NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Math of Space: Surface Area and Volume

Class 9 Ganita Manjari Chapter 14 Quick Links:

• Think and Reflect
• Exercise Set 14.1
• Exercise Set 14.2
• Exercise Set 14.3
• Exercise Set 14.4
• End of Chapter Exercises

• Download Class 9 Maths Offline App to study Class 9 Maths Ganita Manjari Offline.

Class 9 Maths Ganita Manjari Chapter 14 Think and Reflect Solutions

Try to work out for yourself why this model explains the formula for the volume of a cuboid, i.e.,

volume = area of base × height.

Answer:
Imagine the cuboid as a stack of thin rectangular layers, like the playing cards.
For whole-number dimensions, each layer of height 1 unit contains l × w unit cubes. A height of h units gives h such layers.
Total number of unit cubes = l × w × h.
For fractional dimensions, we can use smaller unit cubes and the same grouping idea.
Therefore, volume of a cuboid = area of base × height = lwh.

For a cube of side a:
Surface area = 6a², because it has six square faces of area a² each.
Volume = a × a × a = a³.

Answer:
There is no single choice for every school. I would choose B if the school has enough space and water, because it is shallower and gives more surface space for swimming.

Pool A:
Base area = 50 × 20 = 1000 ft².
Volume = 8 × 50 × 20 = 8000 ft³.
Water needed ≈ 8000 × 28.3 = 2,26,400 litres.

Pool B:
Base area = 100 × 15 = 1500 ft².
Volume = 6 × 100 × 15 = 9000 ft³.
Water needed ≈ 9000 × 28.3 = 2,54,700 litres.

B is 2 ft shallower, but it needs 500 ft² more ground area and about 28,300 litres more water. It is also longer but narrower.

I would consider the available land, students’ swimming ability, depth, water supply, construction cost and maintenance. Being shallower does not by itself make B suitable for beginners. If land and water are limited, A may be more practical, though its greater depth is a concern.

Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.1 Solutions

Answer:
Let the side be a cm.
a³ = 64, so a = 4 cm.
TSA = 6a² = 6 × 4² = 96 cm².
Hence, the total surface area is 96 cm².

Answer:
Side of the box = 2 m = 200 cm.
Volume of box = 200 × 200 × 200 = 80,00,000

Side of small cube = 20 cm.
Volume of small cube = 20 × 20 × 20 = 8000
So, number of cubes in the box = 80,00,000 ÷ 8000 = 1000.
Hence, 1000 small cubes with side 20 cm can be packed tight in the cubical box.

Answer:
Volume of godown = 40 × 25 × 10 = 10000 m³.
Volume of one box = 2 × 1.25 × 1 = 2.5 m³.
Number of boxes = 10000 ÷ 2.5 = 4000.

Answer:
Side of each cube = ∛125 = 5 cm.
The joined cuboid has dimensions 10 cm × 5 cm × 5 cm.
TSA = 2(lw + lh + wh)
= 2(10 × 5 + 10 × 5 + 5 × 5)
= 250 cm².
Hence, the surface area of the resulting cuboid is 250 cm².

(Note that there is no change in volume but a big change in the surface area. This property has major consequences in the biological world.)
Answer:
Original surface area = 6 × 4² = 96 cm².
Number of small cubes = 4³ ÷ 1³ = 64.
Surface area of one small cube = 6 × 1² = 6 cm².
Total surface area of all separate small cubes = 64 × 6 = 384 cm².
Required ratio = 96 : 384 = 1 : 4.

Answer:
The three face areas are lw, wh and lh.
(lw)(wh)(lh) = (lwh)² = 6 × 15 × 10 = 900.
Volume = lwh = √900 = 30 cm³.
Hence, the volume of the cuboid is 30 cm³.

(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?

Answer:
There are 5 small cubes along each edge.

(i) Exactly three faces painted:
These are the corner cubes. A cube has 8 corners.
So, exactly three faces painted = 8 cubes.

(ii) Exactly two faces painted:
These lie along the edges, excluding the corners.
Each edge has 5 − 2 = 3 such cubes. There are 12 edges.
Number = 12 × 3 = 36 cubes.

