RBSE Maths Chapter 15 – Surface area and Volume Mathematics Class 8 Important questions and solutions are available here. The additional, exercise important questions and solutions of Chapter 15, available at BYJU’S, contain detailed explanations. It can be used as a most useful reference for facing the board exams. All these important questions are based on the new pattern designed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 8 solutions.
Chapter 15 of the RBSE Class 8 Maths will help the students to solve problems related to the cuboids, the total surface area of a cube, surface area of a cylinder, volume of standard units, volume of cube and cuboid, the volume of a cylinder.
RBSE Maths Chapter 15: Exercise 15.1 Textbook Important Questions and Solutions
Question 1: On the basis of the given measurements, determine the surface area of a cuboidal wooden log, cuboidal brick and box.
Solution:
(i) Surface area of the cubical wooden log = 6 * (a)^{2}
= 6 * (8)^{2}
= 6 * 64
= 384 cm^{2}
(ii) Surface area of cuboidal brick = 2 * (lb + bh + hl)
= 2 * (9 * 5 + 5 * 3 + 3 * 9)
= 2 * (45 + 15 + 27)
= 2 (87)
= 174 cm^{2}
(iii) Surface area of cuboidal box = 2 * (lb + bh + hl)
= 2 * (150 * 40 + 40 * 70 + 70 * 150)
= 2 (6000 + 2800+ 10500)
= 2 (19300)
= 38600 cm^{2}
Question 2: Determine the side of a cube whose total surface area is 600 square cm.
Solution:
Let a be the length of the side of the cube.
Total surface area of cube = 6a^{2} cm^{2}
According to the question,
6a^{2} = 600
a^{2 }= 600 / 6
a = √100
a = 10
Hence, the required length of the side of the cube is 10 cm.
Question 3: In the given figures below, whose surface area is the greatest?
Solution:
(i) In the first figure (cylinder),
r = [7 / 2] meters
r = 3.5 meters
h = 7 meter
Surface Area = 2πr (r + h)
= 2 * 3.14 * 3.5 (3.5 + 7)
= 21.98 [10.5]
= 230.79m^{2}
(ii) In the second figure (cube),
a = 7 meters
Surface area = 6a^{2}
= 6 * [7]^{2}
= 6 * 49
= 294 m^{2}
Hence, the surface area of the cube is greater than the surface area of the cylinder.
Question 4: Find the area of a curved surface if the area of the base of the cylindrical tank is 176 cm^{2} and height is 30 cm.
Solution:
Let r cm be the radius of the base.
The circumference of base = 2πr cm
According to the question,
2πr = 176
2 * 3.14 * r = 176
6.2831 * r = 176
r = 176 / 6.2831
r = 28.01 cm
Area of curved surface = 2πrh
= 2 * 3.14 * 28.01 * 30
= 5279.76 cm^{2}
Question 5: From a sheet of 8 square meters, a closed cylindrical tank is formed which has onemeter height and 140 cm diameter. How many sheets will be left after making a tank?
Solution:
Diameter of cylindrical tank = 140 cm
Radius (R) = [140] / [2] = 70cm
Height = 1 meter = 100 cm
Surface area of cylindrical tank = 2πr (r + h)
= 2 * 3.14 * 70 * [70 + 100]
= 74769 cm^{2}
= 7.48 m^{2}
The surface area of the metal sheet = 8 m^{2}
Surface area of remaining sheet = 8 – 7.48 = 0.52 m^{2}
Question 6: How many paint tins having the spread capacity of 100cm^{2} will be required to paint the external surface of the box with dimensions 80 cm * 50 cm * 25 cm?
Solution:
Area of outer surface of box = 2 * (lb +bh + lh)
= 2 * (80 * 50 + 50 * 25 + 25 * 80)
= 2 * (4000+ 1250 + 2000)
= 2 * (7250)
= 14500 cm,
Spread capacity of 1 paint box = 100 cm^{2}
Required number of paint boxes = [Area of outer surface of the box] / [capacity of 1 paint box]
= 14500 / 100
= 145 paint tins will be required to paint the external surface of the box.
Question 7: There are 25 cylindrical pillars in a building. Each pillar has a radius of 28 cm and a height of 4 m. Find the expenditure of painting curved surface area of all the pillars at the rate of Rs. 8 per meter square.
Solution:
For one cylindrical pillar, radius (r) = 28 cm
Height (h) = 4 m = 4 * 100 cm = 400 cm
Curved surface area of 1 pillar = 2πrh
= 2 * π * 28 * 400
= 70400 cm^{2}
Curved surface area of 25 cylindrical pillars = 70400 x 25
= 1760000 cm^{2}
= 176 m^{2}
Cost of painting of 25 cylindrical pillars = 176 * 8 = Rs. 1408.
