KSEEB Solutions for Class 9 Science Chapter 11 Work and Energy

Karnataka Board Class 9 Science Chapter 11 Work and Energy

KSEEB Solutions Class 9 Science Chapter 11 Intext Questions

Question 1.
A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that they force acts on the object through the displacement. What is the work done in this case?
Answer:
Work done on an object = 7N × 8m = 56 Nm or 56 J.

Question 2.
When do we say that work is done?
Answer:
Two conditions need to be satisfied for work to be done:

  1. a force should act on an object and
  2. the object must be displaced.

Question 3.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
Let a constant force F act on an object. Let the object be displaced through a distance S in the direction of the force. Let W be work done. We define work to be equal to the product of the force and displacement.
Work done = force × displacement W = F × S.

Question 4.
Define 1 J of work.
Answer:
1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Question 5.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
Answer:
Force exerted, F = 140N.
Displacement, S = 15m
Work done in ploughing the field
W = F × S
= 140 × 15
= 2100J
= 2.1 × 103 J.

Free Kinetic Energy Calculator – calculate kinetic energy step by step.

Question 6.
What is the kinetic energy of an object?
Answer:
Objects in motion possess energy, we call this energy kinetic energy.

Question 7.
Write an expression for the kinetic energy of an object.
Answer:
KE = \(\frac{1}{2}\) mv2.

Question 8.
The kinetic energy of an object of mass, m moving with a velocity of 5 ms-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
Kinetic energy, k = \(\frac{1}{2}\) mv2.
Where m = mass of the object
V = velocity of the object.
Here mass (m) is the same in both cases (∵ object is same)
\(\frac{\mathbf{k}_{1}}{\mathbf{k}_{2}}=\left(\frac{\mathbf{v}_{1}}{\mathbf{v}_{2}}\right)^{2}\)
Initial kinetic energy k1 = 25J
Initial velocity V2 = 5 ms-1
New kinetic Energy K2 = ?
New velocity v2 = 3v1 = 3 × 5 = 10 ms-1
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 1
∴ When velocity is increased three times its K.E. is 225J.

Question 9.
What is power?
Answer:
Power is defined as the rate of doing work or the rate of transfer of energy w
P = \(\frac{w}{t}\)

Question 10.
Define 1 watt of power.
Answer:
Power is 1 W when the rate of consumption of energy is 1 JS-1.

Question 11.
A lamp consumes 1000 J of electrical energy in 10s. What is its power?
Answer:
Power = 1000 J Work w = 1000J
Time = 10 s
Power of lamp, P = w/t
= 1000/10 = 100W
∴ Power of a lamp = 100 KW.

Question 12.
Define average power.
Answer:
We obtain average power by dividing the total energy consumed by the total time taken.

KSEEB Solutions for Class 9 Science Chapter 11 Textbook Exercises

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’

  1. Suma is swimming in a pond.
  2. A donkey is carrying a load on its back.
  3. A windmill is lifting water from a well.
  4. A green plant is carrying out photosynthesis.
  5. An engine is pulling a train.
  6. Food grains are getting dried in the sun.
  7. A sailboat is moving due to wind energy.

Answer:

  1. Work is being done by Seema because she displaces the water by applying the force.
  2. No work is being done by the donkey because the direction of force i.e the load is vertically downward and displacement is along the horizontal. If displacement and force are perpendiculars then no work is done.
  3. Work is done because the windmill is lifting the water i.e., it is changing the position of water.
  4. No work is done because there are no force and displacement.
  5. Work is done because the engine is changing the position of the train.
  6. No work’ is done because there is no force and no displacement.
  7. Work is done because the force acting on the boat is moving it.

Question 2.
An object was thrown at a certain angle to the ground moves in a curved path and falls back to the ground the initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Work done by the force of gravity, W = mgh.
Where h = difference in height of initial and final positions of the object.
According to the question, the initial and final positions of the object lie in the same horizontal line. So h = 0.
∴ Work done W = mg × 0 = 0

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
In the case given in the question, the battery has chemical energy which is converted into energy. Electric energy provided to the bulb further converted into light energy.

