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Naman Academy For Class 9
Chapter 14: Math of Space: Surface Area and VolumePage 144–147All 22 Questions on This Page

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

Comprehensive step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 14 End-of-Chapter Exercises (Questions 1–22, Page 144–147). Covers guesstimations and spatial scaling, ice cream scoops, unit cubes addition, lifetime water consumption and brick estimates, chocolate volume comparisons, global population footprint, milk glass capacity, cone height and volume from sphere area, planetary volume comparisons (Jupiter, Sun, Earth), Archimedean sphere-in-cylinder ratio, equator string puzzle, inscribed pyramid-in-cylinder volume ratio, differential volume changes, cube formulas inversion, crow and marbles water level rise, ballpoint pen ink volume, chapati and coconut edible volume estimates, textbook paper footprint and packing, human population cube, Earth's water layer thickness, cuboid volume-surface area transformations, and house volume estimation.

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Key Concepts for End-of-Chapter Exercises

Jump to Question 1
1

Volume and Surface Area of Cubes and Cuboids

For a cube of side a, volume is V = a³ and total surface area is TSA = 6a². For a cuboid of dimensions l, w, h, volume is V = lwh and total surface area is TSA = 2(lw + wh + hl). Face areas meeting at a corner satisfy V = √(A₁A₂A₃).

V_cube = a³, TSA_cube = 6a², V_cuboid = lwh, TSA_cuboid = 2(lw + wh + hl)
2

Right Circular Cylinder

A cylinder of base radius r and height h has curved surface area CSA = 2πrh, total surface area TSA = 2πr(r + h), and volume V = πr²h. When a sphere of radius r is snugly enclosed by a cylinder (h = 2r), V_sphere = (2/3)V_cylinder.

CSA = 2πrh, TSA = 2πr(r + h), V = πr²h, V_sphere = (2/3)V_enclosing-cylinder
3

Right Circular Cone

For a cone of base radius r, vertical height h, and slant height l = √(r² + h²), curved surface area is CSA = πrl, total surface area is TSA = πr(l + r), and volume is V = (1/3)πr²h.

l² = r² + h², CSA = πrl, TSA = πr(l + r), V = (1/3)πr²h
4

Sphere and Hemisphere

For a sphere of radius r, surface area is S = 4πr² and volume is V = (4/3)πr³. For a hemisphere, curved surface area is 2πr², total surface area is 3πr², and volume is (2/3)πr³.

S_sphere = 4πr², V_sphere = (4/3)πr³, CSA_hemi = 2πr², TSA_hemi = 3πr², V_hemi = (2/3)πr³
5

Spatial Reasoning, Scaling, and Guesstimation

In real-world spatial problems, identify ideal geometric solid models, ensure uniform units (e.g. 1 m³ = 1000 L = 1,000,000 cm³), establish reasonable physical parameters, and calculate bounds using ratios and conservation of matter.

1 m³ = 1000 L, 1 cm³ = 1 mL, Number of units = V_total / V_single
Question 1
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Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.

Solution

Initial guess: About 40 scoops.

Assume each scoop is approximately spherical, with a radius of 2.5 cm.

The volume of a cuboid is

V = l × w × h

Volume of the ice cream container:

Vcontainer = 10 × 15 × 20
= 3000 cm³

The volume of a sphere is

V = (43)πr³

Taking π ≈ 3.14, the volume of one scoop is

Vscoop = (43) × 3.14 × (2.5)³
≈ 65.42 cm³

The estimated number of scoops is

Number of scoops = Vcontainer / Vscoop
= 300065.42
≈ 45.86

The actual number may differ depending on the scoop size, shape and amount of ice cream left in the container.

Answer:Approximately 46 scoops
Question 2
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A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.

Solution

The number of unit cubes in the original cube is

Voriginal = a³

The side length of the new cube is a + 1. Therefore, the number of unit cubes it contains is

Vnew = (a + 1)³

The number of additional unit cubes required is

Additional cubes = (a + 1)³ − a³

Expanding,

(a + 1)³ − a³ = a³ + 3a² + 3a + 1 − a³
= 3a² + 3a + 1
Answer:3a² + 3a + 1 unit cubes
Question 3
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Solve the following: (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.

