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7 | CBSE | Class10 | Health_and_Physical_Education | Points for Discussion
a) What do you think is responsible for Sohan’s habit of smoking and why?
b) Do you think smoking helped reduce Sohan’s stress?
Why?
c) What else could Sohan have done to reduce his stress effectively?
d) What role could Sohan’s parents have played to prevent him from smoking?Consequences of substance abuse
Substance misuse leads to a number of short-term and longterm effects that are detrimental to health.
Short-term effects: These are effects that appear only a few minutes after the intake of substance.
The user feels a false sense of well-being and a pleasant drowsiness.
Some of the short-term effects are distorted vision, hearing and coordination, impaired judgment, bad breath and hangovers.
Long-term effects: Substances having a long-term impact lead to serious damage due to constant and excessive use.
These effects show up over a course of time and are usually caused by progressive damage to different body organs.
Some of the health consequences include irregular eating habits, sleep disorders, poor hygiene resulting in poor health and low immunity.
This predisposes substance abuser to diseases and infections.
Substance abuse in itself is not a cause of HIV/AIDS or STDs but under the influence of drugs, people may engage in risk-behaviours that make them more susceptible to these infections.
Injectable drug users, however, are more prone to HIV in conditions where the users do not sterilise needles and share needles to inject the drug.
Activity 10.6
Discuss how a substance abuser’s life gets affected.
Substance users may even die suddenly from a so-called overdose, when one consumes more than what the body can tolerate.
Death may also occur from long-term damage to the organs of the body.
These substances are especially harmful if consumed during pregnancy.
These substances reach the foetus through blood and harm it by crippling its growth and development or death.
Substance abuse influences not only the individual and family but also the community.
The persons who get addicted often lose interest in other activities be it school, job or any other responsibility.
They are not able to take care of their responsibilities and may become a liability for their families and finally the society.
Furthermore, it is expensive to buy substances or drugs on a regular basis.
Hence, in desperation, addicted individuals may be forced to engage in petty severe crimes.
Prevention of drug abuse
Paying attention to the following may prevent young people from misusing substances —
•
Peer pressure can be managed by being aware of the implications of substance abuse and developing life skills.
One can not only save oneself from adopting risky behaviours under peer pressure but also persuade the peers not to engage in unhealthy behaviours such as substance misuse.
Advocacy with the involvement of local community can be quite useful.
•
The different ways of dealing with negative peer pressure are saying ‘No’ assertively, making an excuse, reversal of pressure, and giving reasons.
•
Young people should also be educated to recognise that a friend is someone who cares, protects and looks after the welfare of their friends rather than force and initiate into unhealthy habits.
•
Young people should be empowered with life skills to know about their body and their life.
They should be taking well-informed and responsible decisions.
•
The family and society should provide the support system for young children in their ‘low’ times.
As the attitudes related to smoking, drinking and misuse of other substances are formed during pre-adolescence and early adolescence, this is an important age to invest in prevention efforts.
The family, community and educational institutions should empower young people with correct and scientific information.
Young children should be encouraged to get engaged in socially productive activities.
For example, one can start a learning centre for the lesser advantaged children in the neighbourhood, one can learn Braille and help at blind school, and one can create awareness about water conservation in the neighbourhood.
This will help them to channelise their energy, being away from the menace of substance abuse and also inculcate social responsibility
• | jehp110.pdf |
0 | CBSE | Class10 | Health_and_Physical_Education | HealtHy Community living
Our way of life, and the utilisation of resources, decide the quality of our personal and community life.
A healthy way of living in the community is ensured not only by the optimal use of resources but also by keeping our environment clean and hygienic.
Let us understand what is meant by healthy community living.
The following figures show community life in three different situations —
1. A village (Fig 11.1)
2. A township in an urban area (Fig 11.2)
3. A slum in an urban area (Fig 11.3)
A village
Mawlynnong Village in East Khasi Hills District of Meghalaya was declared as Asia’s cleanest village in 2003 by Discover
India Magazine.
It is about 75 km away from Shillong, the capital of Meghalaya.
Most of the houses in this village are beautifully decorated with flowers and plants.
As per the Census of 2011, the total population of Mawlynnong is 414.
The main occupation of the villagers is agriculture, but it has also been an age old tradition of ensuring that the surrounding environment is clean.
In fact cleanliness is a collective effort and this practice is ingrained in the people since their childhood.
The people voluntarily sweep the roads and lanes, water the plants in public areas and clean the drains.
A dustbin made out of bamboo is found all along the village.
Everyone makes it a point that dirt and waste are not thrown anywhere.
All the waste from the dustbin is collected and kept in a pit, which the villagers use as manure.
Trees are planted to restore and maintain nature’s balance.
Mawlynnong’s fame as the cleanest village in Asia, is drawing a lot of domestic as well as international tourists, as a result of which tourism is also an important source of employment for the local youths.
Besides, there are many interesting sights to see such as the famous living root bridge in the nearby village of Riwai, which is a fascinating example of indigenous methods of conservation and sustainability.
Local youths are available as enthusiastic and informative guides.
Fig. 11.1: Mawlynnong Village
A township in an urban area
Fig. 11.2: A modern township
This picture is of a modern township derived from an advertisement in a national newspaper.
Many times, catchy slogans such as ‘Get away from noise, pollution, congestion and a cramped life’ are used by builders to sell their apartments.
People are assured a safe and healthy environment with not only the basic amenities, but also other features, such as, shopping arcades, clubs, gymnasiums, gardens, clinics, food market, lots of open space, etc.
A slum in an urban area
Fig. 11.3: A slum area
These pictures characterise life in a slum area in big cities.
As you can see in the picture, people live in overcrowded houses surrounded by stagnant water which is a potential breeding ground for mosquitoes.
We also find railway tracks adjacent to the houses which is dangerous for children and adults too.
In addition, we find heaps of garbage scattered around the houses.
Activity 11.1
After observing the pictures, discuss the following — • Do you think Mawlynnong Village is a successful example of healthy community living?
Give reasons for your answers.
Activity 11.2
You may conduct a survey in your neighborhood and collect information on the following — • Do you think modern townships as depicted in Fig 11.2, can offer its inhabitants a safe and healthy environment, as assured by the builders?
• Make a list of some of the basic amenities and community resources that are collectively shared by members of your community.
• Make a list of some of the community resources that are not available in your locality.
• The pictures shown in Fig 11.3, does not reflect signs of healthy community living.
Do you agree?
Give reasons for your answers.
• On the basis of your observations and survey, complete the following table —
Features appropriate or necessary for healthy community living Features not appropriate or necessary for healthy community living
1. Water supply 1.
Open drains
2. 2.
3. 3.
4. 4.
5. 5.
These activities will enable you to understand various features of healthy community living.
Through these activities, you may have observed that in a community, a group of people live in a particular local area.
Secondly, they share common facilities, which differ from place to place.
In some areas, people have access to facilities, such as, water supply, sanitation, garbage disposal facilities, health care services, recreational facilities, community centre, schools, transportation, etc., but in some places people do not have access to even the most basic amenities.
A community is said to be healthy when its members continuously work together to maintain, improve and expand the available natural resources and avoid their wastage.
The healthy community only be develped when its members recognise their roles and responsibilities.
They strive to inculcate values and attitudes of cooperation, mutual respect, tolerance, kindness, etc.
The role of Panchayati Raj, civil societies and other government institutions is also very important to promote healthy living.
Important features of healthy community living
Maintaining cleanliness of our home and surroundings is an essential feature of healthy community living.
Healthy community is one in which all residents have access to quality education, safe and healthy homes, adequate employment, transportation, physical activity and nutrition.
Living in overcrowded and unhygienic places with excessive noise and pollution, may lead to various forms of illnesses, diseases and stress.
For instance, lack of adequate space, poor ventilation in rooms and toxic fumes in the air, increase the risk of air borne diseases.
Access to basic amenities such as regular water supply, safe drinking water and sanitation, is important for healthy community living.
In the previous classes (Classes VII to IX), it has been learnt that open and unattended garbage dumps are a potential breeding place for flies, cockroaches, insects and so on.
Stagnant water and open drains breed mosquitoes, the carrier of diseases, such as, malaria, dengue and chikungunya.
Water borne diseases, such as, cholera, jaundice, diarrhea and gastroenteritis are also resulted due to water pollution.
Access to basic amenities also includes other provisions such as electricity, hospitals, dispensaries, housing facilities, roads and transportation, schools, colleges, employment, security, recreational facilities and so on.
Imagine a situation when a member in the family falls seriously ill and requires urgent treatment.
What would happen to this person if there are no hospitals in the nearby city, or no transportation facilities to reach there?
To ensure these basic amenities, the role of the local government or panchayats is very important.
They are required to ensure rules and regulations for safety measures, cleanliness, ensure equal access of public amenities to all members, promote adoption of waste management practices, and help ecological restoration and conservation.
However, we cannot depend solely on the government to take the initiative.
The quality of life in a community also depends on how the members of the community work to improve on available resources and ensure that the government provides for these facilities.
You might have seen this logo before.
This is the logo of the Swachh Bharat Abhiyan.
You must have also seen some of the television advertisements through which the government advertises the importance and necessity of staying clean and keeping our environment clean.
In this Clean India campaign — known popularly as the Swachh Bharat Abhiyan, the vision of a Clean and hygienic India, once seen by Mahatma Gandhi, can happen only if members of every community cooperate and accept individual and collective responsibility of keeping themselves, their homes and their surroundings clean.
In order to work collectively, members of a community need to develop values of co-operation, kindness, respect, gratitude, joy, peace and selflessness.
These attributes are important in fostering the principles of collective work and in nurturing healthy social relationships.
The example of Mawlynnong village shows that high income alone is not the only criterion for healthy living.
Cooperative action, responsibility and positive values can also help in improving the quality of life, and healthy living conditions for one and all. | jehp111.pdf |
1 | CBSE | Class10 | Health_and_Physical_Education | A village
Mawlynnong Village in East Khasi Hills District of Meghalaya was declared as Asia’s cleanest village in 2003 by Discover
India Magazine.
It is about 75 km away from Shillong, the capital of Meghalaya.
Most of the houses in this village are beautifully decorated with flowers and plants.
As per the Census of 2011, the total population of Mawlynnong is 414.
The main occupation of the villagers is agriculture, but it has also been an age old tradition of ensuring that the surrounding environment is clean.
In fact cleanliness is a collective effort and this practice is ingrained in the people since their childhood.
The people voluntarily sweep the roads and lanes, water the plants in public areas and clean the drains.
A dustbin made out of bamboo is found all along the village.
Everyone makes it a point that dirt and waste are not thrown anywhere.
All the waste from the dustbin is collected and kept in a pit, which the villagers use as manure.
Trees are planted to restore and maintain nature’s balance.
Mawlynnong’s fame as the cleanest village in Asia, is drawing a lot of domestic as well as international tourists, as a result of which tourism is also an important source of employment for the local youths.
Besides, there are many interesting sights to see such as the famous living root bridge in the nearby village of Riwai, which is a fascinating example of indigenous methods of conservation and sustainability.
Local youths are available as enthusiastic and informative guides.
Fig. 11.1: Mawlynnong Village
A township in an urban area
Fig. 11.2: A modern township
This picture is of a modern township derived from an advertisement in a national newspaper.
Many times, catchy slogans such as ‘Get away from noise, pollution, congestion and a cramped life’ are used by builders to sell their apartments.
People are assured a safe and healthy environment with not only the basic amenities, but also other features, such as, shopping arcades, clubs, gymnasiums, gardens, clinics, food market, lots of open space, etc.
A slum in an urban area
Fig. 11.3: A slum area
These pictures characterise life in a slum area in big cities.
As you can see in the picture, people live in overcrowded houses surrounded by stagnant water which is a potential breeding ground for mosquitoes.
We also find railway tracks adjacent to the houses which is dangerous for children and adults too.
In addition, we find heaps of garbage scattered around the houses.
Activity 11.1
After observing the pictures, discuss the following — • Do you think Mawlynnong Village is a successful example of healthy community living?
Give reasons for your answers.
Activity 11.2
You may conduct a survey in your neighborhood and collect information on the following — • Do you think modern townships as depicted in Fig 11.2, can offer its inhabitants a safe and healthy environment, as assured by the builders?
• Make a list of some of the basic amenities and community resources that are collectively shared by members of your community.
• Make a list of some of the community resources that are not available in your locality.
• The pictures shown in Fig 11.3, does not reflect signs of healthy community living.
Do you agree?
Give reasons for your answers.
• On the basis of your observations and survey, complete the following table —
Features appropriate or necessary for healthy community living Features not appropriate or necessary for healthy community living
1. Water supply 1.
Open drains
2. 2.
3. 3.
4. 4.
5. 5.
These activities will enable you to understand various features of healthy community living.
Through these activities, you may have observed that in a community, a group of people live in a particular local area.
Secondly, they share common facilities, which differ from place to place.
In some areas, people have access to facilities, such as, water supply, sanitation, garbage disposal facilities, health care services, recreational facilities, community centre, schools, transportation, etc., but in some places people do not have access to even the most basic amenities.
A community is said to be healthy when its members continuously work together to maintain, improve and expand the available natural resources and avoid their wastage.
The healthy community only be develped when its members recognise their roles and responsibilities.
They strive to inculcate values and attitudes of cooperation, mutual respect, tolerance, kindness, etc.
The role of Panchayati Raj, civil societies and other government institutions is also very important to promote healthy living.
Important features of healthy community living
Maintaining cleanliness of our home and surroundings is an essential feature of healthy community living.
Healthy community is one in which all residents have access to quality education, safe and healthy homes, adequate employment, transportation, physical activity and nutrition.
Living in overcrowded and unhygienic places with excessive noise and pollution, may lead to various forms of illnesses, diseases and stress.
For instance, lack of adequate space, poor ventilation in rooms and toxic fumes in the air, increase the risk of air borne diseases.
Access to basic amenities such as regular water supply, safe drinking water and sanitation, is important for healthy community living.
In the previous classes (Classes VII to IX), it has been learnt that open and unattended garbage dumps are a potential breeding place for flies, cockroaches, insects and so on.
Stagnant water and open drains breed mosquitoes, the carrier of diseases, such as, malaria, dengue and chikungunya.
Water borne diseases, such as, cholera, jaundice, diarrhea and gastroenteritis are also resulted due to water pollution.
Access to basic amenities also includes other provisions such as electricity, hospitals, dispensaries, housing facilities, roads and transportation, schools, colleges, employment, security, recreational facilities and so on.
Imagine a situation when a member in the family falls seriously ill and requires urgent treatment.
What would happen to this person if there are no hospitals in the nearby city, or no transportation facilities to reach there?
To ensure these basic amenities, the role of the local government or panchayats is very important.
They are required to ensure rules and regulations for safety measures, cleanliness, ensure equal access of public amenities to all members, promote adoption of waste management practices, and help ecological restoration and conservation.
However, we cannot depend solely on the government to take the initiative.
The quality of life in a community also depends on how the members of the community work to improve on available resources and ensure that the government provides for these facilities.
You might have seen this logo before.
This is the logo of the Swachh Bharat Abhiyan.
You must have also seen some of the television advertisements through which the government advertises the importance and necessity of staying clean and keeping our environment clean.
In this Clean India campaign — known popularly as the Swachh Bharat Abhiyan, the vision of a Clean and hygienic India, once seen by Mahatma Gandhi, can happen only if members of every community cooperate and accept individual and collective responsibility of keeping themselves, their homes and their surroundings clean.
In order to work collectively, members of a community need to develop values of co-operation, kindness, respect, gratitude, joy, peace and selflessness.
These attributes are important in fostering the principles of collective work and in nurturing healthy social relationships.
The example of Mawlynnong village shows that high income alone is not the only criterion for healthy living.
Cooperative action, responsibility and positive values can also help in improving the quality of life, and healthy living conditions for one and all. | jehp111.pdf |
2 | CBSE | Class10 | Health_and_Physical_Education | A township in an urban area
Fig. 11.2: A modern township
This picture is of a modern township derived from an advertisement in a national newspaper.
Many times, catchy slogans such as ‘Get away from noise, pollution, congestion and a cramped life’ are used by builders to sell their apartments.
People are assured a safe and healthy environment with not only the basic amenities, but also other features, such as, shopping arcades, clubs, gymnasiums, gardens, clinics, food market, lots of open space, etc.
A slum in an urban area
Fig. 11.3: A slum area
These pictures characterise life in a slum area in big cities.
As you can see in the picture, people live in overcrowded houses surrounded by stagnant water which is a potential breeding ground for mosquitoes.
We also find railway tracks adjacent to the houses which is dangerous for children and adults too.
In addition, we find heaps of garbage scattered around the houses.
Activity 11.1
After observing the pictures, discuss the following — • Do you think Mawlynnong Village is a successful example of healthy community living?
Give reasons for your answers.
Activity 11.2
You may conduct a survey in your neighborhood and collect information on the following — • Do you think modern townships as depicted in Fig 11.2, can offer its inhabitants a safe and healthy environment, as assured by the builders?
• Make a list of some of the basic amenities and community resources that are collectively shared by members of your community.
• Make a list of some of the community resources that are not available in your locality.
• The pictures shown in Fig 11.3, does not reflect signs of healthy community living.
Do you agree?
Give reasons for your answers.
• On the basis of your observations and survey, complete the following table —
Features appropriate or necessary for healthy community living Features not appropriate or necessary for healthy community living
1. Water supply 1.
Open drains
2. 2.
3. 3.
4. 4.
5. 5.
These activities will enable you to understand various features of healthy community living.
Through these activities, you may have observed that in a community, a group of people live in a particular local area.
Secondly, they share common facilities, which differ from place to place.
In some areas, people have access to facilities, such as, water supply, sanitation, garbage disposal facilities, health care services, recreational facilities, community centre, schools, transportation, etc., but in some places people do not have access to even the most basic amenities.
A community is said to be healthy when its members continuously work together to maintain, improve and expand the available natural resources and avoid their wastage.
The healthy community only be develped when its members recognise their roles and responsibilities.
They strive to inculcate values and attitudes of cooperation, mutual respect, tolerance, kindness, etc.
The role of Panchayati Raj, civil societies and other government institutions is also very important to promote healthy living. | jehp111.pdf |
3 | CBSE | Class10 | Health_and_Physical_Education | Fig. 11.3: A slum area
These pictures characterise life in a slum area in big cities.
As you can see in the picture, people live in overcrowded houses surrounded by stagnant water which is a potential breeding ground for mosquitoes.
We also find railway tracks adjacent to the houses which is dangerous for children and adults too.
In addition, we find heaps of garbage scattered around the houses.
Activity 11.1
After observing the pictures, discuss the following — • Do you think Mawlynnong Village is a successful example of healthy community living?
Give reasons for your answers.Activity 11.1
After observing the pictures, discuss the following — • Do you think Mawlynnong Village is a successful example of healthy community living?
Give reasons for your answers.Activity 11.2
You may conduct a survey in your neighborhood and collect information on the following — • Do you think modern townships as depicted in Fig 11.2, can offer its inhabitants a safe and healthy environment, as assured by the builders?
• Make a list of some of the basic amenities and community resources that are collectively shared by members of your community.
• Make a list of some of the community resources that are not available in your locality.
• The pictures shown in Fig 11.3, does not reflect signs of healthy community living.
Do you agree?
Give reasons for your answers.
• On the basis of your observations and survey, complete the following table —
Features appropriate or necessary for healthy community living Features not appropriate or necessary for healthy community living
1. Water supply 1.
Open drains
2. 2.
3. 3.
4. 4.
5. 5.
These activities will enable you to understand various features of healthy community living.
Through these activities, you may have observed that in a community, a group of people live in a particular local area.
Secondly, they share common facilities, which differ from place to place.
In some areas, people have access to facilities, such as, water supply, sanitation, garbage disposal facilities, health care services, recreational facilities, community centre, schools, transportation, etc., but in some places people do not have access to even the most basic amenities.
A community is said to be healthy when its members continuously work together to maintain, improve and expand the available natural resources and avoid their wastage.
The healthy community only be develped when its members recognise their roles and responsibilities.
They strive to inculcate values and attitudes of cooperation, mutual respect, tolerance, kindness, etc.
The role of Panchayati Raj, civil societies and other government institutions is also very important to promote healthy living.Features appropriate or necessary for healthy community living Features not appropriate or necessary for healthy community living
1. Water supply 1.
Open drains
2. 2.
3. 3.
4. 4.
5. 5.
These activities will enable you to understand various features of healthy community living.
Through these activities, you may have observed that in a community, a group of people live in a particular local area.
Secondly, they share common facilities, which differ from place to place.
In some areas, people have access to facilities, such as, water supply, sanitation, garbage disposal facilities, health care services, recreational facilities, community centre, schools, transportation, etc., but in some places people do not have access to even the most basic amenities.
A community is said to be healthy when its members continuously work together to maintain, improve and expand the available natural resources and avoid their wastage.
The healthy community only be develped when its members recognise their roles and responsibilities.
They strive to inculcate values and attitudes of cooperation, mutual respect, tolerance, kindness, etc.
The role of Panchayati Raj, civil societies and other government institutions is also very important to promote healthy living. | jehp111.pdf |
4 | CBSE | Class10 | Health_and_Physical_Education | Prevention of Coronavirus: soCial DistanCing anD Dealing witH stigma
Prevention through social distancing
Social Distancing: Deliberately increasing the physical space between people to avoid spreading illness and staying at least one meter away from other person decreases chances of catching coronavirus (COVID-19).
Social distancing: Dos
•
Stay at home unless absolutely necessary.
•
Keep a distance of at least one meter between yourself and another person.
Social distancing: Don’ts
•
Do not hold events where people have to gather (even if it is a corner meeting with three or four friends, or an evening chat on the chaupal).
•
Do not go to crowded places like markets, shopping, melas, parties.
•
Do not use public transport.
Fig. 11.4: Social distancing dos and don’ts
Courtesy: https://www.mohfw.gov.in/
Dealing with the stigma
What is stigma?
In any epidemic, it is common for individuals to feel stressed and worried because they fear — falling ill and dying.
•
•
approaching health facilities due to fear of becoming infected while in care.
•
losing livelihoods, not being able to work during isolation, and of being dismissed from work.
•
being socially excluded or placed in quarantine because of being associated with the disease.
•
feeling powerless in protecting loved ones and fear of losing loved ones because of the virus or being separated during quarantine.
•
feelings of helplessness, boredom, loneliness and depression due to being isolated and not working towards caring for a dependent.
Stress is caused due to the above fears and being treated as an outcaste or blamed for spreading the disease.
What is the reason behind the stigma?
The level of stigma associated with COVID-19 is based on three main factors — COVID-19 is a new disease about which many things are still being discovered.
•
•
When something is unknown people are worried which leads to fear.
•
Rumours or fake news give wrong information and spreads the fear.
Recognising the stigma
It is very important to recognise stigma and handle it.
Below are four case studies related to these issues.
Read these case studies and recognise the stigma.
CASE 1
You are in the grocery shop, there are several people who are wearing a mask.
You see Babulal, the store owner, going red in his face as he tries to suppress a cough.
CASE 2
Sukhram has come back from Pune where he works as a taxi driver.
They stay in a joint family and you have taken his contact history as advised by your supervisor.
You come to know that Sukhram’s family members have asked him to leave the house.
CASE 3
Beauty works in Delhi as a house maid.
recently she has come back and you have been told that beauty’s employers have asked her to leave as she had a cold.
CASE 4
Surali is a young girl of 11 years.
She and her 8-year-old brother are staying with an aunt as their parents have been asked to go in for isolation.
Surali’s aunt keeps on complaining to you that the children are a big burden on the family’s resources.
What will you feel like if you were Babulal, Rani, Sukhram and Beauty?
•
Babulal has simple cough.
But he is too scared to cough in front of people as he will loose the customers.
•
Sukhram needs family support to help him stay in isolation.
If everyone takes proper precautions the infection need not spread.
•
Beauty has a seasonal cold but she has been asked to leave by her employers.
•
Surali and her brother are two small children who need to be supported and this kind of incident can cause mental stress even in the future.
Child Protection Cell (CPC) should be approached for appropriate measures for helping children in difficult situations.
Fig. 11.5: What does stigma do?
values anD attituDes ConDuCive for HealtHy anD CooPerative living
How to develop values and attitudes conducive for healthy and cooperative living?
These values have to emanate from within a person, but they can also be developed through various other ways.
Camping can be seen as one of these ways which is extremely appropriate for students, particularly during the adolescent period.
What is camping?
Youngsters might have already attended a camp or may have heard about camping activities from friends.
Camping is an outdoor activity, where familiar surroundings are left to spend a night or several nights at an outdoor site.
The location of these sites varies as per the objectives of the trip, time of the year and budget available.
Common camping sites are: visits to villages or rural areas to understand their pattern of living; mountain areas, national parks, lakes, beaches, forests, etc.
One can also plan camping activities in a nearby place, such as, open fields, park, or even in an open space of the school premises.
The basic camping equipment include tents, cooking utensils or gadgets, sleeping bags, first aid kits, ropes, insect repellents,
Fig. 11.6: Camping
11:46:09 AM
etc.
These requirements will vary as per the location of the camping site and objectives of the trip.
However, irrespective of the location, the essence of all camping activities revolves around two main features —
(i) Bonding with nature
(ii) Importance of living in a community.
Importance of camping
Camping gives students a good break, away from the monotony of the classroom.
It gives an opportunity to learn from nature in a joyful, exciting and adventurous manner.
Living in the natural environment with students from different areas, cultures, religions and backgrounds, help then to learn to work as a team in various activities.
These activities differ from adventurous and challenging ones, to the most basic ones, such as, cooking, cleaning, collecting water, etc.
In this process, students learn the importance of self-reliance, teamwork, co-existence, importance of natural resources and organisational skills.
Camping also offers great opportunities for empowerment of the girls.
Fig. 11.7: Trekking Fig. 11.8: Mountaineering Fig. 11.9: Rock climbing
While camping, one can engage in adventure sport, such as, trekking, river rafting, mountaineering, rock climbing, repelling, paragliding, caving.
Adventure sport help to develop courage, self-confidence, leadership qualities and enhance concentration powers.
This is also a great form of physical exercise.
However, these activities should be conducted only in the presence of qualified and professional trainers.
Joining the National Cadet Corps (NCC), National Service Scheme (NSS) or Bharat Scouts and Guides will also give you opportunities to take part in these activities.
11:46:14 AM
Fig. 11.10: Zipline Fig. 11.11: River rafting
A guided walk through the forest helps to explore nature.
It can also complement much of the textual data in your geography and science textbooks through direct experiences.
At the same time, learn to appreciate the beauty and simplicity of nature to make all understand how much disconnected from ‘Mother Earth one has become.’
You realise the importance of nature and how we should take care of it.
Moreover, through interaction with local communities, one learns about edible and medicinal plants, local crafts, firstaid and various other sustainable forms of living.
During the evenings all may gather around campfires which give a great opportunity for social bonding.
Campfires also enable to learn about different cultures through story telling, songs, dances, skits and games.
On many occasions hidden talents of individuals are brought out during these activities.
Activity 11.3
Some of the camping activities have been given in the following table.
Write down the qualities which can be developed through the respective activities.
