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

Solving questions with the help of Ganita Manjari Class 9 Solutions and Part 2 Class 9 Maths Chapter 14 Math of Space Surface Area and Volume NCERT Solutions improves confidence.

Ganita Manjari Class 9 Chapter 14 Solutions

Class 9 Ganita Manjari Chapter 14 Solutions

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

Think and Reflect (NCERT Textbook Page No. 124)

Question 1.
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.
Write the formula for the surface area and the volume of a cube.
Solution:
(a) Thinking of a cuboid as a stack of thin rectangular cards (or thin layers) visually demonstrates how 3D volume builds up from 2D area:
Base Area (l × b): A single rectangular card lying flat has a 2D surface area defined by its length (l) and width (b), which represents the area of the base.
Height (h): As we stack many identical cards on top of one another, the stack grows vertically to a total height (h).

Total Volume (V): The 3D space occupied by the entire stack is equal to the base area of one card multiplied by the height of the stack.
Thus: Volume = area of base × height

(b) Formulae for a Cube
A cube is a special case of a cuboid where the length, width, and height are all equal to side length a.
Curved Surface Area (CSA): 4a² square units
(Since a cube has 4 identical square faces, excluding the top and bottom, each with an area of a²)
Total Surface Area (TSA) = 6a² square units
(Since a cube has 6 identical square faces, each with an area of a²)
Volume (V) = a³ cubic unit (Derived from l × b × h where l = b = h = a

Intext Question (NCERT Textbook Page No. 124)

Question 1.
Observe how these two formulas are special cases of the formulas 2(wl + hl + hw) and whl, when w = h = l.
Solution:
Let the common value of the length, width, and height be a;
that is, w = h = l = a.
Surface area = 2(wl + hl + hw)
= 2(a × a + a × a + a × a)
= 2(a² + a² + a²)
= 2 × 3a²
= 6a²
Volume = whl
= a × a × a
= a³
So the cube formulas 6a² and a³ are obtained from the cuboid formulas by putting w = h = l = a.
A cube is simply a special cuboid with all sides equal.

Think and Reflect (NCERT Textbook Page No. 126)

Question 1.
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 and (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? [1 cubic foot ~ 28.3 litres]
Solution:
Volume of water (capacity).
Pool A: 8 × 50 × 20 = 8000 ft³
= 8000 × 28.3
= 2,26,400 litres

Pool B: 6 × 100 × 15 = 9000 ft³
= 9000 × 28.3
= 2,54,700 litres

Area to be tiled (floor + four walls; the top is open)
Pool A = 50 × 20 + 2(8 × 50 + 8 × 20)
= 1000 + 2(400 + 160)
= 2120 ft²
Pool B = 100 × 15 + 2(6 × 100 + 6 × 15)
= 1500 + 2(600 + 90)
= 2880 ft²
Water surface area (space for swimmers):
A: 50 × 20 = 1000 ft²
B: 100 × 15 = 1500 ft²
A suitable choice for a school: Option B.
Reasons:
Safety: 6 ft is shallower than 8 ft, which is safer for school students who are learning to swim.
More space: B has 1.5 times the water surface, so more students can use it at the same time.
Longer laps: 100 ft is useful for races and practice.
Points against B: it needs about 12.5% more water (about 28,300 litres more), which costs more to fill, filter, and heat, and about 760 ft² more tiling.
Pool A would be preferable if diving is planned (it needs more depth) or if water and money are very limited.
Parameters considered: Depth (safety), surface area of water (number of users), length (laps), volume (water needed and running cost), area of floor and walls (construction cost), and the space available in the school. (This is an open-ended question; other well-reasoned choices are also acceptable.)

Intext Question (NCERT Textbook Page No. 127)

Question 1.
The following figures show different types of cylinders. What is different in each of these cylinders?

Solution:
Both cylinders have two equal, parallel circular ends (bases).
The difference is in the axis — the line joining the centres of the two circular ends.
In the right circular cylinder, the axis is perpendicular to the circular base.
The cylinder stands straight up, and its side is at right angles to the base.
In the oblique circular cylinder, the axis is inclined (slanted) to the base, not perpendicular to it.
The cylinder leans to one side.
In both, a cross-section parallel to the base is a circle of the same radius.

Ex 14.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 14.1 Solutions

Exercise 14.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 14.1 Solutions

Question 1.
The volume of a cube is 64 cm³. What is its total surface area?
Solution:

Let the side of the cube be a cm.
Then, according to the question
Volume of cube = 64 cm³
⇒ a³ = 64
⇒ a = \(\sqrt[3]{64}\)
⇒ a = 4 cm
So the side of the cube is 4 cm.
Now, the total surface area of the cube = 6a²
= 6 × 4²
= 6 × 16
= 96 cm²
Hence, the total surface area of the cube is 96 cm².

Question 2.
How many small cubes with side 20 cm can be packed tightly in a cubical box with side 2 m?
Solution:

First, let us write both lengths in the same unit:
2 m = 2 × 100 cm = 200 cm (∵ 1 m = 100 cm)
Total volume of cubical box with side 2 m = (200 cm)³
Total volume of small cubes = (20 cm)³
Number of small cubes which can fit in cubical box = \(\frac{(200 \mathrm{~cm})^3}{(20 \mathrm{~cm})^3}\) = 1000 cubes

Question 3.
The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m × 1.25 m × 1 m, then find the number of boxes.
Solution:

Here, volume of the godown = l × b × h
= 40 × 25 × 10
= 10000 m³
Volume of one box = 2 × 1.25 × 1 = 2.5 m³
Hence, number of boxes = \(\frac{\text { volume of godown }}{\text { volume of one box }}\)
= \(\frac{10000}{2.5}\)
= 4000

Question 4.
Two cubes, each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:

Let the side of each cube be a cm.
Then volume of cube = a³ = 125
⇒ a = \(\sqrt[3]{125}\) = 5
So, each cube has side 5 cm.
When the two cubes are joined end to end,
the resulting cuboid has length l = 5 + 5 = 10 cm, width w = 5 cm, height h = 5 cm.
∴ Surface area of cuboid = 2(lw + wh + hl)
= 2(10 × 5) + (5 × 5) + (5 × 10)
= 2(50 + 25 + 50)
= 2 × 125
= 250 cm²
Hence, the surface area of the resulting cuboid is 250 cm².

