R. N. Foundation

Little Heart School Affiliated to C.B.S.E., New Delhi

Kedo, Narwaphar (Near Sundernagar), Jamshedpur, East Singhbum, Jharkhand - 832111

LOCKDOWN PERIOD SYLLABUS

As we look at our surroundings, we see a largevariety of things with different shapes, sizesand textures. Everything in this universe ismade up of material which scientists havenamed “matter”. The air we breathe, the foodwe eat, stones, clouds, stars, plants andanimals, even a small drop of water or aparticle of sand – every thing is matter. Wecan also see as we look around that all thethings mentioned above occupy space andhave mass. In other words, they have bothmass* and volume**.

Since early times, human beings havebeen trying to understand their surroundings.Early Indian philosophers classified matter inthe form of five basic elements – the“Panch Tatva”– air, earth, fire, sky and water.According to them everything, living or non-living, was made up of these five basicelements. Ancient Greek philosophers hadarrived at a similar classification of matter.

Modern day scientists have evolved twotypes of classification of matter based on theirphysical properties and chemical nature.

In this chapter we shall learn aboutmatter based on its physical properties.Chemical aspects of matter will be taken upin subsequent chapters.

1.1 Physical Nature of Matter

1.1.1 MATTER IS MADE UP OF PARTICLES

For a long time, two schools of thought prevailedregarding the nature of matter. One schoolbelieved matter to be continuous like a blockof wood, whereas, the other thought that matterwas made up of particles like sand. Let usperform an activity to decide about the natureof matter – is it continuous or particulate?

Activity ______________ 1.1• Take a 100 mL beaker.• Fill half the beaker with water and

mark the level of water.• Dissolve some salt/ sugar with the help

of a glass rod.• Observe any change in water level.• What do you think has happened to

the salt?• Where does it disappear?• Does the level of water change?

In order to answer these questions weneed to use the idea that matter is made upof particles. What was there in the spoon, saltor sugar, has now spread throughout water.This is illustrated in Fig. 1.1.

1.1.2 HOW SMALL ARE THESE PARTICLES

OF MATTER?

Activity ______________ 1.2

• Take 2-3 crystals of potassiumpermanganate and dissolve them in100 mL of water.

Fig. 1.1: When we dissolve salt in water, the particles

of salt get into the spaces between particles

of water.

* The SI unit of mass is kilogram (kg).** The SI unit of volume is cubic metre (m3). The common unit of measuring volume is

litre (L) such that 1L = 1 dm3, 1L = 1000 mL, 1 mL = 1 cm3.

1

MMMMMATTERATTERATTERATTERATTER INININININ O O O O OURURURURUR S S S S SURROUNDINGSURROUNDINGSURROUNDINGSURROUNDINGSURROUNDINGS

Chapter

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• Take out approximately 10 mL of thissolution and put it into 90 mL of clearwater.

• Take out 10 mL of this solution andput it into another 90 mL of clear water.

• Keep diluting the solution like this 5 to8 times.

• Is the water still coloured ?

1.2.2 PARTICLES OF MATTER ARE

CONTINUOUSLY MOVING

Activity ______________ 1.3

• Put an unlit incense stick in a corner

of your class. How close do you have to

go near it so as to get its smell?

• Now light the incense stick. What

happens? Do you get the smell sitting

at a distance?

• Record your observations.

Activity ______________ 1.4

• Take two glasses/beakers filled with

water.

• Put a drop of blue or red ink slowly

and carefully along the sides of the first

beaker and honey in the same way in

the second beaker.

• Leave them undisturbed in your house

or in a corner of the class.

• Record your observations.

• What do you observe immediately after

adding the ink drop?

• What do you observe immediately after

adding a drop of honey?

• How many hours or days does it take

for the colour of ink to spread evenly

throughout the water?

Activity ______________ 1.5

• Drop a crystal of copper sulphate or

potassium permanganate into a glass

of hot water and another containing

cold water. Do not stir the solution.

Allow the crystals to settle at the

bottom.

• What do you observe just above the

solid crystal in the glass?

• What happens as time passes?

• What does this suggest about the

particles of solid and liquid?

• Does the rate of mixing change with

temperature? Why and how?

From the above three activities (1.3, 1.4 and

1.5), we can conclude the following:

Fig. 1.2: Estimating how small are the particles of

matter. With every dilution, though the colour

becomes light, it is still visible.

This experiment shows that just a fewcrystals of potassium permanganate cancolour a large volume of water (about1000 L). So we conclude that there must bemillions of tiny particles in just one crystalof potassium permanganate, which keep ondividing themselves into smaller and smallerparticles.

The same activity can be done using2 mL of Dettol instead of potassiumpermanganate. The smell can be detectedeven on repeated dilution.

The particles of matter are very small –they are small beyond our imagination!!!!

1.2 Characteristics of Particles ofMatter

1.2.1 PARTICLES OF MATTER HAVE SPACE

BETWEEN THEM

In activities 1.1 and 1.2 we saw that particlesof sugar, salt, Dettol, or potassiumpermanganate got evenly distributed in water.Similarly, when we make tea, coffee orlemonade (nimbu paani ), particles of one typeof matter get into the spaces between particlesof the other. This shows that there is enoughspace between particles of matter.

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• If we consider each student as aparticle of matter, then in which groupthe particles held each other with the

maximum force?

Activity ______________ 1.7

• Take an iron nail, a piece of chalk anda rubber band.

• Try breaking them by hammering,cutting or stretching.

• In which of the above threesubstances do you think the particles

are held together with greater force?

Activity ______________ 1.8

• Take some water in a container, trycutting the surface of water with yourfingers.

• Were you able to cut the surface ofwater?

• What could be the reason behind the

surface of water remaining together?

The above three activities (1.6, 1.7 and 1.8)

suggest that particles of matter have force

acting between them. This force keeps the

particles together. The strength of this force of

attraction varies from one kind of matter to

another.

uestions1. Which of the following are

matter?

Chair, air, love, smell, hate,

almonds, thought, cold, lemon

water, smell of perfume.

2. Give reasons for the following

observation:

The smell of hot sizzling food

reaches you several metres

away, but to get the smell from

cold food you have to go close.

3. A diver is able to cut through

water in a swimming pool. Which

property of matter does this

observation show?

4. What are the characteristics of

the particles of matter?

Particles of matter are continuously

moving, that is, they possess what we call

the kinetic energy. As the temperature rises,

particles move faster. So, we can say that with

increase in temperature the kinetic energy of

the particles also increases.

In the above three activities we observethat particles of matter intermix on their ownwith each other. They do so by getting intothe spaces between the particles. Thisintermixing of particles of two different typesof matter on their own is called diffusion. Wealso observe that on heating, diffusionbecomes faster. Why does this happen?

1.2.3 PARTICLES OF MATTER ATTRACT

EACH OTHER

Activity ______________ 1.6

• Play this game in the field— make fourgroups and form human chains assuggested:

• The first group should hold eachother from the back and lock armslike Idu-Mishmi dancers (Fig. 1.3).

Fig. 1.3

• The second group should hold handsto form a human chain.

• The third group should form a chainby touching each other with only theirfinger tips.

• Now, the fourth group of studentsshould run around and try to break thethree human chains one by one intoas many small groups as possible.

• Which group was the easiest to break?Why?

Q

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1.3 States of Matter

Observe different types of matter around you.

What are its different states? We can see thatmatter around us exists in three differentstates– solid, liquid and gas. These states of

matter arise due to the variation in thecharacteristics of the particles of matter.

Now, let us study about the properties of

these three states of matter in detail.

1.3.1 THE SOLID STATE

Activity _____________ 1.9

• Collect the following articles— a pen,a book, a needle and a piece of woodenstick.

• Sketch the shape of the above articlesin your notebook by moving a pencilaround them.

• Do all these have a definite shape,distinct boundaries and a fixed volume?

• What happens if they are hammered,

pulled or dropped?• Are these capable of diffusing into each

other?

• Try compressing them by applyingforce. Are you able to compress them?

All the above are examples of solids. Wecan observe that all these have a definiteshape, distinct boundaries and fixed volumes,

that is, have negligible compressibility. Solidshave a tendency to maintain their shape whensubjected to outside force. Solids may break

under force but it is difficult to change theirshape, so they are rigid.

Consider the following:

(a) What about a rubber band, can itchange its shape on stretching? Is ita solid?

(b) What about sugar and salt? Whenkept in different jars these take theshape of the jar. Are they solid?

(c) What about a sponge? It is a solidyet we are able to compress it. Why?

All the above are solids as:• A rubber band changes shape under

force and regains the same shape when

the force is removed. If excessive force isapplied, it breaks.

• The shape of each individual sugar orsalt crystal remains fixed, whether wetake it in our hand, put it in a plate or ina jar.

• A sponge has minute holes, in whichair is trapped, when we press it, the airis expelled out and we are able tocompress it.

1.3.2 THE LIQUID STATE

Activity _____________1.10

• Collect the following:(a) water, cooking oil, milk, juice, a

cold drink.(b) containers of different shapes. Put

a 50 mL mark on these containersusing a measuring cylinder fromthe laboratory.

• What will happen if these liquids arespilt on the floor?

• Measure 50 mL of any one liquid andtransfer it into different containers oneby one. Does the volume remain thesame?

• Does the shape of the liquid remain thesame ?

• When you pour the liquid from onecontainer into another, does it floweasily?

We observe that liquids have no fixedshape but have a fixed volume. They take upthe shape of the container in which they arekept. Liquids flow and change shape, so theyare not rigid but can be called fluid.

Refer to activities 1.4 and 1.5 where wesaw that solids and liquids can diffuse intoliquids. The gases from the atmospherediffuse and dissolve in water. These gases,especially oxygen and carbon dioxide, areessential for the survival of aquatic animalsand plants.

All living creatures need to breathe forsurvival. The aquatic animals can breatheunder water due to the presence of dissolvedoxygen in water. Thus, we may conclude thatsolids, liquids and gases can diffuse intoliquids. The rate of diffusion of liquids is

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higher than that of solids. This is due to thefact that in the liquid state, particles movefreely and have greater space between eachother as compared to particles in the solidstate.

1.3.3 THE GASEOUS STATE

Have you ever observed a balloon seller fillinga large number of balloons from a singlecylinder of gas? Enquire from him how manyballoons is he able to fill from one cylinder.

Ask him which gas does he have in the cylinder.

Activity _____________1.11

• Take three 100 mL syringes and closetheir nozzles by rubber corks, asshown in Fig.1.4.

• Remove the pistons from all thesyringes.

• Leaving one syringe untouched, fillwater in the second and pieces of chalkin the third.

• Insert the pistons back into thesyringes. You may apply some vaselineon the pistons before inserting theminto the syringes for their smoothmovement.

• Now, try to compress the content bypushing the piston in each syringe.

We have observed that gases are highlycompressible as compared to solids andliquids. The liquefied petroleum gas (LPG)cylinder that we get in our home for cookingor the oxygen supplied to hospitals incylinders is compressed gas. Compressednatural gas (CNG) is used as fuel these daysin vehicles. Due to its high compressibility,large volumes of a gas can be compressedinto a small cylinder and transported easily.

We come to know of what is being cookedin the kitchen without even entering there,by the smell that reaches our nostrils. Howdoes this smell reach us? The particles of thearoma of food mix with the particles of airspread from the kitchen, reach us and evenfarther away. The smell of hot cooked foodreaches us in seconds; compare this with therate of diffusion of solids and liquids. Due tohigh speed of particles and large spacebetween them, gases show the property ofdiffusing very fast into other gases.

In the gaseous state, the particles moveabout randomly at high speed. Due to thisrandom movement, the particles hit eachother and also the walls of the container. Thepressure exerted by the gas is because of thisforce exerted by gas particles per unit areaon the walls of the container.

Fig. 1.4

• What do you observe? In which casewas the piston easily pushed in?

• What do you infer from yourobservations?

Fig.1.5: a, b and c show the magnified schematic

pictures of the three states of matter. The

motion of the particles can be seen and

compared in the three states of matter.

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1.4.1 EFFECT OF CHANGE OF TEMPERATURE

Activity _____________1.12

• Take about 150 g of ice in a beaker and

suspend a laboratory thermometer so

that its bulb is in contact with the ice,as in Fig. 1.6.

uestions1. The mass per unit volume of a

substance is called density.

(density = mass/volume).

Arrange the following in order of

increasing density – air, exhaust

from chimneys, honey, water,

chalk, cotton and iron.

2. (a) Tabulate the differences in

the characterisitcs of states

of matter.

(b) Comment upon the following:

rigidity, compressibility,

fluidity, filling a gas

container, shape, kinetic

energy and density.

3. Give reasons

(a) A gas fills completely the

vessel in which it is kept.

(b) A gas exerts pressure on the

walls of the container.

(c) A wooden table should be

called a solid.

(d) We can easily move our hand

in air but to do the same

through a solid block of wood

we need a karate expert.

4. Liquids generally have lower

density as compared to solids.

But you must have observed that

ice floats on water. Find out why.

1.4 Can Matter Change its State?

We all know from our observation that water

can exist in three states of matter–• solid, as ice,

• liquid, as the familiar water, and

• gas, as water vapour.

What happens inside the matter during

this change of state? What happens to the

particles of matter during the change of

states? How does this change of state take

place? We need answers to these questions,

isn’t it?

Q

(a)

(b)

Fig. 1.6: (a) Conversion of ice to water, (b) conversion

of water to water vapour

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• Start heating the beaker on a low flame.• Note the temperature when the ice

starts melting.

• Note the temperature when all the ice

has converted into water.• Record your observations for this

conversion of solid to liquid state.

• Now, put a glass rod in the beaker and

heat while stirring till the water startsboiling.

• Keep a careful eye on the thermometer

reading till most of the water has

vaporised.• Record your observations for the

conversion of water in the liquid state

to the gaseous state.

On increasing the temperature of solids,the kinetic energy of the particles increases.Due to the increase in kinetic energy, theparticles start vibrating with greater speed.The energy supplied by heat overcomes theforces of attraction between the particles. Theparticles leave their fixed positions and startmoving more freely. A stage is reached whenthe solid melts and is converted to a liquid.The minimum temperature at which a solidmelts to become a liquid at the atmosphericpressure is called its melting point.

The melting point of a solid is an indicationof the strength of the force of attractionbetween its particles.

The melting point of ice is 273.15 K*. Theprocess of melting, that is, change of solidstate into liquid state is also known as fusion.When a solid melts, its temperatureremains the same, so where does the heatenergy go?

You must have observed, during theexperiment of melting, that the temperatureof the system does not change after themelting point is reached, till all the ice melts.This happens even though we continue toheat the beaker, that is, we continue to supplyheat. This heat gets used up in changing the

state by overcoming the forces of attractionbetween the particles. As this heat energy isabsorbed by ice without showing any rise intemperature, it is considered that it getshidden into the contents of the beaker and isknown as the latent heat. The word latentmeans hidden. The amount of heat energythat is required to change 1 kg of a solid intoliquid at atmospheric pressure at its meltingpoint is known as the latent heat of fusion.So, particles in water at 00 C (273 K) havemore energy as compared to particles in iceat the same temperature.

When we supply heat energy to water,particles start moving even faster. At a certaintemperature, a point is reached when theparticles have enough energy to break freefrom the forces of attraction of each other. Atthis temperature the liquid starts changinginto gas. The temperature at which a liquidstarts boiling at the atmospheric pressure isknown as its boiling point. Boiling is a bulkphenomenon. Particles from the bulk of theliquid gain enough energy to change into thevapour state.

For water this temperature is 373 K(100 0C = 273 + 100 = 373 K).

Can you define the latent heat ofvaporisation? Do it in the same way as wehave defined the latent heat of fusion.Particles in steam, that is, water vapour at373 K (1000 C) have more energy than waterat the same temperature. This is becauseparticles in steam have absorbed extra energyin the form of latent heat of vaporisation.

*Note: Kelvin is the SI unit of temperature, 00 C =273.15 K. For convenience, we take 00 C = 273 K afterrounding off the decimal. To change a temperature on the Kelvin scale to the Celsius scale youhave to subtract 273 from the given temperature, and to convert a temperature on the Celsiusscale to the Kelvin scale you have to add 273 to the given temperature.

So, we infer that the state of matter canbe changed into another state by changingthe temperature.

We have learnt that substances aroundus change state from solid to liquid and fromliquid to gas on application of heat. But there

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enclosed in a cylinder? Will the particles comecloser? Do you think that increasing ordecreasing the pressure can change the stateof matter?

are some that change directly from solid stateto gaseous state and vice versa withoutchanging into the liquid state.

Activity _____________1.13

• Take some camphor or ammoniumchloride. Crush it and put it in a chinadish.

• Put an inverted funnel over the chinadish.

• Put a cotton plug on the stem of thefunnel, as shown in Fig. 1.7.

* atmosphere (atm) is a unit of measuring pressure exerted by a gas. The unit of pressure is Pascal (Pa):1 atmosphere = 1.01 × 105 Pa. The pressure of air in atmosphere is called atmospheric pressure. Theatmospheric pressure at sea level is 1 atmosphere, and is taken as the normal atmospheric pressure.

Fig. 1.7: Sublimation of ammonium chloride

Fig. 1.8: By applying pressure, particles of matter can

be brought close together.

Applying pressure and reducingtemperature can liquefy gases.

Have you heard of solid carbon dioxide(CO

2)?

It is stored under high pressure. Solid

CO2 gets converted directly to gaseous state

on decrease of pressure to 1 atmosphere*without coming into liquid state. This is thereason that solid carbon dioxide is also knownas dry ice.

Thus, we can say that pressure andtemperature determine the state of asubstance, whether it will be solid, liquid orgas.

• Now, heat slowly and observe.

• What do you infer from the above

activity?

A change of state directly from solid to gaswithout changing into liquid state is calledsublimation and the direct change of gas tosolid without changing into liquid is calleddeposition.

1.4.2 EFFECT OF CHANGE OF PRESSURE

We have already learnt that the difference invarious states of matter is due to thedifference in the distances between theconstituent particles. What will happen whenwe start putting pressure and compress a gas

Fig. 1.9: Interconversion of the three states of matter

Deposition

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dish and keep it inside a cupboard or

on a shelf in your class.

• Record the room temperature.

• Record the time or days taken for the

evaporation process in the above cases.

• Repeat the above three steps of activity

on a rainy day and record your

observations.

• What do you infer about the effect of

temperature, surface area and wind

velocity (speed) on evaporation?

You must have observed that the rate of

evaporation increases with–

• an increase of surface area:

We know that evaporation is a surface

phenomenon. If the surface area is

increased, the rate of evaporation

increases. For example, while putting

clothes for drying up we spread them out.

• an increase of temperature:

With the increase of temperature, more

number of particles get enough kinetic

energy to go into the vapour state.

• a decrease in humidity:

Humidity is the amount of water vapour

present in air. The air around us cannot

hold more than a definite amount of

water vapour at a given temperature. If

the amount of water in air is already high,

the rate of evaporation decreases.

• an increase in wind speed:

It is a common observation that clothes

dry faster on a windy day. With the

increase in wind speed, the particles of

water vapour move away with the wind,

decreasing the amount of water vapour

in the surrounding.

1.5.2 HOW DOES EVAPORATION CAUSE

COOLING?

In an open vessel, the liquid keeps on

evaporating. The particles of liquid absorb

energy from the surrounding to regain the

energy lost during evaporation. This

absorption of energy from the surroundings

make the surroundings cold.

uestions1. Convert the following

temperature to celsius scale:

a. 300 K b. 573 K.

2. What is the physical state of

water at:

a. 250ºC b. 100ºC ?

3. For any substance, why does the

temperature remain constant

during the change of state?

4. Suggest a method to liquefy

atmospheric gases.

1.5 Evaporation

Do we always need to heat or change pressurefor changing the state of matter? Can youquote some examples from everyday life wherechange of state from liquid to vapour takesplace without the liquid reaching the boilingpoint? Water, when left uncovered, slowlychanges into vapour. Wet clothes dry up.What happens to water in the above twoexamples?

We know that particles of matter arealways moving and are never at rest. At agiven temperature in any gas, liquid or solid,there are particles with different amounts ofkinetic energy. In the case of liquids, a smallfraction of particles at the surface, havinghigher kinetic energy, is able to break awayfrom the forces of attraction of other particlesand gets converted into vapour. Thisphenomenon of change of a liquid intovapours at any temperature below its boilingpoint is called evaporation.

1.5.1 FACTORS AFFECTING EVAPORATION

Let us understand this with an activity.

Activity _____________1.14

• Take 5 mL of water in a test tube and

keep it near a window or under a fan.

• Take 5 mL of water in an open china

dish and keep it near a window orunder a fan.

• Take 5 mL of water in an open china

Q

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What happens when you pour someacetone (nail polish remover) on your palm?The particles gain energy from your palm orsurroundings and evaporate causing thepalm to feel cool.

After a hot sunny day, people sprinklewater on the roof or open ground becausethe large latent heat of vaporisation of waterhelps to cool the hot surface.

Can you cite some more examples fromdaily life where we can feel the effect of coolingdue to evaporation?Why should we wear cotton clothes insummer?

During summer, we perspire morebecause of the mechanism of our body whichkeeps us cool. We know that duringevaporation, the particles at the surface ofthe liquid gain energy from the surroundingsor body surface and change into vapour. Theheat energy equal to the latent heat ofvaporisation is absorbed from the bodyleaving the body cool. Cotton, being a goodabsorber of water helps in absorbing thesweat and exposing it to the atmosphere foreasy evaporation.

Why do we see water droplets on the outersurface of a glass containing ice-coldwater?

Let us take some ice-cold water in atumbler. Soon we will see water droplets onthe outer surface of the tumbler. The watervapour present in air, on coming in contactwith the cold glass of water, loses energy andgets converted to liquid state, which we seeas water droplets.

uestions1. Why does a desert cooler cool

better on a hot dry day?

2. How does the water kept in an

earthen pot (matka) become cool

during summer?

3. Why does our palm feel cold

when we put some acetone or

petrol or perfume on it?

4. Why are we able to sip hot tea or

milk faster from a saucer rather

than a cup?

5. What type of clothes should we

wear in summer?

More

to k

now

Now scientists are talking of five states of matter: Solid, Liquid, Gas, Plasma and Bose-Einstein Condensate.

Plasma: The state consists of super energetic and super excited particles. These particlesare in the form of ionised gases. The fluorescent tube and neon sign bulbs consist ofplasma. Inside a neon sign bulb there is neon gas and inside a fluorescent tube thereis helium gas or some other gas. The gas gets ionised, that is, gets charged whenelectrical energy flows through it. This charging up creates a plasma glowing insidethe tube or bulb. The plasma glows with a special colour depending on the nature ofgas. The Sun and the stars glow because of the presence of plasma in them. The plasmais created in stars because of very high temperature.

Bose-Einstein Condensate: In 1920, Indian physicist Satyendra Nath Bose had donesome calculations for a fifth state of matter. Building on his calculations, Albert Einstein

predicted a new state of matter – the Bose-EinsteinCondensate (BEC). In 2001, Eric A. Cornell, WolfgangKetterle and Carl E. Wieman of USA received the Nobelprize in physics for achieving “Bose-Einsteincondensation”. The BEC is formed by cooling a gas ofextremely low density, about one-hundred-thousandththe density of normal air, to super low temperatures.You can log on to www.chem4kids.com to get moreinformation on these fourth and fifth states of matter.

Q

S.N. Bose

(1894-1974)Albert Einstein

(1879-1955)

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MATTER IN OUR SURROUNDING S 11

Whatyou havelearnt

• Matter is made up of small particles.

• The matter around us exists in three states— solid, liquid

and gas.

• The forces of attraction between the particles are maximum in

solids, intermediate in liquids and minimum in gases.

• The spaces in between the constituent particles and kinetic

energy of the particles are minimum in the case of solids,

intermediate in liquids and maximum in gases.

• The arrangement of particles is most ordered in the case of

solids, in the case of liquids layers of particles can slip and

slide over each other while for gases, there is no order, particles

just move about randomly.

• The states of matter are inter-convertible. The state of matter

can be changed by changing temperature or pressure.

• Sublimation is the change of solid state directly to gaseous

state without going through liquid state.

• Deposition is the change of gaseous state directly to solid

state without going through liquid state.

• Boiling is a bulk phenomenon. Particles from the bulk

(whole) of the liquid change into vapour state.

• Evaporation is a surface phenomenon. Particles from the

surface gain enough energy to overcome the forces of attraction

present in the liquid and change into the vapour state.

• The rate of evaporation depends upon the surface area exposed

to the atmosphere, the temperature, the humidity and the

wind speed.

• Evaporation causes cooling.

• Latent heat of vaporisation is the heat energy required to change

1 kg of a liquid to gas at atmospheric pressure at its boiling

point.

• Latent heat of fusion is the amount of heat energy required

to change 1 kg of solid into liquid at its melting point.

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Exercises1. Convert the following temperatures to the celsius scale.

(a) 293 K (b) 470 K.

2. Convert the following temperatures to the kelvin scale.

(a) 25°C (b) 373°C.

3. Give reason for the following observations.

(a) Naphthalene balls disappear with time without leavingany solid.

(b) We can get the smell of perfume sitting several metresaway.

4. Arrange the following substances in increasing order of forcesof attraction between the particles— water, sugar, oxygen.

5. What is the physical state of water at—

(a) 25°C (b) 0°C (c) 100°C ?

6. Give two reasons to justify—

(a) water at room temperature is a liquid.

(b) an iron almirah is a solid at room temperature.

7. Why is ice at 273 K more effective in cooling than water at thesame temperature?

8. What produces more severe burns, boiling water or steam?

9. Name A,B,C,D,E and F in the following diagram showingchange in its state

Quantity Unit Symbol

Temperature kelvin K

Length metre m

Mass kilogram kg

Weight newton N

Volume cubic metre m3

Density kilogram per cubic metre kg m–3

Pressure pascal Pa

• Some measurable quantities and their units to remember:

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Group Activity

Prepare a model to demonstrate movement of particles in solids,liquids and gases.

For making this model you will need

• A transparent jar

• A big rubber balloon or piece of stretchable rubber sheet

• A string

• Few chick-peas or black gram or dry green peas.

How to make?

• Put the seeds in the jar.

• Sew the string to the centre of the rubber sheet and put sometape to keep it tied securely.

• Stretch and tie the rubber sheet on the mouth of the jar.

• Your model is ready. Now run your fingers up and down thestring by first tugging at it slowly and then rapidly.

Fig. 1.10: A model for converting of solid to liquid and liquid to gas.

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CHAPTER 8

QUADRILATERALS

8.1 Introduction

You have studied many properties of a triangle in Chapters 6 and 7 and you know that

on joining three non-collinear points in pairs, the figure so obtained is a triangle. Now,

let us mark four points and see what we obtain on joining them in pairs in some order.

Fig. 8.1

Note that if all the points are collinear (in the same line), we obtain a line

segment [see Fig. 8.1 (i)], if three out of four points are collinear, we get a triangle

[see Fig. 8.1 (ii)], and if no three points out of four are collinear, we obtain a closed

figure with four sides [see Fig. 8.1 (iii) and (iv)].

Such a figure formed by joining four points in an order is called a quadrilateral.

In this book, we will consider only quadrilaterals of the type given in Fig. 8.1 (iii) but

not as given in Fig. 8.1 (iv).

A quadrilateral has four sides, four angles and four vertices [see Fig. 8.2 (i)].

Fig. 8.2

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In quadrilateral ABCD, AB, BC, CD and DA are the four sides; A, B, C and D are

the four vertices and ∠ A, ∠ B, ∠ C and ∠ D are the four angles formed at the

vertices.

Now join the opposite vertices A to C and B to D [see Fig. 8.2 (ii)].

AC and BD are the two diagonals of the quadrilateral ABCD.

In this chapter, we will study more about different types of quadrilaterals, their

properties and especially those of parallelograms.

You may wonder why should we study about quadrilaterals (or parallelograms)

Look around you and you will find so many objects which are of the shape of a

quadrilateral - the floor, walls, ceiling, windows of your classroom, the blackboard,

each face of the duster, each page of your book, the top of your study table etc. Some

of these are given below (see Fig. 8.3).

Fig. 8.3

Although most of the objects we see around are of the shape of special quadrilateral

called rectangle, we shall study more about quadrilaterals and especially parallelograms

because a rectangle is also a parallelogram and all properties of a parallelogram are

true for a rectangle as well.

8.2 Angle Sum Property of a Quadrilateral

Let us now recall the angle sum property of a

quadrilateral.

The sum of the angles of a quadrilateral is 360º.

This can be verified by drawing a diagonal and dividing

the quadrilateral into two triangles.

Let ABCD be a quadrilateral and AC be a

diagonal (see Fig. 8.4).

