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

Students often refer to Ganita Manjari Class 9 Solutions and Part 2 Class 9 Maths Chapter 14 Math of Space Surface Area and Volume Solutions to verify their answers.

Math of Space Surface Area and Volume Class 9 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:
The model shows a cuboid as a stack of thin rectangular layers placed one above another, similar to a pack of playing cards.
Each layer has a rectangular base with length (l) and width (w).
The area of one layer is l × w.
The number of layers depends on the height (h) of the cuboid.
Therefore, the total space occupied by the cuboid is the area of the base multiplied by the number of layers.
∴ Volume = Area of base × Height
V = (l × w) × h
Thus, the model explains that the volume of a cuboid is found by multiplying its length, width, and height.
A cube has all its sides equal. Let the side of the cube be a.
A cube has 6 square faces.
Area of one face = a × a = a²
Therefore, the surface area of the cube = 6 × a²
Volume of a cube = Length × Width × Height
Since all sides are equal
V = a × a × a = a³
Hence, the formulas are:
Total Surface Area of a cube = 6a²
Volume of a cube = a³

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:
To decide the suitable dimensions of the pool, we compare the volumes of both pools because volume represents the amount of water the pool can hold.
Option (A): Depth = 8 ft, Length = 50 ft, Width = 20 ft
Volume = l × w × 6 = 50 × 20 × 8 = 8000 ft³
Option (B): Depth = 6 ft, Length = 100 ft, Width = 15 ft
Volume = 100 × 15 × 6 = 9000 ft³
Since 9000 ft³ > 8000 ft³, option (B) can hold more water
than Option (A).
Also, while selecting the dimensions of a swimming pool for a school, we should consider the following:

• Safety: A lower depth is safer for students.
• Swimming space: A longer pool provides more area for swimming activities.
• Number of students using the pool: The pool should provide enough space for students.

So, option (B) (6 ft × 100 ft × 15 ft) is more suitable because it provides a larger swimming area and greater water capacity while having a safer depth. The volume, depth, safety, and usability of the pool are considered while making the choice.

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:
Volume of cube = 64 cm³
Let the side of the cube be a.
⇒ a³ = 64
⇒ a = \(\sqrt[3]{64}\) = 4 cm
Total surface area of the cube = 6a² = 6 × (4)² = 96 cm²
Thus, the total surface area of the cube is 96 cm².

Question 2.
How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Solution:
Side of cubical box = 2 m = 200 cm
Side of small cube = 20 cm
Number of small cubes along one edge = \(\frac{200}{20}\) = 10
Total number of cubes = 10 × 10 × 10 = 1000
Therefore, 1000 small cubes can be packed in the cubical box.

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:
Dimensions of godown: Length = 40 m, Width = 25 m, Height = 10 m .
Dimensions of each box: Length = 2 m, Width = 1.25 m, Height = 1 m
Volume of godown = 40 × 25 × 10 = 10000 m³
Volume of one box = 2 × 1.25 × 1 = 2.5 m³
Volume of one box = \(=\frac{\text { Volume of godown }}{\text { Volume of one box }}\)
= \(\frac{10000}{2.5}\) = 4000
Thus, the godown can contain 4000 boxes.

Question 4.
Two cubes each of volume 125 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 125 cm³
Let the side of each cube be a.
⇒ a³ = 125
⇒ a = 5
When two cubes are joined end to end:
Length = 5 + 5 = 10 cm, Width = 5 cm, Height = 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
Therefore, 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:
Side of original cube = 4 cm
Surface area of original cube: = 6a² = 6(4)² = 96 cm²
Number of small cubes = \(\frac{1}{2}\)
= \(\frac{\text { Volume of original cube }}{\text { Volume of a small cube }}\)
= 64
Surface area of one small cube = 6(1)² = 6 cm²
Total surface area of all small cubes = 64 × 6 = 384 cm²
Ratio = 96 : 384 = 1 : 4
Thus, the ratio of the surface areas of the original cube and all the small cubes is 1 : 4.

Question 6.
The surface areas of the three faces of a cuboid that meet at one of the comers of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
Solution:
Let the dimensions of the cuboid be I, w, and h.
Given: lw = 6, wh = 15, lh = 10
Multiplying all three equations, we get
(lw)(wh)(lh) = 6 × 15 × 10
⇒ l²w²h² = 900
⇒ (lwh)² = (30)²
⇒ lwh = 30
Thus, 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:
(i) Only comer cubes have all three faces painted.
A cube has 8 comers.
Therefore, 8 cubes of 1 cm3 have three faces painted.

(ii) Cubes on edges excluding comers have exactly two faces painted.
Number of edges = 12
Cubes on each edge excluding comers = 5 – 2 = 3
Cubes on edges excluding comers = 12 × 3 = 36
Therefore, 36 cubes of 1 cm³ have exactly two faces painted.

