CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Exercise 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the woman between the age 15 – 44 (in years) worldwide, found the following figures (in %)

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of woman’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above is the major cause.
Solution:
(i) The information given above represent graphically as:

(ii) Reproductive health conditions are the major cause of women’s ill health and death worldwide.
(iii) Two factors which play a major role in the cause of women’s ill health and death worldwide are reproductive health condition and neuropsychiatric condition.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society are given below:

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
(i)

(ii) Two conclusions arrive from the graph are:
(a) Number of girls per thousand boys and most in schedule tribe and least in urban.
(b) The average number of girls per thousand boys is less than 950.

Question 3.
Given below are the seats won by different political parties in the polling outcome of state assembly elections.

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
(i)

(ii) Political party ‘A’ won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant is measured correct to one millimeter, and the obtained data is represented in the following table:

(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves is 153 mm long? Why?
Solution:
(i) First we make the continuous class interval frequency distribution table.

(ii) Yes, we can draw frequency polygon to represent the same data.
(iii) Class interval 145 – 153 of lengths are the maximum number of leaves. But it is not concluded that the maximum number of leaves is 153 mm long.

Question 5.
The following table gives the lifetimes of 400 neon lamps.

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more than 700 hours?
Solution:
(i)

(ii) 184 lamps have a lifetime of more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. The two polygons compare the performance of the two sections.
Solution:

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous]
Solution:
First, we make the continuous class interval frequency distribution table.

Question 8.
A random survey of the number of children of various age group playing in a park was found as follows:

Draw a histogram to represent the data bove.
Solution:
To draw a histogram to represent the data first we make data in which we clear the length of the rectangle.

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) Draw a histogram to represent the first we make a data in which we clear the length of the rectangle.

(ii) The class interval in which the maximum number of surnames lie in 6 – 8 letters.

These NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Exercise 14.2

Question 1.
The blood groups of 30 students of class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students.
Solution:
The frequency distribution table of the following data is as:

The most common blood group is O, and the rarest blood group is AB among the students.

Question 2.
The distance (in km) of 40 female engineers from their residence to their place of work was found as follows:

Construct a grouped frequency distribution table with class size 5 for the data given above, taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution:
Upper limit = 32
Lower limit = 2
Range = 32 – 2 = 30

We observe that more than 60% of female engineers’ residence to their place of work were less than 15 km.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88 etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
(i) Upper limit = 99.2
Lower limit = 84.9
Range = 14.3

(ii)The given data is to be taken in the rainy season as the relative humidity is high.
(iii) The range of this data is 99.2 – 84.9 = 14.3

Question 4.
The height of 50 students, measured to the nearest centimeter have been found to be as follows:

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their height from the above table.
Solution:
(i) Upper limit = 173
Lower limit = 150
Range = 173 – 150 = 23

(ii) One conclusion that we can draw from the above table is that more than 50% of students are shorter than 165 cm.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city, The data obtained for 30 days is as follows:

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Solution:
(i) Upper limit = 0.22
Lower limit = 0.01
Range = 0.21

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:

Prepare a frequency distribution table for the data given above.
Solution:
The frequency distribution table of the given data is

Question 7.
The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
(i) Frequency distribution table of the digits after the decimal point is:

(ii) The most frequently occurring digits are 3 and 9. The least occurring digits are 0.

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
(ii) How many children watched television for 15 or more hours a week?
Solution:
Upper limit = 17
Lower limit = 1
Range = 17 – 1 = 16

(ii) Number of children who watched television for 15 or more hours a week is 2 children.

Question 9.
A Company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5:
Solution:
Upper limit = 4.6
Lower limit = 2.2
Range = 2.4

These NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Exercise 14.1

Question 1.
Give five examples of data that you can collect from your day-to-day life.
Solution:
Five examples of data that we can gather from our day to day life are:
(i) Height of students in our class.
(ii) Blood group of students in our class.
(iii) Number of family members in the family in our colony.
(iv) Electricity bills of our house for last 10 months.
(v) Daily temperature in Delhi in the month of June.

Question 2.
Classify the data in Q.1 above as primary or secondary data.
Solution:
Primary data:
(i) Height of students in our class.
(ii) Blood group of students in our class.
(iv) Electricity bills of our house for last 10 months.
Secondary data:
(iii) Number of family members in the family in our colony.
(v) Daily temperature in Delhi in the month of June.

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.9

Question 1.
A wooden bookshelf has external dimension as follows: Height = 110 cm, Depth = 25 cm, breadth = 85 cm (see Fig. 13.31). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.