Question 4.
A certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.
Answer:
Mass, m = 20 kg.
Initial velocity, u = 5 ms-1
Final velocity v = 2ms-1
Work done by the force = change in kinetic energy
= Final kinetic energy – Initial kinetic energy
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 2

Question 5.
A mass of 10kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravitational force W = mgh
Where h = Difference in the heights of initial and final positions of the object.
Here both the initial and final positions are on the same horizontal line.
So there is no difference in height i.e., h = 0.
∴ work done W = mg × 0 = 0.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
The total mechanical energy remains constant / as P.E. of the freely falling object decreases. Its kinetic energy and the kinetic energy increases on account of an increases its velocity) the law of conservation of energy is not violated.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:

  1. Muscular energy into kinetic energy
  2. the kinetic energy of the rotation of the which into kinetic energy of the bicycle.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock and fall to move it. The energy spent in doing so is absorbed by the.rock. This energy is converted into potential energy of the configuration of the rock which results in its deformation. However t his deformation is not visible on account of the huge size of the rock.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in Joules?
Answer:
Energy consumed w = 250 units
= 250 kwh
= 250 × 1000 W × 3600 S
= 250 × 1000 J/S × 3600 S
= 9 × 108 J.

Question 10.
An object of mass 40 kg is raised to a height of 5m above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down.
Answer:
Mass m = 40 kg.
height h = 5 m.
Potential energy PE = mgh = 40 × 9.8 × 5 = 1960J.
KE at half way down = PE at halfway down = \(\mathrm{mg} \frac{\mathrm{h}}{2}\)
= 40 × 9.8 × \(\frac{5}{2}\) = 980 J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 3
The satellite round the earth moves in a circular orbit. Here the force of gravity acts towards the centre of the earth and displacement of the satellite is along the tangent of the circular path that means therefore and displacement are perpendicular to each other.
So, work done, W = F.S cosθ
= F × S cos 90°
= F × S × 0
= 0
That is, no work is done by the force of gravity.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? think. Discuss this question with your friends and teacher.
Answer:
If an object moves with a constant velocity (i.e, there is no acceleration) then no force acts on it. As the object is moving ie it is displaced from one position to another position.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
The person has no movement ie his displacement is zero. So the person had done no work (∵ work is done only when the object is displaced).

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
Time = 10 h = 10 × 60 min
= 10 × 60 × 60 S
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 4
Energy = power × time
= 1500 × 10 × 60 × 60
= 5.4 × 107 J.

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate why does the bob eventually comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
When the pendulum oscillates in the air, the air friction opposes its motion. So some part of the kinetic energy of the pendulum is used to overcome this friction. With the passage of time, the kinetic energy of the pendulum goes on decreasing and finally becomes zero. The kinetic energy of the pendulum is transferred to the atmosphere. So energy is being transferred ie it is converted into one form to another. So here is no violation of the law of conservation of energy.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
The work done on the object to bring the object to rest
= change in kinetic energy
= Final kinetic energy – Initial Kinetic energy
Here final kinetic energy is zero because the object is brought to rest.
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 5

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at the velocity of 60km/h?
Answer:
Mass, m = 1500 kg.
Initial velocity, u = 60 kmh-1
= 60 × \(\frac{5}{18}\) = 16.67 ms-1
Final velocity, v = 0
(∵ the car comes to rest)
Work done to stop the car = change in kinetic energy
= Final kinetic energy – Initial Kinetic energy
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 6

Question 18.
In each of the following a force, F is acting on an object of mass M. The direction of displacement is from west to east shown by the longer arrow. Observe the diagram carefully and state whether the work done by the force is negative, positive or zero.
KSSEB Solutions for Class 9 Science Chapter 11 Work and Energy 7
Answer:
Cas i) The force and displacement are perpendicular to each other. So S = 90°
Work done = FScos θ°
= FScos 90°
= FS × 0 = 0
(∵ cos90° = 0)
Cas ii) The force and displacement are in the same direction so θ = 0°
Work done = FScos θ°
= FScos 90°
= FS × 1 = FS
(∵ cos θ° = 1)
That is the work done is positive.
Cas iii) The force and displacement in opposite direction θ = 180°
Workdone = FScos θ°
= FScos 180°
= FS × -1 = FS
(∵ cos 180° = -1)
That is workdone is negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are acting on it if the resultant of all the forces acting is zero.

Question 20.
Find the energy in kWh consumed in 10 hours by four devices of power 500 w each.
Answer:
Total power p = 500w × 4 = 2000w.
Time, t = 10h
Energy = p × t.
= 2000w × 10h
= 2kwh × 10h = 20kwh.
∴ Energy = 20kwh.