(iii) Cubes in the middle of each face have exactly one face painted.
For each face: (5 – 2)² = 3² = 9
For 6 faces: 6 × 9 = 54
Therefore, 54 cubes of 1 cm³ have exactly one face painted.

(iv) Cubes with no visible surface have no face painted.
Inner cubes = (5 – 2)³ = 3³ = 27
Therefore, 27 cubes have no face painted.

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 integer edge lengths of the cuboid be a, 6, and c in centimetres.
Total surface area of the cuboid = 2(ab + bc + ca) = 100
⇒ ab + bc + ca = 50
Adding a² to both sides of the equation, we get
a² + ab + ac + bc = a² + 50
⇒ a(a + 6) + c(a + b) = a² + 50
⇒ (a + b)(a + c) = a² + 50

Step 1: Determine the possible values of the shortest edge Let a represent the shortest edge of the cuboid.
If all edges were equal to 5 cm (a cube), the surface area would be 6 (5)² = 150 cm²
Since 150 cm² already exceeds the required 100 cm² surface area any cuboid with all edges 5 cm or larger will have a surface area much greater than 100 cm². Therefore, the shortest edge a can only be 1, 2, 3, or 4 because a cube of side 5 cm has a surface area 150 cm².

Step 2: Test each value of a using factor pairs
Case 1: a = 1
(1 +b)(1 + c) = 1² + 50 = 51
The factor pairs of 51 are:

• 1 × 51 ⇒ 1 + b = 1 ⇒ b = 0 (not a valid edge length)
• 3 × 17 ⇒ 1 + b = 3 and 1 + c = 17 ⇒ 6 = 2, c = 16

∴ First Solution: 1 cm × 2 cm × 16 cm

Case 2: a = 2
(2 + 6)(2 + c) = 22 + 50 = 54 The factor pairs of 54 are:

• 1 × 54 ⇒ 2 + 6 = 1 ⇒ 6 = -1 (not possible)
• 2 × 27 ⇒ 2 + 6 = 2 ⇒ 6 = 0 (not possible)
• 3 × 18 ⇒ 2 + 6 = 3 and 2 + c = 18 ⇒ 6 = 1, c = 16 (this is the same 1 × 2 × 16 cuboid)
• 6 × 9 ⇒ 2 + 6 = 6 and 2 + c = 9 ⇒ 6 = 4 and c = 7

Second Solution: 2 cm × 4 cm × 7 cm

Case 3: a = 3
(3 + 6)(3 + c) = 32 + 50 = 59
Since 59 is a prime number, its only factor pair is 1 × 59.
This gives 3 + 6 = 1 ⇒ 6 = -2 (not possible).
Hence, there are no integer solutions for a = 3.

Case 4: a = 4.
(4 + 6)(4 + c) = 42 + 50 = 66
The factor pairs of 66 are 1 × 66, 2 × 33, 3 × 22
all yield negative values for 6.
6 × 11 ⇒ 4 + 6 = 6 and 4 + c = 11 ⇒ 6 = 2, c = 7 (this is the same 2 × 4 × 7 cuboid).
(i) Yes, there are exactly 2 such cuboids.
(ii) The dimensions of the two cuboids are:
1 cm × 2 cm × 16 cm
2 cm × 4 cm × 7 cm
(iii) By expressing the equation as (a + b)(a + c) = a² + 50 and testing all factor pairs for every possible shortest side (a = 1, 2, 3, 4), all possible integer combinations were checked, confirming that no other integer solutions exist.

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 the radius and height of cylinder A be r and h, respectively.
Therefore, the radius of cylinder B = 2r and h
height of cylinder B = \(\frac{h}{2}\)
Curved surface area (CSA) of a cylinder = 2πrh
For cylinder A: CSA_A = 2πrh
For cylinder B: CSA_B = 2π(2r)\(\left(\frac{h}{2}\right)\) = 2πrh
Therefore, CSA_A : CSA_B = 2πrh : 2πrh = 1 : 1
Now, volume of a cylinder (V) = πr²h
For cylinder A: V_A = πr²h
For cylinder B : V_B = π(2r)²\(\left(\frac{h}{2}\right)\) = 2πr²h
Therefore, V_A : V_B = \(\frac{\pi r^2 h}{2 \pi r^2 h}\) = 1 : 2
Thus, the ratio of their curved surfaces = 1:1 and the ratio of their volumes =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 (r₁ and r₂) be 2x and 3x and their heights (h₁ and h₂) be 3y and 2y.
(a) Volume of a cylinder: (V) = πr²h
Therefore, \(\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\pi r_1^2 h_1}{\pi r_2^2 h_2}=\frac{r_1^2 h_1}{r_2^2 h_2}\)
= \(\frac{(2 x)^2(3 y)}{(3 x)^2(2 y)}=\frac{12}{18}=\frac{2}{3}\)
Therefore, the ratio of their volumes is 2 : 3.

