Class 9

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6

(Assume π = \(\frac{22}{7}\) unless stated otherwise)

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm³ = 1 l)
Solution:
The circumference of the base of cylindrical vessel, C = 132 cm.
height, h = 25 cm
C = 2πr = 132

r = 21 cm
∴ Volume of Cylinder, V = πr²h

= 34650 cm³
1000 cubic cm. = 1 litre
∴ 34650 cm³ = … ? …

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution:
Let the inner diameter of a cylindrical vessel be d₁ and outer diameter be d₂.

height of pipe, h = 35 cm.
∴ volume of pipe, V = \(\pi r_{2}^{2} h-\pi r_{1}^{2} h\)

= 110 × 52
∴ V = 5720 cm³
Mass of pipe of 1 ccm is 0.6 gm.
∴ Mass of 5720 cm³. … ? …
= 5720 × 0.6
= 3432 Kg.

Question 3.
A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much ?
Solution:
Length of rectangular tin, l = 5 cm,
breadth, b = 4 cm,
height, h = 15 cm.
∴ Volume of rectangular tin, V = l × b × h
V = 5 × 4 × 15
= 300 cm³
Diameter of plastic cylinder, d = 7 cm,
∴ r = \(\frac{7}{2}\)
Volume of cylinder, v = \(\pi \mathrm{r}^{2} \mathrm{h}\)

= 11 × 7 × 5
= 385 cm³.
∴ Plastic cylinder’s volume is great.
= Volume of cylindrical tin – Volume of rectangular tin.
= 385 – 300
= 85 cm³.
∴ Volume of plastic tin is greater than rectangular tin by 85 cm³.

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base,
(ii) its volume. (Use π = 3.14)
Solution:
Curved surface area of cylinder = 94.2 cm²
height, h = 5 cm.
r = ?
V = ?
(i) Curved Surface area of cylinder, L.S.A.
A= 2πrh
94.2 = 2 × 3.14 × 5 × r

= 3.01 cm.
= 3 cm.
(ii) Volume of cylinder, V = πr²h
= 3.14 × (3)2 × 5
= 3.14 × 9 × 5
= 141.3 cm³.

Question 5.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m², find
(i) inner curved surface area of the vessel.
(ii) radius of the base,
(iii) capacity of the vessel.
Solution:
(i) Cost of painting for cylindrical vessel = Rs. 2200
Cost of painting is at the rate of Rs. 20 per m².
∴ Curved Surface area of vessel = \(\frac{2200}{20}\)
= 110 m²

(ii) Depth of vessel, h = 10 m
r = ?
Curved surface area of vessel, V = 2πrh

(iii) Capacity of the Vessel = Volume of Vessel
V = πr²h

V = 96.25 m³.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Solution:
Volume of a cylindrical vessel = 15.4 litres
1000 c.cm. = 1 litre.

height, h = 1 m,
radius, r = ?
Curved surface area =?
Volume of cylinder, V = πr²h
0. 0154 = \(\frac{22}{7}\) × r² × 1

r = 0.7 m
Now, curved surface area of cylinder vessel,

= 0.44 × 1.07
A = 0.4708 m²

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:
Diameter of pencil, d = 7 mm = 0.7 m.

length of pencil, h = 14 cm
diameter of pencil, d = 1 mm = 0.1 cm

Volume of wood in pencil, V

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
Diameter of pencil, d = 7 cm

height of soup, h = 4 cm
Volume of soup given for 1 patient is 154 cm³
Volume of soup given for 250 patients .. ? ….
= 154 × 250 = 38500 cm³
For 1000 cm³, 1 litre
For 38500 cm³ ….. ? …
= \(\frac{38500}{1000}\)
= 38.5 litres soup.

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.2.

Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 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?
Answer:
The blood groups of 30 students of class VIII are as follows

i) Out of 30 students only three students have blood group AB. This is rarest blood group.
ii) Most common blood group students are 12. Hence ‘O’ is the maximum blood group.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were 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:
Size of the class interval = 5
First class interval is 0 – 5 (5 is not included).

More number of people, i.e. 5 to 15 Km come.

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:
Size of class interval = 2
(i) Frequency distribution table :

(ii) Relative humidity is more, these data are taken during rainy season.
(iii) Maximum humidity = 99.2
Minimum humidity = 84.0
∴ Range = Maximum – Minimum
= 99.2 – 84.9
= 14.3

Question 4.
The heights of 50 students, measured to the nearest centimetres, 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 heights from the table ?
Solution:
Maximum height : 150 cm.
Minimum height : 173 cm.
Size of class interval = 5
(i) Frequence distribution table for grouped data:

(ii) Referring to above table, 50% students have height less than 165 cm.
Height of majority students is 160 – 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 millon ?
Solution:
(i) Grouped frequency distribution table:

(ii) In 8 days the concentration of sulphur dioxide is more than 0.11 parts per million.