What is the sum of angles in ∆ ADC?Fig. 8.4

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QUADRILATERALS 137

You know that

∠ DAC + ∠ ACD + ∠ D = 180° (1)

Similarly, in ∆ ABC,

∠ CAB + ∠ ACB + ∠ B = 180° (2)

Adding (1) and (2), we get

∠ DAC + ∠ ACD + ∠ D + ∠ CAB + ∠ ACB + ∠ B = 180° + 180° = 360°

Also, ∠ DAC + ∠ CAB = ∠ A and ∠ ACD + ∠ ACB = ∠ C

So, ∠ A + ∠ D + ∠ B + ∠ C = 360°.

i.e., the sum of the angles of a quadrilateral is 360°.

8.3 Types of Quadrilaterals

Look at the different quadrilaterals drawn below:

Fig. 8.5

Observe that :

l One pair of opposite sides of quadrilateral ABCD in Fig. 8.5 (i) namely, AB

and CD are parallel. You know that it is called a trapezium.

l Both pairs of opposite sides of quadrilaterals given in Fig. 8.5 (ii), (iii) , (iv)

and (v) are parallel. Recall that such quadrilaterals are called parallelograms.

So, quadrilateral PQRS of Fig. 8.5 (ii) is a parallelogram.

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138 MATHEMATICS

Similarly, all quadrilaterals given in Fig. 8.5 (iii), (iv) and (v) are parallelograms.

l In parallelogram MNRS of Fig. 8.5 (iii), note that one of its angles namely

∠ M is a right angle. What is this special parallelogram called? Try to recall.

It is called a rectangle.

l The parallelogram DEFG of Fig. 8.5 (iv) has all sides equal and we know that

it is called a rhombus.

l The parallelogram ABCD of Fig. 8.5 (v) has ∠ A = 90° and all sides equal; it

is called a square.

l In quadrilateral ABCD of Fig. 8.5 (vi), AD = CD and AB = CB i.e., two pairs

of adjacent sides are equal. It is not a parallelogram. It is called a kite.

Note that a square, rectangle and rhombus are all parallelograms.

l A square is a rectangle and also a rhombus.

l A parallelogram is a trapezium.

l A kite is not a parallelogram.

l A trapezium is not a parallelogram (as only one pair of opposite sides is parallel

in a trapezium and we require both pairs to be parallel in a parallelogram).

l A rectangle or a rhombus is not a square.

Look at the Fig. 8.6. We have a rectangle and a parallelogram with same perimeter

14 cm.

Fig. 8.6

Here the area of the parallelogram is DP × AB and this is less than the area of the

rectangle, i.e., AB × AD as DP < AD. Generally sweet shopkeepers cut ‘Burfis’ in the

shape of a parallelogram to accomodate more pieces in the same tray (see the shape

of the Burfi before you eat it next time!).

Let us now review some properties of a parallelogram learnt in earlier classes.

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8.4 Properties of a Parallelogram

Let us perform an activity.

Cut out a parallelogram from a sheet of paper

and cut it along a diagonal (see Fig. 8.7). You obtain

two triangles. What can you say about these

triangles?

Place one triangle over the other. Turn one around,

if necessary. What do you observe?

Observe that the two triangles are congruent to

each other.

Repeat this activity with some more parallelograms. Each time you will observe

that each diagonal divides the parallelogram into two congruent triangles.

Let us now prove this result.

Theorem 8.1 : A diagonal of a parallelogram divides it into two congruent

triangles.

Proof : Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.8). Observe

that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC

and ∆ CDA. We need to prove that these triangles are congruent.

In ∆ ABC and ∆ CDA, note that BC || AD and AC is a transversal.

So, ∠ BCA = ∠ DAC (Pair of alternate angles)

Also, AB || DC and AC is a transversal.

So, ∠ BAC = ∠ DCA (Pair of alternate angles)

and AC = CA (Common)

So, ∆ ABC ≅ ∆ CDA (ASA rule)

or, diagonal AC divides parallelogram ABCD into two congruent

triangles ABC and CDA.

Now, measure the opposite sides of parallelogram ABCD. What do you observe?

You will find that AB = DC and AD = BC.

This is another property of a parallelogram stated below:

Theorem 8.2 : In a parallelogram, opposite sides are equal.

You have already proved that a diagonal divides the parallelogram into two congruent

Fig. 8.7

Fig. 8.8

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140 MATHEMATICS

triangles; so what can you say about the corresponding parts say, the corresponding

sides? They are equal.

So, AB = DC and AD = BC

Now what is the converse of this result? You already know that whatever is given

in a theorem, the same is to be proved in the converse and whatever is proved in the

theorem it is given in the converse. Thus, Theorem 8.2 can be stated as given below :

If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So

its converse is :

Theorem 8.3 : If each pair of opposite sides of a quadrilateral is equal, then it

is a parallelogram.

Can you reason out why?

Let sides AB and CD of the quadrilateral ABCD

be equal and also AD = BC (see Fig. 8.9). Draw

diagonal AC.

Clearly, ∆ ABC ≅ ∆ CDA (Why?)

So, ∠ BAC = ∠ DCA

and ∠ BCA = ∠ DAC (Why?)

Can you now say that ABCD is a parallelogram? Why?

You have just seen that in a parallelogram each pair of opposite sides is equal and

conversely if each pair of opposite sides of a quadrilateral is equal, then it is a

parallelogram. Can we conclude the same result for the pairs of opposite angles?

Draw a parallelogram and measure its angles. What do you observe?

Each pair of opposite angles is equal.

Repeat this with some more parallelograms. We arrive at yet another result as

given below.

Theorem 8.4 : In a parallelogram, opposite angles are equal.

Now, is the converse of this result also true? Yes. Using the angle sum property of

a quadrilateral and the results of parallel lines intersected by a transversal, we can see

that the converse is also true. So, we have the following theorem :

Theorem 8.5 : If in a quadrilateral, each pair of opposite angles is equal, then

it is a parallelogram.

Fig. 8.9

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There is yet another property of a parallelogram. Let us study the same. Draw a

parallelogram ABCD and draw both its diagonals intersecting at the point O

(see Fig. 8.10).

Measure the lengths of OA, OB, OC and OD.

What do you observe? You will observe that

OA = OC and OB = OD.

or, O is the mid-point of both the diagonals.

Repeat this activity with some more parallelograms.

Each time you will find that O is the mid-point of both the diagonals.

So, we have the following theorem :

Theorem 8.6 : The diagonals of a parallelogram

bisect each other.

Now, what would happen, if in a quadrilateral

the diagonals bisect each other? Will it be a

parallelogram? Indeed this is true.

This result is the converse of the result of

Theorem 8.6. It is given below:

Theorem 8.7 : If the diagonals of a quadrilateral

bisect each other, then it is a parallelogram.

You can reason out this result as follows:

Note that in Fig. 8.11, it is given that OA = OC

and OB = OD.

So, ∆ AOB ≅ ∆ COD (Why?)

Therefore, ∠ ABO = ∠ CDO (Why?)

From this, we get AB || CD

Similarly, BC || AD

Therefore ABCD is a parallelogram.

Let us now take some examples.

Example 1 : Show that each angle of a rectangle is a right angle.

Solution : Let us recall what a rectangle is.

A rectangle is a parallelogram in which one angle is a right angle.

Fig. 8.10

Fig. 8.11

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142 MATHEMATICS

Let ABCD be a rectangle in which ∠ A = 90°.

We have to show that ∠ B = ∠ C = ∠ D = 90°

We have, AD || BC and AB is a transversal

(see Fig. 8.12).

So, ∠ A + ∠ B = 180° (Interior angles on the same

side of the transversal)

But, ∠ A = 90°

So, ∠ B = 180° – ∠ A = 180° – 90° = 90°

Now, ∠ C = ∠ A and ∠ D = ∠ B

(Opposite angles of the parallellogram)

So, ∠ C = 90° and ∠ D = 90°.

Therefore, each of the angles of a rectangle is a right angle.

Example 2 : Show that the diagonals of a rhombus are perpendicular to each other.

Solution : Consider the rhombus ABCD (see Fig. 8.13).

You know that AB = BC = CD = DA (Why?)

Now, in ∆ AOD and ∆ COD,

OA = OC (Diagonals of a parallelogram

bisect each other)

OD = OD (Common)

AD = CD

Therefore, ∆ AOD ≅ ∆ COD

(SSS congruence rule)

This gives, ∠ AOD = ∠ COD (CPCT)

But, ∠ AOD + ∠ COD = 180° (Linear pair)

So, 2∠ AOD = 180°

or, ∠ AOD = 90°

So, the diagonals of a rhombus are perpendicular to each other.

Example 3 : ABC is an isosceles triangle in which AB = AC. AD bisects exterior

angle PAC and CD || AB (see Fig. 8.14). Show that

Fig. 8.12

Fig. 8.13

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QUADRILATERALS 143

(i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram.

Solution : (i) ∆ ABC is isosceles in which AB = AC (Given)

So, ∠ ABC = ∠ ACB (Angles opposite to equal sides)

Also, ∠ PAC = ∠ ABC + ∠ ACB

(Exterior angle of a triangle)

or, ∠ PAC = 2∠ ACB (1)

Now, AD bisects ∠ PAC.

So, ∠ PAC = 2∠ DAC (2)

Therefore,

2∠ DAC = 2∠ ACB [From (1) and (2)]

or, ∠ DAC = ∠ ACB

(ii) Now, these equal angles form a pair of alternate angles when line segments BC

and AD are intersected by a transversal AC.

So, BC || AD

Also, BA || CD (Given)

Now, both pairs of opposite sides of quadrilateral ABCD are parallel.

So, ABCD is a parallelogram.

Example 4 : Two parallel lines l and m are intersected by a transversal p

(see Fig. 8.15). Show that the quadrilateral formed by the bisectors of interior angles

is a rectangle.

Solution : It is given that PS || QR and transversal p intersects them at points A and

C respectively.

The bisectors of ∠ PAC and ∠ ACQ intersect at B and bisectors of ∠ ACR and

∠ SAC intersect at D.

We are to show that quadrilateral ABCD is a

rectangle.

Now, ∠ PAC = ∠ ACR

(Alternate angles as l || m and p is a transversal)

So,1

2∠ PAC =

1

2 ∠ ACR

i.e., ∠ BAC = ∠ ACD

Fig. 8.14

Fig. 8.15

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144 MATHEMATICS

These form a pair of alternate angles for lines AB and DC with AC as transversal and

they are equal also.

So, AB || DC

Similarly, BC || AD (Considering ∠ ACB and ∠ CAD)

Therefore, quadrilateral ABCD is a parallelogram.

Also, ∠ PAC + ∠ CAS = 180° (Linear pair)

So,1

2 ∠ PAC +

1

2 ∠ CAS =

1

2 × 180° = 90°

or, ∠ BAC + ∠ CAD = 90°

or, ∠ BAD = 90°

So, ABCD is a parallelogram in which one angle is 90°.

Therefore, ABCD is a rectangle.

Example 5 : Show that the bisectors of angles of a parallelogram form a rectangle.

Solution : Let P, Q, R and S be the points of

intersection of the bisectors of ∠ A and ∠ B, ∠ B

and ∠ C, ∠ C and ∠ D, and ∠ D and ∠ A respectively

of parallelogram ABCD (see Fig. 8.16).

In ∆ ASD, what do you observe?

Since DS bisects ∠ D and AS bisects ∠ A, therefore,

∠ DAS + ∠ ADS =1

2 ∠ A +

1

2∠ D

=1

2 (∠ A + ∠ D)

=1

2 × 180° (∠ A and ∠ D are interior angles

on the same side of the transversal)

= 90°

Also, ∠ DAS + ∠ ADS + ∠ DSA = 180° (Angle sum property of a triangle)

or, 90° + ∠ DSA = 180°

or, ∠ DSA = 90°

So, ∠ PSR = 90° (Being vertically opposite to ∠ DSA)

Fig. 8.16

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QUADRILATERALS 145

Similarly, it can be shown that ∠ APB = 90° or ∠ SPQ = 90° (as it was shown for

∠ DSA). Similarly, ∠ PQR = 90° and ∠ SRQ = 90°.

So, PQRS is a quadrilateral in which all angles are right angles.

Can we conclude that it is a rectangle? Let us examine. We have shown that

∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°. So both pairs of opposite angles

are equal.

Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and

so, PQRS is a rectangle.

8.5 Another Condition for a Quadrilateral to be a Parallelogram

You have studied many properties of a parallelogram in this chapter and you have also

verified that if in a quadrilateral any one of those properties is satisfied, then it becomes

a parallelogram.

We now study yet another condition which is the least required condition for a

quadrilateral to be a parallelogram.

It is stated in the form of a theorem as given below:

Theorem 8.8 : A quadrilateral is a parallelogram if a pair of opposite sides is

equal and parallel.

Look at Fig 8.17 in which AB = CD and

AB || CD. Let us draw a diagonal AC. You can show

that ∆ ABC ≅ ∆ CDA by SAS congruence rule.

So, BC || AD (Why?)

Let us now take an example to apply this property

of a parallelogram.

Example 6 : ABCD is a parallelogram in which P

and Q are mid-points of opposite sides AB and CD

(see Fig. 8.18). If AQ intersects DP at S and BQ

intersects CP at R, show that:

(i) APCQ is a parallelogram.

(ii) DPBQ is a parallelogram.

(iii) PSQR is a parallelogram.

Fig. 8.17

Fig. 8.18

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146 MATHEMATICS

Solution : (i) In quadrilateral APCQ,

AP || QC (Since AB || CD) (1)

AP =1

2 AB, CQ =

1

2 CD (Given)

Also, AB = CD (Why?)

So, AP = QC (2)

Therefore, APCQ is a parallelogram [From (1) and (2) and Theorem 8.8]

(ii) Similarly, quadrilateral DPBQ is a parallelogram, because

DQ || PB and DQ = PB

(iii) In quadrilateral PSQR,

SP || QR (SP is a part of DP and QR is a part of QB)

Similarly, SQ || PR

So, PSQR is a parallelogram.

EXERCISE 8.1

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the

quadrilateral.

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it

is a rhombus.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right

angles, then it is a square.

6. Diagonal AC of a parallelogram ABCD bisects

∠ A (see Fig. 8.19). Show that

(i) it bisects ∠ C also,

(ii) ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC

bisects ∠ A as well as ∠ C and diagonal BD

bisects ∠ B as well as ∠ D.

8. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that:

(i) ABCD is a square (ii) diagonal BD bisects ∠ B as well as ∠ D.

Fig. 8.19

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9. In parallelogram ABCD, two points P and Q are

taken on diagonal BD such that DP = BQ

(see Fig. 8.20). Show that:

(i) ∆ APD ≅ ∆ CQB

(ii) AP = CQ

(iii) ∆ AQB ≅ ∆ CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

10. ABCD is a parallelogram and AP and CQ are

perpendiculars from vertices A and C on diagonal

BD (see Fig. 8.21). Show that

(i) ∆ APB ≅ ∆ CQD

(ii) AP = CQ

11. In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF

and BC || EF. Vertices A, B and C are joined to

vertices D, E and F respectively (see Fig. 8.22).

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ ABC ≅ ∆ DEF.

12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that

(i) ∠ A = ∠ B

(ii) ∠ C = ∠ D

(iii) ∆ ABC ≅ ∆ BAD

(iv) diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through C

parallel to DA intersecting AB produced at E.]

Fig. 8.20

Fig. 8.21

Fig. 8.22

Fig. 8.23

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148 MATHEMATICS

8.6 The Mid-point Theorem

You have studied many properties of a triangle as well as a quadrilateral. Now let us

study yet another result which is related to the mid-point of sides of a triangle. Perform

the following activity.

Draw a triangle and mark the mid-points E and F of two sides of the triangle. Join

the points E and F (see Fig. 8.24).

Measure EF and BC. Measure ∠ AEF and ∠ ABC.

What do you observe? You will find that :

EF = 1

2 BC and ∠ AEF = ∠ ABC

so, EF || BC

Repeat this activity with some more triangles.

So, you arrive at the following theorem:

Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle

is parallel to the third side.

You can prove this theorem using the following

clue:

Observe Fig 8.25 in which E and F are mid-points

of AB and AC respectively and CD || BA.

∆ AEF ≅ ∆ CDF (ASA Rule)

So, EF = DF and BE = AE = DC (Why?)

Therefore, BCDE is a parallelogram. (Why?)

This gives EF || BC.

In this case, also note that EF = 1

2 ED =

1

2BC.

Can you state the converse of Theorem 8.9? Is the converse true?

You will see that converse of the above theorem is also true which is stated as

below:

Theorem 8.10 : The line drawn through the mid-point of one side of a triangle,

parallel to another side bisects the third side.

Fig. 8.25

Fig. 8.24

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QUADRILATERALS 149

In Fig 8.26, observe that E is the mid-point of

AB, line l is passsing through E and is parallel to BC

and CM || BA.

Prove that AF = CF by using the congruence of

∆ AEF and ∆ CDF.

Example 7 : In ∆ ABC, D, E and F are respectively

the mid-points of sides AB, BC and CA

(see Fig. 8.27). Show that ∆ ABC is divided into four

congruent triangles by joining D, E and F.

Solution : As D and E are mid-points of sides AB

and BC of the triangle ABC, by Theorem 8.9,

DE || AC

Similarly, DF || BC and EF || AB

Therefore ADEF, BDFE and DFCE are all parallelograms.

Now DE is a diagonal of the parallelogram BDFE,

therefore, ∆ BDE ≅ ∆ FED

Similarly ∆ DAF ≅ ∆ FED

and ∆ EFC ≅ ∆ FED

So, all the four triangles are congruent.

Example 8 : l, m and n are three parallel lines

intersected by transversals p and q such that l, m

and n cut off equal intercepts AB and BC on p

(see Fig. 8.28). Show that l, m and n cut off equal

intercepts DE and EF on q also.

Solution : We are given that AB = BC and have

to prove that DE = EF.

Let us join A to F intersecting m at G..

The trapezium ACFD is divided into two triangles;

Fig. 8.26

Fig. 8.27

Fig. 8.28

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150 MATHEMATICS

namely ∆ ACF and ∆ AFD.

In ∆ ACF, it is given that B is the mid-point of AC (AB = BC)

and BG || CF (since m || n).

So, G is the mid-point of AF (by using Theorem 8.10)

Now, in ∆ AFD, we can apply the same argument as G is the mid-point of AF,

GE || AD and so by Theorem 8.10, E is the mid-point of DF,

i.e., DE = EF.

In other words, l, m and n cut off equal intercepts on q also.

EXERCISE 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are

mid-points of the sides AB, BC, CD and DA

(see Fig 8.29). AC is a diagonal. Show that :

(i) SR || AC and SR = 1

2 AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and

DA respectively. Show that the quadrilateral PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA

respectively. Show that the quadrilateral PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.

A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that

F is the mid-point of BC.

Fig. 8.30

Fig. 8.29

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QUADRILATERALS 151

5. In a parallelogram ABCD, E and F are the

mid-points of sides AB and CD respectively

(see Fig. 8.31). Show that the line segments AF

and EC trisect the diagonal BD.

6. Show that the line segments joining the mid-points of the opposite sides of a

quadrilateral bisect each other.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB

and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC (ii) MD ⊥ AC

(iii) CM = MA = 1

2AB

8.7 Summary

In this chapter, you have studied the following points :

1. Sum of the angles of a quadrilateral is 360°.

2. A diagonal of a parallelogram divides it into two congruent triangles.

3. In a parallelogram,

(i) opposite sides are equal (ii) opposite angles are equal

(iii) diagonals bisect each other

4. A quadrilateral is a parallelogram, if

(i) opposite sides are equal or (ii) opposite angles are equal

or (iii) diagonals bisect each other

or (iv)a pair of opposite sides is equal and parallel

5. Diagonals of a rectangle bisect each other and are equal and vice-versa.

6. Diagonals of a rhombus bisect each other at right angles and vice-versa.

7. Diagonals of a square bisect each other at right angles and are equal, and vice-versa.

8. The line-segment joining the mid-points of any two sides of a triangle is parallel to the

third side and is half of it.

9. A line through the mid-point of a side of a triangle parallel to another side bisects the third

side.

10. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order,

is a parallelogram.

Fig. 8.31

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izsepan

izsepan dk tUe lu~ 1880 esa cukjl osQ yegh xk¡o esa gqvk FkkA mudkewy uke èkuir jk; FkkA izsepan dk cpiu vHkkoksa esa chrk vkSj f'k{kkch-,- rd gh gks ikbZA mUgksaus f'k{kk foHkkx esa ukSdjh dh ijarq vlg;ksxvkanksyu esa lfØ; Hkkx ysus osQ fy, ljdkjh ukSdjh ls R;kxi=k ns fn;kvkSj ys[ku dk;Z osQ izfr iwjh rjg lefiZr gks x,A lu~ 1936 esa blegku dFkkdkj dk nsgkar gks x;kA

izsepan dh dgkfu;k¡ ekuljksoj osQ vkB Hkkxksa esa ladfyr gSaAlsoklnu] izsekJe] jaxHkwfe] dk;kdYi] fueZyk] xcu] deZHkwfe]xksnku muosQ izeq[k miU;kl gSaA mUgksaus gal] tkxj.k] ekèkqjh vkfnif=kdkvksa dk laiknu Hkh fd;kA dFkk lkfgR; osQ vfrfjDr izsepan usfucaèk ,oa vU; izdkj dk x| ys[ku Hkh izpqj ek=kk esa fd;kA izsepanlkfgR; dks lkekftd ifjorZu dk l'kDr ekè;e ekurs FksA mUgksaus ftlxk¡o vkSj 'kgj osQ ifjos'k dks ns[kk vkSj ft;k mldh vfHkO;fDr muosQdFkk lkfgR; esa feyrh gSA fdlkuksa vkSj e”knwjksa dh n;uh; fLFkfr] nfyrksadk 'kks"k.k] lekt esa L=kh dh nqnZ'kk vkSj Lokèkhurk vkanksyu vkfn mudhjpukvksa osQ ewy fo"k; gSaA

izsepan osQ dFkk lkfgR; dk lalkj cgqr O;kid gSA mlesa euq"; ghugha i'kq&if{k;ksa dks Hkh vn~Hkqr vkReh;rk feyh gSA cM+h ls cM+h ckr dksljy Hkk"kk esa lhèks vkSj la{ksi esa dguk izsepan osQ ys[ku dh izeq[k

2019-20

4@ f{kfrt

fo'ks"krk gSA mudh Hkk"kk ljy] ltho ,oa eqgkojsnkj gS rFkk mUgksausyksd izpfyr 'kCnksa dk iz;ksx oqQ'kyrkiwoZd fd;k gSA

nks cSyksa dh dFkk osQ ekè;e ls izsepan us Ñ"kd lekt vkSj i'kqvksa osQ HkkokRedlacaèk dk o.kZu fd;k gSA bl dgkuh esa mUgksaus ;g Hkh crk;k gS fd Lora=krk lgtesa ugha feyrh] mlosQ fy, ckj&ckj la?k"kZ djuk iM+rk gSA bl izdkj ijks{k :i ls;g dgkuh vk”kknh osQ vkanksyu dh Hkkouk ls tqM+h gSA blosQ lkFk gh bl dgkuh esaizsepan us iapra=k vkSj fgrksins'k dh dFkk&iajijk dk mi;ksx vkSj fodkl fd;k gSA

2019-20

tkuojksa esa x/k lcls ”;knk cqf¼ghu le>k tkrk gSA ge tc fdlh vkneh dksijys njts dk csoowQI+kQ dguk pkgrs gSa] rks mls x/k dgrs gSaA x/k lpeqpcsoowQI+kQ gS] ;k mlosQ lh/siu] mldh fujkin lfg".kqrk us mls ;g inoh ns nh gS]bldk fu'p; ugha fd;k tk ldrkA xk;sa lhax ekjrh gSa] C;kbZ gqbZ xk; rksvuk;kl gh flaguh dk :i /kj.k dj ysrh gSA oqQÙkk Hkh cgqr xjhc tkuoj gS]ysfdu dHkh&dHkh mls Hkh Øks/ vk gh tkrk gS_ foaQrq x/s dks dHkh Øks/ djrsugha lquk] u ns[kkA ftruk pkgks xjhc dks ekjks] pkgs tSlh [kjkc] lM+h gqbZ ?kkllkeus Mky nks] mlosQ psgjs ij dHkh vlarks"k dh Nk;k Hkh u fn[kkbZ nsxhA oS'kk[kesa pkgs ,dk/ ckj oqQysy dj ysrk gks_ ij geus rks mls dHkh [kq'k gksrs ughans[kkA mlosQ psgjs ij ,d fo"kkn LFkk;h :i ls Nk;k jgrk gSA lq[k&nq[k]gkfu&ykHk] fdlh Hkh n'kk esa mls cnyrs ugha ns[kkA ½f"k;ksa&eqfu;ksa osQ ftrus xq.kgSa os lHkh mlesa ijkdk"Bk dks igq¡p x, gSa_ ij vkneh mls csoowQI+kQ dgrk gSAln~xq.kksa dk bruk vuknj dgha ugha ns[kkA dnkfpr lh/kiu lalkj osQ fy, mi;qDrugha gSA nsf[k, u] Hkkjrokfl;ksa dh vizQhdk esa D;k nqnZ'kk gks jgh gS\ D;ksa vejhdkesa mUgsa ?kqlus ugha fn;k tkrk\ cspkjs 'kjkc ugha ihrs] pkj iSls oqQle; osQ fy,cpkdj j[krs gSa] th rksM+dj dke djrs gSa] fdlh ls yM+kbZ&>xM+k ugha djrs] pkjckrsa lqudj xe [kk tkrs gSa fiQj Hkh cnuke gSaA dgk tkrk gS] os thou osQ vkn'kZdks uhpk djrs gSaA vxj os Hkh b±V dk tokc iRFkj ls nsuk lh[k tkrs rks 'kk;n

nks cSyksa dh dFkk

2019-20

6@ f{kfrt

lH; dgykus yxrsA tkiku dh felky lkeus gSA ,d gh fot; us mls lalkjdh lH; tkfr;ksa esa x.; cuk fn;kA

ysfdu x/s dk ,d NksVk HkkbZ vkSj Hkh gS] tks mlls de gh x/k gS] vkSj oggS ̂ cSy*A ftl vFkZ esa ge x/s dk iz;ksx djrs gSa] oqQN mlh ls feyrs&tqyrs vFkZesa ^cfN;k osQ rkmQ* dk Hkh iz;ksx djrs gSaA oqQN yksx cSy dks 'kk;ncsoowQI+kQksa esa loZJs"B dgsaxs_ exj gekjk fopkj ,slk ugha gSA cSy dHkh&dHkh ekjrkHkh gS] dHkh&dHkh vfM+;y cSYk Hkh ns[kus esa vkrk gSA vkSj Hkh dbZ jhfr;ksa ls viukvlarks"k izdV dj nsrk gS_ vr,o mldk LFkku x/s ls uhpk gSA

>wjh dkNh osQ nksuksa cSyksa osQ uke Fks ghjk vkSj eksrhA nksukas iNkb± tkfr osQFksµns[kus esa lqanj] dke esa pkSdl] Mhy esa mQ¡psA cgqr fnuksa lkFk jgrs&jgrs nksuksaesa HkkbZpkjk gks x;k FkkA nksuksa vkeus&lkeus ;k vkl&ikl cSBs gq, ,d&nwljs lsewd&Hkk"kk esa fopkj&fofue; djrs FksA ,d] nwljs osQ eu dh ckr oSQls le> tkrkFkk] ge ugha dg ldrsA vo'; gh muesa dksbZ ,slh xqIr 'kfDRk Fkh] ftlls thoksaesa Js"Brk dk nkok djus okyk euq"; oafpr gSA nksuksa ,d&nwljs dks pkVdj vkSjlw¡?kdj viuk izse izdV djrs] dHkh&dHkh nksuksa lhax Hkh feyk fy;k djrsFksµfoxzg osQ ukrs ls ugha] osQoy fouksn osQ Hkko ls] vkReh;rk osQ Hkko ls] tSlsnksLrksa esa ?kfu"Brk gksrs gh /kSy&/Iik gksus yxrk gSA blosQ fcuk nksLrh oqQNiqQliqQlh] oqQN gydh&lh jgrh gS] ftl ij ”;knk fo'okl ugha fd;k tk ldrkAftl oDr ;s nksuksa cSy gy ;k xkM+h esa tksr fn, tkrs vkSj xjnu fgyk&fgykdjpyrs] ml oDr gj ,d dh ;gh ps"Vk gksrh Fkh fd ”;knk&ls&”;knk cks> esjh ghxjnu ij jgsA fnu&Hkj osQ ckn nksigj ;k laè;k dks nksuksa [kqyrs] rks ,d&nwljs dkspkV&pwVdj viuh Fkdku feVk fy;k djrsA uk¡n esa [kyh&Hkwlk iM+ tkus osQ cknnksuksa lkFk mBrs] lkFk uk¡n eas eq¡g Mkyrs vkSj lkFk gh cSBrs FksA ,d eq¡g gVk ysrk]rks nwljk Hkh gVk ysrk FkkA

la;ksx dh ckr] >wjh us ,d ckj xksb± dks llqjky Hkst fn;kA cSyksa dks D;k ekywe]os D;ksa Hksts tk jgs gSaA le>s] ekfyd us gesa csp fn;kA viuk ;ksa cspk tkuk mUgsa