Question 5.
A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes? (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.)
Solution:

Original cube (side 4 cm):
∴ Surface area of cube = 6a²
= 6 × 4²
= 6 × 16
= 96 cm²
Number of small cubes = \(\frac{\text { volume of big cube }}{\text { volume of small cube }}\)
= \(\frac{4^3}{1^3}\)
= \(\frac {64}{1}\)
= 64
Surface area of all 64 small cubes (each of side 1 cm) = 64 × 6a² (Here a = 1 cm)
= 64 × (6 × 1²)
= 64 × 6
= 384 cm²
∴ Required ratio = \(\frac{\text { Surface area of original cube }}{\text { Surface area of cut-out cubes }}=\frac{96}{384}=\frac{1}{4}\) = 1 : 4
The ratio is 1 : 4. The total volume stays 64 cm³, but the total surface area becomes four times as large. (This property is useful in gas exchange in lungs, as lungs have millions of tiny air sacs wrapped in capillaries that increase total surface area within chest volume to speed up oxygen and carbon dioxide exchange.)

Question 6.
The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
Solution:
Let the length, width, and height of the cuboid be l, w, and h.
The three faces that meet at a corner have areas lw, wh, and hl.

Here, lw = 6, wh = 15, hl = 10
Now, area of (lw)(wh)(hl) = 6 × 15 × 10
⇒ l²w²h² = 900
⇒ (lwh)² = 900
⇒ lwh = √900
⇒ lwh = \(\sqrt{9 \times 100}\)
⇒ lwh = √9 × √100 = 3 × 10 = 30
So, the volume of the cuboid is 30 cm³.

Question 7.
A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes have

(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
Solution:
As shown in the figure, the cube is cut into 5 × 5 × 5 = 125 small cubes (5 along each edge).

(i) Exactly three faces painted: Only the small cubes at the corners have three faces on the outside.
A cube has 8 comers, so Number = 8 (marked as x in the diagram above)

(ii) Exactly two faces painted: These lie along the edges, but not at the corners.
Each edge has 5 small cubes, of which 2 are corner cubes, leaving 5 – 2 = 3.
A cube has 12 edges, so Number = 12 × 3 = 36 (marked as y in the diagram above)

(iii) Exactly one face painted: These lie on a face, away from all its edges.
On each face, they form a 3 × 3 square, i.e., 9 cubes.
There are 6 faces, so Number = 6 × 9 = 54 (marked as z in the diagram above)

(iv) No face painted: These form the inner cube of side 5 – 2 = 3 cm, hidden completely inside.
Number = 3 × 3 × 3 = 27
Top layer of the 5 × 5 × 5 cube

Hence,

Faces painted	Position	Number of cubes
Exactly three	Corners	8
Exactly two	Edges (not corners)	36
Exactly one	Faces (not edges)	54
None	Inside	27
Total		125

Question 8.
Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm².
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
Solution:
Let the edges be l, w, h (positive integers, in cm) and l ≤ w ≤ h.
Now, total surface area of cuboid = 2(lw + wh + hl) = 100
⇒ lw + wh + hl = 50
Limiting the smallest edge.
Since l ≤ w ≤ h, each of lw, wh, hl is at least l × l = l²
So, 3l² ≤ 50
⇒ l² ≤ 16.6
⇒ l ≤ 4
So we only need to check l = 1, 2, 3, 4.
A useful trick:
Adding l² to both sides lets us factorise,
lw + wh + hl + l² = 50 + l²
⇒ w(l + h) + l(h + l) = 50 + l²
⇒ (w + l) (h + l) = 50 + l²
Case: l = 1
(w + 1)(h + 1) = 51 = 3 × 17 (the only other factorisation, 1 × 51, would give w = 0).
So, w + 1 = 3, h + 1 = 17, i.e., w = 2, h = 16.
Cuboid: 1 cm × 2 cm × 16 cm.
Case: l = 2
(w + 2) (h + 2) = 54, with w + 2 ≥ 4 (because w ≥ 2) and w + 2 ≤ h + 2.
The factor pairs of 54 are 1 × 54, 2 × 27, 3 × 18, 6 × 9.
Only 6 × 9 is allowed, so w = 4, h = 7.

Cuboid: 2 cm × 4 cm × 7 cm.
Case: l = 3
(w + 3) (h + 3) = 59.
Since 59 is prime, it has no factor pair with both factors at least 6.
No cuboid.
Case: l = 4
(w + 4 )(h + 4) = 66, with 8 ≤ w + 4 ≤ h + 4.
Then (w + 4)2 ≤ 66, so w + 4 = 8
but 8 does not divide 66.
No cuboid.
(i) Yes, there is more than one such cuboid.
(ii) There are exactly two: 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm
(iii) The smallest edge can only be 1, 2, 3 or 4 (because 3l² ≤ 50), and for each of these values we listed every possible factor pair of 50 + l².
Only the two cuboids above work, so we have found them all.
(Arranging the same three edges in a different order gives the same cuboid.)

Ex 14.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 14.2 Solutions

Exercise 14.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 14.2 Solutions

Question 1.
Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.
Solution:

Let cylinder A have radius r and height h.
Then cylinder B has radius 2r and height \(\frac {h}{2}\).
Curved surface areas:
CSA of a cylinder = 2πrh
Now, CSA of cylinder A = 2πrh
and, CSA of cylinder B = 2π(2r)(\(\frac {h}{2}\)) = 2πrh
CSA of cylinder A : CSA of cylinder B = 2πrh : 2πrh = 1 : 1
Volumes:
Volume of cylinder A = πr²h
Volume of cylinder B = π(2r)² (\(\frac {h}{2}\))
= π × 4r² × \(\frac {h}{2}\)
= 2πr²h
∴ Volume of cylinder A : Volume of cylinder B = πr²h : 2πr²h = 1 : 2
Hence, CSA of cylinder A : CSA of cylinder B = 1 : 1
Volume of cylinder A : Volume of cylinder B = 1 : 2.

Question 2.
The radii of two cylinders are in the ratio 2 : 3, and their heights are in the ratio 3 : 2. Find (a) the ratio of their volumes, and (b) the ratio of their curved surface areas.
Solution:
Let the radii of the two cylinders be 2x and 3x, and the heights be 3y and 2y.

(a) Ratio of volumes
Volume of cylinder = πr²h
Volume of cylinder 1 = π(2x)²(3y)
and Volume of cylinder 2 = π(3x)²(2y)
Now, ratio of their volumes = \(\frac{V_1}{V_2}\)
= \(\frac{\pi(2 x)^2(3 y)}{\pi(3 x)^2(2 y)}\)
= \(\frac {12}{18}\)
= \(\frac {2}{3}\)
So V₁ : V₂ = 2 : 3
∴ Ratio of their volumes is 2 : 3.

