Class 9

Haryana State Board HBSE 9th Class Science Important Questions Chapter 11 Work and Energy Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 11 Work and Energy

Very Short-Answer Type Questions

Question 1.
What is work ?
Answer:
When force is applied on an object and it moves in the direction of force, it is said to be work done.

Question 2.
On which the quantity of work depends ?
Answer:
The quantity of work depends upon the applied force & distance moved.

Question 3.
What is the formula of work ?
Answer:
Work = Force x Displacement.

Question 4.
What is the unit of work ?
Answer:
Newton-metre (Joule).

Question 5.
What is energy ?
Answer:
The capacity of doing work is called energy.

Question 6.
What are the types of mechanical energy ?
Answer:
Mechanical energy is of two types:
(1) Kinetic energy,
(2) Potential energy.

Question 7.
What is potential energy ?
Answer:
The energy possessed by a body due to its position or change in shape is called potential energy.

Question 8.
Which type of energies are in aircraft flying on height ?
Answer:
Both type of energies-kinetic and potential.

Question 9.
Which type of energy is there in a stone lying on a hill ?
Answer:
Potential energy.

Question 10.
What is mechanical Energy ?
Answer:
The sum of kinetic energy and potential energy of a body is called mechanical energy.

Question 11.
Which type of energy is there when a bullet is fired from a gun ?
Answer:
Kinetic energy.

Question 12.
Write the formula for kinetic energy.
Answer:
Kinetic energy = \(\frac {1}{2}\) mv², where m → mass, v → velocity.

Question 13.
Write the formula for potential energy.
Answer:
Potential energy = mgh, where m → mass, g → acceleration due to gravity and h → height.

Question 14.
What is the law of conservation of energy ?
Answer:
According to this law, energy can neither be created nor be destroyed but can only be transferred from one form to another.

Question 15.
What is the effect on energy when work is done ?
Answer:
When work is done, energy decrease in some extent.

Question 16.
What is the work done by a man when he lifts a luggage of mass 10 kg on his head and moves 50 metres on horizontal road ?
Answer:
zero.

Question 17.
From where, our body gets energy ?
Answer:
From food taken by us.

Question 18.
Which type of energy is used by plants ?
Answer:
Solar energy.

Question 19.
Write an important use of wind energy.
Answer:
Wind energy is used to grind the flour by wind mills.

Question 20.
Write the name of any two natural sources of energy.
Answer:
(i) Sun
(ii) Water.

Question 21.
Write any four types of energy.
Answer:
(i) Mechanical Energy
(ii) Chemical Energy
(iii) Light Energy
(iv) Sound Energy.

Question 22.
What is gravitational potential energy ?
Answer:
The energy stored in a body when work is done against force of gravity is called gravitational potential energy.

Question 23.
What is mean by elastic potential energy ?
Answer:
The energy possessed by an object due to change in its shape is called elastic potential energy.

Question 24.
Why spark is produces when two stones collides with each other ?
Answer:
Because kinetic energy of stone is converted into light energy.

Question 25.
How arrow gets its kinetic energy ?
Answer:
The arrow converts the potential energy from bow into kinetic energy.

Question 26.
Out of energy and acceleration, which is vector ?
Answer:
Acceleration is a vector quantity.

Question 27.
The kinetic energy goes on decreasing when it is thrown vertically upward. What will be the motion at highest point ?
Answer:
Zero.

Question 28.
Write the name of that device which converts electric energy into mechanical energy.
Answer:
Electric motor.

Question 29.
Which energy is converted in other in a hydroelectric power centre ?
Answer:
The kinetic energy of water is converted into electric energy.

Question 30.
What is power ? What is its unit ?
Answer:
The rate of doing work is power. Its unit is watt.

Question 31.
If the velocity of an object is increased three times, what will be its kinetic energy ?
Answer:
The kinetic energy will increase nine times.

Question 32.
Which type of energy is there in a expanded spring ?
Answer:
Potential energy.

Question 33.
Which type of energy is there in a dam ?
Answer:
Potential energy.

Question 34.
Which type of energy is there in the spring of watch ?
Answer:
Potential energy which gives the kinetic energy to the needles of watch. ,

Question 35.
Which type of energy is there in running train ?
Answer:
Kinetic energy.

Question 36.
What will be potential energy of that body whose height is doubled ?
Answer:
Its potential energy will be doubled.

Question 37.
What is the relation between 1 kilowatt and watt ?
Answer:
1 kilowatt = 1000 watt.

Question 38.
Why the palms become hot when rubbed ?
Answer:
Because kinetic energy of palms is converted into heat energy.

Question 39.
How much energy is required to lift a stone of mass 1 kg to a height of 1 metre ?
Answer:
9.8 Joules [∵ W = mgh = 1 x 9.8 x 1 = 9.8 joules]

Question 40.
A person and his son are running with uniform velocity. If the mass of person become two times more than of his son, then what will be the ratio between kinetic energy of both ?
Answer:
2 : 1 [ ∵ K.E. = \(\frac {1}{2}\) mv² and v is equal]

Question 41.
Which type of energy is required to move for a sailboat ?
Answer:
The sailboat moves by the kinetic energy of wind.

Question 42.
What will be the energy a body has a capacity to do a work of 10 joules ?
Answer:
10 Joules.

Question 43.
Write the name of two units of power used in daily life.
Answer:
Kilowatt and megawatt.

Question 44.
How many watts are there in one horse power ?
Answer:
1 horse power = 746 watt.

Question 45.
What is meant by 1 kilowatt hour energy ?
Answer:
The energy consumed by a source of 1 kilowatt power in 1 hour.

Question 46.
What is the relation between kilowatt hour and joule ?
Answer:
1 kilowatt hour = 36,00,000 joules.

Question 47.
Which type of energy is used when a person climbs stairs ?
Answer:
Muscular energy.

Question 48.
Which type of energy is there in our body ?
Answer:
Chemical energy.

Question 49.
The person requires the least amount of energy in which of the following activity ?
(i) In swimming
(ii) In movement
(iii) In sleeping
(iv) In running foot.
Answer:
(iii) In sleeping.

Question 50.
What is simple pendulum ?
Answer:
Simple pendulum is a metalic bob which is hang by weightless string.

Question 51.
Write the name of energy when simple pendulum is at its extreme point.
Answer:
Potential energy.

Question 52.
Write the name of energy when pendulum is in middle position.
Answer:
Kinetic energy.

Short-Answer Type Questions

Question 1.
What is meant by work ? How is it measured ? What is its unit ?
Answer:
Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the
applied force.
Thus, Work done = Force x The distance moved in the direction of force
W = F x s
Work is a vector quantity. Unit of work is newton-metre or joule (J).

Question 2.
What is energy ? Write its unit and definition. What are its various types ?
Answer:
The capacity of doing work is energy. Its unit is also joule.
Joule: Work done is said to be one joule when a force of 1 newton acting on an object and the object is displaced through 1 m in the direction of force.

Various types of energy:
(i) Mechanical energy
(ii) Heat energy
(iii) Chemical energy
(iv) Electric energy
(v) Nuclear energy’
(vi) Light energy
(vii) Atomic energy
(viii) Magnetic energy
(ix) Sound energy,
(x) Solar energy.

Question 3.
Justify with example that moving bodies have capacity to do work ?
Answer:
The examples are following :
(i) A moving marble has a capacity to displace that marble which is in rest.
(ii) The blowing wind works by rotating the wheels of windmill.
(iii) The moving water also takes the objects from one place to another.
(iv) The blowing wind is helpful for sailing boat.
Thus, we can say that moving bodies have capacity to do work.

Question 4.
What is meant by potential energy and kinetic energy ? Give two examples for each.
Answer:
Potential energy’: The energy possessed by a body due to its change in position or shape is called the potential energy.
Examples:
(1) The book lying on the table has potential energy due to its position.
(2) A binded watch has potential energy.

Kinetic energy:
The energy possessed by a body due to its motion is called the kinetic energy.

Examples:
(1) Moving air has kinetic energy.
(2) The running water has kinetic energy.

Question 5.
Is there energy in following substances ? If yes, then what is the type-potential energy or kinetic energy or both ?
(i) The switched off ceiling fan.
(ii) A man climbing up mountain.
(iii) Flying bird.
(iv) The water in a dam.
(v) The stretched bow more than its shape.
(vi) The rubber locket on a table.
(vii) Stretched locket of rubber.
Answer:
Yes, all substances have energy’.
(i) A switched off ceiling fan has potential energy only.
(ii) A man climbing up mountain has both potential and kinetic energy.
(iii) A flying bird has both kinetic and potential energy.
(iv) The water in the dam has potential energy.
(v) A stretched bow has potential energy.
(vi) The rubber locket on the table has potential energy.
(vii) Stretched locket of rubber has potential energy.

Question 6.
Write the difference between potential energy and kinetic energy.
Answer:
The differences between potential energy and kinetic energy are as follows:
Potential Energy (P.E.):
1. The energy possessed by an object due to its position is called potential energy.
2. Potential energy = mgh
3. The potential energy of an object depends upon the height or depth from the surface of earth.

Kinetic Energy (K.E.):
1. The energy possessed by an object due to its motion is called kinetic energy.
2. Kinetic energy = \(\frac {1}{2}\) mv²
3. Kinetic energy depends upon the motion of the object.

Question 7.
Which type of energy is in following ?
(1) Coiled spring of watch.
(2) Running buffallo.
(3) The arrow released from bow.
(4) A table lying on roof.
(5) Compressed air.
(6) A mango on tree.
Answer:
(1) Potential energy
(2) Kinetic energy
(3) Kinetic energy
(4) Potential energy
(5) Potential energy
(6) Potential energy.

Question 8.
When the capacity of doing work by human body decreases ?
Answer:
The capacity of doing work by human body decreases in following circumstances :
(1) When a person is ill.
(2) When a person grows old.
(3) When a person takes food less than required, his muscular energy decreases.

Question 9.
What is the relation between work and energy ? Write their units.
Answer:
The capacity of doing work is called energy. The magnitude of energy is always equal to that work which has a capacity to do that. Energy is measured in joule and the unit of work is also joule.

For example:
To lift a piece of stone of mass 1 kg upto a height of 1 metre requires 9.8 joules energy. Opposite to it, when stone is dropped from one metre, the stone does the same amount of work (9.8 J). Thus, if an object has an energy of 100 joules. It has a capacity of doing work of 100 joules.

Question 10.
How the energy is transferred in key toy ?
Answer:
There is a spring in key toy or key watch which contracts while binding and stored potential energy in it. After some moments, the spring loses slowly and the toy comes in motion. Thus, potential energy is transferred in kinetic energy.

Question 11.
A ball is thrown upward and it returns back. What is the change in kinetic and potential energy of the ball ? Explain.
Answer:
When a ball is thrown upward its kinetic energy changes continuously into potential energy. At highest point its whole kinetic energy changes into potential energy. When the ball returns back its potential energy changes continuously into kinetic energy and whole of its potential energy changes into kinetic energy near the earth but on reaching the earth, whole of its kinetic energy changes into potential energy.

Question 12.
A boy is standing by holding some weight in his expanded hand. In this stage, there seems no change in the position of both boy and weight. Is he doing any work ? If yes, then justify.
Answer:
Yet, there is no change in the position of boy and weight externally, but internally the boy is working. The muscles of boy are stretched due to weight on palm. Thus, their shape get changes. The heart pumps more blood into muscles. The energy of boy wastes in all these chemical activities and he tired soon. If the boy stand more enough in same position, the external change will be observed as sweating.

Question 13.
How much work is done when an artificial satellite revolve round the earth ? Explain in brief. –
Answer:
The artificial satellite revolves round the earth in almost circular track. The gravitational force between the earth and the satellite works along the line joining the earth and satellite and the satallite moves perpendicular to the direction of force of gravity. Thus, the magnitude of displacement of satellite in the direction of force will be zero and the work done by the earth on the satellite will also be zero.

Question 14.
How the kinetic energy of wind is utilised ? Write its important uses.
Answer:
The electric energy can be produced by the energy in the blowing air. It has much waste. Thus, wind mill is used in grinding flour and drawing out water from ponds. A small amount of electric energy is produced from wind mill. Those places where the wind blows throughout the years, wind mill are used to grind flour and drawing out water.
Important uses of mechanical energy of wind :
(1) The fields can be irrigated by drawing water from ponds.
(2) It can be used in grinding the flour.
(3) The mechanical energy of wind can be converted into electric energy by joining the various wind wheels together.

Question 15.
What is water energy ? Write its two important uses.
Answer:
The running water has a capacity to do work due to its kinetic energy, which is known as water energy.

Uses of water energy: Following are the uses of water energy :
(1) Water energy is converted into electric energy and it is used in houses and factories.
(2) Water energy is used to grind flour by windmills.
(3) The heavier wooden pieces is taken from one place to another due to water energy.

Question 16.
How water energy is used to produce electricity ?
Answer:
The kinetic energy of running water is stored as potential energy in dam. The stored water is dropped on the larger blades of water turbines. The shaft attached with running turbine rotates the generator or dyanmo quickly. The rotating coil of dyanmo produces electricity which is called ‘hydroelectric power’. The electricity is produced on Bhakhra dam by this method. This is also known as the principle of hydroelectricity.

Question 17.
Why the nail becomes hot when it is hammered in the sheet of wood ?
Answer:
The nail becomes hot when it is hammered in the sheet of wood because the lifted hammer has potential energy due to its position. When the hammer is dropped on the nail, it comes in rest position, but transfers, its whole energy in the nail. As a result of this, the nail gets some kinetic energy and moves inside the wood. When the nail enters completely and we continuous hammering the nail becomes hot because the mechanical energy of the hammer is transferred in nail and changes into heat energy.

Question 18.
What are the benefits of the solar energy ?
Answer:
The benefits of the solar energy are as follows :
(1) The solar energy is used in solar cooker to prepare food,
(2) Solar energy is used to warm the water by solar heater in hotels, hospitals and industries.
(3) Solar energy is used in the production of electricity.
(4) It is used to melt the metals in solar furnances.

Question 19.
Differentiate among work, energy and power. Write the SI unit for each. Or What is Power? Write its S.I. unit.
Answer:
Work:
When force is applied on a body and it moves in the direction of force, it is said to be work done.
Thus, Work (W) = Force (F) x Distance (S)
SI unit of work is joule.

Energy: t he capacity ot doing work is called energy. SI unit of energy is joule.

Power: The rate of doing work is called power.
Thus,

SI unit of power is watt.

Question 20.
What do you mean by transfer of energy ? Write two examples of transfer of energy.
Answer:
According to the law of conservation of the energy, The energy can neither be created nor destroyed. It can be transferred from one form to another. It is called the transfer of energy. Thus, the total magnitude of energy is never changed.

Examples:
(1) Ice on mountains melts and becomes water. This is the transfer of potential energy into kinetic energy.
(2) The electric energy in bulb changes into light energy and glows.