(iii) Exactly one face painted:
These lie in the middle of each face, away from the edges.
Each face has (5 − 2)² = 9 such cubes. There are 6 faces.
Number = 6 × 9 = 54 cubes.

(iv) No face painted:
These form the inner cube of side 5 − 2 = 3 cm.
Number = 3³ = 27 cubes.

(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
Answer:
Let the positive integer edge lengths be a ≤ b ≤ c.
2(ab + bc + ca) = 100, so ab + bc + ca = 50.

(i) Yes, there is more than one such cuboid.

(ii) The two different sets of dimensions are:
First cuboid: 1 cm, 2 cm, 16 cm.
Surface Area: 2(1 × 2 + 2 × 16 + 1 × 16) = 2(50) = 100 cm².
Second cuboid: 2 cm, 4 cm, 7 cm.
Surface Area: 2(2 × 4 + 4 × 7 + 2 × 7) = 2(50) = 100 cm².

(iii) Proof that these are all:
Since a ≤ b ≤ c, each of ab, ac and bc is at least a².
Thus, 3a² ≤ 50, so a can only be 1, 2, 3 or 4.
Rearrange the equation as (b + a)(c + a) = 50 + a².

• If a = 1: (b + 1)(c + 1) = 51.
  The factor pairs are (1, 51) and (3, 17). Only (3, 17) gives positive b ≥ 1.
  So b = 2 and c = 16.
• If a = 2: (b + 2)(c + 2) = 54.
  The factor pairs are (1, 54), (2, 27), (3, 18) and (6, 9).
  Since b + 2 must be at least 4, only (6, 9) is allowed.
  So b = 4 and c = 7.
• If a = 3: (b + 3)(c + 3) = 59.
  59 is prime, so no pair has both factors at least 6.
• If a = 4: (b + 4)(c + 4) = 66.
  The factor pairs are (1, 66), (2, 33), (3, 22) and (6, 11).
  None has both factors at least 8.

Therefore, exactly two cuboids are possible, apart from rearranging their length, width and height.

Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.2 Solutions

Answer:
Let A have radius r and height h. Then B has radius 2r and height h/2.
CSA of A = 2πrh.
CSA of B = 2π × 2r × h/2 = 2πrh.
Ratio of curved surface areas, A : B = 1 : 1.

Volume of A = πr²h.
Volume of B = π(2r)²(h/2) = 2πr²h.
Ratio of volumes, A : B = 1 : 2.

Answer:
(a) Volume = πr²h.
Ratio of volumes
= (2² × 3) : (3² × 2)
= 12 : 18
= 2 : 3.

(b) CSA = 2πrh.
Ratio of curved surface areas
= (2 × 3) : (3 × 2)
= 1 : 1.

Answer:
The circular base fits exactly inside a square face of the cube.
Its diameter is r, so the cylinder’s radius is r/2. Its height is r.

Volume = π × (r/2)² × r = πr³/4 cm³.
Hence, the volume of the cylinder is πr³/4 cm³, where r is the cube’s edge, not the cylinder’s radius.

Answer:
New radius = 1.1r. New height = (1 − x/100)h.
Using V = πr²h and cancelling πr²h:
(1.1)²(1 − x/100) = 1.
⇒ 1 − x/100 = 1/1.21 = 100/121.
⇒ x/100 = 21/121.
⇒ x = 2100/121 ≈ 17.36.
Hence, the value of x is approximately 17.36.

(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. (π ≈ 22/7)
Answer:
(i) Volume of cube = 12³ = 1728 cm³.

(ii) Volume of one rod = πr²h
= (22/7) × 2² × 12
= 1056/7 ≈ 150.86 cm³.

(iii) Number = 1728 ÷ (1056/7) = 126/11 ≈ 11.45.
Only 11 complete rods can be made. We must not round up, because there is not enough metal for a twelfth complete rod.
Hence, the number of cylindrical rods = 11, with some metal left over.

Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.3 Solutions

For π, use one of the following approximations: π ≈ 22/7 or π ≈ 3.14

Answer:
Radius = 24 ÷ 2 = 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.
Hence, the total surface area of a cone is 396π m² or approximately 1244.57 m².

Answer:
CSA = πrl
= (22/7) × 7 × 10
= 220 cm².
Hence, the curved surface area of a right circular cone is approximately 220 cm².

Answer:
Slant height l = √(h² + r²)
= √(16² + 12²)
= √400
= 20 cm.

CSA = πrl
= π × 12 × 20
= 240π ≈ 754.29 cm².

TSA = πr(l + r)
= π × 12 × 32
= 384π ≈ 1206.86 cm².
Hence, the curved surface area is 754.29 cm² and the total surface area of the cone is 1206.86 cm².

Answer:
Use π ≈ 3.14 here.
V = ⅓πr²h.
⇒ 1570 = ⅓ × 3.14 × r² × 15 = 15.7r².
⇒ r² = 1570 ÷ 15.7 = 100.
⇒ r = 10 cm.
Hence, the radius of the base is 10 cm.

Answer:
(i) CSA = πrl.
⇒ 308 = (22/7) × r × 14 = 44r.
⇒ r = 7 cm.

(ii) TSA = CSA + πr²
= 308 + (22/7) × 7²
= 308 + 154
= 462 cm².
Hence, the radius of the base is 7 cm and the total surface area of the cone is 462 cm², using π ≈ 22/7.

Answer:
l = √(7² + 24²) = √625 = 25 cm.
A cap is open at its base, so use only the curved surface area.
Area for one cap = πrl = (22/7) × 7 × 25 = 550 cm².
Area for 10 caps = 10 × 550 = 5500 cm².
Hence, the area of the sheet required to make 10 caps is approximately 5500 cm².

Answer:
l = √(8² + 6²) = √100 = 10 m.
The tent has no tarpaulin floor, so use CSA = πrl.
Required curved area = π × 6 × 10 = 60π m².
Length of 3 m wide tarpaulin = area ÷ width = 60π ÷ 3 = 20π m.
Extra length = 20 cm = 0.20 m.
Total length = 20π + 0.20 ≈ 63.0 m, using π ≈ 3.14.
Hence, the length of tarpaulin is approximately 63 m.

Answer:
The solid is a right circular cone.
The side about which the triangle rotates is its height: h = 8 cm.
The other perpendicular side becomes its radius: r = 6 cm.
The hypotenuse becomes its slant height: l = 10 cm.

Volume = ⅓πr²h
= ⅓ × π × 6² × 8
= 96π ≈ 301.71 cm³.
CSA = πrl
= π × 6 × 10
= 60π ≈ 188.57 cm².
Hence, the volume is 301.71 cm³ and the curved surface area of the solid is 188.57 cm².

Answer:
The cup is held with its pointed end down, as in the textbook picture.
Let the full cone have radius r and height h.
At half the height, the radius is also halved, because the water forms a smaller cone of the same shape.

Volume of water = ⅓π(r/2)²(h/2) = ⅛ × ⅓πr²h.
Therefore, volume of water ÷ volume of cup = 1/8.
Hence, the fraction of the volume of the cup is occupied by the water is one-eighth of the cup’s volume.

Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.4 Solutions

Answer:
Surface area = 4πr²
= 4 × (22/7) × 0.7²
= 6.16 cm².
Hence, the surface area is approximately 6.16 cm².

Answer:
Since the spheres are made of the same metal, their volumes are in the same ratio as their weights.
Volume ratio = 5920 : 740 = 8 : 1.
Using V = (4/3)πr³, the ratio of radii is ∛8 : ∛1 = 2 : 1.
Smaller radius = 5 ÷ 2 = 2.5 cm.
Larger radius = 2 × 2.5 = 5 cm.
Hence, the radius of the larger sphere is 5 cm.

Answer:
Ratio of radii, moon : earth = 1 : 4.
Surface area = 4πr², so the ratio of areas = 1² : 4² = 1 : 16.
Hence, the ratio of their surface areas is (Moon’s surface area : Earth’s surface area) 1 : 16.