Solution
(i) Could a person drink enough water in a lifetime to fill a classroom?

The answer depends on the classroom's dimensions, daily water consumption and lifespan. Let us make reasonable assumptions.

Assume:

•Classroom length = 8 m
•Classroom width = 6 m
•Classroom height = 3 m
•Water consumed per day = 2 litres
•Lifespan = 75 years

Volume of the classroom:

The volume of a cuboid is

V = l × w × h
Vclassroom = 8 × 6 × 3
= 144 m³

Since 1 m³ = 1000 litres,

144 m³ = 144,000 litres

Water consumed in a lifetime:

Taking 365 days per year,
Total water = 2 × 365 × 75
= 54,750 litres
Comparing the two quantities,
54,750 < 144,000
Therefore, under these assumptions, a person would not drink enough water in a lifetime to fill the classroom.

The estimate will vary with the actual classroom size, daily water consumption and lifespan.

(ii) Estimate the number of bricks used to build the walls of your classroom

We can estimate the number of bricks by calculating the volume of the walls and dividing it by the volume occupied by one brick together with its share of mortar.

Assume:

•Classroom length = 8 m
•Classroom width = 6 m
•Wall height = 3 m
•Wall thickness = 0.2 m
•Total area of doors and windows = 8 m²

Area of the four walls:

A = 2(l + w) × h
= 2(8 + 6) × 3
= 84 m²
Subtracting the area of the doors and windows,
Awalls = 84 − 8
= 76 m²

Volume of the walls:

Vwalls = Wall area × Thickness
= 76 × 0.2
= 15.2 m³

Assume that one brick together with its share of mortar occupies approximately 20 cm × 10 cm × 10 cm.

Converting to metres, its volume is

V_brick_unit = 0.2 × 0.1 × 0.1
= 0.002 m³

The estimated number of bricks is

N = Vwalls / V_brick_unit
= 15.20.002
= 7600

The actual number depends on the classroom dimensions, wall thickness, brick size, mortar and openings.

Answer:(i) No; 54,750 litres < 144,000 litres; (ii) Approximately 7600 bricks
Question 4
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You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?

Solution

The volume of a cube is

V = a³
Option A: One large cube
Side = 50 mm
VA = 50³
= 125,000 mm³
Option B: One hundred small cubes

Side of each small cube = 10 mm

Volume of one small cube:

V = 10³ = 1000 mm³

Volume of 100 small cubes:

VB = 100 × 1000
= 100,000 mm³
Comparing the volumes,
125,000 mm³ > 100,000 mm³
Therefore, Option A gives more chocolate.

The difference is

VA − VB = 125,000 − 100,000
= 25,000 mm³

As a percentage of Option B,

(25,000100,000) × 100% = 25%
Answer:Option A gives 25% more chocolate (125,000 mm³ vs 100,000 mm³)
Question 5
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If all the people in the world are crowded together in one location, how much area would it cover?

Solution

Initial guess: Approximately 2000 km².

Assume that the world population is approximately 8 billion and each person occupies 0.25 m² when standing closely packed.

The total area occupied is

Area = Number of people × Area per person
= 8,000,000,000 × 0.25
= 2,000,000,000 m²

Since 1 km² = 1,000,000 m², the area in square kilometres is

Area = 2,000,000,0001,000,000
= 2000 km²

The answer depends on the assumed population and how tightly people are packed. For example, assuming 0.5 m² per person would give an estimate of 4000 km².

Answer:Estimated area = 2000 km²
Question 6
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A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.

Solution
Given:
Diameter of each glass = 7 cm

Radius, r = 72 = 3.5 cm

Height, h = 12 cm
Number of students = 1600

The volume of a cylinder is

V = πr²h

Taking π ≈ 227, the volume of milk in one glass is

V = (227) × (3.5)² × 12
= (227) × 12.25 × 12
= 462 cm³

The total volume of milk required is

Vtotal = 1600 × 462
= 739,200 cm³

Since 1000 cm³ = 1 litre,

Milk required = 739,2001000
= 739.2 litres
Answer:Milk required = 739.2 litres
Question 7
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The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.