(b) Curved surface area of cylinder (CSA) = 2πrh
Therefore, \(\frac{\mathrm{CSA}_1}{\mathrm{CSA}_2}=\frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}=\frac{r_1 h_1}{r_2 h_2}\)
= \(\frac{(2 x)(3 y)}{(3 x)(2 y)}=\frac{6}{6}\) = 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:
The edge of the cube = r cm
For the largest possible cylinder:
Let R and h be the radius and height of the cylinder respectively.
Diameter of the base of the cylinder = edge of the cube

Fleight of cylinder (h) = r
Volume of cylinder (V) = πR²h
V = π\(\frac{1}{2}\)(r) = π × \(\left(\frac{r^2}{4}\right)\) × r
= \(\frac{\pi r^3}{4}\) cm³
Therefore, the volume of the cylinder is \(\frac{\pi r^3}{4}\)cm³

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 and height of the cylinder be r and h, respectively.
Original volume (V) = πr²h
New radius = r + 10% of r = r + r × \(\left(\frac{10}{100}\right)=\frac{11 r}{10}\)
New height = h – x% of h = h – h × \(\left(\frac{x}{100}\right)=\left(\frac{100-x}{100}\right)\)h

Since volume remains unchanged

∴ πr²h = π\(\left(\frac{11 r}{10}\right)^2\left(\frac{100-x}{100}\right)\)h

⇒ 1 = \(\left(\frac{121}{100}\right) \times\left(\frac{100-x}{100}\right)\)

⇒ 1 = \(\frac{121(100-x)}{10000}\)
⇒ 10000 = 121(100 – x)
⇒ 100 – x = \(\frac{1}{2}\)
⇒ 100 – x ≈ 82.64
⇒ x ≈ 17.36
Therefore, the height must be decreased by approximately 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) Side of cube = 12 cm
Volume of cube (V) = side³ = (12)³ = 1728 cm³
Thus, the volume of the cube is 1728 cm³.

(ii) Radius of cylinder (r) is 2 cm and height (h) is 12 cm
Volume of a cylinder = πr²h = \(\frac{22}{7}\) × (2)² × 12
= \(\frac{1056}{7}\) cm
Thus, the volume of one rod 150.86 cm³.

(iii) Number of complete cylindrical rods
= \(\frac{\text { Volume of cube }}{\text { Volume of one rod }}\)
= \(\frac{1728}{\frac{1056}{7}}=\frac{1728 \times 7}{1056}\) ≈ 11.45
Only complete rods can be formed.
Therefore, the total number of complete cylindrical rods is 11.

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

Question 1.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Given: Slant height (l) = 21 m and diameter = 24 m
Therefore, radius (r) = \(\frac{24}{2}\) = 12 m
TSA of cone = πrl + πr²
= \(\frac{22}{7}\) × 12 × 21 + \(\frac{22}{7}\) × 12²
= 792 + \(\frac{22}{7}\) × 144
= 792 + 452.57
= 1244.57 m²
Thus, the total surface area of the cone is 1244.57 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: Slant height (7) = 10 cm and radius (r) = 7 cm
CSA of cone = πrl = \(\frac{22}{7}\) × 7 × 10 = 220 cm²
Thus, the curved surface area of the cone 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: Height of cone (h) = 16 cm and radius (r) =12 cm
Let slant height of the cone be l.
Using Baudhayana-Pythagoras theorem, we have
l² = h² + r²
= 16² + 12²
= 256 + 144 = 400
⇒ l = 20 cm
Curved Surface Area of cone (CSA)
= πrl = (3.14) × 12 × 20 = 753.6 cm²
Total Surface Area of cone (TSA)
= πrl + πr² = 753.6 + (3.14) × 12 × 12
= 753.6 + 452.16
= 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: Height of the cone (h) = 15 cm and volume (V) = 1570 cm³
V = \(\frac{1}{3}\) πrh
Substituting the value of V and h, we get
1570 = \(\frac{1}{3}\) × 3.14 × r² × 15 = 15.7 × r²
⇒ r² = 100
⇒ r = 10cm
Thus, the radius of the base of the cone 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= 308 cm² and slant height (h) = 14 cm
(i) Let the radius of the base of the cone be r.
CSA = πrl
308 = \(\frac{22}{7}\) × r × 14
⇒ 308 = 44r
⇒ r = 7 cm
Therefore, the radius of the base is 7 cm.