Question 8: Curved surface area of a hollow cylinder is 4224 cm^{2}. A rectangular sheet having width 33 cm is formed, cutting it along its height. Find the perimeter of the sheet.
Solution:
Let r meter be the radius of the base of cylinder and height be h meter.
The surface area of cylinder = 2πrh cm^{2}
According to the question,
2πrh = 4224
2 * π * r * (33) = 4224
2πr = 4224 / 33
2πr = 128
Perimeter of base = 128 cm
Length of sheet (l) = 128 cm
Breadth of sheet (b) = 33 cm
Perimeter of rectangular sheet = 2 * (I + b)
= 2 * (128 + 33)
= 2 * (161)
= 322 cm
Question 9: To make a road plain, a roller has to complete 750 rounds. If the diameter of the roller is 84 cm and length 1 meter, then find the area of the road.
Solution:
Diameter of roller = 84 cm
Radius (r) = d / 2 cm = 84 / 2 = 42 cm
Length (h), 1 meter = 100 cm
Curved surface area = 2πrh
= 2 * π * 42 * 100
= 26400 cm^{2}
Area of road that is made plane in one round = 26400 cm^{2}.
Area of road that is made plane in 750 rounds = 26400 x 750 = 19800000 cm^{2} = 1980m^{2}.
Question 10: A cube is made by arranging 64 cubes having a side of 1 cm, find the total surface area of the cube so formed.
Solution:
Side of one cube = 1 cm
Volume of one cube = a^{3} = (1)^{3} = 1 cm^{3}
Volume of 64 cubes = 64 * 1 = 64 cm^{3}
Let the side of the new cube be x cm.
Volume of new cube = Volume of 64 cubes
=>x^{3} = 64 cm
x = (64)^{1/3}
x = 4cm
Total surface area of new cube = 6a^{2}
= 6 * (4)^{2}
= 6 * 16
= 96 cm^{2}
RBSE Maths Chapter 15: Exercise 15.2 Textbook Important Questions and Solutions
Question 1: The dimensions of a cuboid are 60 cm * 54cm * 30 cm. How many cubes of side 6 cm can be placed in the cuboid?
Solution:
Volume of the cuboid = 60 cm * 54cm * 30 cm = 97200 cm^{3}
Volume of the cube of side 6 cm = (6)^{3} = 216 cm^{3}
The required number of cubes = Volume of cuboid / Volume of one cube
= 97200 / 216
= 450 cubes
Question 2: How many wooden logs of side 6 cm can be cut from a 3 m long, 50 cm broad and 25 cm high wooden pile.
Solution:
Volume of wooden pile = 300 * 50 * 25 cm [1 meter = 100 cm]
= 375000 cm
Volume of 1 cubical block = 25 * 25 * 25 = 15625 cm^{3}
Required number of cubical blocks = [Volume of wooden pile / Volume of 1 block]
= [375000 / 15625]
= 24
Question 3: Cylinder A has a diameter of 14 cm end height of 7 cm. Cylinder B has a diameter 7 cm and height 14 cm. Without any calculation, determine the volume of the cylinder that is the greatest? Verify the answer by calculation.
Solution:
For cylinder A, the radius of base (r), = [14 / 2] cm = 7 cm
Height (h) = 7 cm
Volume of the cylinder = πr^{2}h
= [22 / 7] * (7)^{2} * (7)
= 1078 cm^{3}
For cylinder B, the radius of base r = [7 / 2] cm
Height (h) = 14 cm
Volume of the cylinder = πr^{2}h
= [22 / 7] * [7 / 2]^{2} * 14
= [22 / 7] * [49 / 4] * 14
= 539 cm^{3}
Hence, the volume of cylinder A is 1078 cm^{3} and the volume of cylinder B is 539cm^{3}.
So, the volume of cylinder A > the volume of cylinder B.
Question 4: From a cylindrical milk tanker of radius 1.5 m and length 7 m, how many polythenes of one litre can be packed, (1m^{3 }= 1000 litre)
Solution:
Radius of the tanker = 1.5 meter
Height (h) = 7 meter
Volume of the cylinder = πr^{2}h
= [22 / 7] * (1.5) * (1.5) * 7
= 49.5 m^{3}
= 49.5 x 1000 litre
= 49500 litre
Hence, the required number of polythene packs are 49500.
Question 5. In what time can a tap giving 60 litres of water per minute fill a cylindrical tank of radius 3.5 m and depth 3 m?