Numerical Questions

Question 1.
A boy of mass 40 kg is running with a ve’ocity of 0.5 ms. What is his kinetic energy ?
Solution:
Here,
m = 40 k
v = 0.5 ms^-1
We know that,
Kinetic Energy (K.E.) = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 40 x 0.5 x 0.5 = 5 Joule Answer:

Question 2.
An object of mass 15 kg is moving with a uniform velocity (if 4 ms. What is tue kincth energy possessed by the object?
Solution:
Here
Mass of the body (m) = 1 5 kg
Velocity of the body (v) = 4 rn/s
We know that
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 15 x 4(2)²
= \(\frac {1}{2}\) x 15 x 16
= 120 Jule

Question 3.
A stone of mass 100 gram is kept at a height ofS metre. What kind of energy does it possess and find out its value.
Solution:
The stone has potential energy because of height.
Here. m = 100g = \(\frac {100}{1000}\) = 0.1 kg
g = 9.8 ms^-2
h = 5m
We know that.
Potential Energy (P.E.) m x g x h = 0.1 x 9.8 x 5 = 4.9 Joule

Question 4.
To what height can an object of mass 1 kg be raised by giving it an energy of 980 joule?
Solution:
Here.
m = 1 kg
g = 9.8 ms^-2
h = ?
Potential Energy (RE.) = 980 J
We know that,
Potential Energy (RE.) = m.g.h.
or 980 = 1 x 9.8 x h
or h = \(\frac{980}{1 \times 9.8}\)
= 100 m

Question 5.
What is the increase in kinetic energy of a particle, if its velocity is increased to four times?
Solution:
Suppose the initial velocity of the particle = v
and its final velocity = 4 v
Suppose the mass of the particle = m
∴Initial kinetic energy (K₁) of the particle = \(\frac {1}{2}\) mv² ………….(i)
Final kinetic energy (K₂) of the particle = \(\frac {1}{2}\)m (4v)²
= \(\frac {1}{2}\)m x 16 v² …………..(ii)
From equations (i) and (ii),
\(\frac{\mathrm{K}_1}{\mathrm{~K}_2}=\frac{\frac{1}{2} \mathrm{mv}^2}{\frac{1}{2} \mathrm{~m} \times 16 \mathrm{v}^2}=\frac{1}{16}\)
or K₂ = 16K₁
Thus, the increase in kinetic energy is 16 times the initial kinetic energy.

Question 6.
The kinetic energy of a ball of mass loo g is 20 joule. Calculate its velocity.
Solution:
Here,
m = 100 g = \(\frac {100}{1000}\)kg = 0.1 kg
v = ?
Kinetic Energy (K.E.) = 20 J
We know that, Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
Or 20 = \(\frac {1}{2}\) x 0.1 x v²
v² = \(\frac {20 x 2}{0.1}\) = 400
or v = \(\sqrt{400}\) = 20 ms^-1

Question 7.
A coolie lifts an object of mass 20 kg from the earth to 1.5 m and keep It on his head. Calculate the amount of work done by him on the object.
Solution:
Here,
Mass of the object (m) = 20kg
Displacement (s) = 1 .5m
Work done (W) = F x s = mg x s
= 2okg x 10 ms^-2 x 15 m
= 300kgms^-2 m
= 300Nm= 300J
The amount of work done by the coolie on the object is 30 joules.

Question 8.
A man of mass 50 kg jumps to a height of 1.2 m. What is its potential energy at the highest point?
Solution:
Here,
m = 50kg
g = 9.8 ms^-2
h = 12m
We know that,
Potential Energy (P.E.) m x g x h
= 50 x 9.8 x 1.2
= 588 J
Thus, at the highest point the potential energy of the man is 588 J.

Question 9.
With what velocity should a man of mass 50 kg run so that his kinetic energy becomes equal to 625 J.
SolutIon:
Here,
m = 50kg
v = ?
Kinetic Energy (K.E.) = 625 J
We know that,
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
Or 625 = \(\frac {1}{2}\) x 50 x v²
v² = \(\frac{625 \times 2}{50}\) = 25
or v = \(\sqrt{25}\) 5ms

Question 10.
A man throws a stone of mass 10 kg from a ladder of height 5m. When ¡t reaches the ground, what will be its kinetic energy ? What will be its velocity at the nearest point to the ground?
Solution:
Here,
m = 10kg
h = 5m
v = ?
Kinetic Energy (K.E.) = ?

We know that,
The kinetic energy (K.E.) of the stone = The loss in potential energy due to falling of the stone
= mgh
= 10 kg x 9.8 ms^-2 x 5m = 490J ……………(i)
Because kinetic energy = \(\frac {1}{2}\) mv² .(ii)
From eqns. (i) and (ii),
\(\frac {1}{2}\) mv^-2 = 490J
or \(\frac {1}{2}\) x 10kg x v² = 490J
or v² =\(\frac{490 \times 2}{10}=\frac{980}{10}\) = 98
or v = \(\sqrt{98}\) = 9.9 ms^-1 (approx.)

Question 11.
A car of mass 1000 kg is moving with a velocity of 30 ms^-1. After applying brakes it stops with uniform acceleration at a distance of 50m. Calculate the force and the work done by the brakes?
Solution:
Here,
m = 1000kg
u = 30 ms^-1
v = 0 ms^-1
s = 50m
F = ?
W = ?
We know that, v² – u² = 2as
a = \(\frac{v^2-u^2}{2 s}=\frac{(0)^2-(30)^2}{2 \times 50}=\frac{-900}{100}\) ms^-2
= – 9 ms^-2 So retardation = 9ms^-2
F = m.a.l000kg x 9ms² = 9000 N
The force applied by the brakes on the car (F) = 9000 N
Work done(W) = F x s = 9000 N x 50 m = 450000 J = 450 kJ

Question 12.
A man of mass 50kg climbs a vertical height of 10m on a mountain in 20 seconds. What ¡s his power ? (Given that g 9.8 ms-2)
Solution:
Here,
F = m x g
= 50 x 9.8 = 490N
Now, W = F x S = 490N x 10m = 4,900 Nm
Time (t) = 20 seconds
Power (P) = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{4,900}{20}\) = 245 watt

Question 13.
A boy of mass 40 kg runs up a staircase of 45 steps in 9 sec. If the height of each step is 15 cm, Calculate his power ? (The value of g is 10 ms²)
Solution:
Weight of the boy (mg) 40kg x 10ms² = 400N
Total height of the steps (h) = 45 x 15/100m = 675m
Time taken to climb the steps (t) = 9s
We now that
Power (P) = Time taken = \(\frac{\mathrm{mgh}}{t}\) = \(\frac{400 \mathrm{~N} \times 6.75 \mathrm{~m}}{9 \mathrm{~s}}\) = 300 W
The power of the boy is 300 W.

Question 14.
A man of mass 50 kg climbs 30 steps in 30 sec. If height of each step is 20 cm. Then calculate the power to climb the total steps.
Solution:
Mass of the man (m) = 50 kg
Total number of steps = 30
height of each step = 20 cm = \(\frac {20}{100}\) m = 0.2 m
Total height covered by the man = 30 x 0.2 m = 6m
Acceleration due to gravity (g) = 10 ms^-2
Thus, work done by the man = m.g.h = 50 x 10 x 6J = 3000J
Time taken to climb the steps = 30 s

= \(\frac{3000 \mathrm{~J}}{30 \mathrm{~s}}\)
= 100 Js^-1 = 100 W(watt)

Question 15.
An electric bulb of 100 Wis used for 2 hours, how much electrical energy is consumed?
Solution:
Power of the bulb = 100W
Time = 2 hours 2 x 3600 = 7200 sec.
Energy consumed = Power x Time
= 100 x 7200Ws = 720000 joule

Question 16.
A rocket has a mass of 3 x 10⁶ kg. Calculate (a) its potential energy (b) its kinetic energy, at a height of 25 km moving with a velocity of 1 km/s. (Take g = 10 ms^-2)
Solution:
Here,
Mass of the rocket (m) 3 x 10⁶ kg
Velocity of the rocket (v) = 1 k ms^-1 = 1000 ms^-1
Height of the rocket (h) = 25 km = 25000 m
Acceleration due to gravity (g) = 10 ms^-2

(a) Potential Energy = m.g.h.
3 x 10⁶ x 10 x 25000 J
= 7.5 x 10¹¹

(b) Kinetic Energy = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 3 x 10⁶ x (1000)² J
= 7.5 x 10¹¹J

Question 17.
A horse is pulling a cart moving with a velocity of 18 kmh^-1 with a force of 300 N. Find out the power of the horse in watt and also in horse power?
Solution:
Here,
Velocity of the cart(v) = 18 kmh^-1
= \(\frac{18 \times 1000}{3600}\) ms^-1 = 5 ms^-1
The force exerted by the horse on cart (F) = 300 N
m.g. = 300N
Work done = m.g.h
= 300 x 5 = 1500J

= \(\frac{15000}{1s}\) = 1500 w
We know that, 746 W = 1 horsepower
1 waft = \(\frac {1}{746}\) horsepower
1500 watt = \(\frac {1}{746}\) x 1500 horsepower
= 2 horsepower

Do Yourself

(1) A car is moving with a uniform velocity of 54 kmh^-1 What will be the kinetic energy of a child of mass 40 kg sitting inside it?
Answer:
[4500J]

(2) Calculate the power of that pump which stores the water of loo kg in 25 seconds into a tank kept at a height of 19m. (Let g = 10 ms²)

(3) What will be the energy of an object of mass 10 kg at a height of 40 m from the ground level?
Answer:
[Take g 100 ms²] [40003]

(4) A force of 5 N is acting on a body. The body displaces to a distance of 2m in the direction of the force. Find out the work done on the object.
Answer:
[103)

(5) An object of mass 15 kg is moving uniformly with a velocity of 4 ms². Calculate the kinetic energy of the object.
Answer:
[120J]

(6) What will be the height from the ground level where an object of mass 1 kg will have a potential energy of 1 joule if g= 10 ms²?
Answer:
[0.1 m]

(7) If 4900 joule of energy is utilised in lifting a body of mass 50 kg to a certain height, then what will be the height of the object if g = 9.8 ms²?
Answer:
[10m]

(8) A body of mass 3 kg starts falling down from rest position. After 3 sec., what will be its kinetic energyifgl0m.c2?
Answer:
[1350m]

(9) An object of mass 10 kg is lifted to a height of 6m from the ground level. Calculate the energy present in the object. The value of g is
9.8 ms².
Answer:
[588 J]

(10) An object of mass 12 kg is kept at a certain height. 1f its potential energy is 480 J; calculate its height from the ground level.
Answer:
[Take g = 10 ms²] [4 m]

(11) A man of mass 50kg takes a packet of mass 20kg upto the terrace of a building of height 14m, what is the work done by the man ?
Answer:
[Take g = 10 ms²] [9604 J]

(12) A vehicle of mass 1800 kg is moving with a velocity of 40 ms. After applying brakes it stops with a uniform acceleration at a distance of 200 m. Calculate the force applied and work done by the brakes.
Answer:
[1440 kJ]

(13) What will be the increase in kinetic energy of a particle if its velocity is doubled?
Answer:
[4 times ]

(14) How much amount of work is done on lifting an object of mass 2 kg to a height of 2m if the value of g is 9.8 ms²?
Answer:
[39.2J1

(15) The height of’Kutubminar’ is 70m. What is the work done by a man of mass 50 kg on reaching the top?
Answer:
[Take g 9.8 ms^-2] [3430011

(16) A stone of mass 10g is kept at a height of 5m. What kind of energy is present in it and calculate its value?
Answer:
[Take g 9.8 ms^-2] [0.49J]

(17) Two girls A and B of mass 400 N each, climbs a rope upto a height of 8m. IfA takes 20 sec. to do this work while B takes 50 sec.,
then calculate the individual power of both the girls.
Answer:
[A 160 W, B = 64W]

(18) Find the energy in KWh consumed inS hours by four devices of power 500 W each.
Answer:
[10Kwh]

Essay Type Questions

Question 1.
Define potential energy. Derive an expression for the potential energy of a body.
Answer:
Potential Energy:
The energy possessed by a body due to its position or change in shape is known as potential energy.

Expression for potential energy:
Consider a body of mass m kept on the.ground level at point ‘A’. it is raised to a point ‘B’ by applying a force F on it. The distance between A and B is ‘h’.
As we know that, the force F acting on the body to lift it, is equal and opposite to the gravitational force
F = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) …………(i)

Where, M_e = Mass of the earth
R_e = Radius of the earth
But, g = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) (g = acceleration due to gravity) ………..(ii)
From equations (i) and (ii),
F = mg
The work done in raising the body to a height ‘h’
W = F x S (∵ s = h) = mgh
Because work done is same as the energy. Therefore, at a height ‘h’ the potential energy of the object is,
Potential Energy (P.E.) = mgh

Question 2.
Define kinetic energy. Obtain an expression for it.
Answer:
Kinetic Energy:
The energy possessed by a body due to its motion is called kinetic energy.

Expression:
Let us suppose a body of mass m is dropped from a height h and its velocity becomes V.
on reaching the ground level Then –
Initial velocity of the body (u) = 0
Acceleration (a) = g (acceleration due to gravity)
Final velocity = v
Distance (s) = Height (h)
We know that, v² – u² = 2as
v² – 0 = 2gh
v² = 2gh
h = \(\frac{v^2}{2 g}\) …………. (i)
As the object comes down, its potential energy decreases and kinetic energy increases. When the body strikes to the ground level, its hoIe potential energy is converted into kinetic energy. According to the law of conservation of energy, Energy can neither be increased nr be decreased.”
Thus, kinetic energy at ground level = Potential energy at height h.
Kineiic Energy = mgh
or K.E. = mg x \(\frac{v^2}{2 g}\) (∵ h = \(\frac{v^2}{2 g}\))
K.E. = \(\frac {1}{2}\) mv²

Question 3.
Define Law of conservation of energy. Give an example of conservation of energy.
Answer:
Law of conservation of energy-According to this law, energy can neither be created nor be destroyed. ¡t can only be transfèrred from one form to another. When the energy is released in any form, the same amount of energy appears in other forms. Thus, total energy of the universe remains constant. Suppose, an object of mass m starts falling from a height h, on the ground level.

(a)EnergyatpointA, PotentialEnergy = mgh
Kinetic Energy = 0
∴ Total energy = mgh + 0 = mgh ……………(i)

(b)Energy at point B:
Suppose the object falls from A to B at a distance s, then its height from ground level will be (h-S).
∴ Potential energy = mg (h-s)
and v²-u² = 2gs
v² = 2gs [∵ u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\)m x 2gs = mgs
Total Energy = mg (h-s) + mgh = mgs – mgs + mgs = mgh …………(2)

(c) Energy at point C; v²-u² = 2gh
v² = 2gh [∵u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\) m x 2gh = mgh
and Potential Energy = 0
Total Energy = mgh + 0 = mgh ……………(iii)
in this way we observe from eQuestion (i), (ii) and (iii) that the energy of the object remains constant at every point. There is potential energy in the object at highest point. When it falls litle downs same part of potential energy is converted into kinetic energy and on ground level whole potential energy is converted into kinetic energy.