You may consult other sources for your answers.
| Activities | Qualities developed
| --- | ---
| Hiking, rock climbing, trekking, rope climbing, river rafting, etc. | Confidence, leadership qualities, team- work
| Nature walk Cultural activities — patriotic songs, lectures on Indian culture and heritage through dance, drama, songs, story telling, skits, quizzes, etc.
| Campfire — story telling that emphasize on moral values based on local folk tales, history, games and songs. Interaction with local community members on local crafts, medicines, first aid, food, etc. Yoga
Voluntary services 11:46:15 AM | jehp111.pdf |
5 | CBSE | Class10 | Health_and_Physical_Education | Prevention through social distancing
Social Distancing: Deliberately increasing the physical space between people to avoid spreading illness and staying at least one meter away from other person decreases chances of catching coronavirus (COVID-19).
Social distancing: Dos
•
Stay at home unless absolutely necessary.
•
Keep a distance of at least one meter between yourself and another person.
Social distancing: Don’ts
•
Do not hold events where people have to gather (even if it is a corner meeting with three or four friends, or an evening chat on the chaupal).
•
Do not go to crowded places like markets, shopping, melas, parties.
•
Do not use public transport.
Fig. 11.4: Social distancing dos and don’ts
Courtesy: https://www.mohfw.gov.in/
Dealing with the stigma
What is stigma?
In any epidemic, it is common for individuals to feel stressed and worried because they fear — falling ill and dying.
•
•
approaching health facilities due to fear of becoming infected while in care.
•
losing livelihoods, not being able to work during isolation, and of being dismissed from work.
•
being socially excluded or placed in quarantine because of being associated with the disease.
•
feeling powerless in protecting loved ones and fear of losing loved ones because of the virus or being separated during quarantine.
•
feelings of helplessness, boredom, loneliness and depression due to being isolated and not working towards caring for a dependent.
Stress is caused due to the above fears and being treated as an outcaste or blamed for spreading the disease.
What is the reason behind the stigma?
The level of stigma associated with COVID-19 is based on three main factors — COVID-19 is a new disease about which many things are still being discovered.
•
•
When something is unknown people are worried which leads to fear.
•
Rumours or fake news give wrong information and spreads the fear.
Recognising the stigma
It is very important to recognise stigma and handle it.
Below are four case studies related to these issues.
Read these case studies and recognise the stigma.
CASE 1
You are in the grocery shop, there are several people who are wearing a mask.
You see Babulal, the store owner, going red in his face as he tries to suppress a cough.
CASE 2
Sukhram has come back from Pune where he works as a taxi driver.
They stay in a joint family and you have taken his contact history as advised by your supervisor.
You come to know that Sukhram’s family members have asked him to leave the house.
CASE 3
Beauty works in Delhi as a house maid.
recently she has come back and you have been told that beauty’s employers have asked her to leave as she had a cold.
CASE 4
Surali is a young girl of 11 years.
She and her 8-year-old brother are staying with an aunt as their parents have been asked to go in for isolation.
Surali’s aunt keeps on complaining to you that the children are a big burden on the family’s resources.
What will you feel like if you were Babulal, Rani, Sukhram and Beauty?
•
Babulal has simple cough.
But he is too scared to cough in front of people as he will loose the customers.
•
Sukhram needs family support to help him stay in isolation.
If everyone takes proper precautions the infection need not spread.
•
Beauty has a seasonal cold but she has been asked to leave by her employers.
•
Surali and her brother are two small children who need to be supported and this kind of incident can cause mental stress even in the future.
Child Protection Cell (CPC) should be approached for appropriate measures for helping children in difficult situations.
Fig. 11.5: What does stigma do?Social distancing: Dos
•
Stay at home unless absolutely necessary.
•
Keep a distance of at least one meter between yourself and another person.
Social distancing: Don’ts
•
Do not hold events where people have to gather (even if it is a corner meeting with three or four friends, or an evening chat on the chaupal).
•
Do not go to crowded places like markets, shopping, melas, parties.
•
Do not use public transport.
Fig. 11.4: Social distancing dos and don’ts
Courtesy: https://www.mohfw.gov.in/Social distancing: Don’ts
•
Do not hold events where people have to gather (even if it is a corner meeting with three or four friends, or an evening chat on the chaupal).
•
Do not go to crowded places like markets, shopping, melas, parties.
•
Do not use public transport.
Fig. 11.4: Social distancing dos and don’ts
Courtesy: https://www.mohfw.gov.in/Fig. 11.4: Social distancing dos and don’ts
Courtesy: https://www.mohfw.gov.in/ | jehp111.pdf |
6 | CBSE | Class10 | Health_and_Physical_Education | What is stigma?
In any epidemic, it is common for individuals to feel stressed and worried because they fear — falling ill and dying.
•
•
approaching health facilities due to fear of becoming infected while in care.
•
losing livelihoods, not being able to work during isolation, and of being dismissed from work.
•
being socially excluded or placed in quarantine because of being associated with the disease.
•
feeling powerless in protecting loved ones and fear of losing loved ones because of the virus or being separated during quarantine.
•
feelings of helplessness, boredom, loneliness and depression due to being isolated and not working towards caring for a dependent.
Stress is caused due to the above fears and being treated as an outcaste or blamed for spreading the disease.What is the reason behind the stigma?
The level of stigma associated with COVID-19 is based on three main factors — COVID-19 is a new disease about which many things are still being discovered.
•
•
When something is unknown people are worried which leads to fear.
•
Rumours or fake news give wrong information and spreads the fear.Recognising the stigma
It is very important to recognise stigma and handle it.
Below are four case studies related to these issues.
Read these case studies and recognise the stigma.
CASE 1
You are in the grocery shop, there are several people who are wearing a mask.
You see Babulal, the store owner, going red in his face as he tries to suppress a cough.
CASE 2
Sukhram has come back from Pune where he works as a taxi driver.
They stay in a joint family and you have taken his contact history as advised by your supervisor.
You come to know that Sukhram’s family members have asked him to leave the house.
CASE 3
Beauty works in Delhi as a house maid.
recently she has come back and you have been told that beauty’s employers have asked her to leave as she had a cold.
CASE 4
Surali is a young girl of 11 years.
She and her 8-year-old brother are staying with an aunt as their parents have been asked to go in for isolation.
Surali’s aunt keeps on complaining to you that the children are a big burden on the family’s resources.CASE 1
You are in the grocery shop, there are several people who are wearing a mask.
You see Babulal, the store owner, going red in his face as he tries to suppress a cough.CASE 2
Sukhram has come back from Pune where he works as a taxi driver.
They stay in a joint family and you have taken his contact history as advised by your supervisor.
You come to know that Sukhram’s family members have asked him to leave the house.CASE 3
Beauty works in Delhi as a house maid.
recently she has come back and you have been told that beauty’s employers have asked her to leave as she had a cold.CASE 4
Surali is a young girl of 11 years.
She and her 8-year-old brother are staying with an aunt as their parents have been asked to go in for isolation.
Surali’s aunt keeps on complaining to you that the children are a big burden on the family’s resources.What will you feel like if you were Babulal, Rani, Sukhram and Beauty?
•
Babulal has simple cough.
But he is too scared to cough in front of people as he will loose the customers.
•
Sukhram needs family support to help him stay in isolation.
If everyone takes proper precautions the infection need not spread.
•
Beauty has a seasonal cold but she has been asked to leave by her employers.
•
Surali and her brother are two small children who need to be supported and this kind of incident can cause mental stress even in the future.
Child Protection Cell (CPC) should be approached for appropriate measures for helping children in difficult situations.
Fig. 11.5: What does stigma do? | jehp111.pdf |
7 | CBSE | Class10 | Health_and_Physical_Education | What is camping?
Youngsters might have already attended a camp or may have heard about camping activities from friends.
Camping is an outdoor activity, where familiar surroundings are left to spend a night or several nights at an outdoor site.
The location of these sites varies as per the objectives of the trip, time of the year and budget available.
Common camping sites are: visits to villages or rural areas to understand their pattern of living; mountain areas, national parks, lakes, beaches, forests, etc.
One can also plan camping activities in a nearby place, such as, open fields, park, or even in an open space of the school premises.
The basic camping equipment include tents, cooking utensils or gadgets, sleeping bags, first aid kits, ropes, insect repellents,
Fig. 11.6: Camping
11:46:09 AM
etc.
These requirements will vary as per the location of the camping site and objectives of the trip.
However, irrespective of the location, the essence of all camping activities revolves around two main features —
(i) Bonding with nature
(ii) Importance of living in a community.Fig. 11.6: Camping
11:46:09 AM
etc.
These requirements will vary as per the location of the camping site and objectives of the trip.
However, irrespective of the location, the essence of all camping activities revolves around two main features —
(i) Bonding with nature
(ii) Importance of living in a community.Importance of camping
Camping gives students a good break, away from the monotony of the classroom.
It gives an opportunity to learn from nature in a joyful, exciting and adventurous manner.
Living in the natural environment with students from different areas, cultures, religions and backgrounds, help then to learn to work as a team in various activities.
These activities differ from adventurous and challenging ones, to the most basic ones, such as, cooking, cleaning, collecting water, etc.
In this process, students learn the importance of self-reliance, teamwork, co-existence, importance of natural resources and organisational skills.
Camping also offers great opportunities for empowerment of the girls.
Fig. 11.7: Trekking Fig. 11.8: Mountaineering Fig. 11.9: Rock climbing
While camping, one can engage in adventure sport, such as, trekking, river rafting, mountaineering, rock climbing, repelling, paragliding, caving.
Adventure sport help to develop courage, self-confidence, leadership qualities and enhance concentration powers.
This is also a great form of physical exercise.
However, these activities should be conducted only in the presence of qualified and professional trainers.
Joining the National Cadet Corps (NCC), National Service Scheme (NSS) or Bharat Scouts and Guides will also give you opportunities to take part in these activities.
11:46:14 AM
Fig. 11.10: Zipline Fig. 11.11: River rafting
A guided walk through the forest helps to explore nature.
It can also complement much of the textual data in your geography and science textbooks through direct experiences.
At the same time, learn to appreciate the beauty and simplicity of nature to make all understand how much disconnected from ‘Mother Earth one has become.’
You realise the importance of nature and how we should take care of it.
Moreover, through interaction with local communities, one learns about edible and medicinal plants, local crafts, firstaid and various other sustainable forms of living.
During the evenings all may gather around campfires which give a great opportunity for social bonding.
Campfires also enable to learn about different cultures through story telling, songs, dances, skits and games.
On many occasions hidden talents of individuals are brought out during these activities.
Activity 11.3
Some of the camping activities have been given in the following table.
Write down the qualities which can be developed through the respective activities.
You may consult other sources for your answers.
| Activities | Qualities developed
| --- | ---
| Hiking, rock climbing, trekking, rope climbing, river rafting, etc. | Confidence, leadership qualities, team- work
| Nature walk Cultural activities — patriotic songs, lectures on Indian culture and heritage through dance, drama, songs, story telling, skits, quizzes, etc.
| Campfire — story telling that emphasize on moral values based on local folk tales, history, games and songs. Interaction with local community members on local crafts, medicines, first aid, food, etc. Yoga
Voluntary services 11:46:15 AM | jehp111.pdf |
8 | CBSE | Class10 | Health_and_Physical_Education | assessment
I. Multiple Choice Questions
1. Which of the following is the most important criterion for healthy community living?
a) Cooperative action
b) Kindness
c) Harmony
d) Safety measures
2. Community participation is essential for
a) ecological restoration
b) public hygiene and health
c) social harmony
d) All of the above
II. State whether True or false
a) Maintenance of public hygiene is the sole responsibility of the government.
b) Community health and individual health are closely interlinked
c) Camping is the only means for developing values of cooperative living
d) Overcrowding increases the risk to air borne diseases
e) Interpersonal relationships are as important as our physical environment.
III. Answer the following Questions
1. What is your vision of a healthy community?
Give at least 3 examples.
2. A camping trip has been arranged by your school.
Two girls from your community are not allowed to go for the trip.
What are the points you would highlight to convince the parents on the importance of camping trip for girls.
3. Arrange a collage of pictures showing various activities in a Camping Trip.
4. Give two suggestions for keeping your community healthy, both at the individual and collective level, based on the points enumerated below.
For example,
Record your suggestions regarding cleanliness drive within the community.
The first one has been done for you.
I will ensure that, I do not throw litter around.
We will ensure that, there are no open garbage dumps in the community.
11:46:15 AM
a) Record your suggestions regarding rules and regulations within the community for safety measures.
I will ensure that.
We will ensure that.
b) Record your suggestions for ensuring equal access of public amenities to all members.
I will ensure that.
We will ensure that.
c) Record your suggestions for adopting effective waste management practices.
I will ensure that.
We will ensure that.
d) Record your suggestions for ecological restoration and conservation within your community.
I will ensure that.
We will ensure that.
e) Record your suggestions for ensuring kindness and consideration to the aged and differently abled.
I will ensure that.
We will ensure that.
f) Record your suggestions for counseling provisions for the youth, recreational facilities, etc.
I will arrange for.
We will arrange for.
10.
Imagine that you have been asked to prepare an activity schedule for a day in a camping trip.
First write down the location of the camping site and objectives of the camping trip and include other details in the following table.
Location of camping site.
Objectives of the camping trip.
Table : Schedule for a camping trip
Time Activities Equipment Required No. of participants No. of Teachers/ Instructors Expected Outcome Budget
11:46:16 AMI. Multiple Choice Questions
1. Which of the following is the most important criterion for healthy community living?
a) Cooperative action
b) Kindness
c) Harmony
d) Safety measures
2. Community participation is essential for
a) ecological restoration
b) public hygiene and health
c) social harmony
d) All of the aboveb) Kindnessc) Harmony
d) Safety measures
2. Community participation is essential for
a) ecological restoration
b) public hygiene and health
c) social harmony
d) All of the aboveII. State whether True or false
a) Maintenance of public hygiene is the sole responsibility of the government.
b) Community health and individual health are closely interlinked
c) Camping is the only means for developing values of cooperative living
d) Overcrowding increases the risk to air borne diseases
e) Interpersonal relationships are as important as our physical environment. | jehp111.pdf |
0 | CBSE | Class10 | Health_and_Physical_Education | Social HealtH
You have already learnt that, ‘Health is a state of complete well-being and includes physical, mental, emotional and social health, Many animals show remarkable social behavior.
Humans too are social animals and much of their behaviour is based on social norms laid down by the human societies since time immemorial.
Therefore, social health is a very important issue for all human beings — men-women, young-old, educateduneducated, rich-poor.
Those children who learn to become socially healthy remain so all their lives.
This chapter, will deal with the ways and means of growing up as socially healthy individuals and also helping them to clarify myths related to coronavirus.
Myths and facts related to coronavirus
Statement: A person with coronavirus can recover fully and not be infectious any longer.
Fact: 80 per cent of the people have recovered from the disease without needing special treatment.
But information on the virus treatment is still being researched.
Statement: Eating raw garlic, sesame seeds will protect you against the virus.
Fact: Garlic is a healthy food that has other benefits but does not protect you against the coronavirus.
Statement: The virus can die easily once it is out of the body.
Fact: We do not know about this particular virus as of now. Similar viruses (SARS, MERS) survive from 8 to 24 hours depending on types of surfaces.
Statement: You can get COVID-19 through mosquito bites Fact: The coronavirus cannot be spread through the bite of a mosquito.
It is spread thorough droplets spread when an infected person sneezes or coughs.
After learning this lesson one should be able to explain what is social health, help all to remain socially healthy and spread awareness about benefits of social health.Myths and facts related to coronavirus
Statement: A person with coronavirus can recover fully and not be infectious any longer.
Fact: 80 per cent of the people have recovered from the disease without needing special treatment.
But information on the virus treatment is still being researched.
Statement: Eating raw garlic, sesame seeds will protect you against the virus.
Fact: Garlic is a healthy food that has other benefits but does not protect you against the coronavirus.
Statement: The virus can die easily once it is out of the body.
Fact: We do not know about this particular virus as of now. Similar viruses (SARS, MERS) survive from 8 to 24 hours depending on types of surfaces.
Statement: You can get COVID-19 through mosquito bites Fact: The coronavirus cannot be spread through the bite of a mosquito.
It is spread thorough droplets spread when an infected person sneezes or coughs.
After learning this lesson one should be able to explain what is social health, help all to remain socially healthy and spread awareness about benefits of social health. | jehp112.pdf |
1 | CBSE | Class10 | Health_and_Physical_Education | Activity 12.1
Read the following conversation between an old grandfather, Mr. Das and 12-year-old grandson’s friend Ravi.
Mr. Das : How are you Ravi and how is your grandfather?
Ravi : I am fine but my grandfather always feels angry and unhappy.
He criticises everyone all the time, so no one wants to sit and talk with him. He is boring and not as loving as you are.
Why does he behave like that Dadu?
Mr. Das : It is because he has been unwell for a long time.
Your grandpa is not getting company from any one, this makes him feel lonely and isolated.
In other words, he is suffering physically, as well as mentally due to his old age.
He gets angry because he is unsure of himself now and annoys others by his comments.
He does not think that he is acceptable in society now.
Which is why, he expresses his frustrations by being rude to others without any reason and fault of theirs.
Why don’t you spend time with him? Talk to him about your friends, your day in school, a happy incident, a sad one also and you may observe the change in him within a few days.
Ravi : Thank you, Dadu.
Now I understand that if someone is sick and emotionally unhappy, he would likely be angry and alienate others.
Answer in brief
1. Give reasons for the loneliness of Ravi’s grandfather.
2. State situations that made Ravi think his grandfather was alienating others.
3. Say yes or no
(a) Do you agree that physical, mental and social health are interrelated?
(b) Living with others harmoniously requires a person to be socially healthy.
State the reason for your choice.Activity 12.2
• You are in a tricky situation.
Your best friend had a fight with another friend and you have seen that your best friend is at fault.
You are asked to intervene to bring back peace.
Will you announce that the cause of the fight was your best friend?
This may become a source of annoyance to your best friend but if you did, that is Justice.
• If you help, the two boys become friends again, that is fraternity.
• Your friend Arun noticed that food was being distributed free of charge on the footpath.
A boy wearing tattered dirty clothes was repeatedly being sent to the end of the queue.
Arun held the boy’s hand and insisted that the boy be given the food packet then and there and not at the end.
Arun believed in Equality of opportunity for all.
• If you listen carefully to others opinions on a subject even if they do not match yours and try to understand others’ viewpoints, you believe in liberty.
Read the preamble to the constitution of India given at the beginning of all NCERT text books.
It is reprinted here for you.
(photocopy of preamble of constitution).
Fig. 12.1: Preamble to the Constitution of India
Everyone is a member of a social group and everyone is part of their peer group, family and kin, community, city, region and country, as well as the physical and biological environment.
Would you agree that the country’s constitution provides for Justice, Liberty, Equality and Fraternity which should be adopted in life for living happily and having consideration for all others in the group?
Let us understand these four terms of our constitution which are directly related to “social health of a person or a country”, and then try to define social health.
In other words, if one is socially healthy, than will be able to develop interpersonal relations, through maintaining equality, fraternity and justice.
Let us now try and define social health.Definition of ‘social health’
Social health may be defined as the ability to form satisfying interpersonal relationships with others.
One who is able to make positive relationships and acquires the ability to adapt in different social situations and act appropriately as per the situation concerned, and can be called a socially healthy person.
Need for developing social health
Primitive humans were hunter-gatherers, who lived in small groups or clusters and spent their lives at individual levels.
Around 10,000 years ago, they moved near rivers to grow their own food and began to live together.
With the passage of time, they started to live in a society and developed a language for interaction with each other.
As ‘human society’ progressed, an individual became part of many social groups for example a member of a peer group, a family and kin, a class in school, a native of a region and a citizen of one’s country.
Social changes occur from time to time and many societies lay down norms and values for group living.
Urbanisation has brought about many changes which are different from traditional rural societies.
Social attributes, of people however, remain the same as they are based on interrelationship between members of any society and necessary for accomplishing tasks requiring teamwork.
School is one of the platforms for learning social skills. | jehp112.pdf |
2 | CBSE | Class10 | Health_and_Physical_Education | Role of vaRious institutions
Promotion of social health among children requires collective efforts and appropriate skills.
All people in the field of education have to think about this.
Role of teacher training institutes and other organisations
School teachers need a degree, diploma or certificates from a teacher training institute.
There are a number of organisation concerned with school education.
The syllabus of teacher training courses should include the topics of important issues for children of which health should be one.
Organisations such as NCERT, SCERT, DIET should periodically organise workshops, publish journals and provide inservice training programmes on health issues including social health.
Role of schools in promoting social health
Schools have a very significant role to play in the promotion of health and safety of children.
They spend a lot of time in school in early years.
School environment forms ideal setting for acquiring knowledge of healthy choices of food.
It is in school that they participate in physical activity through sports, games, yoga, gymnastics, exercise and gain benefits of each.
School helps children to learn social skills which assist in establishing lifelong healthy behavior.
Children learn team spirit and training in rules and regulations of social wellbeing just as in the defense services, which are inbuilt in training of defense service personnel, so they work together as a team to protect our nation.
In order to promote social health, the school should have a positive environment where children mingle with teachers, peer group and non-teaching staff without fear and apprehensions.
Teacher have a significant role in this regard.
Role of teachers
It is well known that teachers are the mentors and therefore, the teacher training institutes should make training in leadership and mentorship a part of teacher training.
Also well known saying that ‘example in better than precept’ Students easily learn to be socially healthy if the teacher herself or himself sets an example rather than lecture on social health.
A good teacher ensures that students grow up into physically, mentally and socially healthy individuals.
In doing this the teacher has to make efforts to ensure that students —
(i) eat a proper diet and perform regular exercise and physical activity which are requirements of good physical health.
(ii) inculcate friendship between classmates, indulge in amicable behavior with others in school so that students develop a helpful nature.
These are necessary for mental and emotional health.
3. Teachers should train students in learning life skills like —
a) empathy and self awareness
b) effective communication so that they develop healthy Interpersonal relationships
c) problem-solving and decision-making to learn to be stress free
d) creative and critical thinking
e) coping with stress and emotion.
The above are absolutely essential for developing social skills in order to be socially healthy.
Social skills help a student to have a desirable self image and self esteem and also self confidence.
This makes children acquire ability to live harmoniously in the society.
Role of technology in building social health
Technology has made communication convenient through Mobiles, Skype, Facebook, Twitter, Instagram and E-mail messages.
Knowledge about different media can help to develop the skills to access the appropriate media for accurate information on a specific topic or issues.
More so, because media brings awareness, and provides access to global knowledge and learning.
But refrain from using mobiles and viewing the T.V.
for long periods as that tends to be counter-productive and reduce interaction time with others.
This makes us socially withdrawn.
Although media is a source of information, all of it may not be true or reliable.
It is advisable to seek guidenes from a trusted adult while accessing media and internet.
Moreover, there is a need to understand about real and fake news or information as these affects our attitude and behavior.
Role of students in building habits of social health
Social health comes from social skills.
A few important guidelines for promoting social skills are outlined below.
(i) Building self awareness is an important skill: Practice self-care by developing habits of cleanliness and hygiene, keeping away from substance abuse, engaging in physical activity and regularly consuming a balanced diet.
(ii) Do not be blameful and judgmental: Remember when you point a finger at others, three fingers of yours point towards you.
Hence indulge in knowing yourself better.
It helps to discriminate between a good and a bad act of yours and helps to make friends.
Empathy and self awareness go hand in hand.
When you seea child hitting an animal, tell that child that if they were hit, they would feel the same pain.
That is empathy for others and leads to self awareness.
(iii) Learn to identify your own mistakes: There is no harm in saying sorry for a mistake and rectifying it.
People will have faith in you if you did so.
(iv) Make an effort to reconnect with old relationships and friendships: The socially healthy person makes an effort to contact and meet old friends, to remember enjoyable periods of childhood which can be a good way to beat stress and spend leisure time.
(v) Appreciate yourself and others: But never let your ego rule your behaviour which can sometimes drive you to lose a relationship.
(vi) Try and be respectful, positive and supportive towards the needy, the physically and mentally challenged, the downtrodden and those belonging to faith and cultures other than yours.
Lend an ear to others opinion.
It teaches tolerance.
Tolerance and appreciation are virtues in socially healthy individuals.
Role of family in inculcating social skills in young family members
The importance of family in inculcating social skills among, children is paramount.
Guardians or parents are the prime teachers and caretakers who feed them healthy food and are also their play mates.
They have to be aware themselves to be role model and make children appreciate as they grow.
The benefits of enjoying nutritious food and being aware of a balanced diet is of paramount importance.
Watching of television for longer duration is bad for students.
Parents themselves have to be cautious in what they do in front of children and how long they are in front of the TV.
However, at the same time good TV programmes are an avenue for learning.
Encourage them to indulge in physical activity to build their muscles and bones.
Fit body and fitness depends primarily on proper diet and healthy body and mind.
The parents and elder family members need to guide them in social skills.
Parents and elders have to be accessible to children for necessary guidance.
They form a “safety network’.
They are to guide adolescents through advice, answer their queries and occupy them through activities to keep them away from involvement in addictive substances like tobacco, drugs and alcohol which they might believe are helpful in tiding over stress.
However, these substances have deleterious effects on the growing body. | jehp112.pdf |
3 | CBSE | Class10 | Health_and_Physical_Education | Role of teacher training institutes and other organisations
School teachers need a degree, diploma or certificates from a teacher training institute.
There are a number of organisation concerned with school education.
The syllabus of teacher training courses should include the topics of important issues for children of which health should be one.
Organisations such as NCERT, SCERT, DIET should periodically organise workshops, publish journals and provide inservice training programmes on health issues including social health.Role of schools in promoting social health
Schools have a very significant role to play in the promotion of health and safety of children.
They spend a lot of time in school in early years.
School environment forms ideal setting for acquiring knowledge of healthy choices of food.
It is in school that they participate in physical activity through sports, games, yoga, gymnastics, exercise and gain benefits of each.
School helps children to learn social skills which assist in establishing lifelong healthy behavior.
Children learn team spirit and training in rules and regulations of social wellbeing just as in the defense services, which are inbuilt in training of defense service personnel, so they work together as a team to protect our nation.
In order to promote social health, the school should have a positive environment where children mingle with teachers, peer group and non-teaching staff without fear and apprehensions.
Teacher have a significant role in this regard. | jehp112.pdf |
4 | CBSE | Class10 | Health_and_Physical_Education | Role of technology in building social health
Technology has made communication convenient through Mobiles, Skype, Facebook, Twitter, Instagram and E-mail messages.
Knowledge about different media can help to develop the skills to access the appropriate media for accurate information on a specific topic or issues.
More so, because media brings awareness, and provides access to global knowledge and learning.
But refrain from using mobiles and viewing the T.V.
for long periods as that tends to be counter-productive and reduce interaction time with others.
This makes us socially withdrawn.
Although media is a source of information, all of it may not be true or reliable.
It is advisable to seek guidenes from a trusted adult while accessing media and internet.
Moreover, there is a need to understand about real and fake news or information as these affects our attitude and behavior.Role of students in building habits of social health
Social health comes from social skills.