(ii) TSA = πrl + πr² = 308 + \(\frac{22}{7}\) × 49
= 308 + 154
= 462 cm²
Thus, 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 of cap (r) = 7 cm and height (h) = 24 cm
Let slant height be h.
Using Baudhayana-Pythagoras theorem, we have
l² = h² + r²
= 24² + 7²
= 576 + 49 = 625
⇒ l = 25 cm

Area of sheet required for one cap = CSA of a cap
= πrl = \(\frac{22}{7}\) × 7 × 25
= 550 cm²

Area of sheet required for 10 cups = CSA of 10 caps
= 10 × 550 cm²
= 5500 cm²

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 of tent (h) = 8 m, and radius of base (r) = 6 m
Let the slant height be l.
Using Baudhayana-Pythagoras theorem, we have
l² = h² + r²
= 8² + 6²
= 64 + 36 = 100
⇒ l = 10 m
Area of tarpaulin required = πrl = 3.14 × 6 × 10
= 188.4 m²
Width of tarpaulin = 3 m
Length = \(\frac{188.4}{3}\) = 62.8 m
Adding an extra length of material of 20 cm.
So, the total length of tarpaulin
= 62.8 + 0.2 = 63 m (20 cm = 0.2 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:
Sides of the right-angled triangle are 6 cm, 8 cm, and 10 cm.
When the right-angled triangle is rotated through 360° about the side of 8 cm, the solid generated is a right circular cone, where r = 6cm, h = 8 cm, and l = 10 cm

Volume of the cone (V)
= \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × (3.14) × (6)² × (8)
= 301.44 cm³
Curved Surface Area of the cone (CSA) = πrl
= 3.14 × 6 × 10
= 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:
The volume of the entire conical cup is given by:
V_cup = \(\frac{1}{3}\)πR²h
When the conical cup is filled to half its depth, the water itself forms a smaller similar right circular cone at the vertex of the cup:
Depth of water (h) = \(\frac{H}{2}\)
By using similar triangles formed by the vertical cross-section of the cone:
\(\frac{r}{\mathrm{R}}=\frac{h}{\mathrm{H}}=\frac{1}{2}\)
r = \(\frac{R}{2}\)
where r is the radius of the circular surface of the water. The volume of water (V_water) is the volume of this smaller cone:
V_water = h
Substituting r = \(\frac{R}{2}\) and h = \(\frac{H}{2}\)

Thus, the water occupies one-eighth of the total volume of the cup.

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:
Radius of the spherical ball bearing (r) = 0.7 cm
Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (0.7)²
=4 × 22 × 0.1 × 0.7
= 6.16 cm²
Therefore, 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 larger sphere (W₁) = 5920 g
Weight of smaller sphere (W₂) = 740 g
Diameter of smaller sphere = 5 cm
Radius of smaller sphere (r) = \(\frac{5}{2}\) = 2.5 cm
Let the radius of the larger sphere be R.
Since both spheres are made of the same metal, their density is uniform.
Therefore, the ratio of their weights equals the ratio of their volumes
⇒ \(\frac{W_1}{W_2}=\frac{V_1}{V_2}\)

\(\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\frac{4}{3} \pi \mathrm{R}^3}{\frac{4}{3} \pi r^3}=\left(\frac{\mathrm{R}}{r}\right)^3\)

Substituting the given weights, we get
\(\left(\frac{\mathrm{R}}{r}\right)^3=\frac{5920}{740}\) = 8
Taking the cube root on both sides, we get
\(\frac{\mathrm{R}}{r}=\sqrt[3]{8}\) = 2
Now, substituting r = 2.5 cm, we get
R = 2 × r = 2 × 2.5 cm = 5cm
Thus, 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:
Let the diameter of the Earth be D and its radius be
R = \(\frac{D}{2}\)
Then, the diameter of the Moon is d = \(\frac{1}{4}\)D, and its radius
r = \(\frac{d}{2}=\frac{\frac{1}{4} \mathrm{D}}{2}=\frac{1}{4}\left(\frac{\mathrm{D}}{2}\right)=\frac{1}{4}\)R
Surface area of a sphere = 4πr²
Now, \(\frac{\text { Surface Area of Moon }}{\text { Surface Area of Earth }}=\frac{4 \pi r^2}{4 \pi \mathrm{R}^2}=\left(\frac{r}{\mathrm{R}}\right)^2\)

= \(\left(\frac{\frac{1}{4} R}{R}\right)^2=\left(\frac{1}{4}\right)^2\)

= \(\frac{1}{16}\)
Thus, the ratio of their surface areas is 1: 16.

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

(ii) Total Surface Area (TSA) = 3,tr2
TSA = 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:
Dimensions of the rectangular box: length (l) = 16 cm, width (w) = 8 cm, height (h) = 8 cm
Radius of each metal sphere (r) = 2 cm
Number of spheres packed = 16
Volume of the rectangular box = l × w × h = 16 × 8 × 8
= 1024 cm³
Volume of a sphere = \(\frac{4}{3}\)πr³

Volume of 16 metal spheres = 16 × (\(\frac{4}{3} \times \frac{22}{7}\) × 2³)
= 16 × \(\frac{704}{21}=\frac{11264}{21}\) cm³

Now, volume of preservation liquid
= Volume of the rectangular box – Volume of 16 spheres
= 1024 – \(\frac{11264}{21}=\frac{21504-11264}{21}=\frac{10240}{21}\) ≈ 487.62 cm³
Rounding to the nearest integer gives 488 cm³.
Thus, the volume of the preservative liquid is approximately 488 cm³.