Practical Work

Experiment 1.
Describe by an experiment that the wark done by a force can either be positive or negative.

Procedure:
A child is pulling a toy car parallel to the ground as shown in figure. The child has exerted a force in the direction of displacement of car. In this situation, the work done will be equal to the product of the force and displacement. In this situation the work done by the force is taken as positive. Consider a situation in which an object is being displaced by the action of forces and we identit one of the forces, F acting Opposite to the direction of the displacement s, thus, the angle between the two directions is 1800.

In such a situation, the work ‘ done by the force F is taken as negative and denoted by the minus sign. The work done by the force is F x (-s) or (-F x s). When the force acts opposite to the direction of the displacement, the work done is negative. When the force acts in the direction ofdisplacernen, the work done is positive.

Experimenf 2.
Describe by an activity in which both positive and negative forces are working.

Procedure:
When an object is lifted up, both positive and negative forces work, because the force applied by us, displaces the object upward and called positive force, while the force of gravity applied by earth pulls the object downward which is called negative force.

Experiment 3.
Describe by an experiment that the object at more height has more energy.

Procedure:
(1) Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from height of about 25 cm. The ball creates a depression.
(2) Repeat this activity from heights of 50 cm, I m and 1.5 m.
(3) Ensure that all the depressions are distinctly visible.
(4) Mark the depressions to indicate the height from which the ball was dropped.
(5) Compare their depths.

We will observe that when the ball was dropped from the height of 1.5 m, the depression was the deepest; which proves that the object at more height has more energy which makes the deepest depression.

Expereiient 4.
Describe by an experiment that the kinetic energy of an object increases by increasing its mass.

Procedure:
(1) Set up the apparatus as shown in figure
(2) Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
(3) Place a known mass on the pan so that the trolley starts moving.
(4) The trolley moves forward and hits the wooden block and displaces the block.
(5) Note down the displacement of the block. It is, clear that the work is done on the block by the trolley as the block has gained energy.
(6) Repeat this activity by increasing the mass on the pan.

You will observe that the displacement will increase with the increase in mass, which proves that the kinetic energy increases by increasing the mass.

Quick Review of the Chapter

1. What is required for doing a work?
(A) displacement in object
(B) force should be applied on the object
(C) both (A) and (B)
(D) no displacement in object
Answer:
(C) both (A) and (B)

2. What type of energy is there inside a body kept at height?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) no relation with energy
Answer:
(A) potential

3. The device that converts the electrical energy In heat energy:
(A) Electrical iron
(B) Electrical bulb
(C) Radio
(D) Electrical motor
Answer:
(A) Electrical iron

4. Which type of energy is there in a flying bird?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) neither of the two
Answer:
(C) both (A) and (B)

5. Rado transfer the electrical enegry in:
(A) light energy
(B) sound energy
(C) mechanical energy
(D) heat energy
Answer:
(B) sound energy

6. Which formula is correct for work?
(A) W = \(\frac {F}{s}\)
(B) W = \(\frac {s}{F}\)
(C) W= Fs
(D) W = \(\frac {1}{Fs}\)
Answer:
(C) W= Fs

7. The unit of force Is:
(A) Newton
(B) Newton metre
(C) Joule
(D) both (B) and (C)
Answer:
(A) Newton

8. If 5n force acts on an object, then displacement is 2m, what will be the work done?
(A) lONm
(B) IOJ
(C) lOJm
(D) both(A)and(B)
Answer:
(D) both (A) and (B)

9. A coolie lifts an object of mass iS kg from the earth to 1.5 m keep it on his head. Calculate the
amount of work done by the coolie.
(A) 22.5J
(B) 2.25 J .
(C) 225J
(D) 225N
Answer:
(C) 225 J

10. The potential energy of an object of mass 1kg kept at the height &f 2m will be:
(A) 9.8J
(B) 19.6J
(C) 29.4 J
(D) 39.2 J
Answer:
(B) 19.6 J

11. Whose unit is Newton metre (Nm)?
(A) work
(B) force
(C) acceleration
(D) power
Answer:
(A) work

12. How many watts are in 1 Kilowatt?
(A) 10³
(B) 10⁴
(C) 10⁵
(D) 10⁶
Answer:
(A) 10³

13. How many joules are in 1 Kilowatt?
(A) 36 lakhs
(B) 36 thousands
(C) 36 hundreds
(D) 36 crores
Answer:
(A) 36 lakhs

14. What is another name of newton metre?
(A) Erg
(B) Coulomb
(C) Joule
(D) Hertz
Answer:
(C) Joule

15. ……….. is vector.
(A) potential energy
(B) kinetic energy
(C) work
(D) none of these
Answer:
(D) none of these

16. A man stands for 10 minutes by carrying a weight of 20 kg the work done s ill be:
(A) 200joule
(B) onejoule
(C) zero
(D) 400joule
Answer:
(C) zero

17. The unit of energy is:
(A) Nm
(B) Joule
(C) both (A) and (B)
(D) no unit
Answer:
(C) both (A) and (B)

18. The example of kinetic energy is not:
(A) stone kept on mountain
(B) revolving wheel
(C) blowing air
(D) bullet shot out of a gun
Answer:
(A) stone kept on mountain

19. The example of potential energy is not :
(A) an arrow kept in the bow
(B) a stone on a mountain
(C) falling coconut
(D) a bullet in a gun
Answer:
(C) falling coconut

20. An object of 10 kg mass is carried to a height of 6 m. Calculate the energy of the object. (g = 9.8 mc²)
(A) 98 J
(B) 58.8 J
(C) 588 J
(D) 5880 J
Answer:
(C) 588 J

21. The law of conservation of mass is:
(A) all works require energy
(B) energy can be produced
(C) energy can be destroyed
(D) energy neither can be produced nor destroyed
Answer:
(D) energy neither can be produced nor destroyed

22. Which statement is not correct?
(A) Potential energy + kinetic energy constant
(B) Kinetic energy + potential energy = mechanical energy
(C) As much decrease is there in potential energy at a point so is in kinetic energy
(D) As much decrease is there in potential energy so much increase takes place in kinetic energy.
Answer:
(C) As much decrease is there in potential energy at a point so is in kinetic energy

23. Which formula is correct?
(A) P = W x t
(B) P = \(\frac {W}{t}\)
(C) P = \(\frac {t}{W}\)
(D) P = \(\frac{1}{W \times t}\)
Answer:
(B) P = \(\frac {W}{t}\)

24. What is the unit of power?
(A) Joule
(B) Nm
(C) erg
(D) Watt
Answer:
(D) Watt

25. How much energy will be consumed by a person of 600 newton weight for climbing a height of 10m?
(A) 60 joule
(B) 600 jouie
(C) ó000joule
(D) 588 joule
Answer:
(C) 6000 joule

26. When will be the minimum amount of potential energy in your body ? When you:
(A) are standing
(B) are sitting in a chair
(C) ire lying on the earth
(D) are sitting on the earth
Answer:
(C) are lying on the earth

27. When ¡s work not done?
(A) on fixing a nail in a wood
(B) on sliding a box on the floor
(C) when there is no constituent of parallel force in the direction of motion
(D) a weight hanging on a peg
Answer:
(C) when there is no constituent of parallel force in the direction of motion

28.
10 seconds are taken to fetch a bucket of 5 kg mass from a well of 10 m depth. The power used in the power will be:
(A) 5ONm
(B) 50)
(C) 50 W
(D) 500 W
Answer:
(C) 50 W

29. How much energy will be consumed by 5 fans of 100 watt each in 4 hours?
(A) 20 kwh
(B) 10 kwh
(C) 2 kwh
(D) 1 kwh
Answer:
(C)2 kwh

30. The rate of doing work is called:
(A) Power
(B) Force
(C) Energy
(D) None of these
Answer:
(A) Power

31. What is the momentum of an object of mass m, moving with a velocity u?
(A) (mυ)²
(B) mυ²
(C) \(\frac {1}{2}\) mυ²
(D) mυ²
Answer:
(C) \(\frac {1}{2}\) mυ²

Haryana State Board HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Very Short-Answer Type Questions

Question 1.
What is the meaning of health?
Answer:
The proper shape of the entire organs and systems of the body, their location and functioning is called as health. Health is the normal position of the ability of physical, mental and social life.

Question 2.
What is called an ailment or disease?
Answer:
Any kind of disruption developing in good health is called as a disease. Or in other words, occurring of any disorder in the physical system is called a disease.

Question 3.
Give the name of a factor that influences health.
Answer:
Personal hygiene.

Question 4.
Give a reason that affects the community’s cleanliness.
Answer:
Failure of disposing of garbage scientifically.

Question 5.
How does food affect our health ?
Answer:
Excess of deficiency of nutritious elements in diet both leave ill effects on our health.

Question 6.
How does our environment harm us ?
Answer:
Unwanted materials found in the environment proved to be the cause of diseases for us.

Question 7.
Is it called health that the body is disease-free ?
Answer:
‘No, even if we are disease-free, we cannot consider ourselves healthy.

Question 8.
How is a disease diagnosed ?
Answer:
From the symptoms of the disease.

Question 9.
What are the symptoms of diseases ?
Answer:
The signs that indicate any disorder in our body, are called as the symptoms of disease.

Question 10.
What do the physicians do to confirm a certain disease ?
Answer:
They refer the laboratory’ test,

Question 11.
How many types of diseases are there (on the basis of time duration) ?
Answer:
Two types: Acute diseases, Chronic diseases.

Question 12.
What are called as acute diseases ?
Answer:
Ailments that are short lived are called as acute diseases.

Question 13.
Write down names of two acute diseases.
Answer:
Cough-cold, headache.

Question 14.
What are called as chronic diseases?
Answer:
Chronic diseases-the sustain for long time or throughout life.

Question 15.
Write down the name of chronic disease.
Answer:
Filariasis (elephantiasis).

Question 16.
Can a person fall ill in the inadequacy of diet?
Answer:
Yes, in the inadequacy diet a person can fall ill.

Question 17.
The antibiotic penicillin blocks the processes that build in bacteria.
Answer:
cell wall.

Question 18.
What are called as communicable diseases ?
Answer:
Diseases that spread through contact are called as communicable diseases.

Question 19.
Write down the names of two communicable diseases.
Answer:
Cholera, jaundice.

Question 20.
What are called as non-communicable diseases ?
Answer:
Diseases that do not spread through infection or pathogens are called as non-communicable diseases.

Question 21.
Write down names of two non-communicable diseases.
Answer:
Goitre and Diabetes.

Question 22.
Write names of two diseases caused due to virus.
Answer:
Cough-cold, Influenza, Polio.

Question 23.
What type of disease is cancer ?
Answer:
Non-communicable disease.

Question 24.
Which diseases are caused due to bacteria ?
Answer:
Cholera and Tuberculosis.

Question 25.
Give two examples of the diseases caused due to fungi.
Answer:
Skin diseases (dermatosis)-Ringworm, eczema etc.

Question 26.
Write names of two diseases caused by protozo
Answer:
Malaria and kala-azar.

Question 27.
What are the names of diseases caused by a worm?
Answer:
Elephantiasis (Filariasis), ascariasis.

Question 28.
Write down the name of the factor of acne or pimple.
Answer:
Staphylococci bacteria.

Question 29.
What is the cause of sleeping sickness?
Answer:
Trypanosoma protozoa.

Question 30.
Which disease is caused due to lachmania protozoan?
Answer:
Kala-azar.

Short-Answer Type Questions

Question 1.
What is health ? Write the advantages of good health.
Answer:
Health is a state of being well enough for a person to function well physically, mentally and socially. The structure and functions of all the body parts of a healthy person are normal. He is free from all types of psychological, mental and social tensions. A person free from only physical diseases is not frilly healthy.
Advantages: Following are the advantages of good health:
(1) The personal life of a healthy person is full of happiness.
(2) A healthy person utilises his potential fully as an individual, in the family, in society and in the country and does not become a burden on anyone.

Question 2.
What are the conditions necessary to maintain good health?
Answer:
The following conditions are necessary to maintain good health:
(1) Personal and family hygiene.
(2) Proper exercise and relaxation.
(3) Balanced diet.
(4) Community hygiene.
(5) Good habits and abstinence from intoxicants/drugs.

Question 3.
How does the environment affect health?
Answer:
The health of all persons is based on the surrounding environment. Physical factors constitute the environment. A person can fulfil his basic requirements only from the environment. This is possible only in a hygienic environment. A polluted environment has a bad effect on our health.

Question 4.
What things should be kept in mind while cleaning OUT homes 2
Answer:
Apart from dusting, cleaning cobwebs and disposing the wastes, we should also regularly keep a check on insects like cockroaches, mosquitoes, houseflies, spiders etc. Apart from this, it is necessary to white wash the house to keep it clean and beautiful. We should also keep our immediate surroundings neat and clean so that we don’t catch a disease.

Question 5.
What is a disease? How does disease occur?
Answer:
Any variation in the normal structure or function of a tissue or organ of a body. Or a change in psychological condition is called a disease. A diseased person does not feel comfortable.

Causes of disease:
The disease can occur because of any of the following three causes:
(1) Diseases caused due to deficiency or accumulation of some nutritive components e.g. Beri-beri, Obesity etc.
(2) Through infection (communicable) e.g., cholera, tuberculosis etc.
(3) Non-communicable diseases, like heart diseases, stones in the kidney or gall bladder etc.

Question 6.
What are non-communicable diseases ? Name any two.
Answer:
The diseases which are not caused by infection (and remain confined to the bearer only) are called non-communicable diseases, e.g. heart diseases, non-functioning of kidney, gall bladder or kidney stones etc.

Question 7.
What are hereditary (congenita!) diseases ? Give examples.
Answer:
The diseases which are carried by the parents to the offspring are called hereditary diseases. One such disease is haemophilia, in which the patient’s blood does not clot easily. Generally, the time taken for the clotting of blood increases from 2-8 minutes to 30 minutes-24 hours. A haemophilic person can die because of a continuous flow of blood even from a minor injury. Another such type of disease in sickle cell anaemia. The patient’s RBCs (Red Blood Cells) have defective haemoglobin, which reduces the oxygen-carrying capacity of the RBCs and the person goes on to become weak.

Question 8.
What are antibiotics? Why is their excessive use harmful?
Answer:
Antibiotics are the drugs which are derived from bacteria and are fatal for other bacteria. They do not harm the protoplasm of humans For example; penicillin, Terramycin, streptomycin etc. These save us by stopping the growth of pathogens or by. destroying them. If we use them excessively, then the useful micro-organisms in our alimentary canal will start getting destroyed which will affect our digestive system. .

