Class 9 Maths Part 2 Chapter 14 – Math of Space: Surface Area and Volume from Ganita Manjari with easy Notes and complete Book Solutions. Get clear explanations, important concepts, questions and answers, and revision material to help Class 9 students learn, practise and prepare effectively for examinations.
Math of Space: Surface Area and Volume
Easy Concepts, Notes and Complete Book Solutions
Chapter Introduction
In the previous grade we explored nets of some solids including cubes, cylinders, and cones. We saw that solids have a surface and occupy some space or volume.
In this chapter, we will learn how to determine surface areas and volumes of cuboids, cylinders, cones, pyramids, and spheres, and how we can use these to solve some interesting problems.
14.1 Cuboids and Cubes
A cuboid has six faces, all rectangles. It is characterised by three quantities:
- Length
- Width
- Height
These are denoted by l, w, h respectively.
A useful way of thinking about a cuboid is that it is a three-dimensional (3D) version of a rectangle.
Total Surface Area of a Cuboid
Considering the faces individually, the total surface area of a cuboid is:
Volume of a Cuboid
Volume measures the amount of three-dimensional space an object or a container occupies.
V = lwh
Cube
A cube is a particular and important case of a cuboid, whose length, width, and height are the same.
If the side of a cube is a, then:
Volume of cube = a³
Cube Numbers
Cube numbers are the numbers in the sequence 1, 8, 27, 64, …
These represent the number of unit cubes that can fit inside cubes of side lengths 1, 2, 3, 4, … respectively.
Example 1
Two solid objects are made from the same material:
Cube A has side 6 cm and Cuboid B has dimensions 9 cm × 6 cm × 4 cm.
For Cube A:
Surface area = 6a² = 6 × 6² = 216 cm²
For Cuboid B:
Surface area = 2(lw + wh + hl)
= 2[(9 × 6) + (6 × 4) + (4 × 9)]
= 228 cm²
Thus, the two objects have the same volume but different surface areas. The cuboid has greater surface area than the cube.
Conclusion: For the same storage capacity, a cube requires less covering material, while a cuboid exposes more area to its surroundings. This concept is important in packaging, heat transfer and engineering.
Questions: Cuboids and Cubes
Volume = 64 cm³.
For a cube, a³ = 64.
Therefore, a = 4 cm.
Total surface area = 6a²
= 6 × 4² = 6 × 16
= 96 cm²
Convert 2 m into centimetres:
2 m = 200 cm.
Number along one edge = 200 ÷ 20 = 10.
Total cubes = 10 × 10 × 10.
= 1000 cubes
Volume of godown = 40 × 25 × 10 = 10,000 m³.
Volume of one box = 2 × 1.25 × 1 = 2.5 m³.
Number of boxes = 10,000 ÷ 2.5.
= 4000 boxes
For each cube, a³ = 125.
Therefore, a = 5 cm.
When two cubes are joined end to end, dimensions are:
10 cm × 5 cm × 5 cm.
Surface area = 2(lw + wh + hl)
= 2[(10 × 5) + (5 × 5) + (5 × 10)]
= 2(50 + 25 + 50)
= 250 cm²
Surface area of original cube = 6 × 4² = 96 cm².
Number of small cubes = 4³ = 64.
Surface area of one small cube = 6 × 1² = 6 cm².
Surface area of all small cubes = 64 × 6 = 384 cm².
Required ratio = 96 : 384.
= 1 : 4
Let the dimensions be l, w and h.
Then:
lw = 6, wh = 15, hl = 10.
Multiplying:
(lw)(wh)(hl) = 6 × 15 × 10.
(lwh)² = 900.
Therefore, lwh = 30.
Volume = 30 cm³
There are 5 × 5 × 5 = 125 small cubes.
(i) Exactly three faces painted: Only corner cubes have three painted faces.
8 cubes
(ii) Exactly two faces painted: Each edge has 3 non-corner cubes. There are 12 edges.
12 × 3 = 36 cubes.
(iii) Exactly one face painted: Each face has a 3 × 3 group of cubes inside it, so 9 per face.
6 × 9 = 54 cubes.
(iv) No face painted: Inner cubes form a 3 × 3 × 3 cube.
27 cubes
For dimensions l, w, h:
2(lw + wh + hl) = 100.
Therefore:
lw + wh + hl = 50.