2019-20

çsepan @7

vPNk yxk ;k cqjk] dkSu tkus] ij >wjh osQ lkys x;k dks ?kj rd xksb± ys tkusesa nk¡rksa ilhuk vk x;kA ihNs ls gk¡drk rks nksuksa nk,¡&ck,¡ Hkkxrs] ixfg;k idM+djvkxs ls [khaprk] rks nksuksa ihNs dks ”kksj yxkrsA ekjrk rks nksuksa lhax uhps djosQgq¡dkjrsA vxj bZ'oj us mUgsa ok.kh nh gksrh] rks >wjh ls iwNrsµrqe ge xjhcksa dksD;ksa fudky jgs gks\ geus rks rqEgkjh lsok djus esa dksbZ dlj ugha mBk j[khA vxjbruh esgur ls dke u pyrk Fkk rks vkSj dke ys ysrsA gesa rks rqEgkjh pkdjh esa ejtkuk dcwy FkkA geus dHkh nkus&pkjs dh f'kdk;r ugha dhA rqeus tks oqQN f[kyk;k]og flj >qdkdj [kk fy;k] fiQj rqeus gesa bl ”kkfye osQ gkFk D;ksa csp fn;k\

laè;k le; nksuksa cSy vius u, LFkku ij igq¡psA fnu&Hkj osQ Hkw[ks Fks] ysfdutc uk¡n esa yxk, x,] rks ,d us Hkh mlesa eq¡g u MkykA fny Hkkjh gks jgk FkkAftls mUgksaus viuk ?kj le> j[kk Fkk] og vkt muls NwV x;k FkkA ;g u;k ?kj]u;k xk¡o] u, vkneh] mUgsa csxkuksa&ls yxrs FksA

nksuksa us viuh ewd&Hkk"kk esa lykg dh] ,d&nwljs dks duf[k;ksa ls ns[kk vkSjysV x,A tc xk¡o esa lksrk iM+ x;k] rks nksuksa us ”kksj ekjdj ixgs rqM+k Mkys vkSj?kj dh rjI+kQ pysA ixgs cgqr e”kcwr FksA vuqeku u gks ldrk Fkk fd dksbZ cSymUgsa rksM+ losQxk_ ij bu nksuksa eas bl le; nwuh 'kfDr vk xbZ FkhA ,d&,d >VosQesa jfLl;k¡ VwV xb±A

>wjh izkr%dky lksdj mBk] rks ns[kk fd nksuksa cSy pjuh ij [kM+s gSaA nksuksa dhxjnuksa esa vkèkk&vk/k xjk¡o yVd jgk gSA ?kqVus rd ik¡o dhpM+ ls Hkjs gSa vkSjnksuksa dh vk¡[kksa esa fonzksge; Lusg >yd jgk gSA

>wjh cSyksa dks ns[kdj Lusg ls xn~xn gks x;kA nkSM+dj mUgsa xys yxk fy;kAizsekfyaxu vkSj pqacu dk og n`'; cM+k gh euksgj FkkA

?kj vkSj xk¡o osQ yM+osQ tek gks x, vkSj rkfy;k¡ ctk&ctkdj mudk Lokxrdjus yxsA xk¡o osQ bfrgkl esa ;g ?kVuk vHkwriwoZ u gksus ij Hkh egRoiw.kZ FkhAcky&lHkk us fu'p; fd;k] nksuksa i'kq&ohjksa dks vfHkuanu&i=k nsuk pkfg,A dksbZvius ?kj ls jksfV;k¡ yk;k] dksbZ xqM+] dksbZ pksdj] dksbZ HkwlhA

2019-20

8@ f{kfrt

,d ckyd us dgkµ,sls cSy fdlh osQ ikl u gksaxsAnwljs us leFkZu fd;kµbruh nwj ls nksuksa vosQys pys vk,Arhljk cksykµcSy ugha gSa os] ml tue osQ vkneh gSaAbldk izfrokn djus dk fdlh dks lkgl u gqvkA>wjh dh L=kh us cSyksa dks }kj ij ns[kk] rks ty mBhA cksyhµoSQls uedgjke

cSy gSa fd ,d fnu ogk¡ dke u fd;k_ Hkkx [kM+s gq,A>wjh vius cSyksa ij ;g vk{ksi u lqu ldkµuedgjke D;ksa gSa\ pkjk&nkuk u

fn;k gksxk] rks D;k djrs\L=kh us jksc osQ lkFk dgkµcl] rqEgha rks cSyksa dks f[kykuk tkurs gks] vkSj rks

lHkh ikuh fiyk&fiykdj j[krs gSaA>wjh us fp<+k;kµpkjk feyrk rks D;ksa Hkkxrs\L=kh fp<+hµHkkxs blfy, fd os yksx rqe&tSls cq¼qvksa dh rjg cSyksa dks

lgykrs ughaA f[kykrs gSa] rks jxM+dj tksrrs Hkh gSaA ;s nksuksa Bgjs dkepksj] HkkxfudysA vc ns[kw¡] dgk¡ ls [kyh vkSj pksdj feyrk gS! lw[ks Hkwls osQ flok oqQNu nw¡xh] [kk,¡ pkgsa ejsaA

ogh gqvkA etwj dks dM+h rkdhn dj nh xbZ fd cSyksa dks [kkyh lw[kk Hkwlkfn;k tk,A

cSyksa us uk¡n esa eq¡g Mkyk] rks iQhdk&iQhdkA u dksbZ fpdukgV] u dksbZ jl!D;k [kk,¡\ vk'kk&Hkjh vk¡[kksa ls }kj dh vksj rkdus yxsA

>wjh us etwj ls dgkµFkksM+h&lh [kyh D;ksa ugha Mky nsrk cs\^ekyfdu eq>s ekj gh MkysaxhA*^pqjkdj Mky vkA*^u nknk] ihNs ls rqe Hkh mUgha dh&lh dgksxsA*

2019-20

çsepan @9

2

nwljs fnu >wjh dk lkyk fiQj vk;k vkSj cSyksa dks ys pykA vcdh mlus nksuksadks xkM+h esa tksrkA

nks&pkj ckj eksrh us xkM+h dks lM+d dh [kkbZ esa fxjkuk pkgk_ ij ghjk uslaHkky fy;kA og ”;knk lgu'khy FkkA

laè;k&le; ?kj igq¡pdj mlus nksuksa dks eksVh jfLl;ksa ls ck¡/k vkSj dy dh'kjkjr dk e”kk p[kk;kA fiQj ogh lw[kk Hkwlk Mky fn;kA vius nksuksa cSyksa dks[kyh] pwuh lc oqQN nhA

nksuksa cSyksa dk ,slk vieku dHkh u gqvk FkkA >wjh bUgsa iwQy dh NM+h ls Hkhu Nwrk FkkA mldh fVVdkj ij nksuksa mM+us yxrs FksA ;gk¡ ekj iM+hA vkgr&lEekudh O;Fkk rks Fkh gh] ml ij feyk lw[kk Hkwlk!

uk¡n dh rjiQ vk¡[ksa rd u mBkb±Anwljs fnu x;k us cSyksa dks gy esa tksrk] ij bu nksuksa us tSls ik¡o u mBkus dh

dle [kk yh FkhA og ekjrs&ekjrs Fkd x;k_ ij nksuksa us ik¡o u mBk;kA ,d ckjtc ml funZ;h us ghjk dh ukd ij [kwc MaMs tek,] rks eksrh dk xqLlk dkcw osQckgj gks x;kA gy ysdj HkkxkA gy] jLlh] tqvk] tksr] lc VwV&VkV dj cjkcjgks x;kA xys esa cM+h&cM+h jfLl;k¡ u gksrha] rks nksuksa idM+kbZ esa u vkrsA

ghjk us ewd&Hkk"kk esa dgkµHkkxuk O;FkZ gSAeksrh us mÙkj fn;kµrqEgkjh rks blus tku gh ys yh FkhA^vcdh cM+h ekj iM+sxhA*^iM+us nks] cSy dk tUe fy;k gS] rks ekj ls dgk¡ rd cpsaxs\*^x;k nks vknfe;ksa osQ lkFk nkSM+k vk jgk gSA nksuksa osQ gkFkksa esa ykfB;k¡ gSaA*eksrh cksykµdgks rks fn[kk nw¡ oqQN e”kk eSa HkhA ykBh ysdj vk jgk gSAghjk us le>k;kµugha HkkbZ! [kM+s gks tkvksA^eq>s ekjsxk] rks eSa Hkh ,d&nks dks fxjk nw¡xk!*

2019-20

10@ f{kfrt

^ughaA gekjh tkfr dk ;g /eZ ugha gSA*eksrh fny esa ,saBdj jg x;kA x;k vk igq¡pk vkSj nksuksa dks idM+ dj ys pykA

oqQ'ky gqbZ fd mlus bl oDr ekjihV u dh] ugha rks eksrh Hkh iyV iM+rkA mlosQrsoj ns[kdj x;k vkSj mlosQ lgk;d le> x, fd bl oDr Vky tkuk ghelygr gSA

vkt nksuks a osQ lkeus fiQj ogh lw[kk Hkwlk yk;k x;kA nksuks a pqipki [kM+sjgsA ?kj osQ yksx Hkkstu djus yxsA ml oDr NksVh&lh yM+dh nks jksfV;k¡fy, fudyh] vkSj nksuks a osQ eq¡g esa nsdj pyh xbZA ml ,d jksVh ls budhHkw[k rks D;k 'kkar gksrh_ ij nksuks a osQ ân; dks ekuks Hkkstu fey x;kA ;gk¡Hkh fdlh lTtu dk okl gSA yM+dh HkSjks dh FkhA mldh ek¡ ej pqdh FkhAlkSrsyh ek¡ mls ekjrh jgrh Fkh] blfy, bu cSyks a ls mls ,d izdkj dhvkReh;rk gks xbZ FkhA

2019-20

çsepan @11

nksuksa fnu&Hkj tksrs tkrs] MaMs [kkrs] vM+rsA 'kke dks Fkku ij ck¡/ fn, tkrsvkSj jkr dks ogh ckfydk mUgsa nks jksfV;k¡ f[kyk tkrhA

izse osQ bl izlkn dh ;g cjdr Fkh fd nks&nks xky lw[kk Hkwlk [kkdj Hkh nksuksanqcZy u gksrs Fks] exj nksuksa dh vk¡[kksa esa] jkse&jkse esa fonzksg Hkjk gqvk FkkA

,d fnu eksrh us ewd&Hkk"kk esa dgkµvc rks ugha lgk tkrk] ghjk!^D;k djuk pkgrs gks\*^,dk/ dks lhaxksa ij mBkdj isaQd nw¡xkA*^ysfdu tkurs gks] og I;kjh yM+dh] tks gesa jksfV;k¡ f[kykrh gS] mlh dh

yM+dh gS] tks bl ?kj dk ekfyd gSA ;g cspkjh vukFk u gks tk,xh\*^rks ekyfdu dks u isaQd nw¡A ogh rks ml yM+dh dks ekjrh gSA*^ysfdu vkSjr tkr ij lhax pykuk euk gS] ;g Hkwys tkrs gksA*^rqe rks fdlh rjg fudyus gh ugha nsrsA crkvks] rqM+kdj Hkkx pysaA*^gk¡] ;g eSa Lohdkj djrk gw¡] ysfdu bruh eksVh jLlh VwVsxh oSQls\*^bldk ,d mik; gSA igys jLlh dks FkksM+k&lk pck yksA fiQj ,d >VosQ esa

tkrh gSA*jkr dks tc ckfydk jksfV;k¡ f[kykdj pyh xbZ] nksuksa jfLl;k¡ pckus yxs] ij

eksVh jLlh eq¡g esa u vkrh FkhA cspkjs ckj&ckj ”kksj yxkdj jg tkrs FksAlglk ?kj dk }kj [kqyk vkSj ogh yM+dh fudyhA nksuksa flj >qdkdj mldk

gkFk pkVus yxsA nksuksa dh iw¡Nsa [kM+h gks xb±A mlus muosQ ekFks lgyk, vkSj cksyhµ[kksys nsrh gw¡A pqiosQ ls Hkkx tkvks] ugha rks ;gk¡ yksx ekj MkysaxsA vkt ?kj esalykg gks jgh gS fd budh ukdksa esa ukFk Mky nh tk,A

mlus xjk¡o [kksy fn;k] ij nksuksa pqipki [kM+s jgsAeksrh us viuh Hkk"kk esa iwNkµvc pyrs D;ksa ugha\ghjk us dgkµpysa rks ysfdu dy bl vukFk ij vkiQr vk,xhA lc blh ij

lansg djsaxsA lglk ckfydk fpYykbZµnksuksa iwQiQkokys cSy Hkkxs tk jgs gSaA vks nknk!nknk! nksuksa cSy Hkkxs tk jgs gSa] tYnh nkSM+ksA

2019-20

12@ f{kfrt

x;k gM+cM+kdj Hkhrj ls fudyk vkSj cSyksa dks idM+us pykA os nksukas HkkxsAx;k us ihNk fd;kA vkSj Hkh rs”k gq,A x;k us 'kksj epk;kA fiQj xk¡o osQ oqQNvknfe;ksa dks Hkh lkFk ysus osQ fy, ykSVkA nksuksa fe=kksa dks Hkkxus dk ekSdk feyx;kA lh/s nkSM+rs pys x,A ;gk¡ rd fd ekxZ dk Kku u jgkA ftl ifjfpr ekxZls vk, Fks] mldk ;gk¡ irk u FkkA u,&u, xk¡o feyus yxsA rc nksuksa ,d [ksrosQ fdukjs [kM+s gksdj lkspus yxs] vc D;k djuk pkfg,A

ghjk us dgkµekywe gksrk gS] jkg Hkwy x,A^rqe Hkh csrgk'kk HkkxsA ogha mls ekj fxjkuk FkkA*^mls ekj fxjkrs] rks nqfu;k D;k dgrh\ og viuk /eZ NksM+ ns] ysfdu ge

viuk /eZ D;ksa NksM+sa\*nksuksa Hkw[k ls O;koqQy gks jgs FksA [ksr eas eVj [kM+h FkhA pjus yxsA jg&jgdj

vkgV ys ysrs Fks] dksbZ vkrk rks ugha gSAtc isV Hkj x;k] nksuksa us vk”kknh dk vuqHko fd;k] rks eLr gksdj

mNyus&owQnus yxsA igys nksuksa us Mdkj yhA fiQj lhax feyk, vkSj ,d&nwljs dksBsyus yxsA eksrh us ghjk dks dbZ dne ihNs gVk fn;k] ;gk¡ rd fd og [kkbZesa fxj x;kA rc mls Hkh Øks/ vk;kA laHkydj mBk vkSj fiQj eksrh ls fey x;kAeksrh us ns[kkµ[ksy esa >xM+k gqvk pkgrk gS] rks fdukjs gV x;kA

3vjs! ;g D;k\ dksbZ lk¡M MkSadrk pyk vk jgk gSA gk¡] lk¡M gh gSA og lkeus vkigq¡pkA nksuksa fe=k cxysa >k¡d jgs gSaA lk¡M iwjk gkFkh gSA mlls fHkM+uk tku ls gkFk/ksuk gS_ ysfdu u fHkM+us ij Hkh tku cprh ugha u”kj vkrhA bUgha dh rjI+kQ vkHkh jgk gSA fdruh Hk;adj lwjr gS!

eksrh us ewd&Hkk"kk esa dgkµcqjs iQ¡lsA tku cpsxh\ dksbZ mik; lkspksAghjk us fpafrr Loj esa dgkµvius ?keaM esa Hkwyk gqvk gSA vkj”kw&fourh u lqusxkA^Hkkx D;ksa u pysa\*

2019-20

çsepan @13

^Hkkxuk dk;jrk gSA*^rks fiQj ;gha ejksA cank rks ukS&nks&X;kjg gksrk gSA*^vkSj tks nkSM+k,\*^rks fiQj dksbZ mik; lkspks tYn!*^mik; ;gh gS fd ml ij nksuksa tus ,d lkFk pksV djsa\ eSa vkxs ls jxsnrk

gw¡] rqe ihNs ls jxsnks] nksgjh ekj iM+sxh] rks Hkkx [kM+k gksxkA esjh vksj >iVs] rqecxy ls mlosQ isV esa lhax ?kqlsM+ nsukA tku tksf[ke gS_ ij nwljk mik; ugha gSA*

nksuksa fe=k tku gFksfy;ksa ij ysdj yiosQA lk¡M dks Hkh laxfBr 'k=kqvksa ls yM+usdk rtjck u FkkA og rks ,d 'k=kq ls eYy;q¼ djus dk vknh FkkA T;ksa gh ghjkij >iVk] eksrh us ihNs ls nkSM+k;kA lk¡M mldh rjI+kQ eqM+k] rks ghjk us jxsnkA lk¡Mpkgrk Fkk fd ,d&,d djosQ nksuksa dks fxjk ys_ ij ;s nksuksa Hkh mLrkn FksA mlsog volj u nsrs FksA ,d ckj lk¡M >Yykdj ghjk dk var dj nsus osQ fy, pykfd eksrh us cxy ls vkdj isV esa lhax Hkksad fn;kA lk¡M Øks/ esa vkdj ihNs fiQjkrks ghjk us nwljs igyw esa lhax pqHkk fn;kA vkf[kj cspkjk ”k[eh gksdj Hkkxk vkSjnksuksa fe=kksa us nwj rd mldk ihNk fd;kA ;gk¡ rd fd lk¡M csne gksdj fxj iM+kArc nksuksa us mls NksM+ fn;kA

nksuksa fe=k fot; osQ u'ks esa >wers pys tkrs FksAeksrh us viuh lkaosQfrd Hkk"kk esa dgkµesjk th rks pkgrk Fkk fd cPpk dks

ekj gh Mkyw¡Aghjk us frjLdkj fd;kµfxjs gq, cSjh ij lhax u pykuk pkfg,A^;g lc <ksax gSA cSjh dks ,slk ekjuk pkfg, fd fiQj u mBsA*^vc ?kj oSQls igq¡psaxs] og lkspksA*^igys oqQN [kk ysa] rks lkspsaA*lkeus eVj dk [ksr Fkk ghA eksrh mlesa ?kql x;kA ghjk euk djrk jgk] ij

mlus ,d u lquhA vHkh nks gh pkj xzkl [kk, Fks fd nks vkneh ykfB;k¡ fy, nkSM+iM+s vkSj nksuksa fe=kksa dks ?ksj fy;kA ghjk rks esM+ ij Fkk] fudy x;kA eksrh lhaps gq,

2019-20

14@ f{kfrt

[ksr esa FkkA mlosQ [kqj dhpM+ esa /¡lus yxsA u Hkkx ldkA idM+ fy;kA ghjk usns[kk] laxh ladV esa gSa] rks ykSV iM+kA iQ¡lsaxs rks nksuksa iQ¡lsaxsA j[kokyksa us mls HkhidM+ fy;kA

izkr%dky nksuksa fe=k dkathgkSl esa can dj fn, x,A

4

nksuksa fe=kksa dks thou esa igyh ckj ,slk lkfcdk iM+k fd lkjk fnu chr x;k vkSj[kkus dks ,d frudk Hkh u feykA le> gh esa u vkrk Fkk] ;g oSQlk Lokeh gSAblls rks x;k fiQj Hkh vPNk FkkA ;gk¡ dbZ HkSalsa Fkha] dbZ cdfj;k¡] dbZ ?kksM+s] dbZx/s_ ij fdlh osQ lkeus pkjk u Fkk] lc ”kehu ij eqjnksa dh rjg iM+s FksA dbZrks brus de”kksj gks x, Fks fd [kM+s Hkh u gks ldrs FksA lkjk fnu nksuksa fe=k iQkVddh vksj VdVdh yxk, rkdrs jgs_ ij dksbZ pkjk ysdj vkrk u fn[kkbZ fn;kA rcnksuksa us nhokj dh uedhu fe^h pkVuh 'kq: dh] ij blls D;k r`fIr gksrh\

jkr dks Hkh tc oqQN Hkkstu u feyk] rks ghjk osQ fny esa fonzksg dh Tokykngd mBhA eksrh ls cksykµvc rks ugha jgk tkrk eksrh!

eksrh us flj yVdk, gq, tokc fn;kµeq>s rks ekywe gksrk gS] izk.k fudyjgs gSaA

^bruh tYn fgEer u gkjks HkkbZ! ;gk¡ ls Hkkxus dk dksbZ mik; fudkyukpkfg,A*

^vkvks nhokj rksM+ MkysaA*^eq>ls rks vc oqQN ugha gksxkA*^cl blh cwrs ij vdM+rs Fks!*^lkjh vdM+ fudy xbZA*ckM+s dh nhokj dPph FkhA ghjk e”kcwr rks Fkk gh] vius uqdhys lhax nhokj

esa xM+k fn, vkSj ”kksj ekjk] rks fe^h dk ,d fpIiM+ fudy vk;kA fiQj rks mldk

2019-20

çsepan @15

lkgl c<+kA blus nkSM+&nkSM+dj nhokj ij pksVsa dha vkSj gj pksV esa FkksM+h&FkksM+hfe^h fxjkus yxkA

mlh le; dkathgkSl dk pkSdhnkj ykyVsu ysdj tkuojksa dh gkf”kjh ysus vkfudykA ghjk dk mtaiu ns[kdj mlus mls dbZ MaMs jlhn fd, vkSj eksVh&lhjLlh ls ck¡/ fn;kA

eksrh us iM+s&iM+s dgkµvkf[kj ekj [kkbZ] D;k feyk\^vius cwrs&Hkj ”kksj rks ekj fn;kA*^,slk ”kksj ekjuk fdl dke dk fd vkSj ca/u esa iM+ x,A*^”kksj rks ekjrk gh tkmQ¡xk] pkgs fdrus gh ca/u iM+rs tk,¡A*^tku ls gkFk /ksuk iM+sxkA*^oqQN ijokg ughaA ;ksa Hkh rks ejuk gh gSA lkspks] nhokj [kqn tkrh] rks fdruh

tkusa cp tkrhaA brus HkkbZ ;gk¡ can gSaA fdlh dh nsg esa tku ugha gSA nks&pkj fnuvkSj ;gh gky jgk] rks lc ej tk,¡xsA*

^gk¡] ;g ckr rks gSA vPNk] rks yk] fiQj eSa Hkh ”kksj yxkrk gw¡A*eksrh us Hkh nhokj esa mlh txg lhax ekjkA FkksM+h&lh fe^h fxjh vkSj fiQj

fgEer c<+hA fiQj rks og nhokj esa lhax yxkdj bl rjg ”kksj djus yxk] ekuksfdlh izfr}a}h ls yM+ jgk gSA vkf[kj dksbZ nks ?kaVs dh ”kksj&vk”kekbZ osQ ckn nhokjmQij ls yxHkx ,d gkFk fxj xbZA mlus nwuh 'kfDr ls nwljk /Ddk ekjk] rks vkèkhnhokj fxj iM+hA

nhokj dk fxjuk Fkk fd v/ejs&ls iM+s gq, lHkh tkuoj psr mBsA rhuksa ?kksfM+;k¡ljiV Hkkx fudyhaA fiQj cdfj;k¡ fudyhaA blosQ ckn HkSalsa Hkh f[kld xb±_ ij x/svHkh rd T;ksa&osQ&R;ksa [kM+s FksA

ghjk us iwNkµrqe nksuksa D;ksa ugha Hkkx tkrs\,d x/s us dgkµtks dgha fiQj idM+ fy, tk,¡!^rks D;k gjt gSA vHkh rks Hkkxus dk volj gSA*^gesa rks Mj yxrk gS] ge ;gha iM+s jgsaxsA*

2019-20

16@ f{kfrt

vk/h jkr ls mQij tk pqdh FkhA nksuksa x/s vHkh rd [kM+s lksp jgs Fks fdHkkxsa ;k u Hkkxsa] vkSj eksrh vius fe=k dh jLlh rksM+us esa yxk gqvk FkkA tc oggkj x;k] rks ghjk us dgkµrqe tkvks] eq>s ;gha iM+k jgus nksA 'kk;n dgha HksaVgks tk,A

eksrh us vk¡[kksa esa vk¡lw ykdj dgkµrqe eq>s bruk LokFkhZ le>rs gks] ghjk\ge vkSj rqe brus fnuksa ,d lkFk jgs gSaA vkt rqe foifÙk esa iM+ x,] rks eSa rqEgsaNksM+dj vyx gks tkmQ¡A

ghjk us dgkµcgqr ekj iM+sxhA yksx le> tk,¡xs] ;g rqEgkjh 'kjkjr gSAeksrh xoZ ls cksykµftl vijk/ osQ fy, rqEgkjs xys esa ca/u iM+k] mlosQ fy,

vxj eq> ij ekj iM+s] rks D;k fpark! bruk rks gks gh x;k fd ukS&nl izkf.k;ksa dhtku cp xbZA os lc rks vk'khokZn nsaxsA

;g dgrs gq, eksrh us nksuksa x/ksa dks lhaxksa ls ekj&ekjdj ckM+s osQ ckgjfudkyk vkSj rc vius ca/q osQ ikl vkdj lks jgkA

Hkksj gksrs gh eqa'kh vkSj pkSdhnkj rFkk vU; deZpkfj;ksa esa oSQlh [kycyh eph]blosQ fy[kus dh ”k:jr ughaA cl] bruk gh dkI+kQh gS fd eksrh dh [kwc ejEergqbZ vkSj mls Hkh eksVh jLlh ls ck¡/ fn;k x;kA

5

,d lIrkg rd nksuksa fe=k ogk¡ c¡/s iM+s jgsA fdlh us pkjs dk ,d r`.k Hkh uMkykA gk¡] ,d ckj ikuh fn[kk fn;k tkrk FkkA ;gh mudk vk/kj FkkA nksuksa brusnqcZy gks x, Fks fd mBk rd u tkrk Fkk] BBfj;k¡ fudy vkbZ FkhaA

,d fnu ckM+s osQ lkeus MqXxh ctus yxh vkSj nksigj gksrs&gksrs ogk¡ ipkl&lkBvkneh tek gks x,A rc nksuksa fe=k fudkys x, vkSj mudh ns[kHkky gksus yxhA yksxvk&vkdj mudh lwjr ns[krs vkSj eu iQhdk djosQ pys tkrsA ,sls e`rd cSyksa dkdkSu [kjhnkj gksrk\

2019-20

çsepan @17

lglk ,d nf<+;y vkneh] ftldh vk¡[ksa yky Fkha vkSj eqnzk vR;ar dBksj]vk;k vkSj nksuksa fe=kksa osQ owQYgksa esa m¡xyh xksndj eqa'kh th ls ckrsa djus yxkAmldk psgjk ns[kdj varKkZu ls nksuksa fe=kksa osQ fny dk¡i mBsA og dkSu gS vkSj mUgsaD;ksa VVksy jgk gS] bl fo"k; esa mUgsa dksbZ lansg u gqvkA nksuksa us ,d&nwljs dks Hkhrus=kksa ls ns[kk vkSj flj >qdk fy;kA

ghjk us dgkµx;k osQ ?kj ls ukgd HkkxsA vc tku u cpsxhAeksrh us vJ¼k osQ Hkko ls mÙkj fn;kµdgrs gSa] Hkxoku lcosQ mQij n;k djrs

gSaA mUgsa gekjs mQij D;ksa n;k ugha vkrh\^Hkxoku osQ fy, gekjk ejuk&thuk nksuksa cjkcj gSA pyks] vPNk gh gS] oqQN

fnu mlosQ ikl rks jgsaxsA ,d ckj Hkxoku us ml yM+dh osQ :i esa gesa cpk;kFkkA D;k vc u cpk,¡xs\*