Question 9.
What are the simple and easy precautions to remain free from the diseases caused by microorganisms ?
Answer:
The simple and easy precautions are:
(1) To keep the infected person away from a healthy one.
(2) By vaccination.
(3) To boil water before drinking.
(4) By keeping our surroundings neat and clean.
(5) By keeping check (on the growth) of mosquitoes and houseflies.
(6) By wearing properly washed clothes.
(7) By dumping the waste products in covered places. .
(8) By spraying insecticides so as to keep the surroundings free from harmful insects and pathogens.

Question 10.
What is the difference between bacteria and virus ?
Answer:
Given is the difference between bacteria and virus:
Bacteria:
1. They are living organisms.
2. They have a cellular structure.
3. They can be seen under a microscope.
4. Bacteria are bom from the same type of bacteria and multiply on their own.
5. Bacteria can be harmful as well as useful.

Virus:
1. They possess the features of both living as well as non-living.
2. Their structure is not cellular.
3. They can be seen only with the help of an electron microscope.
4. Viruses can be multiplied only inside a jost cell.
5. Viruses are only harmful.

Question 11.
How do antibiotics act against bacterial infection? Why are they not effective against virus? Explain with reason.
Answer:
Bacteria create a cell wall for their protection. But the antibiotic substances hinder the process of creation of the cell soil due to which bacteria cannot protect themselves and the antibiotics destroy them. Pathogens are destroyed by the same process. Because virus do not create any cell wall so, that antibiotics are not able to effect virus.

Question 12.
How does infection through air take place ?
Answer:
When an infected person exhales, sneezes or coughs, the pathogens spread in the air. When another person inhales the same air, these pathogens enter his body. Common cold, pneumonia and tuberculosis spread through this mode only. The occurence of common cold in crowded places is again due to this.

Question 13.
How does infection take place through water ?
Answer:
When the pathogens from the infected person enter the drinking water, then a healthy person also gets infected when he drinks this water. There is more incidence of water-borne diseases in rainy season. Cholera Spreads through this method.

Question 14.
Why are V.D. and gonorrhoea considered a social evil ?
Answer:
They are both sexual diseases. Their pathogens enter a healthy person’s body through cuts, wounds or genitals. In such a condition, these diseases are spread from a woman to a man or from a man to a woman during sexual contact. As these diseases also spread through sexual activity, so these diseases are considered a social evil because the incidence of these diseases is more in people who establish illegitimate sexual relations. Prostitution is the main cause of this. Social awareness can reduce the incidence of these diseases.

Question 15.
What is the sign inflammation ?
Answer:
During infection, immune system recruits many cells to the affected tissue to kill the disease causing microbes, this process is called imflammation. As a part of this process, there are local effects such as swelling, pain and general effects such as fever.

Question 16.
Why is vaccination of children and expectant mothers done ?
Answer:
Polio, whooping cough, measles diphtheria, typhoid and tetanus are some infective diseases common in the children. To prevent these diseases newly bom babies and children must be got vaccinated or immunised. To immunise all the children and expectant mothers under a plan, sufficient quantity of vaccines have been manufactured and distributed in India. If the expectant mothers are treated with the tetanus vaccine before she delivers the baby, the newly bom baby will have the ability of immunity. Therefore, lacs of lives of children can be saved by vaccination at proper time.

Essay Type Questions

Question 1.
How do pathogens spread diseases ?
Answer:
Those pathogens, that spread diseases are called as pathogens. With the entrance of pathogens into the human body and their reaction a man falls ill.

Pathogens can spread diseases in the following ways:
1. Through Contact: Some pathogens spread diseases through direct contact with the patient or the thing used by the patient; for example, if a healthy person puts on the clothes of a patient or uses his bed, the pathogens get entered into his body and make him ill, like pathogens causing skin diseases.

2. Through eatables: Pathogens of typhoid, cholera etc. the remaining food of the patient enter into our body and spread diseases.

3. Through air: Pathogens of common cold, influenza etc. enter the body of a healthy person through breathing.

4. Through vectors: Houseflies, mosquitoes, mice, bugs etc. are some insects that spread diseases; like malaria spreads through female anopheles mosquito bite. Plague is spread by mice.

5. Through water: Pathogen on mixing into drinking water reach to the healthy person and diseases
like cholera, diarrhoea, typhoid etc. surround the human beings. Therefore, specially during rainy season it is suggested to boil drinking water and then cool it down to drink there after, so that the pathogens might destroy.

Question 2.
Give a brief introduction of the infectious diseases.
Answer:
Infectious diseases are caused due to micro-organisms like:
1. Cholera: The cause of qholera is vibriocoli. It occurs due to drinking dirty water or infected eatables. Pathogens of cholera are spread through housefly. The patient suffers from painless water loss of due to acute vomitting and diarrhoea.

2. Tuberculosis: It’s cause is micro bacterium tuberculosis. This disease affects the lungs. The patient suffers from constant fever, blood in sputum, bodyache, diarrhoea and the neck ribs get swollow.

3. Typhoid: It spreads through salmonela bacterium typall. Its main symptoms are fever, headache, bleeding nose, black circles on the skin, reddish blisters on the chest.

4. Cough and Cold: It is caused by rhino works. Constant flowing of liquid through nostrils, redness in the eyes, irritation and headache are its main symptoms.

5. Chickenpox: It is caused by the pathogen berisola joster. Here, big circles on the patients body
get formed, severe backache, high fever and headache are its other symptoms. Its a effect lasts long for 14 days. .

6. Polio: It is an infectious disease among the children. It spreads through the pathogen of polio mylitis.
Its symptoms are fever, vomiting, bodyache, stiffness and difficulty in controlling muscles. The children develop paralysis. .

7. Rabies: It is caused by the Rabies pathogen. It is caused due to the bite of dog, cat, bat, jackal, fox etc. Patient suffering from Rabies scares water (hydrophobia). It affects the nervous system. Due to this the patient meets a painful death.

8. AIDS: It comes under the category of sexually transmitted diseases which is caused due to virus. A person suffering from HIV-AIDS, Immunisation system becomes weaker, with the result the body fails to face even the minor infection. The patient of HIV runs with pneumonia even with minor cough and cold. Presently, AIDS is an incurabled disease.

Question 3.
Which programmes have been adopted the Govt, of control communicable diseases ?
Answer:
Following programme have been adopted by the Govt, of India to control communicable diseases :

1. National Malaria Eradiction Programme:
This started as a control measure in 1953. In 1975, it was converted from control of the disease to total eradication programme. The changed programmed was started in 1977 which was effectively enforced. The programme include inspection of areas affected by malaria educated the people using suitable medicines, spraying of insecticides to kill mosquitoes, identifying the malaria patients, distribution of antimalarial drugs to keep their accounts etc.

Proper Vaccination List:

2. National Tuberculosis Eradication Programme:

Following stages are there in this programme:
(1) To identify the person suffering from tuberculosis in the very first stage and giving him treatment.
(2) To prevent the diseases by inoculating the children at an early age by BCG vaccine.
(3) Many TB clinics, district centres and education centres started in country.

3. National Cholera Eradication Programme: It includes cleaning the cholera effected area, making fresh drinking water availability and proper disposal of urban faecol waste. To educate people and for immediate treatment, mobile treatment units have been established for it. To inoculate people with cholera injections has been arranged in these.

4. National Leprosy Eradication Programme : A survey was conducted for thistdiseases in India and many treatment centres have been setup. Houses have been constructed for rehabilitation of the infected persons. Tamilnadu Govt, has opened a Leprosy Education and Research Centres, where there is an advisory committee for this diseases. Sulpha drugs are used for the treatment of the disease.

Practical Work

Experiment 1.
To study the effect of possible calamities on the health of people.

Procedure:
There are two effects of natural calamities on the health. Prepare its table with the help of science-teacher.

Experiment 2.
To study the provisions made by local municipal corporation for the supply of clear drinking water.

Procedure:
Prepare a list of provisions made by the local municipal corporation for the supply of clear drinking water.
(1) The cholorination of drinking water is done.
(2) Water storage tank is cleaned regulary.
(3) Tubewells are used to get underground drinking water.
(4) Water treatment plant is setup for canal water.
(5) Water supplies by pipes to avoid pollution.

Experiment 3.
To study the provisions made by local authority to manage garbage.

Procedure:
Prepare a list of provisions related to it from officer related to local conservator.
(1) The garbage is cleaned by landfill.
(2) Organic garbage is treated by roasting.
(3) Covered vehicles are used to throw garbage.
(4) Composte prepared from organic garbage is selling to farmers.
(5) Local water is planned to set up solid waste treatment plant.

Quick Review of the Chapter

1. Cholera is a ………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(A) bacterial

2. Malaria spreads by:
(A) mouse
(B) fly
(C) mosquito
(D) bug
Answer:
(C) mosquito

3. Cholera spreads by :
(A) fly
(B) mouse
(C) mosquito
(D) none of these
Answer:
(A) fly

4. Respiratory diseases spread by :
(A) food
(B) water
(C) air
(D) animal
Answer:
(C) air

5. Communicable disease is :
(A) goitre
(B) beri-beri
(C) obesity
(D) tuberculosis
Answer:
(D) tuberculosis

6. Example of chronic disease is :
(A) cough
(B) cold
(C) elephantiasis
(D) headache
Answer:
(C) elephantiasis

7. Acute disease is:
(A) lungs tuberculosis
(B) Dengue fever
(C) blindness
(D) disentary
Answer:
(D) disentary

8. Which of the following disease is caused by virus ?
(A) Anthrax
(B) Dengue fever
(C) Malaria
(D) Cholera
Answer:
(B) Dengue fever

9. The reason of AIDS is :
(A) bacteria
(B) protozoa
(C) virus
(D) fungi
Answer:
(C) virus.

10. The prostitution in society is the reason which of the following disease
(A) pneumonia
(B) gonorrhoea
(C) giardiasis
(D) influenza
Answer:
(B) gonorrhoea

11. AIDS can spread by:
(A) kiss
(B) embracing
(C) combined used blade of shave
(D) hug
Answer:
(C) combined used blade of shave

12. AIDS cannot spread by:
(A) sexual relations
(B) blood transfusion
(C) from placenta of mother
(D) by shaking hands
Answer:
(D) by shaking hands

13. Which disease can be caused by attack on lever?
(A) diabetes
(B) malaria
(C) jaundice
(D) goitre
Answer:
(C) jaundice

14. Which disease can be caused by fungi?
(A) kala-azar
(B) cholera
(C) malaria
(D) skin disease
Answer:
(D) skin disease

15. Vomiting is related to :
(A) mind
(B) liver
(C) lungs
(D) kidney
Answer:
(A) mind

16. Which fight against microbes :
(A) RBC
(B) WBC
(C) platelets
(D) lymphocytes
Answer:
(B) WBC

17. Which disease happens first and last times ?
(A) rabies
(B) polio
(C) chicken pox
(D) small pox
Answer:
(D) small pox

18. The main symptom of AIDS is:
(A) the loss in resistance power
(B) fever
(C) arthritus
(D) high cough ‘
Answer:
(A) the loss in resistance power

19. Rabies is a …………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(B) viral

20. Before going to fair, there is vaccination of:
(A) tuberculosis
(B) hepatitis
(C) typhoid
(D) cholera
Answer:
(D) cholera

21. Deficiency disease is not:
(A) diarrohea
(B) scurvy
(C) pellagra
(D) xeropthalmia
Answer:
(A) diarrohea

22. Communicable disease is :
(A) intluenza
(B) hemophilia
(C) goitre
(D) anaemia
Answer:
(A) influenza

23. The disease in mother’s placenta is:
(A) polio
(B) AIDS
(C) goitre
(D) scurvy
Answer:
(B) AIDS

24. Vaccination of which of the following is not available?
(A) whooping cough
(B) polio
(C) diphtheria
(D) AIDS
Answer:
(D) AIDS

25. Which disease can be caused by throwing solid waste in open?
(A) elephantiasis
(B) goitre
(C) cholera
(D) AIDS
Answer:
(C) cholera

26. Match the following:

(A) a-iv, b-i,c-ii,d-iii
(B) a-i,b-ii,c-iii,d-iv
(C) a-ii,, b-ic—iii,d-ii,
(D) a-iii. b-Ii, c-i, d’-iv
Answer:
(A) a-iv b-i, c-ii d-III

27. Which disease is caused by viruses?
(A) Common cold
(B) Dengue fever
(C) Influenza
(D) All of these
Answer:
(D) All of these

Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Ex 6.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Exercise 6.2

प्रश्न 1.
आकृति में, x और y के मान ज्ञात कीजिए और फिर दर्शाइए कि AB || CD है।

हल :
यहाँ पर PQ एक रेखा है।

∴ ∠PRA + ∠ARQ = 180° [रैखिक युग्म]
50° + x = 180°
या x = 180° – 50° = 130° उत्तर …………..(i)
अब CD और PQ दो प्रतिच्छेद रेखाएँ S पर प्रतिच्छेद करती हैं।
∴ ∠QSC = ∠DSR [शीर्षाभिमुख कोण]
⇒ 130° = y
या y = 130° उत्तर ………………(ii)
समीकरण (i) व (ii) से
∠x = ∠y [प्रत्येक 130°]
परंतु ये एकांतर कोण हैं
∴ AB || CD [इति सिद्धम]

प्रश्न 2.
आकृति में, यदि AB || CD, CD || EF और y : z = 3 : 7 है, तो x का मान ज्ञात कीजिए। [B.S.E.H. March, 2017]

हल :
क्योंकि CD || EF तथा तिर्यक रेखा PQ उनको क्रमशः S तथा T पर काटती है।

∴ ∠CST = ∠STF [एकांतर कोण]
या 180°- y = z [∵ y + ∠CST = 180° रैखिक युग्म]
या y + z = 180°
परंतु y : z = 3 : 7
अनुपाती योग = 3 + 7 = 10
∴ ∠y = \(\frac {3}{10}\) × 180° = 54°
∠z = \(\frac {7}{10}\) × 180° = 126°
अब क्योंकि AB || CD तथा तिर्यक रेखा PQ उनको क्रमशः R तथा S पर काटती है।
∴ ∠ARS + ∠RSC = 180°
[∵ अभ्यंतर कोण संपूरक होते हैं।]
या x + y = 180°
या x = 180° – y = 180° – 54° = 126°
[∵ y = 54°]
अतः x = 126° उत्तर

प्रश्न 3.
आकृति में, यदि AB || CD, EF ⊥ CD और ∠GED = 126° है, तो ∠AGE, ∠GEF और ∠FGE जात कीजिए।

हल :
क्योंकि AB || CD तथा तिर्यक रेखा GE इन्हें क्रमशः G तथा E पर प्रतिच्छेद करती है ∠AGE = ∠GED [एकांतर कोण]
परंतु ∠GED = 126°
∴ ∠AGE = 126°
आकृति अनुसार
∠GEF = ∠GED – ∠FED
= 126°- 90° = 36°
या ∠FGE = ∠GEC [एकांतर कोण]
⇒ ∠FGE = 90° – ∠GEF [त्रिभुज GEF का तीसरा कोण]
= 90° – 36° = 54°
अतः
∠AGE = 126°, ∠GEF = 36°
∠FGE = 54° उत्तर