Solution:
We have given that
Height of bookshelf = 110 cm
Depth of bookshelf = 25 cm
and Breadth of bookshelf = 85 cm
The thickness of the plank = 5 cm
Total surface area of external face (Shading portion) = 2 (25 × 110) + 2 (85 × 25) + 110 × 85 + 4(5 × 85) + 2(90 × 5)
= 5500 + 4250 + 9350 + 1700 + 900
= 21,700 cm²
Total surface area of internal faces (without shading portion)
= 6(20 × 75) + 2(90 × 20) + 90 × 75
= 9000 + 3600 + 6750
= 19350 cm²
Now, cost of polishing the external faces at the rate of 20 paise/cm² is = 21700 × 0.20 = Rs. 4340
and cost of painting the internal faces at tire rate of 10 paise/cm² is, the bookself = 4340 + 1935 = Rs. 6275

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. 13.32. Eight such spheres are used, for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Solution:
The surface area of one wooden sphere = 4πr²
= \(4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2}\)
So, total surface area of all the 8.
Wooden sphere = \(8 \times 4 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^{2}\) = 11088 cm²
Now, some areas of the sphere can’t be silver printed due to the support.
Support area = \(8 \times \frac{22}{7} \times(1.5)^{2}\) = 56.57 cm²
Required area for silver painting = 11088 – 56.57 = 11031.43 cm²
Cost of silver paint = 11031.43 × 0.25 = Rs. 2757.85
Now, support is to be painted in black colour.
So, curved surface of cylindrical support of:
= 8 × 2 × \(\frac {22}{7}\) × 7 × 15
= 528 cm²
Cost to paint the support in black = 528 × 0.05 = Rs. 26.40
Total cost of paint = 2759.85 + 26.40 = Rs. 2784.25

Question 3.
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution:
Let the diameter of sphere = d

∴ Radius of sphere = \(\frac{d}{2}\)
∴ Curved surface area of sphere = 4πr²
= \(4 \times \pi \times \frac{d}{2} \times \frac{d}{2}\)
= πd²
In second case,
diameter of new sphere = d – 25% of d
= d – \(\frac{25}{100} \times d\)
= \(\frac{3 d}{4}\)
∴ Radius of new sphere = \(\frac{3 d}{8}\)

∴ Curved surface area of new sphere = 4πr²
= \(4 \times \pi \times \frac{3 d}{8} \times \frac{3 d}{8}\)
= \(\frac{9 \pi d^{2}}{16}\)
Therefore, surface area decrease in second case is = πd² – \(\frac{9 \pi d^{2}}{16}\)
= \(\frac{16 \pi d^{2}-9 \pi d^{2}}{16}\)
= \(\frac{7 \pi d^{2}}{16}\)
So, Percent decreasing = \(\frac{7 \pi d^{2}}{\frac{16}{\pi d^{2}} \times 100}\)
= \(\frac{7 \pi d^{2}}{16 \pi d^{2}} \times 100\)
= 43.75%
So, the curved surface area decrease in the second case is 43.75%.

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Solution:
(i) We have radius of sphere = 7 cm

∴ Volume of sphere = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7\)
= 1437 \(\frac {1}{3}\) cm³

(ii) We have
Radius of sphere = 0.63 m

∴ Volume of sphere = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63\)
= 1.0478 m³
= 1.05 m³ (approx)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm
(ii) 0.21 m
Solution:
(i) We have given that

Diameter of spherical ball = 28 cm
∴ Radius of spherical ball = \(\frac {28}{2}\) = 14 cm
Volume of spherical ball = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14\)
= 11498\(\frac {2}{3}\) cm³
Therefore, amount of water displaced by solid sphere is 11498\(\frac {2}{3}\) cm³.

(ii) Diameter of spherical ball = 0.21 m

∴ Radius of spherical a ball = \(\frac {0.21}{2}\) = 0.105 m
So, volume of spherical ball = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 0.105 × 0.105 × 0.105
= 0.004851 m³
Therefore, the amount of water displaced by the solid sphere is 0.004851 m³

Question 3.
The diameter of a metallic ball is 4.2. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?
Solution:

We have given that diameter of metallic ball = 4.2 cm
Radius of metallic ball = \(\frac {4.2}{2}\) = 2.1 cm
volume of metallic ball = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 2.1 × 2.1 × 2.1
= 38.808 cm³
The density of the metal is 8.9 g per cm³.
Mass of the metallic ball = 38.808 × 8.9 = 345.39 g (approx)

Question 4.
The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon.
Solution:
Let the diameter of earth = x

∴ Radius of earth = \(\frac{x}{2}\)
Volume of earth = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \pi \times \frac{x}{2} \times \frac{x}{2} \times \frac{x}{2}\)
= \(\frac{\pi x^{3}}{6}\)
Again, the diameter of the moon is one-fourth of the diameter of the earth.