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:
Frequency distribution table :

Question 7.
The value of n 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:
Frequency distribution table :

(ii) Digits which have more frequency are 3 and 9.
‘0’ is the digit which has less frequency.

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:
(i) Size of Class Interval = 5

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

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:
Size of the class interval = 0.5
Class interval : 2 – 2.5, 2.5 – 3 …….
Grouped Frequency distribution table:

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Statistics Ex 14.2 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Statistics Exercise 14.2, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Solution:
Match box is in the form of cuboid.
Its length, l = 4 cm.
breadth, b = 2.5 cm.
height, h = 1.5 cm.
∴ Volume of Cuboid, V = l × b × h
= 4 × 2.5 × 1.5
V = 15 cm³.
Volume of 1 match box is 15 cm³.
Volume of 12 match boxes …?… .
= 15 × 12
= 180 cm³.

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold ? (1 m³ = 1000 l)
Solution:
Length of cuboidal water tank, l = 6 m
breadth b = 5 m
height, h =4.5 m
∴ Volume of cuboidal water tank, V= l × b × h
= 6 × 5 × 4.5
V= 135 m³.
1 m³ = 1000 l.
135 m³ = ? = 135 × 1000
= 135000 lit.
∴ Number of litres of water tank = 1350000 litres.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid ?
Solution:
Volume of cuboidal vessel, V = 380 m³.
length, l = 10 m
breadth, b = 8 m height, h = ?
Volume of vessel, V= l × b × h
380 = 10 × 8 × h
380 = 80 h

∴ h = 4.75 m.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3.
Solution:
Length of cuboidal pit, l = 8 m
breadth, b = 6 m height, h = 3 m
∴ Volume of pit, V = l × b × h
= 8 × 6 × 3
= 144 m³.
∴ Cost of digging a cuboidal 1 m³ is Rs. 30
Cost of digging a cuboidal pit for 144 m³ … ? …
= 144 × 30
= Rs. 4320.

Question 5.
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
Volume of cuboidal tank, V = 50000 lit.
1 c.c = 1000 litres
∴ Volume of tank = \(\frac{50000}{1000}\) = 50 c.c
length of tank, l = 2.5 m
height, h = 10m
breadth,b = ?
Volume of cuboid, V = l × b × h

∴ b = 2m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last ?
Solution:
Length of cuboid water tank, l = 20 m
breadth, b = 15 m
height, h = 6 m
Volume, V = ?
Volume of water tank, V = l × b × h
= 20 × 15 × 6
= 1800 m³.
1m³ = 1000 litres
∴ 1800m³ = 1800 x 1000
= 1800000 litres.
Water required daily per man= 150 litres
∴ Quantity of water required for 4000 men?
= 4000 × 150
= 600000 litres.
Quantity of water for 1 day = 600000
Number of days for consuming 1800000 litres … ? …

= 3 Days.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown
Solution:
Length of rectangular godown, l₁ = 40 m
breadth, b₁ = 25 m
height, h₁ = 10 m
∴ Volume of godown, V= l₁ × b₁ × h₁
= 40 × 25 × 10
= 10000 m³.
Length of wooden crate, l₂ = 1.5 m
breadth, b₂ = 1.25 m
height, h₂ = 0.5 m
∴ Volume of wooden crate. V = l₂ × b₂ x h₂
= 1.5 × 1.25 × 0.5
= 0.9375 m³.
For 0.9375 m³, 1 crate

= 10666.66

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Each side of a cube, a = 12 cm.
∴ Volume of cube, V = a³
= (12)³
= 1728 cm³.
Solid cube is cut into 8 cubes of equal volume.
Volume of each small cube,

∴ l³ = 216
∴ l = 6 cm.
∴ Each side of small cube is 6 cm.
Ratio of outer area :

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
Depth of river, h = 3m
breadth, b = 40 m
1 Km. = 1000 m
∴ 2 Km. = 2000 m.
Speed of the river per hour is 2 Km.
For 60 minutes, 2000m
For 1 minute …?… = \(\frac{2000}{60}\)

∴ Depth of river in 1 minute, l = \(\frac{100}{3}\) m.
Volume of water flows in 1 minute,
V = l × b × h
= \(\frac{100}{3}\) × 40 × 3
= 4000 m³
∴ In One minute, 4000 m³ water reaches the sea.