प्रश्न 4.
आकृति में, यदि PQ || ST, ∠PQR = 110° और ∠RST = 130° है, तो ∠QRS ज्ञात कीजिए।

हल :
यहाँ पर PQ को M बिंदु पर काटने के लिए SR तक बढ़ाया।

अब PM || ST तथा तिर्यक रेखा SM क्रमशः उनको M तथा S पर काटती है।
∠SMQ = ∠TSM [एकांतर कोण]
परंतु ∠TSM = 130°
अतः
∠SMQ = 130°
⇒ ∠QMR = 180° – 130° = 50°
[∵ SR एक रेखा है]
क्योंकि किरण RQ, PM को Q पर काटती है।
∴ ∠PQR + ∠RQM = 180°
या 110° + ∠RQM = 180°
या ∠RQM = 180° – 110° = 70°
अब ΔQRM में,
∠QRM + ∠RQM + ∠QMR = 180°
या ∠QRS + 70° + 50° = 180°
या ∠QRS + 120° = 180°
या ∠QRS = 180° – 120° = 60° उत्तर

प्रश्न 5.
आकृति में, यदि AB || CD, ∠ARQ = 50° और ∠PRD = 127° है, तो x और y ज्ञात कीजिए।

हल :
आकृति अनुसार AB || CD तथा तिर्यक रेखा PQ इनको क्रमशः P और Q पर प्रतिच्छेद करती है।
∠PQR = ∠APQ [एकांतर कोण]
परंतु ∠PQR = x व ∠APQ = 50°
∴ x = 50°
अब क्योंकि AB || CD तथा तिर्यक रेखा PR इनको क्रमशः P तथा R पर प्रतिच्छेद करती है।
∴ ∠APR = ∠PRD [एकांतर कोण]
या ∠APQ + ∠QPR = 127° [∵ ∠PRD = 127°]
या 50° + y = 127° [∵ ∠APQ = 50°]
या y = 127° – 50° = 77°
इस प्रकार x = 50° तथा y = 77° उत्तर

प्रश्न 6.
आकृति में, PQ और RS दो दर्पण हैं जो एक-दूसरे के समांतर रखे गए हैं। एक आपतन किरण (incident ray) AB, दर्पण PQ से B पर टकराती है और परावर्तित किरण (reflected ray) पथ BC पर चलकर दर्पण Rs से C पर टकराती है तथा पुनः CD के अनुदिश परावर्तित हो जाती है। सिद्ध कीजिए कि AB || CD है।

हल :
यहाँ पर दो समतल दर्पण PQ तथा RS एक-दूसरे के समांतर रखे गए हैं।
∴ PQ || RS [दिया है।

दी गई आपतित किरण AB परावर्तित होने के बाद पथ BC और CD से होकर जाती है।
BN तथा CM समतल दर्पण PQ तथा Rs के अभिलंब खींचे गए हैं।
क्योंकि BN ⊥ PQ, CM ⊥ RS तथा PQ || RS
⇒ BN || CM
इसलिए BN तथा CM दो समांतर रेखाएँ हैं तथा तिर्यक रेखा BC इन्हें क्रमशः B तथा C पर काटती है।
∴ ∠2 = ∠3 [एकांतर कोण]
परंतु परावर्तन के नियम से
∠1 = ∠2 और ∠3 = ∠4
⇒ ∠1 + ∠2 = ∠2 + ∠2
और ∠3 + ∠3 = ∠3 + ∠4
⇒ ∠1 + ∠2 = 2∠2
और ∠3 + ∠4 = 2∠3
परंतु ∠2 = ∠3 या 2∠2 = 2∠3
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠BCD
परंतु ये एकांतर अंतः कोण हैं।
इसलिए AB || CD

Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 बहुपद Ex 2.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 2 बहुपद Exercise 2.5

प्रश्न 1.
उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्नलिखित गुणनफल ज्ञात कीजिए
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (3 – 2x) (3 + 2x)
हल :
(i) (x + 4) (x + 10) = x² + (4 + 10)x + 4 × 10
= x² + 14x + 40 उत्तर
(ii) (x + 8 ) (x – 10) = x² + (8 – 10)x + 8 × (-10)
= x² – 2x – 80 उत्तर
(iii) (3x + 4) (3x – 5) = (3x)² + (4 – 5) × 3x + 4 × (-5)
= 9x² + (-1) × 3x – 20
= 9x² – 3x – 20 उत्तर

(iv) (y² + \(\frac {3}{2}\)) (y² – \(\frac {3}{2}\))
= (y²)² – (\(\frac {3}{2}\))²
= y⁴ – \(\frac {9}{4}\) उत्तर

(v) (3 – 2x) (3 + 2x) = (3)² – (2x)²
= 9 – 4x² उत्तर

प्रश्न 2.
सीधे गुणा किए बिना निम्नलिखित गुणनफलों के मान ज्ञात कीजिए [March, 2020]
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
हल :
(i) 103 × 107 = (100 + 3) (100 + 7)
= (100)² + (3 + 7) × 100 + 3 × 7
= 10000 + 10 × 100 + 21
= 10000 + 1000 + 21 = 11021 उत्तर

(ii) 95 × 96 = (100 – 5)(100 – 4)
= (100)² + (-5 – 4) × 100 + (-5)(-4)
= 10000 – 900 + 20
= 10020 – 900 = 9120 उत्तर

(iii) 104 × 96 = (100 + 4) (100 – 4)
= (100)² – (4)²
= 10000 – 16 = 9984 उत्तर

प्रश्न 3.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित का गुणनखंडन कीजिए
(i) 9x² + 6xy + y²
(ii) 4y² – 4y + 1
(iii) x² – \(\frac{y^2}{100}\) [B.S.E.H. 2019]
हल :
(i) 9x² + 6xy + y² = (3x)² +2 (3x) (y) + (y)²
= (3x + y)² = (3x + y) (3x + y) उत्तर
(ii) 4y² – 4y + 1 = (2y)² – 2 (2y)(1) + (1)²
= (2y – 1)² = (2y – 1) (2y- 1) उत्तर
(iii) x² – \(\frac{y^2}{100}\) = (x)² – (\(\frac {y}{10}\))²
= (x – \(\frac {y}{10}\)) (x + \(\frac {y}{10}\)) उत्तर

प्रश्न 4.
उपयुक्त सर्वसमिकाओं का प्रयोग करके निम्नलिखित में से प्रत्येक का प्रसार कीजिए
(i) (x + 2y + 4z)²
(ii) (2x – y + z)²
(iii) (-2x + 3y + 2z)²
(iv) (3a – 7b – c)²
(v) (-2x + 5y – 3z)²
(vi) [\(\frac {1}{4}\)a – \(\frac {1}{2}\)b + 1]²
हल :
(i) (x + 2y + 4z)² = x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2 (4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx उत्तर

(ii) (2x – y + z)² = [2x + (-y) + z]²
= (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx उत्तर

(iii) (-2x + 3y + 2z) = [(-2x) + 3y + 2z²]
= (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx उत्तर

(iv) (3a – 7b – c)² = [3a + (-7b) + (-c)]² = (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2 (-c)(3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ac उत्तर

(v) (-2x + 5y -3z)² = [(-2x) + 5y + (-3z)]² = (-2x)² + (5y)² + (-3z)² + 2(-2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx उत्तर

प्रश्न 5.
गुणनखंडन कीजिए –
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz
हल:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz = (2x)² + (3y)² + (-4z)² + 2(-4z)(2x)
= (2x + 3y – 4z)²
= (2x + 3y – 4z) (2x + 3y – 4z) उत्तर

(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz = (-\(\sqrt{2}\)x)² + (y)² + (2\(\sqrt{2}\)z)² + 2(\(\sqrt{2}\)z) (-\(\sqrt{2}\)x)
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)\(\sqrt{2}\)
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z) (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z) उत्तर

प्रश्न 6.
निम्नलिखित घनों को प्रसारित रूप में लिखिए
(i) (2x + 1)³
(ii) (2a – 3b)³
(iii) [\(\frac {3}{2}\)x + 1]³
(iv) [x – \(\frac {2}{3}\)y]³ [B.S.E.H. March, 2018]
हल :
(i) (2x + 1)³ = (2x)³ + (1)³ + 3 × 2x × 1 (2x + 1)
[∵ (a + b)³ = a³ + b³ + 3ab(a + b)]
= 8x³ + 1 + 6x (2x + 1)
= 8x³ + 1 + 12x² + 6x
= 8x³ + 12x² + 6x + 1 उत्तर

(ii) (2a – 3b)³ = (2a)³ – (3b)³ – 3 (2a) (3b)(2a – 3b)
[∵ (a – b)³ = a³ – b³ – 3ab(a – b)]
= 8a³ – 27b³ – 18ab (2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab² उत्तर

प्रश्न 7.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के मान ज्ञात कीजिए
(i) (99)³ (ii) (102)³ (iii) (998)³
हल :
(i) (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3 × 100 × 1 (100 – 1)
= 1000000 – 1 – 30000 + 300
= 1000300 – 30001 = 970299 उत्तर

(ii) (102)³ = (100 + 2)³
= (100)³ + (2)³ + 3 × 100 × 2(100 + 2)
= 1000000 + 8 + 600 (100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208 उत्तर

(iii) (998)³ = (1000 – 2)³
= (1000)³ – (2)³ – 3 × 1000 × 2(1000 – 2)
= 1000000000 – 8 – 6000 × (1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 1000012000 – 6000008
= 994011992 उत्तर

प्रश्न 8.
निम्नलिखित में से प्रत्येक का गुणनखंडन कीजिए
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ – 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) 27p³ – \(\frac{1}{216}-\frac{9}{2}\)p² + \(\frac {1}{4}\)p [March 2020]
हल :
(i) 8a³ + b³ + 12a²b + 6ab²
= (2a)³ + (b)³ + 3 (2a) (b) (2a + b)
= (2a + b)³
= (2a + b)(2a + b)(2a + b) उत्तर

(ii) 8a³ – b³ – 12a²b + 6ab²
= (2a)³ – (b)³ – 3 (2a)(b)(2a – b)
= (2a – b)³
= (2a – b)(2a – b)(2a – b) उत्तर

(iii) 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a) (3 – 5a)
= (3 – 5a)³ = (3 – 5a) (3 – 5a) (3 – 5a) उत्तर

(iv) 64a³ – 27b³ – 144a²b + 108aba²
= (4a)³ – (3b)³ – 3 (4a) (3b) (4a – 3b)
= (4a – 3b)³
= (4a – 3b) (4a – 3b) (4a – 3b) उत्तर

प्रश्न 9.
सत्यापित कीजिए
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
हल:
(i) R.H.S. = (x + y) (x² – xy + y²)
= (x) (x² – xy + y²) + y (x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³
= L.H.S.
अतः सत्यापित हुआ।

(ii) R.H.S. = (x – y) (x² + xy + y²)
= x (x² + xy + y²) – y (x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³
= L.H.S.
अतः सत्यापित हुआ।

प्रश्न 10.
निम्नलिखित में से प्रत्येक का गुणनखंडन कीजिए
(i) 27y³ + 125z³ [B.S.E.H. 2019]
(ii) 64m³ – 343n³
हल :
(i) 27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) [(3y)² – 3y × 5z + (5z)²]
= (3y + 5z) [9y² – 15yz + 25z²] उत्तर

(ii) 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n) [(4m)² + 4m × 7n + (7n)²]
= (4m – 7n) [16m² + 28mn + 49n²] उत्तर

प्रश्न 11.
गुणनखंडन कीजिए
27x³ + y³ + z³ – 9xyz [B.S.E.H. March, 2017, 2018]
हल :
हम जानते हैं कि a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
अब 27x³ + y³ +z³ – 9xyz
= (3x)³ + (y)³ + (z)³ -3 (3x) (y) (z)
= (3x + y + z) [(3x)² + (y)² + (z)² – (3x)(y) – (y)(z) – (z)(3x)]
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx) उत्तर

प्रश्न 12.
सत्यापित कीजिए
x³ + y³ + z³ – 3xyz = \(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²]
हल :
यहाँ पर (x – y)² + (y – z)² + (z – x)² = (x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)
= 2x² + 2y² + 2z² – 2xy – 2yz – 2zx
= 2 (x² + y² + z² – xy – yz – zx)
दोनों पक्षों को \(\frac {1}{2}\)(x + y + z) से गुणा करने पर,
\(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²]
= \(\frac {1}{2}\)(x + y + z) × 2 (x² + y² + z² – xy – yz – zx)
= (x + y + z) (x² + y² + z² – xy – yz – zx) (i)
हम जानते हैं कि
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx) (ii)
समीकरण (i) व (ii) से
x³ + y³ + z³ – 3xyz = \(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²] |इति सिद्धम]

प्रश्न 13.
यदि x + y + z = 0 हो, तो दिखाइए कि x³ + y³ + z³ = 3xyz है।
हल :
यहाँ पर
x + y + z = 0
⇒ x + y = – z
दोनों पक्षों को घन करने पर
(x + y)³ = (-z)³
या x³ + y³ + 3xy (x + y) = – z³
या x³ + y³ – 3xyz = – z³ [∵ x + y = – z]
या x³ + y³ + z³ = 3xyz [इति सिद्धम]

प्रश्न 14.
वास्तव में पनों का परिकलन किए बिना निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए –
(i) (-12)³ + (7)³ + (5)³
(ii) (28)³ + (-15)³ + (-13)³ [B.S.E.H. March, 2020]
हल:
(i) माना x = – 12, y = 7 और z = 5
यहाँ पर
x + y + z = – 12 + 7 + 5 = 0
यदि x + y + z = 0 हो तो
x³ + y³ + z³ = 3xyz
अतः
(-12)³ + (7)³ + (5)³ = 3 × (-12) × 7 × 5 उत्तर
= – 1260 उत्तर

(ii) माना x = 28, y = – 15 और z = – 13
यहाँ पर
x + y + z = 28 – 15 – 13 = 0
यदि x + y + z = 0 हो तो
x³ + y³ + z³ = 3xyz
(28)³ + (-15)³ + (-13)³ = 3 × (28) × (-15) × (-13)
= 16380 उत्तर

प्रश्न 15.
नीचे दिए गए आयतों, जिनमें उनके क्षेत्रफल दिए गए हैं, में से प्रत्येक की लंबाई और चौड़ाई के लिए संभव व्यंजक दीजिए
(i) क्षेत्रफल : 25a² – 35a +12
(ii) क्षेत्रफल : 35y² + 13y – 12
हल :
(i) 25a² – 35a + 12 = 25a² – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
हम जानते हैं कि
आयत का क्षेत्रफल = लंबाई × चौड़ाई
⇒ 25a² – 35a + 12 = (5a – 3) × (5a – 4)
अतः आयत की लंबाई 5a-3 तथा चौड़ाई 5a-4 उत्तर