^;g vkneh Nqjh pyk,xkA ns[k ysukA*^rks D;k fpark gS\ ek¡l] [kky] lhax] gah lc fdlh&u&fdlh dke vk

tk,¡xsA*uhyke gks tkus osQ ckn nksuksa fe=k ml nf<+;y osQ lkFk pysA nksuksa dh

cksVh&cksVh dk¡i jgh FkhA cspkjs ik¡o rd u mBk ldrs Fks] ij Hk; osQ ekjsfxjrs&iM+rs Hkkxs tkrs Fks_ D;ksafd og ”kjk Hkh pky /heh gks tkus ij ”kksj ls MaMktek nsrk FkkA

jkg esa xk;&cSyksa dk ,d jsoM+ gjs&gjs gkj esa pjrk u”kj vk;kA lHkh tkuojizlUu Fks] fpdus] piyA dksbZ mNyrk Fkk] dksbZ vkuan ls cSBk ikxqj djrk FkkAfdruk lq[kh thou Fkk budk_ ij fdrus LokFkhZ gSa lcA fdlh dks fpark ugha fdmuosQ nks HkkbZ cf/d osQ gkFk iM+s oSQls nq[kh gSaA

lglk nksuksa dks ,slk ekywe gqvk fd ;g ifjfpr jkg gSA gk¡] blh jkLrs ls x;kmUgsa ys x;k FkkA ogh [ksr] ogh ckx] ogh xk¡o feyus yxsA izfr{k.k mudh pky rs”kgksus yxhA lkjh Fkdku] lkjh nqcZyrk xk;c gks xbZA vkg\ ;g yks! viuk gh gkj vkx;kA blh oqQ,¡ ij ge iqj pykus vk;k djrs Fks] ;gh oqQvk¡ gSA

2019-20

18@ f{kfrt

eksrh us dgkµgekjk ?kj uxhp vk x;kAghjk cksykµHkxoku dh n;k gSA^eSa rks vc ?kj Hkkxrk gw¡A*^;g tkus nsxk\*^bls eSa ekj fxjkrk gw¡A*^ugha&ugha] nkSM+dj Fkku ij pyksA ogk¡ ls ge vkxs u tk,¡xsA*nksuksa mUeÙk gksdj cNM+ksa dh Hkk¡fr oqQysysa djrs gq, ?kj dh vksj nkSM+sA og

gekjk Fkku gSA nksuksa nkSM+dj vius Fkku ij vk, vkSj [kM+s gks x,A nf<+;y HkhihNs&ihNs nkSM+k pyk vkrk FkkA

>wjh }kj ij cSBk /wi [kk jgk FkkA cSyksa dks ns[krs gh nkSM+k vkSj mUgsa ckjh&ckjhls xys yxkus yxkA fe=kksa dh vk¡[kksa ls vkuan osQ vk¡lw cgus yxsA ,d >wjh dkgkFk pkV jgk FkkA

nf<+;y us tkdj cSyksa dh jfLl;k¡ idM+ yhaA>wjh us dgkµesjs cSy gSaA^rqEgkjs cSy oSQls\ eSa eos'kh[kkus ls uhyke fy, vkrk gw¡A*^eSa rks le>k gw¡ pqjk, fy, vkrs gks! pqiosQ ls pys tkvksA esjs cSy gSaA eSa

cspw¡xk rks fcosQaxsA fdlh dks esjs cSy uhyke djus dk D;k vf[r;kj gS\*^tkdj Fkkus esa jiV dj nw¡xkA*^esjs cSy gSaA bldk lcwr ;g gS fd esjs }kj ij [kM+s gSaA*nf<+;y >Yykdj cSyksa dks ”kcjnLrh idM+ ys tkus osQ fy, c<+kA mlh oDRk

eksrh us lhax pyk;kA nf<+;y ihNs gVkA eksrh us ihNk fd;kA nf<+;y HkkxkA eksrhihNs nkSM+kA xk¡o osQ ckgj fudy tkus ij og #dk_ ij [kM+k nf<+;y dk jkLrk ns[kjgk Fkk] nf<+;y nwj [kM+k /efd;k¡ ns jgk Fkk] xkfy;k¡ fudky jgk Fkk] iRFkj isaQdjgk FkkA vkSj eksrh fot;h 'kwj dh Hkk¡fr mldk jkLrk jksosQ [kM+k FkkA xk¡o osQ yksx;g rek'kk ns[krs Fks vkSj g¡lrs FksA

2019-20

çsepan @19

tc nf<+;y gkjdj pyk x;k] rks eksrh vdM+rk gqvk ykSVkAghjk us dgkµeSa Mj jgk Fkk fd dgha rqe xqLls eas vkdj ekj u cSBksA^vxj og eq>s idM+rk] rks eSa cs&ekjs u NksM+rkA*^vc u vk,xkA*^vk,xk rks nwj gh ls [kcj yw¡xkA ns[kw¡] oSQls ys tkrk gSA*^tks xksyh ejok ns\*^ej tkmQ¡xk] ij mlosQ dke rks u vkmQ¡xkA*^gekjh tku dks dksbZ tku gh ugha le>rkA*^blhfy, fd ge brus lh/s gSaA*”kjk nsj esa uk¡nksa esa [kyh] Hkwlk] pksdj vkSj nkuk Hkj fn;k x;k vkSj nksuksa fe=k

[kkus yxsA >wjh [kM+k nksuksa dks lgyk jgk Fkk vkSj chlksa yM+osQ rek'kk ns[k jgs FksAlkjs xk¡o esa mNkg&lk ekywe gksrk FkkA

mlh le; ekyfdu us vkdj nksuksa osQ ekFks pwe fy,A

iz'u&vH;kl

1- dkathgkSl esa oSQn i'kqvksa dh gkf”kjh D;ksa yh tkrh gksxh\

2- NksVh cPph dks cSyksa osQ izfr izse D;ksa meM+ vk;k\

3- dgkuh esa cSyksa osQ ekè;e ls dkSu&dkSu ls uhfr&fo"k;d ewY; mHkj dj vk, gSa\

4- izLrqr dgkuh esa iszepan us x/s dh fdu LoHkkoxr fo'ks"krkvksa osQ vk/kj ij mlosQ izfr :<+vFkZ ^ew[kZ* dk iz;ksx u dj fdl u, vFkZ dh vksj laosQr fd;k gS\

5- fdu ?kVukvksa ls irk pyrk gS fd ghjk vkSj eksrh esa xgjh nksLRkh Fkh\

6- ^ysfdu vkSjr tkr ij lhax pykuk euk gS] ;g Hkwy tkrs gksA* µ ghjk osQ bl dFku osQ ekè;els L=kh osQ izfr çsepan osQ n`f"Vdks.k dks Li"V dhft,A

2019-20

20@ f{kfrt

7- fdlku thou okys lekt esa i'kq vkSj euq"; osQ vkilh laca/kas dks dgkuh esa fdl rjg O;Drfd;k x;k gS\

8- ^bruk rks gks gh x;k fd ukS nl izkf.k;ksa dh tku cp xbZA os lc rks vk'khokZn nsaxsa*µeksrhosQ bl dFku osQ vkyksd esa mldh fo'ks"krk,¡ crkb,A

9- vk'k; Li"V dhft,µ(d) vo'; gh muesa dksbZ ,slh xqIr 'kfDr Fkh] ftlls thokssa esa Js"Brk dk nkok djus

okyk euq"; oafpr gSA([k) ml ,d jksVh ls mudh Hkw[k rks D;k 'kkar gksrh_ ij nksuksa osQ ân; dks ekuks Hkkstu

fey x;kA10- x;k us ghjk&eksrh dks nksuksa ckj lw[kk Hkwlk [kkus osQ fy, fn;k D;ksafdµ

(d) x;k ijk;s cSyksa ij vf/d [kpZ ugha djuk pkgrk FkkA

([k) xjhch osQ dkj.k [kyh vkfn [kjhnuk mlosQ cl dh ckr u FkhA

(x) og ghjk&eksrh osQ O;ogkj ls cgqr nq[kh FkkA

(?k) mls [kyh vkfn lkexzh dh tkudkjh u FkhA

(lgh mÙkj osQ vkxs (ü) dk fu'kku yxkb,A)

jpuk vkSj vfHkO;fDr

11- ghjk vkSj eksrh us 'kks"k.k osQ f[kykI+kQ vkok”k mBkbZ ysfdu mlosQ fy, izrkM+uk Hkh lghAghjk&eksrh dh bl izfrfØ;k ij roZQ lfgr vius fopkj izdV djsaA

12- D;k vkidks yxrk gS fd ;g dgkuh vk”kknh dh yM+kbZ dh vksj Hkh laosQr djrh gS\

Hkk"kk&vè;;u

13- cl bruk gh dkI+kQh gSA

fiQj eSa Hkh ”kksj yxkrk gw¡A

^gh*] ̂ Hkh* okD; esa fdlh ckr ij ”kksj nsus dk dke dj jgs gSaA ,sls 'kCnksa dks fuikr dgrsgSaA dgkuh esa ls ik¡p ,sls okD; Nk¡fV, ftuesa fuikr dk iz;ksx gqvk gksA

2019-20

çsepan @21

14- jpuk osQ vk/kj ij okD; Hksn crkb, rFkk miokD; Nk¡Vdj mlosQ Hkh Hksn fyf[k,µ

(d) nhokj dk fxjuk Fkk fd v/ejs&ls iM+s gq, lHkh tkuoj psr mBsA

([k) lglk ,d nf<+;y vkneh] ftldh vk¡[ks yky Fkha vkSj eqnzk vR;ar dBksj] vk;kA

(x) ghjk us dgkµx;k osQ ?kj ls ukgd HkkxsA

(?k) eSa cspw¡xk] rks fcosaQxsA

(Ä) vxj og eq>s idM+rk rks eSa cs&ekjs u NksM+rkA

15- dgkuh esa txg&txg eqgkojksa dk iz;ksx gqvk gSA dksbZ ik¡p eqgkojs Nk¡fV, vkSj mudk okD;ksaesa iz;ksx dhft,A

ikBsrj lfØ;rk

• i'kq&if{k;ksa ls lacaf/r vU; jpuk,¡ <w¡<+dj if<+, vkSj d{kk esa ppkZ dhft,A

'kCn&laink

fujkin µ lqjf{kr

lfg".kqrk µ lgu'khyrk

iNkb± µ ikyrw i'kqvksa dh ,d uLy

xksb± µ tksM+h

oqQysy µ ØhM+k (dYyksy)

fo"kkn µ mnklh

ijkdk"Bk µ vafre lhek

x.; µ x.kuh;] lEekfur

foxzg µ vyxko

ixfg;k µ i'kq ck¡/us dh jLlh

2019-20

22@ f{kfrt

xjk¡o µ iq¡Qnsnkj jLlh tks cSy vkfn osQ xys esa igukbZ tkrh gSA

izfrokn µ fojks/

fVVdkj µ eq¡g ls fudyus okyk fVd&fVd dk 'kCn

elygr µ fgrdj

jxsnuk µ [knsM+uk

eYy;q¼ µ oqQ'rh

lkfcdk µ okLrk] ljksdkj

dkathgkSl µ eos'kh[kkuk] og ckM+k ftlesa nwljs dk [ksr vkfn [kkus okys ;k (dkbu gkml) ykokfjl pkSik;s can fd, tkrs gSa vkSj oqQN naM ysdj NksM+s

;k uhyke fd, tkrs gSaA

jsoM+ µ i'kqvksa dk >qaM

mUeÙk µ erokyk

Fkku µ i'kqvksa osQ ck¡/s tkus dh txg

mNkg µ mRlo] vkuan

2019-20

Activity _____________13.1

• We have all heard of the earthquakes inLatur, Bhuj, Kashmir etc. or the cyclonesthat lashed the coastal regions. Thinkof as many different ways as possible inwhich people’s health would be affectedby such a disaster if it took place in ourneighbourhood.

• How many of these ways we can think ofare events that would occur when thedisaster is actually happening?

• How many of these health-related eventswould happen long after the actualdisaster, but would still be because of thedisaster?

• Why would one effect on health fall intothe first group, and why would another

fall into the second group?

When we do this exercise, we realise thathealth and disease in human communitiesare very complex issues, with manyinterconnected causes. We also realise thatthe ideas of what ‘health’ and ‘disease’ meanare themselves very complicated. When weask what causes diseases and how we preventthem, we have to begin by asking what thesenotions mean.

We have seen that cells are the basic unitsof organisms. Cells are made of a variety ofchemical substances–proteins, carbo-hydrates,fats or lipids, and so on. Cell is a dynamic place.Something or the other is always happeninginside them. Complex reactions and repair goeson inside cells. New cells are being made. Inour organs or tissues, there are variousspecialised activities going on– the heart isbeating, the lungs are breathing, the kidneyis filtering urine, the brain is thinking.

All these activities are interconnected. Forexample, if the kidneys are not filtering urine,

poisonous substances will accumulate. Undersuch conditions, the brain will not be able tothink properly. For all these interconnectedactivities, energy and raw material are needed.Food is a necessity for cell and tissue functions.Anything that prevents proper functioning ofcells and tissues will lead to a lack of properactivity of the body.

It is in this context that we will look at thenotions of health and disease.

13.1 Health and its Failure

13.1.1 THE SIGNIFICANCE OF ‘HEALTH’

We have heard the word ‘health’ used quitefrequently. We use it ourselves as well, whenwe say things like ‘my grandmother’s healthis not good’. Our teachers use it when theyscold us saying ‘this is not a healthy attitude’.What does the word ‘health’ mean?

If we think about it, we realise that italways implies the idea of ‘being well’. We canthink of this well-being as ef fectivefunctioning. For our grandmothers, being ableto go out to the market or to visit neighboursis ‘being well’, and not being able to do suchthings is ‘poor health’. Being interested infollowing the teaching in the classroom so thatwe can understand the world is called a‘healthy attitude’; while not being interestedis called the opposite. ‘Health’ is therefore astate of being well enough to function wellphysically, mentally and socially.

13.1.2 PERSONAL AND COMMUNITY ISSUES

BOTH MATTER FOR HEALTH

If health means a state of physical, mental and

social well-being, it cannot be something that

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healthy. Social equality and harmony aretherefore necessary for individual health. Wecan think of many other such examples ofconnections between community issues andindividual health.

each one of us can achieve entirely on our own.The health of all organisms will depend on

their surroundings or their environment. Theenvironment includes the physicalenvironment. So, for example, health is at risk

in a cyclone in many ways.Human beings live in societies. Our social

environment, therefore, is an important factor

in our individual health. We live in villages,towns or cities. In such places, even ourphysical environment is decided by our social

environment.Consider what would happen if no agency

is ensuring that garbage is collected and

disposed. What would happen if no one takesresponsibility for clearing the drains andensuring that water does not collect in the

streets or open spaces?So, if there is a great deal of garbage

thrown in our streets, or if there is open drain-

water lying stagnant around where we live,the possibility of poor health increases.Therefore, public cleanliness is important for

individual health.

Activity _____________13.2

• Find out what provisions are made byyour local authority (panchayat/

municipal corporation) for the supplyof clean drinking water.

• Are all the people in your locality able

to access this?

Activity _____________13.3

• Find out how your local authority

manages the solid waste generated inyour neighbourhood.

• Are these measures adequate?

• If not, what improvements would yousuggest?

• What could your family do to reduce

the amount of solid waste generatedduring a day/week?

We need food to be healthy, and this foodwill have to be earned by doing work. For this,the opportunity to do work has to be available.

We need to be happy in order to be trulyhealthy, and if we mistreat each other and areafraid of each other, we cannot be happy or

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13.1.3 DISTINCTIONS BETWEEN ‘HEALTHY’

AND ‘DISEASE-FREE’

If this is what we mean by ‘health’, what do wemean by ‘disease’? The word is actually self-explanatory – we can think of it as ‘disease’ –disturbed ease. Disease, in other words,literally means being uncomfortable. However,the word is used in a more limited meaning.We talk of disease when we can find a specificand particular cause for discomfort. This doesnot mean that we have to know the absolutefinal cause; we can say that someone issuffering from diarrhoea without knowingexactly what has caused the loose motions.

We can now easily see that it is possible tobe in poor health without actually sufferingfrom a particular disease. Simply not beingdiseased is not the same as being healthy.‘Good health’ for a dancer may mean beingable to stretch his body into difficult butgraceful positions. On the other hand, goodhealth for a musician may mean having enoughbreathing capacity in his/her lungs to controlthe notes from his/her flute. To have theopportunity to realise the unique potential inall of us is also necessary for real health.

The Five ‘F’s — What is to be done?

Protect the water source (H)

Treat and store water

safely (S)

Wash hands before preparing and

taking food (H)

Wash hands after defecation (S)

Cover the food (H)

Control flies (S)

Clean vegetables and fruits beforeuse (H)

Avoid open defecation (S)

Proper drainage system (H)

Treatment of water (S)

Prevention of Transmission of Diseases by MaintainingSanitation and Hygiene

Hygiene

Sanitation

Healthy personFaecal matter

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So, we can be in poor health without therebeing a simple cause in the form of anidentifiable disease. This is the reason why,when we think about health, we think aboutsocieties and communities. On the otherhand, when we think about disease, we thinkabout individual sufferers.

uestions1. State any two conditions essential

for good health.

2. State any two conditions essential

for being free of disease.

3. Are the answers to the above

questions necessarily the same or

different? Why?

13.2 Disease and Its Causes

13.2.1 WHAT DOES DISEASE LOOK LIKE?

Let us now think a little more aboutdiseases. In the first place, how do we knowthat there is a disease? In other words, howdo we know that there is something wrongwith the body? There are many tissues inthe body, as we have seen in Chapter 6.These tissues make up physiologicalsystems or organ systems that carry outbody functions. Each of the organ systemshas specific organs as its parts, and it hasparticular functions. So, the digestivesystem has the stomach and intestines, andit helps to digest food taken in from outsidethe body. The musculoskeletal system,which is made up of bones and muscles,holds the body parts together and helps thebody move.

When there is a disease, either thefunctioning of one or more systems of the bodywill change for the worse. These changes giverise to symptoms and signs of disease.Symptoms of disease are the things we feel asbeing ‘wrong’. So we have a headache, we havecough, we have loose motions, we have awound with pus; these are all symptoms.These indicate that there may be a disease, butthey don’t indicate what the disease is. Forexample, a headache may mean just

examination stress or, very rarely, it maymean meningitis, or any one of a dozendifferent diseases.

Signs of disease are what physicians willlook for on the basis of the symptoms. Signswill give a little more definite indication of thepresence of a particular disease. Physicians willalso get laboratory tests done to pinpoint thedisease further.

13.2.2 ACUTE AND CHRONIC DISEASES

The manifestations of disease will be differentdepending on a number of factors. Somediseases last for only very short periods of time,and these are called acute diseases. We all knowfrom experience that the common cold lastsonly a few days. Other ailments can last for along time, even as much as a lifetime, and arecalled chronic diseases. An example is theinfection causing elephantiasis, which is verycommon in some parts of India.

Activity _____________13.4

• Survey your neighbourhood to find out:(1) how many people suffered from

acute diseases during the lastthree months,

(2) how many people developedchronic diseases during this sameperiod,

(3) and finally, the total number ofpeople suffering from chronicdiseases in your neighbourhood.

• Are the answers to questions (1) and(2) different?

• Are the answers to questions (2) and(3) different?

• What do you think could be thereason for these differences? What doyou think would be the effect of thesedifferences on the general health ofthe population?

13.2.3 CHRONIC DISEASES AND POOR

HEALTH

Acute and chronic diseases have differenteffects on our health. Any disease that causespoor functioning of some part of the body willaffect our health. This is because all functionsof the body are necessary for being healthy.

Q

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difference or the poor nourishment alonewould not lead to loose motions. But they dobecome contributory causes of the disease.

Why was there no clean drinking water forthe baby? Perhaps because the public servicesare poor where the baby’s family lives. So,poverty or lack of public services become thirdcause of the baby’s disease.

It will now be obvious that all diseases willhave immediate causes and contributorycauses. Also, most diseases will have manycauses, rather than one single cause.

13.2.5 INFECTIOUS AND NON-INFECTIOUS

CAUSES

As we have seen, it is important to keep publichealth and community health factors in mindwhen we think about causes of diseases. Wecan take that approach a little further. It isuseful to think of the immediate causes ofdisease as belonging to two distinct types. Onegroup of causes is the infectious agents,mostly microbes or micro-organisms.Diseases where microbes are the immediatecauses are called infectious diseases. This isbecause the microbes can spread in thecommunity, and the diseases they cause willspread with them.

Things to ponder

1. Do all diseases spread to peoplecoming in contact with a sick person?

2. What are the diseases that are notspreading?

3. How would a person develop thosediseases that don’t spread by contactwith a sick person?

On the other hand, there are also diseasesthat are not caused by infectious agents. Theircauses vary, but they are not external causeslike microbes that can spread in thecommunity. Instead, these are mostly internal,non-infectious causes.

For example, some cancers are caused bygenetic abnormalities. High blood pressurecan be caused by excessive weight and lackof exercise. You can think of many otherdiseases where the immediate causes will notbe infectious.

But an acute disease, which is over very soon,will not have time to cause major effects ongeneral health, while a chronic disease will doso.

As an example, think about a cough andcold, which all of us have from time to time.Most of us get better and become well within aweek or so. And there are no lasting effects onour health. But if we get infected with a chronicdisease such as tuberculosis of the lungs, thenbeing ill over the years does make us loseweight and feel tired all the time.

We may not go to school for a few days ifwe have an acute disease. But a chronicdisease will make it difficult for us to followwhat is being taught in school and reduce ourability to learn. In other words, we are likelyto have prolonged general poor health if wehave a chronic disease. Chronic diseasestherefore, have very drastic long-term effectson people’s health as compared to acutediseases.

13.2.4 CAUSES OF DISEASES

What causes disease? When we think aboutcauses of diseases, we must remember thatthere are many levels of such causes. Let uslook at an example. If there is a baby sufferingfrom loose motions, we can say that the causeof the loose motions is probably an infection.

But the next question is – where did theinfection come from? Suppose we find that theinfection came through unclean drinkingwater. But many babies must have had thisunclean drinking water. So, why is it that onebaby developed loose motions when the otherbabies did not?

One reason might be that this baby is nothealthy. As a result, it might be more likely tohave disease when exposed to risk, whereashealthier babies would not. Why is the babynot healthy? Perhaps because it is not wellnourished and does not get enough food. So,lack of good nourishment becomes a secondcause of the disease. Further, why is the babynot well nourished? Perhaps because it is froma household which is poor.

It is also possible that the baby has somegenetic difference that makes it more likely tosuffer from loose motions when exposed to apathogen. Without the pathogen, the genetic

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The ways in which diseases spread, and

the ways in which they can be treated and

prevented at the community level would be

different for different diseases. This would

depend a lot on whether the immediate causes

are infectious or non-infectious.

uestions

1. List any three reasons why you

would think that you are sick and

ought to see a doctor. If only one

of these symptoms were present,

would you still go to the doctor?

Why or why not?

2. In which of the following case do

you think the long-term effects on

your health are likely to be most

unpleasant?

• if you get jaundice,

• if you get lice,

• if you get acne.

Why?

13.3 Infectious Diseases

13.3.1 INFECTIOUS AGENTS

We have seen that the entire diversity seen in

the living world can be classified into a few

groups. This classification is based on

common characteristics between different

organisms. Organisms that can cause disease

are found in a wide range of such categories

of classification. Some of them are viruses,

some are bacteria, some are fungi, some are

single-celled animals or protozoans (Fig. 13.1).

Some diseases are also caused by

multicellular organisms, such as worms of

different kinds.

Peptic ulcers and the Nobel prize

For many years, everybody used to thinkthat peptic ulcers, which cause acidity–related pain and bleeding in the stomachand duodenum, were because of lifestylereasons. Everybody thought that a stressfullife led to a lot of acid secretion in thestomach, and eventually caused pepticulcers.

Then two Australians made a discoverythat a bacterium, Helicobacter pylori, wasresponsible for peptic ulcers. Robin Warren(born 1937), a pathologist from Perth,Australia, saw these small curved bacteriain the lower part of the stomach in manypatients. He noticed that signs ofinflammation were always present aroundthese bacteria. Barry Marshall (born 1951),a young clinical fellow, became interestedin Warren’s findings and succeeded incultivating the bacteria from these sources.

In treatment studies, Marshall andWarren showed that patients could be curedof peptic ulcer only when the bacteria werekilled off from the stomach. Thanks to thispioneering discovery by Marshall andWarren, peptic ulcer disease is no longer achronic, frequently disabling condition, buta disease that can be cured by a shortperiod of treatment with antibiotics.

For this achievement, Marshall and

Q

Warren (seen in the picture) received theNobel prize for physiology and medicine in2005.

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Fig. 13.1(b): Picture of staphylococci, the bacteria

which can cause acne. The scale of the

image is indicated by the line at top left,

which is 5 micrometres long.

Fig. 13.1(d): Picture of Leishmania, the protozoan

organism that causes kala-azar. The

organisms are oval-shaped, and each

has one long whip-like structure. One

organism (arrow) is dividing, while a cell

of the immune system (lower right) has

gripped on the two whips of the dividing

organism and is sending cell processes

up to eat up the organism. The immune

cell is about ten micrometres in diameter.

Fig. 13.1(e): Picture of an adult roundworm (Ascaris

lumbricoides) from the small intestine. The

ruler next to it shows four centimetres to

give us an idea of the scale.

Fig. 13.1(a): Picture of SARS viruses coming out (see

arrows for examples) of the surface of

an infected cell. The white scale line

represents 500 nanometres, which is

half a micrometre, which is one-

thousandth of a millimetre. The scale line

gives us an idea of how small the things

we are looking at are.

Courtesy: Emerging Infectious

Deseases, a journal of CDC, U.S.

Fig. 13.1(c): Picture of Trypanosoma, the protozoan

organism responsible for sleeping

sickness. The organism is lying next to

a saucer-shaped red blood cell to give

an idea of the scale.

Copyright: Oregon Health and Science

University, U.S.

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Common examples of diseases caused byviruses are the common cold, influenza,dengue fever and AIDS. Diseases like typhoid

fever, cholera, tuberculosis and anthrax arecaused by bacteria. Many common skininfections are caused by different kinds of

fungi. Protozoan microbes cause manyfamiliar diseases, such as malaria and kala-azar. All of us have also come across intestinal

worm infections, as well as diseases likeelephantiasis caused by diffferent species ofworms.

Why is it important that we think of thesecategories of infectious agents? The answeris that these categories are important factors

in deciding what kind of treatment to use.Members of each one of these groups –viruses, bacteria, and so on – have many

biological characteristics in common.All viruses, for example, live inside host

cells, whereas bacteria very rarely do. Viruses,

bacteria and fungi multiply very quickly, whileworms multiply very slowly in comparison.Taxonomically, all bacteria are closely related

to each other than to viruses and vice versa.This means that many important lifeprocesses are similar in the bacteria group

but are not shared with the virus group. As aresult, drugs that block one of these lifeprocesses in one member of the group is likely

to be effective against many other membersof the group. But the same drug will not workagainst a microbe belonging to a different

group.As an example, let us take antibiotics. They

commonly block biochemical pathwaysimportant for bacteria. Many bacteria, forexample, make a cell-wall to protectthemselves. The antibiotic penicillin blocksthe bacterial processes that build the cell-wall. As a result, the growing bacteria becomeunable to make cell-walls, and die easily.Human cells don’t make a cell-wall anyway,so penicillin cannot have such an effect on us.Penicillin will have this effect on any bacteriathat use such processes for making cell-walls.Similarly, many antibiotics work against manyspecies of bacteria rather than simply workingagainst one.

But viruses do not use these pathways at

all, and that is the reason why antibiotics do

not work against viral infections. If we have a

common cold, taking antibiotics does not

reduce the severity or the duration of the

disease. However, if we also get a bacterial

infection along with the viral cold, taking

antibiotics will help. Even then, the antibiotic

will work only against the bacterial part of

the infection, not the viral infection.

Activity _____________13.5

• Find out how many of you in your

class had cold/cough/fever recently.

• How long did the illness last?

• How many of you took antibiotics (ask

your parents if you had antibiotics)?

• How long were those who took

antibiotics ill?

• How long were those who didn’t take

antibiotics ill?

• Is there a difference between these two

groups?

• If yes, why? If not, why not?

13.3.2 MEANS OF SPREAD

How do infectious diseases spread? Many

microbial agents can commonly move from

an affected person to someone else in a variety

of ways. In other words, they can be

‘communicated’, and so are also called

communicable diseases.

Such disease-causing microbes can spread

through the air. This occurs through the little

droplets thrown out by an infected person who

sneezes or coughs. Someone standing close by

can breathe in these droplets, and the microbes

get a chance to start a new infection. Examples

of such diseases spread through the air are

the common cold, pneumonia and

tuberculosis.

We all have had the experience of sitting

near someone suffering from a cold and

catching it ourselves. Obviously, the more

crowded our living conditions are, the more

likely it is that such airborne diseases will

spread.

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Diseases can also be spread throughwater. This occurs if the excreta from someonesuffering from an infectious gut disease, suchas cholera, get mixed with the drinking waterused by people living nearby. The cholera-causing microbes will enter a healthy personthrough the water they drink and causedisease in them. Such diseases are much morelikely to spread in the absence of safe suppliesof drinking water.