KSEEB Solutions for Class 9 Maths

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4

(Assume π = \(\frac{22}{7}\), unless stated otherwise)

Question 1.
Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm.
Solution:

Surface area of the sphere = \(4 \pi r^{2}\)

= 22 × 3 × 21
= 1386 cm².

(ii) r = 5.6 cm. = \(\frac{56}{10}\)cm.
Surface area of the sphere = \(4 \pi r^{2}\)

= 394.24 cm².

(iii) r = 14 cm.
Surface area of the sphere = \(4 \pi r^{2}\)

= 88 × 28
= 2464 cm².

Question 2.
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m.
Solution:
(i) Diameter, d = 14 cm.

Surface area of a sphere = \(4 \pi r^{2}\)

= 616 cm².
(ii) Diameter, d = 21 cm.

Surface area of a sphere = \(4 \pi r^{2}\)

= 1386 cm².
(iii) Diameter, d = 3.5 m. = \(\frac{7}{2}\)

Surface area of a sphere = \(4 \pi r^{2}\)

= 38.5 m².

Question 3.
Find the total surface area of a hemisphere of radius 10 cm.
(Use π = 3.14 )
Solution:
r = 10 cm.
Total surface area of a hemisphere = \(3 \pi r^{2}\)

= 942 cm².

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Radius of spherical balloons be r₁ and r₂, r₁ = 7 cm. and r₂ = 14 cm.
Ratio of surface area


∴ Ratio of surface areas of the balloons = 1 : 4.

Question 5.
A hemispherical bowl made of brass has an inner diameter of 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm².
Solution:
Inner diameter of hemispherical bowl = 10.5 cm.

Curved surface area of the bowl = \(2 \pi r^{2}\)

= 173.25 cm².
Cost of tin-plating for 100 cm² is Rs. 16.
Cost of tin-plating for 173.25 sq.cm. …?…

= Rs. 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Total surface area of a sphere= 154 cm²
Radius, r =?

∴ r = 3.5 cm

Question 7.
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the diameter of earth be d’ unit
∴ Diameter of Moon = \(\frac{d}{4}\)
radius of the earth = \(\frac{\mathrm{d}}{2}\)
radius of the moon = \(\frac{1}{2} \times \frac{\mathrm{d}}{4}=\frac{\mathrm{d}}{8}\)
Ratio of surface area,

= 1 : 16

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Thickness of the hemispherical bowl = 0.25 cm.
inner radius, r = 5 cm.
∴ Outer Surfce area= 5 + 0.25 = 5.25 cm.
Outer Curved Surface area = 2πr²

= 173.25 cm².

Question 9.
A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder
(iii) ratio of the areas obtained in (i) & (ii).

Solution:
(i) If radius of a sphere is ‘r’ cm, its Surface area = 4πr²
(ii) height of cylinder, h = diameter of sphere
h = r + r
∴ h = 2r
∴ Curved surface area of cylinder = 2πrh
= 2πr × h
= 2πr × 2r
= 4πr²

= 1 : 1

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.1.

Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.1

Question 1.
Give five examples of data that you can collect from your day-to-day life.
Answer:
Some of the examples of data that we can collect from our day-to-day life are as follows :

1. Number of TV viewers in the city.
2. Number of Colleges in the city.
3. Number of sugar factories in the city.
4. Measuring the height of students in the classroom.
5. Number of children below 15 years in India.

Question 2.
classify the data in Q.1 above as primary or secondary data.
Answer:
i) Secondary Data
Because in example (iv) to find our heights of students, the investigator is in contact with the student.
ii) E.g., (i), (ii), (iii) and (v) are primary data. Because these are information collected from a source.

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Probability Exercise 15.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 15 Probability Exercise 15.1.

Karnataka Board Class 9 Maths Chapter 15 Probability Ex 15.1.

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Total number of hittings of ball = 30
∴ n(S) = 30.
Total number of boundary = 6
Total number of No boundaries = 30 – 6 = 24
∴ n(E) = 24
Probability of No boundary hits =

Question 2.
1500 families with 2 children were selected randomly, and the following data were recorded :

Number of girls in a family	2	1	0
Number of families	475	814	211

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl,
(iii) No girl
Also check whether the sum of these probabilities is 1.
Solution:
Total number of families =1500
(i) Family with 2 girls = 475
∴ P₁(Family with 2 girls)

(ii) Family with 1 girl = 814
∴ P₂ (Family with 1 girl)

(iii) Family without girls = 211
∴ P₃ (Family without girls)

Sum of probabilities

∴ Sum of probabilities is 1.