Solution
Given:
Radius of the sphere = 5 cm

Radius of the cone, r = 4 cm

Let the slant height of the cone be l.

The surface area of a sphere is

S = 4πr²
Therefore, the surface area of the sphere is

S = 4π × 5² = 100π cm²

The curved surface area of a cone is

CSA = πrl

According to the question,

Surface area of sphere = 5 × CSA of cone

Substituting the values,

100π = 5 × π × 4 × l

100π = 20πl
l = 5 cm
Finding the height of the cone

The relation between the height, radius and slant height is

l² = h² + r²
Therefore,
5² = h² + 4²
25 = h² + 16
h² = 9
h = 3 cm
Finding the volume of the cone

The volume of a cone is

V = (13)πr²h

V = (13) × π × 4² × 3

= 16π cm³
Taking π ≈ 3.14,

V ≈ 16 × 3.14 = 50.24 cm³

Answer:Height of cone = 3 cm; Volume of cone = 16π cm³ ≈ 50.24 cm³
Question 8
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Take Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius 695,700 km. Compute approximately: (i) the ratio of the volume of Jupiter to the volume of Earth; (ii) the ratio of the volume of the Sun to the volume of Earth. (Calculator can be used.)

  • (i)Compute approximately the ratio of the volume of Jupiter to the volume of Earth.
  • (ii)Compute approximately the ratio of the volume of the Sun to the volume of Earth.
Solution

The volume of a sphere is

V = (43)πr³

Since (43)π is common to all three spheres, the ratio of their volumes is the ratio of the cubes of their radii.

(i) Ratio of Jupiter's volume to Earth's volume
Given:
Radius of Jupiter = 69,900 km
Radius of Earth = 6370 km
VJ / VE = (69,9006370)³
≈ 1321.34
(ii) Ratio of the Sun's volume to Earth's volume
Given:
Radius of the Sun = 695,700 km
Radius of Earth = 6370 km
VS / VE = (695,7006370)³
≈ 1,302,709.90
Answer:VJ : VE ≈ 1321 : 1; VS : VE ≈ 1,302,710 : 1
Question 9
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Show that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder which encloses it.

Enclosing Cylinder of a Sphere

Geometric relationship: Smallest cylinder enclosing a sphere of radius r has height h = 2r and base radius r.

Solution

Let the radius of the sphere be r.

The smallest cylinder that encloses the sphere has:

•Radius = r
•Height = 2r, equal to the diameter of the sphere.

The volume of the sphere is

Vsphere = (43)πr³

The volume of a cylinder is

Vcylinder = πr²h
Substituting h = 2r,
Vcylinder = πr²(2r)
= 2πr³
Comparing the volumes,
Vsphere / Vcylinder = ((43)πr³) / (2πr³)
= 23
Thus, the volume of a sphere is 23 of the volume of the smallest cylinder that encloses it.
Answer:Vsphere = (23) Vcylinder
Question 10
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Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere. 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?

Solution

Let the radius of the Earth be R metres and the height of the second string above the ground be h metres.

The circumference of a circle is

C = 2πr

The length of the original string is

C₁ = 2πR

The second string is 1 metre longer, and its radius is R + h. Therefore,

C₂ = 2π(R + h)

Since the second string is 1 metre longer,

C₂ − C₁ = 1
2π(R + h) − 2πR = 1
2πR + 2πh − 2πR = 1
2πh = 1
h = 1 / (2π) m
Taking π ≈ 3.14,
h ≈ 16.28
≈ 0.159 m
≈ 15.9 cm

The same calculation applies to all three objects because the original radius cancels out of the equation.

•Moon: approximately 15.9 cm
•Jupiter: approximately 15.9 cm
•Volleyball: approximately 15.9 cm

Are you surprised? Yes, the result is surprising because we might expect a 1-metre increase in string length to produce a much smaller gap around a large object than around a small one. However, the increase in radius depends only on the extra circumference, not on the object's original size.

Answer:Height above the ground ≈ 15.9 cm (identical for Earth, Moon, Jupiter, and a volleyball)
Question 11
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We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.

Square Pyramid Inscribed in a Cylinder

Geometric arrangement: Square base inscribed in circular base of radius r has diagonal 2r, so base area s² = 2r².