A few important guidelines for promoting social skills are outlined below.
(i) Building self awareness is an important skill: Practice self-care by developing habits of cleanliness and hygiene, keeping away from substance abuse, engaging in physical activity and regularly consuming a balanced diet.
(ii) Do not be blameful and judgmental: Remember when you point a finger at others, three fingers of yours point towards you.
Hence indulge in knowing yourself better.
It helps to discriminate between a good and a bad act of yours and helps to make friends.
Empathy and self awareness go hand in hand.
When you seea child hitting an animal, tell that child that if they were hit, they would feel the same pain.
That is empathy for others and leads to self awareness.
(iii) Learn to identify your own mistakes: There is no harm in saying sorry for a mistake and rectifying it.
People will have faith in you if you did so.
(iv) Make an effort to reconnect with old relationships and friendships: The socially healthy person makes an effort to contact and meet old friends, to remember enjoyable periods of childhood which can be a good way to beat stress and spend leisure time.
(v) Appreciate yourself and others: But never let your ego rule your behaviour which can sometimes drive you to lose a relationship.
(vi) Try and be respectful, positive and supportive towards the needy, the physically and mentally challenged, the downtrodden and those belonging to faith and cultures other than yours.
Lend an ear to others opinion.
It teaches tolerance.
Tolerance and appreciation are virtues in socially healthy individuals.Role of family in inculcating social skills in young family members
The importance of family in inculcating social skills among, children is paramount.
Guardians or parents are the prime teachers and caretakers who feed them healthy food and are also their play mates.
They have to be aware themselves to be role model and make children appreciate as they grow.
The benefits of enjoying nutritious food and being aware of a balanced diet is of paramount importance.
Watching of television for longer duration is bad for students.
Parents themselves have to be cautious in what they do in front of children and how long they are in front of the TV.
However, at the same time good TV programmes are an avenue for learning.
Encourage them to indulge in physical activity to build their muscles and bones.
Fit body and fitness depends primarily on proper diet and healthy body and mind.
The parents and elder family members need to guide them in social skills.
Parents and elders have to be accessible to children for necessary guidance.
They form a “safety network’.
They are to guide adolescents through advice, answer their queries and occupy them through activities to keep them away from involvement in addictive substances like tobacco, drugs and alcohol which they might believe are helpful in tiding over stress.
However, these substances have deleterious effects on the growing body. | jehp112.pdf |
5 | CBSE | Class10 | Health_and_Physical_Education | These are —
•
Empathy: Is the ability to understand another’s feelings in a particular situation.
•
Self Awareness: Is the recognition of one’s character, strengths, areas of growth, beliefs and values.
•
Effective Communication: Is having skills of communication that facilitate interaction with people in positive ways.
•
Interpersonal Relationships: Building relationships of friendship and goodwill with all others.
•
Problem solving: Is the ability to resolve challenges.
•
Decision Making: Is the quality of analyzing problem to find and act to reach an appropriate solution.
•
Creative thinking: Is the ability to do something in a novel manner.
•
Critical Thinking: Is the capacity to analyse multiple perspectives and objectively evaluate the same.
•
Coping with Stress and emotional distress: These refer to management and regulation of one’s emotions and moments of stress.
All these life skills help to develop desirable social health and live happily in a society.
1. The teacher needs to play an active role in discouraging her students from engaging in socially unhealthy practices such as vindictive attitudes, selfishness, jealousy and culture of hatred.
For this, teacher has to sacrifice time and energy.
It is however, necessary for a teacher and school authorities to understand that more than finishing the syllabus and passing exams, it is the teacher’s responsibility to build good human beings.
Home has a large role to play but it has been the mission of teachers to contribute towards grooming students into socially healthy adults
2. Another role of a teacher in inculcating ‘social health’ is to have a friendly, stress-free atmosphere in class.
This can happen if students are engaged in ‘group activities’, especially activities for ‘experiential learning’.
Members of groups are reshuffled from time to time for team activities so that students may understand that it takes all kinds of people to make the world and the socially healthy groups can live in harmony despite differing in opinions.
Group activities build team spirit and remove boredom.iDeas foR PRomoting social health of stuDents
1. If there is a canteen, permit sale of healthy food like fruits or fruit juices.
School authorities should be aware as to what is being sold for consumption immediately outside school and permit only those selling healthy eatables or low fat and healthy snacks.
2. Provide for safe drinking water.
3. Provide for examination and treatment of students with poor health conditions, bad teeth and weak vision.
Organise health checks for them.
4. Organise variety of co-curricular activities, and ensure participation of maximum number of students, and training them leadership as well as team spirit.
5. Organise interclass sport and games competitions and finally, a sport day and prize distribution for encouragement.
6. Include a games period every working day in the time table.
7. Seek and insist on cooperation of parents so that even guardians who have not learnt social skills and have unhealthy habits may learn through their participation.
It shall be community service by the school and a means to encourage everyone to undertake healthy living.
8. Have events for teaching and non teaching staff and allow voluntary participation in games using school equipment.
9. Have experts for teaching Yoga, Taekwondo, Judo and Karate for self-defense and utilise school premises for the same.
10.
Schools can have activities involving children to learn to love and respect all living beings, plants and animals.
11.
Encourage team games, ‘House system’ should be adopted, which helps children to interact with students of other classes too.
12.
Organise periodic slide shows or movies with valuable messages.
In this manner students, teachers, parents and community will realise the benefits of being physically, mentally and socially healthy.
And what a great way it will be to build a healthy nation!assessment
I. Answer the following Questions
1. Define
a) Health
b) Mental health
c) Social health
2. What do you mean by life skills?
Enlist them.
3. What are the characteristics of a socially healthy person?
4. Suggest three ways in which social health can be promoted in children by each of the following —
a) School
b) Teachers
c) Family
5. Write a short story of your choice to express life skills.
6. What kind of society do you envisage if majority of its members are socially healthy?I. Answer the following Questions
1. Define
a) Health
b) Mental health
c) Social health
2. What do you mean by life skills?
Enlist them.
3. What are the characteristics of a socially healthy person?
4. Suggest three ways in which social health can be promoted in children by each of the following —
a) School
b) Teachers
c) Family
5. Write a short story of your choice to express life skills.
6. What kind of society do you envisage if majority of its members are socially healthy? | jehp112.pdf |
0 | CBSE | Class10 | Health_and_Physical_Education | Agencies And AwArds Promoting HeAltH, sPorts And YogA 13
Awards and rewards are the essential parts of performance enhancing efforts in health, sport, games and Yoga.
It is managed and monitored by specific agencies.
There are a number of agencies that promote health, physical education, including sport and yoga in our country.
In India there are provisions for various kinds of awards.
The chapter shall discuss those agencies and awards that promote health, physical education, sport and yoga. | jehp113.pdf |
1 | CBSE | Class10 | Health_and_Physical_Education | NCERT
National Council of Educational Research and Training (NCERT) is an apex autonomous organisation set up in 1961 by the Government of India for quality improvement of school education and teacher education.
It functions in the areas of educational research, training and development of curriculum and instructional materials for school education.
It’s another critical role is to assist and advise the Central and State governments on policies and programmes in school education.
The major constituent units of the NCERT are —
Activity 13.1
Where is the headquarter of NCERT located?
1. National Institute of Education (NIE) undertakes research and development activities related to pedagogical aspects of curriculum, instructional materials and supplementary materials.
It prepares national curricular policy documents, develops database and various types of materials and organises in-service training for different target groups.
2. Central Institute of Educational Technology CIET is concerned with development of educational technology, design and production of media software.
It holds programmes to build competencies of media personnel and need-based researches.
It evaluates activities undertaken and studies carried out to assess the effectiveness of materials and programmes.
3. Pandit Sunderlal Sharma Central Institute of Vocational Education (PSSCIVE) is located at Bhopal and organises research, development, training and extension programmes related to Work education and Vocational education.Activity 13.1
Where is the headquarter of NCERT located?
1. National Institute of Education (NIE) undertakes research and development activities related to pedagogical aspects of curriculum, instructional materials and supplementary materials.
It prepares national curricular policy documents, develops database and various types of materials and organises in-service training for different target groups.
2. Central Institute of Educational Technology CIET is concerned with development of educational technology, design and production of media software.
It holds programmes to build competencies of media personnel and need-based researches.
It evaluates activities undertaken and studies carried out to assess the effectiveness of materials and programmes.
3. Pandit Sunderlal Sharma Central Institute of Vocational Education (PSSCIVE) is located at Bhopal and organises research, development, training and extension programmes related to Work education and Vocational education.Where is the headquarter of NCERT located?
1. National Institute of Education (NIE) undertakes research and development activities related to pedagogical aspects of curriculum, instructional materials and supplementary materials.
It prepares national curricular policy documents, develops database and various types of materials and organises in-service training for different target groups.
2. Central Institute of Educational Technology CIET is concerned with development of educational technology, design and production of media software.
It holds programmes to build competencies of media personnel and need-based researches.
It evaluates activities undertaken and studies carried out to assess the effectiveness of materials and programmes.
3. Pandit Sunderlal Sharma Central Institute of Vocational Education (PSSCIVE) is located at Bhopal and organises research, development, training and extension programmes related to Work education and Vocational education.Activity 13.2
Find out how many clusters are taking part is CBSE competitive sport.
4. Regional Institutes of Education (RIEs) are located at Ajmer, Bhopal, Bhubaneswar and Mysore.
The RIEs cater to the needs of school education and teacher education (pre-service and in-service education) including those teachers educators in the States and UTs under their respective jurisdictions.
Besides these, yet another regional institute, known as NorthEastern Regional Institute of Education (NERIE) is located at Shillong.
Do You Know?
CBSE has started competitive sports for all private schools affiliated to CBSE since 1988-89.
5. NCERT: Role in Health and Physical Education NCERT as an apex body includes functioning of Health and Physical Education like all other subject areas.
The National Curriculum Framework (NCF-2005) prepared by NCERT considers health and physical education a compulsory subject from Class I to X and optional subject at Classes XI and XII.
As a follow up of NCF2005, NCERT has prepared syllabus on health and physical education which has been approved by the National Steering Committee set by the Government of India.
Health and Physical Education components have also been included in the pre-service training courses running at each RIE’s. It contributes to the policy formulation process of the Central and State governments related to Health and Physical Education. | jehp113.pdf |
2 | CBSE | Class10 | Health_and_Physical_Education | CBSE
Central Board of Secondary Education (CBSE) is the first board of education that was set up in 1921 under jurisdiction of Rajputana, Central India and Gwalior.
Government of India decided to set up a Joint Board in 1929 and it was named as the ‘Board of High School and Intermediate Education.
Later in 1952, the constitution of the Board was amended and it was named as ‘Central Board of Secondary Education’.
In 1962 the Board was reconstituted once again with the objectives: (i) to serve the educational institutions more effectively (ii) to be responsive to the educational needs of those students whose parents were employed in the Central Government and had frequently transferable jobs.
The major functions of the CBSE have been to develop curriculum for all the subjects at the secondary and higher secondary levels, to conduct evaluation and examination activities, to organise teacher training workshops, to develop resource materials for teachers and students, to publish some text books for secondary and senior secondary classes and to monitor various academic projects.
The CBSE has been preparing syllabi on Health and Physical Education, conducting competitive sport activities for schools affiliated to it and promoting the transaction of Health and Physical Education at secondary and higher secondary levels.
It has also been ensuring that the Comprehensive and Continuous Evaluation (CCE) is focused on health and physical education activities.School education agencies in states
We have found that there are government agencies at the state level for preparing curriculam, training teachers and other functionaries and evaluating the performance of students.
There is a State Council of Educational Research and Training (SCERT) in almost all the major States.
This institution is responsible for preparation of syllabi and textbooks for all the classes at primary and upper primary stages.
In some of the States and Union Territories, the State Institutes of Education (SIE) or the Directorates of Education perform these roles.
All these institutions at the state level perform these roles for the subject area of Health and Physical Education also.
Then there are State Boards of Education, that are responsible for preparation of Syllabi and textbooks and evaluation of students of all classes at secondary and higher secondary stages.
These institutions also conduct in-service teacher training for all the subjects including the subject of Health and Physical Education.
Under SCERTs there are District Institutes of Education and Training (DIETs) that are responsible for pre-service teacher education at the elementary stage (primary and upper primary stages).SAI
Sport Authority of India (SAI) was set up by the Government of India in 1984, aimed to promote sport in India and developing excellence by upgrading the skills of the Indian sport persons.
Some of the prestigious institutes run by the SAI are:
(i) Netaji Subhash National Institute of Sport, (Chandigarh, Sonipat, Lucknow, Guwahati, Imphal, Bangaluru, Madurai, Kolkata, Patiala and Gandhinagar), and (ii) Laxmibai National College of Physical Education (Thiruvananthapuram).
Activity 13.3
Education is a concurrent subject of the state government.
Activity 13.4
Where is the head office of SAI.
The main objective to establish SAI was to upgrade the skills of the budding sport talents in India.
In order to attain this objective 23 training centers spread over the entire country are functioning.
Through various schemes formulated for sub-junior, junior and senior levels, it ensures that the enthusiasm for sport is widened among different age groups of people.
SAI has also provided competitive exposures to the talented sportpersons.
Some of the SAI schemes formulated for the promotion of sport in India include National Sport Talent Contest.
The Sport Projects Development Area and the Sport Hostel Scheme.
Besides, the Army Boys Sport Company (ABSC) in association with the Indian Army authorities is also run by SAI.
SAI provides facilities like sport equipment for the trainees, kit, stipend as well as coaches.
Currently, there are eight ABSCs all over India.
Another scheme proposed by SAI is called SAI Training Centers (STC).
This Scheme has been successful to a great extent, in fulfilling SAI’s objective of spotting and nurturing sport talents.
Another ambicious schemes run by SAI are - Special Area Games (SAG), and Centre of excellence (COX), producing high level National/ International sport.Activity 13.3
Education is a concurrent subject of the state government. | jehp113.pdf |
3 | CBSE | Class10 | Health_and_Physical_Education | NSNIS
After independence, on May 7, 1961 the National Institute of Sport (NIS) was set up by the Government of India for the development of sport at the Motibagh Palace of the then Maharaja of Patiala.
With the objective of developing sport in the country on scientific lines and to train the coaches in different sport disciplines.
On January 23, 1973, it was renamed as Netaji Subhas National Institute of Sport (NSNIS).
Presently, NSNIS Patiala is Asia’s one of the best Sport Institute and is popularly known as the “Mecca” of Indian Sport.
It has produced coaches of high caliber and significantly contributed in rendering their expertise and assistance in the preparation of the national teams for various International competitions.
The Institute is producing high caliber coaches in sport disciplines through its Diploma in Sport Coaching and Master Course in Sport Coaching.
The Institute is presently conducting 6 courses pertaining to sport.
(i) Master Degree in Sport Coaching (ii) Post Graduate Diploma in Sport Medicine (iii) Diploma Course in Sport Coaching (iv) Diploma Course in Sport Coaching (for candidates from North-East Region and Andaman and Nicobar) (v) Refresher Course and (vi)Certificate Course in Sport Coaching.
(www.nsnis.org) Other Government Institutions promoting health, physical education, sport and Yoga are Laxmibai National Institute of Physical Education, Gwalior, Indira Gandhi Institute of physical education and Sport Science (University Delhi) New Delhi, Government College of Physical Education, Patiala etc.
Besides these there are many other colleges, Faculties and Departments of Physical Education and Sport under various state and central Universities, which promotes sport, games and physical education.
Further, a detailed information on various institutions conducting Bachelor’s Degree, Postgraduate Degree, Master of Philosophy in Physical Education and Doctorate degree programs could be obtained from their respective websites.Sport schools
Another important agency is Motilal Nehru School of Sport, RAI which was founded in July 1973 by the Government of Haryana to provide excellent educational facilities with extra emphasis on sport to deserving students.
It is a fully residential and coeducation school.
Besides this there are other sport schools such as: G.V.Raja Sport School, Trivandrum; Maharana Pratap Sport College, Dehradun, Sport College, Lucknow, Sport School Jalandhar etc.
that cater to the development of sport in the school education sector.
There are also provisions for sport scholarships and sport hostels which prospective students can avail.
The information can be obtained from concerned institutions or through internet.Yoga institutions
There are a number of institutions devoted to promotion of yoga, providing everything from in depth courses to flexible drop-in classes.
As the style of yoga and approach to teaching varies at each center, it is important to give proper thought to your needs before approaching the Institute.
Some of the Institutes that offer yoga education are as follows:
1. Ramamani Iyengar Memorial Yoga Institue, Pune
2. Krishnamacharya Yoga Mandiram, Chennai
3. Bihar School of Yoga, Munger
4. The Yoga Institute, Mumbai
5. Kaivalyadham Shriram Mahadevji Yoga Samiti, Lonavala, Pune
6. Morarji Desai National Institute of Yoga, New Delhi
7. Swami Vivekananda Yoga Anusandhan Samsthana (SVYASA), Bengaluru
8. Patanjali Yoga Pith, Haridwar
9. Malviya Toga Sansthan BHU, Varanasi
10.
Uttarakhand University of Sanskrit and Yogic Sciences, Haridwar Additional information about above Yoga institutions could be obtained from their respective websites.1. Ramamani Iyengar Memorial Yoga Institue, Pune2. Krishnamacharya Yoga Mandiram, Chennai3. Bihar School of Yoga, Munger4. The Yoga Institute, Mumbai
5. Kaivalyadham Shriram Mahadevji Yoga Samiti, Lonavala, Pune6. Morarji Desai National Institute of Yoga, New Delhi7. Swami Vivekananda Yoga Anusandhan Samsthana (SVYASA), Bengaluru8. Patanjali Yoga Pith, Haridwar9. Malviya Toga Sansthan BHU, Varanasi
10.
Uttarakhand University of Sanskrit and Yogic Sciences, Haridwar Additional information about above Yoga institutions could be obtained from their respective websites. | jehp113.pdf |
4 | CBSE | Class10 | Health_and_Physical_Education | Rajiv Gandhi Khel Ratna Award
The Rajiv Gandhi Khel Ratna award was instituted in the year 1991-
Fig. 13.1: Rajiv Gandhi Khel Ratna
92.
It is India’s highest honour given for achievement in sport.
The words “Khel Ratna” literally mean “Sport Gem” in Hindi.
The award is named after the late Rajiv Gandhi, former Prime Minister of India.
It carries a medal, a scroll of honour and a cash component of ` 7,50,000.
The Khel Ratna was devised to be an overarching honour, conferred for outstanding sporting performance, whether by an individual or a team, across all sporting disciplines in a given year.Fig. 13.1: Rajiv Gandhi Khel Ratna
92.
It is India’s highest honour given for achievement in sport.
The words “Khel Ratna” literally mean “Sport Gem” in Hindi.
The award is named after the late Rajiv Gandhi, former Prime Minister of India.
It carries a medal, a scroll of honour and a cash component of ` 7,50,000.
The Khel Ratna was devised to be an overarching honour, conferred for outstanding sporting performance, whether by an individual or a team, across all sporting disciplines in a given year.Dronacharya Award
Fig. 13.2: Dronacharya award
Dronacharya Award was instituted in 1985 to honour eminent coaches who have done ‘outstanding and meritorious’ work consistently with a singularity of purpose for raising the standards of sportpersons to highest performance in National and International events.
As the best sportperson award is named Arjuna Award, it is appropriate that the coaching award is named after Dronacharya, as he was the Guru of Arjuna.
The award comprises a plaque (bronze statuette of Dronacharya), a scroll of honour and a cash prize of ` 5,00,000 (Rupees five lakh).
Activity 13.5
| Gather | information
| regarding the benefits you can get from different institutions/agencies working in the areas of games and sport, health and physical education and sport training from different sources, such as by discussing with your teachers, sport teachers, and relevant publications or through internet. Identify the types of help you can get from each one of them to promote your abilities in games and sport.
12:27:36 PM
Arjuna Award
The Arjuna Award was instituted in 1961 by the Government of India to recognise outstanding achievement in games and sport.
The award carries a cash prize of ` 500,000 a bronze statuette of Arjuna and a scroll.
Over the years the scope of the award has been expanded and a large number of sportpersons who belonged to the pre-Arjuna Award era were also included in the list.
Further, the number of disciplines for which the award is given was increased to include indigenous games and the differently abled category.
The Government has recently revised the scheme for the Arjuna Award.
According to the revised guidelines, to be eligible for the Award, a sportperson should not only have had good performance consistently for the previous three years at the international level with excellence for the year for which the Award is recommended, but should also have shown qualities of leadership, sportpersonship and a sense of discipline.
Fig. 13.3: Arjuna award
From the year 2001, the award is given only in disciplines falling under the following categories:
•
Olympic Games/Asian Games/Commonwealth Games / World Cup/World Championship Disciplines and Cricket
•
Indigenous Games
•
Sport for the Physically Challenged
Dhyan Chand Award
The Dhyan Chand Award is a Life Time Achievement given to the veteran sportpersons of India for their achievements in their respective fields of sport.
It is named after Dhyan Chand the legendary Indian hockey player.
This is a new award instituted by the Government of India in the year 2002.
The award carries a cash prize of ` 5,00,000/- (Rupees five lakh), a statuette and scroll of honour.
The main objective of the award is to bestow honour on those sportpersons who have contributed a lot to their respective sport by their performance and who still continue to contribute to the promotion of sport after their retirement from the active sporting career.
Fig. 13.4: Dhyan Chand award
Maulana Abul Kalam Azad (MAKA) Trophy
Government instituted MAKA Trophy in 1956-57 as a tool for promoting the competitive sport amongst colleges and Universities.
The top overall performing University in the InterUniversity tournaments in India is given the MAKA Trophy, which is a rolling trophy along with cash prize.
The cash prize has also been recently enhanced to Rs.10 lakh, Rs.
5 lakh and Rs.3 lakh for the Universities securing first, second and third position respectively in Inter-University tournaments.
Fig. 13.5: Maka trophy
12:27:36 PM | jehp113.pdf |
5 | CBSE | Class10 | Health_and_Physical_Education | Activity 13.5
| Gather | information
| regarding the benefits you can get from different institutions/agencies working in the areas of games and sport, health and physical education and sport training from different sources, such as by discussing with your teachers, sport teachers, and relevant publications or through internet. Identify the types of help you can get from each one of them to promote your abilities in games and sport.
12:27:36 PM
Arjuna Award
The Arjuna Award was instituted in 1961 by the Government of India to recognise outstanding achievement in games and sport.
The award carries a cash prize of ` 500,000 a bronze statuette of Arjuna and a scroll.
Over the years the scope of the award has been expanded and a large number of sportpersons who belonged to the pre-Arjuna Award era were also included in the list.
Further, the number of disciplines for which the award is given was increased to include indigenous games and the differently abled category.
The Government has recently revised the scheme for the Arjuna Award.
According to the revised guidelines, to be eligible for the Award, a sportperson should not only have had good performance consistently for the previous three years at the international level with excellence for the year for which the Award is recommended, but should also have shown qualities of leadership, sportpersonship and a sense of discipline.
Fig. 13.3: Arjuna award
From the year 2001, the award is given only in disciplines falling under the following categories:
•
Olympic Games/Asian Games/Commonwealth Games / World Cup/World Championship Disciplines and Cricket
•
Indigenous Games
•
Sport for the Physically Challenged
Dhyan Chand Award
The Dhyan Chand Award is a Life Time Achievement given to the veteran sportpersons of India for their achievements in their respective fields of sport.
It is named after Dhyan Chand the legendary Indian hockey player.
This is a new award instituted by the Government of India in the year 2002.
The award carries a cash prize of ` 5,00,000/- (Rupees five lakh), a statuette and scroll of honour.
The main objective of the award is to bestow honour on those sportpersons who have contributed a lot to their respective sport by their performance and who still continue to contribute to the promotion of sport after their retirement from the active sporting career.
Fig. 13.4: Dhyan Chand award
Maulana Abul Kalam Azad (MAKA) Trophy
Government instituted MAKA Trophy in 1956-57 as a tool for promoting the competitive sport amongst colleges and Universities.
The top overall performing University in the InterUniversity tournaments in India is given the MAKA Trophy, which is a rolling trophy along with cash prize.
The cash prize has also been recently enhanced to Rs.10 lakh, Rs.
5 lakh and Rs.3 lakh for the Universities securing first, second and third position respectively in Inter-University tournaments.
Fig. 13.5: Maka trophy
12:27:36 PMArjuna Award
The Arjuna Award was instituted in 1961 by the Government of India to recognise outstanding achievement in games and sport.
The award carries a cash prize of ` 500,000 a bronze statuette of Arjuna and a scroll.
Over the years the scope of the award has been expanded and a large number of sportpersons who belonged to the pre-Arjuna Award era were also included in the list.
Further, the number of disciplines for which the award is given was increased to include indigenous games and the differently abled category.
The Government has recently revised the scheme for the Arjuna Award.
According to the revised guidelines, to be eligible for the Award, a sportperson should not only have had good performance consistently for the previous three years at the international level with excellence for the year for which the Award is recommended, but should also have shown qualities of leadership, sportpersonship and a sense of discipline.
Fig. 13.3: Arjuna award
From the year 2001, the award is given only in disciplines falling under the following categories:
•
Olympic Games/Asian Games/Commonwealth Games / World Cup/World Championship Disciplines and Cricket
•
Indigenous Games
•
Sport for the Physically ChallengedOlympic Games/Asian Games/Commonwealth Games / World Cup/World Championship Disciplines and Cricket
•Indigenous Games
•Sport for the Physically Challenged | jehp113.pdf |
6 | CBSE | Class10 | Health_and_Physical_Education | Maulana Abul Kalam Azad (MAKA) Trophy
Government instituted MAKA Trophy in 1956-57 as a tool for promoting the competitive sport amongst colleges and Universities.
The top overall performing University in the InterUniversity tournaments in India is given the MAKA Trophy, which is a rolling trophy along with cash prize.
The cash prize has also been recently enhanced to Rs.10 lakh, Rs.
5 lakh and Rs.3 lakh for the Universities securing first, second and third position respectively in Inter-University tournaments.
Fig. 13.5: Maka trophy
12:27:36 PMFig. 13.5: Maka trophy
12:27:36 PMRashtriya Khel Protsahan Puraskar
The RKPP award was instituted in 2009 by the Govt.
of India to recognise the contribution in sport by entities other than sport persons and coaches.
The objective of the award is to encourage and promote corporate involvement in the promotion and development of sport.
The award carries a citation and a trophy in each categories, like
Fig. 13.6: Rashtriya Khel Protsahan Puraskar
1. The community sport identification and nurturing of budding young talent
2. Financial support for sport excellence.
3. Establishment and management of sport academics of excellence.
4. Employment of sport persons and sport welfare measure.Fig. 13.6: Rashtriya Khel Protsahan Puraskar
1. The community sport identification and nurturing of budding young talent
2. Financial support for sport excellence.
3. Establishment and management of sport academics of excellence.
4. Employment of sport persons and sport welfare measure.Tenzing Norgay National Adventure Award
The TNNAA was instituted in 1993 by the Govt.
of India.
It is named after Tenzing Norgay, who is one of the first two individual to reach the summit of Mount Everest in 1953 along with Sir Edmund Percival Hillary.