Solution:
Radius of the tank = 3.5 meter
Depth of the tank (h) = 3 meter
Volume of the cylinder = πr^{2}h
= [22 / 7] * (3.5) * (3.5) * 3
= 115.5 m^{3}
= 115.5 x 1000 litre
= 115500 litre
Required time = [Volume of the tank] / [Volume of water flowing in a minute]
= [115500] / [60]
= 1925 minutes
Question 6: Dimensions of a cuboidal ice is 50 cm * 30 cm * 20 cm. Find its weight in kilograms if the weight of 1000 cm^{3 } ice is 900 gram.
Solution:
Volume of the ice plate = 50 * 30 * 20
= 30000 cm^{3}
Weight of the ice plate = 30000 * [900 / 1000]
= 27000 grams
= 27000 / 1000 [1kg = 1000 grams]
= 27 kg
Question 7: If the side of a cube is doubled then,
(i) How many times will its surface area increase?
(ii) How many times will its volume increase?
Solution:
(i) Let a cm be the side of the cube.
Surface area S_{1} = 6a^{2} cm^{2}
New side = 2a cm
New surface area S_{2} = 6 * (2a)^{2} = 24 a^{2} cm^{2}
S_{2} / S_{1} = 24a^{2} / 6a^{2 }= 4
S_{2} = 4S_{1}
So, the surface area of the new cube will increase by 4 times.
(ii) Volume V_{1 }= a^{3 }cm^{3}
New volume = V_{2} = (2a)^{3} cm^{3}
= 8a^{3} cm^{3}
V_{2} / V_{1} = 8a^{3} / a^{3} = 8
V_{2} = 8V_{1}
So, the volume increases by 8 times.
Question 8: Volume of a cylindrical tank having 7meter diameter is 770 cubic meters. Find the height of the tank.
Solution:
The diameter of the tank [d] = 7 meters
The radius of the tank = d / 2 = 7 / 2 m = 3.5 m
Let the height of the tank be ‘h’ m.
The volume of the cylindrical tank = πr^{2}h
770 = [22 / 7] * (3.5) * (3.5) * h
770 = 38.5 * h
770 / 38.5 = h
h = 20 m
The height of the tank is 20 meters.
RBSE Maths Chapter 15: Additional Questions and Solutions
Question 1: If the cuboid is formed by sticking 3 sides of a cube ‘x’, then what will be the dimensions of the cuboid?
Solution:
Condition I
Length = 3x
Breadth = x
Height = x
Condition II
Length = x
Breadth = x
Height = 3x
Question 2: Find the surface area of a cube having side 3 cm. What will be the surface area of 5 such cubes?
Solution:
Surface area of cube of side 3 cm = 6a^{2}
= 6 * (3)^{2}
= 6 * 9
= 54 cm
Surface area of 5 cubes = 54 * 5 = 270 cm^{2}
Question 3: What will be the total surface area of 5 cubes having side 3 cm.? If they are stocked one after one, then how much surface area will be decreased?
Solution:
Surface area of cube of side 3 cm = 6a^{2}
= 6 * (3)^{2}
= 6 * 9
= 54 cm
Surface area of 5 cubes = 54 * 5 = 270 cm^{2}
If 5 cubes are joined one after one, then it becomes a cuboid of length [l]
= 3 cm + 3 cm + 3 cm + 3 cm + 3 cm
= 15cm
Breadth (b) = 3 cm
Height (h) = cm
Surface area = 2 * (lb + bh + lh)
= 2 * (15 * 3 + 3 * 3 + 3 * 15)
= 2 * (45 + 9 + 45)
= 198 cm^{2}
Hence, the surface area will be reduced and the difference = 270 cm^{2} – 198 = 72 cm^{2}.
Question 4: Find the lateral surface area of the cuboid given below.
Solution:
Length of the cuboid = 10 cm.
Breadth of the cuboid = 4cm
Height of the cuboid = 4 cm
The lateral surface area of cuboid = 2 * (l + b) *h
= 2 * (10 + 4) * (4)
= 2 * 14 * 4
= 112 cm^{2}
Question 5: Determine the total surface area of the following cylinders.
Solution:
[i] In the figure [i], the radius of the cylinder = diameter / 2 = 7 / 2 = 3.5 cmHeight of cylinder [h] = 10cm
Total surface area of the cylinder = 2πr (r + h)
= 2 * [22 / 7] * [3.5] * [3.5 + 10]
= 297 cm^{2}
[ii] In the figure [ii], the radius of the cylinder = diameter / 2 = 7 / 2 = 3.5 cmHeight of cylinder [h] = 10.5 cm
Total surface area of the cylinder = 2πr (r + h)
= 2 * [22 / 7] * [3.5] * [3.5 + 10.5]
= 308 cm^{2}
Question 6: By arranging the cubes of unit length, find the volume of the cubes so obtained.
