NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 14 Math of Space Surface Area and Volume Examples

Examples

Example 1. Two solid objects are made from the same material: Cube A has side 6 cm and Cuboid B has dimensions 9 cm × 6 cm × 4 cm. Compare their volumes and surface areas and determine which object has a greater surface area. How is this useful in a real-life situation?
Answer:

Cube A (side a = 6 cm):
Volume = a³ = 6³ = 216 cm³
Surface area = 6a² = 6 × 36 = 216 cm²
Cuboid B (l = 9 cm, w = 6 cm, h = 4 cm):
Volume = l × w × h = 9 × 6 × 4 = 216 cm³
Surface area = 2(lw + wh + hl) = 2(54 + 24 + 36) = 2 × 114 = 228 cm²
Comparing: both solids have the same volume, 216 cm³, but the surface areas are 216 cm² (cube) and 228 cm² (cuboid).
Cuboid B has the greater surface area, by 228 − 216 = 12 cm².
Real-life use: for the same capacity, a cube-shaped box needs the least wrapping or packing material, so boxes close to a cube save material. On the other hand, when we want something to heat up or cool down quickly (radiator fins, cooling trays), we choose a shape with more surface area for the same volume.

Example 2. Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope (see Fig. 14.6). What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take π = \( \frac{22}{7} \).
Answer:

Radius of the base r = 3.5 cm; length (height) of the kaleidoscope h = 25 cm.
The chart paper forms only the curved surface, so we need the curved surface area.
Area of chart paper = 2πrh = 2 × \( \frac{22}{7} \) × 3.5 × 25 = 2 × 22 × 0.5 × 25 = 550 cm²
Check: opened out, the curved surface is a rectangle 2πr = 22 cm long and 25 cm wide, and 22 × 25 = 550 cm².
Savitri needs 550 cm² of chart paper.

Example 3. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is rotated through 360° about the side with length 12 cm. Find the volume of the solid so obtained.
Answer:

First check the triangle: 5² + 12² = 25 + 144 = 169 = 13². By the converse of the Baudhāyana–Pythagoras theorem it is right-angled, with the right angle between the 5 cm and 12 cm sides.
Turning a right triangle about one of its perpendicular sides traces out a right circular cone. Here the side of 12 cm is the axis, so
height h = 12 cm, radius r = 5 cm (and slant height l = 13 cm).
V = \( \frac{1}{3} \) πr²h = \( \frac{1}{3} \) × π × 25 × 12 = 100π cm³
With π ≈ 3.14, V ≈ 314 cm³.
The volume of the cone is 100π cm³ ≈ 314 cm³.

Example 4. A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Answer:

The water the bowl can hold fills a hemisphere of radius r = 3.5 cm.
V = \( \frac{2}{3} \) πr³ = \( \frac{2}{3} \) × \( \frac{22}{7} \) × 3.5 × 3.5 × 3.5
= \( \frac{2}{3} \) × 22 × 0.5 × 3.5 × 3.5 = \( \frac{2}{3} \) × 134.75 ≈ 89.8 cm³
The bowl can hold about 89.8 cm³ of water.

Example 5. Lallan’s golgappa (a small hollow ball made of wheat) stall has a cylindrical vessel with pani (spiced water) in it. The vessel is of radius 10 cm and it has pani up to a height of 50 cm. Estimate the number of golgappas he can serve assuming all golgappas are spheres. Before calculating, make a guess. Can you guess what number this could be — is it in the hundreds, thousands, or more?
Answer:

Guess: the vessel holds roughly 15 litres and each golgappa only a spoonful or two, so the answer should be somewhere between several hundred and two thousand.
Number of golgappas = \( \frac{\text{pani in the vessel}}{\text{pani in one golgappa}} \)
Pani in the vessel (r = 10 cm, height 50 cm): V = πr² × height = π × 10 × 10 × 50 = 5000π cm³ ≈ 15708 cm³ (about 15.7 litres).
Two students made different assumptions:
Student 1: radius of a golgappa 1.5 cm; volume = \( \frac{4}{3} \) π(1.5)³ ≈ 14.14 cm³; filled 60%, so pani per golgappa ≈ 14.14 × 0.6 ≈ 8.5 cm³; number ≈ 15708 ÷ 8.5 ≈ 1848, i.e. about 1850 golgappas.
Student 2: radius of a golgappa 1.75 cm; volume = \( \frac{4}{3} \) π(1.75)³ ≈ 22.45 cm³ (the book prints 22.43); filled 80%, so pani per golgappa ≈ 18 cm³; number ≈ 15708 ÷ 18 ≈ 873, i.e. about 870 golgappas.
So Lallan can serve somewhere between about 870 and 1850 golgappas — in the high hundreds to under two thousand. The exact estimate depends on the assumed size of a golgappa and how full it is filled.

Example 6. Anirban saw a few tennis balls lying around. He wondered, “How many tennis balls can fit in an empty classroom?”. The classroom is of dimensions 30 ft × 30 ft × 15 ft. Estimate how many tennis balls of radius 3 cm fit in this classroom. Before calculating, make a guess. Also make an estimate based on the second and third configurations of Fig. 14.17.
Answer:
Guess: somewhere between a few lakh and a few tens of lakhs of balls.
Volume of one ball = \( \frac{4}{3} \) × π × 3³ ≈ 113 cm³.
Using 1 ft ≈ 30 cm, 1 ft³ ≈ 28000 cm³ (exactly, 1 ft = 30.48 cm and 1 ft³ ≈ 28317 cm³).
Volume of the room = 30 × 30 × 15 = 13500 ft³ ≈ 13500 × 28000 = 37,80,00,000 cm³.
Upper bound: 37,80,00,000 ÷ 113 ≈ 33,45,132. Balls always leave gaps between them, so the real number must be smaller than this.
Configuration A (balls stacked straight on top of one another): diameter = 6 cm; 30 ft ≈ 900 cm and 15 ft ≈ 450 cm.
Along the length: 900 ÷ 6 = 150; along the width: 150; along the height: 450 ÷ 6 = 75.
Number of balls = 150 × 150 × 75 = 16,87,500.

Configuration B (each ball sits in the hollow of four balls below): the four centres below form a square of side 6 cm. The upper centre is above the middle of this square, at a horizontal distance of half the diagonal, 3√2 cm, from each of them. The centres are 6 cm apart, so the vertical gap between layers is √(6² − (3√2)²) = √18 ≈ 4.24 cm.
Number of layers = (444 ÷ 4.24, rounded down) + 1 = 104 + 1 = 105 (the lowest centres are 3 cm above the floor and the highest must be at least 3 cm below the ceiling).
The layers alternate between 150 × 150 = 22500 balls and 149 × 149 = 22201 balls, so the total is 53 × 22500 + 52 × 22201 = 11,92,500 + 11,54,452 = 23,46,952 ≈ 23.5 lakh.
Configuration C (rows fitted into hollows, each ball on three balls below): inside a layer, the rows are √(6² − 3²) = √27 ≈ 5.20 cm apart, so there are (894 ÷ 5.196, rounded down) + 1 = 172 + 1 = 173 rows, alternately of 150 and 149 balls: 87 × 150 + 86 × 149 = 25864 balls per layer.
A ball of the next layer sits above the centre of an equilateral triangle of side 6 cm, which is \( \frac{6}{\sqrt{3}} \) = 2√3 cm from each vertex, so the layers are √(36 − 12) = √24 ≈ 4.90 cm apart. Number of layers = (444 ÷ 4.90, rounded down) + 1 = 90 + 1 = 91.
Total ≈ 91 × 25864 = 23,53,624 ≈ 23.5 lakh.
Summary: upper bound 33,45,132; A: 16,87,500; B: about 23,47,000; C: about 23,54,000.
With configuration A about 16.9 lakh balls fit; with B or C about 23.5 lakh balls fit — roughly 39% more than A, but still well below the upper bound of about 33 lakh. Spheres always leave gaps, which is why cubes, not spheres, are used as units of volume.