Solution

The radius of the cylinder is r, and its height is h.

The square base of the pyramid is inscribed in the circular base of the cylinder. Thus, the diagonal of the square equals the diameter of the circle.

Diagonal of square = 2r

Let the side of the square be s.

Using the Pythagorean theorem,

s² + s² = (2r)²
2s² = 4r²
s² = 2r²
Therefore, the area of the square base is
Base area = s² = 2r²

The pyramid's height is h, the same as the cylinder's height.

The volume of a pyramid is

V = (13) × Base area × Height

Hence,
Vpyramid = (13) × 2r² × h
= (23)r²h

The volume of the cylinder is

Vcylinder = πr²h
Therefore,
Vpyramid / Vcylinder = ((23)r²h) / (πr²h)
= 2 / (3π)
Answer:Vpyramid : Vcylinder = 2 : 3π
Question 12
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What is the change in volume when: (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?

Solution
(i) Length of a cuboid increased by 1 unit

The original volume is

V₁ = lwh

The new length is l + 1. Therefore, the new volume is

V₂ = (l + 1)wh

The increase in volume is

V₂ − V₁ = (l + 1)wh − lwh

= lwh + wh − lwh
= wh
Increase in volume = wh cubic units

Correct option: (c)

(ii) Radius of a cylinder increased by 1 unit

The original volume is

V₁ = πr²h

The new radius is r + 1. Therefore, the new volume is

V₂ = π(r + 1)²h

The increase in volume is

V₂ − V₁ = π(r + 1)²h − πr²h

= πh[(r + 1)² − r²]
= πh(r² + 2r + 1 − r²)
= 2πrh + πh

Increase in volume = (2πrh + πh) cubic units

Correct option: (e)

(iii) Radius of a sphere decreased by 1 unit

Let the original radius be r units, where r ≥ 1.

The original volume is

V₁ = (43)πr³

The new radius is r − 1. Therefore, the new volume is

V₂ = (43)π(r − 1)³

The decrease in volume is original volume minus new volume:

V₁ − V₂ = (43)π[r³ − (r − 1)³]

= (43)π[r³ − (r³ − 3r² + 3r − 1)]
= (43)π(3r² − 3r + 1)
= 4πr² − 4πr + (43)π

The expression is in cubic units.

Answer:(i) wh cubic units (Option c); (ii) 2πrh + πh cubic units (Option e); (iii) Decrease = 4πr² − 4πr + (43)π cubic units
Question 13
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Given a cube with volume V, express its total surface area S in terms of V.

Solution

Let the side of the cube be a.

The volume of a cube is

V = a³
Therefore,
a = ∛V

The total surface area of a cube is

S = 6a²
Substituting a = ∛V,
S = 6(∛V)²
= 6∛(V²)
Answer:S = 6∛(V²)
Question 14
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Given a cube with total surface area S, express its volume V in terms of S.

Solution

Let the side of the cube be a.

The total surface area of a cube is

S = 6a²
Therefore,
a² = S / 6
Taking the positive square root,
a = √(S / 6)

The volume of a cube is

V = a³
Substituting the value of a,
V = (√(S / 6))³
= (S / 6)√(S / 6)
Answer:V = (S / 6)√(S / 6)
Question 15
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A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm. A crow wants to drink water from this glass. The water must be at a height of 20 cm for the crow to reach it. There are some marbles lying around. How many marbles of radius 1 cm should the crow drop into the glass to make the water reach the required height?

Solution
Given:
Radius of the glass = 4 cm
Initial water height = 16 cm
Required water height = 20 cm
Radius of each marble = 1 cm

The water level must rise by

20 − 16 = 4 cm

The volume needed to fill this extra height in the cylindrical glass is

Vrequired = πr²h
= π × 4² × 4
= 64π cm³

The volume of one spherical marble is

Vmarble = (43)πr³
Putting r = 1 cm,
Vmarble = (43)π cm³
Therefore, the number of marbles required is
N = Vrequired / Vmarble
= 64π / ((43)π)
= 64 × (34)
= 48

This assumes the marbles sink and are fully submerged in the water.

