Ã— 6m Ã— 3.5m

Where, Length = 7m, Breadth = 6m, Height = 3.5m

Area of four walls (including doors & windows) = 2 (lh + bh) m^{2}

= 2 (7Ã—3.5 + 6Ã—3.5)

= 91m^{2}

Area of four walls (without doors & windows) =

Area including doors & windows â€“ area occupied by doors & windows

= 91 â€“ 17 = 74 m^{2}

Then,

Cost for white washing 1m^{2}Â area of walls = Rs 1.50

âˆ´Â Total cost for white washing the walls = 74 Ã— 1.50 = Rs 111

**17. The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3m Ã—1.5m and 10 windows each of size 1.5mÃ— 1m. If the cost of white washing the walls of the hall at the rate of Rs 1.20 per m ^{2}Â is Rs 2385.60, find the breadth of the hall.**

**Solution:**

Given details are,

Dimensions of central hall of a school = Length = 80 m , height = 8m

Let breadth of hall be â€˜bâ€™ m

So,

Area of each door = 3m Ã— 1.5m = 4.5m^{2}

Area of 10 doors = 10 Ã— 4.5 = 45m^{2}

Area of each window = 1.5m Ã— 1m = 1.5 m^{2}

Area of 10 windows = 10 Ã— 1.5 = 15m^{2}

Area occupied by doors and windows = 45 + 15 = 60 m^{2}

Area of the walls of the hall including doors and windows = 2 (lh + bh) m^{2}

= 2 (80Ã—8 + bÃ—8)

= 2(640+8b) m^{2}

Then,

Area of only walls = area of walls including doors & windows â€“ area occupied by doors & windows

= 2(640+8b) â€“ 60

= 1280 + 16b â€“ 60

= (1220 + 16b) m^{2}

Given, Total cost for white washing = Rs 2385.60

Rate of white washing = Rs 1.20 per m^{2}

So,

Total cost = Rate Ã— (areas of walls only)

2385.60 = 1.20 Ã— (1220 + 16b)

2385.60 / 1.20 = (1220 + 16b)

1988 = 1220 + 16b

16b = 1988 â€“ 1220

= 768

b = 768/16

= 48

âˆ´ Breadth of hall is 48 m