The award is given to individual for Excelling Adventure Activities on land, Sea and Air.
It carries a statuette, Certificate and a cash prize of 5 Lakhs.Other sports awards in India
Apart from the awards that have been discussed so far, there are some other special awards that are given by the Government of India to the medal winners in the international sport events.
Some State governments also confer these awards.
Generally, the Governments give away some cash awards for winning medals or cups in the international championships, Olympic Games, the World cup or World Championships, the Asian and Commonwealth Games or Championships.
The players who become victorious in the game of Chess and Billiards or Snooker as well as the junior sportpersons who win medals in the World, Asian and Commonwealth Championships, are also given these awards.
Fig. 13.7: TNNA Award
Activity 13.5
Gather information of the current year’s award winners in the field of Games.
Do You Know?
Tenzing Norgay was the first person from India to climb Mount Everest in 29, May 1953.Activity 13.5
Gather information of the current year’s award winners in the field of Games.
Do You Know?
Tenzing Norgay was the first person from India to climb Mount Everest in 29, May 1953.Do You Know?
Tenzing Norgay was the first person from India to climb Mount Everest in 29, May 1953. | jehp113.pdf |
0 | CBSE | Class10 | Health_and_Physical_Education | Health and Physical EducationTextbook for Class X
1:15:10 PM
First Edition
August 2020 Shravana 1942
Reprinted
October 2021 Ashwin 1943 December 2021 Agrahayana 1943
PD 50T RSP
© National Council of Educational Research and Training, 2020
Publication Team
Head, Publication : Anup Kumar Rajput
Division Chief Editor : Shveta Uppal
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Publication Team
Head, Publication : Anup Kumar Rajput
Division Chief Editor : Shveta Uppal
Chief Production : Arun Chitkara
Officer
Chief Business : Vipin Dewan
Manager Production Assistant : Sunil Kumar
Cover and Layout
DTP Cell, PD
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No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher.
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The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable.
OFFICES OF THE PUBLICATION DIVISION, NCERT
NCERT Campus Sri Aurobindo Marg
| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
| CWC Campus Kolkata 700 114 | Phone : 033-25530454
| CWC Complex Guwahati 781 021 | Phone : 0361-2674869
Publication Team
Head, Publication : Anup Kumar Rajput
Division Chief Editor : Shveta Uppal
Chief Production : Arun Chitkara
Officer
Chief Business : Vipin Dewan
Manager Production Assistant : Sunil Kumar
Cover and Layout
DTP Cell, PD
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No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher.
This book is sold subject to the condition that it shall not, by way of trade, be lent, re-sold, hired out or otherwise disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published.
The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable.
OFFICES OF THE PUBLICATION DIVISION, NCERT
NCERT Campus Sri Aurobindo Marg
| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
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Head, Publication : Anup Kumar RajputDivision Chief Editor : Shveta UppalChief Production : Arun Chitkara
OfficerChief Business : Vipin Dewan | jehp1ps.pdf |
1 | CBSE | Class10 | Health_and_Physical_Education | Cover and Layout
DTP Cell, PD
ALL RIGHTS RESERVED
No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher.
This book is sold subject to the condition that it shall not, by way of trade, be lent, re-sold, hired out or otherwise disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published.
The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable.
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| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
| CWC Campus Kolkata 700 114 | Phone : 033-25530454
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No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher.
This book is sold subject to the condition that it shall not, by way of trade, be lent, re-sold, hired out or otherwise disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published.
The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable.
OFFICES OF THE PUBLICATION DIVISION, NCERT
NCERT Campus Sri Aurobindo Marg
| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
| CWC Campus Kolkata 700 114 | Phone : 033-25530454
| CWC Complex Guwahati 781 021 | Phone : 0361-2674869
ALL RIGHTS RESERVED
No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher.
This book is sold subject to the condition that it shall not, by way of trade, be lent, re-sold, hired out or otherwise disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published.
The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable.
OFFICES OF THE PUBLICATION DIVISION, NCERT
NCERT Campus Sri Aurobindo Marg
| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
| CWC Campus Kolkata 700 114 | Phone : 033-25530454
| CWC Complex Guwahati 781 021 | Phone : 0361-2674869
OFFICES OF THE PUBLICATION DIVISION, NCERT
NCERT Campus Sri Aurobindo Marg
| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
| CWC Campus Kolkata 700 114 | Phone : 033-25530454
| CWC Complex Guwahati 781 021 | Phone : 0361-2674869
NCERT Campus Sri Aurobindo Marg
| New Delhi 110 016 | Phone : 011-26562708
| --- | ---
| 108, 100 Feet Road Bengaluru 560 085 | Phone : 080-26725740
| Navjivan Trust Building Ahmedabad 380 014 | Phone : 079-27541446
| CWC Campus Kolkata 700 114 | Phone : 033-25530454
| CWC Complex Guwahati 781 021 | Phone : 0361-2674869
ISBN 978-93-5292-335-91077 – HealtH aNd PHySIcal educatIoN
Textbook for Class X
160.00
Printed on 80 GSM paper
Published at the Publication Division by the Secretary, National Council of Educational Research and Training, Sri Aurobindo Marg, New Delhi 110016 and printed at International PrintO-Pac Ltd., C-4 to C-9, Housery Complex, Ph.
II Extn., Noida 201305 (U.P.)Textbook for Class X
160.00Printed on 80 GSM paper
Published at the Publication Division by the Secretary, National Council of Educational Research and Training, Sri Aurobindo Marg, New Delhi 110016 and printed at International PrintO-Pac Ltd., C-4 to C-9, Housery Complex, Ph.
II Extn., Noida 201305 (U.P.) | jehp1ps.pdf |
2 | CBSE | Class10 | Health_and_Physical_Education | aCknowledgemenTs
The National Council of Educational Research and Training (NCERT) acknowledges the valuable contribution of the following individuals and organisations for reviewing and refining the manuscript of this book.
Amit Sharma, TGT Physical Education, Kendriya Vidyalaya No.-1, Delhi Cantt.
; Asha Arora, Science Teacher, CIE, Department of Education, University of Delhi; Bhagwati Yadav, Yoga Teacher, KV NSG, Manesar, Gurugram, Haryana; Guru Dutt Ghai, Professor, Department of Health and Physical Education, Laxmibai Institute of Physical Education, Gwalior, Madhya Pradesh; Indu Sharma, Assistant Professor, Yoga, Morarji Desai National Institute of Yoga, New Delhi; Namrata Khanna, Consultant, Food Safety and Standards Authority of India (FSSAI), FDA Bhawan, New Delhi; Neetu Sharma, Yoga Therapist, Morarji Desai National Institute of Yoga, New Delhi; N.K.
Dutta, Associate Professor and Head, Physical Education, Ichchhanath Surat, Gujarat; Preeti Shukla, TGT Physical Education, Kendriya Vidyalaya, Tagore Garden, New Delhi; Priti Verma, TGT, Physical Education, Kendriya Vidyalaya, Shivgarh, Raibareli, Uttar Pradesh; Rajveer Singh, Teacher, Physical Education, Jawahar Navodaya Vidyalaya, Baghra, Muzaffarnagar Uttar Pradesh; Sadhana Arya, Former Guest Faculty, Morarji Desai National Institute of Yoga, New Delhi; Samiran Chakrabarty, Associate Professor, IGIPESS, University of Delhi, B-Block, Vikas Puri, New Delhi; Shivani Bansal Moghe, Consultant, Food Safety and Standards Authority of India (FSSAI), New Delhi; Vikram Singh, Professor, Department of Physical Education & Sports, JNU, New Delhi; Abhimanyu Singh, Professor and Head, Department of Physical Education, Banaras Hindu University, Varanasi; Archana Chahal, Professor, Physical Education University of Allahabad, Prayagraj; Ashwani Bhati, D.M.
School, R.I.E.
Ajmer; Atul Dubey, Faculty of Physical Education, I.T.O., Ajmer Azad Khan, Head, Sportz Village Pvt.
Ltd, Bangalore; Bijaya Kumar Malik, Assistant Professor, NCERT, New Delhi; Deepak Lakhera, DMS, RIE Bhopal; Eram S. Rao, Associate Professor, Department of Food Technology, Bhaskaracharya College of Applied Science, University of Delhi; Jitendra Verma, TGT, Health and Physical Education, Kendriya Vidyalaya Sangthan, Sec-II, Rohini, New Delhi; Kamlesh Soni, DMS, R.I.E.
Bhubaneswar; Kusum Yadav, Rao Birender Singh College of Education, Rewari; Mahendra Barua, T.G.T., Physical Education, R.I.E.
Bhopal; Ravipal Singh, K.V.S.
Shakurbasti (KVS), Delhi;
Sujit Panigrahi, Founder and C.E.O., Education Startup Fitness 365, Sohna Road, Gurugram, Haryana;
We acknowledge the contribution of Blue Fish Designs Pvt.
Ltd. for making the illustrations for this textbook, Azra Khatoon, Shefali Sharma, and Aakansha, Junior Project Fellow, NCERT, New Delhi for going through the material, Rani Sharma (PA) for creating a CRC, Mohd.
Kashif, Ruby Malik and Ravi Sharma for typing the material and Bittoo, D.T.P.
Operator for laying out the text.
The contributions of Soumma Chandra, Assistant Editor (Contractual), and Rajshree Saini, DTP Operator (Contractual) are also acknowledged for editing and preparing the layout this book.
(viii)
The Council gratefully acknowledges the valuable suggestions and feedback received during try-out by teachers.
How To use THis book
It is expected that the transaction of this book titled Health and Physical Education for Class X will enable the teachers and students to transform theoretical knowledge into action.
Further the teachers will be empowered to bring out the positive change through modification during learning and practicing various games and sports among students.
In schools, where infrastructural facilities are not available, the teacher may improvise them with the help of students like playground, equipments, rules of the games, etc.
This will also develop creativity among children.
Playing games and sports in all conditions important for holistic health.
The book contains 13 chapters.
The first chapter deals with Physical Education and its relationship with other subjects.
The second chapter explains the effects of physical activities on human body.
The third chapter discusses the issues related to growth and development during adolescence.
Chapters 4, 5, 6, and 7 discuss the details of individual and team games.
Efforts have been made to include latest rules and regulations of various games and sports and other related areas.
Since the rules and regulations keep on changing, teachers and students may visit the official website of the recognised Federation or Association of the different games and sports.
The names of some websites are given separately.
The teachers and students may also see some of the videos of different games and sports given in the official websites related to a particular federation for understanding the rules and learning the skills.
Yoga for Healthy Living is the eighth chapter.
Chapter 9 explains dietary considerations and food quality.
Safety for healthy living, healthy community living and social health are discussed in Chapters 10, 11 and 12 respectively.
The last chapter provides details about the agencies and awards promoting health, sport and yoga.
The book also includes information about corona virus, more specifically known as COVID-19 causing massive loss of human lives and creating panic across the world.
It is important to prevent and deal with this challenge and to empower the younger generation to face this challenge.
The book has provided links as well as, added the relevant information at suitable places.
(x)
The overall objective of this book is to make games and sports joyfull and thereby making childern physically fit, mentally alert and emotionally strong.
The teacher may also encourage to select games or sports of their choice by maintaining social distancing
for practical experience.
Yogic practices need to be more focussed and considered as an integral part of their everyday life in the present context.
The teacher should also emphasise on developing life skills, such as, understanding self, social awareness, team building, cooperation, empathy, communication skill, creative thinking and also development of values.
The students can correlate the given activities in the textbook with their day-to-day life.
As a teacher, you have to understand that this textbook is different from other subject textbooks in the sense that its contents need to be understood well and applied throughout life for one’s own well being and that of others.
Its use should not, therefore, be solely examination driven.
Even a general discussion from time to time would be useful.
Activities included in it, are of practical nature and enjoyable and one can make sure that the concepts are clarified by involving students in experiential learning.
Physical activities and exercises of all kinds have been given to ensure the development of fitness and questioning skills, including life skills.
We would welcome your feedback on this book in terms of — How did you like this textbook?
What are your experiences in organising or being a part of various activities?
What were the dificulties faced by you?
What changes would you like to see in the next version of this book?
Do write to us on all these and other matters related to this textbook.
You could be a parent, a teacher, a student or just a casual reader.
You can send your feedback in the form given at the end of this book to the undersigned.
We sincerely hope you enjoy this book and learn more than it offers.
Saroj Yadav
| | Professor and Dean (Academic)
| --- | ---
| New Delhi | National Council of Educational
| May 2020 | Research and Training
| jehp1ps.pdf |
3 | CBSE | Class10 | Health_and_Physical_Education | (viii)
The Council gratefully acknowledges the valuable suggestions and feedback received during try-out by teachers.How To use THis book
It is expected that the transaction of this book titled Health and Physical Education for Class X will enable the teachers and students to transform theoretical knowledge into action.
Further the teachers will be empowered to bring out the positive change through modification during learning and practicing various games and sports among students.
In schools, where infrastructural facilities are not available, the teacher may improvise them with the help of students like playground, equipments, rules of the games, etc.
This will also develop creativity among children.
Playing games and sports in all conditions important for holistic health.
The book contains 13 chapters.
The first chapter deals with Physical Education and its relationship with other subjects.
The second chapter explains the effects of physical activities on human body.
The third chapter discusses the issues related to growth and development during adolescence.
Chapters 4, 5, 6, and 7 discuss the details of individual and team games.
Efforts have been made to include latest rules and regulations of various games and sports and other related areas.
Since the rules and regulations keep on changing, teachers and students may visit the official website of the recognised Federation or Association of the different games and sports.
The names of some websites are given separately.
The teachers and students may also see some of the videos of different games and sports given in the official websites related to a particular federation for understanding the rules and learning the skills.
Yoga for Healthy Living is the eighth chapter.
Chapter 9 explains dietary considerations and food quality.
Safety for healthy living, healthy community living and social health are discussed in Chapters 10, 11 and 12 respectively.
The last chapter provides details about the agencies and awards promoting health, sport and yoga.
The book also includes information about corona virus, more specifically known as COVID-19 causing massive loss of human lives and creating panic across the world.
It is important to prevent and deal with this challenge and to empower the younger generation to face this challenge.
The book has provided links as well as, added the relevant information at suitable places.
(x)
The overall objective of this book is to make games and sports joyfull and thereby making childern physically fit, mentally alert and emotionally strong.
The teacher may also encourage to select games or sports of their choice by maintaining social distancing
for practical experience.
Yogic practices need to be more focussed and considered as an integral part of their everyday life in the present context.
The teacher should also emphasise on developing life skills, such as, understanding self, social awareness, team building, cooperation, empathy, communication skill, creative thinking and also development of values.
The students can correlate the given activities in the textbook with their day-to-day life.
As a teacher, you have to understand that this textbook is different from other subject textbooks in the sense that its contents need to be understood well and applied throughout life for one’s own well being and that of others.
Its use should not, therefore, be solely examination driven.
Even a general discussion from time to time would be useful.
Activities included in it, are of practical nature and enjoyable and one can make sure that the concepts are clarified by involving students in experiential learning.
Physical activities and exercises of all kinds have been given to ensure the development of fitness and questioning skills, including life skills.
We would welcome your feedback on this book in terms of — How did you like this textbook?
What are your experiences in organising or being a part of various activities?
What were the dificulties faced by you?
What changes would you like to see in the next version of this book?
Do write to us on all these and other matters related to this textbook.
You could be a parent, a teacher, a student or just a casual reader.
You can send your feedback in the form given at the end of this book to the undersigned.
We sincerely hope you enjoy this book and learn more than it offers.
Saroj Yadav
| | Professor and Dean (Academic)
| --- | ---
| New Delhi | National Council of Educational
| May 2020 | Research and Training
| jehp1ps.pdf |
0 | CBSE | Class10 | Mathematics | REAL NUMBERS
1.1 Introduction
In Class IX, you began your exploration of the world of real numbers and encountered irrational numbers.
We continue our discussion on real numbers in this chapter.
We begin with two very important properties of positive integers in Sections 1.2 and 1.3, namely the Euclid’s division algorithm and the Fundamental Theorem of Arithmetic.
Euclid’s division algorithm, as the name suggests, has to do with divisibility of integers.
Stated simply, it says any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b. Many of you probably recognise this as the usual long division process.
Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers.
We touch upon a few of them, and use it mainly to compute the HCF of two positive integers.
The Fundamental Theorem of Arithmetic, on the other hand, has to do something with multiplication of positive integers.
You already know that every composite number can be expressed as a product of primes in a unique way— this important fact is the Fundamental Theorem of Arithmetic.
Again, while it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics.
We use the Fundamental Theorem of Arithmetic for two main applications.
First, we use it to prove the irrationality of many of the numbers you studied in Class IX, such as 2, 3 and 5.
Second, we apply this theorem to explore when exactly the decimal expansion of a rational number, say (0)
p q
q
, is terminating and when it is non- terminating repeating.
We do so by looking at the prime factorisation of the denominator q of p q. You will see that the prime factorisation of q will completely reveal the nature
of the decimal expansion of p q.
So let us begin our exploration.
1.2 The Fundamental Theorem of Arithmetic
In your earlier classes, you have seen that any natural number can be written as a product of its prime factors.
For instance, 2 = 2, 4
=
2 × 2, 253 = 11 × 23, and so on.
Now, let us try and look at natural numbers from the other direction.
That is, can any natural number be obtained by multiplying prime numbers?
Let us see.
Take any collection of prime numbers, say 2, 3, 7, 11 and 23.
If we multiply some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers (In fact, infinitely many).
Let us list a few :
| 7 × 11 × 23 = 1771 | 3 × 7 × 11 × 23 = 5313
| 2 × 3 × 7 × 11 × 23 = 10626 | 23 × 3 × 73 = 8232
22 ×
3
× 7 × 11 × 23 = 21252 and so on.
Now, let us suppose your collection of primes includes all the possible primes.
What is your guess about the size of this collection?
Does it contain only a finite number of integers, or infinitely many?
Infact, there are infinitely many primes.
So, if we combine all these primes in all possible ways, we will get an infinite collection of numbers, all the primes and all possible products of primes.
The question is – can we produce all the composite numbers this way?
What do you think?
Do you think that there may be a composite number which is not the product of powers of primes?
Before we answer this, let us factorise positive integers, that is, do the opposite of what we have done so far.
We are going to use the factor tree with which you are all familiar.
Let us take some large number, say, 32760, and factorise it as shown.
An equivalent version of Theorem 1.2 was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic.
However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae.
Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton.
He has made fundamental contributions to both mathematics and science.
So we have factorised 32760 as 2
×
2 × 2
×
3 × 3
×
5 × 7 × 13 as a product of primes, i.e., 32760 = 23 × 32 ×
5
× 7 × 13 as a product of powers of primes.
Let us try another number, say, 123456789.
This can be written as 32 × 3803 × 3607.
Of course, you have to check that 3803 and 3607 are primes!
(Try it out for several other natural numbers yourself.) This leads us to a conjecture that every composite number can be written as the product of powers of primes.
In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers.
Let us now formally state this theorem.
Theorem 1.1 (Fundamental Theorem of Arithmetic) : Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Carl Friedrich Gauss (1777 – 1855)
The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes.
Actually it says more.
It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur.
That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur.
So, for example, we regard 2
×
3 × 5
×
7 as the same as 3
×
5 × 7 × 2, or any other possible order in which these primes are written.
This fact is also stated in the following form: The prime factorisation of a natural number is unique, except for the order of its factors.
| | In general, given a composite number x, we factorise it as x = p | 1 p 2 | p | n , where
| --- | --- | --- | --- | ---
| 1, | p 2,, | p n | are | primes | and | written | in | ascending | order, | i.e., | p 1 | | p 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| p
. .. p n. If we combine the same primes, we will get powers of primes.
For example, 32760 =
2
× 2 ×
2
× 3 ×
3
× 5 ×
7
× 13 = 23 × 32 ×
5
× 7 × 13 Once we have decided that the order will be ascending, then the way the number is factorised, is unique.
The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields.
Let us look at some examples.
Example 1 : Consider the numbers 4n, where n is a natural number.
Check whether there is any value of n for which 4n ends with the digit zero.
Solution : If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5.
That is, the prime factorisation of 4n would contain the prime 5.
This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2.
So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n.
So, there is no natural number n for which 4n ends with the digit zero.
You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it!
This method is also called the prime factorisation method.
Let us recall this method through an example.
Example 2 : Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution : We have :
6
= 21 × 31 and 20 =
2
× 2 ×
5
= 22 × 51.
You can find HCF(6, 20) = 2 and LCM(6, 20) =
2
× 2 ×
3
× 5 = 60, as done in your earlier classes.
Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers.
From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20) =
6
× 20.
In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) =
a
× b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.
Example 3: Find the HCF of 96 and 404 by the prime factorisation method.
Hence, find their LCM.
Solution : The prime factorisation of 96 and 404 gives :
96 = 25 × 3, 404 = 22 × 101 Therefore, the HCF of these two integers is 22 = 4.
| | |
| --- | --- | ---
| Also, | LCM (96, 404) = | 96 | 404 | 96 | 404 9696
| --- | --- | --- | --- | --- | ---
| | | HCF(96, 404) | | | 4
Example 4 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Solution : We have :
6
=
2
× 3, 72 = 23 × 32, 120 = 23 ×
3
× 5 Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively.
| So, | HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6
| 23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively
| involved in the three numbers.
| So, | LCM (6, 72, 120) = 23 × 32 × 51 = 360
Remark : Notice, 6 × 72 × 120 HCF (6, 72, 120) × LCM (6, 72, 120).
So, the product of three numbers is not equal to the product of their HCF and LCM. | jemh101.pdf |
1 | CBSE | Class10 | Mathematics | 1.2 The Fundamental Theorem of Arithmetic
In your earlier classes, you have seen that any natural number can be written as a product of its prime factors.
For instance, 2 = 2, 4
=
2 × 2, 253 = 11 × 23, and so on.
Now, let us try and look at natural numbers from the other direction.
That is, can any natural number be obtained by multiplying prime numbers?
Let us see.
Take any collection of prime numbers, say 2, 3, 7, 11 and 23.
If we multiply some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers (In fact, infinitely many).
Let us list a few :
| 7 × 11 × 23 = 1771 | 3 × 7 × 11 × 23 = 5313
| 2 × 3 × 7 × 11 × 23 = 10626 | 23 × 3 × 73 = 8232
22 ×
3
× 7 × 11 × 23 = 21252 and so on.
Now, let us suppose your collection of primes includes all the possible primes.
What is your guess about the size of this collection?
Does it contain only a finite number of integers, or infinitely many?
Infact, there are infinitely many primes.
So, if we combine all these primes in all possible ways, we will get an infinite collection of numbers, all the primes and all possible products of primes.
The question is – can we produce all the composite numbers this way?
What do you think?
Do you think that there may be a composite number which is not the product of powers of primes?
Before we answer this, let us factorise positive integers, that is, do the opposite of what we have done so far.
We are going to use the factor tree with which you are all familiar.
Let us take some large number, say, 32760, and factorise it as shown.
An equivalent version of Theorem 1.2 was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic.
However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae.
Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton.
He has made fundamental contributions to both mathematics and science.
So we have factorised 32760 as 2
×
2 × 2
×
3 × 3
×
5 × 7 × 13 as a product of primes, i.e., 32760 = 23 × 32 ×
5
× 7 × 13 as a product of powers of primes.
Let us try another number, say, 123456789.
This can be written as 32 × 3803 × 3607.
Of course, you have to check that 3803 and 3607 are primes!
(Try it out for several other natural numbers yourself.) This leads us to a conjecture that every composite number can be written as the product of powers of primes.
In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers.
Let us now formally state this theorem.
Theorem 1.1 (Fundamental Theorem of Arithmetic) : Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Carl Friedrich Gauss (1777 – 1855)
The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes.
Actually it says more.
It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur.
That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur.
So, for example, we regard 2
×
3 × 5
×
7 as the same as 3
×
5 × 7 × 2, or any other possible order in which these primes are written.
This fact is also stated in the following form: The prime factorisation of a natural number is unique, except for the order of its factors.
| | In general, given a composite number x, we factorise it as x = p | 1 p 2 | p | n , where
| --- | --- | --- | --- | ---
| 1, | p 2,, | p n | are | primes | and | written | in | ascending | order, | i.e., | p 1 | | p 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| p
. .. p n. If we combine the same primes, we will get powers of primes.
For example, 32760 =
2
× 2 ×
2
× 3 ×
3
× 5 ×
7
× 13 = 23 × 32 ×
5
× 7 × 13 Once we have decided that the order will be ascending, then the way the number is factorised, is unique.
The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields.
Let us look at some examples.
Example 1 : Consider the numbers 4n, where n is a natural number.
Check whether there is any value of n for which 4n ends with the digit zero.
Solution : If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5.
That is, the prime factorisation of 4n would contain the prime 5.
This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2.
So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n.
So, there is no natural number n for which 4n ends with the digit zero.
You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it!
This method is also called the prime factorisation method.
Let us recall this method through an example.
Example 2 : Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution : We have :
6
= 21 × 31 and 20 =
2
× 2 ×
5
= 22 × 51.
You can find HCF(6, 20) = 2 and LCM(6, 20) =
2
× 2 ×
3
× 5 = 60, as done in your earlier classes.
Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers.
From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20) =
6
× 20.
In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) =
a
× b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.
Example 3: Find the HCF of 96 and 404 by the prime factorisation method.
Hence, find their LCM.
Solution : The prime factorisation of 96 and 404 gives :
96 = 25 × 3, 404 = 22 × 101 Therefore, the HCF of these two integers is 22 = 4.
| | |
| --- | --- | ---
| Also, | LCM (96, 404) = | 96 | 404 | 96 | 404 9696
| --- | --- | --- | --- | --- | ---
| | | HCF(96, 404) | | | 4
Example 4 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Solution : We have :
6
=
2
× 3, 72 = 23 × 32, 120 = 23 ×
3
× 5 Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively.
| So, | HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6
| 23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively
| involved in the three numbers.
| So, | LCM (6, 72, 120) = 23 × 32 × 51 = 360
Remark : Notice, 6 × 72 × 120 HCF (6, 72, 120) × LCM (6, 72, 120).
So, the product of three numbers is not equal to the product of their HCF and LCM. | jemh101.pdf |
2 | CBSE | Class10 | Mathematics | EXERCISE 1.1
1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
| | (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
| 3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
| | (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
5. Check whether 6n can end with the digit 0 for any natural number n.
6. Explain why 7 × 11 × 13 + 13 and 7
×
6 × 5
×
4 × 3
×
2 × 1
+
5 are composite numbers.
7. There is a circular path around a sports field.
Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same.
Suppose they both start at the
same point and at the same time, and go in the same direction.
After how many minutes will they meet again at the starting point?
1.3 Revisiting Irrational Numbers
In Class IX, you were introduced to irrational numbers and many of their properties.
You studied about their existence and how the rationals and the irrationals together made up the real numbers.
You even studied how to locate irrationals on the number line.
However, we did not prove that they were irrationals.
In this section, we will prove that 2, 3, 5 and, in general, p is irrational, where p is a prime.
One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
Recall, a number ‘s’ is called irrational if it cannot be written in the form, where p and q are integers and q ¹ 0.