Answer:Number of marbles required = 48
Question 16
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Find the volume of ink in a new ballpoint pen. Take the necessary measurements and make approximations, as needed.

Solution

This is an estimation question. The answer depends on the actual dimensions of the ink-filled tube.

For an example, assume:

•Length of the ink-filled tube = 10 cm
•Inner diameter of the tube = 1.4 mm = 0.14 cm
Therefore, the inner radius is

r = 0.142 = 0.07 cm

Approximating the ink-filled part as a cylinder, its volume is

V = πr²h
Taking π ≈ 22/7,
V = (227) × (0.07)² × 10
= (227) × 0.0049 × 10
= 0.154 cm³

Since 1 cm³ = 1 mL,

Estimated volume of ink ≈ 0.15 mL

This is only an example estimate. For a more accurate answer, measure the length and inner diameter of the ink-filled tube in an actual new pen and use those values in the formula.

Answer:Estimated volume of ink ≈ 0.15 mL (or 0.154 cm³)
Question 17
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Solve the following estimation problems: (i) Estimate how many chapatis can be made from a ball of chapati dough of radius 6 cm. (ii) Cut a coconut or muskmelon in half and estimate the volume of edible flesh or fruit by taking the necessary measurements.

Solution
(i) Estimate how many chapatis can be made from dough ball of radius 6 cm

Initial guess: About 25 chapatis.

Assume the ball of dough is spherical. After rolling, suppose each chapati has an approximate radius of 7.5 cm and thickness of 0.2 cm.

The volume of a sphere is

V = (43)πr³

Volume of the dough ball:

Vdough = (43) × 3.14 × 6³
≈ 904.32 cm³

Approximating each chapati as a thin cylinder, its volume is

V = πr²h
Vchapati = 3.14 × (7.5)² × 0.2
≈ 35.33 cm³

Estimated number of chapatis:

N = Volume of doughVolume of one chapati

= 904.3235.33
≈ 25.6

The actual number depends on the size and thickness of the chapatis and the amount of dough lost during preparation.

(ii) Cut a coconut or muskmelon in half and estimate the volume of edible flesh

Example using half a coconut:

Suppose the measured outer radius of the edible coconut flesh is approximately 5 cm, and its inner hollow cavity has a radius of 4 cm. The thickness of the edible flesh is therefore about 1 cm.

We can approximate the edible portion as a hemispherical shell.

The volume of a hemisphere is

V = (23)πr³
Hence, the volume of edible flesh is approximately
Vflesh = (23)π(R³ − r³)

where R = 5 cm and r = 4 cm.

Taking π ≈ 3.14,

Vflesh = (23) × 3.14 × (5³ − 4³)

= (23) × 3.14 × (125 − 64)
= (23) × 3.14 × 61
≈ 127.7 cm³

This is an illustrative estimate. The answer will vary with the actual measurements of the coconut. A similar method can be used for a muskmelon, with suitable measurements of its edible portion and seed cavity.

Answer:(i) Approximately 26 chapatis; (ii) Estimated edible flesh ≈ 128 cm³
Question 18
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Estimation using the Ganita Manjari textbook: (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?

Solution
(i) Side-by-side pages on classroom floor

Suppose the textbook has approximately 148 pages, each measuring 20 cm × 28 cm.

Area of one page:

A = 20 × 28
= 560 cm²
= 0.056 m²

Approximate total area of all pages:

Apages = 148 × 0.056
= 8.288 m²

Assume the classroom floor measures 8 m × 6 m.

Aclassroom = 8 × 6
= 48 m²

Since 8.288 m² < 48 m², the pages would not cover the entire classroom floor under these assumptions.

(ii) Maximum number of textbooks in a storeroom (15 ft × 20 ft × 30 ft)

Assume each textbook measures approximately:

•Length = 28 cm
•Width = 20 cm
•Thickness = 1 cm
Convert the storeroom dimensions into centimetres, using 1 ft = 30.48 cm:
15 ft = 457.2 cm
20 ft = 609.6 cm
30 ft = 914.4 cm

Arrange the textbooks neatly, with the 28 cm side along the 30 ft side, the 20 cm side along the 20 ft side and the 1 cm thickness along the 15 ft side.