(b) Ratio of curved surface areas
CSA of cylinder = 2πrh
CSA of cylinder 1 = 2π(2x)(3y)
and CSA of cylinder 2 = 2π(3x)(2y)
∴ \(\frac{\mathrm{CSA}_1}{\mathrm{CSA}_2}=\frac{2 \pi(2 x)(3 y)}{2 \pi(3 x)(2 y)}=\frac{6 x y}{6 x y}=1\)
So CSA₁ : CSA₂ = 1 : 1
Therefore, the ratio of their curved surface areas is 1 : 1.

Question 3.
The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)?
Solution:

Here, the diameter of the base = edge of the cube = r,
so radius = \(\frac {r}{2}\) cm
and height of the cylinder = edge of the cube = r cm
Volume, V = π × (radius)² × height
Now, V = \(\frac{22}{7} \times\left(\frac{r}{2}\right)^2 \times r=\frac{22}{7} \times \frac{r^2}{4} \times r=\frac{11 r^3}{14} \mathrm{~cm}^3\)
Hence, the volume of the cylinder is \(\frac{11 r^3}{14}\) cm, where r is the cube’s edge.

Question 4.
The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x.
Solution:
Let the original radius be r, and the original height be h.
Original volume = πr²h
When the radius is increased by 10%
New radius = r + 10% of r
= r + \(\frac {10}{100}\) × r
= r + \(\frac {r}{10}\)
= \(\frac{10 r+r}{10}\)
= \(\frac {11}{10}\)r
= 1.1r
and when height is decreased by x%
New height = h – x% of h
= h – \(\frac {x}{100}\) × h
= h – \(\frac {xh}{100}\)
= h(1 – \(\frac {x}{100}\))
Then volume of new cylinder = πr²h = \(\pi \times(1.1 r)^2 \times\left(1-\frac{x}{100}\right) h\)
The volume is unchanged, so the volume of the new cylinder = the volume of the original cylinder.

Hence, the value of x is approx. 17.36.

Question 5.
A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find:
(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 ≈ \(\frac {22}{7}\))
Solution:
(i) Volume of the cube
Volume of a cube = (side)³
Vcube = 12³
= 12 × 12 × 12
= 1728 cm³

(ii) Volume of one rod (r = 2 cm, h = 12 cm)
Volume of a cylinder = πr²h
V_rod = πr²h
= \(\frac {22}{7}\) × 2² × 12
= \(\frac {22}{7}\) × 4 × 12
= \(\frac {22}{7}\) × 48
= \(\frac {1056}{7}\)
= 150\(\frac {6}{7}\) cm³
= 150.85 cm³

(iii) Number of rods
When metal is melted and recast, the volume does not change.
Number of rods = \(\frac{\text { Volume of cube }}{\text { Volume of one rod }}\)
= \(\frac{12 \times 12 \times 12}{\pi \times 4 \times 12}\)
= \(\frac{12 \times 3 \times 7}{22}\)
≈ 11.4
Since we can make only complete rods, we take the integer part; hence the number of rods formed = 11.
Another way to calculate:
Number of rods = \(\frac{\text { Volume of cube }}{\text { Volume of one rod }}=\frac{1728 \mathrm{~cm}^3}{150.85 \mathrm{~cm}^3}=\frac{1728}{150.85}\) = 11.4
But the calculation given above is easy.

Ex 14.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 14.3 Solutions

Exercise 14.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 14.3 Solutions

For π, use one of the following approximations: π ≈ \(\frac {22}{7}\) or π ≈ 3.14

Question 1.
Find the total surface area of a cone if its slant height is 21 m and the diameter of its base is 24 m.
Solution:
Given: Slant height (l) = 21 m
radius (r) = \(\frac {24}{2}\) = 12 m

TSA of cone = πr(l + r)
= \(\frac {22}{7}\) × 12(21 + 12)
= \(\frac {8712}{7}\)
= 1244.57 m²
The total surface area is about 1244.57 m² (using π = 3.14 it is 1243.44 m²).

Question 2.
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution:
Given: r = 7, l = 10 cm
CSA of right circular cone = πrl
= \(\frac {22}{7}\) × 7 × 10
= 220 cm2
The curved surface area is 220 cm2.

Question 3.
The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone.
Solution:
Given: h = 16 cm, r = 12 cm.
Now, the slant height (l):
l² = h² + r²
= 16² + 12²
= 256 + 144
= 400
l = √400 = 20 cm
Using π = 3.14
CSA of a cone = πrl
= 3.14 × 12 × 20
= 240π
= 753.6 cm²
TSA of a cone = πr(l + r)
= 3.14 × 12 × (20 + 12)
= 1205.76 cm²
Therefore, CSA = 753.6 cm²
TSA = 1205.76 cm².

Question 4.
A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
Solution:
Given: h = 15 cm, V = 1570 cm³
Take π = 3.14

Volume of the cone = \(\frac{1}{3} \pi r^2 h\)
⇒ 1570 = \(\frac {1}{3}\) × 3.14 × r² × 15
⇒ 1570 = 15.7r²
⇒ \(\frac {1570}{15.7}\) = r²
⇒ 100 = r²
⇒ r = √100 = 10
∴ The radius of the base is 10 cm.

Question 5.
The curved surface area of a cone is 308 cm², and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
Solution:
Given: CSA of the cone = 308 cm²
Slant height (l) = 14 cm
(i) Radius of the base
CSA of the cone, πrl = 308 cm²
⇒ \(\frac {22}{7}\) × r × 14 = 308
⇒ 44r = 308
⇒ r = \(\frac {308}{44}\) = 7 cm
Hence, the radius of the cone is 7 cm.

(ii) Total surface area of the cone.
TSA of the cone = CSA + πr²
= 308 + \(\frac {22}{7}\) × 7²
= 308 + 154
= 462 cm²
So, the total surface area of the cone is 462 cm².

Question 6.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Given: Radius (r) = 7 cm, height (h) = 24 cm
Now, slant height (l):
l² = h² + r²
= 24² + 7²
= 576 + 49
= 625
l = √625 = 25 cm
A cap is open at the bottom, so only the curved surface is made from the sheet.
Sheet required for one cap (CSA) = πrl
= \(\frac {22}{7}\) × 7 × 25
= 550 cm²
Sheet required for 10 caps = 10 × 550 = 5500 cm²
5500 cm² of sheet is required to make 10 caps.