Solution:
Sl.no 
Cuboid Volume 
Length [L] 
Breadth [B] 
Height [H] 
L * B * H [cubic unit] 
[i] 
21 cubic unit 
7 units 
3 units 
1 unit 
7 * 3 * 1 = 21 
[ii] 
42 cubic unit 
7 units 
3 units 
2 units 
7 * 3 * 2 = 42 
[iii] 
63 cubic unit 
7 units 
3 units 
3 units 
7 * 3 * 3 = 63 
Question 7: Find the volume of the given cuboids on the basis of the given measures.
Solution:
(i) Length of the cuboid [L] = 8 cm
Breadth of the cuboid [B] = 3 cm
Height of the cuboid [H] = 2 cm
Volume of the cuboid = L * B * H = 8 * 3 * 2 = 48 cm^{3}
(ii) Length of the cuboid [L] = 1 cm
Breadth of the cuboid [B] = 1 cm
Height of the cuboid [H] = 7 cm
Volume of the cuboid = L * B * H = 1 * 1 * 7 = 7 cm^{3}
Question 8: Complete the following table.
Solution:
Sl.no 
Cube 
Number of unit cubes 
Side 
Side * Side * Side 
(Side)^{3} 
[i] 

8 cubic units 
2 unit 
2 * 2 * 2 
8 cubic units 
[ii] 

27 cubic units 
3 
3 * 3 * 3 
27 cubic units 
[iii] 

64 cubic units 
4 
4 * 4 * 4 
64 cubic units 
Question 9: Determine the volume of the cube having side.
[i] 1.5cm
[ii] 4m
Solution:
(i) 1.5 cm
Here, a = 1.5 cm
Volume of cube = a^{3}
= (1.5)^{3}
= 1.5 * 1.5 * 1.5
= 3.375 cm^{3}
(ii) 4 cm
Here a = 4 cm
Volume of cube = a^{3}
= (4)^{3}
= 4 * 4 * 4
= 64 cm^{3}
Question 10: Match the following.
Question 
Answer Choices 
1] Number of vertices of a cube 
A] 6m 
2] Number of edges of a cube 
B] 2 
3] Number of edges of the cylinder 
C] 12 
4] Side of a cube of volume 216 m^{3} 
D] 8 
Solution:
1] – D]
2] – C]
3] – B]
4] – A]
Question 11: Is the below figure a cylinder? Justify your answer.
Solution:
A cylinder consists of two edges and circles which are congruent. In the above figure, the circles are not congruent. Hence, it is not a cylinder.
Question 12: The ratio of two cylindrical poles is 3:2, and the ratio of their heights is 2:3. Find the ratio of the curved surface area of poles.
Solution:
S_{1} / S_{2} = 2πr_{1}h_{1} / 2πr_{2}h_{2}
= [r_{1} / r_{2}] * [h_{1} / h_{2}]
= [3 / 2] * [2 / 3]
= 1:1
Question 13: The area of three adjacent faces of a cuboid are p, q and r. The volume of the cuboid is V. Prove that V^{2 }= pqr.
Solution:
Let Iength [l], breadth [b] and height [h] be the dimensions of the cuboid respectively.
Then, p = lb, q = bh, r = lh
p* q * r = (Ib) (bh) (lh)
= l^{2}b^{2}h^{2}
= V^{2}
So, V^{2} = pqr
Question 14: If V is the volume of a cuboid of dimensions a, b and c, surface area is S, then prove that [1 / V] = [2 / s] * [(1 / a) + (1 / b) + (1 / c)].
Solution:
LHS = [1 / V] = [1 / abc] —– (1)
RHS = [2 / s] * [(1 / a) + (1 / b) + (1 / c)]
= [2 / 2 [ab + bc + ca]] * [(bc + ca + ab) / abc)]
= 1 / abc —– (2)
From (1) and (2), LHS = RHS
[1 / V] = [2 / s] * [(1 / a) + (1 / b) + (1 / c)]Question 15: The population of a village is 2000. Every citizen needs 150 litres of water per day. The dimensions of the water tank are 20 m * 15 m * 6 m in the village. For how many days is this tank sufficient for the village?