(ii) 35y² + 13y – 12 = 35y² + 28y – 15y – 12
= 7y (5y + 4) – 3 (5y + 4)
= (5y + 4)(7y – 3)
हम जानते हैं कि
आयत का क्षेत्रफल = लंबाई × चौड़ाई
⇒ 35y² + 13y – 12 = (7y – 3) × (5y + 4)
अतः आयत की लंबाई 7y – 3 तथा चौड़ाई 5y + 4 उत्तर

प्रश्न 16.
बनाभों (cuboids), जिनके आयतन नीचे दिए गए हैं, की विमाओं के लिए संभव व्यंजक क्या हैं ?
(i) आयतन : 3x² – 12x
(ii) आयतन : 12ky² + 8ky – 20k
हल :
(i) 3x² – 12x = 3x(x – 4)
= 3 × x × (x – 4)
हम जानते हैं कि
घनाभ का आयतन = लंबाई × चौड़ाई × ऊँचाई
अतः दिए गए व्यंजक की लंबाई, चौड़ाई एवं ऊँचाई क्रमशः 3, x तथा (x – 4) है। उत्तर

(ii) 12ky² + 8ky – 20k = 4k [3y² + 2y – 5]
= 4k [3y² + 5y – 3y – 5]
= 4k [y(3y + 5) – 1(3y + 5)]
= 4k (3y + 5)(y – 1)
हम जानते हैं कि
घनाभ का आयतन = लंबाई × चौड़ाई × ऊँचाई
अतः दिए गए व्यंजक की लंबाई, चौड़ाई एवं ऊँचाई क्रमशः 4k, (3y + 5) तया (y – 1) है। उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Exercise 9.1

प्रश्न 1.
निम्नलिखित आकृतियों में से कौन-सी आकृतियां एक ही आधार और एक ही समांतर रेखाओं के बीच स्थित हैं? ऐसी स्थिति में, उभयनिष्ठ आधार और दोनों समांतर रेखाएं लिखिए।

हल :
(i) ΔPDC व समांतर चतुर्भुज ABCD एक ही आधार DC तथा समांतर रेखाओं DC व AB के मध्य में स्थित हैं।
(ii) चतुर्भुज SRNM व SRQP एक आधार SR पर हैं परंतु एक ही समांतर रेखाओं के बीच नहीं हैं।
(iii) ΔTQR व समांतर चतुर्भुज PQRS एक ही आधार QR तथा समांतर रेखाओं QR व PS के मध्य में स्थित हैं।
(iv) ΔRPQ व || चतुर्भुज ABCD एक ही आधार पर नहीं हैं परंतु एक ही समांतर रेखाओं BC व AD के मध्य में स्थित हैं।
(v) समांतर चतुर्भुज ABCD व समांतर चतुर्भुज APQD एक ही आधार AD तथा एक ही समांतर रेखाओं AD व BQ के मध्य में स्थित हैं।
(vi) चतुर्भुज PSDA, PSCB व PSRQ एक ही आधार PS पर हैं परन्तु एक ही समांतर रेखाओं के बीच नहीं हैं।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 5 युक्लिड के ज्यामिति का परिचय Exercise 5.1

प्रश्न 1.
निम्नलिखित कथनों में से कौन-से कथन सत्य हैं और कौन-से कथन असत्य हैं ? अपने उत्तरों के लिए कारण दीजिए
(i) एक बिंदु से होकर केवल एक ही रेखा खींची जा सकती है।
(ii) दो भिन्न बिंदुओं से होकर जाने वाली असंख्य रेखाएँ हैं।
(iii) एक सांत रेखा दोनों ओर अनिश्चित रूप से बढ़ाई जा सकती है।
(iv) यदि दो वृत्त बराबर हैं, तो उनकी त्रिज्याएँ बराबर होती हैं।
(v) आकृति में, यदि AB = PQ और PQ = XY है, तो AB = XY होगा।

हल :
(i) यह कथन असत्य है, क्योंकि एक बिंद से होकर एक नहीं, बल्कि अनेक रेखाएँ खींची जा सकती हैं। जैसे बिंद O से दर्शाई गई हैं।

(ii) यह कथन असत्य है, क्योंकि दो बिंदुओं से होकर एक ओर केवल एक ही रेखा खींची जा सकती है, जैसे बिंदु X और Y से होती हुई केवल एक ही रेखा खींची जा सकती है।

(iii) यह कथन सत्य है, क्योंकि अभिगृहीत 2 अनुसार एक सांत रेखा (Terminated Line) को अनिश्चित रूप से बढ़ाया जा सकता है। जैसे AB को बढ़ाया हुआ दिखाया गया है।

(iv) यह कथन सत्य है, क्योंकि यदि एक वृत्त द्वारा घेरे गए क्षेत्र को दूसरे वृत्त पर अध्यारोपित किया जाए तो वे संपाती होंगे। अतः इनके केंद्र और परिसीमाएँ संपाती होंगी। इस कारण इनकी त्रिज्याएँ संपाती होंगी।
(v) यह कथन सत्य है, क्योंकि यूक्लिड के प्रथम अभिगृहीत अनुसार वे बस्तुएँ जो एक ही वस्तु के बराबर हों एक-दूसरे के बराबर होती हैं।

प्रश्न 2.
निम्नलिखित पदों में से प्रत्येक की परिभाषा दीजिए। क्या इनके लिए कुछ ऐसे पद हैं, जिन्हें परिभाषित करने की आवश्यकता है? वे क्या हैं और आप इन्हें कैसे परिभाषित कर पाएँगे ?
(i) समांतर रेखाएँ[B.S.E.H, 2018]
(ii) लंब रेखाएँ
(iii) रेखाखंड [B.S.E.H. 2018]
(iv) वृत्त की त्रिज्या
(v) वर्ग
हल :
उपरोक्त पदों को परिभाषित करने के लिए हमें निम्नलिखित पदों की आवश्यकता पड़ेगी
(a) बिंदु – एक बिंदु केवल कल्पना है जिसकी न कोई लंबाई, न कोई चौड़ाई और न कोई मोटाई होती है अर्थात एक बिंदु वह है जिसका कोई भाग नहीं होता।
(b) रेखा – एक रेखा चौड़ाई रहित लंबाई होती है। इसे दोनों ओर अनिश्चित रूप से बढ़ाया जा सकता है।
(c) तल – तल की लंबाई तथा चौड़ाई होती है। इसकी कोई मोटाई नहीं होती; जैसे कागज।
(d) किरण – एक निश्चित बिंदु से चलकर अंत तक जाने वाली रेखा किरण कहलाती है अर्थात इसका आरंभिक बिंदु होता है। परंतु अंत बिंदु नहीं होता; जैसे किरण AB दशाई गई है।

(e) कोण – एक उभयनिष्ठ बिंदु वाली दो किरणों का सम्मिलन कोण कहलाता है; जैसे \(\overline{AB}\) और \(\overline{AC}\) के बीच का क्षेत्र ∠BAC कहलाता है।

(f) वृत्त – किसी तल में किसी निश्चित बिंदु से उसी तल में दी गई समान अचर दूरी पर स्थित बिंदुओं का समुच्चय वृत्त कहलाता है। निश्चित बिंदु को वृत्त का केंद्र तथा निश्चित दूरी को वृत्त की त्रिज्या कहते हैं।

(i) समांतर रेखाएँ-वे दो रेखाएँ समांतर रेखाएँ कहलाती हैं, जब (a) वे प्रतिच्छेद न करें, (b) जब वे समतलीय हों।
आकृति में l₁ तथा l₂ दो समांतर रेखाएँ दर्शाई गई हैं।

(ii) लंब रेखाएँ – दो रेखाएँ AB तथा CD जो एक ही तल में स्थित हों, लंब रेखाएँ कहलाती हैं, यदि वे समकोण बनाएँ; जैसे AB ⊥ CD.

(iii) रेखाखंड – एक रेखाखंड रेखा का वह भाग होता है, जब दो विभिन्न बिंदु A तथा B एक रेखा पर दिए गए हैं, तब रेखा का भाग जिसके अंतःबिंदु A तथा B हों रेखाखंड कहलाती है।

इसे AB कहा जाता है। AB तथा BA एक ही रेखाखंड को दर्शाते हैं।

(iv) वृत्त की त्रिज्या – केंद्र से वृत्त की परिधि पर एक बिंदु की दूरी वृत्त की त्रिज्या कहलाती है।

(v) वर्ग-वह चतुर्भुज जिसकी चारों भुजाएँ बराबर हों तथा प्रत्येक कोण समकोण हो, वर्ग कहलाता है।

प्रश्न 3.
नीचे दी हुई दो अभिधारणाओं पर विचार कीजिए
(i) दो भिन्न बिंदु A और B दिए रहने पर, एक तीसरा बिंदु C ऐसा विद्यमान है, जो A और B के बीच स्थित होता है।
(ii) यहाँ कम-से-कम ऐसे तीन बिंदु विद्यमान हैं कि वे एक रेखा पर स्थित नहीं हैं।
क्या इन अभिधारणाओं में कोई अपरिभाषित शब्द हैं ? क्या ये अभिधारणाएँ अविरोधी हैं ? क्या ये यूक्लिड की अभिधारणाओं से प्राप्त होती हैं ? स्पष्ट कीजिए।
हल :
ऐसे अनेक अपरिभाषित शब्द हैं जिनकी जानकारी छात्र को होनी चाहिए। ये संगत होते हैं, क्योंकि इनमें दो अलग-अलग स्थितियों का अध्ययन किया जाता है अर्थात
(i) यदि दो बिंदु A और B दिए हुए हों, तो उनके बीच में स्थित एक बिंद होता है।
(ii) यदि A और B दिए हुए हों, तो आप एक ऐसा बिंद ले सकते हैं जो A और B से होकर जाने वाली रेखा पर स्थित नहीं होता।
ये अभिगृहीत यूक्लिड की अभिगृहीतों का अनुसरण नहीं करते। फिर भी ये अभिगृहीत 5.1 का अनुसरण करते हैं क्योंकि दो दिए विभिन्न बिंदुओं से केवल एक रेखा ही गुजर सकती है।

प्रश्न 4.
यदि दो बिंदुओं A और B के बीच एक बिंदु C ऐसा स्थित है कि AC = BC है, तो सिद्ध कीजिए कि AC = \(\frac {1}{2}\) AB है। एक आकृति खींचकर इसे स्पष्ट कीजिए। [B.S.E.H. March 2017, 2019]
हल :
यहाँ पर एक बिंदु C है जोकि दो बिंदुओं A तथा B के बीच में इस प्रकार स्थित है कि AC = BC. दोनों ओर AC जोड़ने पर,

AC + AC = AC + BC
⇒ 2AC = AB [∵ AC + CB, AB के संपाती है।]
या AC = \(\frac {1}{2}\)AB
इति सिद्धम

प्रश्न 5.
प्रश्न 4 में, C रेखाखंड AB का एक मध्य-बिंदु कहलाता है। सिद्ध कीजिए कि एक रेखाखंड का एक और केवल एक ही मध्य-बिंदु होता है।
हल :

माना AB का अन्य मध्य-बिंदु D है.
AD = DB ……..(i)
लेकिन दिया गया है C, AB का मध्य-बिंदु है।
⇒ AC = CB ……..(ii)
समीकरण (ii) में से समीकरण (i) को घटाने पर
AC – AD = CB – DB
या DC = – DC
या DC + DC = 0
या 2DC = 0
या DC = \(\frac {0}{2}\) = 0
इस प्रकार C और D संपाती बिंदु हैं।
अतः प्रत्येक रेखाखंड का एक और केवल एक ही मध्य-बिंदु होता है।

प्रश्न 6.
आकृति में, यदि AC = BD है, तो सिद्ध कीजिए कि AB = CD है। [B.S.E.H. March, 2020]

हल :
यहाँ पर दिया है
AC = BD
आकृति अनुसार
AB + BC = BC + CD
[∵ AC = AB + BC तथा BD = BC + CD]
या AB + BC – BC = CD
या AB = CD [इति सिद्धम]

प्रश्न 7.
यूक्लिड की अभिगृहीतों की सूची में दिया हुआ अभिगृहीत 5 एक सर्वव्यापी सत्य क्यों माना जाता है ? (ध्यान दीजिए कि यह प्रश्न पाँचवीं अभिधारणा से संबंधित नहीं है।)
हल :
क्योंकि यूक्लिड की अभिगृहीतों की सूची में दिया हुआ अभिगृहीत 5 ब्रह्मांड की प्रत्येक चीज के लिए सत्य है। इसलिए यह सदैव सत्य है।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 12 हीरोन का सूत्र Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 12 हीरोन का सूत्र Exercise 12.1

प्रश्न 1.
एक यातायात संकेत बोर्ड पर ‘आगे स्कूल है’ लिखा है और यह भुजा ‘a’ वाले एक समबाहु त्रिभुज के आकार का है। हीरोन के सूत्र का प्रयोग करके इस बोर्ड का क्षेत्रफल ज्ञात कीजिए। यदि संकेत बोर्ड का परिमाप 180 cm है, तो इसका क्षेत्रफल क्या होगा ?
हल :
यहाँ समबाहु त्रिभुज की प्रत्येक भुजा = a अर्थात a = a, b = a, c = a
∴ s = \(\frac{a+a+a}{2}=\frac{3 a}{2}\)

a + a + a = 180 cm
3a = 180
a = \(\frac{180}{3}\) = 60cm
∵ वांछित क्षेत्रफल = \(\frac{\sqrt{3}}{4}\) × (60)² cm²
= \(\frac{\sqrt{3}}{4}\) × 3600 = 900\(\sqrt{3}\) cm² उत्तर

प्रश्न 2.
किसी फ्लाईओवर (flyover) की त्रिभुजाकार दीवार को विज्ञापनों के लिए प्रयोग किया जाता है। दीवार की भुजाओं की लंबाइयाँ 122 m, 22 m और 120 m हैं (देखिए आकृति)। इस विज्ञापन से प्रति वर्ष ₹ 5000 प्रति m2 की प्राप्ति होती है। एक कंपनी ने एक दीवार को विज्ञापन देने के लिए 3 महीने के लिए किराए पर लिया। उसने कुल कितना किराया दिया ?