Answer:
(i) CSA = 2πr²
= 2 × (22/7) × 21²
= 2772 cm².

(ii) TSA = 3πr²
= 3 × (22/7) × 21²
= 4158 cm².
The total area includes the flat circular base.
Hence, the curved surface area is 2772 cm² and the total surface area is 4158 cm².

Answer:
Volume of box = 16 × 8 × 8 = 1024 cm³.
Volume of one sphere = (4/3)π × 2³ = 32π/3 cm³.
Volume of 16 spheres = 512π/3 cm³.
Volume of liquid = 1024 − 512π/3.
Using π ≈ 22/7 gives approximately 487.62 cm³.
Rounded to the nearest integer, the volume is 488 cm³.
Hence, the volume of this liquid is 488 cm³.

Answer:
New radius = 1.1r.
Old volume = (4/3)πr³.
New volume = (4/3)π(1.1r)³ = 1.331 × old volume.
Increase = (1.331 − 1) × old volume = 0.331 × old volume.
Percentage increase = 0.331 × 100 = 33.1%.
Hence proved.

Answer:
New volume ÷ old volume = 1 + 72.8/100 = 1.728.
Since volume is proportional to r³:
(1 + x/100)³ = 1.728 = 1.2³.
⇒ 1 + x/100 = 1.2.
⇒ x = 20.
Hence, the value of x is 20.

Answer:
2πr = 35.2.
r = 35.2 ÷ (2 × 22/7) = 5.6 m.
Only the curved surface of the dome is painted.
CSA = 2πr² = 2 × (22/7) × 5.6² = 197.12 m².

1 m² = 10000 cm².
Area = 197.12 × 10000 = 19,71,200 cm².
Cost = (19,71,200 ÷ 100) × ₹10 = ₹1,97,120.
Hence, the cost of painting is ₹1,97,120.

Class 9 Maths Ganita Manjari Chapter 14 End-of-Chapter Exercises Solutions

For guesstimate problems, make a guess first before solving.

Answer:
My first guess: About 100 scoops.
Assume a full container, no wastage and each scoop approximately a sphere of radius 2 cm.
Container volume = 10 × 15 × 20 = 3000 cm³.
Volume of one scoop = (4/3)π × 2³ = 32π/3 ≈ 33.52 cm³.
Number of scoops ≈ 3000 ÷ 33.52 ≈ 89.5.
Hence, estimate is about 90 scoops. The answer will change with the scoop size.

Answer:
Original number of unit cubes = a³.
New number = (a + 1)³.
Number to add = (a + 1)³ − a³
= a³ + 3a² + 3a + 1 − a³
= 3a² + 3a + 1.
Hence, the added unit cubes are 3a² + 3a + 1.

(i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom?
(ii) Estimate the number of bricks used to build the walls of your classroom.

Answer:
(i) My first guess: No, for a normal-sized classroom.
For this estimate, assuming a person drinks 2 litres daily for 70 years.
Water drunk = 2 × 365 × 70 = 51100 litres = 51.1 m³.

Assuming the classroom measures 8 m × 6 m × 3 m.
Room volume = 8 × 6 × 3 = 144 m³ = 144000 litres.
Since 51100 < 144000, the water would not fill this classroom.
It would fill about 51100/144000 ≈ 35.5% of it.

(ii) My first guess: About 8000 bricks.
Assuming internal dimensions 8 m × 6 m, wall height 3 m and wall thickness 0.20 m.
Outer dimensions = 8.4 m × 6.4 m.
Volume of all four walls before subtracting openings:
[(8.4 × 6.4) − (8 × 6)] × 3 = 17.28 m³.

Assuming one door of size 1 m × 2 m and two windows of size 1.5 m × 1.2 m.
Volume of openings = [1 × 2 + 2 × 1.5 × 1.2] × 0.20 = 1.12 m³.
Net wall volume = 17.28 − 1.12 = 16.16 m³.