Question 3.
Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.
Solution:
Total number of students in IX class = 40
∴ n(S) = 40
Number of students born in August = 6
∴ n(E) = 6

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes :

Outcome	3 Heads	2 Heads	1 Head	No head
Frequency	23	72	77	28

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Total number of events tossing 3 coins simultaneously = 200
∴ n(S) = 200
Frequency in which two heds up = 72
∴ n(E) = 72

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below :

Monthly Income (in Rs.)	Vehicles per family
	0	1	2	Above 2
Less than 7000	10	160	25	0
7,000 – 10,000	0	305	27	2
10,000- 13,000	1	535	29	1
13,000- 16,000	2	469	59	25
16,000 or more	1	579	82	88

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs. 7000 per month and does not own any vehicle.
(iv) earning Rs. 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Total number of families surveyed = 2400
∴ n(S) = 2400
(i) earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles = 29
∴ n(A) = 29

(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579
∴ n(B) = 579

(iii) earning less than Rs. 7000 per month and does not own any vehicle =10
∴ n(C) = 10

(iv) earning Rs. 13000 – 16000 per month and owning more than 2 vehicles = 25
∴ n(D) = 25

(v) owning not more than 1 vehicle 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062
∴ n(E) = 2062

Question 6.
Refer to Table 14.7, Chapter 14.
(i) Find the probability that a student obtained less than 20% in the Mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution:
(i) Number of students obtained less than 20% in the Mathematics test = 7
n(A) = 7
Total number of students = 90
n(S) = 90

(ii) Number of students obtained marks 60 or above
= 15 + 8 = 23 ∴ n(B) = 23
Total Number of students = 90
n(S) = 90

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

Opinion	No. of students
like dislike	135 65

Find the probability that a student chosen at random.
(i) likes statistics,
(ii) does not like it.
Solution:
Total number of students = 135 + 65
= 200
n(S) = 200
(i) Number of students who like Statistics
= 135
n(A) = 135

(ii) Number of students who does not like Statistics = 65
n(B) = 65

Question 8.
Refer to Q. 2, Exercise 14.2. What is the empirical probability that an engineer lives :
(i) less than 7 km from her place of work ?
(ii) more than or equal to 7 km from her place of work ?
(iii) within \(\frac{1}{2}\) km from her place of work.
Solution:
Referring to Q.2 Exercise 14.2
(i) Total Number of engineers = 40
n(S) = 40
Number of engineers living less than 7 km from her place of work = 9
n(A) = 9

(ii) Number of engineers living more than or equal to 7 km from her place of work.
= 40 – 9 = 31
n(B) = 31

(iii) Number of engineers living within \(\frac{1}{2}\) Km from her place of work = 0
n(C) = 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four- wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg.)
4.97 5.05 5.08 5.03 5.00 5.06
5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Total number of bags of wheat flour = 11
n(S) = 11
Number of bags which have flour more than 5 kg. = 7
n(A) = 0

Question 12.
In Q. 5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12- 0.16 on any of these days.
Solution:
In Q.5, Exercixe 14.2 of the Text Book,
Number of days recorded = 30
n(S) = 30
Concentration of Sulphur Dioxide (SO₂) in the interval 0.12 – 0.16 = 2
n(A) = 2

Question 13.
In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution:
In Q. 1, Exercise 14.2 of the Text Book,
Total Number of students = 30
n(S) = 30
Number of students having Blood group AB = 3
n(A) = 3

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Probability Exercise 15.1, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3

(If values are not given for ‘n’ Assume n = \(\frac{22}{7}\).)

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
Diameter of the base of a cone, d = 10.5 cm.
height h = 10 cm.
Curved Surface Area, C.S.A. = ?
d = 10.5 cm, \(=10 \frac{1}{2}=\frac{21}{2}\) cm.
h = l = 10 cm.
∴ Curved Surface Area of a cone = πrl

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Diameter of base of a cone, d = 24 m.
height, h = 21 m.
Total Surface Area, T.S.A. = ?
d = 24 m, ∴ r = \(\frac{\mathrm{d}}{2}=\frac{24}{2}=12\)
h = l = 21 m.
T.S.A of a cone = πr(r + l)

= 1244.57 m²

Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base, and
(ii) total surface area of the cone.
Solution:
If Curved surface area of a cone, πrl = 308 cm².
height, h = l = 14 cm.
Then, (i) r = ?, (ii) T.S.A. = ?
(i) CSA of a Cone = πrl

∴ r = 7 cm
(ii) TSA of a cone= πr(r + l)

= 22 × 21
= 462 cm²

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70.
Solution:
(i) ABC is a conical tent,
height, h = 10 m.
radius, r = 24 m.
Slant height of the tent, l = ?