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
P.E. of the configuration of the body and the ground (the body may be deformed and the ground may at the place of collision).
This process energy in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy charges is called dissipation of energy.

KSEEB Solutions for Class 9 Science Chapter 11 Additional Questions and Answer

Fill in the blanks:
Question 1.
Work done = ___________
Answer:
force, displacement

Question 2.
An object having the capability to do work is said to possess __________
Answer:
energy.

Question 3.
__________ is the energy possessed by an object due to its motion.
Answer:
Kinetic energy

Question 4.
P.E
Answer:
mgh.

Question 5.
Larger unit of energy is called
Answer:
kilojoule.

Answer the following questions:

Question 1.
Can there be displacement of an object in the absence of any force acting on it?
Answer :
In the absence of any force on the object i.e., F = 0, ma, (as F = ma) since m ≠ 0 and a = 0. In such a case the object is either at rest or in a state of uniform motion in a straight line in the latter case there is a displacement . of the object without any force acting on it.

Question 2.
How does a bullet pierce a target?
Answer :
A bullet moves with large velocity and as such possesses a lot of kinetic energy the work in piercing the target is derived from the kinetic energy of the bullet.

Question 3.
Why do some engines require fuels like petrol and diesel?
Answer :
Internal combustion heat engines use the chemical energy of fossil fuel (petrol and diesel) for their operation. These engines first convert the chemical energy of the fuels into heat energy. Which is later on converted into mechanical energy.

Question 4.
Calculate the work done by a body. By the force of 5 N makes to move through a distance of 12 m.
Answer :
W = F x s = 5 x 12 = 60 joules

Question 5.
When force 6 N applied on a wall, the wall remains in the same position, calculate the work done.
Answer :
W = F x s
W – 6 x s = 6 x 0 = 0 Joules No work is done on the body

Question 6.
Electricity is the most convenient form of energy. Why?
Answer :
It can be converted into other forms of energy easily.
It can be produced by different means. It is ecofriendly

Question 7.
Calculate the power experienced by a source that can do work of 50 joules in 5 seconds.
Answer :
Power = \($\frac{\text { Workdone }}{\text { Time taken }}$\)
\(=\frac{50}{5} \frac{\text { Joules }}{\text { Second }}\) = 10 J/s = 10 watts

KSEEB Solutions for Class 9 Science

 

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KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
cosec2 A – cot2 A = 1
cosec2 A = 1 + cot2 A
cosec2 A = cot2 A + 1
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 1

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
i) sin2 A + cos2 A = 1
sin2 A = 1 – cos2 A
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 2
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 3

Question 3.
Evaluate :
i) \(\frac{\sin ^{2} 63^{\circ}+\sin 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 4

ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° + cos 25°
= sin2 25° + cos2 25°
= 1. [∵ cos2 θ + sin2 θ = 1]

Question 4.
Choose the correct option. Justify your choice.
i) 9 sec2 A – 9 tan2 A.
A) 1
B) 9
C) 8
D) 0
Solution:
B) 9
9 sec2 A – 9 tan2 A
= 9(sec2 A – tan2 A)
= 9 × 1
= 9

ii) (1+tan θ + sec θ) (1+ cot θ- cosec θ) =
A) 0
B) 1
C) 2
D) -1
Solution:
C) 2
(1+tan θ + sec θ) (1+ cot θ- cosec θ)
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 5
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 6

iii) (sec A + tan A) (1 – sin A) =
A) sec A
B) sin A
C) cosec A
D) cos A
Solution:
D) cos A
(sec A + tan A) (1 – sin A)
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 7

iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) =
A) sec2 A
B) -1
C) cot2 A
D) tan2 A
Solution:
D) tan2 A
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 8

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 9
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 10

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 11
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 12

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 13
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 14
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 15

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 16
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 17

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 18
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 19
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 20
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 21
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 22
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 23
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 24
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 25
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 26
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 27
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 28
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 29
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 30
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 31
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 32

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KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2

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Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2

Question 1.
Evaluate the following :
i) sin 60° cos 30° + sin 30° cos 60°
ii) 2 tan2 45° + cos2 30° – sin2 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 1
Solution:
i) sin 60° cos 30° + sin 30° cos 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 2

ii) 2 tan2 45° + cos2 30° – sin260°
= 2(tan 45°)2 + (cos 30°)2 – (sin 60°)2
= 2 (1)2 + \(\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= 2 × 1
= 2
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 3
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 4
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 5