Positive integer dimensions give the following unordered solutions:
- 1 cm × 1 cm × 24 cm
- 1 cm × 2 cm × 16 cm
- 1 cm × 3 cm × 11 cm
- 1 cm × 4 cm × 8 cm
- 1 cm × 5 cm × 6 cm
- 2 cm × 3 cm × 7 cm
- 2 cm × 4 cm × 6 cm
- 3 cm × 4 cm × 5 cm
Thus there is more than one such cuboid. These are the integer-sided cuboids satisfying the condition.
Think and Reflect: Swimming Pool
Suppose a swimming pool is being built in your school. Two choices are available for the dimensions of the pool:
(A) 8 ft (depth) × 50 ft × 20 ft
(B) 6 ft (depth) × 100 ft × 15 ft
Which option would be suitable/appropriate for your school? Why? What parameters/aspects did you consider to make the choice?
For comparison, calculate:
B: 6 × 100 × 15 = 9000 ft³
The choice depends on the school’s requirements such as available space, depth, capacity, cost, safety and intended use.
14.2 Right Circular Cylinders
A right circular cylinder has a circular cross-section and its axis is perpendicular to its circular base.
An oblique circular cylinder has an axis that is inclined rather than perpendicular to the base.
We restrict ourselves to the study of the right circular cylinder.
Cross-Section
Given a three-dimensional object, if we take an extremely thin slice of it, the shape we get is called a cross-section.
For a cylinder:
- A slice parallel to the base gives a circular cross-section.
- A slice parallel to the axis gives a rectangular cross-section.
Surface Area of a Cylinder
Let the radius of the circular base be r and height be h.
For a cylinder closed at both ends:
= 2πr(h + r)
For a cylinder closed at one end:
= πr(2h + r)
Volume of cylinder:
Example 2
Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope.
Radius = 3.5 cm
Height = 25 cm
= 2πrh
= 2 × 22/7 × 3.5 × 25
= 550 cm²
Therefore, 550 cm² of chart paper is required.
Questions: Cylinders
Let cylinder A have radius r and height h.
Then B has radius 2r and height h/2.
CSA A = 2πrh.
CSA B = 2π(2r)(h/2) = 2πrh.
CSA ratio A : B = 1 : 1.
Volume A = πr²h.
Volume B = π(2r)²(h/2) = 2πr²h.
Volume ratio A : B = 1 : 2.
Let r₁:r₂ = 2:3 and h₁:h₂ = 3:2.
(a) Volumes:
V₁:V₂ = r₁²h₁ : r₂²h₂
= 2² × 3 : 3² × 2
= 12 : 18
= 2 : 3
(b) Curved surface areas:
CSA₁:CSA₂ = r₁h₁:r₂h₂
= 2 × 3 : 3 × 2
= 1 : 1
The largest cylinder has height r and diameter r.
Therefore radius = r/2.
= πr³/4
Volume = πr³/4 cm³
Taking the intended increase as 10%, the new radius is 1.1r.
Let the new height be (1 − x/100)h.
Volume remains unchanged:
(1.1r)²(1 − x/100)h = r²h
1.21(1 − x/100) = 1
1 − x/100 = 1/1.21
x ≈ 17.36.
Therefore, x ≈ 17.36%.
(i) Volume of cube:
12³ = 1728 cm³.
(ii) Volume of one rod:
πr²h = π × 2² × 12 = 48π cm³.
Using π ≈ 22/7, volume ≈ 150.86 cm³.
(iii) Number of complete rods:
1728 ÷ 150.86 ≈ 11.46.
Therefore, 11 complete cylindrical rods can be formed.
14.3 Cones
A right circular cone is generated when a right triangle is rotated about one of its legs.
The relationship among the slant length l, height h and radius r is:
Curved Surface Area
Total Surface Area
TSA = πr(l + r)
Volume
The volume of a cone is one-third the volume of a cylinder having the same base radius and the same height.
Example 3
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is rotated through 360° about the side with length 12 cm.
The perpendicular sides are 5 cm and 12 cm and the hypotenuse is 13 cm.
When the triangle is rotated about the side of length 12 cm, it forms a right circular cone.
Radius = 5 cm and height = 12 cm.
= 1/3 × π × 5² × 12
= 100π cm³
Volume = 100π cm³.
Exercise 14.3
Diameter = 24 m, so radius = 12 m.
Slant height l = 21 m.
TSA = πr(l + r)
= π × 12 × (21 + 12)
= 396π m².
Using π = 22/7, TSA = 1244.57 m² approximately.
CSA = πrl.
= π × 7 × 10
= 70π cm².
Using π = 22/7, CSA = 220 cm².
h = 16 cm and r = 12 cm.
Slant height:
l = √(h² + r²) = √(16² + 12²) = √400 = 20 cm.