In ⊥∆AOC, ∠O = 90°
As per Pythagoras theorem,
AC² = AO² + OC²
l² = (01)² +(24)²
= 100 + 576
l² = 676
l = \(\sqrt{676}\)
Area of canvas required to prepare tent, C.S.A. = ?
C.S.A.= πrl

Cost of 1 m² canvas is Rs. 70,

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14.)
Solution:
ABC is a conical tent.
Height, AO = h = 8 m.
base radius, OC = r = 6 cm
In ⊥∆AOC,
AC² = AO² + OC²
l² = (8)² +(6)²
l² = 64 + 36
l² = 100
∴ l = \(\sqrt{100}\)
l = 10 m
Curved surface Area of cone C.S.A. = πrl
= 3.14 × 6 × 10
= 188.4 m²
Let the length of tarpaulin required be 1 m.
In that 20 cm. (0.2 m) is wastage means remaining tarpaulin is (1 – 0.2 m)
breadth, b = 3m
∴ Area of tarpaulin = curved surface area of tent.
(l – 0.2 m) × 3 = 188.4 m²

l – 0.2 = 62.8
∴ l = 62.8 + 0.2
∴ l = 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m².
Solution:
Slant height of a cone, l = 25 m
Diameter, 2r = 14m ∴ r = 7 m
Curved Surface Area = πrl

= 22 × 25
= 550 m².
Cost of white-washing is Rs. 210 per m².
Cost of white-washing for 550 m² … ? …

= Rs. 1155.

Question 7.
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:
A joker’s cap is in the form of a right circular cone.
radius of base, r = 7 cm.
height, h = 24 cm.
slant height, 1 = ?
Curved surface area, C.S.A.= ?

In ⊥∆AOC, ∠AOC = 90°
∴ AC² = AO² + OC²

∴ Curved surface area= πrl

= 550 m²
Area of 1 cap is 550 m²
Area of 10 caps…,v … ? …
= 550 × 10
= 5500 m².

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m², what will be the cost of painting all these cones ?
(Use π = 3. 14 and take \(\sqrt{1.04}\) = 1.02)
Solution:
Diameter of a cone, d = 40 cm = 0.40 m.

height, h = l m.

Curved surface area of cone = πrl

= 0.64056 m²
Area of 1 cone = 0.64056 m²
Area of 50 cone = ?
= 50 × 0.64056
= 32.028 m²
Cost of painting for 1 sq.m, is Rs. 12.
Cost of painting for 32.028 sq.m. …? …
= 32.028 × 12
= Rs.384.37.

KSEEB Solutions for Class 9 Maths

Students can Download Maths Chapter 9 Commercial Arithmetic Ex 9.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.5

Question 1.
Find the simple interest on Rs. 2500 for 4 years at 6 \(\frac { 1 }{ 4 }\) % per annum.
Answer:
P = Rs. 2500
T = 4 years

Simple interest is Rs. 625.

Question 2.
Find the simple interest on Rs. 3500 at the rate of 214% per annum for 165 days.
Answer:
P= Rs. 3500

Simple interest = Rs. 39.55.

Question 3.
In what period will Rs. 5200 amounts to Rs. 7,384 at 12% per annum simple interest?
Answer:
P = Rs. 5200
Amount = Rs. 7384
R = 12%
T = ?
Simple interest = Amount – principal = 7384 – 5200 = Rs. 2184

If amounts to Rs. 7,384 in 3.5 years

Question 4.
Ramya borrowed a loan from a bank for buying a computer. After 4 years she paid Rs. 26,640 and settled the account. If the rate of interest is 12% per annum what was the sum she borrowed?
Answer:
T = 4 years
R = 12%
After 4 years she paid Rs. 26,640.
This is nothing but amount.
Let the principal borrowed be R Amount = principal + S.I.
∴ Simple interest = Amount – principal S.I. = 26,640 – P

The principal borrowed – Rs. 18,000

Question 5.
A sum of money triples itself in 8 years. Find the rate of interest.
Answer:
Let the principal be x. It amounts to 3x in 8 years.
Amount = principal + simple interest
∴ Simple interest = Amount – principal = 3x – x = 2x

R = 25%

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.2

(If values are not given for ‘π’ Assume π = \(\frac{22}{7}\))

Question 1.
The curved surface area of a right circular cylinder of a height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Solution:
Lateral surface area of cylinder which has height 14 cm is 88 cm².
Diameter of bottom of cylinder, d = ?
h = 14 cm, L.S.A. = 88 cm², d = ?
Lateral surface area of cylinder, L.S.A.
L.S.A.= 2πrh = 88

88 × r = 88
∴ r = \(\frac{88}{88}\) ∴ r = 1 cm.
∴ Diameter, d = 2r =2 × 1 = 2 cm.