Question 2.
Choose the correct option and justify your choice :
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) =
A) sin 60°
B) cos 60°
C) tan 60°
D) sin 30°
Solution:
A) sin 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 6

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) =
A) tan 90°
B) 1
C) sin 45°
D) 0
Solution:
D) 0
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 7

iii) sin 2A = 2 sin A is true when A ;
A) 0°
B) 30°
C) 45°
D) 60°
Solution:
A) 0°
LHS = sin 2A = sin 0° = 0
RHS = 2sin A = 2.sin0° = 0

iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) =
A) cos 60°
B) sin 60°
C) tan 60°
D) sin 30°
Solution:
C) tan 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 8
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 9

Question 3.
If tan (A + B) =\(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) 0° < A + B ≤ 90°; A > B. find A and B.
Solution:
tan (A + B) = \(\sqrt{3}\)
tan (A + B) = tan 60°
A + B = 60°
tan (A – B) = \(\frac{1}{\sqrt{3}}\) = tan 30°
tan(A – B) = tan 30°
A – B = 30° → (2)
Adding (1) and (2)
A + B + A – B = 60 + 30
2A = 90
A = \(\frac{90}{2}\) = 45°
A = 45°
Put A = 45° in eqn (1)
A + B = 60
B = 60 – A= 60 – 45°
B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
i) sin (A + B) = sin A + sin B
Solution:
False
Take A = 45° and B = 45°
LHS: sin (45° + 45°) = sin 90° = 1
RHS: sin 45 + sin 45 = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}\)
LHS ≠ RHS.

ii) The value of sin θ increases as θ increases.
Solution:
True.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 10
∴ The value of sin θ increases as θ increases.

iii) The value of cos θ increases as θ increases.
Solution:
False.
cos 0° = 1 cos 30° = \(\frac{\sqrt{3}}{2}\) =0.87
cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 cos 60° = \(\frac{1}{2}=\) = 0.5
cos 90° = 1
∴ The value of cos 0 increases as 0 increases – the statement is False.

iv) sin θ = cos θ for all values of θ.
Solution:
False.
sin 30° = \(\frac{1}{2}\) cos 30° = \(\frac{\sqrt{3}}{2}\)
sin 30° ≠ cos 30°
But sin 45° = cos 45° = \(\frac{1}{\sqrt{2}}\)

v) cot A is not defined for A = 0°.
Solution:
True.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 11
∴ When A = 0°, cot A is not defined.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1

Question 1.
In ∆ABC, right-angled at B, AB = 24 cm., BC = 7 cm. Determine:
i) sin A, cos A
ii) sin C, cos C
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 1
In ⊥∆ABC, ∠B = 90°
As per Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7) 2
= 576 + 49
AC2 = 625
∴ AC = 25 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 2

Question 2.
In the given figure, find tan P – cot R.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 3
In ⊥∆PQR, ∠Q = 90°
∴ PQ2 + QR2 = PR2
(12)2 + QR2 = (13)2
144 + QR2 = 169
QR2 = 169 – 144
QR2 = 25
∴ QR = 5 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 4

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 5
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 6

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 7

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios
Solution;
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 8

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
In ⊥∆ABC, ∠A and ∠B are acute angles. CD ⊥ AB is drawn.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 9
CD is common.
S.S.S. Postulate :
∴ ∆ADC ~ ∆ADB
∴ ∠A = ∠B
∵ “Angles of similar triangles are equiangular.”

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate :
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot2θ
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 10
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 11

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos2 A – sin2 A or not
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 12
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 13

Question 9.
In ∆ABC, right-angled at B, if tan A =\(\frac{1}{\sqrt{3}}\) find the value of:
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 14
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 15

Question 10.
In ∆PQR, right-angled at Q, PR + QR = 25 cm. and PQ = 5 cm. Determine the values of sin P, cos P and tan P
Solution:
PQ = 5 cm
PR + QR = 25 cm
∴ PR = 25 – QR
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 16
PR2 = PQ2 + QR2
QR2 = PR2 – PQ2
= (25 – QR)2 – (5)2
QR2 = 625 – 50QR + QR2 – 25
50QR = 600
∴ QR = 12 cm.
∴ PR = 25 – QR = 25 – 12 = 13 cm.
∴ QR = 12 cm
∴ PR = 25 – QR = 25 – 12 = 13 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 17

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
Answer:
False. 60° = \(\sqrt{3}\) > 1

(ii) If sec A = \(\frac{12}{5}\) for some value of angle A.
Answer:
True. because sec A > 1.