Assuming each brick together with its share of mortar occupies 0.20 m × 0.10 m × 0.10 m = 0.002 m³.
Number of bricks ≈ 16.16 ÷ 0.002 = 8080.
Hence, the number of bricks used to build the walls of your classroom is about 8100.

Answer:
Volume in A = 50³ = 125000 mm³.
Volume in B = 100 × 10³ = 100000 mm³.
Difference = 25000 mm³.
Hence, option A gives more chocolate. It gives 25% more than B.

Answer:
My first guess: A few thousand square kilometres.
For a simple guesstimate, assume a rounded population of 8 billion people. This is an assumed number for the exercise, not an exact current population count.
Allow each person a ground area of 0.5 m × 0.5 m = 0.25 m².

Total area ≈ 8,000,000,000 × 0.25 = 2,000,000,000 m².
Since 1 km² = 1,000,000 m², the area is about 2000 km².
Hence, the estimate is about 2000 km², without additional space for paths or facilities.

Answer:
Radius of each glass = 7/2 = 3.5 cm.
Milk in one glass = πr²h
= (22/7) × 3.5² × 12
= 462 cm³.

Milk for 1600 students
= 462 × 1600
= 739200 cm³.
In litres = 739200 ÷ 1000 = 739.2 litres.
Hence, 739.2 litres of milk needed for 1600 students.

Answer:
Sphere’s surface area = 4π × 5² = 100π cm².
Let the cone’s slant height be l.
100π = 5 × π × 4 × l = 20πl.
Therefore, l = 5 cm.

Height h = √(l² − r²) = √(5² − 4²) = 3 cm.
Volume = ⅓πr²h = ⅓ × π × 4² × 3 = 16π cm³ ≈ 50.29 cm³.
Hence, the height is 3 cm and volume is 16π cm³ or approximately 50.29 cm³.

Compute approximately:
(i) the ratio of the volume of Jupiter to the volume of the Earth;
(ii) the ratio of the volume of the Sun to the volume of the Earth.
(Calculator can be used.)
Answer:
For spheres, V = (4/3)πr³. The common factor (4/3)π cancels in a ratio.

(i) Jupiter : Earth = (69900/6370)³ : 1 ≈ 1321.34 : 1.
Thus, Jupiter’s volume is about 1321 times Earth’s volume.

(ii) Sun : Earth = (695700/6370)³ : 1 ≈ 1302709.90 : 1.
Thus, the Sun’s volume is about 13,02,710 times Earth’s volume.

Answer:
Let the sphere’s radius be r.
The smallest enclosing cylinder has radius r and height 2r.

Volume of sphere = (4/3)πr³.
Volume of cylinder = πr² × 2r = 2πr³.
Ratio = [(4/3)πr³] ÷ [2πr³] = 2/3.
Hence, the sphere’s volume is two-thirds of the cylinder’s volume.

How high above the ground is the second string?
Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?

Answer:
Let the original radius be R metres and the extra height be x metres.
Original string length = 2πR.

New string length = 2π(R + x).
Difference = 1 metre, so:
2π(R + x) − 2πR = 1.
⇒ 2πx = 1.
⇒ x = 1/(2π) m ≈ 0.159 m ≈ 15.9 cm.

Answer for Earth: About 15.9 cm.
(i) Moon: About 15.9 cm.
(ii) Jupiter: About 15.9 cm.
(iii) Volleyball: About 15.9 cm.

At first this seems surprising because the objects have very different sizes. But R cancels from the calculation. The gap depends only on the extra string length, not on the original radius.

Answer:
The diagonal of the square base is the diameter of the circle, 2r.
Let the square’s side be a. By the Baudhayana–Pythagoras theorem:
a² + a² = (2r)².
2a² = 4r², so a² = 2r².
Thus, the square’s area is 2r².

The pyramid’s perpendicular height is h.
Volume of pyramid = ⅓ × base area × height = (2/3)r²h.
Volume of cylinder = πr²h.
Ratio, pyramid : cylinder = (2/3)r²h : πr²h = 2 : 3π.
Hence, the ratio of the volume of this pyramid to the volume of the cylinder is 2 : 3π or approximately 7 : 33 if π ≈ 22/7.