Some examples of irrational numbers, with which you are already familiar, are :
p q
2,2, 3, 15,, 0.10110111011110 3
, etc.
Before we prove that 2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.
Theorem 1.2 : Let p be a prime number.
If p divides a2, then p divides a, where a is a positive integer.
Proof : Let the prime factorisation of a be as follows :
a = p
1 p 2.
.. p n, where p 1 ,p 2 ,.
.., p n are primes, not necessarily distinct.
Therefore, a2 = (p
| 1 p 2 | . p n
| --- | ---
| )( p | 1 p 2 | . p | n ) = p2 1 p2 2 | . p2 n .
| --- | --- | --- | --- | ---
Now, we are given that p divides a2.
Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2.
However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p
| | 1 , p 2 , . . ., p n . So p is one of p | 1 , p | 2 , . . ., p | n .
| --- | --- | --- | --- | ---
| Now, since a = p | 1 p 2 . . . p n , p divides a. | | |
We are now ready to give a proof that 2 is irrational.
The proof is based on a technique called ‘proof by contradiction’.
(This technique is discussed in some detail in Appendix 1).
Theorem 1.3 : 2 is irrational.
Proof : Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (¹ 0) such that 2
=
r s.
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get,2 a
b
where a and b are coprime.
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2.
Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that 2 is irrational.
Example 5 : Prove that 3 is irrational.
Solution : Let us assume, to the contrary, that 3 is rational.
That is, we can find integers a and b (¹ 0) such that 3
=
a b Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, 3b a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that 3 is irrational.
In Class IX, we mentioned that :
the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational.
| We prove some particular cases here. |
| --- | ---
| Example 6 : Show that 5 – | 3 is irrational.
| Solution : Let us assume, to the contrary, that 5 – | 3 is rational.
That is, we can find coprime a and b (b 0) such that 5
3
a b
| Therefore, 5 | 3 a b
| Rearranging this equation, we get | 3 | 5 – a | 5 b | a | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | b b | b | | | | | | |
| Since a and b are integers, we get | 5 – | a b is rational, and so | | | 3 is rational. | | | | |
| But this contradicts the fact that | 3 is irrational. | | | | | | | | |
| This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational.
| So, we conclude that 5 3 is irrational. | | | | | | | | | |
Example 7 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b 0) such that 3
2
a b
Rearranging, we get 2
3
a
b
a
b is rational, and so 2 is rational.
Since 3, a and b are integers, But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational. | jemh101.pdf |
3 | CBSE | Class10 | Mathematics | 1.3 Revisiting Irrational Numbers
In Class IX, you were introduced to irrational numbers and many of their properties.
You studied about their existence and how the rationals and the irrationals together made up the real numbers.
You even studied how to locate irrationals on the number line.
However, we did not prove that they were irrationals.
In this section, we will prove that 2, 3, 5 and, in general, p is irrational, where p is a prime.
One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
Recall, a number ‘s’ is called irrational if it cannot be written in the form, where p and q are integers and q ¹ 0.
Some examples of irrational numbers, with which you are already familiar, are :
p q
2,2, 3, 15,, 0.10110111011110 3
, etc.
Before we prove that 2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.
Theorem 1.2 : Let p be a prime number.
If p divides a2, then p divides a, where a is a positive integer.
Proof : Let the prime factorisation of a be as follows :
a = p
1 p 2.
.. p n, where p 1 ,p 2 ,.
.., p n are primes, not necessarily distinct.
Therefore, a2 = (p
| 1 p 2 | . p n
| --- | ---
| )( p | 1 p 2 | . p | n ) = p2 1 p2 2 | . p2 n .
| --- | --- | --- | --- | ---
Now, we are given that p divides a2.
Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2.
However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p
| | 1 , p 2 , . . ., p n . So p is one of p | 1 , p | 2 , . . ., p | n .
| --- | --- | --- | --- | ---
| Now, since a = p | 1 p 2 . . . p n , p divides a. | | |
We are now ready to give a proof that 2 is irrational.
The proof is based on a technique called ‘proof by contradiction’.
(This technique is discussed in some detail in Appendix 1).
Theorem 1.3 : 2 is irrational.
Proof : Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (¹ 0) such that 2
=
r s.
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get,2 a
b
where a and b are coprime.
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2.
Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that 2 is irrational.
Example 5 : Prove that 3 is irrational.
Solution : Let us assume, to the contrary, that 3 is rational.
That is, we can find integers a and b (¹ 0) such that 3
=
a b Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, 3b a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that 3 is irrational.
In Class IX, we mentioned that :
the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational.
| We prove some particular cases here. |
| --- | ---
| Example 6 : Show that 5 – | 3 is irrational.
| Solution : Let us assume, to the contrary, that 5 – | 3 is rational.
That is, we can find coprime a and b (b 0) such that 5
3
a b
| Therefore, 5 | 3 a b
| Rearranging this equation, we get | 3 | 5 – a | 5 b | a | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | b b | b | | | | | | |
| Since a and b are integers, we get | 5 – | a b is rational, and so | | | 3 is rational. | | | | |
| But this contradicts the fact that | 3 is irrational. | | | | | | | | |
| This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational.
| So, we conclude that 5 3 is irrational. | | | | | | | | | |
Example 7 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b 0) such that 3
2
a b
Rearranging, we get 2
3
a
b
a
b is rational, and so 2 is rational.
Since 3, a and b are integers, But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational. | jemh101.pdf |
4 | CBSE | Class10 | Mathematics | p q
2,2, 3, 15,, 0.10110111011110 3
, etc.
Before we prove that 2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.
Theorem 1.2 : Let p be a prime number.
If p divides a2, then p divides a, where a is a positive integer.
Proof : Let the prime factorisation of a be as follows :
a = p
1 p 2.
.. p n, where p 1 ,p 2 ,.
.., p n are primes, not necessarily distinct.
Therefore, a2 = (p
| 1 p 2 | . p n
| --- | ---
| )( p | 1 p 2 | . p | n ) = p2 1 p2 2 | . p2 n .
| --- | --- | --- | --- | ---
Now, we are given that p divides a2.
Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2.
However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p
| | 1 , p 2 , . . ., p n . So p is one of p | 1 , p | 2 , . . ., p | n .
| --- | --- | --- | --- | ---
| Now, since a = p | 1 p 2 . . . p n , p divides a. | | |
We are now ready to give a proof that 2 is irrational.
The proof is based on a technique called ‘proof by contradiction’.
(This technique is discussed in some detail in Appendix 1).
Theorem 1.3 : 2 is irrational.
Proof : Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (¹ 0) such that 2
=
r s.
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get,2 a
b
where a and b are coprime.
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2.
Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that 2 is irrational.
Example 5 : Prove that 3 is irrational.
Solution : Let us assume, to the contrary, that 3 is rational.
That is, we can find integers a and b (¹ 0) such that 3
=
a b Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, 3b a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that 3 is irrational.
In Class IX, we mentioned that :
the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational.
| We prove some particular cases here. |
| --- | ---
| Example 6 : Show that 5 – | 3 is irrational.
| Solution : Let us assume, to the contrary, that 5 – | 3 is rational.
That is, we can find coprime a and b (b 0) such that 5
3
a b
| Therefore, 5 | 3 a b
| Rearranging this equation, we get | 3 | 5 – a | 5 b | a | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | b b | b | | | | | | |
| Since a and b are integers, we get | 5 – | a b is rational, and so | | | 3 is rational. | | | | |
| But this contradicts the fact that | 3 is irrational. | | | | | | | | |
| This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational.
| So, we conclude that 5 3 is irrational. | | | | | | | | | |
Example 7 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b 0) such that 3
2
a b
Rearranging, we get 2
3
a
b
a
b is rational, and so 2 is rational.
Since 3, a and b are integers, But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational. | jemh101.pdf |
5 | CBSE | Class10 | Mathematics | So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2.
Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that 2 is irrational.
Example 5 : Prove that 3 is irrational.
Solution : Let us assume, to the contrary, that 3 is rational.
That is, we can find integers a and b (¹ 0) such that 3
=
a b Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.So, 3b a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that 3 is irrational.
In Class IX, we mentioned that :
the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational.
| We prove some particular cases here. |
| --- | ---
| Example 6 : Show that 5 – | 3 is irrational.
| Solution : Let us assume, to the contrary, that 5 – | 3 is rational.
That is, we can find coprime a and b (b 0) such that 5
3
a b
| Therefore, 5 | 3 a b
| Rearranging this equation, we get | 3 | 5 – a | 5 b | a | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | b b | b | | | | | | |
| Since a and b are integers, we get | 5 – | a b is rational, and so | | | 3 is rational. | | | | |
| But this contradicts the fact that | 3 is irrational. | | | | | | | | |
| This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational.
| So, we conclude that 5 3 is irrational. | | | | | | | | | |
Example 7 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b 0) such that 3
2
a b
In Class IX, we mentioned that :
the sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational.
| We prove some particular cases here. |
| --- | ---
| Example 6 : Show that 5 – | 3 is irrational.
| Solution : Let us assume, to the contrary, that 5 – | 3 is rational.
That is, we can find coprime a and b (b 0) such that 5
3
a b
| Therefore, 5 | 3 a b
| Rearranging this equation, we get | 3 | 5 – a | 5 b | a | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | b b | b | | | | | | |
| Since a and b are integers, we get | 5 – | a b is rational, and so | | | 3 is rational. | | | | |
| But this contradicts the fact that | 3 is irrational. | | | | | | | | |
| This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational.
| So, we conclude that 5 3 is irrational. | | | | | | | | | |
Example 7 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b 0) such that 3
2
a b
Rearranging, we get 2
3
a
b
a
b is rational, and so 2 is rational.
Since 3, a and b are integers, But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational.EXERCISE 1.2
1. Prove that 5 is irrational.
2. Prove that 3 2 5 is irrational.
3. Prove that the following are irrationals :
(i) 1 2 (ii) 7 5 (iii) 6 2
1.4 Summary
In this chapter, you have studied the following points:
1. The Fundamental Theorem of Arithmetic :
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
2. If p is a prime and p divides a2, then p divides a, where a is a positive integer.
3. To prove that 2, 3 are irrationals. | jemh101.pdf |
0 | CBSE | Class10 | Mathematics | POLYNOMIALS
2.1 Introduction
In Class IX, you have studied polynomials in one variable and their degrees.
Recall that if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).
For example, 4x + 2 is a polynomial in the variable x of degree 1, 2y2 – 3y + 4 is a polynomial in the variable y of degree 2, 5x3 – 4x2 +
x
– 2
| is a polynomial in the variable x of degree 3 and 7u6 – | 4 | 4 | 23 | 8
| --- | --- | --- | --- | ---
| 2 u | u | u | | is a polynomial
| in the variable u of degree 6. Expressions like 1 | 1x , | 2x , 2 | 1 2 3x x | etc., are
| not polynomials. A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3,
| 3 5,x 2y, 2 11 x , 3z + 4, 2 1 3 u , etc., are all linear polynomials. Polynomials | | | |
| such as 2x + 5 – x2, x3 + 1, etc., are not linear polynomials.
| A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’
| has been derived from the word ‘quadrate’, which means ‘square’. | 2 | 2 | ,2 | 3 5 x x
| u u v v z | | are | some | examples of
y2 – 2, 22 3,x x
2
2 22 1 2 5, 5, 4
3
3
7 quadratic polynomials (whose coefficients are real numbers).
More generally, any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a 0.
A polynomial of degree 3 is called a cubic polynomial.
Some examples of a cubic polynomial are 2 – x3, x3, 32,x 3 – x2 + x3, 3x3 – 2x2 +
x
– 1.
In fact, the most general form of a cubic polynomial is ax3 + bx2 + cx + d, where, a, b, c, d are real numbers and a 0.
Now consider the polynomial p(x) = x2 – 3x – 4.
Then, putting x
=
2 in the polynomial, we get p(2) = 22 –
3
× 2 –
4
= – 6.
The value ‘– 6’, obtained by replacing x by 2 in x2 – 3x – 4, is the value of x2 – 3x – 4 at x = 2.
Similarly, p(0) is the value of p(x) at x = 0, which is – 4.
If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).
What is the value of p(x) = x2 –3x – 4 at x = –1?
We have : p(–1) = (–1)2 –{3 × (–1)} –
4
= 0 Also, note that p(4) = 42 – (3 4) –
4
= 0.
As p(–1) = 0 and p(4) = 0, –1 and 4 are called the zeroes of the quadratic polynomial x2 – 3x – 4.
More generally, a real number k is said to be a zero of a polynomial p(x), if p(k) = 0.
You have already studied in Class IX, how to find the zeroes of a linear polynomial.
For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us 2k +
3
= 0, i.e., k
=
3 2
In general, if k is a zero of p(x) = ax + b, then p(k) = ak +
b
= 0, i.e.,.
b k
| So, the zero of the linear polynomial ax + b is | | (Constant term) Coefficient of
| --- | --- | ---
| | b a | x
Thus, the zero of a linear polynomial is related to its coefficients.
Does this happen in the case of other polynomials too?
For example, are the zeroes of a quadratic polynomial also related to its coefficients?
In this chapter, we will try to answer these questions.
We will also study the division algorithm for polynomials.
2.2 Geometrical Meaning of the Zeroes of a Polynomial
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0.
But why are the zeroes of a polynomial so important?
To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.
Consider first a linear polynomial ax + b, a 0.
You have studied in Class IX that the graph of y = ax + b is a straight line.
For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7).
| x | –2 | 2
| --- | --- | ---
| y = 2x + 3 | –1 | 7
From Fig. 2.1, you can see
that the graph of y = 2x + 3 intersects the x - axis mid-way between x = –1 and x
=
– 2, that is, at the point 3, 0 2.
You also know that the zero of
2x + 3 is 3 2 .
Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis.
Fig. 2.1
Table 2.1
In general, for a linear polynomial ax + b, a 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely,, 0 b a.
Therefore, the linear polynomial ax + b, a 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x2 – 3x – 4.
Let us see what the graph* of y = x2 – 3x – 4 looks like.
Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table 2.1. y = x2 – 3x –
4
6 0 –
4
– 6 –
6
– 4 0 6
* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated.
x
–
2 –1 0
1
2 3 4 5
If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.
In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a
>
0 or a < 0.
(These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial.
Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Fig. 2.2 Case (i) : Here, the graph cuts x-axis at two distinct points A and A.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.
From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
The x-coordinates of A and A are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see Fig. 2.3).
Fig. 2.3
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points.
So, the two points A and A of Case (i) coincide here to become one point A (see Fig. 2.4).
Fig. 2.4
The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case.
Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis.
So, it does not cut the x-axis at any point (see Fig. 2.5).
Fig. 2.5
So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero.
This also means that a polynomial of degree 2 has atmost two zeroes.
Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be?
Let us find out.
Consider the cubic polynomial x3 – 4x.
To see what the graph of y = x3 – 4x looks like, let us list a few values of y corresponding to a few values for x as shown in Table 2.2.
Table 2.2
| x | –2 | –1 | 0 | 1 | 2
| --- | --- | --- | --- | --- | ---
| y = x3 – 4x | 0 | 3 | 0 | –3 | 0
Locating the points of the table on a graph paper and drawing the graph, we see that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6. We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x.
Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis.
Since the curve meets the x -axis in only these 3 points, their x -coordinates are the only zeroes of the polynomial.
Let us take a few more examples.
Consider the cubic polynomials x3 and x3 – x2.
We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively.
Fig. 2.6
Fig. 2.7
Fig. 2.8
Note that 0 is the only zero of the polynomial x3.
Also, from Fig. 2.7, you can see that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis.
Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2.
Also, from Fig. 2.8, these values are the x -coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis.
From the examples above, we see that there are at most 3 zeroes for any cubic polynomial.
In other words, any polynomial of degree 3 can have at most three zeroes.
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at atmost n points.
Therefore, a polynomial p(x) of degree n has at most n zeroes.
Example 1 : Look at the graphs in Fig. 2.9 given below.
Each is the graph of y = p(x), where p(x) is a polynomial.
For each of the graphs, find the number of zeroes of p(x).
Fig. 2.9
Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3.
(Why?)
(iv) The number of zeroes is 1.
(Why?)
(v) The number of zeroes is 1.
(Why?)
(vi) The number of zeroes is 4.
(Why?) | jemh102.pdf |
1 | CBSE | Class10 | Mathematics | 2.2 Geometrical Meaning of the Zeroes of a Polynomial
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0.
But why are the zeroes of a polynomial so important?
To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.
Consider first a linear polynomial ax + b, a 0.
You have studied in Class IX that the graph of y = ax + b is a straight line.
For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7).
| x | –2 | 2
| --- | --- | ---
| y = 2x + 3 | –1 | 7
From Fig. 2.1, you can see
that the graph of y = 2x + 3 intersects the x - axis mid-way between x = –1 and x
=
– 2, that is, at the point 3, 0 2.
You also know that the zero of
2x + 3 is 3 2 .
Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis.
Fig. 2.1
Table 2.1
In general, for a linear polynomial ax + b, a 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely,, 0 b a.
Therefore, the linear polynomial ax + b, a 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x2 – 3x – 4.
Let us see what the graph* of y = x2 – 3x – 4 looks like.
Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table 2.1. y = x2 – 3x –
4
6 0 –
4
– 6 –
6
– 4 0 6
* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated.
x
–
2 –1 0
1
2 3 4 5
If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.
In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a
>
0 or a < 0.
(These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial.
Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Fig. 2.2 Case (i) : Here, the graph cuts x-axis at two distinct points A and A.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.
From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
The x-coordinates of A and A are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see Fig. 2.3).
Fig. 2.3
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points.
So, the two points A and A of Case (i) coincide here to become one point A (see Fig. 2.4).
Fig. 2.4
The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case.
Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis.
So, it does not cut the x-axis at any point (see Fig. 2.5).
Fig. 2.5
So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero.
This also means that a polynomial of degree 2 has atmost two zeroes.
Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be?
Let us find out.
Consider the cubic polynomial x3 – 4x.
To see what the graph of y = x3 – 4x looks like, let us list a few values of y corresponding to a few values for x as shown in Table 2.2.
Table 2.2
| x | –2 | –1 | 0 | 1 | 2
| --- | --- | --- | --- | --- | ---
| y = x3 – 4x | 0 | 3 | 0 | –3 | 0
Locating the points of the table on a graph paper and drawing the graph, we see that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6. We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x.
Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis.
Since the curve meets the x -axis in only these 3 points, their x -coordinates are the only zeroes of the polynomial.
Let us take a few more examples.
Consider the cubic polynomials x3 and x3 – x2.
We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively.
Fig. 2.6
Fig. 2.7
Fig. 2.8
Note that 0 is the only zero of the polynomial x3.
Also, from Fig. 2.7, you can see that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis.
Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2.
Also, from Fig. 2.8, these values are the x -coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis.
From the examples above, we see that there are at most 3 zeroes for any cubic polynomial.
In other words, any polynomial of degree 3 can have at most three zeroes.
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at atmost n points.
Therefore, a polynomial p(x) of degree n has at most n zeroes.
Example 1 : Look at the graphs in Fig. 2.9 given below.
Each is the graph of y = p(x), where p(x) is a polynomial.
For each of the graphs, find the number of zeroes of p(x).
Fig. 2.9
Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3.
(Why?)
(iv) The number of zeroes is 1.
(Why?)
(v) The number of zeroes is 1.
(Why?)
(vi) The number of zeroes is 4.
(Why?) | jemh102.pdf |
2 | CBSE | Class10 | Mathematics | From Fig. 2.1, you can see
that the graph of y = 2x + 3 intersects the x - axis mid-way between x = –1 and x
=
– 2, that is, at the point 3, 0 2.
You also know that the zero of
2x + 3 is 3 2 .
Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis.
Fig. 2.1
Table 2.1
In general, for a linear polynomial ax + b, a 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely,, 0 b a.
Therefore, the linear polynomial ax + b, a 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x2 – 3x – 4.
Let us see what the graph* of y = x2 – 3x – 4 looks like.
Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table 2.1. y = x2 – 3x –
4
6 0 –
4
– 6 –
6
– 4 0 6
* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated.
x
–
2 –1 0
1
2 3 4 5
If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.
In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a
>
0 or a < 0.
(These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial.
Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Fig. 2.2 Case (i) : Here, the graph cuts x-axis at two distinct points A and A.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.
From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
The x-coordinates of A and A are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see Fig. 2.3).
Fig. 2.3
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points.
So, the two points A and A of Case (i) coincide here to become one point A (see Fig. 2.4).
Fig. 2.4
The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case.
Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis.
So, it does not cut the x-axis at any point (see Fig. 2.5).
Fig. 2.5
So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero.
This also means that a polynomial of degree 2 has atmost two zeroes.
Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be?
Let us find out.
Consider the cubic polynomial x3 – 4x.
To see what the graph of y = x3 – 4x looks like, let us list a few values of y corresponding to a few values for x as shown in Table 2.2.
Table 2.2
| x | –2 | –1 | 0 | 1 | 2
| --- | --- | --- | --- | --- | ---
| y = x3 – 4x | 0 | 3 | 0 | –3 | 0
Locating the points of the table on a graph paper and drawing the graph, we see that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6. We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x.
Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis.
Since the curve meets the x -axis in only these 3 points, their x -coordinates are the only zeroes of the polynomial.
Let us take a few more examples.
Consider the cubic polynomials x3 and x3 – x2.
We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively.
Fig. 2.6
Fig. 2.7
Fig. 2.8
Note that 0 is the only zero of the polynomial x3.
Also, from Fig. 2.7, you can see that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis.
Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2.
Also, from Fig. 2.8, these values are the x -coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis.
From the examples above, we see that there are at most 3 zeroes for any cubic polynomial.
In other words, any polynomial of degree 3 can have at most three zeroes.
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at atmost n points.
Therefore, a polynomial p(x) of degree n has at most n zeroes.
Example 1 : Look at the graphs in Fig. 2.9 given below.
Each is the graph of y = p(x), where p(x) is a polynomial.
For each of the graphs, find the number of zeroes of p(x).
Fig. 2.9
Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3.
(Why?)
(iv) The number of zeroes is 1.
(Why?)
(v) The number of zeroes is 1.
(Why?)
(vi) The number of zeroes is 4.
(Why?) | jemh102.pdf |
3 | CBSE | Class10 | Mathematics | Fig. 2.1
Table 2.1
In general, for a linear polynomial ax + b, a 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely,, 0 b a.
Therefore, the linear polynomial ax + b, a 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x2 – 3x – 4.
Let us see what the graph* of y = x2 – 3x – 4 looks like.
Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table 2.1. y = x2 – 3x –
4
6 0 –
4
– 6 –
6
– 4 0 6
* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated.
x
–
2 –1 0
1
2 3 4 5
If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.
In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a
>
0 or a < 0.
(These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial.
Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Fig. 2.2 Case (i) : Here, the graph cuts x-axis at two distinct points A and A.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.
From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
The x-coordinates of A and A are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see Fig. 2.3).Table 2.1
In general, for a linear polynomial ax + b, a 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely,, 0 b a.
Therefore, the linear polynomial ax + b, a 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x2 – 3x – 4.
Let us see what the graph* of y = x2 – 3x – 4 looks like.
Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table 2.1. y = x2 – 3x –
4
6 0 –
4
– 6 –
6
– 4 0 6
* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students, nor is to be evaluated.
x
–
2 –1 0
1
2 3 4 5
If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.
In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a
>
0 or a < 0.
(These curves are called parabolas.) You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial.
Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Fig. 2.2 Case (i) : Here, the graph cuts x-axis at two distinct points A and A.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.
From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
The x-coordinates of A and A are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see Fig. 2.3).Fig. 2.3
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points.
So, the two points A and A of Case (i) coincide here to become one point A (see Fig. 2.4).Fig. 2.4
The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case.
Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis.
So, it does not cut the x-axis at any point (see Fig. 2.5). | jemh102.pdf |
4 | CBSE | Class10 | Mathematics | Table 2.2
| x | –2 | –1 | 0 | 1 | 2
| --- | --- | --- | --- | --- | ---
| y = x3 – 4x | 0 | 3 | 0 | –3 | 0
Locating the points of the table on a graph paper and drawing the graph, we see that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6. We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x.
Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis.
Since the curve meets the x -axis in only these 3 points, their x -coordinates are the only zeroes of the polynomial.
Let us take a few more examples.
Consider the cubic polynomials x3 and x3 – x2.
We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively.
Fig. 2.6
Fig. 2.7
Fig. 2.8
Note that 0 is the only zero of the polynomial x3.
Also, from Fig. 2.7, you can see that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis.
Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2.
Also, from Fig. 2.8, these values are the x -coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis.
From the examples above, we see that there are at most 3 zeroes for any cubic polynomial.
In other words, any polynomial of degree 3 can have at most three zeroes.
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at atmost n points.
Therefore, a polynomial p(x) of degree n has at most n zeroes.
Example 1 : Look at the graphs in Fig. 2.9 given below.
Each is the graph of y = p(x), where p(x) is a polynomial.
For each of the graphs, find the number of zeroes of p(x).Fig. 2.9
Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3.
(Why?)
(iv) The number of zeroes is 1.
(Why?)
(v) The number of zeroes is 1.
(Why?)
(vi) The number of zeroes is 4.
(Why?)Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3.
(Why?)
(iv) The number of zeroes is 1.
(Why?)
(v) The number of zeroes is 1.
(Why?)
(vi) The number of zeroes is 4.
(Why?) | jemh102.pdf |
5 | CBSE | Class10 | Mathematics | Fig. 2.102.3 Relationship between Zeroes and Coefficients of a Polynomial
You have already seen that zero of a linear polynomial ax + b is b a .
We will now try to answer the question raised in Section 2.1 regarding the relationship between zeroes and coefficients of a quadratic polynomial.
For this, let us take a quadratic polynomial, say p(x) = 2x2 – 8x + 6.
In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term.
So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is 6 × 2x2 = 12x2.
So, we write 2x2 – 8x +
6
= 2x2 – 6x – 2x +
6
= 2x(x – 3) – 2(x – 3) = (2x – 2)(x – 3) = 2(x – 1)(x – 3) So, the value of p(x) = 2x2 – 8x + 6 is zero when x
–
1 = 0 or x
–
3 = 0, i.e., when x
=
1 or x = 3.
So, the zeroes of 2x2 – 8x + 6 are 1 and 3.
Observe that :
3x2 + 5x –
2
= 3x2 + 6x –
x
– 2 = 3x(x + 2) –1(x + 2)
when x
=
1 Sum of its zeroes = Product of its zeroes =
(8) (Coefficient of) 1
3
4 2 Coefficient of x
x
6 Constant term 1
3
3 2 Coefficient of x 3 1
2 2
= (3x – 1)(x + 2) 2 Let us take one more quadratic polynomial, say, p(x) = 3x2 + 5x – 2.
By the method of splitting the middle term, Hence, the value of 3x2 + 5x – 2 is zero when either 3x –
1
= 0 or x
+
2 = 0, i.e., or x = –2.
So, the zeroes of 3x2 + 5x – 2 are 3 and – 2.