Number of books along the 30 ft side:

⌊914.428⌋ = 32

Number of books along the 20 ft side:

⌊609.620⌋ = 30

Number of layers along the 15 ft side:

⌊457.21⌋ = 457
Therefore, the number of textbooks in this arrangement is
N = 32 × 30 × 457
= 438,720

This is an estimate for the stated textbook dimensions and aligned arrangement. The actual maximum may vary with the dimensions of the books and how they can be packed.

Answer:(i) No; pages cover approx 8.3 m² < 48 m²; (ii) Approximately 438,720 textbooks
Question 19
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If the entire human population decided to climb into one giant cube, how long would the side have to be?

Solution

Initial guess: A cube with side length of around 1 km.

Assume:

•World population = 8 billion = 8 × 10⁹ people
•Average body volume of one person = 0.07 m³

The total volume of all people combined is

Vtotal = 8 × 10⁹ × 0.07
= 5.6 × 10⁸ m³

The volume of a cube is

V = a³
Therefore, the required side length of the cube is
a = ∛(5.6 × 10⁸)
≈ 824 m

In kilometres, this is

824 m ≈ 0.82 km

This calculation assumes that people can be packed tightly together without gaps. A real cube containing people would need to be larger because bodies cannot fit together perfectly.

Answer:Side of the cube ≈ 824 m (approximately 0.82 km)
Question 20
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The Earth's surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering Earth's surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth's radius is approximately 6371 km. (i) Write an expression that gives the thickness of this water layer. (ii) Simplify the expression using a calculator.

Solution
(i) Write an expression for the thickness of the water layer

Let the thickness of the water layer be t km.

Radius of Earth:

R = 6371 km

The outer radius, including the water layer, is 6371 + t.

The volume of a sphere is

V = (43)πr³

The volume of the water layer equals the volume of the outer sphere minus the volume of Earth:

(43)π[(6371 + t)³ − 6371³] = 1.38 × 10⁹

Rearranging,

(6371 + t)³ = 6371³ + (3(1.38 × 10⁹)) / (4π)
Taking the cube root and subtracting 6371,

t = ∛(6371³ + (3(1.38 × 10⁹)) / (4π)) − 6371

(ii) Simplify the expression using a calculator
Evaluating the expression,
t ≈ 2.705 km
Therefore, the thickness is approximately 2.71 km.
Answer:(i) t = ∛(6371³ + (3(1.38 × 10⁹)) / (4π)) − 6371; (ii) Thickness ≈ 2.71 km (approx 2705 m)
Question 21
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Give the dimension of a cuboid whose volume is halved when its surface area is doubled.

Solution

One valid 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

The volume of a cuboid is

V = lwh
V₁ = 8 × 14 × 14
= 1568 cm³

The total surface area is

S = 2(lw + wh + hl)

S₁ = 2(8 × 14 + 14 × 14 + 14 × 8)

= 2(112 + 196 + 112)
= 840 cm²
New cuboid
V₂ = 1 × 28 × 28
= 784 cm³

Since 784 = 15682,

the volume is halved.

Its total surface area is

S₂ = 2(1 × 28 + 28 × 28 + 28 × 1)

= 2(28 + 784 + 28)
= 1680 cm²

Since 1680 = 2 × 840,

the surface area is doubled.

Note
This is one valid pair of cuboids; the question does not specify a unique answer. Multiple pairs of dimensions are possible.
Answer:1 cm × 28 cm × 28 cm (from original cuboid 8 cm × 14 cm × 14 cm)
Question 22
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Find the volume of your house, making necessary approximations. Present how you solved it.

Solution

For a simple estimate, suppose the house can be approximated by a cuboid with the following dimensions:

•Length = 10 m
•Width = 8 m
•Height = 3 m

The volume of a cuboid is

V = lwh
Therefore,
V = 10 × 8 × 3
= 240 m³

This is only an example. For your own house, measure its approximate length, width and height. If it has several floors or sections of different sizes, estimate each section's volume separately and add them.

Answer:Estimated volume of the house = 240 m³ (for 10 m × 8 m × 3 m model)

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