(i) the length of a cuboid with dimensions l, w, h is increased by 1 unit?
(a) 1 cubic unit
(b) (lwh + 1) cubic unit
(c) wh cubic units
(d) lw cubic units
(e) lh cubic units
(ii) the radius of a cylinder with dimensions r, h is increased by 1 unit?
(a) 1 cubic unit
(b) πr²h cubic units
(c) πr² cubic units
(d) 2πrh + 2πh cubic units
(e) 2πrh + πh cubic units
(iii) the radius of a sphere is decreased by 1 unit?
Answer:
(i) Old volume = lwh.
New volume = (l + 1)wh.
Increase = (l + 1)wh − lwh = wh cubic units.
Correct option: (c).

(ii) Old volume = πr²h.
New volume = π(r + 1)²h.
Increase = πh[(r + 1)² − r²]
= πh(2r + 1)
= 2πrh + πh cubic units.
Correct option: (e).

(iii) Let the original radius be r units, with r ≥ 1.
Old volume = (4/3)πr³.
New volume = (4/3)π(r − 1)³.
Decrease = (4/3)π[r³ − (r − 1)³]
= (4/3)π(3r² − 3r + 1) cubic units.
Equivalently, the change “new volume − old volume” is the negative of this quantity.

Answer:
If its side is a, then V = a³, so a = ∛V.
S = 6a² = 6(∛V)².
Hence, the total surface area S in terms of V is given by 6V^(2/3) or S = 6(∛V)².

Answer:
S = 6a², so a = √(S/6).
V = a³ = [√(S/6)]³.
Hence, the volume V in terms of S is given by (S/6)^(3/2) or V = [√(S/6)]³.

Answer:
Required rise = 20 − 16 = 4 cm.
The marbles must displace a water volume equal to the cylindrical space for this rise.

Required displacement = π × 4² × 4 = 64π cm³.
Volume of one marble = (4/3)π × 1³ = 4π/3 cm³.
Number of marbles = 64π ÷ (4π/3) = 48.
Hence, the number of marbles are 48.

Answer:
My first guess: About 0.3 mL.
So measurements: Assume the ink-filled part of the refill is a cylinder of internal diameter 2 mm and length 8 cm.
Internal radius = 1 mm = 0.1 cm.
Ink volume ≈ πr²h = π × 0.1² × 8 = 0.08π ≈ 0.251 cm³.
Since 1 cm³ = 1 mL, this is about 0.25 mL.
Estimate: About 0.25 mL for these assumed measurements, ignoring the small tip region.

(i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it?

(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
Answer:
(i) My first guess: About 40 small chapatis.
Assume each uncooked chapati is a thin cylinder of radius 6 cm and thickness 0.2 cm, with no dough wasted.
Volume of dough ball = (4/3)π × 6³ = 288π cm³.
Volume of one chapati = π × 6² × 0.2 = 7.2π cm³.
Number ≈ 288π ÷ 7.2π = 40.
Estimate: About 40 chapatis of these assumed dimensions.

(ii) My first guess: About 130 cm³ of flesh in one half of a coconut.
Sample measurements: Model the half coconut as a hemispherical layer. Assume the outer radius of the edible flesh, just inside the hard shell, is R = 5 cm and the radius of the empty inner cavity is r = 4 cm.

Volume of edible flesh in one half:
V ≈ (2/3)π(R³ − r³)
= (2/3)π(5³ − 4³)
= 122π/3 ≈ 127.81 cm³.
Estimate: About 128 cm³ in one half or about 256 cm³ in two similar halves, using π ≈ 22/7.

(i) If all the pages of this textbook were laid out side-by-side on the floor would they cover the entire classroom floor?
(ii) What is the maximum number of textbooks that can fit in an empty storeroom of dimensions 15 ft × 20 ft × 30 ft?
Answer:
(i) My first guess: No.
Assume each paper sheet is 25 cm × 20 cm and the book has 200 numbered pages, which are 100 physical sheets. Ignoring the covers.
Area of one sheet = 25 × 20 = 500 cm².
Area of all sheets = 100 × 500 = 50000 cm² = 5 m².
Assume the classroom floor is 8 m × 6 m = 48 m².
Since 5 < 48, the sheets will not cover the classroom floor.