The sexual act is one of the closest physicalcontact two people can have with each other.Not surprisingly, there are microbial infectionssuch as syphilis or AIDS that are transmittedby sexual contact from one partner to theother. However, such sexually transmitteddiseases are not spread by casual physicalcontact. Casual physical contacts includehandshakes or hugs or sports, like wrestling,or by any of the other ways in which we toucheach other socially. Other than the sexualcontact, the virus causing AIDS (HIV) can alsospread through blood-to-blood contact withinfected people or from an infected mother toher baby during pregnancy or through breastfeeding.

We live in an environment that is full ofmany other creatures apart from us. It isinevitable that many diseases will betransmitted by other animals. These animalscarry the infecting agents from a sick personto another potential host. These animals arethus the intermediaries and are called vectors.

The commonest vectors we all know aremosquitoes. In many species of mosquitoes,the females need highly nutritious food in theform of blood in order to be able to lay matureeggs. Mosquitoes feed on many warm-bloodedanimals, including us. In this way, they cantransfer diseases from person to person(Fig. 13.2).

13.3.3 ORGAN-SPECIFIC AND TISSUE-SPECIFIC MANIFESTATIONS

The disease-causing microbes enter the bodythrough these different means. Where do theygo then? The body is very large whencompared to the microbes. So there are manypossible places, organs or tissues, where theycould go. Do all microbes go to the same tissueor organ, or do they go to different ones?

Different species of microbes seem to haveevolved to home in on different parts of thebody. In part, this selection is connected totheir point of entry. If they enter from the airvia the nose, they are likely to go to the lungs.This is seen in the bacteria causingtuberculosis. If they enter through the mouth,they can stay in the gut lining like typhoid-causing bacteria. Or they can go to the liver,like the viruses that cause jaundice.

But this needn’t always be the case. Aninfection like HIV, that comes into the bodyvia the sexual organs, will spread to lymphnodes all over the body. Malaria-causingmicrobes, entering through a mosquito bite,will go to the liver, and then to the red bloodcells. The virus causing Japanese encephalitis,or brain fever, will similarly enter through amosquito bite. But it goes on to infect the brain.

The signs and symptoms of a disease willthus depend on the tissue or organ which themicrobe targets. If the lungs are the targets,then symptoms will be cough andbreathlessness. If the liver is targeted, there willbe jaundice. If the brain is the target, we willobserve headaches, vomiting, fits orunconsciousness. We can imagine what thesymptoms and signs of an infection will be ifwe know what the target tissue or organ is,and the functions that are carried out by thistissue or organ.

Fig. 13.2: Common methods of transmission of

diseases.

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In addition to these tissue-specific effectsof infectious disease, there will be other commoneffects too. Most of these common effects dependon the fact that the body’s immune system isactivated in response to infection. An activeimmune system recruits many cells to theaffected tissue to kill off the disease-causingmicrobes. This recruitment process is calledinflammation. As a part of this process, thereare local effects such as swelling and pain, andgeneral effects such as fever.

In some cases, the tissue-specificity of theinfection leads to very general-seeming effects.For example, in HIV infection, the virus goesto the immune system and damages itsfunction. Thus, many of the effects of HIV-AIDSare because the body can no longer fight offthe many minor infections that we faceeveryday. Instead, every small cold can becomepneumonia. Similarly, a minor gut infectioncan produce major diarrhoea with blood loss.Ultimately, it is these other infections that killpeople suffering from HIV-AIDS.

It is also important to remember that theseverity of disease manifestations depend onthe number of microbes in the body. If thenumber of microbes is very small, the diseasemanifestations may be minor or unnoticed.But if the number is of the same microbe large,the disease can be severe enough to be life-threatening. The immune system is a majorfactor that determines the number of microbessurviving in the body. We shall look into thisaspect a little later in the chapter.

13.3.4 PRINCIPLES OF TREATMENT

What are the steps taken by your family whenyou fall sick? Have you ever thought why yousometimes feel better if you sleep for some time?When does the treatment involve medicines?

Based on what we have learnt so far, itwould appear that there are two ways to treatan infectious disease. One would be to reducethe effects of the disease and the other to killthe cause of the disease. For the first, we canprovide treatment that will reduce thesymptoms. The symptoms are usuallybecause of inflammation. For example, we cantake medicines that bring down fever, reducepain or loose motions. We can take bed rest so

that we can conserve our energy. This willenable us to have more of it available to focuson healing.

But this kind of symptom-directedtreatment by itself will not make the infectingmicrobe go away and the disease will not becured. For that, we need to be able to kill offthe microbes.

How do we kill microbes? One way is touse medicines that kill microbes. We have seenearlier that microbes can be classified intodifferent categories. They are viruses, bacteria,fungi or protozoa. Each of these groups oforganisms will have some essentialbiochemical life process which is peculiar tothat group and not shared with the othergroups. These processes may be pathways forthe synthesis of new substances or respiration.

These pathways will not be used by useither. For example, our cells may make newsubstances by a mechanism different from thatused by bacteria. We have to find a drug thatblocks the bacterial synthesis pathway withoutaffecting our own. This is what is achieved bythe antibiotics that we are all familiar with.Similarly, there are drugs that kill protozoasuch as the malarial parasite.

One reason why making anti-viralmedicines is harder than making anti-bacterial medicines is that viruses have fewbiochemical mechanisms of their own. Theyenter our cells and use our machinery for theirlife processes. This means that there arerelatively few virus-specific targets to aim at.Despite this limitation, there are now effectiveanti-viral drugs, for example, the drugs thatkeep HIV infection under control.

13.3.5 PRINCIPLES OF PREVENTION

All of what we have talked about so far dealswith how to get rid of an infection in someonewho has the disease. But there are threelimitations of this approach to dealing withinfectious disease. The first is that oncesomeone has a disease, their body functionsare damaged and may never recovercompletely. The second is that treatment willtake time, which means that someonesuffering from a disease is likely to be

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an infectious microbe does not necessarily

mean developing noticeable disease.

So, one way of looking at severe infectiousdiseases is that it represents a lack of success

of the immune system. The functioning of the

immune system, like any other system in our

body, will not be good if proper and sufficientnourishment and food is not available.

Therefore, the second basic principle of

prevention of infectious disease is the

availability of proper and sufficient food foreveryone.

Activity _____________13.6

• Conduct a survey in your locality.

Talk to ten families who are well-off

and ten who are very poor (in your

estimation). Both sets of families

should have children who are below

five years of age. Measure the heights

of these children. Draw a graph of the

height of each child against its age

for both sets of families.

• Is there a difference between the

groups? If yes, why?

• If there is no difference, do you think

that your findings mean that being

well-off or poor does not matter for

health?

These are the general ways of preventinginfections. What are the specific ways? Theyrelate to a peculiar property of the immunesystem that usually fights off microbialinfections. Let us cite an example to try andunderstand this property.

These days, there is no smallpox anywherein the world. But as recently as a hundredyears ago, smallpox epidemics were not at alluncommon. In such an epidemic, people usedto be very afraid of coming near someonesuffering from the disease since they wereafraid of catching the disease.

However, there was one group of peoplewho did not have this fear. These people wouldprovide nursing care for the victims ofsmallpox. This was a group of people who hadhad smallpox earlier and survived it, althoughwith a lot of scarring. In other words, if youhad smallpox once, there was no chance of

bedridden for some time even if we can giveproper treatment. The third is that the personsuffering from an infectious disease can serveas the source from where the infection mayspread to other people. This leads to themultiplication of the above difficulties. It isbecause of such reasons that prevention ofdiseases is better than their cure.

How can we prevent diseases? There aretwo ways, one general and one specific to eachdisease. The general ways of preventinginfections mostly relate to preventingexposure. How can we prevent exposure toinfectious microbes?

If we look at the means of their spreading,we can get some easy answers. For airbornemicrobes, we can prevent exposure byproviding living conditions that are notovercrowded. For water-borne microbes, wecan prevent exposure by providing safedrinking water. This can be done by treatingthe water to kill any microbial contamination.For vector-borne infections, we can provideclean environments. This would not, forexample, allow mosquito breeding. In otherwords, public hygiene is one basic key to theprevention of infectious diseases.

In addition to these issues that relate tothe environment, there are some other generalprinciples to prevent infectious diseases. Toappreciate those principles, let us ask aquestion we have not looked at so far. Normally,we are faced with infections everyday. Ifsomeone is suffering from a cold and cough inthe class, it is likely that the children sittingaround will be exposed to the infection. Butall of them do not actually suffer from thedisease. Why not?

This is because the immune system of ourbody is normally fighting off microbes. We havecells that specialise in killing infectingmicrobes. These cells go into action each timeinfecting microbes enter the body. If they aresuccessful, we do not actually come down withany disease. The immune cells manage to killoff the infection long before it assumes majorproportions. As we noted earlier, if the numberof the infecting microbes is controlled, themanifestations of disease will be minor. In otherwords, becoming exposed to or infected with

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Q

suffering from it again. So, having the diseaseonce was a means of preventing subsequentattacks of the same disease.

This happens because when the immunesystem first sees an infectious microbe, itresponds against it and then remembers itspecifically. So the next time that particularmicrobe, or its close relatives enter the body,the immune system responds with even greatervigour. This eliminates the infection even morequickly than the first time around. This is thebasis of the principle of immunisation.

to the infecting microbe from turning intoactual disease.

Many such vaccines are now available forpreventing a whole range of infectious diseases,and provide a disease-specific means ofprevention. There are vaccines against tetanus,diphtheria, whooping cough, measles, polioand many others. These form the public healthprogramme of childhood immunisation forpreventing infectious diseases.

Of course, such a programme can beuseful only if such health measures areavailable to all children. Can you think ofreasons why this should be so?

Some hepatitis viruses, which causejaundice, are transmitted through water. Thereis a vaccine for one of them, hepatitis A, in themarket. But the majority of children in manyparts of India are already immune to hepatitisA by the time they are five years old. This isbecause they are exposed to the virus throughwater. Under these circumstances, would youtake the vaccine?

Activity _____________13.7

• Rabies virus is spread by the bite ofinfected dogs and other animals. Thereare anti-rabies vaccines for both humansand animals. Find out the plan of yourlocal authority for the control of rabiesin your neighbourhood. Are thesemeasures adequate? If not, whatimprovements would you suggest?

uestions1. Why are we normally advised to

take bland and nourishing foodwhen we are sick?

2. What are the different means bywhich infectious diseases arespread?

3. What precautions can you take inyour school to reduce theincidence of infectious diseases?

4. What is immunisation?5. What are the immunisation

programmes available at thenearest health centre in yourlocality? Which of these diseasesare the major health problems inyour area?

Immunisation

Traditional Indian and Chinese medicinalsystems sometimes deliberately rubbed theskin crusts from smallpox victims into theskin of healthy people. They thus hoped toinduce a mild form of smallpox that wouldcreate resistance against the disease.

Famously, two centuries ago, an Englishphysician named Edward Jenner, realised

that milkmaidswho had hadcowpox did notcatch smallpoxeven duringe p i d e m i c s .Cowpox is a verymild disease.Jenner trieddeliberately givingcowpox to people(as he can be seendoing in the

picture), and found that they were nowresistant to smallpox. This was because thesmallpox virus is closely related to thecowpox virus. ‘Cow’ is ‘vacca’ in Latin, andcowpox is ‘vaccinia’. From these roots, theword ‘vaccination’ has come into our usage.

We can now see that, as a general principle,we can ‘fool’ the immune system intodeveloping a memory for a particular infectionby putting something, that mimics the microbewe want to vaccinate against, into the body.This does not actually cause the disease butthis would prevent any subsequent exposure

2019-2020

WHY DO WE FALL ILL? 187

Whatyou havelearnt

• Health is a state of physical, mental and social well-being.

• The health of an individual is dependent on his/her physicalsurroundings and his/her economic status.

• Diseases are classified as acute or chronic, depending ontheir duration.

• Disease may be due to infectious or non-infectious causes.

• Infectious agents belong to different categories of organismsand may be unicellular and microscopic or multicellular.

• The category to which a disease-causing organism belongsdecides the type of treatment.

• Infectious agents are spread through air, water, physicalcontact or vectors.

• Prevention of disease is more desirable than its successfultreatment.

• Infectious diseases can be prevented by public health hygienemeasures that reduce exposure to infectious agents.

• Infectious diseases can also be prevented by usingimmunisation.

• Effective prevention of infectious diseases in the communityrequires that everyone should have access to public hygieneand immunisation.

Exercises1. How many times did you fall ill in the last one year? What

were the illnesses?

(a) Think of one change you could make in your habits inorder to avoid any of/most of the above illnesses.

(b) Think of one change you would wish for in yoursurroundings in order to avoid any of/most of the aboveillnesses.

2. A doctor/nurse/health-worker is exposed to more sick peoplethan others in the community. Find out how she/he avoidsgetting sick herself/himself.

3. Conduct a survey in your neighbourhood to find out whatthe three most common diseases are. Suggest three stepsthat could be taken by your local authorities to bring downthe incidence of these diseases.

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SCIENCE188

4. A baby is not able to tell her/his caretakers that she/he issick. What would help us to find out

(a) that the baby is sick?

(b) what is the sickness?

5. Under which of the following conditions is a person most likelyto fall sick?

(a) when she is recovering from malaria.

(b) when she has recovered from malaria and is taking care ofsomeone suffering from chicken-pox.

(c) when she is on a four-day fast after recovering frommalaria and is taking care of someone suffering fromchicken-pox.

Why?

6. Under which of the following conditions are you most likelyto fall sick?

(a) when you are taking examinations.

(b) when you have travelled by bus and train for two days.

(c) when your friend is suffering from measles.

Why?

2019-2020

R. N. FOUNDATION LITTLE HEART SCHOOL KEDO, NARWAPAHAR, Affiliated to C.B.S.E., NEW DELHI

(Students are requested to do the work in any notebook neatly)

ASSIGNMENT FOR STD – IX

SUBJECT: BIOLOGY

Qla. Classify infectious agents into different categories and also mention the diseases caused by

them.

Qlb. What are the different means by which infectious diseases are spread?

SUBJECT: CHEMISTRY

TOPIC:MATTER

1. Describe in brief:

* Physical nature of matter

* States of matter and change of states of matter

* Evaporation

* Uses of evaporation in daily life (Draw pictures/diagram as required. OR stick pictures)

SUBJECT: MATHEMATICS

1.Complete chapter from text on Real, Rational and Irrational numbers.

2.Complete chapter on quadrilaterals from the text book. 3.Complete chapter on Triangles from the text book.

SUBJECT: ENGLISH

I. Write an Article on the following topics in about l50 words:-

l. Hard work is the key to success.

2. Education for All.

3. National Integration. The need of the Hour.

4. The need to save water

5. Terrorism – A Big Threat to Humanity

6. Inculcating Hard Values in Children

II. Write 4 original stories in about l50 – 200 words.

SUBJECT: PHYSICS

TOPIC: “MOTION”

SUB TOPIC: l. Distance and Displacement

2. Speed

3. Velocity

4. Uniform and Non-Uniform motion along a straight line

5. Acceleration

6. Distance – time and velocity – time graphs for uniform motion and uniformly accelerated motion.

Explain and write the definition, draw the diagrams, give some examples related to the topics etc.,

also write the mathematical formula and its S.I .units.

SUBJECT: HINDI

In everyday life, we see some objects at rest

and others in motion. Birds fly, fish swim,

blood flows through veins and arteries, and

cars move. Atoms, molecules, planets, stars

and galaxies are all in motion. We often

perceive an object to be in motion when its

position changes with time. However, there

are situations where the motion is inferred

through indirect evidences. For example, we

infer the motion of air by observing the

movement of dust and the movement of leaves

and branches of trees. What causes the

phenomena of sunrise, sunset and changing

of seasons? Is it due to the motion of the

earth? If it is true, why don’t we directly

perceive the motion of the earth?

An object may appear to be moving for

one person and stationary for some other. For

the passengers in a moving bus, the roadside

trees appear to be moving backwards. A

person standing on the road–side perceives

the bus alongwith the passengers as moving.

However, a passenger inside the bus sees his

fellow passengers to be at rest. What do these

observations indicate?

Most motions are complex. Some objects

may move in a straight line, others may take

a circular path. Some may rotate and a few

others may vibrate. There may be situations

involving a combination of these. In this

chapter, we shall first learn to describe the

motion of objects along a straight line. We

shall also learn to express such motions

through simple equations and graphs. Later,

we shall discuss ways of describing circular

motion.

Activity ______________ 8.1

• Discuss whether the walls of yourclassroom are at rest or in motion.

Activity ______________ 8.2

• Have you ever experienced that thetrain in which you are sitting appearsto move while it is at rest?

• Discuss and share your experience.

Think and Act

We sometimes are endangered by the

motion of objects around us, especially

if that motion is erratic and

uncontrolled as observed in a flooded

river, a hurricane or a tsunami. On the

other hand, controlled motion can be a

service to human beings such as in the

generation of hydro-electric power. Do

you feel the necessity to study the

erratic motion of some objects and

learn to control them?

8.1 Describing Motion

We describe the location of an object by

specifying a reference point. Let us

understand this by an example. Let us

assume that a school in a village is 2 km north

of the railway station. We have specified the

position of the school with respect to the

railway station. In this example, the railway

station is the reference point. We could have

also chosen other reference points according

to our convenience. Therefore, to describe the

position of an object we need to specify a

reference point called the origin.

8

MMMMMOTIONOTIONOTIONOTIONOTION

Chapter

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8.1.1 MOTION ALONG A STRAIGHT LINE

The simplest type of motion is the motionalong a straight line. We shall first learn todescribe this by an example. Consider themotion of an object moving along a straightpath. The object starts its journey from Owhich is treated as its reference point(Fig. 8.1). Let A, B and C represent theposition of the object at different instants. Atfirst, the object moves through C and B andreaches A. Then it moves back along the samepath and reaches C through B.

displacement, are used to describe the overallmotion of an object and to locate its finalposition with reference to its initial position

at a given time.

Activity ______________ 8.3

• Take a metre scale and a long rope.• Walk from one corner of a basket-ball

court to its oppposite corner along itssides.

• Measure the distance covered by youand magnitude of the displacement.

• What difference would you noticebetween the two in this case?

Activity ______________ 8.4

• Automobiles are fitted with a devicethat shows the distance travelled. Sucha device is known as an odometer. Acar is driven from Bhubaneshwar toNew Delhi. The difference between thefinal reading and the initial reading ofthe odometer is 1850 km.

• Find the magnitude of the displacementbetween Bhubaneshwar and New Delhiby using the Road Map of India.

The total path length covered by the objectis OA + AC, that is 60 km + 35 km = 95 km.This is the distance covered by the object. Todescribe distance we need to specify only thenumerical value and not the direction ofmotion. There are certain quantities whichare described by specifying only theirnumerical values. The numerical value of aphysical quantity is its magnitude. From thisexample, can you find out the distance of thefinal position C of the object from the initialposition O? This difference will give you thenumerical value of the displacement of theobject from O to C through A. The shortestdistance measured from the initial to the finalposition of an object is known as thedisplacement.

Can the magnitude of the displacementbe equal to the distance travelled by anobject? Consider the example given in(Fig. 8.1). For motion of the object from O toA, the distance covered is 60 km and themagnitude of displacement is also 60 km.During its motion from O to A and back to B,the distance covered = 60 km + 25 km = 85 km

Fig. 8.1: Positions of an object on a straight line path

while the magnitude of displacement = 35 km.Thus, the magnitude of displacement (35 km)is not equal to the path length (85 km).Further, we will notice that the magnitude ofthe displacement for a course of motion maybe zero but the corresponding distancecovered is not zero. If we consider the objectto travel back to O, the final position concideswith the initial position, and therefore, thedisplacement is zero. However, the distancecovered in this journey is OA + AO = 60 km +60 km = 120 km. Thus, two different physicalquantities — the distance and the

MOTION 99

2019-2020

SCIENCE100

uestions1. An object has moved through a

distance. Can it have zero

displacement? If yes, support

your answer with an example.

2. A farmer moves along the

boundary of a square field of side

10 m in 40 s. What will be the

magnitude of displacement of the

farmer at the end of 2 minutes 20

seconds from his initial position?

3. Which of the following is true for

displacement?

(a) It cannot be zero.

(b) Its magnitude is greater than

the distance travelled by the

object.

8.1.2 UNIFORM MOTION AND NON-UNIFORM MOTION

Consider an object moving along a straight

line. Let it travel 5 m in the first second,

5 m more in the next second, 5 m in the

third second and 5 m in the fourth second.

In this case, the object covers 5 m in each

second. As the object covers equal distances

in equal intervals of time, it is said to be in

uniform motion. The time interval in this

motion should be small. In our day-to-day

life, we come across motions where objects

cover unequal distances in equal intervals

of time, for example, when a car is moving

on a crowded street or a person is jogging

in a park. These are some instances of

non-uniform motion.

Activity ______________ 8.5

• The data regarding the motion of two

different objects A and B are given in

Table 8.1.

• Examine them carefully and state

whether the motion of the objects is

uniform or non-uniform.

Q

(a)

(b)

Fig. 8.2

Table 8.1

Time Distance Distance

travelled by travelled by

object A in m object B in m

9:30 am 10 12

9:45 am 20 19

10:00 am 30 23

10:15 am 40 35

10:30 am 50 37

10:45 am 60 41

11:00 am 70 44

8.2 Measuring the Rate of Motion

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MOTION 101

Look at the situations given in Fig. 8.2. Ifthe bowling speed is 143 km h–1 in Fig. 8.2(a)

what does it mean? What do you understandfrom the signboard in Fig. 8.2(b)?

Different objects may take different

amounts of time to cover a given distance.Some of them move fast and some move

slowly. The rate at which objects move can

be different. Also, different objects can moveat the same rate. One of the ways of

measuring the rate of motion of an object isto find out the distance travelled by the objectin unit time. This quantity is referred to as

speed. The SI unit of speed is metre persecond. This is represented by the symbolm s–1 or m/s. The other units of speed include

centimetre per second (cm s–1) and kilometreper hour (km h–1). To specify the speed of anobject, we require only its magnitude. The

speed of an object need not be constant. Inmost cases, objects will be in non-uniformmotion. Therefore, we describe the rate of

motion of such objects in terms of theiraverage speed. The average speed of an objectis obtained by dividing the total distance

travelled by the total time taken. That is,

average speed = Total distance travelled

Total time taken

If an object travels a distance s in time t thenits speed v is,

v = s

t(8.1)

Let us understand this by an example. Acar travels a distance of 100 km in 2 h. Itsaverage speed is 50 km h–1. The car mightnot have travelled at 50 km h–1 all the time.Sometimes it might have travelled faster andsometimes slower than this.

Example 8.1 An object travels 16 m in 4 s

and then another 16 m in 2 s. What isthe average speed of the object?

Solution:

Total distance travelled by the object =16 m + 16 m = 32 mTotal time taken = 4 s + 2 s = 6 s

Average speed = Total distance travelled

Total time taken

= 32 m

6 s = 5.33 m s–1

Therefore, the average speed of the objectis 5.33 m s–1.

8.2.1 SPEED WITH DIRECTION

The rate of motion of an object can be more

comprehensive if we specify its direction of

motion along with its speed. The quantity that

specifies both these aspects is called velocity.

Velocity is the speed of an object moving in a

definite direction. The velocity of an object

can be uniform or variable. It can be changed

by changing the object’s speed, direction of

motion or both. When an object is moving

along a straight line at a variable speed, we

can express the magnitude of its rate of

motion in terms of average velocity. It is

calculated in the same way as we calculate

average speed.

In case the velocity of the object is

changing at a uniform rate, then average

velocity is given by the arithmetic mean of

initial velocity and final velocity for a given

period of time. That is,

average velocity =

initial velocity + final velocity

2

Mathematically, vav

= u + v

2(8.2)

where vav

is the average velocity, u is the initial

velocity and v is the final velocity of the object.

Speed and velocity have the same units,

that is, m s–1 or m/s.

Activity ______________ 8.6

• Measure the time it takes you to walk

from your house to your bus stop or

the school. If you consider that your

average walking speed is 4 km h–1,

estimate the distance of the bus stop

or school from your house.

2019-2020

SCIENCE102

= 50km 1000m 1h

× ×h 1km 3600s

= 13.9 m s–1

The average speed of the car is50 km h–1 or 13.9 m s–1.

Example 8.3 Usha swims in a 90 m longpool. She covers 180 m in one minute

by swimming from one end to the otherand back along the same straight path.Find the average speed and average

velocity of Usha.Solution:

Total distance covered by Usha in 1 minis 180 m.

Displacement of Usha in 1 min = 0 m

Average speed = Total distance covered

Total timetaken

=180m 180 m 1 min

= ×1min 1min 60s

= 3 m s-1

Average velocity =

Displacement

Total timetaken

=

0m

60 s

= 0 m s–1

The average speed of Usha is 3 m s–1

and her average velocity is 0 m s–1.

8.3 Rate of Change of Velocity

During uniform motion of an object along a

straight line, the velocity remains constant

with time. In this case, the change in velocity

of the object for any time interval is zero.

However, in non-uniform motion, velocity

varies with time. It has different values at

different instants and at different points of

the path. Thus, the change in velocity of the

object during any time interval is not zero.

Can we now express the change in velocity of

an object?

Activity ______________ 8.7

• At a time when it is cloudy, there maybe frequent thunder and lightning. Thesound of thunder takes some time toreach you after you see the lightning.

• Can you answer why this happens?• Measure this time interval using a

digital wrist watch or a stop watch.• Calculate the distance of the nearest

point of lightning. (Speed of sound inair = 346 m s-1.)

uestions1. Distinguish between speed and

velocity.

2. Under what condition(s) is the

magnitude of average velocity of

an object equal to its average

speed?

3. What does the odometer of an

automobile measure?

4. What does the path of an object

look like when it is in uniform

motion?

5. During an experiment, a signal

from a spaceship reached the

ground station in five minutes.

What was the distance of the

spaceship from the ground

station? The signal travels at the

speed of light, that is, 3 × 108

m s–1.

Example 8.2 The odometer of a car reads

2000 km at the start of a trip and2400 km at the end of the trip. If thetrip took 8 h, calculate the average

speed of the car in km h–1 and m s–1.

Solution:

Distance covered by the car,s = 2400 km – 2000 km = 400 kmTime elapsed, t = 8 hAverage speed of the car is,

vav

= 400 km

8 h=

s

t

= 50 km h–1

Q

2019-2020

MOTION 103

To answer such a question, we have tointroduce another physical quantity calledacceleration, which is a measure of the

change in the velocity of an object per unittime. That is,

acceleration =change in velocity

time taken

If the velocity of an object changes froman initial value u to the final value v in time t,

the acceleration a is,

v – ua =

t(8.3)

This kind of motion is known asaccelerated motion. The acceleration is taken

to be positive if it is in the direction of velocityand negative when it is opposite to thedirection of velocity. The SI unit of

acceleration is m s–2 .If an object travels in a straight line and

its velocity increases or decreases by equal

amounts in equal intervals of time, then theacceleration of the object is said to beuniform. The motion of a freely falling body

is an example of uniformly acceleratedmotion. On the other hand, an object cantravel with non-uniform acceleration if its

velocity changes at a non-uniform rate. Forexample, if a car travelling along a straightroad increases its speed by unequal amounts

in equal intervals of time, then the car is saidto be moving with non-uniform acceleration.

Activity ______________ 8.8

• In your everyday life you come across

a range of motions in which(a) acceleration is in the direction of

motion,

(b) acceleration is against thedirection of motion,

(c) acceleration is uniform,

(d) acceleration is non-uniform.

• Can you identify one example each

for the above type of motion?

Example 8.4 Starting from a stationary

position, Rahul paddles his bicycle to

attain a velocity of 6 m s–1 in 30 s. Thenhe applies brakes such that the velocity

of the bicycle comes down to 4 m s-1 inthe next 5 s. Calculate the acceleration

of the bicycle in both the cases.

Solution:

In the first case:

initial velocity, u = 0 ;final velocity, v = 6 m s–1 ;time, t = 30 s .

From Eq. (8.3), we have

v – ua =

t

Substituting the given values of u,v andt in the above equation, we get

( )–1 –16m s – 0m s

=30 s

a

= 0.2 m s–2

In the second case:initial velocity, u = 6 m s–1;final velocity, v = 4 m s–1;

time, t = 5 s.

Then, ( )–1 –14m s – 6m s

=5 s

a

= –0.4 m s–2 .

The acceleration of the bicycle in the

first case is 0.2 m s–2 and in the secondcase, it is –0.4 m s–2.

uestions1. When will you say a body is in

(i) uniform acceleration? (ii) non-

uniform acceleration?

2. A bus decreases its speed from

80 km h–1 to 60 km h–1 in 5 s.

Find the acceleration of the bus.

3. A train starting from a railway

station and moving with uniform

acceleration attains a speed

40 km h–1 in 10 minutes. Find its

acceleration.