Question 6.
The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Solution:
Let the radius of the original sphere be r.
Volume = \(\frac{4}{3}\) πr³
Now, the radius is increased by 10%:
The new radius (r’) = r + r = 1.1 r
New volume = \(\frac{4}{3}\) π(r’)³ = \(\frac{4}{3}\) π(1.1r)³ = (1.1)³ × (\(\frac{4}{3}\)πr³)
= 1.331 × volume of original sphere

Increase in volume = New volume – Original volume
= 0.331 × Original volume

Percentage increase in volume
= \(\left(\frac{0.331 \times \text { original volume }}{\text { original volume }}\right)\) × 100%
= 0.331 × 100% = 33.1%
Hence, the volume increases by 33.1%.

Question 7.
The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value ofx.
Solution:
Let initial radius be r
Then, initial volume (V) = \(\frac{4}{3}\)πr³
adius increased by x%.

The new radius (r’) = r(1 + \(\frac{x}{100}\))
Volume increase = 72.8%
Then, the new volume (V’) = V + 0.728V = 1.728V
Since volume is directly proportional to the cube of the radius:
\(\frac{1}{2}\)
Substituting the values, we get
1.728 = (1 + \(\frac{x}{100}\))
⇒ \(\sqrt[3]{1.728}\) = 1 + \(\frac{x}{100}\)
⇒ 1.2 = 1 + \(\frac{x}{100}\)
⇒ \(\frac{x}{100}\) = 1.2 – 1 = 0.2
⇒ x = 0.2 × 100 = 20
Thus, x = 20 (i.e., the radius is increased by 20%).

Question 8.
The hemispherical dome of a building needs to be painted (see Figure).

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:
Circumference of circular base = 2πr = 35.2 m
⇒ 2 × \(\frac{22}{7}\) × r = 35.2
⇒ \(\frac{44}{7}\) × r = 35.2
⇒ r = \(\frac{35.2 \times 7}{44}\) = 0.8 × 7 = 5.6m
Since only the outer surface of the dome needs painting, we find its curved surface area
CSA = 2πr
= 2 × \(\frac{22}{7}\) × (5.6)²
= 2 × \(\frac{22}{7}\) × 5.6 × 5.6 7
= 2 × 22 × 0.8 × 5.6
= 197.12 m²
1 m = 100 cm
⇒ 1 m² = 10,000 cm²
Area in cm2 = 197.12 × 10,000 = 19,71,200 cm²
Rate of painting = ₹ 10 per 100 cm².
Rate per cm = ₹ \(\frac{10}{100}\) = 0.1
Total Cost = Area × rate
= 19,71,200× \(\frac{10}{100}\)
= ₹ 1,97,120
Thus, the cost of painting the hemispherical 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

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:
Volume of the cuboidal container:
V_container = lwh = 10 × 15 × 20
= 3000 cm³
An average hemispherical or spherical scoop has a diameter of about 5 cm
Then, radius (r) = 2.5 cm
Treating one standard scoop as a sphere:
V_scoop = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) × 3.14 × (2.5)³
= \(\frac{4}{3}\) × 3.14 × 15.625 ≈ 65.42 cm³

Number of scoops = \(\frac{\mathrm{V}_{\text {container }}}{\mathrm{V}_{\text {scoop }}} \approx \frac{3000}{65.42}\) ≈ 45.9
So, about 40 to 50 scoops of ice cream can be obtained.

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 volume of the original cube of side a is V₁ = a³ unit cubes.
The volume of the new cube of side a + 1 is V₂ = (a + 1 )³ unit cubes.
Number of unit cubes to be added
= V₂ – V₁ = (a + 1 )³ – a³
= (a³ + 3a² + 3a + 1) – a³ = 3a² + 3a + 1
Therefore, the required expression is 3a² + 3a + 1.

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) Assume: Length of classroom = 6 m
Width of classroom = 5 m
Height of classroom = 3 m
Volume of classroom = 6 × 5 × 3 = 90 m³
Since 1 m³ = 1000 litres
90 m³ = 90 × 1000 = 90000 litres
Average water consumed by a person in one day ≈ 3 litres
For a lifetime of 70 years = 70 × 365 × 3 = 76650 litres
Since 90000 > 76650
Therefore, a person cannot drink enough water in a lifetime to fill an entire classroom.