(iii) cos A is the abbreviation used for the cosecant of angle A.
Answer:
False. Because cos A is simplified as cos.

(iv) cot A is the product of cot and A.
Answer:
False. cot is ∠A or meaningless. Here,
cot A = \(\frac{\text { Adjacent side }}{\text { Opposite side }}\)

(v) sin θ = \(\frac{4}{3}\) for some angle θ
Answer:
False. because sin θ ≯ 1

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

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

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Solution:
(i) r = 7 cm, V = ?
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 1

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm.
(ii) 0.21 m.
Solution:
(i) Diameter of solid spherical ball, diameter, d = 28 cm,
∴ radius, r = \(\frac{28}{2}\) = 14 cm.
Volume of solid spherical ball, V = \(\frac{4}{3}\) πr3.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 2
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 3
Amount of water displaced by the ball = 0.00485 m3.

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3 ?
Solution:
Diameter of metallic ball, d = 4.2 cm. density = 8.9 gm/cm3
mass, m = ? d = 4.2 cm.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 4
∴ V = 38.808 cm3.
∴ Mass = Density × Volume
= 8.9 × 38.808
= 345.40 gm.

Question 4.
The diameter of the moon is approximately one-fourth of 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 be ‘d’ unit.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 5
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 6
∴ Volume of Moon = \(\frac{1}{64}\) × Volume of Earth

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Diameter of a hemispherical bowl, d = 10.5 cm
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 7

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:
In a hemispherical tank,
Inner radius, r1 = 1 m
Outer radius, r2 = 1 + 0.01 = 1.01 m (∵ 1 cm = 0.01)
∴ Volume of hemispherical tank, V
= Volume of outer diameter – Volume of inner diameter.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 8
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 9

Question 7.
Find the volume of sphere whose surface area is 154 cm2.
Solution:
Surface area of sphere, 4πr2 =154 cm2.
Volume of Sphere, V = ?
A = 4πr2 = 154
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 10

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is Rs. 2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
(i) Cost of white-washing dome is Rs. 498.96
Cost of white-washing is Rs. 2 per sq. metre.
∴ Surface Area = \(\frac{498.96}{2}\) = 249.48 m2.

(ii) Surface Area of hemisphere, V = 2πr2
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 11
∴ Inner volume of hemisphere dome, V
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 12

Question 9.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S1. Find the
(i) radius r of the new sphere,
(ii) ratio of S and S1
Solution:
(i) Volume of 1 sphere, V = \(\frac{4}{3}\)πr3
Volume of 27 solid sphere
= 27 × \(\frac{4}{3}\)πr3
Let r1is the radius of the new sphere.
Volume of new sphere = Volume of 27 solid sphere
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 13

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:
Diameter of capsule of medicine, d
d = 3.5 mm = \(\frac{7}{2}\) mm.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 14

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows:
Height = 110 cm, Depth = 25 cm, Breadth = 85 cm.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 1
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:
Outer length, l = 25 cm.
outer breadth, b = 85 cm.
outer height, h = 110 cm.
Outer surface area = lh + 2lb + 2bh = lh + 2(lb + bh)
= 85 × 110 + 2(85 × 25 + 25 × 110)
= 9350 + 9750
= 19100 cm2.
Area of front face,
= [85 × 110 – 75 × 100 + 2(75 × 5)]
= (9350 – 7500 + 2(375)]
= 9350 – 7500 + 750
= 11000 – 7500
= 3500 cm2.
Area to be polished,
= 19100 + 3500
= 22600 cm2.
Cost of polishing 1 cm3 is Rs. 0.20.
Cost of polishing 22600 cm3 … ? …
= 22600 × 0.20 = Rs. 4520.
Length of horizontal shelf, l = 75 cm.
breadth, b = 20 cm.
height, h = 30 cm.
Area of horizontal shelf
= 2(l + h)b + lh
= [2(75 + 30) × 20 + 75 × 30]
= (4200 + 2250] cm2.
= 6450 cm2.
∴ Area of painting 3 horizontal rows
= 3 × 6450
= 19350 cm2.
Cost of painting for 1 cm3 is Rs. 0.10.
∴ Cost of painting 19350 cm3 … ?
= 19350 × 0.10 = Rs. 1935.
∴ Total cost of polish and painting
= Rs. 4520 + Rs. 1935
= Rs. 6455.