Q

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SCIENCE104

8.4 Graphical Representation of

Motion

Graphs provide a convenient method topresent basic information about a variety ofevents. For example, in the telecast of aone-day cricket match, vertical bar graphsshow the run rate of a team in each over. Asyou have studied in mathematics, a straightline graph helps in solving a linear equationhaving two variables.

To describe the motion of an object, wecan use line graphs. In this case, line graphsshow dependence of one physical quantity,such as distance or velocity, on anotherquantity, such as time.

8.4.1 DISTANCE–TIME GRAPHS

The change in the position of an object withtime can be represented on the distance-timegraph adopting a convenient scale of choice.In this graph, time is taken along the x–axisand distance is taken along the y-axis.Distance-time graphs can be employed undervarious conditions where objects move withuniform speed, non-uniform speed, remainat rest etc.

Fig. 8.3: Distance-time graph of an object moving

with uniform speed

We know that when an object travels equaldistances in equal intervals of time, it moveswith uniform speed. This shows that the

distance travelled by the object is directlyproportional to time taken. Thus, for uniformspeed, a graph of distance travelled againsttime is a straight line, as shown in Fig. 8.3.The portion OB of the graph shows that thedistance is increasing at a uniform rate. Notethat, you can also use the term uniformvelocity in place of uniform speed if you takethe magnitude of displacement equal to thedistance travelled by the object along they-axis.

We can use the distance-time graph todetermine the speed of an object. To do so,consider a small part AB of the distance-timegraph shown in Fig 8.3. Draw a line parallelto the x-axis from point A and another lineparallel to the y-axis from point B. These twolines meet each other at point C to form atriangle ABC. Now, on the graph, AC denotesthe time interval (t

2 – t

1) while BC corresponds

to the distance (s2 – s

1). We can see from the

graph that as the object moves from the pointA to B, it covers a distance (s

2 – s

1) in time

(t2 – t

1). The speed, v of the object, therefore

can be represented as

v = 2 1

2 1

–

–

s s

t t(8.4)

We can also plot the distance-time graphfor accelerated motion. Table 8.2 shows thedistance travelled by a car in a time intervalof two seconds.

Table 8.2: Distance travelled by acar at regular time intervals

Time in seconds Distance in metres

0 0

2 1

4 4

6 9

8 16

10 25

12 36

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MOTION 105

The distance-time graph for the motionof the car is shown in Fig. 8.4. Note that theshape of this graph is different from the earlierdistance-time graph (Fig. 8.3) for uniformmotion. The nature of this graph shows non-linear variation of the distance travelled bythe car with time. Thus, the graph shown inFig 8.4 represents motion with non-uniformspeed.

8.4.2 VELOCITY-TIME GRAPHS

The variation in velocity with time for anobject moving in a straight line can berepresented by a velocity-time graph. In thisgraph, time is represented along the x-axis

Fig. 8.4: Distance-time graph for a car moving with

non-uniform speed

Fig. 8.5: Velocity-time graph for uniform motion of

a car

and the velocity is represented along they-axis. If the object moves at uniform velocity,the height of its velocity-time graph will notchange with time (Fig. 8.5). It will be a straightline parallel to the x-axis. Fig. 8.5 shows thevelocity-time graph for a car moving withuniform velocity of 40 km h–1.

We know that the product of velocity andtime give displacement of an object movingwith uniform velocity. The area enclosed byvelocity-time graph and the time axis will beequal to the magnitude of the displacement.

To know the distance moved by the carbetween time t

1 and t

2 using Fig. 8.5, draw

perpendiculars from the points correspondingto the time t

1 and t2 on the graph. The velocity

of 40 km h–1 is represented by the height ACor BD and the time (t

2 – t

1) is represented by

the length AB.So, the distance s moved by the car in

time (t2 – t

1) can be expressed as

s = AC × CD= [(40 km h–1) × (t

2 – t

1) h]

= 40 (t2– t

1) km

= area of the rectangle ABDC (shadedin Fig. 8.5).

We can also study about uniformlyaccelerated motion by plotting its velocity–time graph. Consider a car being driven alonga straight road for testing its engine. Supposea person sitting next to the driver records itsvelocity after every 5 seconds by noting thereading of the speedometer of the car. Thevelocity of the car, in km h–1 as well as inm s–1, at different instants of time is shownin table 8.3.

Table 8.3: Velocity of a car atregular instants of time

Time Velocity of the car (s) (m s–1) (km h

–1)

0 0 0

5 2.5 9

10 5.0 18

15 7.5 27

20 10.0 36

25 12.5 45

30 15.0 54

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SCIENCE106

In this case, the velocity-time graph for themotion of the car is shown in Fig. 8.6. Thenature of the graph shows that velocitychanges by equal amounts in equal intervalsof time. Thus, for all uniformly acceleratedmotion, the velocity-time graph is astraight line.

Fig. 8.6: Velocity-time graph for a car moving with

uniform accelerations.

You can also determine the distancemoved by the car from its velocity-time graph.The area under the velocity-time graph givesthe distance (magnitude of displacement)moved by the car in a given interval of time.If the car would have been moving withuniform velocity, the distance travelled by itwould be represented by the area ABCDunder the graph (Fig. 8.6). Since themagnitude of the velocity of the car ischanging due to acceleration, the distance stravelled by the car will be given by the areaABCDE under the velocity-time graph(Fig. 8.6).That is,

s = area ABCDE= area of the rectangle ABCD + area of

the triangle ADE

= AB × BC +

1

2(AD × DE)

In the case of non-uniformly acceleratedmotion, velocity-time graphs can have anyshape.

Fig. 8.7: Velocity-time graphs of an object in non-

uniformly accelerated motion.

Fig. 8.7(a) shows a velocity-time graphthat represents the motion of an object whosevelocity is decreasing with time whileFig. 8.7 (b) shows the velocity-time graphrepresenting the non-uniform variation ofvelocity of the object with time. Try to interpretthese graphs.

Activity ______________ 8.9

• The times of arrival and departure of atrain at three stations A, B and C andthe distance of stations B and C fromstation A are given in table 8.4.

Table 8.4: Distances of stations Band C from A and times of arrivaland departure of the train

Station Distance Time of Time offrom A arrival departure(km) (hours) (hours)

A 0 08:00 08:15B 120 11:15 11:30C 180 13:00 13:15

• Plot and interpret the distance-time

graph for the train assuming that itsmotion between any two stations isuniform.

Velo

cit

y (km

h–1)

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MOTION 107

8.5 Equations of Motion byGraphical Method

When an object moves along a straight linewith uniform acceleration, it is possible torelate its velocity, acceleration during motionand the distance covered by it in a certaintime interval by a set of equations known asthe equations of motion. For convenience, aset of three such equations are given below:

v = u + at (8.5)s = ut + ½ at2 (8.6)

2 a s = v2 – u2 (8.7)where u is the initial velocity of the objectwhich moves with uniform acceleration a fortime t, v is the final velocity, and s is thedistance travelled by the object in time t.Eq. (8.5) describes the velocity-time relationand Eq. (8.6) represents the position-timerelation. Eq. (8.7), which represents therelation between the position and the velocity,can be obtained from Eqs. (8.5) and (8.6) byeliminating t. These three equations can bederived by graphical method.

8.5.1 EQUATION FOR VELOCITY-TIME

RELATION

Consider the velocity-time graph of an objectthat moves under uniform acceleration as

Activity _____________8.10

• Feroz and his sister Sania go to schoolon their bicycles. Both of them start atthe same time from their home but takedifferent times to reach the schoolalthough they follow the same route.Table 8.5 shows the distance travelledby them in different times

Table 8.5: Distance covered by Ferozand Sania at different times ontheir bicycles

Time Distance Distancetravelled travelledby Feroz by Sania

(km) (km)

8:00 am 0 0

8:05 am 1.0 0.8

8:10 am 1.9 1.6

8:15 am 2.8 2.3

8:20 am 3.6 3.0

8:25 am – 3.6

Q

• Plot the distance-time graph for theirmotions on the same scale andinterpret.

uestions1. What is the nature of the

distance-time graphs for uniform

and non-uniform motion of an

object?

2. What can you say about the

motion of an object whose

distance-time graph is a straight

line parallel to the time axis?

3. What can you say about the

motion of an object if its speed-

time graph is a straight line

parallel to the time axis?

4. What is the quantity which is

measured by the area occupied

below the velocity-time graph?Fig. 8.8: Velocity-time graph to obtain the equations

of motion

2019-2020

SCIENCE108

= OA × OC + 1

2 (AD × BD) (8.10)

Substituting OA = u, OC = AD = t and BD = at,

we get

s = u × t + 1

( )2

t ×at

or s = u t + 1

2a t 2

8.5.3 EQUATION FOR POSITION–VELOCITY

RELATION

From the velocity-time graph shown in

Fig. 8.8, the distance s travelled by the object

in time t, moving under uniform acceleration

a is given by the area enclosed within the

trapezium OABC under the graph. That is,

s = area of the trapezium OABC

= ( )OA + BC ×OC

2

Substituting OA = u, BC = v and OC = t,

we get

( )2

=u + v t

s (8.11)

From the velocity-time relation (Eq. 8.6),

we get

( )v – u

t =a

(8.12)

Using Eqs. (8.11) and (8.12) we have

( ) ( )×v + u v - us=

2a

or 2 a s = v2 – u2

Example 8.5 A train starting from rest

attains a velocity of 72 km h–1 in

5 minutes. Assuming that the

acceleration is uniform, find (i) the

acceleration and (ii) the distance

travelled by the train for attaining this

velocity.

shown in Fig. 8.8 (similar to Fig. 8.6, but nowwith u ≠ 0). From this graph, you can see thatinitial velocity of the object is u (at point A)and then it increases to v (at point B) in timet. The velocity changes at a uniform rate a.In Fig. 8.8, the perpendicular lines BC andBE are drawn from point B on the time andthe velocity axes respectively, so that theinitial velocity is represented by OA, the finalvelocity is represented by BC and the timeinterval t is represented by OC. BD = BC –CD, represents the change in velocity in timeinterval t.

Let us draw AD parallel to OC. From thegraph, we observe that

BC = BD + DC = BD + OASubstituting BC = v and OA = u,

we get v = BD + u

or BD = v – u (8.8)From the velocity-time graph (Fig. 8.8),

the acceleration of the object is given by

a = Change in velocity

time taken

=BD BD

=AD OC

Substituting OC = t, we get

a = BD

t

or BD = at (8.9)

Using Eqs. (8.8) and (8.9) we getv = u + at

8.5.2 EQUATION FOR POSITION-TIME

RELATION

Let us consider that the object has travelleda distance s in time t under uniformacceleration a. In Fig. 8.8, the distancetravelled by the object is obtained by the areaenclosed within OABC under the velocity-timegraph AB.

Thus, the distance s travelled by the objectis given by

s = area OABC (which is a trapezium)= area of the rectangle OADC + area of

the triangle ABD

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MOTION 109

Solution:

We have been givenu = 0 ; v = 72 km h–1 = 20 m s-1 andt = 5 minutes = 300 s.(i) From Eq. (8.5) we know that

( )v – ua =

t

–1 –1

–2

20 m s – 0 m s=

300s

1= m s

15

(ii) From Eq. (8.7) we have2 a s = v2 – u2 = v2 – 0Thus,

2

–1 2

–2

=2

(20 m s )=

2×(1/15) m s

vs

a

= 3000 m

= 3 km

The acceleration of the train is 1

15m s– 2

and the distance travelled is 3 km.

Example 8.6 A car accelerates uniformly

from 18 km h–1

to 36 km h–1

in 5 s.

Calculate (i) the acceleration and (ii) the

distance covered by the car in that time.

Solution:

We are given thatu = 18 km h–1 = 5 m s–1

v = 36 km h–1 = 10 m s–1 andt = 5 s .

(i) From Eq. (8.5) we have

v – ua =

t

=

-1 -110 m s – 5 m s

5s

= 1 m s–2

(ii) From Eq. (8.6) we have

s = u t + 1

2a t 2

= 5 m s–1 × 5 s + 1

2× 1 m s–2 × (5 s)2

= 25 m + 12.5 m

= 37.5 m

The acceleration of the car is 1 m s–2

and the distance covered is 37.5 m.

Example 8.7 The brakes applied to a car

produce an acceleration of 6 m s-2 in

the opposite direction to the motion. If

the car takes 2 s to stop after the

application of brakes, calculate the

distance it travels during this time.

Solution:

We have been givena = – 6 m s–2 ; t = 2 s and v = 0 m s–1.From Eq. (8.5) we know that

v = u + at

0 = u + (– 6 m s–2) × 2 s or u = 12 m s–1 .

From Eq. (8.6) we get

s = u t + 1

2 a t 2

= (12 m s–1 ) × (2 s) + 1

2 (–6 m s–2 ) (2 s)2

= 24 m – 12 m= 12 m

Thus, the car will move 12 m before itstops after the application of brakes.Can you now appreciate why driversare cautioned to maintain somedistance between vehicles whiletravelling on the road?

uestions

1. A bus starting from rest moves

with a uniform acceleration of

0.1 m s-2 for 2 minutes. Find (a)

the speed acquired, (b) the

distance travelled.Q2019-2020

SCIENCE110

2. A train is travelling at a speedof 90 km h–1. Brakes are applied

so as to produce a uniformacceleration of – 0.5 m s-2. Findhow far the train will go before itis brought to rest.

3. A trolley, while going down aninclined plane, has anacceleration of 2 cm s-2. What willbe its velocity 3 s after the start?

4. A racing car has a uniformacceleration of 4 m s-2. Whatdistance will it cover in 10 s afterstart?

5. A stone is thrown in a verticallyupward direction with a velocityof 5 m s-1. If the acceleration of thestone during its motion is 10 m s–2

in the downward direction, whatwill be the height attained by thestone and how much time will ittake to reach there?

8.6 Uniform Circular Motion

When the velocity of an object changes, wesay that the object is accelerating. The changein the velocity could be due to change in itsmagnitude or the direction of the motion orboth. Can you think of an example when anobject does not change its magnitude ofvelocity but only its direction of motion?

Let us consider an example of the motion

of a body along a closed path. Fig 8.9 (a) shows

the path of an athlete along a rectangular

track ABCD. Let us assume that the athlete

runs at a uniform speed on the straight parts

AB, BC, CD and DA of the track. In order to

keep himself on track, he quickly changes

his speed at the corners. How many times

will the athlete have to change his direction

of motion, while he completes one round? It

is clear that to move in a rectangular track

once, he has to change his direction of motion

four times.

Now, suppose instead of a rectangular

track, the athlete is running along a

hexagonal shaped path ABCDEF, as shown

in Fig. 8.9(b). In this situation, the athlete

will have to change his direction six times

while he completes one round. What if the

track was not a hexagon but a regular

octagon, with eight equal sides as shown by

ABCDEFGH in Fig. 8.9(c)? It is observed that

as the number of sides of the track increases

the athelete has to take turns more and more

often. What would happen to the shape of

the track as we go on increasing the number

of sides indefinitely? If you do this you will

notice that the shape of the track approaches

the shape of a circle and the length of each of

the sides will decrease to a point. If the athlete

moves with a velocity of constant magnitude

along the circular path, the only change in

his velocity is due to the change in the

direction of motion. The motion of the athlete

moving along a circular path is, therefore, an

example of an accelerated motion.

We know that the circumference of a circle

of radius r is given by π2 r . If the athlete takes

t seconds to go once around the circular path

of radius r, the speed v is given by

π2 rv =

t(8.13)

When an object moves in a circular path

with uniform speed, its motion is called

uniform circular motion.

(a) Rectangular track (b) Hexagonal track

(d) A circular track(c) Octagonal shaped track

Fig. 8.9: The motion of an athlete along closed tracks

of different shapes.

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MOTION 111

If you carefully note, on being released

the stone moves along a straight line

tangential to the circular path. This is

because once the stone is released, it

continues to move along the direction it has

been moving at that instant. This shows that

the direction of motion changed at every point

when the stone was moving along the circular

path.

When an athlete throws a hammer or a

discus in a sports meet, he/she holds the

hammer or the discus in his/her hand and

gives it a circular motion by rotating his/her

own body. Once released in the desired

direction, the hammer or discus moves in the

direction in which it was moving at the time

it was released, just like the piece of stone in

the activity described above. There are many

more familiar examples of objects moving

under uniform circular motion, such as the

motion of the moon and the earth, a satellite

in a circular orbit around the earth, a cyclist

on a circular track at constant speed

and so on.

Activity _____________8.11

• Take a piece of thread and tie a small

piece of stone at one of its ends. Movethe stone to describe a circular pathwith constant speed by holding thethread at the other end, as shown inFig. 8.10.

Fig. 8.10: A stone describing a circular path with

a velocity of constant magnitude.

• Now, let the stone go by releasing the

thread.

• Can you tell the direction in which the

stone moves after it is released?

• By repeating the activity for a few times

and releasing the stone at differentpositions of the circular path, checkwhether the direction in which thestone moves remains the same or not.

Whatyou havelearnt• Motion is a change of position; it can be described in terms of

the distance moved or the displacement.

• The motion of an object could be uniform or non-uniformdepending on whether its velocity is constant or changing.

• The speed of an object is the distance covered per unit time,and velocity is the displacement per unit time.

• The acceleration of an object is the change in velocity per unittime.

• Uniform and non-uniform motions of objects can be shownthrough graphs.

• The motion of an object moving at uniform acceleration can bedescribed with the help of the following equations, namely

v = u + at

s = ut + ½ at2

2as = v2 – u2

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SCIENCE112

where u is initial velocity of the object, which moves withuniform acceleration a for time t, v is its final velocity and s isthe distance it travelled in time t.

• If an object moves in a circular path with uniform speed, itsmotion is called uniform circular motion.

Exercises1. An athlete completes one round of a circular track of diameter

200 m in 40 s. What will be the distance covered and thedisplacement at the end of 2 minutes 20 s?

2. Joseph jogs from one end A to the other end B of a straight300 m road in 2 minutes 30 seconds and then turns aroundand jogs 100 m back to point C in another 1 minute. What areJoseph’s average speeds and velocities in jogging (a) from A toB and (b) from A to C?

3. Abdul, while driving to school, computes the average speed forhis trip to be 20 km h–1. On his return trip along the sameroute, there is less traffic and the average speed is30 km h–1. What is the average speed for Abdul’s trip?

4. A motorboat starting from rest on a lake accelerates in a straightline at a constant rate of 3.0 m s–2 for 8.0 s. How far does theboat travel during this time?

5. A driver of a car travelling at 52 km h–1 applies the brakes andaccelerates uniformly in the opposite direction. The car stopsin 5 s. Another driver going at 3 km h–1 in another car applieshis brakes slowly and stops in 10 s. On the same graph paper,plot the speed versus time graphs for the two cars. Which ofthe two cars travelled farther after the brakes were applied?

6. Fig 8.11 shows the distance-time graph of three objects A,Band C. Study the graph and answer the following questions:

Fig. 8.11

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MOTION 113

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

7. A ball is gently dropped from a height of 20 m. If its velocityincreases uniformly at the rate of 10 m s-2, with what velocitywill it strike the ground? After what time will it strike theground?

8. The speed-time graph for a car is shown is Fig. 8.12.

Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds.Shade the area on the graph that represents the distancetravelled by the car during the period.

(b) Which part of the graph represents uniform motion ofthe car?

9. State which of the following situations are possible and givean example for each of these:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving with an acceleration but with uniformspeed.

(c) an object moving in a certain direction with anacceleration in the perpendicular direction.

10. An artificial satellite is moving in a circular orbit of radius42250 km. Calculate its speed if it takes 24 hours to revolvearound the earth.

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NUMBER SYSTEMS 1

CHAPTER 1

NUMBER SYSTEMS

1.1 Introduction

In your earlier classes, you have learnt about the number line and how to represent

various types of numbers on it (see Fig. 1.1).

Fig. 1.1 : The number line

Just imagine you start from zero and go on walking along this number line in the

positive direction. As far as your eyes can see, there are numbers, numbers and

numbers!

Fig. 1.2

Now suppose you start walking along the number line, and collecting some of the

numbers. Get a bag ready to store them!

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2 MATHEMATICS

3

-40

166

22-75 2

1 9

0Z3

40

16

74

5

2601

422

58

0

-3-757

-66-21

-40

31

71

340

16

745 2

601 2958

0W

940

16

74

5

2601

4652

58

031

1

7110

N

You might begin with picking up only natural

numbers like 1, 2, 3, and so on. You know that this list

goes on for ever. (Why is this true?) So, now your

bag contains infinitely many natural numbers! Recall

that we denote this collection by the symbol N.

Now turn and walk all the way back, pick up

zero and put it into the bag. You now have the

collection of whole numbers which is denoted by

the symbol W.

Now, stretching in front of you are many, many negative integers. Put all the

negative integers into your bag. What is your new collection? Recall that it is the

collection of all integers, and it is denoted by the symbol Z.

Are there some numbers still left on the line? Of course! There are numbers like

1 3,

2 4, or even

2005

2006

−. If you put all such numbers also into the bag, it will now be the

Z comes from the

German word

“zahlen”, which means

“to count”.

Q

–672112

13

–19

8116 1

4

20052006 –12

13

9

14

–6625

-65

60

19

19

999

0–6

727 58

20052006

3

–5

16

60

999

4

–8–6625

58

0

27

71

17981

–1213

89

–67

2

3

9

14

–

Why Z ?

2019-2020

NUMBER SYSTEMS 3

collection of rational numbers. The collection of rational numbers is denoted by Q.

‘Rational’ comes from the word ‘ratio’, and Q comes from the word ‘quotient’.

You may recall the definition of rational numbers:

A number ‘r’ is called a rational number, if it can be written in the form p

q,

where p and q are integers and q ≠ 0. (Why do we insist that q ≠ 0?)

Notice that all the numbers now in the bag can be written in the form p

q, where p

and q are integers and q ≠ 0. For example, –25 can be written as 25

;1

− here p = –25

and q = 1. Therefore, the rational numbers also include the natural numbers, whole

numbers and integers.

You also know that the rational numbers do not have a unique representation in

the form p

q, where p and q are integers and q ≠ 0. For example,

1

2 =

2

4 =

10

20 =

25

50

= 47

94, and so on. These are equivalent rational numbers (or fractions). However,

when we say that p

q is a rational number, or when we represent

p

q on the number

line, we assume that q ≠ 0 and that p and q have no common factors other than 1

(that is, p and q are co-prime). So, on the number line, among the infinitely many

fractions equivalent to 1

2, we will choose

1

2 to represent all of them.

Now, let us solve some examples about the different types of numbers, which you

have studied in earlier classes.

Example 1 : Are the following statements true or false? Give reasons for your answers.

(i) Every whole number is a natural number.

(ii) Every integer is a rational number.

(iii) Every rational number is an integer.

Solution : (i) False, because zero is a whole number but not a natural number.

(ii) True, because every integer m can be expressed in the form 1

m, and so it is a

rational number.

2019-2020

4 MATHEMATICS

(iii) False, because 3

5 is not an integer.

Example 2 : Find five rational numbers between 1 and 2.

We can approach this problem in at least two ways.

Solution 1 : Recall that to find a rational number between r and s, you can add r and

s and divide the sum by 2, that is 2

r s+ lies between r and s. So,

3

2 is a number

between 1 and 2. You can proceed in this manner to find four more rational numbers

between 1 and 2. These four numbers are 5 11 13 7 ., , and4 8 8 4

Solution 2 : The other option is to find all the five rational numbers in one step. Since

we want five numbers, we write 1 and 2 as rational numbers with denominator 5 + 1,

i.e., 1 = 6

6 and 2 =

12

6. Then you can check that

7

6,

8

6,

9

6,

10

6 and

11

6 are all rational

numbers between 1 and 2. So, the five numbers are 7 4 3 5 11

,, , and6 3 2 3 6

.

Remark : Notice that in Example 2, you were asked to find five rational numbers

between 1 and 2. But, you must have realised that in fact there are infinitely many

rational numbers between 1 and 2. In general, there are infinitely many rational

numbers between any two given rational numbers.

Let us take a look at the number line again. Have you picked up all the numbers?

Not, yet. The fact is that there are infinitely many more numbers left on the number

line! There are gaps in between the places of the numbers you picked up, and not just

one or two but infinitely many. The amazing thing is that there are infinitely many

numbers lying between any two of these gaps too!

So we are left with the following questions:

1. What are the numbers, that are left on the number

line, called?

2. How do we recognise them? That is, how do we

distinguish them from the rationals (rational

numbers)?

These questions will be answered in the next section.

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NUMBER SYSTEMS 5

EXERCISE 1.1

1. Is zero a rational number? Can you write it in the form p

q, where p and q are integers

and q ≠ 0?

2. Find six rational numbers between 3 and 4.

3. Find five rational numbers between 3

5 and

4

5.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

1.2 Irrational Numbers

We saw, in the previous section, that there may be numbers on the number line that

are not rationals. In this section, we are going to investigate these numbers. So far, all

the numbers you have come across, are of the form p

q, where p and q are integers

and q ≠ 0. So, you may ask: are there numbers which are not of this form? There are

indeed such numbers.

The Pythagoreans in Greece, followers of the famous

mathematician and philosopher Pythagoras, were the first

to discover the numbers which were not rationals, around

400 BC. These numbers are called irrational numbers

(irrationals), because they cannot be written in the form of

a ratio of integers. There are many myths surrounding the

discovery of irrational numbers by the Pythagorean,

Hippacus of Croton. In all the myths, Hippacus has an

unfortunate end, either for discovering that 2 is irrational

or for disclosing the secret about 2 to people outside the

secret Pythagorean sect!

Let us formally define these numbers.

A number ‘s’ is called irrational, if it cannot be written in the form p

q, where p

and q are integers and q ≠ 0.

Pythagoras

(569 BCE – 479 BCE)

Fig. 1.3

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6 MATHEMATICS

20052006

3

–5

16

60

999

4

–8

–6625

58

0

27

71

17981

–12

13

89

–67

2

3

9

14

-65–6626

-45

036

19

R

You already know that there are infinitely many rationals. It turns out that there

are infinitely many irrational numbers too. Some examples are:

2, 3, 15,, π, 0.10110111011110...

Remark : Recall that when we use the symbol , we assume that it is the

positive square root of the number. So 4 = 2, though both 2 and –2 are square

roots of 4.

Some of the irrational numbers listed above are familiar to you. For example, you

have already come across many of the square roots listed above and the number π.

The Pythagoreans proved that 2 is irrational. Later in approximately 425 BC,

Theodorus of Cyrene showed that 3, 5, 6, 7, 10, 11, 12, 13, 14, 15

and 17 are also irrationals. Proofs of irrationality of 2 , 3 , 5 , etc., shall be

discussed in Class X. As to π, it was known to various cultures for thousands of

years, it was proved to be irrational by Lambert and Legendre only in the late 1700s.

In the next section, we will discuss why 0.10110111011110... and π are irrational.

Let us return to the questions raised at the end of

the previous section. Remember the bag of rational

numbers. If we now put all irrational numbers into

the bag, will there be any number left on the number

line? The answer is no! It turns out that the collection

of all rational numbers and irrational numbers together

make up what we call the collection of real numbers,

which is denoted by R. Therefore, a real number is either rational or irrational. So, we

can say that every real number is represented by a unique point on the number

line. Also, every point on the number line represents a unique real number.

This is why we call the number line, the real number line.

In the 1870s two German mathematicians,

Cantor and Dedekind, showed that :

Corresponding to every real number, there is a

point on the real number line, and corresponding

to every point on the number line, there exists a

unique real number.

G. Cantor (1845-1918)

Fig. 1.5R. Dedekind (1831-1916)

Fig. 1.4

2019-2020

NUMBER SYSTEMS 7

Let us see how we can locate some of the irrational numbers on the number line.

Example 3 : Locate 2 on the number line.

Solution : It is easy to see how the Greeks might have discovered

2 . Consider a square OABC, with each side 1 unit in length (see

Fig. 1.6). Then you can see by the Pythagoras theorem that

OB = 2 21 1 2+ = . How do we represent 2 on the number line?

This is easy. Transfer Fig. 1.6 onto the number line making sure that the vertex O

coincides with zero (see Fig. 1.7).

Fig. 1.7

We have just seen that OB = 2 . Using a compass with centre O and radius OB,

draw an arc intersecting the number line at the point P. Then P corresponds to 2 on

the number line.

Example 4 : Locate 3 on the number line.

Solution : Let us return to Fig. 1.7.

Fig. 1.8

Construct BD of unit length perpendicular to OB (as in Fig. 1.8). Then using the

Pythagoras theorem, we see that OD = ( )2

22 1 3+ = . Using a compass, with

centre O and radius OD, draw an arc which intersects the number line at the point Q.

Then Q corresponds to 3 .

Fig. 1.6

2019-2020

8 MATHEMATICS

In the same way, you can locate n for any positive integer n, after 1n − has been

located.

EXERCISE 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form m , where m is a natural number.

