Exercise 21.3 deals with the surface area of a cuboid and a cube. We shall discuss the surface area of the walls of a room. Solutions given here are formulated by our subject experts with utmost care. Students who aim to clear their doubts and improve their problem-solving skills can make use of RD Sharma Class 8 Solutions. These are the best resources students can access and prepare for their exams. Students can download the PDF easily that is provided below.
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 (10×12 + 12×14 + 14×10)
= 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 (6×8 + 8×10 + 10×6)
= 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 (2×4 + 4×5 + 5×2)
= 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 (32×30 + 30×25 + 25×32)
= 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 × side2
= 6 × 1.22
= 6 × 1.44
= 8.64 m2
(ii) Given,
Edge of cube = 27 cm
We know that,
Surface area of cube = 6 × side2
= 6 × 272
= 6 × 729
= 4374 cm2
(iii) Given,
Edge of cube = 3 cm
We know that,
Surface area of cube = 6 × side2
= 6 × 32
= 6 × 9
= 54 cm2
(iv) Given,
Edge of cube = 6 m
We know that,
Surface area of cube = 6 × side2
= 6 × 62
= 6 × 36
= 216 m2
(v) Given,
Edge of cube = 2.1 m
We know that,
Surface area of cube = 6 × side2
= 6 × 2.12
= 6 × 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 × 5cm × 4cm
We know that,
Surface area of cuboid = 2 (lb + bh + hl) cm2
= 2 (5×5 + 5×4 + 4×5)
= 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 = 3√(343) = 7m
We know that,
Surface area of cube = 6 × side2
= 6 × 72
= 6 × 49
= 294 m2
(ii) Given details are,
Volume of cube = 216 dm3
Side of cube a = 3√(216) = 6dm
We know that,
Surface area of cube = 6 × side2
= 6 × 62
= 6 × 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 × side2 = 96cm2
Side2 = 96/6
= 16
Side = √16 = 4cm
∴ Volume of a cube = 43 = 64cm3
(ii) Given details are,
Surface area of cube = 150 m2
6 × side2 = 150cm2
Side2 = 150/6
= 25
Side = √25 = 5cm
∴ 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 × 3x × 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 = √9
= 3
∴ 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 × 0.5m × 15cm = 25cm × 50cm × 15cm
We know that,
Area of cardboard required = 2 (lb + bh + hl) cm2
= 2 (25×50 + 50×15 + 15×25)
= 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 × side2
= 6 × 122
= 6 × 144
= 864 cm2
9. The dimensions of an oil tin are 26 cm× 26 cm× 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 × 26cm × 45cm
Then,
Area of tin sheet required for making one oil tin = total surface area of oil tin
= 2 (lb + bh + hl) cm2
= 2 (26×26 + 26×45 + 45×26)
= 2 (676 + 1170 + 1170)
= 2 (3016)
= 6032 cm2
Area of tin sheet required for 20 oil tins = 20 × 6032
= 120640 cm2
= 12.064 m2
Given, Cost of 1 m2 tin sheet = Rs 10
So, Cost of 12.064 m2 tin sheet = 10 × 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 × 8m × 5m
Where, Length = 11m, Breadth = 8m, Height = 5m
We know,
Area of floor = length × breadth
= 11 × 8
= 88 m2
Area of four walls (including doors & windows) = 2 (lh + bh) cm2
= 2 (11×5 + 8×5)
= 2 (55 + 40)
= 2 (95)
= 190m2
∴ 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 × 15m ×3m
Where, Length = 20m , Breadth = 15m , Height = 3m
We know,
Area of floor = length × breadth
= 20 × 15
= 300 m2
Area of walls of swimming pool = 2 (lh + bh) cm2
= 2 (20×3 + 15×3)
= 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 1m2 area = Rs 25
∴ Cost for repairing 510 m2 = 510 × 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
∴ 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 = 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 × b = lb cm2
Then,
Product of areas of two adjacent walls = (l×h) × (b×h) = lbh2 cm4
Product of areas of floor and two adjacent walls = lb × lbh2 cm6
= l2 × b2 × 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
Breadth of wall = 3m
Height of wall = 350cm = 350/100 = 3.5m
Area of ceiling = l × b
= 4.5 × 3
= 13.5 m2
Area of walls = 2 (lh + bh) m2
= 2 (4.5×3.5 + 3×3.5)
= 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 1m2 area = Rs 8
∴ Cost for plastering 66 m2 area = 66 × 8 = Rs 528
15. A cuboid has total surface area of 50 m2 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
∴ 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 × 6m × 3.5m
Where, Length = 7m, Breadth = 6m, Height = 3.5m
Area of four walls (including doors & windows) = 2 (lh + bh) m2
= 2 (7×3.5 + 6×3.5)
= 91m2
Area of four walls (without doors & windows) =
Area including doors & windows – area occupied by doors & windows
= 91 – 17 = 74 m2
Then,
Cost for white washing 1m2 area of walls = Rs 1.50
∴ Total cost for white washing the walls = 74 × 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 ×1.5m and 10 windows each of size 1.5m× 1m. If the cost of white washing the walls of the hall at the rate of Rs 1.20 per m2 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 ‘b’ m
So,
Area of each door = 3m × 1.5m = 4.5m2
Area of 10 doors = 10 × 4.5 = 45m2
Area of each window = 1.5m × 1m = 1.5 m2
Area of 10 windows = 10 × 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 (80×8 + b×8)
= 2(640+8b) m2
Then,
Area of only walls = area of walls including doors & windows – area occupied by doors & windows
= 2(640+8b) – 60
= 1280 + 16b – 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 × (areas of walls only)
2385.60 = 1.20 × (1220 + 16b)
2385.60 / 1.20 = (1220 + 16b)
1988 = 1220 + 16b
16b = 1988 – 1220
= 768
b = 768/16
= 48
∴ Breadth of hall is 48 m
Comments