Question 7.
What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
Solution:
Given: Height (h) = 8 m, radius (r) = 6 m
Now, slant height (l):
l² = h² + r²
= 8² + 6²
= 64 + 36
= 100
l = √100 = 10 m
The tarpaulin forms the curved surface of the tent (the floor is not curved).
CSA of the conical tent = πrl
= 3.14 × 6 × 10
= 188.4 m²
The tarpaulin is 3 cm wide
Area of the tarpaulin = width × length
So, length needed for the tent = \(\frac{\text { area }}{\text { width }}\)
= \(\frac {188.4}{3}\)
= 62.8
Adding the extra 20 cm = 0.2 m for stitching margins and wastage,
Total length = 62.8 + 0.2 = 63 m
63 m of tarpaulin is required. (With π = \(\frac {22}{7}\) the length is about 63.06 m).

Question 8.
A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
Solution:
Since 6² + 8² = 36 + 64 = 100 = 10²
the triangle is right-angled with perpendicular sides 6 cm and 8 cm.

Rotating about the 8 cm side gives a right circular cone with height h = 8 cm, radius r = 6 cm, slant height l = 10 cm.
Now, volume of the cone, V = \(\frac{1}{3} \pi r^2 h\)
= \(\frac {1}{3}\) × 3.14 × 6² × 8
= 3.14 × 96
= 301.44 cm³
and the curved surface area of the cone
CSA = πrl
= 3.14 × 6 × 10
= 60π
= 188.4 cm²
Therefore, volume = 96π = 301.44 cm³
curved surface area = 60π = 188.4 cm².

Question 9.
Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone.

What fraction of the volume of the cup is occupied by the water?
Solution:
Let the cup (a cone standing on its vertex O) have radius r at the top and depth (height) h.

The water also takes the shape of a cone with the same vertex O and height \(\frac {h}{2}\).
In the cross-section, triangles OPQ and OAB are similar (PQ || AB and they share the angle at O).
So, \(\frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{\mathrm{OP}}{\mathrm{OA}}=\frac{h / 2}{h}=\frac{1}{2}\)
Hence, the radius of the water surface is \(\frac {r}{2}\).
Volume of the cup = \(\frac{1}{3} \pi r^2 h\)
Volume of water = \(\frac{1}{3} \pi\left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right)\)
= \(\frac{1}{3} \pi \times \frac{r^2}{4} \times \frac{h}{2}\)
= \(\frac{1}{8} \times \frac{1}{3} \pi r^2 h\)
\(\frac{\text { Volume of water }}{\text { Volume of cup }}=\frac{1}{8}\)
So, the fraction of the volume of the cup occupied by the water is \(\frac {1}{8}\) of the cup’s volume.

Ex 14.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 14.4 Solutions

Exercise 14.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 14.4 Solutions

Question 1.
A ball bearing has a radius of 0.7 cm. Find its surface area.
Solution:
Given: Radius (r) = 0.7 cm.
Surface area of the bearing ball = 4πr²
= 4 × \(\frac {22}{7}\) × 0.7 × 0.7
= 4 × 22 × 0.1 × 0.7
= 6.16 cm²
∴ The surface area of the ball bearing is 6.16 cm².

Question 2.
Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Solution:
Given: Weight of the larger sphere = 5920 g
and the weight of the smaller sphere = 740 g
Radius of the smaller sphere r = \(\frac {5}{2}\) = 2.5 cm
Let the radius of the larger sphere be R.
Both spheres are made of the same metal, so their weights are in the same ratio as their volumes.
∴ \(\frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3}=\frac{5920}{740}\)
⇒ \(\frac{R^3}{r^3}\) = 8
⇒ \(\frac{R}{r}=\sqrt[3]{8}\) = 2
⇒ R = 2r = 2 × 2.5 = 5 cm [∵ r = 2.5 cm]
∴ The radius of the larger sphere is 5 cm.

Question 3.
The diameter of the moon is approximately one fourth the diameter of the Earth. Given that the moon and Earth are both roughly spherical, find the ratio of their surface areas.
Solution:
The diameter of the Moon is approximately one-fourth of the diameter of the Earth.
Since the Moon and Earth are spherical in shape, the surface area of a sphere of radius r = 4πr²
Let the radius of the Earth be R, and the radius of the Moon be r.
Diameter of the moon ≈ \(\frac {1}{4}\) × diameter of the earth (Given)
Thus, the radius of the moon = \(\frac {1}{4}\) × radius of the earth
[Since radius = 2 × Diameter]
⇒ r = \(\frac {1}{4}\) × R
⇒ \(\frac{r}{\mathrm{R}}=\frac{1}{4}\) …….(1)
Now, the surface area of the moon = 4πr²
and the surface area of the Earth = 4πR²
Ratio of their surface areas = \(\frac{4 \pi r^2}{4 \pi R^2}\)
= \(\frac{r^2}{R^2}\)
= \(\left(\frac{r}{R}\right)^2\)
= \(\left(\frac{1}{4}\right)^2\)
= \(\frac {1}{16}\) [From equation (1)]
∴ The ratio of their surface areas = 1 : 16.

Question 4.
Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Solution:
Given: Radius (r) = 21 cm
(i) Curved surface area (CSA) of a hemisphere = 2πr²
= 2 × \(\frac {22}{7}\) × 21 × 21
= 2 × 22 × 3 × 21
= 2772 cm²

(ii) Total surface area (TSA) of a hemisphere = 2πr² + πr²
= 3πr²
= 3 × \(\frac {22}{7}\) × 21 × 21
= 3 × 22 × 3 × 21
= 4158 cm²

Question 5.
Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
Solution:
Given: Radius of each metal sphere = 2 cm
Dimension of cuboid = 16 cm × 8 cm × 8 cm
Each sphere has a diameter of 4 cm, so the box holds spheres exactly
Volume of the box = lbh
= 16 × 8 × 8
= 1024 cm³
Volume of 16 spheres = \(16 \times \frac{4}{3} \pi r^3\)
= \(16 \times \frac{4}{3} \pi \times 2^3\)
= \(\frac{512}{3} \pi\)
= \(\frac {512}{3}\) × 3.1416
≈ 536.17 cm³
Volume of liquid = volume of box – volume of spheres
= 1024 – 536.17
= 487.83 cm³
Rounding to the nearest integer, the volume of the liquid is 488 cm³.
(The answer is also 488 with π = 3.14 or π = \(\frac {22}{7}\))

Question 6.
The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Solution:
Let the original radius be r.
Original volume, V = \(\frac{4}{3} \pi r^3\)
New radius = r + \(\frac {10}{100}\)r = 1.1r
New volume = \(\frac{4}{3} \pi(1.1 r)^3\)
= \(1.331 \times \frac{4}{3} \pi r^3\)
= 1.331 V
Percentage increase = \(\frac{\text { New volume }- \text { Original volume }}{\text { Original volume }} \times 100 \%\)
= \(\frac{1.331 V-V}{V} \times 100 \%\)
= 0.331 × 100%
= 33.1%
Hence, the volume increases by about 33.1%.
(Volume depends on the cube of the radius, so a 10% increase in radius gives more than a 33% increase in volume).