Question 2.
It is required to make a closed cylinderical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same ?
Solution:
Height of cylindrical water tank, h = 1 m
Diameter of bottom, d = 140 cm.
h = 1 m, d = 140 cm,

r = 0.7 m.
Area of metal sheet required to prepare water tank, A
A = 2πr (r + h)

= 44 × 0.1 × 1.7
= 4.4 × 1.7
A = 7.48 m².

Question 3.
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Solution:
Length of metal pipe, l = 77 cm,
h = 77 cm.
Inner diameter = 4 cm, ∴ r₁ = 2 cm.
Outer diameter= 4.4 cm. ∴ r₂ = 2.2 cm.
(i) Inner surfce area of pipe (C.S.A) = 2πr₁h

= 2 × 22 × 2 × 11
= 968 cm².
(ii) Outer surface area of pipe (C.S.A) = 2πr₂h

= 2 × 22 × 2.2 × 11
= 44 × 24.2
= 1064.8 cm².
(iii) Total Surface area of Pipe (T.S.A.) = Inner CSA + Outer CSA + Area of circular ring.
= 2πr₁h + 2πr₂h + (2πr₂²h – 2πr₁²h)

= 2032.8 + 5.28
∴ T.S.A. = 2038.08 cm².

Question 4.
The diameter of a roller is 84 cm and its length is 120 m. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
Solution:
Roller is in the form of cylinder.
Diameter, d = 84 cm,

l = h = 120 cm ⇒ 1.2 m.
∴ Curved surface area of Roller = 2 πrh

= 44 × 0.072
= 3.168 m²
∴ Area of 1 round of roller = 3.168 m².
Area of 500 rounds … ? …
= 3.168 × 500
= 1584 m².
∴ Total area of a playground is 1584 m².

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m².
Solution:
Diameter of cylindrical pole, d = 50 cm.

∴ r = 0.25 m
height of pole, h = 3.5 m.
Curved surface area of pole = 2πrh

= 44 × 0.25 × 0.5
= 5.5 m².
Cost of painting for 1 m2 is Rs. 12.50,
Cost of painting for 5.5 m² is … ? …
= 5.5 × 12.50
= Rs.68.75.

Question 6.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
Curved surface arc of cylinder = 4.4 m²
radius, r = 0.7 m height, h = ?
C.S.A. of cylinder, A= 2πrh
2πrh = 4.4

44 × 0.1 × h = 4.4
4.4 h = 4.4

∴ h = 1 m.

Question 7.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m².
Solution:
Inner diameter of circular well, d = 3.5 m.
Radius, r = \(\frac{3.5}{2}\) m
Depth of the well – height, h = 10 m.
(i) Inner surfacer area of circular well, C.S.A. = 2πrh
C.S.A. = 2πrh

= 22 × 0.5 × 10
= 110 m²
(ii) Cost of plastering of area 1 m² is Rs. 40
Cost of plastering 110 m²… ? …
= 110 × 40
= Rs. 4400.

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
Diameter of cylindrical pipe, d = 5 cm
= 0.5 m

Length of pipe = height, h = 28 m.
Curved surface area of pipe = ?
C.S.A.= 2πrh

Question 9.
Find
(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \(\frac{1}{12}\) of the steel actually used was wasted in making the tank.
Solution:
(i) Petrol tank is in the form of cylinder.
Diameter, d = 4.2 m

h = 4.5 m
TSA of petrol tank = 2πr(r + h)

= 44 × 0.3 × 6.6
= 87.12 m²
(ii) ∴ Area of iron required for manufacturing tanker = 87.12 m².
If iron used is 1 sq. m then there is a loss of \(\frac{1}{12}\).
Iron required except loss = \(1-\frac{1}{12}\)
= \(\frac{11}{12}\) m²
Area of \(\frac{11}{12}\) m² is 87.12 m².
Area of 1 m² … ? …

= 95.04 m².

Question 10.
In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:
Diameter of cylindrical frame, d= 20 cm,
∴ r = 10 cm.
height, h = 30 cm.
For folding 2.5 cm is required.
∴ Total height = 30 + 2.5 + 2.5
= 35 cm.
Curved surface area of cloth covered, A
A= 2πrh

= 44 × 50
= 2200 cm².

