# RD Sharma Solutions for Class 8 Maths Chapter 21 Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube) Exercise 21.3

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.

## Download the Pdf of RD Sharma Solutions for Class 8 Maths Exercise 21.3 Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)

### 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) cm2

= 2 (10Ã—12 + 12Ã—14 + 14Ã—10)

= 2 (120 + 168 + 140)

= 2 (428)

= 856 cm2

(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) cm2

= 2 (6Ã—8 + 8Ã—10 + 10Ã—6)

= 2 (48 + 80 + 60)

= 2 (188)

= 376 dm2

(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) cm2

= 2 (2Ã—4 + 4Ã—5 + 5Ã—2)

= 2 (8 + 20 + 10)

= 2 (38)

= 76 m2

(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) cm2

= 2 (32Ã—30 + 30Ã—25 + 25Ã—32)

= 2 (960 + 750 + 800)

= 2 (2510)

= 5020 dm2

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— side2

= 6 Ã— 1.22

= 6 Ã— 1.44

= 8.64 m2

(ii) Given,

Edge of cube = 27 cm

We know that,

Surface area of cube = 6 Ã— side2

= 6 Ã— 272

= 6 Ã— 729

= 4374 cm2

(iii) Given,

Edge of cube = 3 cm

We know that,

Surface area of cube = 6 Ã— side2

= 6 Ã— 32

= 6 Ã— 9

= 54 cm2

(iv) Given,

Edge of cube = 6 m

We know that,

Surface area of cube = 6 Ã— side2

= 6 Ã— 62

= 6 Ã— 36

= 216 m2

(v) Given,

Edge of cube = 2.1 m

We know that,

Surface area of cube = 6 Ã— side2

= 6 Ã— 2.12

= 6 Ã— 4.41

= 26.46 m2

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) cm2

= 2 (5Ã—5 + 5Ã—4 + 4Ã—5)

= 2 (25 + 20 + 20)

= 2 (65)

= 130 cm2

4. Find the surface area of a cube whose volume is
(i) 343 m3
(ii) 216 dm3

Solution:

(i) Given details are,

Volume of cube = 343 m3

Side of cube, a = 3âˆš(343) = 7m

We know that,

Surface area of cube = 6 Ã— side2

= 6 Ã— 72

= 6 Ã— 49

= 294 m2

(ii) Given details are,

Volume of cube = 216 dm3

Side of cube a = 3âˆš(216) = 6dm

We know that,

Surface area of cube = 6 Ã— side2

= 6 Ã— 62

= 6 Ã— 36

= 216 dm2

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

(i) 96 cm2
(ii) 150 m2

Solution:

(i) Given details are,

Surface area of cube = 96 cm2

6 Ã— side2 = 96cm2

Side2 = 96/6

= 16

Side = âˆš16 = 4cm

âˆ´ Volume of a cube = 43 = 64cm3

(ii) Given details are,

Surface area of cube = 150 m2

6 Ã— side2 = 150cm2

Side2 = 150/6

= 25

Side = âˆš25 = 5cm

âˆ´ Volume of a cube = 53 = 125m3

6. The dimensions of a cuboid are in the ratio 5: 3: 1 and its total surface area is 414 m2. Find the dimensions.

Solution:

Given details are,

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

Total surface area of cuboid = 414 m2

The dimensions are = 5x Ã— 3x Ã— x

Surface area of cuboid = 414 m2

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm2

2 (lb + bh + hl) cm2 = 414

2 (15x2 + 3x2 + 5x2) = 414

2 (23x2) = 414

46x2 = 414

x2 = 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) cm2

= 2 (25Ã—50 + 50Ã—15 + 15Ã—25)

= 2 (1250 + 750 + 375)

= 2 (2375)

= 4750 cm2

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— side2

= 6 Ã— 122

= 6 Ã— 144

= 864 cm2

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) cm2

= 2 (26Ã—26 + 26Ã—45 + 45Ã—26)

= 2 (676 + 1170 + 1170)

= 2 (3016)

= 6032 cm2

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

= 120640 cm2

= 12.064 m2

Given, Cost of 1 m2Â tin sheet = Rs 10

So, Cost of 12.064 m2Â 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 m2

Area of four walls (including doors & windows) = 2 (lh + bh) cm2

= 2 (11Ã—5 + 8Ã—5)

= 2 (55 + 40)

= 2 (95)

= 190m2

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

= 88 + 190

= 278 m2

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 m2

Area of walls of swimming pool = 2 (lh + bh) cm2

= 2 (20Ã—3 + 15Ã—3)

= 2 (60 + 45)

= 2 (105)

= 210m2

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

= 300 + 210

= 510 m2

Given, Cost for repairing 1m2Â area = Rs 25

âˆ´Â Cost for repairing 510 m2Â = 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) m2

= 2h (l+b)

= 2 (3) (15)

= 90m2

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 cm2

Then,

Product of areas of two adjacent walls = (lÃ—h) Ã— (bÃ—h) = lbh2 cm4

Product of areas of floor and two adjacent walls = lb Ã— lbh2 cm6

= l2 Ã— b2 Ã— h2 cm6

= (lbh)2 cm6

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

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

Area of ceiling = l Ã— b

= 4.5 Ã— 3

= 13.5 m2

Area of walls = 2 (lh + bh) m2

= 2 (4.5Ã—3.5 + 3Ã—3.5)

= 2 (15.75 + 10.5)

= 52.5 m2

Sum of Area of ceiling + area of walls = 13.5m2 + 52.5m2

= 66m2

Given, Cost for plastering 1m2Â area = Rs 8

âˆ´Â Cost for plastering 66 m2Â area = 66 Ã— 8 = Rs 528

15. A cuboid has total surface area of 50 m2Â and lateral surface area is 30 m2. Find the area of its base.

Solution:

Given details are,

Total surface area of cuboid = 50 m2

Lateral surface area of cuboid = 30 m2

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 m2

âˆ´ Area of base is 10m2

16. A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m2. What is the cost of white washing the walls at the rate of Rs 1.50 per m2?

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) m2

= 2 (7Ã—3.5 + 6Ã—3.5)

= 91m2

Area of four walls (without doors & windows) =

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

= 91 â€“ 17 = 74 m2

Then,

Cost for white washing 1m2Â 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 m2Â 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 = 4.5m2

Area of 10 doors = 10 Ã— 4.5 = 45m2

Area of each window = 1.5m Ã— 1m = 1.5 m2

Area of 10 windows = 10 Ã— 1.5 = 15m2

Area occupied by doors and windows = 45 + 15 = 60 m2

Area of the walls of the hall including doors and windows = 2 (lh + bh) m2

= 2 (80Ã—8 + bÃ—8)

= 2(640+8b) m2

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) m2

Given, Total cost for white washing = Rs 2385.60

Rate of white washing = Rs 1.20 per m2

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