Question 7.
The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
Solution:
Given: The volume of the sphere increased by 72.8%.
Let the original radius be r.
New radius = \(\left(1+\frac{x}{100}\right) r\)
New volume = Original volume + increased volume
Let V be the original volume
So, new volume = V + \(\frac {72.8}{100}\) × V
= \(\left(\frac{100+72.8}{100}\right) V\)
= \(\frac{V(172.8)}{100}\)
= 1.728 V
∴ New volume = 1.728 × original volume
⇒ \(\frac{4}{3} \pi\left(1+\frac{x}{100}\right)^3 r^3=1.728 \times \frac{4}{3} \pi r^3\)
⇒ \(\left(1+\frac{x}{100}\right)^3\) = 1.728
⇒ \(1+\frac{x}{100}=\sqrt[3]{1.728}\) = 1.2 (since 1.2³ = 1.728)
⇒ \(\frac {x}{100}\) = 1.2 – 1
⇒ \(\frac {x}{100}\) = 0.2
⇒ x = 20
∴ x = 20, i.e., the radius is increased by 20%.

Question 8.
The hemispherical dome of a building needs to be painted (see Fig. below). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹ 10 per 100 cm².

Solution:
Given: Circumference of the base of the dome = 35.2 m
Cost of painting = ₹ 10 per 100 cm².
Let the radius of the base of the dome be r.
Since the circumference of the dome = 35.2 m
⇒ 2πr = 35.2
⇒ 2 × \(\frac {22}{7}\) × r = 35.2
⇒ r = \(\frac{35.2 \times 7}{44}\) = 5.6 m
So, the radius of the dome is 5.6 m.
The curved surface area (CSA) of a dome (hemisphere) is 2πr².
That needs to be painted.
∴ CSA = 2πr²
= 2 × \(\frac {22}{7}\) × 5.6 × 5.6
= 2 × 22 × 0.8 × 5.6
= 197.12 m²
Change to cm²:
1 m² = 100 × 100 cm² = 10000 cm²
So CSA = 197.12 m²
= 197.12 × 10000 cm²
= 1971200 cm²
Number of 100 cm² units = \(\frac {1971200}{100}\) = 19712
The cost is ₹ 10 for every 100 cm²
Cost = ₹ 19712 × 10 = ₹ 197120
So, the cost of painting the dome is ₹ 1,97,120.

Ganita Manjari Class 9 Maths Chapter 14 End of Chapter Exercise Solutions

Math of Space Surface Area and Volume End of Chapter Exercise Solutions

For guesstimate problems, make a guess first before solving.

Question 1.
Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.

Solution:
Given: Volume of ice cream in the container = 10 × 15 × 20 = 3000 cm³
Assumption: A scoop of ice cream is roughly a sphere of diameter 5 cm, i.e., radius 2.5 cm.
∴ Volume of one scoop = \(\frac{4}{3} \pi r^3\)
= \(\frac {4}{3}\) × 3.14 × 2.5³
= \(\frac {4}{3}\) × 3.14 × 15.625
= 65.4 cm³
Number of scoops = \(\frac{\text { Volume of ice cream container }}{\text { Volume of one scoop }}\)
= \(\frac {3000}{65.4}\)
≈ 45.9
About 45 full scoops can be obtained. (With a smaller scoop of radius 2 cm, the volume is about 33.5 cm3 and we get about 89 scoops; the estimate depends on the size of the scoop assumed.)

Question 2.
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 cube of side a contains a³ unit cubes, and the cube of side a + 1 contains (a + 1)³ unit cubes.
Number of unit cubes = cube of side (a + 1) – cube of side a
= (a + 1)³ – a³
= (a + 1) (a + 1) (a + 1) – a³
= (a² + 2a + 1) (a + 1) – a³
= a(a² + 2a + 1) + 1(a² + 2a + 1) – a³
= a³ + 2a² + a + a² + 2a + 1 – a³
= a³ + 3a² + 3a + 1 – a³
= 3a² + 3a + 1
which is the required expression.

Question 3.
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) Water a person drinks in a lifetime vs. volume of a classroom
Guess: Probably not
Assumptions : Classroom 8 m × 6 m × 3.5 m
a person drinks about 2.5 litres of water a day and lives for about 80 years.
Volume of the classroom = l × b × h
= 8 × 6 × 3.5
= 168 m³
Since 1 m³ = 1000 litres, the classroom holds 168 × 1000 = 168000 litres.
Water drunk in a lifetime ≈ 2.5 × 365 × 80
= 73000 litres
= \(\frac {73000}{168000}\)
≈ 0.43
No. A person drinks only about 73,000 litres in a lifetime, which fills less than half (about 43%) of such a classroom.
Even drinking 5 litres a day would give only about 1,46,000 litres.

(ii) Number of bricks in the walls of the classroom
Guess: Several thousand bricks.
Assumptions: Classroom 8 m × 6 m with walls 3.5 m high and 20 cm thick;
one door 1 m × 2.1 m and three windows 1.5 m × 1.2 m;
a standard brick with its share of mortar occupies 20 cm × 10 cm × 10 cm.
Area of the four walls = 2(8 + 6) × 3.5 = 98 m³
Area of door and windows = 1 × 2.1 + 3 × (1.5 × 1.2)
= 2.1 + 5.4
= 7.5 m²
Area of brickwork = 98 – 7.5 = 90.5 m²
Volume of brickwork = 90.5 × 0.2 = 18.1 m³
Volume of one brick with mortar = 0.2 × 0.1 × 0.1 = 0.002 m³
Number of bricks = \(\frac {18.1}{0.002}\) = 9050
About 9,000 bricks (a little less, because the corners are shared between two walls).