(ii) My first guess: About 2 lakh books.
Assume one closed textbook measures 25 cm × 20 cm × 2 cm.
For this estimate, use 1 ft ≈ 30 cm.
Storeroom dimensions ≈ 450 cm × 600 cm × 900 cm.

Arrange books with the 25 cm side along 450 cm, the 20 cm side along 600 cm and the 2 cm thickness along 900 cm.
Number = (450 ÷ 25) × (600 ÷ 20) × (900 ÷ 2)
= 18 × 30 × 450
= 2,43,000.

This arrangement fills the assumed cuboid without gaps. It also equals the room-volume/book-volume upper limit, so it is the maximum under these idealised dimensions.
Estimate: About 2.43 lakh textbooks, with no space left for movement. The actual result requires actual book dimensions and more precise room measurements.

Answer:
My first guess: A few kilometres.
Assume a rounded population of 8 billion and allow 1 m³ of space per person, including some empty space. Ignore walls and floors in this volume estimate.
Required volume ≈ 8,000,000,000 m³.
If the cube’s side is a metres:
a³ = 8,000,000,000.
a = ∛(8 × 10⁹) = 2000 m = 2 km.
Estimate: A cube about 2 km on each side, under these assumptions.

(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.
Answer:
My first guess: A few kilometres.
Let Earth’s radius be R = 6371 km and the water thickness be t km.
Water volume W = 1.38 × 10⁹ km³.
The outer radius is R + t.

(i) Subtract the volume of Earth from the volume enclosed by the outer water surface:
W = (4/3)π[(R + t)³ − R³].
(R + t)³ = R³ + 3W/(4π).

Required expression:
t = ∛[R³ + 3W/(4π)] − R
= ∛[6371³ + (3 × 1.38 × 10⁹)/(4π)] − 6371 km.

(ii) Using a calculator’s value of π:
t ≈ 2.704392 km.
Estimate: About 2.70 km or about 2704 m.

Answer:
One possible example is to change a cuboid of dimensions 8 cm × 14 cm × 14 cm into a cuboid of dimensions 1 cm × 28 cm × 28 cm.

Original cuboid:
Volume = 8 × 14 × 14 = 1568 cm³.
TSA = 2(8 × 14 + 8 × 14 + 14 × 14)
= 2(112 + 112 + 196) = 840 cm².

New cuboid:
Volume = 1 × 28 × 28 = 784 cm³ = 1568/2 cm³.
TSA = 2(1 × 28 + 1 × 28 + 28 × 28)
= 2(28 + 28 + 784) = 1680 cm² = 2 × 840 cm².

Thus, the new volume is half and the new surface area is double. This is one example; the dimensions are not unique.

Answer:
Aim: Estimate the internal volume of a single-storey house.
My first guess: About 150 m³.

Method:

1. Divide the house into non-overlapping rooms and passages.
2. Measure each space’s internal length, width and height.
3. Treat each space as a cuboid and calculate l × w × h.
4. Add the volumes. Count every space only once.

The following are sample assumed measurements, not measurements of your house:

Living room: 4 m × 3 m × 3 m = 36 m³.
Bedroom 1: 4 m × 3 m × 3 m = 36 m³.
Bedroom 2: 3 m × 3 m × 3 m = 27 m³.
Kitchen: 3 m × 2 m × 3 m = 18 m³.
Bathroom: 2 m × 1.5 m × 3 m = 9 m³.
Passage: 2 m × 3 m × 3 m = 18 m³.
Total internal volume = 36 + 36 + 27 + 18 + 9 + 18 = 144 m³.

Conclusion: The estimated internal volume of this sample house is 144 m³, close to my initial guess of 150 m³.
Assumptions: All ceilings are flat and 3 m high. Internal wall volumes are excluded. Furniture is not subtracted because the aim is to find the rooms’ total internal space.