(ii) Assume dimensions of the classroom:
Length = 6 m, Width = 5 m, and Height = 3 m
Area of four walls = 2(1 + w) × h
= 2(6 + 5) × 3
= 66 m²
Assume: Area of doors and windows = 8 m2
⇒ Net wall area = 66 – 8 = 58 m²
Assume: Wall thickness = 0.2 m
Volume of walls = 58 × 0.2 = 11.6 m³
Assume brick size: 20 cm × 10 cm × 6 cm
Volume of one brick = 20 × 10 × 6 cm
= 1200 cm³
= 0.0012 m³
Number of bricks = \(\frac{11.6}{0.0012}\) ≈ 9667
Thus, 9667 bricks are required.

Question 4.
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:
Option (A):
Side of one large cube = 50 mm
Volume = (50 mm)³ = 1,25,000 mm³

Option (B):
Side of one small cube = 10 mm
Volume of 1 small cube = (10 mm)³ = 1000 mm³
Volume of 100 small cubes = 100 × 1000 mm³
= 1,00,000 mm³
Since, 1,25,000 mm³ > 1,00,000 mm³,
Therefore, option (A) gives more chocolate.

Question 5.
If all the people in the world are crowded together in one location, how much area would it cover?
Solution:
Assume the world population is approximately 8 billion people (8 × 10⁹).
In a tight crowd, one person requires approximately
0. 25 m² (a square of 50 cm × 50 cm).
Total area required = 8 × 10⁹ × 0.25 = 2 × 109 m²
Since 1 km² = 10⁶ m²
km² = 2000 km²
Area = \(\frac{2 \times 10^9}{10^6}\) km² = 2000 km²
All the people in the world would occupy approximately 2000 km² of the area.

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:
Diameter of glass = 7 cm ⇒ radius (r) = \(\frac{7}{2}\) =3.5 cm
Height of milk (h) = 12 cm
Volume of milk in one glass (V) = πr²h
= \(\frac{22}{7} \times\left(\frac{7}{2}\right) \times\left(\frac{7}{2}\right)\)
= \(\frac{22}{7} \times \frac{49}{4}\) × 12
= 22 × 7 × 3
= 462 cm³
Total volume of milk for 1600 students
= 1600 × 462 cm³
= 739200 cm³
Since 1000 cm³ = 1 litre
Therefore, 739.2 litres of milk is needed to serve 1600 students.

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) = 5 cm
Radius of cone (r) = 4 cm
Surface Area of Sphere = 4πR² = 4π(5)² = 100π cm²
Curved Surface Area of Cone = πrl = π × 4 × l
Surface Area of Sphere = 5 × Curved Surface Area of Cone
⇒ 100π = 5 × 4πl
⇒ 100π = 20πrl
⇒ l = = 5 cm
Now, using l² = r² + h²
5² = 4² + h²
⇒ 25 = 16 + h²
⇒ h² = 9
⇒ h = 3 cm

Volume of the cone (V) = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 4² × 3
= 1671 cm³
=16 × 3.14
= 50.24 cm³
Thus, the height of the cone is 3 cm and the volume of the cone is 50.24 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. (Calculator can be used.)
Solution:
Since V is directly proportional to r³, the ratio of volumes is \(\frac{1}{2}\)
(i) Ratio of volume of Jupiter to volume of Earth:
Ratio = \(\left(\frac{r_{\text {Jupiter }}}{r_{\text {Earth }}}\right)^3=\left(\frac{69,900}{6370}\right)^3\)
≈ \(\left(\frac{10.9733}{1}\right)^3 \approx \frac{1321}{1}\) = 1321:1

(ii) Ratio of volume of Sun to volume of Earth:
Ratio = \(\left(\frac{r_{\text {Sun }}}{r_{\text {Earth }}}\right)^3=\left(\frac{6,95,700}{6370}\right)^3\)
≈ \(\left(\frac{109.215}{1}\right)^3 \approx \frac{13,02,707}{1}\) = 13,02,707 : 1

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 volume of the sphere is:
V_sphere = \(\frac{4}{3}\)πr³
The smallest cylinder that encloses this sphere tightly must have
Radius = r and
Height (h) = 2r (diameter of the sphere)
The volume of this cylinder = πr²h = πr²(2r) = 2πr³
\(\frac{\text { Volume of Sphere }}{\text { Volume of cylinder }}=\frac{\frac{4}{3} \pi r^3}{2 \pi r^3}=\frac{\frac{4}{3}}{2}=\frac{4}{6}=\frac{2}{3}\)
Volume of Sphere = \(\frac{2}{3}\) × Volume of cylinder
Hence proved.

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 initial radius be R and the original string length (equator circumference) be C₁ = 2πR.
The second string has length C₂ = C₁ + 1 m and forms a concentric circle of radius R + h, where h is the height above the ground
⇒ C₂ = 2π(R + h) = 2πR + 2πh

Substituting C₂ = 2πR + 1, we get
2πR + 1 = 2πR + 2πh
⇒ 2πh = 1 m
⇒ h = \(\frac{1}{2 \pi}\)m
⇒ h = \(\frac{1}{2 \times 3.14}\) m ≈ 0.159 m = 15.9 cm
We notice that in expression h = \(\frac{1}{2 \pi}\) the height (h) does not depend on the radius(R) of the sphere.
Therefore, the same height is obtained for all spheres For (i) the Moon, (ii) Jupiter, and (iii) a volleyball: the string is approximately 15.9 cm above the surface of the sphere (the ground).