Question 4.
You are given two options: (A) one chocolate cube of side 50 mm, (B) one hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?
Solution:
Option A: Cube of one chocolate cube
50³ = 50 × 50 × 50 = 125000 mm³
Option B: Cube of 100 chocolate cubes
100 × 10³ = 100 × 1000 = 100000 mm³
Since 125000 > 100000, option A has more chocolate.
In fact, one 50 mm cube is equal to \(\frac {125000}{1000}\) = 125 small cubes of side 10 mm.
Option (A), the single cube of side 50 mm, gives more chocolate (25% more).

Question 5.
If all the people in the world are crowded together in one location, how much area would it cover?

Solution:
Guess: A few thousand square kilometres.
Assumptions: The world population is about 8 billion = 8 × 109 people
In a tightly packed standing crowd, each person occupies about 0.5 m × 0.5 m = 0.25 m².
Area ≈ 8 × 10⁹ × 0.25 m² = 2 × 10⁹ m²
Since 1 km² = 1000 × 1000 = 10⁶ m²
2 × 10⁹ m² = \(\frac{2 \times 10^9}{10^6}\) km² = 2000 km²
This is a square with a side of about √2000 ≈ 45 km.
About 2000 km² — roughly 1.35 times the area of Delhi (about 1,480 km²).
All of humanity could stand together in a square about 45 km on each side.

Question 6.
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:
We have, radius of a glass r = \(\frac {7}{2}\) = 3.5 cm
Height of milk filled, h = 12 cm
Milk in one glass = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 12
= 22 × 0.5 × 3.5 × 12
= 462 cm³
Milk for 1600 students = 1600 × 462 cm³
= 739200 cm³
= \(\frac {739200}{1000}\) litres
= 739.2 litres
Therefore, 739.2 litres of milk are needed.

Question 7.
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 sphere (\(r_l^3\)) = 5 cm
Radius of cone (\(r_c^3\)) = 4 cm
CSA of sphere = 5 × CSA of cone.
Surface area of the sphere = \(4 \pi r_c^3\)
= 4π × 5²
= 100π cm²
Let the slant height of the cone be l.
Its curved surface area = π × r × l = 4πl
Now, CSA of sphere = 5 × CSA of cone
⇒ 100π = 5 × 4πl
⇒ 100π = 20πl
⇒ l = 5 cm
Height of the cone:
l² = r² + h²
⇒ h² = l² – r²
⇒ h = \(\sqrt{l^2-r^2}\)
⇒ h = \(\sqrt{5^2-4^2}\)
⇒ h = \(\sqrt{25-16}\)
⇒ h = √9
⇒ h = 3 cm
Volume of the cone:
V = \(\frac{1}{3} \pi r^2 h\)
= \(\frac {1}{3}\) × π × 4² × 3
= 16π
≈ 16 × \(\frac {22}{7}\)
≈ 50.29 cm³
Hence, height = 3 cm; volume = 16π ≈ 50.29 cm³.

Question 8.
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 6,95,700 km. 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.
(A calculator can be used.)
Solution:
For two spheres of radii R₁ and R₂,
\(\frac{V_1}{V_2}=\frac{\frac{4}{3} \pi R_1^3}{\frac{4}{3} \pi R_2^3}\) = \(\left(\frac{R_1}{R_2}\right)^3\)
(i) Jupiter : Earth
\(\frac{V_J}{V_E}=\left(\frac{69900}{6370}\right)^3\)
≈ (10.973)³
≈ 1321

(ii) Sun : Earth
\(\frac{V_S}{V_E}=\left(\frac{695700}{6370}\right)^3\)
≈ (109.22)³
≈ 1302170.6
≈ 13 lakh

Question 9.
Show that the volume of a sphere is equal to \(\frac {2}{3}\) of the volume of the smallest cylinder which encloses it.
Solution:
Let the radius of the sphere be r.
The smallest cylinder that encloses the sphere touches it all around and at the top and bottom,
so the radius of the cylinder = r,
height of the cylinder = 2r (the diameter of the sphere).
Volume of the cylinder = πr² × 2r = 2πr³
∴ \(\frac{\text { Volume of sphere }}{\text { Volume of cylinder }}=\frac{\frac{4}{3} \pi r^3}{2 \pi r^3}\)
⇒ \(\frac{\text { Volume of the sphere }}{\text { Volume of cylinder }}=\frac{4}{3} \times \frac{1}{2}=\frac{2}{3}\)
Hence, the volume of the sphere = \(\frac {2}{3}\) × volume of the smallest cylinder that encloses it.
Hence shown.

Question 10.
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 let the second string be x metres above the ground everywhere.
Then the second string is a circle of radius R + x.
Length of the second string – length of the first string = 1
⇒ 2π(R + x) – 2πR = 1
⇒ 2πR + 2πx – 2πR = 1
⇒ 2πx = 1
⇒ x = \(\frac{1}{2 \pi}\)
⇒ x = \(\frac{1}{2 \times 3.141}\)
⇒ x = \(\frac {1}{6.282}\)
⇒ x = 0.159 m
So the second string is about 15.9 cm above the ground.
The value of R is cancelled out, so the answer does not depend on the size of the sphere:

Sphere	Height of the second string
Earth	\(\frac{1}{2 \pi}\) m = 15.9 cm
(i) the Moon	= 15.9 cm
(ii) Jupiter	= 15.9 cm
(iii) a volleyball	= 15.9 cm

Question 11.
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.
Solution:
The corners A, B, C, D of the square base lie on the circle of radius r, so the diagonal AC of the square is a diameter of the circle:
AC = 2r
If the side of the square is s, then by the Baudhaayana-Pythagoras theorem in triangle ABC,
⇒ s² + s² = (2r)²
⇒ 2s² = 4r²
⇒ s² = 2r²
So the area of the square base is 2r² cm².
The apex is on the top of the cylinder, so the height of the pyramid is h.
Volume of pyramid = \(\frac {1}{3}\) × area of base × height
= \(\frac {1}{3}\) × 2r² × h
= \(\frac{2 r^2 h}{3}\)
Volume of cylinder = πr²h
∴ \(\frac{\text { Volume of pyramid }}{\text { Volume of cylinder }}=\frac{\frac{2 r^2 h}{3}}{\pi r^2 h}=\frac{2}{3 \pi}\)
The ratio is 2 : 3π, which is about 0.21 (the pyramid takes up about 21% of the cylinder).