हल :
यहाँ त्रिभुजाकार दीवार के लिए
a = 122 m
b = 22 m
c = 120 m

1 वर्ग मी० क्षेत्रफल का 1 वर्ष का विज्ञापन किराया = ₹ 5000
1320 वर्ग मी० क्षेत्रफल का \(\frac{1}{4}\) वर्ष का किराया = 5000 × 1320 × \(\frac{1}{4}\)
= ₹ 16,50,000 उत्तर

प्रश्न 3.
किसी पार्क में एक फिसल पट्टी (slide) बनी हुई है। इसकी पार्वीय दीवारों (side walls) में से एक दीवार पर किसी रंग से पेंट किया गया है और उस पर “पार्क को हरा-भरा और साफ रखिए” लिखा हुआ है (देखिए आकृति)। यदि इस दीवार की विमाएँ 15 m, 11 m और 6 m हैं, तो रंग से पेंट हुए भाग का क्षेत्रफल ज्ञात कीजिए।

हल :

आकृति यहाँ पेंट की गई दीवार की विमाएँ
a = 15 m
b = 11 m
c = 6 m

प्रश्न 4.
उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसकी दो भुजाएँ 18 cm और 10 cm हैं तथा उसका परिमाप 42 cm है।
हल :
यहाँ पर
a = 18 cm
b = 10 cm
a+ b + c = 42 cm
या 18 + 10 + c = 42
या c = 42 – 28 = 14 cm

प्रश्न 5.
एक त्रिभुज की भुजाओं का अनुपात 12 : 17 : 25 है और उसका परिमाप 540 cm है। इस त्रिभुज का क्षेत्रफल ज्ञात कीजिए। [B.S.E.H. March, 2019]
हल :
यहाँ पर त्रिभुज की भुजाओं का अनुपात = a : b : c = 12 : 17 : 25
अनुपाती योग = 12 + 17 + 25 = 54
∴ a = \(\frac{12}{54}\) × 540 = 120cm
b = \(\frac{17}{54}\) × 540 = 170cm
c = \(\frac{25}{54}\) × 540 = 250cm

प्रश्न 6.
एक समद्विबाहु त्रिभुज का परिमाप 30 cm है और उसकी बराबर भुजाएँ 12 cm लंबाई की हैं। इस त्रिभुज का क्षेत्रफल ज्ञात कीजिए।
हल :
यहाँ पर
a = 12 cm
b = 12 cm
c = 30 – (a + b)
= 30 – (12 + 12) = 30 – 24 = 6 cm

= \(\sqrt{15 \times 3 \times 3 \times 9}\) cm²
= \(\sqrt{15 \times 9 \times 9}\) cm²
= 9\(\sqrt{15}\) cm²

Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 त्रिभुज Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 7 त्रिभुज Exercise 7.1

प्रश्न 1.
चतुर्भुज ACBD में, AC = AD है और AB कोण A को समद्विभाजित करता है (देखिए आकृति)। दर्शाइए कि ΔABC ≅ ΔABD है।
BC और BD के बारे में आप क्या कह सकते हैं?
हल :
यहाँ पर ΔABC और ΔABD में,

AC = AD [दिया है]
∠CAB = ∠BAD [∵ AB कोण Aको समद्विभाजित करता है।]
AB = AB [उभवनिष्ठ]
अतः ΔABC ≅ ΔABD [भुजा-कोण-भुजा सर्वांगसमता]
⇒ BC = BD [∵ सर्वांगसम त्रिभुजों के संगत भाग] [इति सिद्धम]

प्रश्न 2.
ABCD एक चतुर्भुज है, जिसमें AD = BC और ∠DAB = ∠CBA है (देखिए आकृति)। सिद्ध कीजिए कि
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC

हल :
(i) यहाँ पर ΔABD और ΔBAC में,
AD = BC [दिया है]
∠DAB = ∠CBA [दिया है]
AB = AB [उभयनिष्ठ]
अतः ΔABD ≅ ΔBAC [भुजा-कोण-भुजा सर्वांगसमता]

(ii) क्योंकि ΔABD ≅ ΔBAC ∴ BD = AC [∵ सर्वांगसम त्रिभुजों के संगत भाग]

(iii) क्योंकि ΔABD ≅ ΔBAC ∴ ∠ABD = ∠BAC [∵ सर्वांगसम त्रिभुजों के संगत भाग]
[इति सिद्धम]

प्रश्न 3.
एक रेखाखंड AB पर AD और RC दो बराबर लंब रेखाखंड हैं (देखिए आकृति)। दर्शाइए कि CD, रेखाखंड AB को समद्विभाजित करता है।

हल :
यहाँ पर AB और CD, O पर प्रतिच्छेद करते हैं।
∴ ∠AOD = ∠BOC [शीर्षाभिमुख कोण]
ΔAOD और ΔBOC में,
∠AOD = ∠BOC [प्रमाणित]
∠DAO = ∠OBC
AD = BC [दिया है]
∴ ΔAOD ≅ ΔBOC [कोण-कोण-भुजा सर्वांगसमता]
⇒ OA = OB
[∵ सर्वागसम त्रिभुजों के संगत भाग]
अतः CD, रेखाखंड AB को समद्विभाजित करता है। [इति सिद्धम]

प्रश्न 4.
l और m दो समांतर रेखाएं हैं जिन्हें समांतर रेखाओं p और q का एक अन्य युग्म प्रतिच्छेदित करता है (देखिए आकृति)। दर्शाइए कि ΔABC ≅ ΔCDA है।

हल :
क्योंकि l तथा m समांतर रेखाएँ दूसरी समांतर रेखाओं p और q व द्वारा प्रतिच्छेदित की जाती हैं। अतः AD || BC तथा AB || CD
⇒ ABCD समांतर चुतर्भुज है।
⇒ AB = CD और BC = AD [∵ समांतर चतुर्भुज की सम्मुख भुजाएँ बराबर होती हैं।]
अब ΔARC और ΔCDA में,
AB = CD [प्रमाणित]
BC = AD [प्रमाणित]
AC = AC [उभयनिष्ठ]
ΔABC ≅ ΔCDA [भुजा-भुजा-भुजा सर्वांगसमता] [इति सिद्धम]

प्रश्न 5.
रेखा l कोण A को समद्विभाजित करती है और B रेखा l पर स्थित कोई बिंदु है। BP और BQ कोण A की भुजाओं पर B से डाले गए लंब हैं (देखिए आकृति)। दर्शाइए कि –
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ है अर्थात बिंदु B कोण की भुजाओं से समदूरस्थ है।

हल :
(i) यहाँ पर ΔAPB तथा ΔAQB में,
∠APB = ∠AQB [प्रत्येक = 90°]
∠PAB = ∠QAB [क्योंकि AB, ∠PAQ का समद्विभाजक है।]
AB = AB [उभयनिष्ठ]
⇒ ΔAPB ≅ ΔAQB
[कोण-कोण भुजा सर्वांगसमता]

(ii) ∵ ΔAPB ≅ ΔAQB
∴ BP = BQ [सर्वांगसम त्रिभुजों के संगत भाग]
अर्थात बिंदु B, कोण A की भुजाओं से समदूरस्थ है। [इति सिद्धम]

प्रश्न 6.
आकृति में, AC = AE, AB = AD और ∠BAD = ∠EAC है। दर्शाइए कि BC = DE है।

हल :
ΔBAD और ΔEAC में,
∠BAD = ∠EAC [दिया है।]
दोनों ओर ∠DAC जोड़ने पर,
∠BAD + ∠DAC = ∠EAC + ∠DAC
या ∠BAC = ∠EAD
अब ΔBAC और ΔEAD में,
AB = AD [दिया है]
∠BAC = ∠EAD [प्रमाणित]
AC = AE [दिया है]
अतः
ΔBAC ≅ ΔEAD [भुजा-कोण-भुजा सर्वांगसमता]
⇒ BC = DE [सर्वांगसम त्रिभुजों के संगत भाग] [इति सिद्धम]

प्रश्न 7.
AB एक रेखाखंड है और Pइसका मध्य-बिंदु है। D और E रेखाखंड AB के एक ही ओर स्थित दो बिंदु इस प्रकार हैं कि ∠BAD = ∠ABE और ∠EPA = ∠DPB है। (देखिए आकृति)। दर्शाइए कि
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

हल :
(i) ∠EPA = ∠DPB [दिया है|
दोनों ओर ∠DPE जोड़ने पर,
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
अब, ΔEBP तथा ΔDAP में,
∠EPB = ∠DPA [प्रमाणित]
BP = AP [दिया है।]
और ∠EBP = ∠DAP [दिया है]
अतः ΔEBP ≅ ΔDAP [कोण-भुजा-कोण सर्वांगसमता]

(ii) क्योंकि ΔEBP ≅ ΔDAP
AD = BE
[∵ सर्वांगसम त्रिभुजों के संगत भाग]

प्रश्न 8.
एक समकोण त्रिभुज ABC में, जिसमें कोण समकोण है, M कर्ण AB का मध्य-बिंद्ध है। C को M से मिलाकर बिंदु D तक इस प्रकार बढ़ाया गया है कि DM = CM है। बिंदु D को बिंदु B से मिला दिया जाता है (देखिए आकृति)। दर्शाइए कि –

(i) ΔAMC ≅ ΔBMD
(ii) ∠DRC एक समकोण है।
(iii) ΔDBC ≅ ΔACB
(iv) CM = \(\frac {1}{2}\)AB
हल :
(i) ΔAMC तथा ΔBMD में,
AM = BM [∵ M, AB का मध्य-बिंदु है]
∠AMC = ∠BMD [शीर्षाभिमुख कोण]
और CM = MD [दिया गया है]
अतः ΔAMC ≅ ΔBMD [भुजा-कोण-भुजा सर्वांगसमता]
[इति सिद्धम]

(ii) क्योंकि
ΔAMC ≅ ΔBMD
∠BDM = ∠ACM [∵ सर्वांगसम त्रिभुजों के संगत भाग]
परंतु ये एकांतर कोण है, अतः BD || CA
या ∠CBD + ∠BCA = 180° [∵ तिर्यक रेखा के एक ही ओर के आंतरिक कोण]
या ∠CBD + 90° = 180° [∵ ∠BCA = 90°]
या ∠DBC = 90° [इति सिद्धम]

(iii) अब ΔDBC तथा ΔACB में,
BD = CA [सर्वांगसम ΔBMD व ΔAMC के भाग]
∠DBC = ∠ACB [प्रत्येक = 90°]
BC = BC [उभयनिष्ठ]
ΔDBC ≅ ΔACB [भुजा-कोण-भुजा सर्वांगसमता] [इति सिद्धम]

(iv) क्योंकि
CD = AB [∵ सर्वांगसम त्रिभुजों के संगत भाग]
\(\frac {1}{2}\)CD = \(\frac {1}{2}\)AB
या CM = \(\frac {1}{2}\) AB
[इति सिद्धम]

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.9 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Exercise 13.9

प्रश्न 1.
एक लकड़ी के बुकशेल्फ (book-shell) की बाहरी विमाएं निम्न हैं : ऊंचाई = 110 सें०मी०, गहराई = 25 सें०मी०, चौड़ाई = 85 सें०मी० (देखिए आकृति)। प्रत्येक स्थान पर तख्तों की मोटाई 5 सें०मी० है। इसके बाहरी फलकों पर पालिश कराई जाती है और आंतरिक फलकों पर पेंट किया जाना है। यदि पालिश कराने की दर 20 पैसे प्रति सें०मी० है और पेंट कराने की दर 10 पैसे प्रति सें०मी० है, तो इस बुक-शैल्फ पर पालिश और पेंट कराने का कुल व्यय ज्ञात कीजिए।

हल :
यहां पर, पॉलिश वाले तल का क्षेत्रफल = (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5) सें०मी०²
= (9350 + 4250 + 5500 + 1500 + 1100) सें०मी०²
= 21700 सें०मी०²
इस पर पॉलिश कराने का खर्च = 21700 × \(\frac{20}{100}\) = ₹ 4340
पेंट वाले तल का क्षेत्रफल = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90) सें०मी०²
= (9000 + 3600 + 6750) सें०मी०² = 19350 सें०मी०²
इस पर पेंट कराने का खर्च = 19350 × \(\frac{10}{100}\) = ₹ 1935
पॉलिश व पेंट कराने का कुल खर्च = 4340 + 1935 = ₹ 6275 उत्तर

प्रश्न 2.
किसी घर के कंपाउंड के सामने की दीवार को 21 सें०मी० व्यास वाले लकड़ी के गोलों को छोटे आधारों पर टिका कर सजाया जाता है, जैसा कि आकृति में दिखाया गया है। इस प्रकार के आठ गोलों का प्रयोग इस कार्य के लिए किया जाना है और इन गोलों को चांदी वाले रंग में पेंट करवाना है। प्रत्येक आधार 1.5 सें०मी० त्रिज्या और ऊंचाई 7 सें०मी० का एक बेलन है तथा इन्हें काले रंग से पेंट करवाना है। यदि चांदी के रंग के पेंट करवाने की दर 25 पैसे प्रति से०मी० है तथा काले रंग के पेंट करवाने की दर 5 पैसे प्रति सें०मी० हो, तो पेंट करवाने का कुल व्यय ज्ञात कीजिए।

हल :
माना
लकड़ी के गोले का व्यास (d) = 21 सें०मी०
लकड़ी के गोले की त्रिज्या = \(\frac{21}{2}\) सें०मी०
∴ गोले का पृष्ठीय क्षेत्रफल = 4πr2
= \(4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\) = 1386 सें०मी०²
गोले के जितने तल पर चांदी वाला रंग होगा = \(\left[1386-\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2}\right]\) सें०मी०²
= \(\left[1386-\frac{99}{14}\right]\) सें०मी०²
= [1386 – 7.07] सें०मी०²
= 1378.93 सें०मी०²
8 गोलों में जितने तल पर चांदी वाला रंग होगा = 1378.93 × 8 सें०मी०² = 11031.44 सें०मी०²
चांदी का पेंट करने की दर = 25 पैसे प्रति सें०मी०²
∴ चांदी वाले रंग पर व्यय = \(\frac{11031.44 \times 25}{100}\)
= ₹ 2757.86
बेलन की त्रिज्या (r) = 1.5 = \(\frac{15}{10}=\frac{3}{2}\) सें०मी०
बेलन की ऊंचाई (h) = 7 सें०मी०
बेलन का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= \(2 \times \frac{22}{7} \times \frac{3}{2} \times 7\) = 66 सें०मी०²
8 बेलनों का वक्र पृष्ठीय क्षेत्रफल = 66 × 8 = 528 सें०मी०²
काले रंग का पेंट करवाने की दर = 5 पैसे प्रति सें०मी०²
बेलनों पर काला रंग कराने पर व्यय = \(\frac{528 \times 5}{100}\) = ₹ 26.40
दोनों रंगों के पेंट पर कुल व्यय = 2757.86 + 26.40
= ₹ 2784.26 उत्तर

प्रश्न 3.
एक गोले के व्यास में 25% की कमी हो जाती है। उसका वक्र पृष्ठीय क्षेत्रफल कितने प्रतिशत कम हो गया है?
हल :
माना,
गोले का पहला व्यास = 2x मी०
गोले की पहली त्रिज्या = x मी०
गोले का पहला पृष्ठीय क्षेत्रफल = 4πr² = 4πx² मी०²
25% कमी करने के पश्चात्