Solution:
Volume of the tank = 20 m * 15 m * 6 m
= 1800 m^{3}
= 1800 * 1000 litres
Water requirement of village per day = 150 * 2000 = 300,000 litres
Required number of days = Volume of the tank / Requirement of people per day
= [1800 * 1000] / [300000]
= 6 days
Question 16: A closed iron tank of dimensions 12 m in length, 9 m in breadth and 4 m in depth is to be made. The iron sheet whose cost is Rs 50 per meter and which is 2 m wide is used. Find the total cost of the iron sheet required for the iron tank.
Solution:
Surface area of tank = 2 * (lb + bh + lh)
= 2 * (12 * 9 + 9 * 4 + 4 * 12)
= 2 * (108 + 36 + 48) = 2 (192)
= 384 m^{3}
Width of iron sheet = 2 meter
Length of iron sheet = 384 / 2 = 192 meter
Cost of iron sheet = 192 * 50 = Rs 9600.
Question 17: A cylinder, whose height is 3 m, is open from the top. The circumference of its base is 22 m. Find the total surface area.
Solution:
Let the radius of the base be ‘r’ m.
2𝛑r = 22
[2] * [22 / 7] x r = 226.286 * r = 22
r = 3.5 cm
Total Surface Area = 2𝛑rh + 𝛑r^{2}
= 2 * [22 / 7] * [3.5] * 3 + [22 / 7] * [3.5]^{2}
= 66 + 38.5
= 104.5 m^{2}
Question 18: A cylindrical pipe, which is open from both sides is made of an iron sheet of thickness 2 cm. If the external diameter is 16 cm and height is 100 cm, then find the total iron used in making the pipe.
Solution:
External diameter = 16 cm.
External radius (R) = [d / 2] = [16 / 2] cm = 8 cm
Thickness of the sheet = 2 cm
Inner radius (r) = 8 – 2 = 6 cm
Height (h) = 100 cm
Used iron = Extemal volume – Internal Volume
= 𝛑R^{2}h – 𝛑r^{2}h
= 𝛑 * [R^{2} – r^{2}] * h
= [22 / 7] * [8^{2} – 6^{2}] * 100
= [22 / 7] * [64 – 36] * 100
= 8800 cm^{3}
Question 19: Find the length of the longest rod, which can be put in a room of dimensions 12 m * 9 m * 8 m.
Solution:
The given room is of
Length = 12m
Breadth = 9 m
Height = 8m
Length of the longest rod = Length of the diagonal = √l^{2 }+ b^{2} + h^{2}
= √12^{2} + 9^{2} + 8^{2}
= √144 + 81 + 64
= √289
= 17 m
Question 20: The curved surface area of a cylindrical tank is 440m^{2}. Its height is 4 m. Find the volume of the task.
Solution:
Height of cylinder (h) = 4 m
Curved surface area of cylinder = 440 m^{2}
Volume of the cylinder has to be found.
2𝛑rh = 440
2 * [22 / 7] * r * 4 = 440
25.143 * r = 440
r = 440 / 25.143
r = 17.5 m
Volume of the cylinder = 𝛑r^{2}h
= [22 / 7] * [17.5]^{2} * 4
= 962.5 * 4
= 3850 m^{2}
Question 21: Determine the side of a cube whose total surface area is 1014 cm^{2}. Find its volume also.
Solution:
Let the side of the cube be ‘a’ cm.
The total surface area of cube = 6a^{2}
According to question,
6a^{2} = 1014
a^{2 }= 1014 / 6
a^{2} = 169
a = 13 cm
Volume of the cube = (side)^{3 }= [13 * 13 * 13]^{3} = 2197 cm^{3}
Question 22: Radius and total surface area of a cylinder are 7 cm and 968 cm^{2} respectively. Find its height.
Solution:
Given, radius of the cylinder (r) = 7 cm
Total surface area = 968 cm^{2}
Height (h) = ?
Total surface area of cylinder = 2πr (r + h)
968 = 2 * [22 / 7] * 7 (h + 7)
968 = 6.286 * [7h + 49]
968 – 308.014 = 44.002 * h
659.986 / 44.002 = h
h = 15cm
Height of cylinder = 15 cm.
Question 23: Curved surface area of a cylinder is 880 m^{2} whose height is 10 m. Find the volume of a cylinder.
Solution:
The curved surface area of cylinder = 880 m^{2} and height (h) = 10 meters.
2πrh = 880 m
2 * [22 / 7] * r * 10 = 880
62.86 * r = 880
r = 880 / 62.86
r = 14
Volume of the cylinder = πr^{2}h
= [22 / 7] * [14]^{2} * 10
= 6160 m^{3}