Question 12.
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) Cuboid, length is increased by 1 unit
The volume of a cuboid = lwh
New volume after 1 unit increase in length = (l + 1)wh
Change in volume = New volume – Old volume
= (l + 1)wh – lwh
= lwh + wh – lwh
= wh
The volume increases by wh cubic units (one extra slab of thickness 1 unit on the face of area wh).
∴ The correct option is (c).

(ii) Cylinder, radius is increased by 1 unit
Volume of cylinder = πr²h
New volume = π(r + 1)²h
Now, change in volume = New volume – Old volume
= πh(r + 1)² – πr²h
= πh[(r + 1)² – r²]
= πh[(r² + 2r + 1 – r²]
= πh(2r + 1)
= 2πrh + πh
The volume increases by (2πrh + πh) cubic units.
∴ The correct option is (e).

(iii) Sphere: radius is decreased by 1 unit
Volume of sphere = \(\frac{4}{3} \pi r^3\)
New volume = \(\frac{4}{3} \pi(r-1)^3\)
Now, change in volume = Old volume – New volume.

The volume decreases by \(\left(4 \pi r^2-4 \pi r+\frac{4}{3} \pi\right)\) cubic units, i.e., \(\frac{4}{3} \pi\left(3 r^2-3 r+1\right)\) cubic units.

Question 13.
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.
Then the volume of the cube is V = a³
⇒ a = \(\sqrt[3]{V}=V^{\frac{1}{3}}\)
Now, the total surface area of the cube
S = 6a² [∵ a = \(V^{\frac{1}{3}}\)]
= \(6\left(V^{\frac{1}{3}}\right)^2\)
= \(6 V^{\frac{2}{3}}\)
So, total surface area, S = \(6 V^{\frac{2}{3}}=6 \sqrt[3]{V^2}\)

Question 14.
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.
Then the total surface area, S = 6a²
⇒ a² = \(\frac {S}{6}\)
⇒ a = \(\sqrt{\frac{S}{6}}\)
Now, the volume of the cube, V = a³
= \(\left(\sqrt{\frac{S}{6}}\right)^3\) [∵ a = \(\sqrt{\frac{S}{6}}\)]
= \(\left(\frac{S}{6}\right)^{\frac{3}{2}}\)
So, volume, V = \(\left(\frac{S}{6}\right)^{\frac{3}{2}}=\frac{S}{6} \sqrt{\frac{S}{6}}=\frac{S \sqrt{S}}{6 \sqrt{6}}\)

Question 15.
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 glass = 4 cm
Radius of marbles = 1 cm
The marbles sink and push the water up.
The volume of the marbles dropped equals the volume of the rise in water.
Rise needed = 20 – 16 = 4 cm
(The glass is 25 cm tall, so the water does not overflow).
Volume of the rise = πr²h
= π × 4² × 4
= 64π cm³
Volume of one marble = \(\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times 1^3=\frac{4}{3} \pi \mathrm{~cm}^3\)
Number of marbles = \(\frac{64 \pi}{\frac{4}{3} \pi}\) = 48
Therefore, the crow should drop 48 marbles.

Question 16.
Find the volume of ink in a new ballpoint pen. Take the necessary measurements and make approximations, as needed.
Solution:
Guess: Less than one millilitre.
Measurements (sample): The ink is in the refill, a thin cylindrical tube.
In a new pen, the ink column is about 9 cm long, and the inner diameter of the tube is about 2 mm,
so the inner radius is r = 1 mm = 0.1 cm.
Volume, V = πr²h
= 3.14 × 0.1² × 9
= 3.14 × 0.01 × 9
= 0.28 cm³
Since 1 cm³ = 1 millilitre, this is about 0.28 mL.
A new ballpoint pen holds roughly 0.3 cm³ (about a quarter of a millilitre) of ink.
Our answer will depend on the pen we measure.

Question 17.
Solve:
(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.

Solution:
(i) Number of chapatis
Guess: Approximately 18 chapatis
Volume of dough = \(\frac{4}{3} \pi r^3\)
= \(\frac{4}{3} \pi \times 6^3\)
= 288π cm³
≈ 904.3 cm³
Assumption: A rolled chapati is a thin cylinder (disc) of diameter 16 cm (radius 8 cm) and thickness 2 mm = 0.2 cm.
Volume of dough in one chapati = πr²h
= π × 8² × 0.2
= 12.8π cm³
≈ 40.2 cm³
Number of chapatis = \(\frac{288 \pi}{12.8 \pi}\) = 22.5
About 22 chapatis can be made.

(ii) Volume of edible coconut flesh/muskmelon fruit
Coconut (sample measurements after cutting it into two halves): Outer radius of the shell is about 6 cm, shell thickness is about 0.4 cm, thickness of the white flesh is about 1 cm.
The flesh is a thick spherical shell between radius (R) = 6 – 0.4 = 5.6 cm
and radius (r) = 5.6 – 1 = 4.6 cm.
Volume = \(\frac{4}{3} \pi \mathrm{R}^3-\frac{4}{3} \pi r^3\)
= \(\frac{4}{3} \pi\left(\mathrm{R}^3-r^3\right)\)
= \(\frac{4}{3} \pi\left(5.6^3-4.6^3\right)\)
= \(\frac {4}{3}\) × 3.14 × (175.616 – 97.336)
= \(\frac {4}{3}\) × 3.14 × 78.28
≈ 328 cm³
Each half has about 164 cm3 of flesh, about 328 cm³ in all.
Muskmelon (sample measurements): Radius (R) is about 7 cm, rind about 0.5 cm thick, seed cavity (hollow middle) of radius (r) is about 3 cm.
Volume, V = \(\frac{4}{3} \pi \mathrm{R}^3-\frac{4}{3} \pi r^3\)
= \(\frac{4}{3} \pi\left(\mathrm{R}^3-r^3\right)\)
= \(\frac{4}{3} \pi\left(6.5^3-3^3\right)\)
= \(\frac {4}{3}\) × 3.14 × (274.625 – 27)
≈ 1037 cm³
Roughly 330 cm³of coconut flesh, or about 1000 cm³ (1 litre) of muskmelon fruit.
The answers depend on the fruit we measure.

Question 18.
Looking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders:
(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:
Assumptions: The book has about 160 pages, i.e., 80 sheets (leaves);
each page measures about 21 cm × 27.5 cm; the closed book is about 1 cm thick.
The classroom floor is about 8 m × 6 m.
(i) Pages laid out on the floor
Area of one page = 21 × 27.5
= 577.5 cm²
= 0.05775 m²
If the 80 sheets are laid side-by-side,
Area covered = 80 × 0.05775 = 4.6 m²
Even if each of the 160 pages were laid separately (for example, as photocopies), the area would only be about 9.2 m².
Area of classroom floor = 8 × 6 = 48 m²
No. The pages would cover only about 4.6 m² (at most about 9.2 m²), which is less than one-fifth of the classroom floor.