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 comers 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 square base is inscribed in a circle of radius r.
The diagonal of the square = 2r.
Area of square = \(\frac{1}{2}\) × d₁ × d₂ = \(\frac{1}{2}\)(2r)(2r) = 2r²
Height of the pyramid = h
Volume of the pyramid = \(\frac{1}{2}\) × base area × height
= \(\frac{1}{3}\)(2 r²)h = \(\frac{2}{3}\)r2h
Volume of the cylinder = πr²h
\(\frac{\text { Volume of pyramid }}{\text { Volume of cylinder }}=\frac{\frac{2}{3} r^2 h}{\pi r^2 h}=\frac{2}{3 \pi}\)
Volume of cylinder So, the ratio of the volume of the pyramid to the volume of the cylinder is 2 : 3π.

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
Solution:
Initial volume = lwh
New volume = (l + 1)wh = lwh + wh
Change in volume = (lwh + wh) – lwh = wh cubic units
So, the correct option is (c).

(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
Solution:
Initial volume = πr²h
New volume = π(r + 1 )²h = π(r + 2r+ 1 )h
= πr²h + 2πrh + πh
Change in volume = 2πrh + πh cubic units
So, the correct option is (e).

(iii) the radius of a sphere is decreased by 1 unit?
Solution:
Initial volume = \(\frac{4}{3}\) πr³.

New volume = \(\frac{4}{3}\) π (r – 1)³
= \(\frac{4}{3}\)π(r³ – 3r² + 3r – 1)
Decrease in volume
= \(\frac{4}{3}\)πr – \(\frac{4}{3}\)π(r³ – 3r² + 3r – 1)
= \(\frac{4}{3}\) π(3r² – 3r + 1)
= 4πr² – 4πr + \(\frac{4}{3}\) π cubic units

Question 13.
Given a cube with volume V, express its total surface area S in terms ofV.
Solution:
Let the side of the cube be a.
Volume(V) = a³ ⇒ a = V^{\(\frac{1}{3}\)} = \(\sqrt[3]{\mathrm{V}}\)
The total surface area (S) = 6a²
Substituting a = V3, we get
S = 6(V^{\(\frac{1}{3}\)})²
= 6V^{\(\frac{2}{3}\)} (or 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.
Surface area (S) = 6a²
⇒ a² = \(\frac{S}{6}\)
a = \(\sqrt{\frac{S}{6}}\)
The volume (V) = a³
Substituting a = \(\sqrt{\frac{S}{6}}\)
V = \(\left(\sqrt{\frac{S}{6}}\right)^3=\left(\frac{S}{6}\right)^{\frac{3}{2}}=\frac{S \sqrt{S}}{6 \sqrt{6}}\)

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:
Radius of glass (r) = 4 cm
Increase in water height required = 20 – 16 = 4 cm
Volume of water to be displaced = πr²h = π × 4² × 4
= 64π cm³
Radius of a marble = 1 cm
Volume of one marble = \(\frac{1}{2}\) πr³ = \(\frac{4}{3}\)π(1)³ = \(\frac{4}{3}\)π cm³
Number of marbles required = \(\frac{64 \pi}{\frac{4}{3} \pi}\) = 64 × \(\frac{3}{4}\) = 48
Thus, the crow should drop 48 marbles into the glass.

Question 16.
Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
Solution:
Assume the ink container is cylindrical.
Let radius of ink tube (r) = 0.15 cm and
Length of ink tube (h) = 10 cm
V = πr²h = π(0.15)² (10) ≈ 3.14 × 0.0225 × 10
≈ 0.71 cm³
Therefore, the volume of ink in a new ball point pen is approximately 0.71 cm³. (Answer may vary)

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) Volume of the big dough ball:
V = \(\frac{4}{3}\) π(6) = \(\frac{4}{3}\) π(216)
= 288π cm³
Let us assume that one small dough used for a single chapati has a diameter of 4 cm
⇒ radius of 2 cm Volume of one small dough ball used for a chapati
= \(\frac{4}{3}\)π(2) = \(\frac{32}{3}\)π cm
Number of chapatis = \(\frac{288 \pi}{\frac{32}{3} \pi}\) = 27
Therefore, approximately 27 chapatis can be made from it. (Answer may vary)

(ii) A halved coconut or muskmelon is a hemispherical shell.
Let the outer radius be ft and the inner radius be r.
Volume of edible flesh = \(\frac{2}{3}\) πR³ – \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\)π(R³ – r³)
Let outer radius of coconut or muskmelon R = 6 cm and inner radius r = 4.5 cm
Volume = \(\frac{2}{3} \times \frac{22}{7}\) × (216 – 91.125)
= \(\frac{44}{21}\) × 124.875 ≈ 262 cm³
So, the volume of edible fruit in a half coconut or muskmelon is approximately 262 cm³.
(Answer may vary)

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:
This is an estimation-based question.
(i) Step 1: Count the number of pages and dimensions of one page.
Step 2: Calculate the area of one page (Area of one page = Length × Breadth)
Step 3: Calculate the total area of all pages (Total area = Number of pages × Area of one page)
Step 4: Convert the total area into suitable unit.
Step 5: Calculate the area of the classroom floor.
Step 6: Compare the area covered by pages with the classroom floor area.