Question 11.
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using card-board. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution:
Penholder is in the shape of a cylinder, which one of the edges is open and another is closed.
radius, r = 3 cm, height, h = 10.5 cm.
Total cardboard required by one Competitor: = Curved Surface Area + Area of bottom = 2πrh + πr²


∴ Total cardboard required for 35 competitors is :

KSEEB Solutions for Class 9 Maths

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.6

Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:
Data: Two circles having centres A and B, intersect at C and D.
To Prove: ∠ACB = ∠ADB.
Construction: Join A and B.
Proof: In ∆ABC and ∆ABD,
AC = AD (∵ radii of same circle are equal)
BC = DD
AB is common.
∴ ∆ABC ≅ ∆ABD (SSS Postulate.)
∴ ∠ACB = ∠ADB.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm., find the radius of the circle.

Solution:
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre.
Distance between AB and CD is 6 cm.
To Prove: Radius of the circle, OP =?
Construction: Join OP and OQ, OB, and OD.
Proof: Chord AB || Chord CD.
AB = 5 cm, and CD =11 cm.
OP ⊥ AB
∴ BP = AP = \(\frac{5}{2}\) = 2.5 cm.
OQ⊥CD
∴ CQ = QD = \(\frac{11}{2}\) = 5.5 cm.
PQ = 6 cm. (Data)
Let OQ = 2 cm then, OP = (6 – x) cm.
In ∆BPO, ∠P = 90°
As per Pythagoras theorem,
OB² = BP² + PO²
= (2.5)² + (6 – x)²
= 6.25 + 36 – 12x + x²
OB² = x²– 12x + 42.25 …………….. (i)
In ∆OQD, ∠Q= 90°
∴ OD² = OQ² + QD²
= (x)² + (5.5)²
OD² = x² + 30.25 ……………….. (ii)
OB = OD (∵ radii of same circle)
From (i) and (ii).
x² – 12x + 42.25 = x² + 30.25
-12x = 30.25 – 42.25
-12x = -12
12x = 12
∴ x = \(\frac{12}{12}\)
∴ x = 1 cm.
From (ii),
OD² = x² + 30.25
= (1)² + 30.25
= 1 + 30.25
∴ OD² = 31.25
OD = \(\sqrt{31.25}\)
∴ OD = 5.59 cm.
∴ Radius of circle OP = OD = 5.59 cm.

Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm, from the centre, what is the distance of the other chord from the center?

Solution:
Data: Chords of a circle are 6 cm. and 8 cm. are parallel. The smaller chord is at distance of 4 cm. from the centre.
To Prove: Distance between bigger chord and centre =?
Construction: Join OA and OC.
Proof: AB || CD, AB = 6 cm, CD = 8 cm.
OP⊥CD, OQ⊥CD.
In ∆OPA, ∠P = 90°
∴ OA² = OP² + PA² (According to Pythagoras theorem)
= (4)² + (3)² = 16 + 9
OA² = 25
∴ OA = 5 cm.
OA = OC = 5 cm. (radii of the same circle.)
Now, in ∆OQC,
OC² = OQ² + QC²
(5)² = x² + (4)²
25 = x² + 16
x² = 25 – 16 = 9
∴ x = \(\sqrt{9}\) ∴ x = 3 cm.
∴ Bigger chord is at a distance of 3 cm from the centre.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:
Data: The vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle.
To Prove: ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. OR
∠ABC= \(\frac{1}{2}\) [∠DOE – ∠AOC].
Construction: OA, OC, OE, OD are joined.
Proof: In ∆AOD and ∆COE,
OA = OC, OD = OE radii of same circle.
AD = CE (Data)
∴ ∆AOD ≅ ∆COE (SSS Postulate)
∠OAD = ∠OCE ……….. (i)
∴∠ODA = ∠OEC …………. (ii)
OA = OD
∴ ∠OAD = ∠ODA …………. (iii)
From (i) and (ii),
∠OAD = ∠OCE = ∠ODA = ∠OEC = x°.
In ∆ODE, OD = OE
∠ODE = ∠OED = y°.
ADEC is a cyclic quadrilateral.
∴ ∠CAD + ∠DEC = 180°
x + a + x + y = 180
2x + a + y = 180
y = 180 – 2x – a ……….. (iv)
But, ∠DOE = 180 – 2y
∠AOC = 180 – 2a