(iii) Every real number is an irrational number.

2. Are the square roots of all positive integers irrational? If not, give an example of the

square root of a number that is a rational number.

3. Show how 5 can be represented on the number line.

4. Classroom activity (Constructing the ‘square root

spiral’) : Take a large sheet of paper and construct

the ‘square root spiral’ in the following fashion. Start

with a point O and draw a line segment OP1 of unit

length. Draw a line segment P1P

2 perpendicular to

OP1 of unit length (see Fig. 1.9). Now draw a line

segment P2P

3 perpendicular to OP

2. Then draw a line

segment P3P

4 perpendicular to OP

3. Continuing in

this manner, you can get the line segment Pn–1

Pn by

drawing a line segment of unit length perpendicular to OPn–1

. In this manner, you will

have created the points P2, P

3,...., P

n,... ., and joined them to create a beautiful spiral

depicting 2, 3, 4, ...

1.3 Real Numbers and their Decimal Expansions

In this section, we are going to study rational and irrational numbers from a different

point of view. We will look at the decimal expansions of real numbers and see if we

can use the expansions to distinguish between rationals and irrationals. We will also

explain how to visualise the representation of real numbers on the number line using

their decimal expansions. Since rationals are more familiar to us, let us start with

them. Let us take three examples : 10 7 1

, ,3 8 7

.

Pay special attention to the remainders and see if you can find any pattern.

Fig. 1.9 : Constructing

square root spiral

2019-2020

NUMBER SYSTEMS 9

Example 5 : Find the decimal expansions of 10

3,

7

8 and

1

7.

Solution :

3.333... 0.875 0.142857...

3 10 8 7.0 7 1.0

9 64 7

10 60 30

9 56 28

10 40 20

9 40 14

10 0 60

9 56

1 40

35

50

49

1

Remainders : 1, 1, 1, 1, 1... Remainders : 6, 4, 0 Remainders : 3, 2, 6, 4, 5, 1,

Divisor : 3 Divisor : 8 3, 2, 6, 4, 5, 1,...

Divisor : 7

What have you noticed? You should have noticed at least three things:

(i) The remainders either become 0 after a certain stage, or start repeating themselves.

(ii) The number of entries in the repeating string of remainders is less than the divisor

(in 10

3 one number repeats itself and the divisor is 3, in

1

7 there are six entries

326451 in the repeating string of remainders and 7 is the divisor).

(iii) If the remainders repeat, then we get a repeating block of digits in the quotient

(for 10

3, 3 repeats in the quotient and for

1

7, we get the repeating block 142857

in the quotient).

2019-2020

10 MATHEMATICS

Although we have noticed this pattern using only the examples above, it is true for all

rationals of the form p

q (q ≠ 0). On division of p by q, two main things happen – either

the remainder becomes zero or never becomes zero and we get a repeating string of

remainders. Let us look at each case separately.

Case (i) : The remainder becomes zero

In the example of 7

8, we found that the remainder becomes zero after some steps and

the decimal expansion of 7

8 = 0.875. Other examples are

1

2 = 0.5,

639

250 = 2.556. In all

these cases, the decimal expansion terminates or ends after a finite number of steps.

We call the decimal expansion of such numbers terminating.

Case (ii) : The remainder never becomes zero

In the examples of 10

3 and

1

7, we notice that the remainders repeat after a certain

stage forcing the decimal expansion to go on for ever. In other words, we have a

repeating block of digits in the quotient. We say that this expansion is non-terminating

recurring. For example, 10

3 = 3.3333... and

1

7 = 0.142857142857142857...

The usual way of showing that 3 repeats in the quotient of 10

3 is to write it as 3.3 .

Similarly, since the block of digits 142857 repeats in the quotient of 1

7, we write

1

7 as

0.142857 , where the bar above the digits indicates the block of digits that repeats.

Also 3.57272... can be written as 3.572 . So, all these examples give us non-terminating

recurring (repeating) decimal expansions.

Thus, we see that the decimal expansion of rational numbers have only two choices:

either they are terminating or non-terminating recurring.

Now suppose, on the other hand, on your walk on the number line, you come across a

number like 3.142678 whose decimal expansion is terminating or a number like

1.272727... that is, 1.27 , whose decimal expansion is non-terminating recurring, can

you conclude that it is a rational number? The answer is yes!

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NUMBER SYSTEMS 11

We will not prove it but illustrate this fact with a few examples. The terminating cases

are easy.

Example 6 : Show that 3.142678 is a rational number. In other words, express 3.142678

in the form p

q, where p and q are integers and q ≠ 0.

Solution : We have 3.142678 = 3142678

1000000, and hence is a rational number.

Now, let us consider the case when the decimal expansion is non-terminating recurring.

Example 7 : Show that 0.3333... = 0 3. can be expressed in the form p

q, where p and

q are integers and q ≠ 0.

Solution : Since we do not know what 0 3. is , let us call it ‘x’ and so

x = 0.3333...

Now here is where the trick comes in. Look at

10 x = 10 × (0.333...) = 3.333...

Now, 3.3333... = 3 + x, since x = 0.3333...

Therefore, 10 x = 3 + x

Solving for x, we get

9x = 3, i.e., x = 1

3

Example 8 : Show that 1.272727... = 1 27. can be expressed in the form p

q, where p

and q are integers and q ≠ 0.

Solution : Let x = 1.272727... Since two digits are repeating, we multiply x by 100 to

get

100 x = 127.2727...

So, 100 x = 126 + 1.272727... = 126 + x

Therefore, 100 x – x = 126, i.e., 99 x = 126

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12 MATHEMATICS

i.e., x =126 14

99 11=

You can check the reverse that 14

11 = 1 27. .

Example 9 : Show that 0.2353535... = 0 235. can be expressed in the form p

q,

where p and q are integers and q ≠ 0.

Solution : Let x = 0 235. . Over here, note that 2 does not repeat, but the block 35

repeats. Since two digits are repeating, we multiply x by 100 to get

100 x = 23.53535...

So, 100 x = 23.3 + 0.23535... = 23.3 + x

Therefore, 99 x = 23.3

i.e., 99 x =233

10, which gives x =

233

990

You can also check the reverse that 233

990 = 0 235. .

So, every number with a non-terminating recurring decimal expansion can be expressed

in the form p

q (q ≠ 0), where p and q are integers. Let us summarise our results in the

following form :

The decimal expansion of a rational number is either terminating or non-

terminating recurring. Moreover, a number whose decimal expansion is

terminating or non-terminating recurring is rational.

So, now we know what the decimal expansion of a rational number can be. What

about the decimal expansion of irrational numbers? Because of the property above,

we can conclude that their decimal expansions are non-terminating non-recurring.

So, the property for irrational numbers, similar to the property stated above for rational

numbers, is

The decimal expansion of an irrational number is non-terminating non-recurring.

Moreover, a number whose decimal expansion is non-terminating non-recurring

is irrational.

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NUMBER SYSTEMS 13

Recall s = 0.10110111011110... from the previous section. Notice that it is non-

terminating and non-recurring. Therefore, from the property above, it is irrational.

Moreover, notice that you can generate infinitely many irrationals similar to s.

What about the famous irrationals 2 and π? Here are their decimal expansions up

to a certain stage.

2 = 1.4142135623730950488016887242096...

π = 3.14159265358979323846264338327950...

(Note that, we often take 22

7 as an approximate value for π, but π ≠

22

7.)

Over the years, mathematicians have developed various techniques to produce more

and more digits in the decimal expansions of irrational numbers. For example, you

might have learnt to find digits in the decimal expansion of 2 by the division method.

Interestingly, in the Sulbasutras (rules of chord), a mathematical treatise of the Vedic

period (800 BC - 500 BC), you find an approximation of 2 as follows:

2 = 1 1 1 1 1 1

1 1 41421563 4 3 34 4 3

.

+ + × − × × =

Notice that it is the same as the one given above for the first five decimal places. The

history of the hunt for digits in the decimal expansion of π is very interesting.

The Greek genius Archimedes was the first to compute

digits in the decimal expansion of π. He showed 3.140845

< π < 3.142857. Aryabhatta (476 – 550 C.E.), the great

Indian mathematician and astronomer, found the value

of π correct to four decimal places (3.1416). Using high

speed computers and advanced algorithms, π has been

computed to over 1.24 trillion decimal places!

Now, let us see how to obtain irrational numbers.

Example 10 : Find an irrational number between 1

7 and

2

7.

Solution : We saw that 1

7 = 0142857. . So, you can easily calculate

20 285714

7.= .

To find an irrational number between 1

7 and

2

7, we find a number which is

Archimedes (287 BCE – 212 BCE)

Fig. 1.10

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14 MATHEMATICS

non-terminating non-recurring lying between them. Of course, you can find infinitely

many such numbers.

An example of such a number is 0.150150015000150000...

EXERCISE 1.3

1. Write the following in decimal form and say what kind of decimal expansion each

has :

(i)36

100(ii)

1

11(iii)

14

8

(iv)3

13(v)

2

11(vi)

329

400

2. You know that 1

7 = 0142857. . Can you predict what the decimal expansions of

2

7,

3

7,

4

7,

5

7,

6

7 are, without actually doing the long division? If so, how?

[Hint : Study the remainders while finding the value of 1

7 carefully.]

3. Express the following in the form p

q, where p and q are integers and q ≠ 0.

(i) 0 6. (ii) 0 47. (iii) 0 001.

4. Express 0.99999 .... in the form p

q. Are you surprised by your answer? With your

teacher and classmates discuss why the answer makes sense.

5. What can the maximum number of digits be in the repeating block of digits in the

decimal expansion of 1

17? Perform the division to check your answer.

6. Look at several examples of rational numbers in the form p

q (q ≠ 0), where p and q are

integers with no common factors other than 1 and having terminating decimal

representations (expansions). Can you guess what property q must satisfy?

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

8. Find three different irrational numbers between the rational numbers 5

7 and

9

11.

9. Classify the following numbers as rational or irrational :

(i) 23 (ii) 225 (iii) 0.3796

(iv) 7.478478... (v) 1.101001000100001...

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NUMBER SYSTEMS 15

1.4 Representing Real Numbers on the Number Line

In the previous section, you have seen that any

real number has a decimal expansion. This helps

us to represent it on the number line. Let us see

how.

Suppose we want to locate 2.665 on the

number line. We know that this lies between 2

and 3.

So, let us look closely at the portion of the

number line between 2 and 3. Suppose we divide

this into 10 equal parts and mark each point of

division as in Fig. 1.11 (i). Then the first mark to

the right of 2 will represent 2.1, the second 2.2, and so on. You might be finding some

difficulty in observing these points of division between 2 and 3 in Fig. 1.11 (i). To have

a clear view of the same, you may take a magnifying glass and look at the portion

between 2 and 3. It will look like what you see in Fig. 1.11 (ii). Now, 2.665 lies between

2.6 and 2.7. So, let us focus on the portion between 2.6 and 2.7 [See Fig. 1.12(i)]. We

imagine to divide this again into ten equal parts. The first mark will represent 2.61, the

next 2.62, and so on. To see this clearly, we magnify this as shown in Fig. 1.12 (ii).

Fig. 1.12

Again, 2.665 lies between 2.66 and 2.67. So, let us focus on this portion of the

number line [see Fig. 1.13(i)] and imagine to divide it again into ten equal parts. We

magnify it to see it better, as in Fig. 1.13 (ii). The first mark represents 2.661, the next

one represents 2.662, and so on. So, 2.665 is the 5th mark in these subdivisions.

Fig. 1.11

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16 MATHEMATICS

Fig. 1.13

We call this process of visualisation of representation of numbers on the number line,

through a magnifying glass, as the process of successive magnification.

So, we have seen that it is possible by sufficient successive magnifications to visualise

the position (or representation) of a real number with a terminating decimal expansion

on the number line.

Let us now try and visualise the position (or representation) of a real number with a

non-terminating recurring decimal expansion on the number line. We can look at

appropriate intervals through a magnifying glass and by successive magnifications

visualise the position of the number on the number line.

Example 11 : Visualize the representation of 5 37. on the number line upto 5 decimal

places, that is, up to 5.37777.

Solution : Once again we proceed by successive magnification, and successively

decrease the lengths of the portions of the number line in which 5 37. is located. First,

we see that 5 37. is located between 5 and 6. In the next step, we locate 5 37.

between 5.3 and 5.4. To get a more accurate visualization of the representation, we

divide this portion of the number line into 10 equal parts and use a magnifying glass to

visualize that 5 37. lies between 5.37 and 5.38. To visualize 5 37. more accurately, we

again divide the portion between 5.37 and 5.38 into ten equal parts and use a magnifying

glass to visualize that 5 37. lies between 5.377 and 5.378. Now to visualize 5 37. still

more accurately, we divide the portion between 5.377 an 5.378 into 10 equal parts, and

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NUMBER SYSTEMS 17

visualize the representation of 5 37. as in Fig. 1.14 (iv). Notice that 5 37. is located

closer to 5.3778 than to 5.3777 [see Fig 1.14 (iv)].

Fig. 1.14

Remark : We can proceed endlessly in this manner, successively viewing through a

magnifying glass and simultaneously imagining the decrease in the length of the portion

of the number line in which 5 37. is located. The size of the portion of the line we

specify depends on the degree of accuracy we would like for the visualisation of the

position of the number on the number line.

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18 MATHEMATICS

You might have realised by now that the same procedure can be used to visualise a

real number with a non-terminating non-recurring decimal expansion on the number

line.

In the light of the discussions above and visualisations, we can again say that every

real number is represented by a unique point on the number line. Further, every

point on the number line represents one and only one real number.

EXERCISE 1.4

1. Visualise 3.765 on the number line, using successive magnification.

2. Visualise 4 26. on the number line, up to 4 decimal places.

1.5 Operations on Real Numbers

You have learnt, in earlier classes, that rational numbers satisfy the commutative,

associative and distributive laws for addition and multiplication. Moreover, if we add,

subtract, multiply or divide (except by zero) two rational numbers, we still get a rational

number (that is, rational numbers are ‘closed’ with respect to addition, subtraction,

multiplication and division). It turns out that irrational numbers also satisfy the

commutative, associative and distributive laws for addition and multiplication. However,

the sum, difference, quotients and products of irrational numbers are not always

irrational. For example, ( ) ( )6 6+ − , ( ) ( ) ( ) ( )2 2 3 3,− ⋅ and 17

17 are

rationals.

Let us look at what happens when we add and multiply a rational number with an

irrational number. For example, 3 is irrational. What about 2 3+ and 2 3 ? Since

3 has a non-terminating non-recurring decimal expansion, the same is true for

2 3+ and 2 3 . Therefore, both 2 3+ and 2 3 are also irrational numbers.

Example 12 : Check whether 7 5 , 7

2 21 25

, ,+ π − are irrational numbers or

not.

Solution : 5 = 2.236... , 2 = 1.4142..., π = 3.1415...

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NUMBER SYSTEMS 19

Then 7 5 = 15.652..., 7

5 =

7 5 7 5

55 5= = 3.1304...

2 + 21 = 22.4142..., π – 2 = 1.1415...

All these are non-terminating non-recurring decimals. So, all these are irrational numbers.

Now, let us see what generally happens if we add, subtract, multiply, divide, take

square roots and even nth roots of these irrational numbers, where n is any natural

number. Let us look at some examples.

Example 13 : Add 2 2 5 3+ and 2 3 3– .

Solution : ( ) ( )2 2 5 3 2 3 3–+ + = ( ) ( )2 2 2 5 3 3 3–+ +

= (2 + 1) 2 (5 3) 3 3 2 2 3+ − = +

Example 14 : Multiply 6 5 by 2 5 .

Solution : 6 5 × 2 5 = 6 × 2 × 5 × 5 = 12 × 5 = 60

Example 15 : Divide 8 15 by 2 3 .

Solution : 8 3 5

8 15 2 3 4 52 3

×÷ = =

These examples may lead you to expect the following facts, which are true:

(i) The sum or difference of a rational number and an irrational number is irrational.

(ii) The product or quotient of a non-zero rational number with an irrational number is

irrational.

(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or

irrational.

We now turn our attention to the operation of taking square roots of real numbers.

Recall that, if a is a natural number, then a b= means b2 = a and b > 0. The same

definition can be extended for positive real numbers.

Let a > 0 be a real number. Then a = b means b2 = a and b > 0.

In Section 1.2, we saw how to represent n for any positive integer n on the number

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20 MATHEMATICS

line. We now show how to find x for any given positive real number x geometrically.

For example, let us find it for x = 3.5, i.e., we find 3 5. geometrically.

Fig. 1.15

Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such

that AB = 3.5 units (see Fig. 1.15). From B, mark a distance of 1 unit and mark the

new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle

with centre O and radius OC. Draw a line perpendicular to AC passing through B and

intersecting the semicircle at D. Then, BD = 3.5 .

More generally, to find x , for any positive real

number x, we mark B so that AB = x units, and, as in

Fig. 1.16, mark C so that BC = 1 unit. Then, as we

have done for the case x = 3.5, we find BD = x

(see Fig. 1.16). We can prove this result using the

Pythagoras Theorem.

Notice that, in Fig. 1.16, ∆ OBD is a right-angled triangle. Also, the radius of the circle

is 1

2

x + units.

Therefore, OC = OD = OA = 1

2

x + units.

Now, OB = 1 1

2 2

x xx

+ − − = ⋅

So, by the Pythagoras Theorem, we have

BD2 = OD2 – OB2 =

2 21 1 4

2 2 4

x x xx

+ − − = =

.

This shows that BD = x .

Fig. 1.16

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NUMBER SYSTEMS 21

This construction gives us a visual, and geometric way of showing that x exists for

all real numbers x > 0. If you want to know the position of x on the number line,

then let us treat the line BC as the number line, with B as zero, C as 1, and so on.

Draw an arc with centre B and radius BD, which intersects the number line in E

(see Fig. 1.17). Then, E represents x .

Fig. 1.17

We would like to now extend the idea of square roots to cube roots, fourth roots,

and in general nth roots, where n is a positive integer. Recall your understanding of

square roots and cube roots from earlier classes.

What is 3 8 ? Well, we know it has to be some positive number whose cube is 8, and

you must have guessed 3 8 = 2. Let us try 5 243 . Do you know some number b such

that b5 = 243? The answer is 3. Therefore, 5 243 = 3.

From these examples, can you define n a for a real number a > 0 and a positive

integer n?

Let a > 0 be a real number and n be a positive integer. Then n a = b, if bn = a and

b > 0. Note that the symbol ‘ ’ used in 32, 8, n a , etc. is called the radical sign.

We now list some identities relating to square roots, which are useful in various

ways. You are already familiar with some of these from your earlier classes. The

remaining ones follow from the distributive law of multiplication over addition of real

numbers, and from the identity (x + y) (x – y) = x2 – y2, for any real numbers x and y.

Let a and b be positive real numbers. Then

(i) ab a b= (ii)a a

b b=

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22 MATHEMATICS

(iii) ( ) ( )a b a b a b+ − = − (iv) ( ) ( ) 2a b a b a b+ − = −

(v) ( ) ( )a b c d ac ad bc bd+ + = + + +

(vi) ( )2

2a b a ab b+ = + +

Let us look at some particular cases of these identities.

Example 16 : Simplify the following expressions:

(i) ( ) ( )5 7 2 5+ + (ii) ( ) ( )5 5 5 5+ −

(iii) ( )2

3 7+ (iv) ( ) ( )11 7 11 7− +

Solution : (i) ( ) ( )5 7 2 5 10 5 5 2 7 35+ + = + + +

(ii) ( ) ( ) ( )2

25 5 5 5 5 5 25 5 20–+ − = − = =

(iii) ( ) ( ) ( )2 2 2

3 7 3 2 3 7 7 3 2 21 7 10 2 21+ = + + = + + = +

(iv) ( ) ( ) ( ) ( )2 2

11 7 11 7 11 7 11 7 4− + = − = − =

Remark : Note that ‘simplify’ in the example above has been used to mean that the

expression should be written as the sum of a rational and an irrational number.

We end this section by considering the following problem. Look at 1

2⋅ Can you tell

where it shows up on the number line? You know that it is irrational. May be it is easier

to handle if the denominator is a rational number. Let us see, if we can ‘rationalise’ the

denominator, that is, to make the denominator into a rational number. To do so, we

need the identities involving square roots. Let us see how.

Example 17 : Rationalise the denominator of 1

2⋅

Solution : We want to write 1

2 as an equivalent expression in which the denominator

is a rational number. We know that 2 . 2 is rational. We also know that multiplying

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NUMBER SYSTEMS 23

1

2 by

2

2 will give us an equivalent expression, since

2

2 = 1. So, we put these two

facts together to get

1 1 2 2

22 2 2= × = ⋅

In this form, it is easy to locate 1

2 on the number line. It is half way between 0

and 2 .

Example 18 : Rationalise the denominator of 1

2 3⋅

+

Solution : We use the Identity (iv) given earlier. Multiply and divide 1

2 3+ by

2 3− to get 1 2 3 2 3

2 34 32 3 2 3

− −× = = −

−+ −.

Example 19 : Rationalise the denominator of 5

3 5⋅

−

Solution : Here we use the Identity (iii) given earlier.

So,5

3 5− =

( )( )

5 3 55 3 5 53 5

3 5 23 5 3 5

++ − × = = +

−− +

Example 20 : Rationalise the denominator of 1

7 3 2⋅

+

Solution : 1 1 7 3 2 7 3 2 7 3 2

49 18 317 3 2 7 3 2 7 3 2

− − −= × = = −+ + −

So, when the denominator of an expression contains a term with a square root (or

a number under a radical sign), the process of converting it to an equivalent expression

whose denominator is a rational number is called rationalising the denominator.

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24 MATHEMATICS

EXERCISE 1.5

1. Classify the following numbers as rational or irrational:

(i) 2 5− (ii) ( )3 23 23+ − (iii)2 7

7 7

(iv)1

2(v) 2π

2. Simplify each of the following expressions:

(i) ( ) ( )3 3 2 2+ + (ii) ( ) ( )3 3 3 3+ −

(iii) ( )2

5 2+ (iv) ( ) ( )5 2 5 2− +

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter

(say d). That is, π = c

d⋅ This seems to contradict the fact that π is irrational. How will

you resolve this contradiction?

4. Represent 9 3. on the number line.

5. Rationalise the denominators of the following:

(i)1

7(ii)

1

7 6−

(iii)1

5 2+(iv)

1

7 2−

1.6 Laws of Exponents for Real Numbers

Do you remember how to simplify the following?

(i) 172 . 175 = (ii) (52)7 =

(iii) 10

7

23

23 = (iv) 73 . 93 =

Did you get these answers? They are as follows:

(i) 172 . 175 = 177 (ii) (52)7 = 514

(iii)10

3

7

2323

23= (iv) 73 . 93 = 633

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NUMBER SYSTEMS 25

To get these answers, you would have used the following laws of exponents,

which you have learnt in your earlier classes. (Here a, n and m are natural numbers.

Remember, a is called the base and m and n are the exponents.)

(i) am . an = am + n (ii) (am)n = amn

(iii)

mm n

n

aa , m n

a

−= > (iv) ambm = (ab)m

What is (a)0? Yes, it is 1! So you have learnt that (a)0 = 1. So, using (iii), we can

get 1

.n

na

a

−= We can now extend the laws to negative exponents too.

So, for example :

(i)2 –5 –3

3

117 17 17

17⋅ = = (ii) 2 –7 –14(5 ) 5=

(iii)

–10–17

7

2323

23= (iv) –3 –3 –3(7) (9) (63)⋅ =

Suppose we want to do the following computations:

(i)2 1

3 32 2⋅ (ii)

41

53

(iii)

1

5

1

3

7

7

(iv)1 1

5 513 17⋅

How would we go about it? It turns out that we can extend the laws of exponents

that we have studied earlier, even when the base is a positive real number and the

exponents are rational numbers. (Later you will study that it can further to be extended

when the exponents are real numbers.) But before we state these laws, and to even

make sense of these laws, we need to first understand what, for example 3

24 is. So,

we have some work to do!

In Section 1.4, we defined n a for a real number a > 0 as follows:

Let a > 0 be a real number and n a positive integer. Then n a = b, if bn = a and

b > 0.

In the language of exponents, we define n a =

1

na . So, in particular, 1

3 32 2= .

There are now two ways to look at 3

24 .

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26 MATHEMATICS

3

24 =

31

324 2 8

= =

3

24 = ( ) ( )1 1

3 2 24 64 8= =

Therefore, we have the following definition:

Let a > 0 be a real number. Let m and n be integers such that m and n have no

common factors other than 1, and n > 0. Then,

m

na = ( )m

n mn a a=

We now have the following extended laws of exponents:

Let a > 0 be a real number and p and q be rational numbers. Then, we have

(i) ap . aq = ap+q (ii) (ap)q = apq

(iii)

pp q

q

aa

a

−= (iv) apbp = (ab)p

You can now use these laws to answer the questions asked earlier.

Example 21 : Simplify (i)2 1

3 32 2⋅ (ii)

41

53

(iii)

1

5

1

3

7

7

(iv)1 1

5 513 17⋅

Solution :

(i)

2 12 1 3

13 33 3 32 2 2 2 2 2

+

⋅ = = = = (ii)

41 4

5 53 3

=

(iii)

11 1 3 5 255 3 15 15

1

3

77 7 7

7

− −−

= = = (iv)1 1 1 1

5 5 5 513 17 (13 17) 221⋅ = × =

EXERCISE 1.6

1. Find : (i)1

264 (ii)1

532 (iii)1

3125

2. Find : (i)3

29 (ii)2

532 (iii)3

416 (iv)1

3125−

3. Simplify : (i)2 1

3 52 2⋅ (ii)7

3

1

3

(iii)

1

2

1

4

11

11

(iv)1 1

2 27 8⋅

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NUMBER SYSTEMS 27

1.7 Summary

In this chapter, you have studied the following points:

1. A number r is called a rational number, if it can be written in the form p

q, where p and q are

integers and q ≠ 0.

2. A number s is called a irrational number, if it cannot be written in the form p

q, where p and

q are integers and q ≠ 0.

3. The decimal expansion of a rational number is either terminating or non-terminating recurring.

Moreover, a number whose decimal expansion is terminating or non-terminating recurring

is rational.

4. The decimal expansion of an irrational number is non-terminating non-recurring. Moreover,

a number whose decimal expansion is non-terminating non-recurring is irrational.

5. All the rational and irrational numbers make up the collection of real numbers.

6. There is a unique real number corresponding to every point on the number line. Also,

corresponding to each real number, there is a unique point on the number line.

7. If r is rational and s is irrational, then r + s and r – s are irrational numbers, and rs and r

s are

irrational numbers, r ≠ 0.

8. For positive real numbers a and b, the following identities hold:

(i) ab a b= (ii)a a

b b=

(iii) ( ) ( )a b a b a b+ − = − (iv) ( ) ( ) 2a b a b a b+ − = −

(v) ( )2

2a b a ab b+ = + +

9. To rationalise the denominator of 1

,a b+

we multiply this by ,a b

a b

−

− where a and b are

integers.

10. Let a > 0 be a real number and p and q be rational numbers. Then

(i) ap . aq = ap + q (ii) (ap)q = apq

(iii)

pp q

q

aa

a

−= (iv) apbp = (ab)p

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CHAPTER 7

TRIANGLES

7.1 Introduction

You have studied about triangles and their various properties in your earlier classes.

You know that a closed figure formed by three intersecting lines is called a triangle.

(‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For

example, in triangle ABC, denoted as ∆ ABC (see Fig. 7.1); AB, BC, CA are the three

sides, ∠ A, ∠ B, ∠ C are the three angles and A, B, C are three vertices.

In Chapter 6, you have also studied some properties

of triangles. In this chapter, you will study in details

about the congruence of triangles, rules of congruence,

some more properties of triangles and inequalities in

a triangle. You have already verified most of these

properties in earlier classes. We will now prove some

of them.

7.2 Congruence of Triangles

You must have observed that two copies of your photographs of the same size are

identical. Similarly, two bangles of the same size, two ATM cards issued by the same

bank are identical. You may recall that on placing a one rupee coin on another minted

in the same year, they cover each other completely.

Do you remember what such figures are called? Indeed they are called congruent

figures (‘congruent’ means equal in all respects or figures whose shapes and sizes

are both the same).

Now, draw two circles of the same radius and place one on the other. What do

you observe? They cover each other completely and we call them as congruent circles.

Fig. 7.1

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Repeat this activity by placing one

square on the other with sides of the same

measure (see Fig. 7.2) or by placing two

equilateral triangles of equal sides on each

other. You will observe that the squares are

congruent to each other and so are the

equilateral triangles.

You may wonder why we are studying congruence. You all must have seen the ice

tray in your refrigerator. Observe that the moulds for making ice are all congruent.