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.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 2
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:
Diameter of a wooden frame, d = 21 cm.
Radius r = \(\frac{21}{2}\)
Outer surface area of wooden spheres,
A = 4πr2
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 3
∴ A = 1386cm2
Radius of cylinder, r1 = 1.5 cm
height, h = 7 cm
The curved surface area of cylinder support,
A = 2πrh
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 4
Area of silver painted
= [8 × (1386 – 7.07)] cm2
= 8 × 1378.93
= 11031.44 cm2
Cost of silver paint = 11031.44 × 0.25
= Rs. 2757.86
Area of black paint = (8 × 66) cm2
= 528 cm2
Cost of black paint = 528 × 0.05
= Rs. 26.40
Total cost of silver and black paint.
= Rs. 2757.76 + Rs. 26.40
= Rs. 2784.26.

Question 3.
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution:
Let the diametrer of sphere be ‘d’
Radius of sphere, r1 = \(\frac{\mathrm{d}}{2}\)
Radius of outer sphere, r2 = \(\frac{\mathrm{d}}{2}\left(1-\frac{25}{100}\right)\)
∴ r2 = \(\frac{3}{8}\)d
Outer Area of Sphere, S1 = 4πr12
= 4π\(\left(\frac{\mathrm{d}}{2}\right)^{2}\)
S1 = πd2
The diameter of sphere is decreased by 25%. Then its outer surface area,
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 5

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1.

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Data: Constructing an angle of 90° at the initial point of a given ray.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 1
Steps of Construction:

  1. Draw AB straight line.
  2. Taking A as centre and any radius draw an arc and intersect AB at C.
  3. Taking C as centre with the same radius draw an arc which intersects at D.
  4. With the same radius taking D as centre, it intersects at E.
  5. With centres E and D with the same radius draw two arcs which meet both at F.
  6. If AF is joined, AB straight line is the line in point A with 90°, i.e., AF.
    In the above construction ∠DAB = 60° and ∠DAE = 60°.
    Angular bisector of ∠DAE is AF.
    ∴ ∠DAF = ∠EAF = 30°
    ∴∠BAF = ∠DAB + ∠DAE = 60°+ 30°
    ∴∠BAF = 90°.

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Data: To construct an angle of 45° at the initial point of a given ray.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 2
Steps of Construction :

  1. Draw a straight line with any measurement.
  2. Taking A as centre with any radius draw an arc, which meets AB at C.
  3. With C centre, with the same radius draw an arc which meets at D.
  4. With D centre, with the same radius, draw an arc which meets at D.
  5. With centres E and D, with the previous radius draw two arcs which meet at F.
  6. Now AF is joined. Now AP1AF at A. Hence ∠BAF = 90°.
  7. Now, AG which is the angular bisector of ∠BAF is drawn.
  8. ∠BAG = 45° is constructed.
    ∠FAB = 90°
    AG is the angular bisector of ∠FAB
    ∴ ∠FAG = ∠GAB = 45°
    ∴ ∠GAB = 45°.

Question 3.
Construct the angles of the following measurements :
(i) 30°
(ii) \(22 \frac{1}{2}^{\circ}\)
(iii) 15°.
Solution:
(i) To construct 30° angle :
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 3
Steps of Construction :

  1. Draw PQ straight line.
  2. With P as centre with any radius draw \(\frac{1}{2}\) arc- intersects PQ at A.
  3. With A as centre with the same radius draw an arc which intersects at B. Join PB and produced.
  4. With A and B centres, with radius more than half of AB draw two arcs which intersect at C. Join PC.
  5. ∠BPC = ∠CPQ = 30°.