Haryana State Board HBSE 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Exercise 9.2

प्रश्न 1.
आकृति में, ABCD एक समांतर चतुर्भुज है, AE ⊥ DC और CF ⊥ AD है। यदि AB = 16 सें०मी०, AE = 8 सें०मी० और CF = 10 सें०मी० है, तो AD ज्ञात कीजिए। [B.S.E.H. March, 2018]

हल :
यहाँ पर आधार (AB) = 16 सें०मी०
शीर्षलंब (AE) = 8 सें०मी०
∴ समांतर चतुर्भुज ABCD का क्षेत्रफल = आधार × शीर्षलंब
= 16 सें०मी० × 8 सें०मी०
= 128 सें०मी०²
दूसरी अवस्था में
शीर्षलंब (CF) = 10 सें०मी०

प्रश्न 2.
यदि E, F, G और H क्रमशः समांतर चतुर्भुज ABCD की भुजाओं के मध्य-बिंदु हैं, तो दर्शाइए कि ar (EFGH) = \(\frac{1}{2}\)ar (ABCD) है।
हल :

दिया है : समांतर चतुर्भुज ABCD में E, F, G व H क्रमशः भुजाओं AB, BC, CD व DA के मध्य बिंदु हैं।
इन्हें मिलाने पर चतुर्भुज EFGH प्राप्त होता है।
सिद्ध करना है : ar (EFGH) = \(\frac{1}{2}\)ar (ABCD)
रचना : H व F को मिलाओ।
प्रमाण : ΔHGF और समांतर चतुर्भुज HDCF समान आधार HF और समान समांतर रेखाओं HF और DC के मध्य स्थित हैं।
∴ ar (ΔHGF) = \(\frac{1}{2}\)ar (HDCF) …..(i)
इसी प्रकार, ΔHEF और समांतर चतुर्भुज ABFH समान आधार HF और समान समांतर रेखाओं HF और AB के मध्य स्थित है।
∴ ar (ΔHEF) = \(\frac{1}{2}\)ar (ABFH) …..(ii)
समीकरण (i) एवं (ii) को जोड़ने पर,.
ar (ΔHGF) + ar (ΔHEF) = \(\frac{1}{2}\)[ar (HDCF) + ar (ABFH)]
⇒ ar (EFGH) = \(\frac{1}{2}\)ar (ABCD) [इति सिद्धम]

प्रश्न 3.
P और Q क्रमशः समांतर चतुर्भुज ABCD की भुजाओं DC और AD पर स्थित बिंदु हैं। दर्शाइए कि ar (APB) = ar (BQC) है।
हल :

दिया है :
समांतर चतुर्भुज ABCD की भुजाओं DC और AD पर क्रमशः P व Q दो बिंदु स्थित हैं।
सिद्ध करना है : ar (ΔAPB) = ar (ΔBQC)
प्रमाण : यहाँ पर ΔAPB तथा || चतुर्भुज ABCD एक ही आधार AB तथा समांतर रेखाओं AB व CD के मध्य में हैं।
∴ ar (ΔAPB) = \(\frac{1}{2}\)ar (|| चतुर्भुज ABCD) …..(i)
इसी प्रकार ΔBQC तथा || चतुर्भुज ABCD एक ही आधार BC तथा समांतर रेखाओं AD व BC के मध्य में हैं।
∴ ar (ΔBQC) = \(\frac{1}{2}\)ar (॥ चतुर्भुज ABCD) …..(ii)
समीकरण (i) व (ii) की तुलना में,
ar (ΔAPB) = ar (ΔBQC) [इति सिद्धम]

प्रश्न 4.
आकृति में, P समांतर चतुर्भुज ABCD के अभ्यंतर में स्थित कोई बिंदु है। दर्शाइए कि
(i) ar (APB) + ar (PCD) = \(\frac{1}{2}\)ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

हल :

दिया है : समांतर चतुर्भुज ABCD के अभ्यंतर में स्थित कोई बिंदु P है।
सिद्ध करना है : (i) ar (ΔAPB) + ar (ΔPCD) = \(\frac{1}{2}\)ar (ABCD)
(ii) ar (ΔAPD) + ar (ΔPBC) = ar (ΔAPB) + ar (ΔPCD)
रचना : P से EPF समांतर AB या DC तथा GPH समांतर AD या BC खींचिए।
प्रमाण : क्योंकि AB || EF m(रचना से)
तथा AE || BF [|| चतुर्भुज की सम्मुख भुजाएं]
∴ AEFB एक || चतुर्भुज है। इसी प्रकार EDCF भी एक || चतुर्भुज है।
ΔAPB तथा || चतुर्भुज AEFB एक ही आधार AB तथा समांतर रेखाओं AB तथा EF के मध्य में स्थित हैं।
∴ ar (ΔAPB) = \(\frac{1}{2}\)ar (|| चतुर्भुज AEFB) …..(i)
इसी प्रकार ar (ΔPCD) = \(\frac{1}{2}\)ar (|| चतुर्भुज EDCF) …..(ii)
समीकरण (i) व (ii) को जोड़ने पर,
ar (ΔAPB + ar (ΔPCD) = \(\frac{1}{2}\)ar (|| चतुर्भुज AEFB) + \(\frac{1}{2}\)ar (|| चतुर्भुज EDCF)
ar (ΔAPB) + ar (ΔPCD) = \(\frac{1}{2}\){ar (|| चतुर्भुज AEFB) + ar (|| चतुर्भुज EDCF)}
∴ ar (ΔAPB) + ar (ΔPCD) = \(\frac{1}{2}\)ar (|| चतुर्भुज ABCD) …..(iii)
इसी प्रकार हम सिद्ध कर सकते हैं कि,
ar (ΔAPD) + ar (ΔPBC) = \(\frac{1}{2}\)ar (|| चतुर्भुज ABCD) …..(iv)
समीकरण (iii) व (iv) की तुलना से,
ar (ΔAPB) + ar (ΔPCD) = ar (ΔAPD) + ar (ΔPBC) [इति सिद्धम]

प्रश्न 5.
आकृति में, PQRS और ABRS समांतर चतुर्भुज हैं तथा X भुजा BR पर स्थित कोई बिंदु है। दर्शाइए कि
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = \(\frac{1}{2}\)ar (PQRS)

हल :
दिया है : PQRS और ABRS समांतर चतुर्भुज एक ही आधार SR तथा दो समांतर रेखाओं SR व PB के बीच स्थित हैं तथा X भुजा BR पर स्थित कोई बिंदु है।
सिद्ध करना है : (i) ar (PQRS) = ar (ABRS)
(ii) ar (ΔAXS) = \(\frac{1}{2}\)ar (PQRS)
प्रमाण : (i) समांतर चतुर्भुज PQRS और समांतर चतुर्भुज ABRS समान आधार RS और एक ही समांतर रेखाओं SR तथा PB के बीच स्थित हैं।
∴ ar (PQRS) = ar (ABRS) [इति सिद्धम]
(ii) ΔAXS और समांतर चतुर्भुज ABRS समान आधार AS और एक ही समांतर रेखाओं AS बीच RB के बीच स्थित है।
∴ ar (ΔAXS) = \(\frac{1}{2}\)ar (ABRS)
या ar (ΔAXS) = \(\frac{1}{2}\)ar (PQRS) [भाग (i) से] [इति सिद्धम]

प्रश्न 6.
एक किसान के पास समांतर चतुर्भुज PQRS के रूप का एक खेत था। उसने RS पर स्थित कोई बिंदु A लिया और उसे P और Q से मिला दिया। खेत कितने भागों में विभाजित हो गया है? इन भागों के आकार क्या हैं? वह किसान खेत में गेहूँ और दालें बराबर-बराबर भागों में अलग-अलग बोना चाहता है। वह ऐसा कैसे करें?
हल :

इस प्रकार खेत तीन भागों में बँट जाता है तथा तीनों भाग त्रिभुज के आकार में हैं।
(i) ΔAPQ (ii) ΔASP (iii) ΔARQ
क्योंकि ΔAPQ तथा || चतुर्भुज PQRS एक ही आधार PQ तथा एक ही समांतर रेखाओं PQ तथा Rs के मध्य में स्थित है।
∴ ar (ΔAPQ) = \(\frac{1}{2}\)ar (|| चतुर्भुज PORS)
⇒ ar (ΔAPQ) = ar (ΔAPS) + ar (ΔAQR)
किसान को या तो गेहूँ ΔAPQ में तथा दालें अन्य दो त्रिभुजों में बोनी चाहिएं या दालें ΔAPQ में तथा गेहूँ अन्य दो त्रिभुजों में बोनी चाहिए।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Ex 6.3 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Exercise 6.3

प्रश्न 1.
आकृति में, ΔPQR की भुजाओं QP और RQ को क्रमशः बिंदुओं 5 और T तक बढ़ाया गया है। यदि ∠SPR = 135° है और ∠PQT = 110° है, तो ∠PRQ ज्ञात कीजिए।

हल :
क्योंकि QS एक रेखा है।
∴ ∠QPR + ∠SPR = 180° [रैखिक युग्म]
या ∠QPR + 135° = 180°
या ∠QPR = 180° – 135° = 45°
अब ∠TQP = ∠QPR + ∠PRQ [∵ बाह्य कोण, अंतः अभिमुख कोणों के योग के बराबर होता है]
या 110° = 45° + ∠PRQ
या ∠PRQ = 110° – 45° = 65°
अतः ∠PRQ = 65° उत्तर

प्रश्न 2.
आकृति में, ∠X = 62° और ∠XYZ = 54° है। यदि YO और ZO क्रमशः ΔXYZ के ∠XYZ और ∠XZY के समद्विभाजक हैं, तो ∠OZY और ∠YOZ ज्ञात कीजिए।

हल :
ΔXYZ में,
∠YXZ + ∠XYZ + ∠XZY = 180°
या 62° + 54° + ∠XZY = 180°
[∵ ∠YXZ = 62°, ∠XYZ = 54°]
या ∠XZY = 180° – 62° – 54° = 180° – 116° = 64°
क्योंकि YO तथा ZO, ∠XYZ तथा ∠XZY के समद्विभाजक हैं, इसलिए
∠OYZ = \(\frac {1}{2}\) × ∠XYZ = \(\frac {1}{2}\) × 54° = 27°
तथा
∠OZY = \(\frac {1}{2}\) × ∠XZY = \(\frac {1}{2}\) × 64° = 32°
अब ΔOYZ में,
∠YOZ + ∠OYZ + ∠OZY = 180° .
या ∠YOZ + 27° + 32° = 180°
या ∠YOZ + 59° = 180°
या ∠YOZ = 180° – 59° = 121°
अतः ∠OZY = 32° व ∠YOZ = 121° उत्तर

प्रश्न 3.
आकृति में, यदि AB || DE, ∠BAC = 35° और ∠CDE = 53° है, तो ∠DCE ज्ञात कीजिए।

हल :
आकृति में, AB || DE तथा तिर्यक रेखा AE इन्हें क्रमशः A तथा E पर काटती है।
∴ ∠DEA = ∠BAE [एकांतर कोण]
परंतु ∠BAE = 35°
⇒ ∠DEA = 35° या ∠DEC = 35°
ΔDEC में,
∠DCE + ∠DEC + ∠CDE = 180°
या ∠DCE + 35° + 53° = 180°
⇒ ∠DCE = 180° – 35° – 53° = 180° – 88° = 92°
अतः ∠DCE = 92° उत्तर

प्रश्न 4.
आकृति में, यदि रेखाएँ PQ और Rs बिंदु T पर इस प्रकार प्रतिच्छेद करती हैं कि ∠PRT = 40°, ∠RPT = 95° और ∠TSQ = 75° है, तो ∠SQT ज्ञात कीजिए।

हल :
ΔPRT में,
∠PRT + ∠RTP + ∠TPR = 180°
या 40° + ∠RTP + 95° = 180°
⇒ ∠RTP = 180° – 40° – 95° = 180° – 135° = 45°
परंतु ∠STQ = ∠RTP [शीर्षाभिमुख कोण]
⇒ ∠STQ = 45°

अब ΔTQS में,
∠SQT + ∠STQ + ∠TSQ = 180°
या ∠SQT + 45° + 75° = 180°
या ∠SQT = 180° – 45° – 75°
= 180° – 120° = 60°
अतः SQT = 60° उत्तर

प्रश्न 5.
आकृति में, यदि PQ ⊥ PS, PQ || SR, ∠SQR = 28° और ∠QRT = 65° है, तो x और y के मान ज्ञात कीजिए।

हल :
हम जानते हैं कि किसी त्रिभुज का बाय कोण अंतः अभिमुख कोणों के योग के बराबर होता है।
∴ ΔSRQ में,
∠QRT = ∠QSR + ∠RQS
या 65° = ∠QSR + 28°
या ∠QSR = 65° – 28° = 37°
अब क्योंकि PQ || SR तथा तिर्यक रेखा PS इन्हें क्रमशः P और S पर प्रतिच्छेदित करती है।
∴ ∠PSR + ∠SPQ = 180° [अंतः कोण युग्म]
या (∠PSQ + ∠QSR) + 90° = 180°
या y + 37° + 90° = 180°
या y + 127° = 180°
या y = 180° – 127° = 53°
समकोण ΔPQS में,
∠PQS = 180° – ∠SPQ – ∠PSQ
या x = 180° – 90° = 53°
x = 180° – 143° = 37°
अतः x = 37° व y = 53° उत्तर

प्रश्न 6.
आकृति में, ΔPQR की भुजा QR को बिंदु S तक बढ़ाया गया है। यदि ∠PQR और ∠PRS के समद्विभाजक बिंदु T पर मिलते हैं, तो सिद्ध कीजिए कि ∠QTR = \(\frac {1}{2}\)∠QPR है।

हल :
ΔPOR में, बाह्य ∠PRS = ∠P + ∠Q
दोनों ओर 2 से भाग करने पर
⇒ \(\frac {1}{2}\) बाह्य ∠PRS = \(\frac {1}{2}\)∠P + \(\frac {1}{2}\)∠Q ……(i)
क्योंकि. TR, ∠PRS को समद्विभाजित करता है।
∴ ∠PRS = 2∠TRS
इसी प्रकार TQ, ∠Q को समद्विभाजित करता है।
∴ ∠Q = 2∠TQR
∠PRS व ∠Q का मान समीकरण (i) में रखने पर
\(\frac {1}{2}\)(2∠TRS) = \(\frac {1}{2}\)∠P + \(\frac {1}{2}\)(2∠TQR)
या ∠TRS = \(\frac {1}{2}\)∠P + ∠TQR
परंतु ΔTQR में, ∠TRS = ∠TQR + ∠QTR
अतः ∠TQR + ∠QTR = \(\frac {1}{2}\)∠P + ∠TQR
या ∠QTR = \(\frac {1}{2}\)∠P
या ∠QTR = \(\frac {1}{2}\)∠QPR [इति सिद्धम]