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Solution:
Data: ABCD is a rhombus. Circle is drawn taking side CD diameter. Let the diagonals AC and BD intersect at ‘O’.
To Prove: Circle passes through the point ‘O’ of the intersection of its diagonals.
Proof: ∠DOC = 90° (Angle in the semicircle) and diagonals of rhombus bisect at right angles at ‘O’.
∴ ∠DOC = ∠COB = ∠BOA = ∠AOD = 90°
∴ Circle passes the point of intersection of its diagonal through ‘O’.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Solution:
Data: ABCD is a parallelogram. The circle through A, B, and C intersect CD at E. AE is joined.
To Prove: AE = AD
Proof: ∠AEC + ∠AED = 180° …………. (i) (linear pair)
ABCE is a cyclic quadrilateral.
∴ ∠ABC + ∠AEC = 180° ………….(ii) (opposite angles)
Comparing (i) and (ii),
∠AEC = ∠AED = ∠ABC + ∠AEC
∠AED = ∠ABC ………….. (iii)
But, ∠ABC = ∠ADE (Opposite angles of quadrilateral)
Substituting in equation (iii),
∠AED = ∠ADE
∴ AE = AD.

Question 7.
AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
Solution:
Data : AC and BD are chords of a circle bisect each other at ‘O’.
To Prove:
i) AC and BD are diameters.
ii) ABCD is a rectangle.
Construction: AB, BC, CD and DA are joined.

Proof: In ∆AOB and ∆COD,
AO = OC (Data)
BO = OD (Data)
∠AOB = ∠COD (vertically opposite angles)
∴ ∆AOB ≅ ∆COD (SAS postulate)
∴ ∠OAB = ∠OCD
These are pair of alternate angles.
∴ AB || CD and AB = CD.
∴ ABCD is a parallelogram.
∴ ∠BAD = ∠BCD (Opposite angles of parallelogram)
But. ∠BAD + ∠BCD = 180 (∵ Angles of cyclic quadrilateral)
∠BAD + ∠BAD = 180
2(∠BAD) = 180
∴ ∠BAD = \(\frac{180}{2}\)
∴ ∠BAD = 90°.
If angles of a quadrilateral are right angles it is rectangle. ABCD is a recrtangle.
∠BAD = 90°
∠BAD is separated from chord BD.
∴ This is the angl in semicircle.
∴ Chord BD is a diameter.
Similarly, ∠ADC = 90°
∴ Chord AC is a diameter.

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – \(\frac{1}{2}\) A, 90° – \(\frac{1}{2}\) B and 90° – \(\frac{1}{2}\) C.
Solution:
Data: AD, BE and CF are angular bisectors of angles A, B and C of ∆ABC intersects its circumference at D, E and F respectively.

To prove: Angles of ∆DEF are 90° – \(\frac{1}{2}\) A, 90° – \(\frac{1}{2}\) B and 90° – \(\frac{1}{2}\) C.
Proof: AD, BE and CF are angular bisectors of angles A, B and C of ∆ABC.
∴ ∠BAD = ∠CAD = \(\frac{\angle \mathrm{A}}{2}\)

Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lies on the two circles.
Prove that BP = BQ.

Solution:
Data : Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.
To Prove: BP = BQ
Construction: Join AB.
Proof: Two congruent triangles with centres O and O’ intersects at A and B. Through A segment PAQ is drawn so that P, Q lie on the two circles.
Similarly, ∠AQB= 70° in circle subtended by chord AB. Because Angles subtended by circumference by same chord.
∴ ∠APB = ∠AQB = 70°.
Now, in ∆PBQ, ∠QPB = ∠PQB.
∴ Sides opposite to each other are equal.
∴ BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Solution:
Data: In ∆ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect each other. O is the centre of the circle.
To Prove: Angle bisector of ∠A and perpendicular bisector of BC intersect at D.
Construction: Join OB, OC.
Proof: Angle subtended at the Centre
= 2 × angles subtended in the circumference.
∠BOC = 2 × ∠BAC
In ∆BOE and ∆COE,
∠OEB = ∠OEC = 90° (∵ OE⊥BC)
∴ BO = OC (radii)
OE is common.
∴ ∆BOE ≅ ∆COE (RHS postulate)
But, ∠BOE + ∠COE = ∠BOC
∠BOE + ∠BOE = ∠BOC
2∠BOE = ∠BOC
2∠BOE = 2∠BAC
∴ ∠BOE = ∠BAC
But, ∠BOE = ∠COE = ∠BAC
∠BAD = \(\frac{1}{2}\) ∠BAC
∠BAD = \(\frac{1}{2}\) ∠BOE
∠BAD = \(\frac{1}{2}\) ∠BOD
∴ ∠BOD = 2∠BAD
∴ The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference.
∴ Angle bisector of ∠A and perpendicular bisector of BC intersect at D.

KSEEB Solutions for Class 9 Maths