(ii) Step 1: Measure the dimensions of one textbook.
Step 2: Convert the storeroom dimensions from feet to centimetres.
Step 3: Find the volume of the storeroom.
(Volume = l × w × h)
Step 4: Find the volume of one textbook.
Step 5: Calculate the maximum number of textbooks.
(Number of textbooks = \(\frac{\text { Volume of storeroom }}{\text { Volume of one textbook }}\))
Step 6: Give the estimated number of textbooks that can fit in the storeroom.

Question 19.
If the entire human population decided to climb into one giant cube, how long would the side have to be?
Solution:
Human population ≈ 8 billion = 8 × 10⁹ people
Volume occupied by an average human body = 0.07 m³.
Accounting for packing space/air, assume each person takes a volume of 0.125 m3 (a cube of side 0.5 m).
Total volume required (V) = 8 x 10⁹ × 0.125 m³ = 10⁹ m³ = 1 km³
Side of the cube = a³ = 10⁹ m³
⇒ a = \(\sqrt[3]{10^9 \mathrm{~m}^3}\) = 1000 m = 1 km
So, the side of the giant cube would have to be about 1 km long.
A realistic estimate would place the side length of the cube within the range of 1 to 3 km, depending on how tightly or comfortably people are packed.

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:
(i) Let the thickness of water layer be t.
Earth’s radius R = 6371 km.
The volume of the water layer is approximately equal to the surface area of the Earth multiplied by the thickness
⇒ V = 4πR² × t
⇒ t = \(\frac{V}{4 \pi R^2}\)

(ii) V = 1.38 × 10⁹ km³,
R = 6371 km
Surface area of Earth:
4πR² = 4π × (63 71)²
⇒ 4 × (3.14) × 40,589,641 ≈ 509805891 km²

Thickness (t) = \(\frac{1.38 \times 10^9}{509805891}\) ≈ 2.7069 km
≈ 2707 metres
Thus, the thickness is approximately 2.71 km or 2707 m

Question 21.
Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
Solution:
Let the dimensions of the cuboid be l, w, and h.
Volume of cuboid (V) = lwh
Surface area of cuboid (S) = 2(lw + wh + lh)
We need a cuboid whose volume becomes half and surface area becomes double.
To get half the volume, half only one dimension \(\frac{l}{2}\) ,w,h
New volume:
V’= \(\frac{l}{2}\) × w × h = \(\frac{l}{2}\) Iwh = \(\frac{V}{2}\)
Now check the surface area:
Original surface area (S) = 2(lw + wh + lh)
New surface area (S’) = 2(\(\frac{lw}{2}\) + wh + \(\frac{lh}{2}\)) = lw + 2wh + lh
For S’ = 2S, we need:
lw + 2 wh + lh = 4 (lw + wh + lh)
⇒ 3 lw + 2 wh + 3lh = 0
This condition is not satisfied for a general cuboid.
So, changing only one dimension (or scaling the whole shape) can never halve the volume and double the surface area. To halve the volume and double the surface area, you must change at least two sides, and not by the same scale factor. For example, consider the cuboid of dimensions: 2 × 3 × 4 (V = 24 cubic units, S = 52 sq. units) becomes the cuboid with dimensions 12 × 0.25 × 4 (V = 12 cubic units, S = 104 sq. units).

Question 22.
Project: Find the volume of your house making necessary approximations. Present how you solved it.
Solution:
Assume the house is approximately a cuboid.
Let length of the house (l) = 12 m
width of the house (w) = 8 m
height of the house (h) = 3 m
Volume of cuboid (V) = l × w × h
Substituting the values, we get
V= 12 × 8 × 3 = 288 m³
Therefore, the approximate volume of the house is 288 m³
Note: Actual dimensions may vary according to the house. The volume can be calculated by dividing the house into simple shapes such as cuboids and adding their volumes. Methodology: Divide the house into identifiable 3D geometrical shapes (mainly rectangular cuboids for each room, plus a triangular prism or pyramid if the roof is sloped).
Sample Presentation:

• Floor Area: Measure or take the built-up area of the flat/ house.
• Ceiling Height: Measure the average floor-to-ceiling height.
• Volume Calculation: Total Volume ≈ Floor Area × Height