CSA = πrl = π × 12 × 20 = 240π cm².
TSA = πr(l + r) = π × 12 × 32 = 384π cm².
CSA = 240π cm²; TSA = 384π cm².
V = 1570 cm³ and h = 15 cm.
V = 1/3 πr²h.
1570 = 1/3 × π × r² × 15.
Using π = 3.14:
1570 = 15.7r².
r² = 100.
r = 10 cm.
CSA = πrl.
308 = 22/7 × r × 14.
Therefore r = 7 cm.
(i) Radius = 7 cm.
(ii) TSA = πr(l + r)
= 22/7 × 7 × (14 + 7)
= 462 cm².
r = 7 cm, h = 24 cm.
Slant height l = √(7² + 24²) = √625 = 25 cm.
Sheet required for one cap = CSA = πrl.
= 22/7 × 7 × 25 = 550 cm².
For 10 caps = 550 × 10.
= 5500 cm².
r = 6 m, h = 8 m.
l = √(6² + 8²) = 10 m.
Curved surface area = πrl = π × 6 × 10 = 60π m².
Tarpaulin width = 3 m.
Length required = 60π ÷ 3 = 20π ≈ 62.86 m.
Extra length = 20 cm = 0.20 m.
Approximately 63.06 m of tarpaulin is required.
The triangle forms a cone.
Radius = 6 cm and height = 8 cm.
Slant height = 10 cm.
Volume = 1/3 × π × 6² × 8 = 96π cm³.
CSA = πrl = π × 6 × 10 = 60π cm².
If the water reaches half the height, the smaller cone has all corresponding dimensions half of the original cone.
Volume changes as the cube of the linear dimension.
Therefore:
The water occupies 1/8 of the volume.
14.4 Pyramidal Shapes
A pyramidal shape has a base that may be a triangle, square or any other closed shape. All points in the base are connected to a single point called the apex.
The same formula is applicable whether the base is triangular, square or any other shaped flat surface.
The surface area of a pyramidal shape is found by adding the areas of all the polygonal faces.
A cone can be thought of as a pyramid with a circular base.
14.5 Spheres and Hemispheres
A sphere is the three-dimensional equivalent of a circle.
A sphere is the collection of all points in 3D space that are at a given distance r from a given point O, the centre of the sphere.
Surface Area of a Sphere
Volume of a Sphere
The volume of a sphere can also be expressed as:
14.5.2 Hemisphere
A hemisphere is a three-dimensional solid formed when a sphere is cut into two equal halves by a plane passing through its centre.
A hemisphere has one curved surface and one circular base.
Total surface area = 3πr²
Volume = 2/3 πr³
Example 4
A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
= 2/3 × 22/7 × 3.5³
≈ 89.8 cm³
The bowl can contain approximately 89.8 cm³ of water.
Questions: Spheres and Hemispheres
Surface area = 4πr².
= 4 × 22/7 × 0.7 × 0.7.
= 6.16 cm².
For spheres made of the same metal, weight is proportional to volume.
Therefore:
V₂/V₁ = 7409/5920.
Since volume is proportional to r³:
(r₂/r₁)³ = 7409/5920.
Smaller diameter = 5 cm, so r₁ = 2.5 cm.
Therefore:
r₂ = 2.5 × ∛(7409/5920).
r₂ ≈ 2.58 cm.
Surface area is proportional to the square of radius.
Diameter ratio Moon : Earth = 1 : 4.
Therefore radius ratio = 1 : 4.
Surface area ratio = 1² : 4².
= 1 : 16.
r = 21 cm.
(i) CSA = 2πr² = 2 × 22/7 × 21².
= 2772 cm².
(ii) TSA = 3πr² = 3 × 22/7 × 21².
= 4158 cm².
Volume of box = 16 × 8 × 8 = 1024 cm³.
Volume of one sphere = 4/3π(2³) = 32π/3 cm³.
Volume of 16 spheres = 16 × 32π/3 ≈ 536.17 cm³.
Volume of liquid = 1024 − 536.17.
≈ 488 cm³.
New volume = 172.8% of original volume.
Since volume is proportional to r³:
(1 + x/100)³ = 1.728.
∛1.728 = 1.2.
Therefore:
1 + x/100 = 1.2.
x = 20%.
Circumference = 2πr = 35.2 m.
Using π = 22/7:
r = 5.6 m.
Curved surface area = 2πr².
= 2 × 22/7 × 5.6²
= 197.12 m².
Convert to cm²:
197.12 × 10,000 = 1,971,200 cm².