Diameter of moon = \(\frac{x}{4}\)
Radius of moon = \(\frac{x}{8}\)
∴ Volume of moon = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \pi \times \frac{x}{8} \times \frac{x}{8} \times \frac{x}{8}\)
= \(\frac{x^{3} \pi}{384}\)
Therefore, fraction of the volume of the moon and the volume of earth is = \(\frac{\frac{\pi x^{3}}{384}}{\frac{\pi x^{3}}{6}}=\frac{6}{384}=\frac{1}{64}\)

Question 5.
How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold.
Solution:
We have a diameter of hemispherical bowl = 10.5 cm

∴ Radius of hemispherical = \(\frac{10.5}{2}\) = 5.25 cm
∴ Volume of hemisphere = \(\frac{2}{3} \pi r^{3}\)
= \(\frac{2}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 5.25
= 303.187 cm³
We know that 1000 cm³ = 1 litre
∴ 303.18 cm³ = 0.303 litre (approx)
Therefore, a hemispherical bowl of diameter 10.2 cm holds 0.303 liters of milk.

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick if the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
We have given that
The inner radius of hemispherical tank = 1 m
and thickness of iron sheet = 1 cm = 0.01 m
The outer radius of hemispherical tank = 1.01 m

So, volume of iron used to make the tank = Volume of out spherical tank – Volume of inner spherical tank
= \(\left.\frac{2}{3} \pi(1.01)^{3}\right)-\frac{2}{3} \pi(1)^{3}\)
= \(\frac{2}{3}\) × π × 1.030301 – \(\frac{2}{3}\) × π × 1
= \(\frac{2}{3}\) × π × (1.030301 – 1)
= \(\frac{2}{3} \times \frac{22}{7} \times 0.030301\)
= 0.06348 m³ (approx)
Therefore, the volume of the iron used to make the tank is 0.06348 m³ (approx)

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
We have given that
The surface area of a sphere is 154 cm²
but, the surface area of sphere = 4πr²

Volume of sphere = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 3.5 × 3.5 × 3.5
= 179\(\frac{2}{3}\) cm³

Question 8.
A dome of a building is in the form of a hemisphere. From inside it was while washed at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2.00 per square meter find the:
(i) Inside the surface area of the dome.
(ii) Volume of the air inside the dome.
Solution:
(i) We have given that Rs. 2 is the rate of whitewashing of 1 m² area
Rs. 1 is the rate of white-washing \(\frac{1}{2}\) m² area

Rs. 498.96 is the rate of white-washing = \(\frac{1}{2}\) × 498.96 = 249.48 m² area.
Therefore, according to question, inside surface area of dome = 249.48 m²

(ii) We have,
surface area of hemispherical dome = 249.48 m²
but surface area of hemispherical dome = 2πr²
⇒ 2πr² = 249.48
⇒ r² = \(\frac{249.48 \times 7}{2 \times 22}\)
⇒ r² = 39.69
⇒ r = √39.69 = 6.3 m
Therefore,
Volume of hemispherical dome = \(\frac{2}{3} \pi r^{3}\)
= \(\frac{2}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3
= 523.908 m³
Therefore, the volume of the air inside the dome is 523.9 m³ (approx).

Question 9.
Twenty-seven solid iron spheres, each of radius r and surface area s are melted to form a sphere with surface area s’. Find the:
(i) radius r’ of the new sphere
(ii) ratio of s and s’.
Solution:
We have given that

The radius of sphere = r
Surface area(s) = 4πr²
and volume of sphere = \(\frac{4}{3} \pi r^{3}\)
Volume of 27 such speres = \(\frac{4}{3} \times \pi r^{3} \times 27\) = 36πr³

According to question Volume of 27 small sphere = volume of big sphere
or, 36πr³ = \(\frac{4}{3} \pi\left(r^{\prime}\right)^{3}\) (∵ radius = r’)
⇒ (r’)³ = \(\frac{36 \pi r^{3} \times 3}{4 \times \pi}\)
(∴ volume of big sphere = volume of 27 small sphere)
⇒ (r’)³ = 27r³
⇒ r’ = 3r
Therefore, radius r’ of new sphere is equal to 3r.

(ii) Surface area of new-sphere (s’) =4π(r’)²
and surface area of small sphere = 4πr²
∴ Ratio between them = \(\left(\frac{4 \pi r^{2}}{4 \pi\left(r^{\prime}\right)^{2}}\right)\)
= \(\left(\frac{4 \pi r^{2}}{4 \pi \times 3 r \times 3 r}\right)\) (∴ r’ = 3r)
= \(\frac{1}{9}\)
= 1 : 9

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule.
Solution:
We have given

Diameter of spherical capsule = 3.5 mm
∴ Radius of spherical capsule = \(\frac{3.5}{2}\) = 1.75 mm
∴ Volume of spherical capsule = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 1.75
= 22.4583 mm³ (approx)
Therefore, medicine is needed to fill the capsule is 22.4583 mm³ (approx)

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm.
(ii) radius 3.5 cm, height 12 cm.
Solution:
(i) We have given
radius of right circular cone = 6 cm
and height of right circular cone = 7 cm.