(ii) Maximum number of textbooks in the storeroom
We know that 1 ft = 30.48 cm
So, 15 ft = 457.2 cm, 20 ft = 609.6 cm, 30 ft = 914.4 cm
Upper bound using volumes:
\(\frac{\text { volume of storeroom }}{\text { volume of a book }} \approx \frac{457.2 \times 609.6 \times 914.4}{21 \times 27.5 \times 1}\)
≈ \(\frac {254850000}{577.5}\)
≈ 441300
Actual packing: Books cannot be cut, so count how many fit along each direction.
The best arrangement found is: thickness (1 cm) along the 15 ft side, 21 cm along the 20 ft side, 27.5 cm along the 30 ft side.
Along 15 ft: \(\frac {457.2}{1}\) → 457 books
Along 20 ft: \(\frac {609.6}{21}\) ≈ 29.03 → 29 books
Along 30 ft: \(\frac {914.4}{27.5}\) ≈ 33.25 → 33 books
Number of textbooks = 457 × 29 × 33 = 437349
About 4.37 lakh textbooks (4,37,349 in this arrangement; the volume of upper bound is about 4.41 lakh).

Question 19.
If the entire human population decided to climb into one giant cube, how long would the side have to be?

Solution:
Guess: About a kilometre.
Assumptions: Population about 8 × 109 people.
The average human body has a volume of about 0.065 m³ (a mass of about 65 kg with a density close to that of water, 1000 kg/m³).
If people could be packed with no gaps at all:
Total volume = 8 × 109 × 0.065 = 5.2 × 108 m³
Side = \(\sqrt[3]{5.2 \times 10^8}\) ≈ 804 m
More realistically, people stand in layers: each person needs about 0.5 m × 0.5 m of floor and a layer about 2 m high, i.e., 0.5 m³ of space.
Total volume = 8 × 10⁹ × 0.5 = 4 × 10⁹ m³
Side = \(\sqrt[3]{4 \times 10^9}\) ≈ 1587 m ≈ 1.6 km
The cube would need a side of roughly 0.8 km (tightly packed) to about 1.6 km (people standing in floors) – only about a kilometre or so!

Question 20.
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 the Earth’s surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth’s radius is ~ 6371 km.
(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.
Solution:
Let R = 6371 km be the radius of the Earth,
V = 1.38 × 10⁹ km³ the volume of water,
and t be the thickness of the layer.
(i) Expression for the thickness
The water fills the space between two spheres, of radii R and R + t:

Because the layer is very thin compared with the Earth, the water is almost like a flat sheet spread over the surface area 4πR².
So, approximately,
volume ≈ surface area × thickness
⇒ t ≈ \(\frac{V}{4 \pi R^2}\)

(ii) Using a calculator
4πR² = 4 × 3.1416 × 6371²
≈ 4 × 3.1416 × 40589641
≈ 510064000 km² (≈ 5.1 × 108 km²)
t ≈ \(\frac{1.38 \times 10^9}{5.1006 \times 10^8}\) ≈ 2.71 km
(The exact expression gives t ≈ 2.70 km, almost the same.)
The water layer would be about 2.7 km thick.

Question 21.
Give the dimensions of a cuboid whose volume is halved when its surface area is doubled.
Solution:
1. The Algebraic Conditions
Let the original dimensions of the cuboid be x, y, and z.
Original Volume (V): xyz
Original Surface Area (S): 2(xy + yz + zx)
If we transform this cuboid into a new one with dimensions x’, y’, and z’ such that its volume is halved and its surface area is doubled, the new dimensions must satisfy two conditions:
(i) Volume Condition:
x’y’z’ = \(\frac {1}{2}\)xyz
(ii) Surface Area Condition:
2(x’y’ + y’z’ + z’x’) = 4(xy + yz + zx)
or x’y’ + y’z’ + z’x’ = 2(xy + yz + zx)
Since we have 3 unknown dimensions (x’, y’, z’) but only 2 equations, infinitely many combinations of dimensions can satisfy these conditions depending on how the individual sides are altered.

2. Why Uniform Scaling Fails
If we attempt to change all dimensions of the cuboid by the same scale factor k:
The new volume scales by k³, meaning
k³ = \(\frac {1}{2}\)
⇒ k = \(\left(\frac{1}{2}\right)^{1 / 3}\) ≈ 0.7937
The new surface area scales by k², meaning
k² = 2
⇒ k = √2 ≈ 1.4142
Because \(\left(\frac{1}{2}\right)^{1 / 3} \neq \sqrt{2}\), uniform scaling is impossible.
We cannot shrink or stretch all sides uniformly to meet both criteria simultaneously.

3. How It Can Be Achieved (Non-Uniform Scaling)
To satisfy the conditions, the dimensions must be altered non-uniformly (e.g., shrinking one or two dimensions significantly while stretching or altering the others).
For example, if we keep one dimension constant or scale them independently (k₁, k₂, k₃), we can find valid proportional relationships that satisfy both equations, meaning the new dimensions depend entirely on the specific starting proportions of the original cuboid.

Question 22.
Project: Find the volume of your house, making necessary approximations. Present how you solved it.
Solution:
Method: Model each room as a cuboid, measure (or estimate) its length, width, and height, find each volume with V = l × w × h, and add.
Ignore the thickness of walls and small items.
A sample for a two-bedroom flat with 3 m high ceilings:

Room	Length (m)	Width (m)	Height (m)	Volume (m3)
Living room	5	4	3	60
Bedroom 1	4	3.5	3	42
Bedroom 2	3.5	3	3	31.5
Kitchen	3	2.5	3	22.5
Two bathrooms	2	1.5	3	2 × 9 = 18
Passage	4	1	3	12
Total				186

Volume of the house = 60 + 42 + 31.5 + 22.5 + 18 + 12 = 186 m³
Since 1 m³ = 1000 litres, the house encloses about 1,86,000 litres of space.
Presentation: Draw a rough floor plan, write the measured dimensions of each room, show the calculation table as above, and state the approximations (rooms treated as cuboids, walls, cupboards, and the balcony ignored). Our own measurements will give a different answer.

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