In-text Questions and Activities

Think and Reflect (Page 124). (a) Try to work out for yourself why this model (a cuboid thought of as a pack of cards) explains the formula for the volume of a cuboid, i.e., volume = area of base × height.
(b) Write the formula for the surface area and the volume of a cube.
Answer:
(a)

Imagine the cuboid as a pile of n identical thin sheets, like playing cards, each of length l, width w and very small thickness t.
Each sheet has area l × w, the same as the base of the cuboid, and the height of the pile is h = n × t.
Volume of one sheet = lw × t, so the volume of the whole pile is
V = n × (lw × t) = lw × (n × t) = lw × h
Since lw is the area of the base, volume = area of base × height. Adding more sheets only makes the pile taller; every layer still has the area of the base.
(b) A cube is a cuboid with l = w = h = a, and its six faces are equal squares of area a².

Total surface area of a cube = 6a²; volume of a cube = a × a × a = a³.

In-text Question (Page 124). Observe how these two formulas are special cases of the formulas 2(wl + hl + hw) and whl, when w = h = l.
Answer:
Put w = h = l = a (the side of the cube).
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 come straight out of the cuboid formulas when all three dimensions are equal — a cube is just a special cuboid.

Think and Reflect (Page 126). 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]
Answer:

Water needed (volume):
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, top open):
Pool A: 50 × 20 + 2(8 × 50 + 8 × 20) = 1000 + 1120 = 2120 ft²
Pool B: 100 × 15 + 2(6 × 100 + 6 × 15) = 1500 + 1380 = 2880 ft²
Water surface (room for swimmers): A: 50 × 20 = 1000 ft²; B: 100 × 15 = 1500 ft².
A suitable choice for a school is Option B.
1. Safety: 6 ft is shallower than 8 ft, which is safer for students who are still learning to swim.
2. More room: B has 1.5 times the water surface, so more students can swim at once.
3. Longer laps: a 100 ft length is better for races and practice.
Against B: it needs about 12.5% more water (about 28,300 litres more) to fill, filter and heat, and about 760 ft² more tiling. Pool A would be better if diving is planned (it needs depth) or if water and money are very limited.
Parameters considered: depth (safety), water surface area (number of users), length (laps), volume (water and running cost), tiled area (building cost) and the space available in the school. This is an open-ended question, so other well-argued choices are also acceptable.

In-text Question (Page 127). The following figures show different types of cylinders. What is different in each of these cylinders?
Answer:

Both cylinders have two equal, parallel circular ends. The difference lies in the axis, the line joining the centres of the two circular ends.
1. In the right circular cylinder the axis is perpendicular to the base, so the cylinder stands straight up.
2. In the oblique circular cylinder the axis is slanted (not perpendicular) to the base, so the cylinder leans to one side.
In both, every cross-section parallel to the base is a circle of the same radius. In this chapter we work only with right circular cylinders.

Activity (Page 141). Find the approximate volumes of a gas cylinder, a bucket, any cylindrical vessel, a cupboard, a water tank that you can find near you (use 1000 cm³ = 1 litre).
Answer:
Measure each object with a tape and model it by the nearest simple solid. Sample measurements (yours will differ), with π ≈ 3.14:
1. LPG gas cylinder — cylinder, diameter 30 cm, height 60 cm: V ≈ 3.14 × 15² × 60 = 42390 cm³ ≈ 42 litres.
2. Bucket — cylinder using the average radius (top diameter 30 cm, bottom 24 cm, so average radius 13.5 cm), height 30 cm: V ≈ 3.14 × 13.5² × 30 ≈ 17168 cm³ ≈ 17 litres.
3. Cylindrical vessel (patila) — radius 12 cm, height 20 cm: V ≈ 3.14 × 12² × 20 ≈ 9043 cm³ ≈ 9 litres.
4. Cupboard (almirah) — cuboid 90 cm × 50 cm × 180 cm: V = 8,10,000 cm³ = 810 litres.
5. Cylindrical water tank on the roof — diameter 110 cm, height 110 cm: V ≈ 3.14 × 55² × 110 ≈ 10,44,835 cm³ ≈ 1045 litres.
To change to litres, divide the volume in cm³ by 1000; for example, the gas cylinder holds 42390 ÷ 1000 ≈ 42 litres.