The cast used for moulding in the tray also has congruent depressions (may be all are

rectangular or all circular or all triangular). So, whenever identical objects have to be

produced, the concept of congruence is used in making the cast.

Sometimes, you may find it difficult to replace the refill in your pen by a new one

and this is so when the new refill is not of the same size as the one you want to

remove. Obviously, if the two refills are identical or congruent, the new refill fits.

So, you can find numerous examples where congruence of objects is applied in

daily life situations.

Can you think of some more examples of congruent figures?

Now, which of the following figures are not congruent to the square in

Fig 7.3 (i) :

Fig. 7.3

The large squares in Fig. 7.3 (ii) and (iii) are obviously not congruent to the one in

Fig 7.3 (i), but the square in Fig 7.3 (iv) is congruent to the one given in Fig 7.3 (i).

Let us now discuss the congruence of two triangles.

You already know that two triangles are congruent if the sides and angles of one

triangle are equal to the corresponding sides and angles of the other triangle.

Fig. 7.2

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Now, which of the triangles given below are congruent to triangle ABC in

Fig. 7.4 (i)?

Fig. 7.4

Cut out each of these triangles from Fig. 7.4 (ii) to (v) and turn them around and

try to cover ∆ ABC. Observe that triangles in Fig. 7.4 (ii), (iii) and (iv) are congruent

to ∆ ABC while ∆ TSU of Fig 7.4 (v) is not congruent to ∆ ABC.

If ∆ PQR is congruent to ∆ ABC, we write ∆ PQR ≅ ∆ ABC.

Notice that when ∆ PQR ≅ ∆ ABC, then sides of ∆ PQR fall on corresponding

equal sides of ∆ ABC and so is the case for the angles.

That is, PQ covers AB, QR covers BC and RP covers CA; ∠ P covers ∠ A,

∠ Q covers ∠ B and ∠ R covers ∠ C. Also, there is a one-one correspondence

between the vertices. That is, P corresponds to A, Q to B, R to C and so on which is

written as

P ↔ A, Q ↔ B, R ↔ C

Note that under this correspondence, ∆ PQR ≅ ∆ ABC; but it will not be correct to

write ∆QRP ≅ ∆ ABC.

Similarly, for Fig. 7.4 (iii),

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FD ↔ AB, DE ↔ BC and EF ↔ CA

and F ↔ A, D ↔ B and E ↔ C

So, ∆ FDE ≅ ∆ ABC but writing ∆ DEF ≅ ∆ ABC is not correct.

Give the correspondence between the triangle in Fig. 7.4 (iv) and ∆ ABC.

So, it is necessary to write the correspondence of vertices correctly for writing of

congruence of triangles in symbolic form.

Note that in congruent triangles corresponding parts are equal and we write

in short ‘CPCT’ for corresponding parts of congruent triangles.

7.3 Criteria for Congruence of Triangles

In earlier classes, you have learnt four criteria for congruence of triangles. Let us

recall them.

Draw two triangles with one side 3 cm. Are these triangles congruent? Observe

that they are not congruent (see Fig. 7.5).

Fig. 7.5

Now, draw two triangles with one side 4 cm and one angle 50° (see Fig. 7.6). Are

they congruent?

Fig. 7.6

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See that these two triangles are not congruent.

Repeat this activity with some more pairs of triangles.

So, equality of one pair of sides or one pair of sides and one pair of angles is not

sufficient to give us congruent triangles.

What would happen if the other pair of arms (sides) of the equal angles are also

equal?

In Fig 7.7, BC = QR, ∠ B = ∠ Q and also, AB = PQ. Now, what can you say

about congruence of ∆ ABC and ∆ PQR?

Recall from your earlier classes that, in this case, the two triangles are congruent.

Verify this for ∆ ABC and ∆ PQR in Fig. 7.7.

Repeat this activity with other pairs of triangles. Do you observe that the equality

of two sides and the included angle is enough for the congruence of triangles? Yes, it

is enough.

Fig. 7.7

This is the first criterion for congruence of triangles.

Axiom 7.1 (SAS congruence rule) : Two triangles are congruent if two sides

and the included angle of one triangle are equal to the two sides and the included

angle of the other triangle.

This result cannot be proved with the help of previously known results and so it is

accepted true as an axiom (see Appendix 1).

Let us now take some examples.

Example 1 : In Fig. 7.8, OA = OB and OD = OC. Show that

(i) ∆ AOD ≅ ∆ BOC and (ii) AD || BC.

Solution : (i) You may observe that in ∆ AOD and ∆ BOC,

OA = OB(Given)

OD = OC

Fig. 7.8

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Also, since ∠ AOD and ∠ BOC form a pair of vertically opposite angles, we have

∠ AOD = ∠ BOC.

So, ∆ AOD ≅ ∆ BOC (by the SAS congruence rule)

(ii) In congruent triangles AOD and BOC, the other corresponding parts are also

equal.

So, ∠ OAD = ∠ OBC and these form a pair of alternate angles for line segments

AD and BC.

Therefore, AD || BC.

Example 2 : AB is a line segment and line l is its perpendicular bisector. If a point P

lies on l, show that P is equidistant from A and B.

Solution : Line l ⊥ AB and passes through C which

is the mid-point of AB (see Fig. 7.9). You have to

show that PA = PB. Consider ∆ PCA and ∆ PCB.

We have AC = BC (C is the mid-point of AB)

∠ PCA = ∠ PCB = 90° (Given)

PC = PC (Common)

So, ∆ PCA ≅ ∆ PCB (SAS rule)

and so, PA = PB, as they are corresponding sides of

congruent triangles.

Now, let us construct two triangles, whose sides are 4 cm and 5 cm and one of the

angles is 50° and this angle is not included in between the equal sides (see Fig. 7.10).

Are the two triangles congruent?

Fig. 7.10

Fig. 7.9

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Notice that the two triangles are not congruent.

Repeat this activity with more pairs of triangles. You will observe that for triangles

to be congruent, it is very important that the equal angles are included between the

pairs of equal sides.

So, SAS congruence rule holds but not ASS or SSA rule.

Next, try to construct the two triangles in which two angles are 60° and 45° and

the side included between these angles is 4 cm (see Fig. 7.11).

Fig. 7.11

Cut out these triangles and place one triangle on the other. What do you observe?

See that one triangle covers the other completely; that is, the two triangles are congruent.

Repeat this activity with more pairs of triangles. You will observe that equality of two

angles and the included side is sufficient for congruence of triangles.

This result is the Angle-Side-Angle criterion for congruence and is written as

ASA criterion. You have verified this criterion in earlier classes, but let us state and

prove this result.

Since this result can be proved, it is called a theorem and to prove it, we use the

SAS axiom for congruence.

Theorem 7.1 (ASA congruence rule) : Two triangles are congruent if two angles

and the included side of one triangle are equal to two angles and the included

side of other triangle.

Proof : We are given two triangles ABC and DEF in which:

∠ B = ∠ E, ∠ C = ∠ F

and BC = EF

We need to prove that ∆ ABC ≅ ∆ DEF

For proving the congruence of the two triangles see that three cases arise.

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Case (i) : Let AB = DE (see Fig. 7.12).

Now what do you observe? You may observe that

AB = DE (Assumed)

∠ B = ∠ E (Given)

BC = EF (Given)

So, ∆ ABC ≅ ∆ DEF (By SAS rule)

Fig. 7.12

Case (ii) : Let if possible AB > DE. So, we can take a point P on AB such that

PB = DE. Now consider ∆ PBC and ∆ DEF (see Fig. 7.13).

Fig. 7.13

Observe that in ∆ PBC and ∆ DEF,

PB = DE (By construction)

∠ B = ∠ E (Given)

BC = EF (Given)

So, we can conclude that:

∆ PBC ≅ ∆ DEF, by the SAS axiom for congruence.

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Since the triangles are congruent, their corresponding parts will be equal.

So, ∠ PCB = ∠ DFE

But, we are given that

∠ ACB = ∠ DFE

So, ∠ ACB = ∠ PCB

Is this possible?

This is possible only if P coincides with A.

or, BA = ED

So, ∆ ABC ≅ ∆ DEF (by SAS axiom)

Case (iii) : If AB < DE, we can choose a point M on DE such that ME = AB and

repeating the arguments as given in Case (ii), we can conclude that AB = DE and so,

∆ ABC ≅ ∆ DEF.

Suppose, now in two triangles two pairs of angles and one pair of corresponding

sides are equal but the side is not included between the corresponding equal pairs of

angles. Are the triangles still congruent? You will observe that they are congruent.

Can you reason out why?

You know that the sum of the three angles of a triangle is 180°. So if two pairs of

angles are equal, the third pair is also equal (180° – sum of equal angles).

So, two triangles are congruent if any two pairs of angles and one pair of

corresponding sides are equal. We may call it as the AAS Congruence Rule.

Now let us perform the following activity :

Draw triangles with angles 40°, 50° and 90°. How many such triangles can you

draw?

In fact, you can draw as many triangles as you want with different lengths of

sides (see Fig. 7.14).

Fig. 7.14

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Observe that the triangles may or may not be congruent to each other.

So, equality of three angles is not sufficient for congruence of triangles. Therefore,

for congruence of triangles out of three equal parts, one has to be a side.

Let us now take some more examples.

Example 3 : Line-segment AB is parallel to another line-segment CD. O is the

mid-point of AD (see Fig. 7.15). Show that (i) ∆AOB ≅ ∆DOC (ii) O is also the

mid-point of BC.

Solution : (i) Consider ∆ AOB and ∆ DOC.

∠ ABO = ∠ DCO

(Alternate angles as AB || CD

and BC is the transversal)

∠ AOB = ∠ DOC

(Vertically opposite angles)

OA = OD (Given)

Therefore, ∆AOB ≅ ∆DOC (AAS rule)

(ii) OB = OC (CPCT)

So, O is the mid-point of BC.

EXERCISE 7.1

1. In quadrilateral ACBD,

AC = AD and AB bisects ∠ A

(see Fig. 7.16). Show that ∆ ABC ≅ ∆ ABD.

What can you say about BC and BD?

Fig. 7.15

Fig. 7.16

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2. ABCD is a quadrilateral in which AD = BC and

∠ DAB = ∠ CBA (see Fig. 7.17). Prove that

(i) ∆ ABD ≅ ∆ BAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC.

3. AD and BC are equal perpendiculars to a line

segment AB (see Fig. 7.18). Show that CD bisects

AB.

4. l and m are two parallel lines intersected by

another pair of parallel lines p and q

(see Fig. 7.19). Show that ∆ ABC ≅ ∆ CDA.

5. Line l is the bisector of an angle ∠ A and B is any

point on l. BP and BQ are perpendiculars from B

to the arms of ∠ A (see Fig. 7.20). Show that:

(i) ∆ APB ≅ ∆ AQB

(ii) BP = BQ or B is equidistant from the arms

of ∠ A.

Fig. 7.17

Fig. 7.18

Fig. 7.19

Fig. 7.20

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6. In Fig. 7.21, AC = AE, AB = AD and

∠ BAD = ∠ EAC. Show that BC = DE.

7. AB is a line segment and P is its mid-point. D and

E are points on the same side of AB such that

∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB

(see Fig. 7.22). Show that

(i) ∆ DAP ≅ ∆ EBP

(ii) AD = BE

8. In right triangle ABC, right angled at C, M is

the mid-point of hypotenuse AB. C is joined

to M and produced to a point D such that

DM = CM. Point D is joined to point B

(see Fig. 7.23). Show that:

(i) ∆ AMC ≅ ∆ BMD

(ii) ∠ DBC is a right angle.

(iii) ∆ DBC ≅ ∆ ACB

(iv) CM = 1

2 AB

7.4 Some Properties of a Triangle

In the above section you have studied two criteria for congruence of triangles. Let us

now apply these results to study some properties related to a triangle whose two sides

are equal.

Fig. 7.22

Fig. 7.23

Fig. 7.21

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Perform the activity given below:

Construct a triangle in which two sides are

equal, say each equal to 3.5 cm and the third side

equal to 5 cm (see Fig. 7.24). You have done such

constructions in earlier classes.

Do you remember what is such a triangle

called?

A triangle in which two sides are equal is called

an isosceles triangle. So, ∆ ABC of Fig. 7.24 is

an isosceles triangle with AB = AC.

Now, measure ∠ B and ∠ C. What do you observe?

Repeat this activity with other isosceles triangles with different sides.

You may observe that in each such triangle, the angles opposite to the equal sides

are equal.

This is a very important result and is indeed true for any isosceles triangle. It can

be proved as shown below.

Theorem 7.2 : Angles opposite to equal sides of an isosceles triangle are equal.

This result can be proved in many ways. One of

the proofs is given here.

Proof : We are given an isosceles triangle ABC

in which AB = AC. We need to prove that

∠ B = ∠ C.

Let us draw the bisector of ∠ A and let D be

the point of intersection of this bisector of

∠ A and BC (see Fig. 7.25).

In ∆ BAD and ∆ CAD,

AB = AC (Given)

∠ BAD = ∠ CAD (By construction)

AD = AD (Common)

So, ∆ BAD ≅ ∆ CAD (By SAS rule)

So, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles.

So, ∠ B = ∠ C

Fig. 7.24

Fig. 7.25

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Is the converse also true? That is:

If two angles of any triangle are equal, can we conclude that the sides opposite to

them are also equal?

Perform the following activity.

Construct a triangle ABC with BC of any length and ∠ B = ∠ C = 50°. Draw the

bisector of ∠ A and let it intersect BC at D (see Fig. 7.26).

Cut out the triangle from the sheet of paper and fold it along AD so that vertex C

falls on vertex B.

What can you say about sides AC and AB?

Observe that AC covers AB completely

So, AC = AB

Repeat this activity with some more triangles.

Each time you will observe that the sides opposite

to equal angles are equal. So we have the

following:

Theorem 7.3 : The sides opposite to equal angles of a triangle are equal.

This is the converse of Theorem 7.2.

You can prove this theorem by ASA congruence rule.

Let us take some examples to apply these results.

Example 4 : In ∆ ABC, the bisector AD of ∠ A is perpendicular to side BC

(see Fig. 7.27). Show that AB = AC and ∆ ABC is isosceles.

Solution : In ∆ABD and ∆ACD,

∠ BAD = ∠ CAD (Given)

AD = AD (Common)

∠ ADB = ∠ ADC = 90° (Given)

So, ∆ ABD ≅ ∆ ACD (ASA rule)

So, AB = AC (CPCT)

or, ∆ ABC is an isosceles triangle.

Fig. 7.26

Fig. 7.27

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Example 5 : E and F are respectively the mid-points

of equal sides AB and AC of ∆ ABC (see Fig. 7.28).

Show that BF = CE.

Solution : In ∆ ABF and ∆ ACE,

AB = AC (Given)

∠ A = ∠ A (Common)

AF = AE (Halves of equal sides)

So, ∆ ABF ≅ ∆ ACE (SAS rule)

Therefore, BF = CE (CPCT)

Example 6 : In an isosceles triangle ABC with AB = AC, D and E are points on BC

such that BE = CD (see Fig. 7.29). Show that AD = AE.

Solution : In ∆ ABD and ∆ ACE,

AB = AC (Given) (1)

∠ B = ∠ C

(Angles opposite to equal sides) (2)

Also, BE = CD

So, BE – DE = CD – DE

That is, BD = CE (3)

So, ∆ ABD ≅ ∆ ACE

(Using (1), (2), (3) and SAS rule).

This gives AD = AE (CPCT)

EXERCISE 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect

each other at O. Join A to O. Show that :

(i) OB = OC (ii) AO bisects ∠ A

2. In ∆ ABC, AD is the perpendicular bisector of BC

(see Fig. 7.30). Show that ∆ ABC is an isosceles

triangle in which AB = AC.

Fig. 7.28

Fig. 7.29

Fig. 7.30

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3. ABC is an isosceles triangle in which altitudes

BE and CF are drawn to equal sides AC and AB

respectively (see Fig. 7.31). Show that these

altitudes are equal.

4. ABC is a triangle in which altitudes BE and CF to

sides AC and AB are equal (see Fig. 7.32). Show

that

(i) ∆ ABE ≅ ∆ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the

same base BC (see Fig. 7.33). Show that

∠ ABD = ∠ ACD.

6. ∆ABC is an isosceles triangle in which AB = AC.

Side BA is produced to D such that AD = AB

(see Fig. 7.34). Show that ∠ BCD is a right angle.

7. ABC is a right angled triangle in which ∠ A = 90°

and AB = AC. Find ∠ B and ∠ C.

8. Show that the angles of an equilateral triangle

are 60° each.

Fig. 7.32

Fig. 7.33

Fig. 7.34

Fig. 7.31

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7.5 Some More Criteria for Congruence of Triangles

You have seen earlier in this chapter that equality of three angles of one triangle to

three angles of the other is not sufficient for the congruence of the two triangles. You

may wonder whether equality of three sides of one triangle to three sides of another

triangle is enough for congruence of the two triangles. You have already verified in

earlier classes that this is indeed true.

To be sure, construct two triangles with sides 4 cm, 3.5 cm and 4.5 cm

(see Fig. 7.35). Cut them out and place them on each other. What do you observe?

They cover each other completely, if the equal sides are placed on each other. So, the

triangles are congruent.

Fig. 7.35

Repeat this activity with some more triangles. We arrive at another rule for

congruence.

Theorem 7.4 (SSS congruence rule) : If three sides of one triangle are equal to

the three sides of another triangle, then the two triangles are congruent.

This theorem can be proved using a suitable construction.

You have already seen that in the SAS congruence rule, the pair of equal angles

has to be the included angle between the pairs of corresponding pair of equal sides and

if this is not so, the two triangles may not be congruent.

Perform this activity:

Construct two right angled triangles with hypotenuse equal to 5 cm and one side

equal to 4 cm each (see Fig. 7.36).

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Fig. 7.36

Cut them out and place one triangle over the other with equal side placed on each

other. Turn the triangles, if necessary. What do you observe?

The two triangles cover each other completely and so they are congruent. Repeat

this activity with other pairs of right triangles. What do you observe?

You will find that two right triangles are congruent if one pair of sides and the

hypotenuse are equal. You have verified this in earlier classes.

Note that, the right angle is not the included angle in this case.

So, you arrive at the following congruence rule:

Theorem 7.5 (RHS congruence rule) : If in two right triangles the hypotenuse

and one side of one triangle are equal to the hypotenuse and one side of the

other triangle, then the two triangles are congruent.

Note that RHS stands for Right angle - Hypotenuse - Side.

Let us now take some examples.

Example 7 : AB is a line-segment. P and Q are

points on opposite sides of AB such that each of them

is equidistant from the points A and B (see Fig. 7.37).

Show that the line PQ is the perpendicular bisector

of AB.

Solution : You are given that PA = PB and

QA = QB and you are to show that PQ ⊥ AB and

PQ bisects AB. Let PQ intersect AB at C.

Can you think of two congruent triangles in this figure?

Let us take ∆ PAQ and ∆ PBQ.

In these triangles,

Fig. 7.37

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AP = BP (Given)

AQ = BQ (Given)

PQ = PQ (Common)

So, ∆ PAQ ≅ ∆ PBQ (SSS rule)

Therefore, ∠ APQ = ∠ BPQ (CPCT).

Now let us consider ∆ PAC and ∆ PBC.

You have : AP = BP (Given)

∠ APC = ∠ BPC (∠ APQ = ∠ BPQ proved above)

PC = PC (Common)

So, ∆ PAC ≅ ∆ PBC (SAS rule)

Therefore, AC = BC (CPCT) (1)

and ∠ ACP = ∠ BCP (CPCT)

Also, ∠ ACP + ∠ BCP = 180° (Linear pair)

So, 2∠ ACP = 180°

or, ∠ ACP = 90° (2)

From (1) and (2), you can easily conclude that PQ is the perpendicular bisector of AB.

[Note that, without showing the congruence of ∆ PAQ and ∆ PBQ, you cannot show

that ∆ PAC ≅ ∆ PBC even though AP = BP (Given)

PC = PC (Common)

and ∠ PAC = ∠ PBC (Angles opposite to equal sides in

∆APB)

It is because these results give us SSA rule which is not always valid or true for

congruence of triangles. Also the angle is not included between the equal pairs of

sides.]

Let us take some more examples.

Example 8 : P is a point equidistant from two lines l and m intersecting at point A

(see Fig. 7.38). Show that the line AP bisects the angle between them.

Solution : You are given that lines l and m intersect each other at A. Let PB ⊥ l,

PC ⊥ m. It is given that PB = PC.

You are to show that ∠ PAB = ∠ PAC.

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Let us consider ∆ PAB and ∆ PAC. In these two

triangles,

PB = PC (Given)

∠ PBA = ∠ PCA = 90° (Given)

PA = PA (Common)

So, ∆ PAB ≅ ∆ PAC (RHS rule)

So, ∠ PAB = ∠ PAC (CPCT)

Note that this result is the converse of the result proved in Q.5 of Exercise 7.1.

EXERCISE 7.3

1. ∆ ABC and ∆ DBC are two isosceles triangles on

the same base BC and vertices A and D are on the

same side of BC (see Fig. 7.39). If AD is extended

to intersect BC at P, show that

(i) ∆ ABD ≅ ∆ ACD

(ii) ∆ ABP ≅ ∆ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ∠ A.

3. Two sides AB and BC and median AM

of one triangle ABC are respectively

equal to sides PQ and QR and median

PN of ∆ PQR (see Fig. 7.40). Show that:

(i) ∆ ABM ≅ ∆ PQN

(ii) ∆ ABC ≅ ∆ PQR

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence

rule, prove that the triangle ABC is isosceles.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that

∠ B = ∠ C.

Fig. 7.38

Fig. 7.39

Fig. 7.40

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7.6 Inequalities in a Triangle

So far, you have been mainly studying the equality of sides and angles of a triangle or

triangles. Sometimes, we do come across unequal objects, we need to compare them.

For example, line-segment AB is greater in length as compared to line segment CD in

Fig. 7.41 (i) and ∠ A is greater than ∠ B in Fig 7.41 (ii).

Fig. 7.41

Let us now examine whether there is any relation between unequal sides and

unequal angles of a triangle. For this, let us perform the following activity:

Activity : Fix two pins on a drawing board say at B and C and tie a thread to mark a

side BC of a triangle.

Fix one end of another thread at C and tie a pencil

at the other (free) end . Mark a point A with the

pencil and draw ∆ ABC (see Fig 7.42). Now, shift

the pencil and mark another point A′ on CA beyond

A (new position of it)

So, A′C > AC (Comparing the lengths)

Join A′ to B and complete the triangle A′BC.

What can you say about ∠ A′BC and ∠ ABC?

Compare them. What do you observe?

Clearly, ∠ A′BC > ∠ ABC

Continue to mark more points on CA (extended) and draw the triangles with the

side BC and the points marked.

You will observe that as the length of the side AC is increased (by taking different

positions of A), the angle opposite to it, that is, ∠ B also increases.

Let us now perform another activity :

Fig. 7.42

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Activity : Construct a scalene triangle (that is a triangle in which all sides are of

different lengths). Measure the lengths of the sides.

Now, measure the angles. What do you

observe?

In ∆ ABC of Fig 7.43, BC is the longest side

and AC is the shortest side.

Also, ∠ A is the largest and ∠ B is the smallest.

Repeat this activity with some other triangles.

We arrive at a very important result of inequalities in a triangle. It is stated in the

form of a theorem as shown below:

Theorem 7.6 : If two sides of a triangle are unequal, the angle opposite to the

longer side is larger (or greater).

You may prove this theorem by taking a point P

on BC such that CA = CP in Fig. 7.43.

Now, let us perform another activity :

Activity : Draw a line-segment AB. With A as centre

and some radius, draw an arc and mark different

points say P, Q, R, S, T on it.

Join each of these points with A as well as with B (see Fig. 7.44). Observe that as

we move from P to T, ∠ A is becoming larger and larger. What is happening to the

length of the side opposite to it? Observe that the length of the side is also increasing;

that is ∠ TAB > ∠ SAB > ∠ RAB > ∠ QAB > ∠ PAB and TB > SB > RB > QB > PB.

Now, draw any triangle with all angles unequal

to each other. Measure the lengths of the sides

(see Fig. 7.45).

Observe that the side opposite to the largest angle

is the longest. In Fig. 7.45, ∠ B is the largest angle

and AC is the longest side.

Repeat this activity for some more triangles and

we see that the converse of Theorem 7.6 is also true.

In this way, we arrive at the following theorem:

Fig. 7.43

Fig. 7.44

Fig. 7.45

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Theorem 7.7 : In any triangle, the side opposite to the larger (greater) angle is

longer.

This theorem can be proved by the method of contradiction.

Now take a triangle ABC and in it, find AB + BC, BC + AC and AC + AB. What

do you observe?

You will observe that AB + BC > AC,

BC + AC > AB and AC + AB > BC.

Repeat this activity with other triangles and with this you can arrive at the following

theorem :

Theorem 7.8 : The sum of any two sides of a

triangle is greater than the third side.

In Fig. 7.46, observe that the side BA of ∆ ABC has

been produced to a point D such that AD = AC. Can you

show that ∠ BCD > ∠ BDC and BA + AC > BC? Have

you arrived at the proof of the above theorem.

Let us now take some examples based on these results.

Example 9 : D is a point on side BC of ∆ ABC such that AD = AC (see Fig. 7.47).

Show that AB > AD.

Solution : In ∆ DAC,

AD = AC (Given)

So, ∠ ADC = ∠ ACD

(Angles opposite to equal sides)

Now, ∠ ADC is an exterior angle for ∆ABD.

So, ∠ ADC > ∠ ABD

or, ∠ ACD > ∠ ABD

or, ∠ ACB > ∠ ABC

So, AB > AC (Side opposite to larger angle in ∆ ABC)

or, AB > AD (AD = AC)

Fig. 7.46

Fig. 7.47

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EXERCISE 7.4

1. Show that in a right angled triangle, the

hypotenuse is the longest side.

2. In Fig. 7.48, sides AB and AC of ∆ ABC are

extended to points P and Q respectively. Also,

∠ PBC < ∠ QCB. Show that AC > AB.

3. In Fig. 7.49, ∠ B < ∠ A and ∠ C < ∠ D. Show that

AD < BC.

4. AB and CD are respectively the smallest and

longest sides of a quadrilateral ABCD

(see Fig. 7.50). Show that ∠ A > ∠ C and

∠ B > ∠ D.

5. In Fig 7.51, PR > PQ and PS bisects ∠ QPR. Prove

that ∠ PSR > ∠ PSQ.

Fig. 7.49

Fig. 7.48

Fig. 7.50

Fig. 7.51

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6. Show that of all line segments drawn from a given point not on it, the perpendicular

line segment is the shortest.

EXERCISE 7.5 (Optional)*

1. ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all

the vertices of ∆ ABC.

2. In a triangle locate a point in its interior which is equidistant from all the sides of the

triangle.

3. In a huge park, people are concentrated at three

points (see Fig. 7.52):

A : where there are different slides and swings

for children,

B : near which a man-made lake is situated,

C : which is near to a large parking and exit.

Where should an icecream parlour be set up so

that maximum number of persons can approach

it?

(Hint : The parlour should be equidistant from A, B and C)

4. Complete the hexagonal and star shaped Rangolies [see Fig. 7.53 (i) and (ii)] by filling

them with as many equilateral triangles of side 1 cm as you can. Count the number of

triangles in each case. Which has more triangles?

(i) (ii)

Fig. 7.53

Fig. 7.52

*These exercises are not from examination point of view.

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7.7 Summary

In this chapter, you have studied the following points :

1. Two figures are congruent, if they are of the same shape and of the same size.

2. Two circles of the same radii are congruent.

3. Two squares of the same sides are congruent.

4. If two triangles ABC and PQR are congruent under the correspondence A ↔ P,

B ↔ Q and C ↔ R, then symbolically, it is expressed as ∆ ABC ≅ ∆ PQR.

5. If two sides and the included angle of one triangle are equal to two sides and the included

angle of the other triangle, then the two triangles are congruent (SAS Congruence Rule).

6. If two angles and the included side of one triangle are equal to two angles and the

included side of the other triangle, then the two triangles are congruent (ASA Congruence

Rule).

7. If two angles and one side of one triangle are equal to two angles and the corresponding

side of the other triangle, then the two triangles are congruent (AAS Congruence Rule).

8. Angles opposite to equal sides of a triangle are equal.

9. Sides opposite to equal angles of a triangle are equal.

10. Each angle of an equilateral triangle is of 60°.

11. If three sides of one triangle are equal to three sides of the other triangle, then the two

triangles are congruent (SSS Congruence Rule).

12. If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse

and one side of other triangle, then the two triangles are congruent (RHS Congruence

Rule).

13. In a triangle, angle opposite to the longer side is larger (greater).

14. In a triangle, side opposite to the larger (greater) angle is longer.

15. Sum of any two sides of a triangle is greater than the third side.

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