Cost = 1,971,200 ÷ 100 × ₹10.
Cost = ₹197,120.
14.6 Areas and Volumes Around Us
We learned about different 3D solids. Identify and list objects or structures around you that are cubical, cuboidal, cylindrical, conical, spherical, or pyramidal in shape, or closely resemble these shapes.
Activity: Find the approximate volumes of a gas cylinder, a bucket, any cylindrical vessel, a cupboard, a water tank that you can find near you.
Use:
14.6.1 Guesstimate: Did You Ever Wonder?
Guesstimate questions are problems in which you make a reasonable estimate for a quantity when exact data are not available.
Instead of finding a perfectly accurate answer, you use assumptions, approximations, logic, and everyday knowledge to model a given situation and arrive at a sensible estimate.
Example 5: Golgappas
Lallan’s golgappa stall has a cylindrical vessel with pani in it. The vessel is of radius 10 cm and it has pani up to a height of 50 cm. Estimate the number of golgappas he can serve assuming all golgappas are spheres.
Volume of pani in vessel:
= π × 10² × 50
= 5000π cm³
The answer depends on the assumptions made about the size of the golgappa and the amount of pani inside it.
This shows that in guesstimate problems, different people may get different answers, but the reasoning used matters the most.
Example 6: Tennis Balls in a Classroom
Anirban saw a few tennis balls lying around. He wondered, “How many tennis balls can fit in an empty classroom?”. The classroom is of dimensions 30 ft × 30 ft × 15 ft. Estimate how many tennis balls of radius 3 cm fit in this classroom.
The volume of the classroom is:
We cannot simply divide the volume of the classroom by the volume of one tennis ball because spheres leave gaps between them.
Using a simple ordered arrangement gives an approximate number of balls.
Along length: 900 ÷ 6 = 150.
Along width: 150.
Along height: 450 ÷ 6 = 75.
This is an approximate value for that particular arrangement.
Guesstimate Exercises
Volume of container = 10 × 15 × 20 = 3000 cm³.
The exact number depends on the size of one scoop.
If one scoop is approximately 50 cm³:
3000 ÷ 50 = 60.
Approximately 60 scoops.
Original number of unit cubes = a³.
New cube contains (a + 1)³ unit cubes.
Number to be added:
(a + 1)³ − a³
= a³ + 3a² + 3a + 1 − a³.
= 3a² + 3a + 1.
(i) Estimate classroom volume and compare it with the total amount of water a person drinks in a lifetime. The classroom volume is generally much larger than a person’s lifetime drinking amount, so the answer is no.
(ii) Estimate the total volume of the walls and divide it by the approximate volume of one brick. The result gives an approximate number of bricks.
The exact answer depends on the classroom dimensions and brick size.
(A) Volume = 50³ = 125,000 mm³.
(B) One cube = 10³ = 1000 mm³.
100 cubes = 100 × 1000 = 100,000 mm³.
Option A gives more chocolate.
This is a guesstimate. Assume approximately 0.25 m² per person in a very crowded standing arrangement.
Multiply this by the world’s population to estimate the total area.
The exact answer depends on the assumed space per person.
Radius = 7/2 = 3.5 cm.
Volume of one glass = πr²h.
= 22/7 × 3.5² × 12
≈ 462 cm³.
For 1600 students:
462 × 1600 = 739,200 cm³.
Convert to litres:
739,200 ÷ 1000.
= 739.2 litres.
Sphere surface area = 4π × 5² = 100π.
It is five times cone CSA.
Therefore cone CSA = 20π.
πrl = 20π.
π × 4 × l = 20π.
l = 5 cm.
Using l² = h² + r²:
25 = h² + 16.
h² = 9.
h = 3 cm.
Volume = 1/3πr²h.
= 1/3 × π × 16 × 3.
Volume = 16π cm³.
Volume of a sphere is proportional to r³.
(i) Jupiter : Earth = (69900/6370)³.
Approximately 1325 : 1.
(ii) Sun : Earth = (695700/6370)³.
Approximately 1,300,000 : 1.
The smallest enclosing cylinder has radius r and height 2r.
Volume of cylinder = πr² × 2r = 2πr³.
Volume of sphere = 4/3πr³.
Therefore:
Sphere volume ÷ cylinder volume = (4/3πr³)/(2πr³) = 2/3.
Thus the volume of the sphere is 2/3 of the volume of the smallest enclosing cylinder.
Original circumference = 2πr.
New circumference = 2π(r + h).
Difference = 1 m.