(ii) To draw an angle of \(22 \frac{1}{2}^{\circ}\)
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 4
Steps of Construction:

  1. Draw a straight line AB.
  2. With centre ‘A’, taking a convenient radius to draw an arc which intersect AB at P.
  3. With P as centre with the same radius draw an arc at Q, with Q as centre with the same radius draw an arc which intersects at R.
  4. With R and Q centres with the same radius draw two arcs which intersect at S. Join AS, ∠BAS = 90°.
  5. Now construct AT which is the angular bisector of ∠BAS, and joined, ∠TAB= 45°.
  6. Now AU which is the angular bisector of ∠TAB, AU is joined. Now, ∠UAB = \(22 \frac{1}{2}^{\circ}\)

(iii) To construct an angle of 15° :
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 5
Steps of Construction :

  1. Draw a straight line AB.
  2. With A as centre with a convenient radius draw an arc which intersect AB at C.
  3. With C as centre, with the same radius, draw an arc which intersects at D.
    Now. ∠DAB = 60°
  4. Construct AE, the bisector of ∠DAB. Join then, ∠EAB = 30°.
  5. Construct AF, the bisector of ∠EAB. Join then, ∠FAB = 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor :
(i) 75°
(ii) 105°
(iii) 135°.
Solution:
(i) To construct angle 75° :
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 6
Steps of Construction :

  1. Draw AB with any measurement.
  2. With A as centre, with convenient scale draw an arc AB which intersects at C.
  3. With C as centre with the same radius draw an arc which intersects at D. ∠DAB = 60°.
  4. With D as centre with the same radius draw an arc which intersects at E.
  5. With Centres E and D, by taking more than half of ED draw two arcs which meet at E. AF is joined.
    Now, ∠FAB = 90°.
  6. Draw AG bisector of ∠EAD, AG joined. ∠GAD = 15°.
  7. ∠GAD + ∠DAB = 15 + 60 = 75°
    ∴ ∠GAB = 75°.

(ii) To construct angle 105° :
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 7
Steps of Construction :

  1. Draw PQ with any measurement.
  2. With P as centre and with any radius, draw an arc which intersects PQ at A.
  3. With A as centre, with the same radius draw an arc which intersect at B.
  4. With B as centre with same radius draw another arc, it intersects at C.
  5. With centres C and B if two arcs are drawn these two meet at D. Join PD. ∠DPQ = 90°. Straight-line PD intersects the arc CB at E.
  6. Now taking radius half of CE, if the line is drawn from C to E it intersects at F. FP is joined. Now ∠FPD = 15°.
  7. ∠FPD + ∠DPQ = 15 + 90 = 105°.
    ∴ ∠FPQ = 105° is constructed.

(iii) To construct an angle of 135°
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 8
Steps of Construction :

  1. Draw a straight line AB.
  2. With A as centre, a semicircle is drawn which meets AB at C and it intersects the produced BA line at G.
  3. With C as centre with the same radius and centre CD, draw a radius of arc DE.
  4. With centres E and D by taking a radius more than half draw two arcs which meet at F. Join AF. ∠FAB = 90°. AF intersects ED at H.
  5. With G and H centres by taking radius more than \(\frac{1}{2}\) of the radius GH draw arcs which meet at I. AI is joined. Now, ∠IAF = 45°.
  6. ∠IAF + ∠FAB = 45° + 90° = 135°
    ∴ ∠IAB = 135° is constructed.

Question 5.
Construct an equilateral triangle, given its side, and justify the construction.
Solution:
Data: Construct an equilateral triangle, given its side.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 9
Steps of Construction :

  1. Construct AB = 6 cm.
  2. With A as centre and convenient scale draw an arc, it meets AB at P.
  3. With P as centre, with same radius draw an arc, it intersects the first arc at Q, AQ is joined and produced. ∠QAB = 60°.
  4. With A as centre and taking 6 cm radius draw an arc. It intersects at C which is the produced line of AQ.
  5. Next, BC is joined, ABC is an equilateral triangle.

In ∆ABC, ∠A = ∠B = ∠C = 60°
AB = BC = CA = 6 cm.
In ∆ABC, AB = AC = 6 cm. ∠A = 60°
∠B = 60°
∠A + ∠B + ∠C = 180°
60 + ∠B + ∠B =180 (∵ ∠B = ∠C)
60 + 2∠B = 180
2∠B = 180 – 60
2∠B = 120°
∴ ∠B = \(\frac{120}{2}\)
∴ ∠B = 60°
∴∠B = ∠C = 60°
∴∠A = ∠B = ∠C = 60°
AC = BC (Opposite sides of equal angles)
But, AB = AC (known)
∴ AB = BC = CA = 6 cm.

We hope the KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6

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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 1
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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 2
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?
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 3
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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 4
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
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 5

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 6
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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 7
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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 8
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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 9
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 = \(\frac{\angle \mathrm{A}}{2}\)
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 10
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 11

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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 12
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.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 13
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

 

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