Volume of right circular cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7} 6 \times 6 \times 7\)
= 264 cm³

(ii) We have given
Radius of cone = 3.5 cm
and height of cone = 12 cm.

Volume of cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 3.5 × 3.5 × 12
= 154 cm³

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm.
(ii) height 12 cm and slant height 13 cm.
Solution:
(i) We have given that,
Radius of conical vessel = 7 cm
and slant height = 25 cm
By Pythagoras theorem,

∴ Volume of conical vessel = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 7 × 7 × 24
= 1232 cm³
We know that
1000 cm³ = 1 l
∴ 1232 cm³ = 1.232 l
Therefore, capacity of conical versel is 1.232 litres.

(ii) We have given that
Height of conical vessel (h) = 12 cm
Slant height of conical vessel (l) = 13 cm
By Pythagoras theorem

∴ Volume of conical vessel = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 5 × 5 × 12
= \(\frac{2200}{7}\) cm³
Since, 1000 cm³ = 1 l
\(\frac{2200}{7}\) cm³ = \(\frac{2.2}{7}\) l
Therefore, the capacity of conical vessel is \(\frac{2.2}{7}\) litres.

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base.
Solution:
We have given
Height of cone (h) = 15 cm
and volume of cone = 1570 cm³
We know that
Volume of cone = \(\frac{1}{3} \pi r^{2} h\)

Therfore, radius of required cone = 10 cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
We have given that
Height of cone = 9 cm
and volume of cone = 48π
But we know that volume of cone = \(\frac{1}{3} \pi r^{2} h\)

Therefore diameter of required right circular cone is 2 × 4 = 8 cm.

Question 5.
A conical pit top diameter 3.5 is 12 m deep. What is its capacity in kilo liters.
Solution:
We have given that
diameter of conical pit = 3.5m.
radius of conical pit = \(\frac{3.5}{2}\) m = 1.75 m
and height of conical pit = 12 m

∴ Volume of conical pit = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 12
= 38.5 m³
We know that
1 m³ = 10001
38.5 m³ = 1000 × 38.5
= 38500 litres
= 38.5 kilolitres.

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find.
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
Solution:
(i) We have given that
diameter of cone = 28 cm
Radius of cone = 14 cm
and volume of cone = 9856 cm³

We know that,
Volume of cone = \(\frac{1}{3} \pi r^{2} h\)
9856 = \(\frac{1}{3} \times \frac{22}{7}\) × 14 × 14 × h
h = \(\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}\) = 48 cm

(ii) We have,
height of the cone = 48 cm
and radius of cone = 14 cm
By Pathagoras theorem,
AC = \(\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(48)^{2}+(14)^{2}}\)
= \(\sqrt{2304+196}\)
= 50 cm.
Therefore slant height of the cone = 50 cm

(iii) We know that,
Curved surface area of cone = πrl
= \(\frac {22}{7}\) × 14 × 50
= 2200 cm²
Hence, curved surface area of the cone is 2200 cm²

Question 7.
A right triangle ABC with side 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
When we revolved the right ∆ABC about the side 12 cm, then it form a right circular cone of radius 6 cm and height 12 cm.

So, volume of such cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac {1}{3}\) × π × 5 × 5 × 12
= 100π cm³
Therefore, volume of the solid formed by the revolved a right angle triangle about the side 12 cm is 100π cm³.

Question 8.
If the triangle ABC in the question 7 is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solid obtained.
Solution:
If we revolved the right angle triangle ABC about the side 5 cm is formed a right circular cone.

So, volume of the such cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac {1}{3}\) × π × 12 × 12 × 5
= 240π
Therefore, volume of the solid formed by the revolved a right angle triangle about the side 5 cm is 240π cm³.
Ratio of the volumes of the two solid obtained by the revolved a right angle triangle is = \(\frac{100 \pi}{240 \pi}\) = 5 : 12

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it form rain. Find the area of the cavas required.
Solution:
We have given that
Diameter of right circular cone = 10.5 cm
Radius of right circular cone = 3.25 m
and height of the cone = 3m

∴ Volume of the cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 3
= 86.625 m³
Again,
By Pythagoars theorem.
AC = \(\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(3)^{3}+(5.25)^{2}}\)
= \(\sqrt{9+27.5625}\)
= 6.04 m (approx)
Therefore,
Canvas required to covers it = Curved surface area of cone
Now, Curved surface area of cone = πrl
= \(\frac {22}{7}\) × 5.25 × 6.04
= 99.66 m²
Hence, canvas required to cover the conical shape heap of wheat is 99.66 m².