Therefore:
2πh = 1.
h = 1/(2π).
h ≈ 0.159 m = 15.9 cm.
The answer is the same for Earth, Moon, Jupiter and a volleyball because the radius cancels out.
The square base is inscribed in the circular base.
The diagonal of the square is 2r.
If side = a, then a√2 = 2r.
Therefore a² = 2r².
Pyramid volume = 1/3 × 2r² × h.
Cylinder volume = πr²h.
Ratio = (2r²h/3)/(πr²h).
Ratio = 2 : 3π.
(i) New volume = (l + 1)wh.
Change = (l + 1)wh − lwh = wh.
Answer: (c) wh cubic units.
(ii) New volume = π(r + 1)²h.
Change = π[(r + 1)² − r²]h.
= π(2r + 1)h.
= 2πrh + πh cubic units.
(iii) Original volume = 4/3πr³.
New volume = 4/3π(r − 1)³.
The change is the new volume minus the original volume.
V = a³.
Therefore a = V1/3.
S = 6a².
S = 6V2/3.
S = 6a².
a² = S/6.
a = √(S/6).
V = a³.
V = (S/6)3/2.
Required rise in water level = 20 − 16 = 4 cm.
Volume needed = πr²h = π × 4² × 4 = 64π cm³.
Volume of one marble = 4/3π(1³) = 4π/3 cm³.
Number of marbles = 64π ÷ (4π/3).
= 48.
48 marbles are required.
This is a measurement and estimation question.
Measure the approximate radius and length of the cylindrical ink tube.
Use:
Then convert the answer into millilitres using 1 cm³ = 1 mL.
(i) This is an estimation question. Find the volume of the dough ball:
V = 4/3π(6³) = 288π ≈ 905 cm³.
Estimate the volume of dough needed for one chapati and divide the total volume by that amount.
(ii) Measure the approximate dimensions of the edible portion and model it as a suitable solid such as a hemisphere or cylinder. Then use the corresponding volume formula.
(i) Measure the area of one page, multiply by the number of pages, and compare it with the area of the classroom floor.
(ii) Find the volume of the storeroom:
15 × 20 × 30 = 9000 ft³.
Estimate the volume of one textbook and divide the storeroom volume by it. The result is only an upper estimate because books cannot normally fill every gap.
This is a guesstimate question.
Estimate the volume occupied by one person and multiply it by the population.
If total volume is V, the side of the giant cube is:
The exact result depends on the assumptions used for the volume occupied by each person.
Let the thickness of the water layer be t.
Earth’s original volume = 4/3πR³.
New volume = 4/3π(R + t)³.
Water volume is the difference:
Put R = 6371 km and solve for t using a calculator.
For a thin layer, approximately:
Surface area ≈ 4πR².
This gives a thickness of approximately 4.3 km.
This question does not give enough information to determine a unique cuboid.
Let the original dimensions be l, w, h. Its volume is lwh and surface area is 2(lw + wh + hl).
A particular numerical answer requires an additional condition.
Therefore, the dimensions cannot be uniquely determined from the given information alone.
Divide the house into simple shapes such as cuboids, cylinders or other suitable solids.
Measure or estimate the length, width and height of each part.
Calculate the volume of each part.
Add all the volumes.
This gives an approximate volume of the house.
Chapter Summary
| 3D Shape | Surface Area | Volume |
|---|---|---|
| Cuboid | 2(lw + wh + hl) | lwh |
| Cube | 6a² | a³ |
| Right Circular Cylinder | CSA = 2πrh TSA, both ends = 2πr(h + r) | πr²h |
| Right Circular Cone | CSA = πrl TSA = πr(l + r) | 1/3πr²h |
| Pyramid | Sum of areas of polygonal faces | 1/3 × area of base × height |
| Sphere | 4πr² | 4/3πr³ |
| Hemisphere | CSA = 2πr² TSA = 3πr² | 2/3πr³ |
Quick Revision
- Volume tells us how much three-dimensional space a solid occupies.
- Surface area tells us the total area of the outer surfaces.
- For a cuboid, volume = length × width × height.
- For a cylinder, volume = πr²h.
- For a cone, volume = 1/3πr²h.
- For a sphere, volume = 4/3πr³.
- For a hemisphere, volume = 2/3πr³.
- The curved surface area of a cylinder is 2πrh.
- The curved surface area of a cone is πrl.
- The surface area of a sphere is 4πr².
- Always check that all measurements use the same units before calculating.
prepare more with the given Worksheet of chapter 14, which will help you to improve your knowledge .
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