Observe that :
Sum of its zeroes =
Product of its zeroes =
| 1 | 5 | (Coefficient of | )
| --- | --- | --- | ---
| 3 | | | 3 x | x | Coefficient of x | x | x
| --- | --- | --- | --- | --- | --- | --- | ---
| (2)
| 1 | | | 2 | | Constant term | |
| 3 | (2) | | 3 | | Coefficient of x | |
2
In general, if * and * are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, a 0, then you know that x
–
and x
–
are the factors of p(x).
Therefore, ax2 + bx +
c
= k(x – ) (x – ), where k is a constant
Comparing the coefficients of x2, x and constant terms on both the sides, we get a = k, b = k( + ) and c = k = k[x2 – ( + )x + ] = kx2 – k( + )x + k
This gives + = –b a,
c a
=
* , are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively.
We will use later one more letter ‘’ pronounced as ‘gamma’.
i.e., 2 (Coefficient
product of zeroes = =
Let us consider some examples.
Example 2 : Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Solution : We haveThis gives + = –b a,
c a
=
* , are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively.
We will use later one more letter ‘’ pronounced as ‘gamma’.
i.e., 2 (Coefficient
product of zeroes = =
Let us consider some examples.
Example 2 : Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Solution : We havec a
=
* , are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively.
We will use later one more letter ‘’ pronounced as ‘gamma’.
i.e., 2 (Coefficient
product of zeroes = =
Let us consider some examples.
Example 2 : Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Solution : We havex2 + 7x + 10 = (x + 2)(x + 5)
So, the value of x2 + 7x + 10 is zero when x
+
2 = 0 or x
+
5 = 0, i.e., when x
=
– 2 or sum of zeroes = product of zeroes = Example 3 : Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.
x = –5.
Therefore, the zeroes of x2 + 7x + 10 are – 2 and – 5.
Now,
| | b | | | | x |
| --- | --- | --- | --- | --- | --- | ---
| 2 (Coefficient
| | a | x | x | x | | ,
| Coefficient of
| c a | x | x Constant term | . | | |
Coefficient of
| | of of |
| --- | --- | ---
| sum of zeroes = + = | 2 (Coefficient |
| | of | )
2 (Coefficient
2
x
2
x
(7) – (Coefficient of) ,– 2 (–5) –
(7) 1 Coefficient of x
2 2
x
10 Constant term (2) (5) 10 1 Coefficient of x
3x x
Solution : Recall the identity a2 – b2 = (a – b)(a + b).
Using it, we can write:
x2 –
3
= 3
So, the value of x2 – 3 is zero when x
=
3 or x
=
– 3
Therefore, the zeroes of x2 – 3 are 3 and 3
Now,
sum of zeroes =
product of zeroes = Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are
| | (Coefficient of) ,3 3 0 Coefficient of
| --- | ---
| Constant term 3 3 | Constant term 3 – 3 | 3 Constant term 3 | Constant term 3 |
| --- | --- | --- | --- | ---
| 2
| | | 1 | |
– 3 and 2, respectively.
Solution : Let the quadratic polynomial be ax2 + bx + c, and its zeroes be and .
We have ,
+
= 3
=
b and | jemh102.pdf |
6 | CBSE | Class10 | Mathematics | a
=
2
= c
a.
If a = 1, then b
=
3 and c = 2.
So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2.
You can check that any other quadratic polynomial that fits these conditions will be of the form k(x2 + 3x + 2), where k is real.
Let us now look at cubic polynomials.
Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients?
Let us consider p(x) = 2x3 – 5x2 – 14x + 8. Since p(x) can have atmost three You can check that p(x) = 0 for x = 4, – 2, 1 2 zeroes, these are the zeores of 2x3 – 5x2 – 14x + 8. Now, 2 sum of the zeroes = product of the zeroes = However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have
1 5 (5) (Coefficient of) 4 (2)
3
| 2 | 2 | 2 | Coefficient of x x | | | | | | | | ,
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | | 1 | | | | 8 | – Constant term | | |
| 4 | (2) | | | | | 4 | | | | |
| 3
| | | | 2 | | | | 2 | Coefficient of x | | . |
| 4 | (2) (2) | 4 |
| --- | --- | --- | ---
| | 1 1 | |
| | | |
| | 2 2 | |
|
| | | |
| | = 14 – 8 1 2 | | 7
Coefficient of
x
x.
In general, it can be proved that if , , are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then
+ + =
–b a,
+ + =
c a,
=
– d a. Let us consider an example.
Example 5* : Verify that 3, –1,
1 3
are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get a = 3, b
=
– 5, c = –11, d
=
– 3.
Further p(3) =
3
× 33 – (5 × 32) – (11 × 3) –
3
= 81 – 45 – 33 –
3
= 0, p(–1) =
3
× (–1)3 –
5
× (–1)2 – 11 × (–1) –
3
= –3 –
5
+ 11 –
3
= 0,,
| | 3 | 2
| --- | --- | ---
| 1 | 1 | 1 | 1
| --- | --- | --- | ---
| | 3 | 5 | 11 | 3
| --- | --- | --- | --- | ---
| 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 p |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = 1 | 5 | 11 | 2 | 2 – | 3 | – | 0
| --- | --- | --- | --- | --- | --- | --- | ---
| 9 | 9 | 3 | 3 | 3 | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
Therefore, 3, –1 and 1 are the zeroes of 3x3 – 5x2 – 11x – 3.
So, we take = 3, = –1 and
=
1 3,,.
Now,
1
1
5 (5)
| | 3 | (1) | 2
| --- | --- | --- | ---
| | | | | | | | | 3 | | | 3 | | 3 | | 3 b a | b a |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 1 | 1 | 1 | 11
| --- | --- | --- | --- | ---
| | 3 | (1) | (1) | 3 | 3 | 1
| --- | --- | --- | --- | --- | --- | ---
| | | | | | | | | | | 3 | | | | | 3 | | | | 3 | | | | 3 | c a
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 1 | (3)
| --- | --- | ---
| 3 | (1) | 1
| --- | --- | ---
| | | | | | 3 | d a | 3 d a | d a |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
* Not from the examination point of view. | jemh102.pdf |
7 | CBSE | Class10 | Mathematics | –b a,
+ + =
c a,
=
– d a. Let us consider an example.
Example 5* : Verify that 3, –1,
1 3
are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get a = 3, b
=
– 5, c = –11, d
=
– 3.
Further p(3) =
3
× 33 – (5 × 32) – (11 × 3) –
3
= 81 – 45 – 33 –
3
= 0, p(–1) =
3
× (–1)3 –
5
× (–1)2 – 11 × (–1) –
3
= –3 –
5
+ 11 –
3
= 0,,
| | 3 | 2
| --- | --- | ---
| 1 | 1 | 1 | 1
| --- | --- | --- | ---
| | 3 | 5 | 11 | 3
| --- | --- | --- | --- | ---
| 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 p |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = 1 | 5 | 11 | 2 | 2 – | 3 | – | 0
| --- | --- | --- | --- | --- | --- | --- | ---
| 9 | 9 | 3 | 3 | 3 | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
Therefore, 3, –1 and 1 are the zeroes of 3x3 – 5x2 – 11x – 3.
So, we take = 3, = –1 and
=
1 3,,.
Now,
1
1
5 (5)
| | 3 | (1) | 2
| --- | --- | --- | ---
| | | | | | | | | 3 | | | 3 | | 3 | | 3 b a | b a |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 1 | 1 | 1 | 11
| --- | --- | --- | --- | ---
| | 3 | (1) | (1) | 3 | 3 | 1
| --- | --- | --- | --- | --- | --- | ---
| | | | | | | | | | | 3 | | | | | 3 | | | | 3 | | | | 3 | c a
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 1 | (3)
| --- | --- | ---
| 3 | (1) | 1
| --- | --- | ---
| | | | | | 3 | d a | 3 d a | d a |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
* Not from the examination point of view.c a,
=
– d a. Let us consider an example.
Example 5* : Verify that 3, –1,
1 3
are the zeroes of the cubic polynomial p(x) = 3x3 – 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get a = 3, b
=
– 5, c = –11, d
=
– 3.
Further p(3) =
3
× 33 – (5 × 32) – (11 × 3) –
3
= 81 – 45 – 33 –
3
= 0, p(–1) =
3
× (–1)3 –
5
× (–1)2 – 11 × (–1) –
3
= –3 –
5
+ 11 –
3
= 0,,
| | 3 | 2
| --- | --- | ---
| 1 | 1 | 1 | 1
| --- | --- | --- | ---
| | 3 | 5 | 11 | 3
| --- | --- | --- | --- | ---
| 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 3 p | 3 p | 3 p | 3 p | 3 p | 3 p | 3 p |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = 1 | 5 | 11 | 2 | 2 – | 3 | – | 0
| --- | --- | --- | --- | --- | --- | --- | ---
| 9 | 9 | 3 | 3 | 3 | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
Therefore, 3, –1 and 1 are the zeroes of 3x3 – 5x2 – 11x – 3.
So, we take = 3, = –1 and
=
1 3,,.
Now,
1
1
5 (5)
| | 3 | (1) | 2
| --- | --- | --- | ---
| | | | | | | | | 3 | | | 3 | | 3 | | 3 b a | b a |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 1 | 1 | 1 | 11
| --- | --- | --- | --- | ---
| | 3 | (1) | (1) | 3 | 3 | 1
| --- | --- | --- | --- | --- | --- | ---
| | | | | | | | | | | 3 | | | | | 3 | | | | 3 | | | | 3 | c a
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 1 | (3)
| --- | --- | ---
| 3 | (1) | 1
| --- | --- | ---
| | | | | | 3 | d a | 3 d a | d a |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
* Not from the examination point of view. | jemh102.pdf |
0 | CBSE | Class10 | Mathematics | PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
3
3.1 Introduction
You must have come across situations like the one given below :
Akhila went to a fair in her village.
She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely, you get it).
The number of times she played Hoopla is half the number of rides she had on the Giant Wheel.
If each ride costs ` 3, and a game of Hoopla costs ` 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent ` 20.
May be you will try it by considering different cases.
If she has one ride, is it possible?
Is it possible to have two rides?
And so on.
Or you may use the knowledge of Class IX, to represent such situations as linear equations in two variables.
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she played Hoopla by y. Now the situation can be represented by the two equations:
| y = 1 2 x | (1)
| 3x + 4y = 20 | (2)
Can we find the solutions of this pair of equations?
There are several ways of finding these, which we will study in this chapter.
3.2 Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear equations.
A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.
A pair of linear equations which are equivalent has infinitely many distinct common solutions.
Such a pair is called a dependent pair of linear equations in two variables.
Note that a dependent pair of linear equations is always consistent.
We can now summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point.
In this case, the pair of equations has a unique solution (consistent pair of equations).
(ii) the lines may be parallel.
In this case, the equations have no solution (inconsistent pair of equations).
(iii) the lines may be coincident.
In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
(ii) 2x + 3y –
9
= 0 and 4x + 6y – 18 = 0 (The lines coincide)
(iii) x + 2y –
4
= 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of 1
1
1 22 2, and a
b
c ca b in all the 1 and a 2, b 2, c 2 denote the coefficents of equations given in the general form in Section 3.2. three examples.
Here, a, b, c
1 1
Table 3.1
| | 1 2 | 1 2 | 1 2
| --- | --- | --- | ---
| 1. | x – 2y = 0 | 1 3 | 2 4 − | 0 20− | 1 | 1
| --- | --- | --- | --- | --- | --- | ---
| a a | b b | c c | Compare the | Graphical Algebraic No. | ratios | representation interpretation
| Sl Pair of lines
| 3x + 4y – 20 = 0
| | | | | | a | b
| | | | | | a b ≠ |
| | | | | | |
| | | | | | 2 | 2
| | | | | | a | b
| | | | | | a c = | b
| | | | | | 2 | 2
| | | | | | a | b
| | | | 2 4 | − − 4 12 | |
| 1 | 1 | 1
| --- | --- | ---
| a | b | c = | ≠ | Parallel lines | No solution
| --- | --- | --- | --- | --- | ---
2
| | | | 2 | 2 | 2 | |
| --- | --- | --- | --- | --- | --- | --- | ---
| 2. 2x + 3y – 9 = 0 2 4 3 6 9 18 − − 1 1 1 lines many solutions 4x + 6y – 18 = 0
| 3. x + 2y – 4 = 0 2x + 4y – 12 = 0 | 1 | | | | | |
| From the table above, you can observe that if the lines represented by the equation
| | a + b y + c 1 1 | 1 = 0 | | | | |
| | + b y + c 2 2 | 2 = 0 | | | | |
| | | a b | | | | |
| | | a b ≠ ⋅ | | | | |
| | | 2 2 | | | | |
| | a | b c | | | | |
| | a | b c = = | ⋅ | | | |
| | 2 | 2 | 2 | | | |
| | a | b c | | | | |
| | a | b c = ≠ ⋅ | | | | |
| | 2 | 2 2 | | | | |
| and | a
| --- | ---
| (i) intersecting, then | | 1 | | 1 |
| --- | --- | --- | --- | --- | ---
| (ii) coincident, then | 1 | | 1 | | 1
| (iii) parallel, then 1 | | 1 | | 1 |
In fact, the converse is also true for any pair of lines.
You can verify them by considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example 1 : Check graphically whether the pair of equations
| | x + 3y = 6 | (1)
| --- | --- | ---
| and | 2x – 3y = 12 | (2)
is consistent.
If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2).
For this, we find two solutions of each of the equations, which are given in Table 3.2
Table 3.2
x
0
6
x 0 3
y =
| 6 3 x−
| ---
| 2 | 0 | y = 2 | 12
| --- | --- | --- | ---
3
x −
– 4 –2
Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ as shown in Fig. 3.1. We observe that there is a point B (6, 0) common to both the lines AB and PQ.
So, the solution of the pair of linear equations is x
=
6 and y = 0, i.e., the given pair of equations is consistent.
Fig. 3.1
Example 2 : Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
| 5x – 8y + 1 = 0 | (1)
| --- | ---
| 3x – | 24 5 y + 3 5 | = | 0 | (2)
| --- | --- | --- | --- | ---
| Solution : Multiplying Equation (2) by | | 5 , 3 we get | |
5x – 8y +
1
= 0 But, this is the same as Equation (1).
Hence the lines represented by Equations (1) and (2) are coincident.
Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts.
When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased.
Also, the number of skirts is four less than four times the number of pants purchased”.
Help her friends to find how many pants and skirts Champa bought.
Solution : Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x – 2 (1) and y = 4x – 4 (2) Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.
They are given in Table 3.3.
| Table 3.3
| ---
| x | 2 | 0
| --- | --- | ---
| y = 2x – 2 | 2 | – 2
| x | 0 | 1
| y = 4x – 4 | – 4 | 0
Fig. 3.2
Plot the points and draw the lines passing through them to represent the equations, as shown in Fig. 3.2.
The two lines intersect at the point (1, 0).
So, x = 1, y
=
0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Verify the answer by checking whether it satisfies the conditions of the given problem. | jemh103.pdf |
1 | CBSE | Class10 | Mathematics | 3.2 Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear equations.
A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.
A pair of linear equations which are equivalent has infinitely many distinct common solutions.
Such a pair is called a dependent pair of linear equations in two variables.
Note that a dependent pair of linear equations is always consistent.
We can now summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point.
In this case, the pair of equations has a unique solution (consistent pair of equations).
(ii) the lines may be parallel.
In this case, the equations have no solution (inconsistent pair of equations).
(iii) the lines may be coincident.
In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)
(ii) 2x + 3y –
9
= 0 and 4x + 6y – 18 = 0 (The lines coincide)
(iii) x + 2y –
4
= 0 and 2x + 4y – 12 = 0 (The lines are parallel)
Let us now write down, and compare, the values of 1
1
1 22 2, and a
b
c ca b in all the 1 and a 2, b 2, c 2 denote the coefficents of equations given in the general form in Section 3.2. three examples.
Here, a, b, c
1 1
Table 3.1
| | 1 2 | 1 2 | 1 2
| --- | --- | --- | ---
| 1. | x – 2y = 0 | 1 3 | 2 4 − | 0 20− | 1 | 1
| --- | --- | --- | --- | --- | --- | ---
| a a | b b | c c | Compare the | Graphical Algebraic No. | ratios | representation interpretation
| Sl Pair of lines
| 3x + 4y – 20 = 0
| | | | | | a | b
| | | | | | a b ≠ |
| | | | | | |
| | | | | | 2 | 2
| | | | | | a | b
| | | | | | a c = | b
| | | | | | 2 | 2
| | | | | | a | b
| | | | 2 4 | − − 4 12 | |
| 1 | 1 | 1
| --- | --- | ---
| a | b | c = | ≠ | Parallel lines | No solution
| --- | --- | --- | --- | --- | ---
2
| | | | 2 | 2 | 2 | |
| --- | --- | --- | --- | --- | --- | --- | ---
| 2. 2x + 3y – 9 = 0 2 4 3 6 9 18 − − 1 1 1 lines many solutions 4x + 6y – 18 = 0
| 3. x + 2y – 4 = 0 2x + 4y – 12 = 0 | 1 | | | | | |
| From the table above, you can observe that if the lines represented by the equation
| | a + b y + c 1 1 | 1 = 0 | | | | |
| | + b y + c 2 2 | 2 = 0 | | | | |
| | | a b | | | | |
| | | a b ≠ ⋅ | | | | |
| | | 2 2 | | | | |
| | a | b c | | | | |
| | a | b c = = | ⋅ | | | |
| | 2 | 2 | 2 | | | |
| | a | b c | | | | |
| | a | b c = ≠ ⋅ | | | | |
| | 2 | 2 2 | | | | |
| and | a
| --- | ---
| (i) intersecting, then | | 1 | | 1 |
| --- | --- | --- | --- | --- | ---
| (ii) coincident, then | 1 | | 1 | | 1
| (iii) parallel, then 1 | | 1 | | 1 |
In fact, the converse is also true for any pair of lines.
You can verify them by considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example 1 : Check graphically whether the pair of equations
| | x + 3y = 6 | (1)
| --- | --- | ---
| and | 2x – 3y = 12 | (2)
is consistent.
If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2).
For this, we find two solutions of each of the equations, which are given in Table 3.2
Table 3.2
x
0
6
x 0 3
y =
| 6 3 x−
| ---
| 2 | 0 | y = 2 | 12
| --- | --- | --- | ---
3
x −
– 4 –2
Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ as shown in Fig. 3.1. We observe that there is a point B (6, 0) common to both the lines AB and PQ.
So, the solution of the pair of linear equations is x
=
6 and y = 0, i.e., the given pair of equations is consistent.
Fig. 3.1
Example 2 : Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
| 5x – 8y + 1 = 0 | (1)
| --- | ---
| 3x – | 24 5 y + 3 5 | = | 0 | (2)
| --- | --- | --- | --- | ---
| Solution : Multiplying Equation (2) by | | 5 , 3 we get | |
5x – 8y +
1
= 0 But, this is the same as Equation (1).
Hence the lines represented by Equations (1) and (2) are coincident.
Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts.
When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased.
Also, the number of skirts is four less than four times the number of pants purchased”.
Help her friends to find how many pants and skirts Champa bought.
Solution : Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x – 2 (1) and y = 4x – 4 (2) Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.
They are given in Table 3.3.
| Table 3.3
| ---
| x | 2 | 0
| --- | --- | ---
| y = 2x – 2 | 2 | – 2
| x | 0 | 1
| y = 4x – 4 | – 4 | 0
Fig. 3.2
Plot the points and draw the lines passing through them to represent the equations, as shown in Fig. 3.2.
The two lines intersect at the point (1, 0).
So, x = 1, y
=
0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Verify the answer by checking whether it satisfies the conditions of the given problem. | jemh103.pdf |
2 | CBSE | Class10 | Mathematics | Table 3.1
| | 1 2 | 1 2 | 1 2
| --- | --- | --- | ---
| 1. | x – 2y = 0 | 1 3 | 2 4 − | 0 20− | 1 | 1
| --- | --- | --- | --- | --- | --- | ---
| a a | b b | c c | Compare the | Graphical Algebraic No. | ratios | representation interpretation
| Sl Pair of lines
| 3x + 4y – 20 = 0
| | | | | | a | b
| | | | | | a b ≠ |
| | | | | | |
| | | | | | 2 | 2
| | | | | | a | b
| | | | | | a c = | b
| | | | | | 2 | 2
| | | | | | a | b
| | | | 2 4 | − − 4 12 | |
| 1 | 1 | 1
| --- | --- | ---
| a | b | c = | ≠ | Parallel lines | No solution
| --- | --- | --- | --- | --- | ---
2
| | | | 2 | 2 | 2 | |
| --- | --- | --- | --- | --- | --- | --- | ---
| 2. 2x + 3y – 9 = 0 2 4 3 6 9 18 − − 1 1 1 lines many solutions 4x + 6y – 18 = 0
| 3. x + 2y – 4 = 0 2x + 4y – 12 = 0 | 1 | | | | | |
| From the table above, you can observe that if the lines represented by the equation
| | a + b y + c 1 1 | 1 = 0 | | | | |
| | + b y + c 2 2 | 2 = 0 | | | | |
| | | a b | | | | |
| | | a b ≠ ⋅ | | | | |
| | | 2 2 | | | | |
| | a | b c | | | | |
| | a | b c = = | ⋅ | | | |
| | 2 | 2 | 2 | | | |
| | a | b c | | | | |
| | a | b c = ≠ ⋅ | | | | |
| | 2 | 2 2 | | | | |
| and | a
| --- | ---
| (i) intersecting, then | | 1 | | 1 |
| --- | --- | --- | --- | --- | ---
| (ii) coincident, then | 1 | | 1 | | 1
| (iii) parallel, then 1 | | 1 | | 1 |
In fact, the converse is also true for any pair of lines.
You can verify them by considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example 1 : Check graphically whether the pair of equations
| | x + 3y = 6 | (1)
| --- | --- | ---
| and | 2x – 3y = 12 | (2)
is consistent.
If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2).
For this, we find two solutions of each of the equations, which are given in Table 3.2Table 3.2
x
0
6
x 0 3
y =
| 6 3 x−
| ---
| 2 | 0 | y = 2 | 12
| --- | --- | --- | ---
3x −
– 4 –2
Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ as shown in Fig. 3.1. We observe that there is a point B (6, 0) common to both the lines AB and PQ.
So, the solution of the pair of linear equations is x
=
6 and y = 0, i.e., the given pair of equations is consistent.
Fig. 3.1
Example 2 : Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:
| 5x – 8y + 1 = 0 | (1)
| --- | ---
| 3x – | 24 5 y + 3 5 | = | 0 | (2)
| --- | --- | --- | --- | ---
| Solution : Multiplying Equation (2) by | | 5 , 3 we get | |
5x – 8y +
1
= 0 But, this is the same as Equation (1).
Hence the lines represented by Equations (1) and (2) are coincident.
Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Example 3 : Champa went to a ‘Sale’ to purchase some pants and skirts.
When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased.
Also, the number of skirts is four less than four times the number of pants purchased”.
Help her friends to find how many pants and skirts Champa bought.
Solution : Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x – 2 (1) and y = 4x – 4 (2) Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.
They are given in Table 3.3.
| Table 3.3
| ---
| x | 2 | 0
| --- | --- | ---
| y = 2x – 2 | 2 | – 2
| x | 0 | 1
| y = 4x – 4 | – 4 | 0
| jemh103.pdf |
3 | CBSE | Class10 | Mathematics | EXERCISE 3.1
1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz.
If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together cost ` 46.
Find the cost of one pencil and that of one pen.
| | a | b | c
| --- | --- | --- | ---
| 2. | On comparing the ratios | 1 | 1 | 1
| --- | --- | --- | --- | ---
| | | a b | | c, find out whether the lines representing the
| | | 2 2, | 2 2, and |
following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y +
8
= 0 (ii) 9x + 3y + 12 = 0
7x + 6y –
9
= 0 18x + 6y + 24 = 0 (iii) 6x – 3y + 10 = 0
2x –
y
+ 9 = 0, find out whether the following pair of linear equations are consistent, or inconsistent.
3. On comparing the ratios 1 1
2 2
,a b a b
and 1 2
c c
(i) 3x + 2y =
5
; 2x – 3y = 7 (ii) 2x – 3y =
8
; 4x – 6y = 9
(iii) 3 5 7
2
3
x y ; 9x – 10y = 14 (iv) 5x – 3y = 11 ; 10x + 6y = –22
(v) 4 2 8
3
x y ; 2x + 3y = 12
4. Which of the following pairs of linear equations are consistent/inconsistent?
If consistent, obtain the solution graphically:
(i) x
+
y = 5, 2x + 2y = 10
(ii) x
–
y = 8, 3x – 3y = 16
(iii) 2x +
y
– 6 = 0, 4x – 2y –
4
= 0
(iv) 2x – 2y –
2
= 0, 4x – 4y –
5
= 0
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
6. Given the linear equation 2x + 3y –
8
= 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines
(iii) coincident lines
7. Draw the graphs of the equations x
–
y + 1
=
0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
3.3 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically.
The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like ()3
2 7, (–1.75, 3.3), 4 1, 13 19
, etc.
There is every possibility of making mistakes while reading such coordinates.
Is there any alternative method of finding the solution?
There are several algebraic methods, which we shall now discuss.
3.3.1 Substitution Method : We shall explain the method of substitution by taking some examples.
| Example 4 : Solve the following pair of equations by substitution method: | 7x – 15y = 2 | (1)
| --- | --- | ---
| | x + 2y = 3 | (2)
| Solution : Step 1 : We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) :
| x + 2y = 3
| and write it as | x = 3 – 2y | (3)
| Step 2 : Substitute the value of x in Equation (1). We get 7(3 – 2y) – 15y = 2
| i.e., | 21 – 14y – 15y = 2 |
| i.e., | – 29y = –19 |
| Therefore, | y = | 19
29
Step 3 : Substituting this value of y in Equation (3), we get =
=
3 – 19 2
29 49 29 49 29, y =
Therefore, the solution is x = Putting this value of t in Equation (2), we get s
=
3 (12) +
6
= 42 So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.
19 29.
Verification : Substituting x = 49 29 and y = 19 29, you can verify that both the Equations
(1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:
Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved.
Sometimes, as in Examples 9 and 10 below, you can get statements with no variable.
If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions.
If the statement is false, then the pair of linear equations is inconsistent.
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.
Remark : We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations.
That is why the method is known as the substitution method.
Example 5 : Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then.
Also, three years from now, I shall be three times as old as you will be.”
(Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.
Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively.
Then, the pair of linear equations that represent the situation is s
–
7 = 7 (t – 7), i.e., s – 7t + 42 = 0 (1) and s
+
3 = 3 (t + 3), i.e., s – 3t = 6 (2) Using Equation (2), we get s = 3t + 6.
Putting this value of s in Equation (1), we get (3t + 6) – 7t + 42 = 0,
| i.e., | 4t = 48, which gives t = 12. |
| --- | --- | ---
| Example 6 : Solution : The pair of linear equations formed were: 2x + 3y = 9 | In a shop the cost of 2 pencils and 3 erasers is `9 and the cost of 4 pencils and 6 erasers is `18. Find the cost of each pencil and each eraser. | (1)
| | 4x + 6y = 18 | (2)
We first express the value of x in terms of y from the equation 2x + 3y = 9, to get
| = 9 3 2 y− | (3)
| Now we substitute this value of x in Equation (2), to get
| | 4(9 3) 2 y− + 6y = 18
| i.e., | 18 – 6y + 6y = 18
| i.e., | 18 = 18
This statement is true for all values of y. However, we do not get a specific value of y as a solution.
Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same.
Therefore, Equations (1) and (2) have infinitely many solutions.
We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Example 7 : Two rails are represented by the equations x + 2y –
4
= 0 and 2x + 4y – 12 = 0.
Will the rails cross each other?
Solution : The pair of linear equations formed were:
| x + 2y – 4 = 0 | (1)
| 2x + 4y – 12 = 0 | (2)
We express x in terms of y from Equation (1) to get =
4
– 2y Now, we substitute this value of x in Equation (2) to get 2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0 i.e., 4
=
0 which is a false statement.
Therefore, the equations do not have a common solution.
So, the two rails will not cross each other. | jemh103.pdf |
4 | CBSE | Class10 | Mathematics | c c
(i) 3x + 2y =
5
; 2x – 3y = 7 (ii) 2x – 3y =
8
; 4x – 6y = 9
(iii) 3 5 7
2
3
x y ; 9x – 10y = 14 (iv) 5x – 3y = 11 ; 10x + 6y = –22
(v) 4 2 8
3
x y ; 2x + 3y = 12
4. Which of the following pairs of linear equations are consistent/inconsistent?
If consistent, obtain the solution graphically:
(i) x
+
y = 5, 2x + 2y = 10
(ii) x
–
y = 8, 3x – 3y = 16
(iii) 2x +
y
– 6 = 0, 4x – 2y –
4
= 0
(iv) 2x – 2y –
2
= 0, 4x – 4y –
5
= 0
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
6. Given the linear equation 2x + 3y –
8
= 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines
(iii) coincident lines
7. Draw the graphs of the equations x
–
y + 1
=
0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
3.3 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically.
The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like ()3
2 7, (–1.75, 3.3), 4 1, 13 19
, etc.
There is every possibility of making mistakes while reading such coordinates.
Is there any alternative method of finding the solution?
There are several algebraic methods, which we shall now discuss.
3.3.1 Substitution Method : We shall explain the method of substitution by taking some examples.
| Example 4 : Solve the following pair of equations by substitution method: | 7x – 15y = 2 | (1)
| --- | --- | ---
| | x + 2y = 3 | (2)
| Solution : Step 1 : We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) :
| x + 2y = 3
| and write it as | x = 3 – 2y | (3)
| Step 2 : Substitute the value of x in Equation (1). We get 7(3 – 2y) – 15y = 2
| i.e., | 21 – 14y – 15y = 2 |
| i.e., | – 29y = –19 |
| Therefore, | y = | 19
29
Step 3 : Substituting this value of y in Equation (3), we get =
=
3 – 19 2
29 49 29 49 29, y =
Therefore, the solution is x = Putting this value of t in Equation (2), we get s
=
3 (12) +
6
= 42 So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.
19 29.
Verification : Substituting x = 49 29 and y = 19 29, you can verify that both the Equations
(1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:
Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved.
Sometimes, as in Examples 9 and 10 below, you can get statements with no variable.
If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions.
If the statement is false, then the pair of linear equations is inconsistent.
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.
Remark : We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations.
That is why the method is known as the substitution method.
Example 5 : Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then.
Also, three years from now, I shall be three times as old as you will be.”
(Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.
Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively.
Then, the pair of linear equations that represent the situation is s
–
7 = 7 (t – 7), i.e., s – 7t + 42 = 0 (1) and s
+
3 = 3 (t + 3), i.e., s – 3t = 6 (2) Using Equation (2), we get s = 3t + 6.
Putting this value of s in Equation (1), we get (3t + 6) – 7t + 42 = 0,
| i.e., | 4t = 48, which gives t = 12. |
| --- | --- | ---
| Example 6 : Solution : The pair of linear equations formed were: 2x + 3y = 9 | In a shop the cost of 2 pencils and 3 erasers is `9 and the cost of 4 pencils and 6 erasers is `18. Find the cost of each pencil and each eraser. | (1)
| | 4x + 6y = 18 | (2)
We first express the value of x in terms of y from the equation 2x + 3y = 9, to get
| = 9 3 2 y− | (3)
| Now we substitute this value of x in Equation (2), to get
| | 4(9 3) 2 y− + 6y = 18
| i.e., | 18 – 6y + 6y = 18
| i.e., | 18 = 18
This statement is true for all values of y. However, we do not get a specific value of y as a solution.
Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same.
Therefore, Equations (1) and (2) have infinitely many solutions.
We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Example 7 : Two rails are represented by the equations x + 2y –
4
= 0 and 2x + 4y – 12 = 0.
Will the rails cross each other?
Solution : The pair of linear equations formed were:
| x + 2y – 4 = 0 | (1)
| 2x + 4y – 12 = 0 | (2)
We express x in terms of y from Equation (1) to get =
4
– 2y Now, we substitute this value of x in Equation (2) to get 2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0 i.e., 4
=
0 which is a false statement.
Therefore, the equations do not have a common solution.
So, the two rails will not cross each other. | jemh103.pdf |
5 | CBSE | Class10 | Mathematics | 3.3 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically.
The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like ()3
2 7, (–1.75, 3.3), 4 1, 13 19
, etc.
There is every possibility of making mistakes while reading such coordinates.
Is there any alternative method of finding the solution?
There are several algebraic methods, which we shall now discuss.
3.3.1 Substitution Method : We shall explain the method of substitution by taking some examples.
| Example 4 : Solve the following pair of equations by substitution method: | 7x – 15y = 2 | (1)
| --- | --- | ---
| | x + 2y = 3 | (2)
| Solution : Step 1 : We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2) :
| x + 2y = 3
| and write it as | x = 3 – 2y | (3)
| Step 2 : Substitute the value of x in Equation (1). We get 7(3 – 2y) – 15y = 2
| i.e., | 21 – 14y – 15y = 2 |
| i.e., | – 29y = –19 |
| Therefore, | y = | 19
29
Step 3 : Substituting this value of y in Equation (3), we get =
=
3 – 19 2
29 49 29 49 29, y =
Therefore, the solution is x = Putting this value of t in Equation (2), we get s
=
3 (12) +
6
= 42 So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.
19 29.
Verification : Substituting x = 49 29 and y = 19 29, you can verify that both the Equations
(1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:
Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved.
Sometimes, as in Examples 9 and 10 below, you can get statements with no variable.
If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions.
If the statement is false, then the pair of linear equations is inconsistent.
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.
Remark : We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations.
That is why the method is known as the substitution method.
Example 5 : Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then.
Also, three years from now, I shall be three times as old as you will be.”
(Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.
Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively.
Then, the pair of linear equations that represent the situation is s
–
7 = 7 (t – 7), i.e., s – 7t + 42 = 0 (1) and s
+
3 = 3 (t + 3), i.e., s – 3t = 6 (2) Using Equation (2), we get s = 3t + 6.
Putting this value of s in Equation (1), we get (3t + 6) – 7t + 42 = 0,
| i.e., | 4t = 48, which gives t = 12. |
| --- | --- | ---
| Example 6 : Solution : The pair of linear equations formed were: 2x + 3y = 9 | In a shop the cost of 2 pencils and 3 erasers is `9 and the cost of 4 pencils and 6 erasers is `18. Find the cost of each pencil and each eraser. | (1)
| | 4x + 6y = 18 | (2)
We first express the value of x in terms of y from the equation 2x + 3y = 9, to get
| = 9 3 2 y− | (3)
| Now we substitute this value of x in Equation (2), to get
| | 4(9 3) 2 y− + 6y = 18
| i.e., | 18 – 6y + 6y = 18
| i.e., | 18 = 18
This statement is true for all values of y. However, we do not get a specific value of y as a solution.
Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same.
Therefore, Equations (1) and (2) have infinitely many solutions.
We cannot find a unique cost of a pencil and an eraser, because there are many common solutions, to the given situation.
Example 7 : Two rails are represented by the equations x + 2y –
4
= 0 and 2x + 4y – 12 = 0.
Will the rails cross each other?
Solution : The pair of linear equations formed were:
| x + 2y – 4 = 0 | (1)
| 2x + 4y – 12 = 0 | (2)
We express x in terms of y from Equation (1) to get =
4
– 2y Now, we substitute this value of x in Equation (2) to get 2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0 i.e., 4
=
0 which is a false statement.
Therefore, the equations do not have a common solution.
So, the two rails will not cross each other. | jemh103.pdf |
6 | CBSE | Class10 | Mathematics | 3.3.2 Elimination Method
Now let us consider another method of eliminating (i.e., removing) one variable.
This is sometimes more convenient than the substitution method.
Let us see how this method works.
Example 8 : The ratio of incomes of two persons is 9
:
7 and the ratio of their expenditures is 4 : 3.
If each of them manages to save ` 2000 per month, find their monthly incomes.
Solution : Let us denote the incomes of the two person by ` 9x and ` 7x and their expenditures by ̀ 4y and ̀ 3y respectively.
Then the equations formed in the situation is given by :
| | 9x – 4y = 2000 | (1)
| --- | --- | ---
| and | 7x – 3y = 2000 | (2)
| Step 1 : Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then we get the equations:
| | 27x – 12y = 6000 | (3)
| | 28x – 12y = 8000 | (4)
Step 2 : Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same.
So, we get (28x – 27x) – (12y – 12y) = 8000 – 6000
| i.e., | x = 2000
| Step 3 : Substituting this value of x in (1), we get
| 9(2000) – 4y = 2000
| i.e., | y = 4000
So, the solution of the equations is x = 2000, y = 4000.
Therefore, the monthly incomes of the persons are ` 18,000 and ` 14,000, respectively.
Verification : 18000 : 14000 =
9
: 7.
Also, the ratio of their expenditures = 18000 – 2000 : 14000 – 2000 = 16000 : 12000 =
4
: 3Remarks :
1. The method used in solving the example above is called the elimination method, because we eliminate one variable first, to get a linear equation in one variable.
In the example above, we eliminated y. We could also have eliminated x. Try doing it that way.
2. You could also have used the substitution, or graphical method, to solve this problem.
Try doing so, and see which method is more convenient.
Let us now note down these steps in the elimination method :
Step 1 : First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
Step 2 : Then add or subtract one equation from the other so that one variable gets eliminated.
If you get an equation in one variable, go to Step 3.
If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step 3 : Solve the equation in one variable (x or y) so obtained to get its value.
Step 4 : Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
Now to illustrate it, we shall solve few more examples.
Example 9 : Use elimination method to find all possible solutions of the following pair of linear equations :
| 2x + 3y = 8 | (1)
| 4x + 6y = 7 | (2)
Solution :
Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal.
Then we get the equations as :
| 4x + 6y = 16 | (3)
| 4x + 6y = 7 | (4)
Step 2 : Subtracting Equation (4) from Equation (3), (4x – 4x) + (6y – 6y) = 16 – 7 i.e., 0 = 9, which is a false statement.
Therefore, the pair of equations has no solution.
Example 10 : The sum of a two-digit number and the number obtained by reversing the digits is 66.
If the digits of the number differ by 2, find the number.
How many such numbers are there?
Solution : Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the first number may be written as 10 x
+
y in the expanded form (for example, 56 = 10(5) + 6).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit.
This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5).
According to the given condition.
| (10x + y) + (10y + x) = 66
| ---
| i.e., | 11(x + y) = 66 |
| --- | --- | ---
| i.e., | x + y = 6 | (1)
| We are also given that the digits differ by 2, therefore,
| either | x – y = 2 | (2)
| or | y – x = 2 | (3)
If x
–
y = 2, then solving (1) and (2) by elimination, we get x
=
4 and y = 2.
In this case, we get the number 42.
If y
–
x = 2, then solving (1) and (3) by elimination, we get x
=
2 and y = 4.
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Verification : Here 42 + 24 = 66 and 4
–
2 = 2. Also 24 + 42 = 66 and 4
–
2 = 2. | jemh103.pdf |
0 | CBSE | Class10 | Mathematics | QUADRATIC EQUATIONS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials.
One type was the quadratic polynomial of the form ax2 + bx + c, a 0.
When we equate this polynomial to zero, we get a quadratic equation.
Quadratic equations come up when we deal with many real-life situations.
For instance, suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth.
What should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres.
Then, its length should be (2x + 1) metres.
We can depict this information pictorially as shown in Fig. 4.1.
Fig. 4.1
| Now, | area of the hall = (2x + 1). x m2 = (2x2 + x) m2
| So, | 2x2 + x = 300 (Given)
| Therefore, | 2x2 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x2 +
x
– 300 = 0 which is a quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation of the form x2 – px +
q
= 0.
Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present day terminology, are solutions of quadratic equations.
Solving of quadratic equations, in general form, is often credited to ancient Indian mathematicians.
In fact, Brahmagupta (C.E.598–665) gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later, Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula, (as quoted by Bhaskara II) for solving a quadratic equation by the method of completing the square.
An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied quadratic equations of different types.
Abraham bar Hiyya Ha-Nasi, in his book ‘Liber embadorum’ published in Europe in C.E.
1145 gave complete solutions of different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding their roots.
You will also see some applications of quadratic equations in daily life situations.
4.2 Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax2 + bx +
c
= 0, where a, b, c are real numbers, a 0.
For example, 2x2 +
x
– 300 = 0 is a quadratic equation.
Similarly, 2x2 – 3x +
1
= 0, 4x – 3x2 +
2
= 0 and 1 – x2 + 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation.
But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation.
That is, ax2 + bx +
c
= 0, a
0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in different fields of mathematics.
Let us consider a few examples.
Example 1 : Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles.
Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day.
The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day.
On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles =
x
– 5 The number of marbles left with Jivanti, when she lost 5 marbles = 45 –
x
– 5 = 40 – x Therefore, their product = (x – 5) (40 – x) = 40x – x2 – 200 + 5x
| | = | x2 + 45x – 200
| --- | --- | ---
| So, | x2 + 45x – 200 = 124 | (Given that product = 124)
| i.e., | x2 + 45x – 324 = 0 |
| i.e., | x2 – 45x + 324 = 0 |
Therefore, the number of marbles John had, satisfies the quadratic equation x2 – 45x + 324 = 0
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x So, the total cost of production (in rupees) that day = x (55 – x)
| Therefore, | x (55 – x) = 750
| i.e., | 55x – x2 = 750
| i.e., | x2 + 55x – 750 = 0
| --- | ---
| i.e., | x2 – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation x2 – 55x + 750 = 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) (x – 2)2 +
1
= 2x – 3 (ii) x(x + 1) +
8
= (x + 2) (x – 2)
(iii) x (2x + 3) = x2 + 1 (iv) (x + 2)3 = x3 – 4
Solution :
(i) LHS = (x – 2)2 +
1
= x2 – 4x +
4
+ 1 = x2 – 4x + 5
Therefore, (x – 2)2 +
1
= 2x – 3 can be rewritten as x2 – 4x +
5
= 2x – 3
i.e., x2 – 6x +
8
= 0 It is of the form ax2 + bx +
c
= 0.
Therefore, the given equation is a quadratic equation.
(ii) Since x(x + 1) +
8
= x2 +
x
+ 8 and (x + 2)(x – 2) = x2 – 4
| | Therefore, | x2 + x + 8 = | x2 – 4
| --- | --- | --- | ---
| | i.e., | x + 12 = 0 |
| It is not of the form ax2 + bx + c = 0.
| Therefore, the given equation is not a quadratic equation.
| (iii) Here, LHS = x (2x + 3) = 2x2 + 3x
| | So, | x (2x + 3) = | x2 + 1 can be rewritten as
| 2x2 + 3x = x2 + 1
| | Therefore, we get | x2 + 3x – 1 = 0 |
| It is of the form ax2 + bx + c = 0.
| So, the given equation is a quadratic equation.
| (iv) Here, LHS = (x + 2)3 = x3 + 6x2 + 12x + 8
| | Therefore, | (x + 2)3 = | x3 – 4 can be rewritten as
| x3 + 6x2 + 12x + 8 = x3 – 4
| i.e., | 6x2 + 12x + 12 = 0 | or, | x2 + 2x + 2 = 0
It is of the form ax2 + bx +
c
= 0.
So, the given equation is a quadratic equation.
Remark : Be careful!
In (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3) and not a quadratic equation.
But it turns out to be a quadratic equation.
As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not. | jemh104.pdf |
1 | CBSE | Class10 | Mathematics | 4.1 Introduction
In Chapter 2, you have studied different types of polynomials.
One type was the quadratic polynomial of the form ax2 + bx + c, a 0.
When we equate this polynomial to zero, we get a quadratic equation.
Quadratic equations come up when we deal with many real-life situations.
For instance, suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth.
What should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres.
Then, its length should be (2x + 1) metres.
We can depict this information pictorially as shown in Fig. 4.1.
Fig. 4.1
| Now, | area of the hall = (2x + 1). x m2 = (2x2 + x) m2
| So, | 2x2 + x = 300 (Given)
| Therefore, | 2x2 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x2 +
x
– 300 = 0 which is a quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation of the form x2 – px +
q
= 0.
Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present day terminology, are solutions of quadratic equations.
Solving of quadratic equations, in general form, is often credited to ancient Indian mathematicians.
In fact, Brahmagupta (C.E.598–665) gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later, Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula, (as quoted by Bhaskara II) for solving a quadratic equation by the method of completing the square.
An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied quadratic equations of different types.
Abraham bar Hiyya Ha-Nasi, in his book ‘Liber embadorum’ published in Europe in C.E.
1145 gave complete solutions of different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding their roots.
You will also see some applications of quadratic equations in daily life situations.Fig. 4.1
| Now, | area of the hall = (2x + 1). x m2 = (2x2 + x) m2
| So, | 2x2 + x = 300 (Given)
| Therefore, | 2x2 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x2 +
x
– 300 = 0 which is a quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation of the form x2 – px +
q
= 0.
Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present day terminology, are solutions of quadratic equations.
Solving of quadratic equations, in general form, is often credited to ancient Indian mathematicians.
In fact, Brahmagupta (C.E.598–665) gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later, Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula, (as quoted by Bhaskara II) for solving a quadratic equation by the method of completing the square.
An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied quadratic equations of different types.
Abraham bar Hiyya Ha-Nasi, in his book ‘Liber embadorum’ published in Europe in C.E.
1145 gave complete solutions of different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding their roots.
You will also see some applications of quadratic equations in daily life situations. | jemh104.pdf |
2 | CBSE | Class10 | Mathematics | Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles =
x
– 5 The number of marbles left with Jivanti, when she lost 5 marbles = 45 –
x
– 5 = 40 – x Therefore, their product = (x – 5) (40 – x) = 40x – x2 – 200 + 5x
| | = | x2 + 45x – 200
| --- | --- | ---
| So, | x2 + 45x – 200 = 124 | (Given that product = 124)
| i.e., | x2 + 45x – 324 = 0 |
| i.e., | x2 – 45x + 324 = 0 |
Therefore, the number of marbles John had, satisfies the quadratic equation x2 – 45x + 324 = 0
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x So, the total cost of production (in rupees) that day = x (55 – x)
| Therefore, | x (55 – x) = 750
| i.e., | 55x – x2 = 750
| i.e., | x2 + 55x – 750 = 0
| --- | ---
| i.e., | x2 – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation x2 – 55x + 750 = 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) (x – 2)2 +
1
= 2x – 3 (ii) x(x + 1) +
8
= (x + 2) (x – 2)
(iii) x (2x + 3) = x2 + 1 (iv) (x + 2)3 = x3 – 4Solution :
(i) LHS = (x – 2)2 +
1
= x2 – 4x +
4
+ 1 = x2 – 4x + 5
Therefore, (x – 2)2 +
1
= 2x – 3 can be rewritten as x2 – 4x +
5
= 2x – 3
i.e., x2 – 6x +
8
= 0 It is of the form ax2 + bx +
c
= 0.
Therefore, the given equation is a quadratic equation.
(ii) Since x(x + 1) +
8
= x2 +
x
+ 8 and (x + 2)(x – 2) = x2 – 4
| | Therefore, | x2 + x + 8 = | x2 – 4
| --- | --- | --- | ---
| | i.e., | x + 12 = 0 |
| It is not of the form ax2 + bx + c = 0.
| Therefore, the given equation is not a quadratic equation.
| (iii) Here, LHS = x (2x + 3) = 2x2 + 3x
| | So, | x (2x + 3) = | x2 + 1 can be rewritten as
| 2x2 + 3x = x2 + 1
| | Therefore, we get | x2 + 3x – 1 = 0 |
| It is of the form ax2 + bx + c = 0.
| So, the given equation is a quadratic equation.
| (iv) Here, LHS = (x + 2)3 = x3 + 6x2 + 12x + 8
| | Therefore, | (x + 2)3 = | x3 – 4 can be rewritten as
| x3 + 6x2 + 12x + 8 = x3 – 4
| i.e., | 6x2 + 12x + 12 = 0 | or, | x2 + 2x + 2 = 0
It is of the form ax2 + bx +
c
= 0.
So, the given equation is a quadratic equation.
Remark : Be careful!
In (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3) and not a quadratic equation.
But it turns out to be a quadratic equation.
As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not. | jemh104.pdf |
3 | CBSE | Class10 | Mathematics | 4.3 Solution of a Quadratic Equation by Factorisation
Consider the quadratic equation 2x2 – 3x +
1
= 0.
If we replace x by 1 on the LHS of this equation, we get (2 × 12) – (3 × 1) +
1
= 0 = RHS of the equation.
We say that 1 is a root of the quadratic equation 2x2 – 3x +
1
= 0.
This also means that 1 is a zero of the quadratic polynomial 2x2 – 3x + 1.
In general, a real number is called a root of the quadratic equation ax2 + bx +
c
= 0, a
0 if a 2 + b +
c
= 0. We also say that x = is a solution of the quadratic equation, or that satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx +
c
= 0 are the same.
You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes.
So, any quadratic equation can have atmost two roots.
You have learnt in Class IX, how to factorise quadratic polynomials by splitting their middle terms.
We shall use this knowledge for finding the roots of a quadratic equation.
Let us see how.
Example 3 : Find the roots of the equation 2x2 – 5x +
3
= 0, by factorisation.
Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = 6x2 = (2x2) × 3].
So, 2x2 – 5x +
3
= 2x2 – 2x – 3x +
3
= 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1) Now, 2x2 – 5x +
3
= 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x2 – 5x +
3
= 0 are the same for which (2x – 3)(x – 1) = 0, i.e., either 2x –
3
= 0 or x
–
1 = 0.
Now, 2x –
3
= 0 gives
3 2
x and x
–
1 = 0 gives x = 1.
So,
3
2
3 2
x and x
=
1 are the solutions of the equation.
In other words, 1 and are the roots of the equation 2x2 – 5x +
3
= 0.
Verify that these are the roots of the given equation.
Note that we have found the roots of 2x2 – 5x +
3
= 0 by factorising 2x2 – 5x + 3 into two linear factors and equating each factor to zero.
Example 4 : Find the roots of the quadratic equation 6x2 –
x
– 2 = 0.
Solution : We have 6x2 –
x
– 2 = 6x2 + 3x – 4x –
2
= 3x (2x + 1) – 2 (2x + 1) = (3x – 2)(2x + 1) The roots of 6x2 –
x
– 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0 Therefore, 3x –
2
= 0 or 2x +
1
= 0,
| i.e., | | | | | x = 2 3 or | | x = | | 1 2 | | | | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, the roots of 6x2 – x – 2 = 0 are 2 | | | | | | | | | | 1.and – | | | | | | | | | |
| | | | | | | | 3 | | | 2 | | | | | | | | | |
| We verify the roots, by checking that 2 | | | | | | | | | | | 1 and 2 satisfy 6x2 – x – 2 = 0. | | | | | | | | |
| Example 5 : Find the roots of the quadratic equation 23 2 6 2 0x x .
| Solution : 23 2 6 2x x = 23 6 6 2x x x
| = 3
| | | | | | 3 | | 2 | | | 2 | 3 x | | | 2x x | | | | | |
| | | | | | = | | 3 | x | | 2x | 3 2 | | | | | | | | |
| So, the roots of the equation are the values of x for which
| | | | | | 3 x | | 2 | | 0x | 3 2 | | | | | | | | | |
| Now, 3 | 2 0x | for | | x | 2 3 . | | | | | | | | | | | | | | |
So, this root is repeated twice, one for each repeated factor 3 2x .
Therefore, the roots of 23 2
6
2 0x x are 2
3
, 2 3.
Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1.
Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x satisfies the equation 2x2 +
x
– 300 = 0.
Applying the factorisation method, we write this equation as 2x2 – 24x + 25x – 300 = 0
| | 2x (x – 12) + 25 (x – 12) = 0
| i.e., | (x – 12)(2x + 25) = 0
| So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is 12 m. Its length = 2x +
1
= 25 m. | jemh104.pdf |
4 | CBSE | Class10 | Mathematics | EXERCISE 4.2
1. Find the roots of the following quadratic equations by factorisation:
| (i) | x2 – 3x – 10 = 0 | (ii) 2x2 + x – 6 = 0
| --- | --- | ---
| (iii) | 22 | 7 | x | 5 | 2 0x | (iv) 2x2 – x + 1 8 | = 0
| --- | --- | --- | --- | --- | --- | --- | ---
(v) 100 x2 – 20x +
1
= 0
2. Solve the problems given in Example 1.
3. Find two numbers whose sum is 27 and product is 182.
4. Find two consecutive positive integers, sum of whose squares is 365.
5. The altitude of a right triangle is 7 cm less than its base.
If the hypotenuse is 13 cm, find the other two sides.
6. A cottage industry produces a certain number of pottery articles in a day.
It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day.
If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article.
4.4 Nature of Roots
The equation ax2 + bx +
c
= 0 are given by
=
2– 4 2
b b ac a
If b2 – 4ac > 0, we get two distinct real roots If b2 – 4ac = 0, then x = 0
| | | | | | | | | 2 | | 2 2 b | 4 acb |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | | | | | | | a | a a | a | a | and
| 2 4
| – 2 2 b acb
| a a | a | a | a | a | a | | | . | | | |
b
a
, i.e., or –
2
2 b b
x
a a
2
So, the roots of the equation ax2 + bx +
c
= 0 are both
2
b a
Therefore, we say that the quadratic equation ax2 + bx +
c
= 0 has two equal real roots in this case.
If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac.
Therefore, there are no real roots for the given quadratic equation in this case.
Since b2 – 4ac determines whether the quadratic equation ax2 + bx +
c
= 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax2 + bx +
c
= 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.
Let us consider some examples.
Example 7: Find the discriminant of the quadratic equation 2x2 – 4x +
3
= 0, and hence find the nature of its roots.
Solution : The given equation is of the form ax2 + bx +
c
= 0, where a = 2, b
=
– 4 and c = 3.
Therefore, the discriminant
b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.
Example 8
:
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres.
Is it possible to do so?
If yes, at what distances from the two gates should the pole be erected?
Solution : Let us first draw the diagram (see Fig. 4.2).
Let P be the required location of the pole.
Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP – BP (or, BP AP) = 7 m. Therefore, AP = (x + 7) m.
Fig. 4.2
Now, AB = 13m, and since AB is a diameter, APB = 90° (Why?) Therefore, AP2 + PB2 = AB2 (By Pythagoras theorem) b2 – 4ac = 72 –
4
× 1 × (– 60) = 289 > 0.
So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park.
Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get
| i.e., | (x + 7)2 + x2 = 132
| --- | ---
| i.e., | x2 + 14x + 49 + x2 = 169
| i.e., | 2x2 + 14x – 120 = 0
So, the distance ‘x’ of the pole from gate B satisfies the equation x2 + 7x – 60 = 0
So, it would be possible to place the pole if this equation has real roots.
To see if this is so or not, let us consider its discriminant.
The discriminant is
7 289
| = | 2 | 7 | 17
| --- | --- | --- | ---
| | = | 2 |
Therefore, x
=
5 or – 12.
Example 9 : Find the discriminant of the equation 3x2 – 2x +
1
3 = 0 and hence find the nature of its roots.
Find them, if they are real.
Solution : Here a = 3, b
=
– 2 and 1
3
c .
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x
=
– 12 will have to be ignored.
So, x = 5.
Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.
Therefore, discriminant b2 – 4ac = (– 2)2 –
4
× 3 ×
1
3 = 4
–
4 = 0.
Hence, the given quadratic equation has two equal real roots.
| The roots are | | | | | 2 1,,, | 2 1,,, | 1,,, , | 1,,, i.e., | 1 1,,, | 1,,, , | | .i.e.,
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 2 | 3 b a | 2 | 3 b b a | 6 3 b | 6 3 b | 3 b | 3 b | 3 3 b | 3 b | 3 b |
| jemh104.pdf |
5 | CBSE | Class10 | Mathematics | b a
Therefore, we say that the quadratic equation ax2 + bx +
c
= 0 has two equal real roots in this case.
If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac.
Therefore, there are no real roots for the given quadratic equation in this case.
Since b2 – 4ac determines whether the quadratic equation ax2 + bx +
c
= 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax2 + bx +
c
= 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.
Let us consider some examples.
Example 7: Find the discriminant of the quadratic equation 2x2 – 4x +
3
= 0, and hence find the nature of its roots.
Solution : The given equation is of the form ax2 + bx +
c
= 0, where a = 2, b
=
– 4 and c = 3.
Therefore, the discriminant
b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.
Example 8
:
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres.
Is it possible to do so?
If yes, at what distances from the two gates should the pole be erected?
Solution : Let us first draw the diagram (see Fig. 4.2).
Let P be the required location of the pole.
Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP – BP (or, BP AP) = 7 m. Therefore, AP = (x + 7) m.
Fig. 4.2
Now, AB = 13m, and since AB is a diameter, APB = 90° (Why?) Therefore, AP2 + PB2 = AB2 (By Pythagoras theorem) b2 – 4ac = 72 –
4
× 1 × (– 60) = 289 > 0.
So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park.
Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get
| i.e., | (x + 7)2 + x2 = 132
| --- | ---
| i.e., | x2 + 14x + 49 + x2 = 169
| i.e., | 2x2 + 14x – 120 = 0
So, the distance ‘x’ of the pole from gate B satisfies the equation x2 + 7x – 60 = 0
So, it would be possible to place the pole if this equation has real roots.
To see if this is so or not, let us consider its discriminant.
The discriminant is
7 289
| = | 2 | 7 | 17
| --- | --- | --- | ---
| | = | 2 |
Therefore, x
=
5 or – 12.
Example 9 : Find the discriminant of the equation 3x2 – 2x +
1
3 = 0 and hence find the nature of its roots.
Find them, if they are real.
Solution : Here a = 3, b
=
– 2 and 1
3
c .
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x
=
– 12 will have to be ignored.
So, x = 5.
Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.
Therefore, discriminant b2 – 4ac = (– 2)2 –
4
× 3 ×
1
3 = 4
–
4 = 0.
Hence, the given quadratic equation has two equal real roots.
| The roots are | | | | | 2 1,,, | 2 1,,, | 1,,, , | 1,,, i.e., | 1 1,,, | 1,,, , | | .i.e.,
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | 2 | 3 b a | 2 | 3 b b a | 6 3 b | 6 3 b | 3 b | 3 b | 3 3 b | 3 b | 3 b |
| jemh104.pdf |
0 | CBSE | Class10 | Mathematics | ARITHMETIC PROGRESSIONS
5ARITHMETIC PROGRESSIONS
5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone, etc.
We now look for some patterns which occur in our day-to-day life.
Some such examples are :
(i) Reena applied for a job and got selected.
She has been offered a job with a starting monthly salary of ` 8000, with an annual increment of ` 500 in her salary.
Her salary (in ̀) for the 1st, 2nd, 3rd,.
.. years will be, respectively
8000, 8500, 9000,.
.. .
(ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see Fig. 5.1).
The bottom rung is 45 cm in length.
The lengths (in cm) of the 1st, 2nd, 3rd,.
.., 8th rung from the bottom to the top are, respectively
Fig. 5.1
45, 43, 41, 39, 37, 35, 33, 31 (iii) In a savings scheme, the amount becomes 5 4 times of itself after every 3 years.
The maturity amount (in `) of an investment of ` 8000 after 3, 6, 9 and 12 years will be, respectively :
10000, 12500, 15625, 19531.25
(iv) The number of unit squares in squares with side 1, 2, 3,.
.. units (see Fig. 5.2) are, respectively
12, 22, 32,.
.. .
Fig. 5.2
(v) Shakila puts ` 100 into her daughter’s money box when she was one year old and increased the amount by ̀ 50 every year.
The amounts of money (in ̀) in the
box on the 1st, 2nd, 3rd, 4th,.
.. birthday were 100, 150, 200, 250,.
.., respectively.
(vi) A pair of rabbits are too young to produce in their first month.
In the second, and
every subsequent month, they produce a new pair.
Each new pair of rabbits produce a new pair in their second month and in every subsequent month (see Fig. 5.3).
Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd,.
.., 6th month, respectively are :
1, 1, 2, 3, 5, 8
Fig. 5.3
In the examples above, we observe some patterns.
In some, we find that the succeeding terms are obtained by adding a fixed number, in other by multiplying with a fixed number, in another we find that they are squares of consecutive numbers, and so on.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms.
We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone, etc.
We now look for some patterns which occur in our day-to-day life.
Some such examples are :
(i) Reena applied for a job and got selected.
She has been offered a job with a starting monthly salary of ` 8000, with an annual increment of ` 500 in her salary.
Her salary (in ̀) for the 1st, 2nd, 3rd,.
.. years will be, respectively
8000, 8500, 9000,.
.. .
(ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see Fig. 5.1).
The bottom rung is 45 cm in length.
The lengths (in cm) of the 1st, 2nd, 3rd,.
.., 8th rung from the bottom to the top are, respectively
Fig. 5.1
45, 43, 41, 39, 37, 35, 33, 31 (iii) In a savings scheme, the amount becomes 5 4 times of itself after every 3 years.
The maturity amount (in `) of an investment of ` 8000 after 3, 6, 9 and 12 years will be, respectively :
10000, 12500, 15625, 19531.25
(iv) The number of unit squares in squares with side 1, 2, 3,.
.. units (see Fig. 5.2) are, respectively
12, 22, 32,.
.. .
Fig. 5.2
(v) Shakila puts ` 100 into her daughter’s money box when she was one year old and increased the amount by ̀ 50 every year.
The amounts of money (in ̀) in the
box on the 1st, 2nd, 3rd, 4th,.
.. birthday were 100, 150, 200, 250,.
.., respectively.
(vi) A pair of rabbits are too young to produce in their first month.
In the second, and
every subsequent month, they produce a new pair.
Each new pair of rabbits produce a new pair in their second month and in every subsequent month (see Fig. 5.3).
Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd,.
.., 6th month, respectively are :
1, 1, 2, 3, 5, 8 | jemh105.pdf |
1 | CBSE | Class10 | Mathematics | 5.2 Arithmetic Progressions
Consider the following lists of numbers :
(i) 1, 2, 3, 4,.
..
(ii) 100, 70, 40, 10,.
..
(iii) – 3, –2, –1, 0,
(iv) 3, 3, 3, 3,.
..
(v) 1.0, –1.5, –2.0, –2.5,.
..
Each of the numbers in the list is called a term.
Given a term, can you write the next term in each of the lists above?
If so, how will you write it?
Perhaps by following a pattern or rule.
Let us observe and write the rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to the term preceding it.
In (iv), all the terms in the list are 3, i.e., each term is obtained by adding (or subtracting) 0 to the term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it.
In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms.
Such list of numbers is said to form an Arithmetic Progression ( AP).
So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference of the AP.
Remember that it can be positive, negative or zero.
Let us denote the first term of an AP by a, second term by a,.
.., nth term by, a,.
.., a
1
2
n and the common difference by d. Then the AP becomes a
1
2 , a 3
a
n.
So, a
2
–
a
1 = a
3
– a 2 =.
.. =
a
n – a
n
– 1 = d. Some more examples of AP are:
(a) The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149,.
.., 157.
(b) The minimum temperatures (in degree celsius) recorded for a week in the month of January in a city, arranged in ascending order are
3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
(c) The balance money ( in ` ) after paying 5 % of the total loan of ` 1000 every month is 950, 900, 850, 800,.
.., 50.
(d) The cash prizes ( in ` ) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350,.
.., 750.
(e) The total savings (in `) after every month for 10 months when ` 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
It is left as an exercise for you to explain why each of the lists above is an AP.
You can see that a, a + d, a + 2d, a + 3d,.
..
represents an arithmetic progression where a is the first term and d the common difference.
This is called the general form of an AP.
Note that in examples (a) to (e) above, there are only a finite number of terms.
Such an AP is called a finite AP.
Also note that each of these Arithmetic Progressions (APs) has a last term.
The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions.
Such APs do not have a last term.
Now, to know about an AP, what is the minimum information that you need?
Is it enough to know the first term?
Or, is it enough to know only the common difference?
You will find that you will need to know both – the first term a and the common difference d.
For instance if the first term a is 6 and the common difference d is 3, then the AP is 6, 9,12, 15,.
..
and if a is 6 and d is – 3, then the AP is 6, 3, 0, –3,.
..
Similarly, when a
=
– 7, d
=
– 2, the AP is 7, – 9, – 11, – 13,.
.. a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3,.
..
a = 0, d = 1
1
| 2, | the AP is 0, 1 1 2, 3, 4 1 2 | , 6, . . .
| --- | --- | ---
| a = 2, | d = 0, | the AP is 2, 2, 2, 2, . . .
So, if you know what a and d are, you can list the AP.
What about the other way round?
That is, if you are given a list of numbers can you say that it is an AP and then find a and d? Since a is the first term, it can easily be written.
We know that in an AP, every succeeding term is obtained by adding d to the preceding term.
So, d found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP.
For example, for the list of numbers :
6, 9, 12, 15,.
.. , We have a
2 – a 1 = 9 6 = 3, a 3 – a 2 = 12 – 9 = 3, a 4 – a 3 = 15 – 12 = 3
Here the difference of any two consecutive terms in each case is 3.
So, the given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers : 6, 3, 0, – 3,.
.., a
| 2 – a | 1 = 3 – 6 = – 3
| a
| 3 – a | 2 = 0 – 3 = – 3
| a
| 4 – a | 3 = –3 – 0 = –3
Similarly this is also an AP whose first term is 6 and the common difference is –3.
In general, for an AP a
1
,
a 2
,. .., a, we have d = a
n
k
+
1
– a k where a
k
+ 1 and a k are the (
k
+ 1)th and the kth terms respectively.
To obtain d in a given AP, we need not find all of a,.
.. .
2
–
a
1 , a
3
– a 2
,
a 4 – a 3
It is enough to find only one of them.
Consider the list of numbers 1, 1, 2, 3, 5,.
.. .
By looking at it, you can tell that the difference between any two consecutive terms is not the same.
So, this is not an AP.
Note that to find d in the AP : 6, 3, 0, – 3,.
.., we have subtracted 6 from 3 and not 3 from 6, i.e., we should subtract the kth term from the (k + 1) th term even if the (k + 1) th term is smaller.
Let us make the concept more clear through some examples.
Example 1 : For the AP :
3
2
1
2 3 2 3
1 2, –, –,., write the first term a and the common difference d. Solution : Here, a =
2, d =
| 1 | 3
| 2 – | 2 = – 1.
Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP.
Example 2 : Which of the following list of numbers form an AP?
If they form an AP, write the next two terms :
| (i) 4, 10, 16, 22, . . . | (ii) 1, – 1, – 3, – 5, . . .
| --- | ---
| Solution : | (i) We have a | 2 – a 1 = 10 – 4 = | 6
| --- | --- | --- | ---
| (iii) 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .
| | a | 3 – a 2 = 16 – 10 = 6 a |
| | | 4 – a 3 = 22 – 16 = 6 |
| i.e., a k + 1 – a | k is the same every time. | |
| So, the given list of numbers forms an AP with the common difference d = 6.
| The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.
| (ii) a
| 2 – a | 1 = 1 – 1 = – 2 a | |
| 3 – a | 2 = 3 – ( –1 ) = – 3 + 1 = – 2 a | |
| 4 – a | 3 = – 5 – ( –3 ) = – 5 + 3 = – 2 | |
| i.e., a k + 1 – a | k is the same every time. | |
| So, the given list of numbers forms an AP with the common difference d = – 2.
| The next two terms are:
| | 5 + (– 2 ) = – 7 | and | 7 + (– 2 ) = – 9
| (iii) a
| 1 = 2 – (– 2) = 2 + 2 = 4 a 2 – a | | |
| 3 – a | 2 = – 2 – 2 = – 4 | |
2, the given list of numbers does not form an AP.
As a
2
–
a
1 a
3
– a
(iv) a
2
–
a
1 3 – a 2
=
1
–
1 = 0
a
= 1 –
1
= 0 a
| 4 – a 3 = 2 – 1 = 1 | |
| --- | --- | ---
| Here, a
| 2 – a 1 | |
| | = a 3 | – a
| 2 a 4 | – a 3.
So, the given list of numbers does not form an AP. | jemh105.pdf |
2 | CBSE | Class10 | Mathematics | So, a
2
–
a
1 = a
3
– a 2 =.
.. =
a
n – a
n
– 1 = d. Some more examples of AP are:
(a) The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149,.
.., 157.
(b) The minimum temperatures (in degree celsius) recorded for a week in the month of January in a city, arranged in ascending order are
3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
(c) The balance money ( in ` ) after paying 5 % of the total loan of ` 1000 every month is 950, 900, 850, 800,.
.., 50.
(d) The cash prizes ( in ` ) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350,.
.., 750.
(e) The total savings (in `) after every month for 10 months when ` 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
It is left as an exercise for you to explain why each of the lists above is an AP.
You can see that a, a + d, a + 2d, a + 3d,.
..
represents an arithmetic progression where a is the first term and d the common difference.
This is called the general form of an AP.
Note that in examples (a) to (e) above, there are only a finite number of terms.
Such an AP is called a finite AP.
Also note that each of these Arithmetic Progressions (APs) has a last term.
The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions.
Such APs do not have a last term.
Now, to know about an AP, what is the minimum information that you need?
Is it enough to know the first term?
Or, is it enough to know only the common difference?
You will find that you will need to know both – the first term a and the common difference d.
For instance if the first term a is 6 and the common difference d is 3, then the AP is 6, 9,12, 15,.
..
and if a is 6 and d is – 3, then the AP is 6, 3, 0, –3,.
..
Similarly, when a
=
– 7, d
=
– 2, the AP is 7, – 9, – 11, – 13,.
.. a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3,.
..
a = 0, d = 1
1
| 2, | the AP is 0, 1 1 2, 3, 4 1 2 | , 6, . . .
| --- | --- | ---
| a = 2, | d = 0, | the AP is 2, 2, 2, 2, . . .
So, if you know what a and d are, you can list the AP.
What about the other way round?
That is, if you are given a list of numbers can you say that it is an AP and then find a and d? Since a is the first term, it can easily be written.
We know that in an AP, every succeeding term is obtained by adding d to the preceding term.
So, d found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP.
For example, for the list of numbers :
6, 9, 12, 15,.
.. , We have a
2 – a 1 = 9 6 = 3, a 3 – a 2 = 12 – 9 = 3, a 4 – a 3 = 15 – 12 = 3
Here the difference of any two consecutive terms in each case is 3.
So, the given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers : 6, 3, 0, – 3,.
.., a
| 2 – a | 1 = 3 – 6 = – 3
| a
| 3 – a | 2 = 0 – 3 = – 3
| a
| 4 – a | 3 = –3 – 0 = –3
Similarly this is also an AP whose first term is 6 and the common difference is –3. | jemh105.pdf |
3 | CBSE | Class10 | Mathematics | EXERCISE 5.1
1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ̀ 15 for the first km and ̀ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1 4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ̀ 150 for the first metre and rises by ` 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum.
2. Write first four terms of the AP, when the first term a and the common difference d are
given as follows:
(i) a = 10, d = 10 (ii) a = –2, d = 0
(iii) a = 4, d
=
– 3 (iv) a
=
– 1, d = 1
2
(v) a
=
– 1.25, d
=
– 0.25
3. For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3,.
.. (ii) 5, – 1, 3, 7,.
..
(iii) 1
5
9 13,,,,
3
3
3 3
.. .
(iv) 0.6, 1.7, 2.8, 3.9,.
..
4. Which of the following are APs ?
If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16,.
.. (ii)
| 5 | | 72,
| --- | --- | ---
| , 3,,
| 2 2
| . . .
(iii) 1.2, – 3.2, – 5.2, – 7.2,.
.. (iv) 10, – 6, – 2, 2,.
..
(v) 3, 3 2, 3 2 2, 3 3 2,.. .
(vi) 0.2, 0.22, 0.222, 0.2222,.
..
| (vii) 0, – 4, – 8, –12, . . . | (viii) –
1
2 , –
These numbers are in AP.
(Why?) Now, looking at the pattern formed above, can you find her monthly salary for
1
2
1
2 1 2
, –,.
(ix) 1, 3, 9, 27,.
.. (x) a, 2a, 3a, 4a,.
..
(xi) a, a2, a3, a4,.
.. (xii) 2, 8, 18, 32,.. .
(xiii) 3, 6, 9, 12,.. .
(xiv) 12, 32, 52, 72,.
..
(xv) 12, 52, 72, 73,.
..
5.3 nth Term of an AP
Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected.
She has been offered the job with a starting monthly salary of ` 8000, with an annual increment of ̀ 500.
What would be her monthly salary for the fifth year?
To answer this, let us first see what her monthly salary for the second year would be.
It would be ` (8000 + 500) = ` 8500.
In the same way, we can find the monthly salary for the 3rd, 4th and 5th year by adding ` 500 to the salary of the previous year.
So, the salary for the 3rd year = ` (8500 + 500) = ` (8000 + 500 + 500) = ` (8000 +
2
× 500)
| | = ` [8000 + (3 – 1) × 500] | (for the 3rd year)
| --- | --- | ---
| = ` 9000
| Salary for the 4th year = ` (9000 + 500)
| = ` (8000 + 500 + 500 + 500)
| | (for the 4th year) | (for the 4th year)
| = ` 9500
| Salary for the 5th year = ` (9500 + 500)
| = ` (8000+500+500+500 + 500)
| | (for the 5th year) | (for the 5th year)
= ` 10000 Observe that we are getting a list of numbers 8000, 8500, 9000, 9500, 10000,.
..
the 6th year?
The 15th year?
And, assuming that she will still be working in the job, what about the monthly salary for the 25th year?
You would calculate this by adding ` 500 each time to the salary of the previous year to give the answer.
Can we make this process shorter?
Let us see.
You may have already got some idea from the way we have obtained the salaries above.
Salary for the 15th year = Salary for the 14th year + ` 500 =
= ` [8000 + 14 × 500] = ` [8000 + (15 – 1) × 500] = ` 15000
| | | 2 = | a + d = a + (2 – 1) d
| --- | --- | --- | ---
| i.e., | First salary + (15 – 1) × Annual increment.
| In the same way, her monthly salary for the 25th year would be | ` [8000 + (25 – 1) × 500] = ` 20000
| = First salary + (25 – 1) × Annual increment
| This example would have given you some idea about how to write the 15th term, or the 25th term, and more generally, the nth term of the AP.
| Let a 1, a 2, a 3, be an AP whose first term a 1 is a and the common difference is d.
| Then,
| the second term a
| the third term | a | 3 = | a 2 + d = (a + d) + d = a + 2d = a + (3 – 1) d
| the fourth term | a | 4 = | a 3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d
Looking at the pattern, we can say that the nth term a
n
= a + (n – 1) d. So, the nth term an of the AP with first term a and common difference d is given by an =
a
+ (n – 1) d.
an is also called the general term of the AP.
If there are m terms in the AP, then am represents the last term which is sometimes also denoted by l.
Let us consider some examples.
Example 3 : Find the 10th term of the AP : 2, 7, 12,.
..
| Solution : Here, a = 2, | d = 7 – 2 = 5 | and | n = 10.
| --- | --- | --- | ---
| We have a n = a + (n – 1) d | | |
| So, a 10 = 2 + (10 – 1) × 5 = 2 + 45 = 47 | | |
Therefore, the 10th term of the given AP is 47.
| Example 4 : Which term of the AP : 21, 18, 15, . . . is | | 81? Also, is any term 0? Give
| --- | --- | ---
| reason for your answer.
| Solution : Here, a = 21, | d = 18 – 21 = – 3 and a | n = – 81, and we have to find n.
| As a n = a + ( n – 1) d, | |
| we have 81 = 21 + (n – 1)(– 3) | |
| 81 = 24 – 3n
| 105 = 3n | |
| So, | n = | 35
| Therefore, the 35th term of the given AP is – 81.
| Next, we want to know if there is any n for which a | | n = 0. If such an n is there, then
| 21 + (n – 1) (–3) = 0,
| i.e., | 3(n – 1) = 21 |
| i.e., | n = 8 |
So, the eighth term is 0.
Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution : We have a
| | 3 = a + (3 – 1) d = | a + 2d = 5 | (1)
| --- | --- | --- | ---
| and | a 7 = a + (7 – 1) d = | a + 6d = 9 | (2)
Solving the pair of linear equations (1) and (2), we get a = 3, d
=
1 Hence, the required AP is 3, 4, 5, 6, 7,.
..
Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23,.
..
Solution : We have : a
2
–
a
1 = 11 –
5
= 6, a
3
– a 2 = 17 – 11 = 6, a
4
– a 3 = 23 – 17 = 6
| As a | k + 1 – a | k is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
| --- | --- | ---
| Now, | a = 5 | and | d = 6.
| --- | --- | --- | ---
Let 301 be a term, say, the nth term of this AP.
| We know that | a | n = | a + (n – 1) d
| --- | --- | --- | ---
| So, | 301 = 5 + (n – 1) × 6 | |
| i.e., | 301 = 6n – 1 | |
| So, | n = | 302 | 151
| --- | --- | --- | ---
| Is this an AP? Yes it is. Here, | a = 12, | d = 3, a | n = 99.
| --- | --- | --- | ---
| 6 3 But n should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.
| Example 7 : How many two-digit numbers are divisible by 3? Solution : The list of two-digit numbers divisible by 3 is :
| 12, 15, 18, . . . , 99
| As | a | n = | a + (n – 1) d,
| we have | 99 = 12 + (n – 1) × 3 | |
| i.e., | 87 = (n – 1) × 3 | |
| i.e., | n – 1 = | | 87 3 = 29
| i.e., | n = 29 + 1 = 30 | |
So, there are 30 two-digit numbers divisible by 3.
Example 8 : Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4,.
.., – 62.
Solution : Here, a = 10, d
=
7 – 10 = – 3, l
=
– 62, where l
=
a + (n – 1) d To find the 11th term from the last term, we will find the total number of terms in the AP.
| So, | 62 = 10 + (n – 1)(–3)
| i.e., | 72 = (n – 1)(–3)
| i.e., | n – 1 = 24
| or | n = 25
| So, there are 25 terms in the given AP.
| The 11th term from the last term will be the 15th term. (Note that it will not be the 14th term. Why?)
| So, | a 15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32
i.e., the 11th term from the last term is – 32.
Alternative Solution :
If we write the given AP in the reverse order, then a
=
– 62 and d
=
3 (Why?) So, the question now becomes finding the 11th term with these a and d.
So, a 11 = 62 + (11 – 1) ×
3
= – 62 + 30 = – 32 So, the 11th term, which is now the required term, is – 32.
Example 9
:
A sum of ̀ 1000 is invested at 8% simple interest per year.
Calculate the interest at the end of each year.
Do these interests form an AP?
If so, find the interest at the end of 30 years making use of this fact.
Solution : We know that the formula to calculate simple interest is given by
Simple Interest = P× R ×T
100
So, the interest at the end of the 1st year = 1000 × 8×1 100 `
=
` 80
The interest at the end of the 2nd year = 1000 × 8× 2 100 `
=
` 160
| The interest at the end of the | 3rd year = | | 1000 × 8× 3 |
| --- | --- | --- | --- | ---
| Now, | | | | | a 30 = | a + (30 – 1) d = 80 + 29 × 80 = 2400 | | | | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| 100 ` = ` 240 Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
| So, the interest (in `) at the end of the 1st, 2nd, 3rd, . . . years, respectively are 80, 160, 240, . . .
| It is an AP as the difference between the consecutive terms in the list is 80, i.e., d = 80. Also, a = 80.
| So, to find the interest at the end of 30 years, we shall find a.
| 30
| So, the interest at the end of 30 years will be ` 2400.
| Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
| Solution : The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
| 23, 21, 19, . . ., 5
| It forms an AP (Why?). Let the number of rows in the flower bed be n.
| Then | a = 23, | | | | d = 21 – 23 = – 2, a | | | | n = 5 | | | | | | | |
| As, | | | | | a n = | a + (n – 1) d | | | | | | | | | | |
| We have, | | | | | 5 = 23 + (n – 1)(– 2) | | | | | | | | | | | |
| i.e., | 18 = (n – 1)(– 2)
| --- | ---
| i.e., | n = 10
So, there are 10 rows in the flower bed. | jemh105.pdf |
4 | CBSE | Class10 | Mathematics | 5.3 nth Term of an AP
Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected.
She has been offered the job with a starting monthly salary of ` 8000, with an annual increment of ̀ 500.
What would be her monthly salary for the fifth year?
To answer this, let us first see what her monthly salary for the second year would be.
It would be ` (8000 + 500) = ` 8500.
In the same way, we can find the monthly salary for the 3rd, 4th and 5th year by adding ` 500 to the salary of the previous year.
So, the salary for the 3rd year = ` (8500 + 500) = ` (8000 + 500 + 500) = ` (8000 +
2
× 500)
| | = ` [8000 + (3 – 1) × 500] | (for the 3rd year)
| --- | --- | ---
| = ` 9000
| Salary for the 4th year = ` (9000 + 500)
| = ` (8000 + 500 + 500 + 500)
| | (for the 4th year) | (for the 4th year)
| = ` 9500
| Salary for the 5th year = ` (9500 + 500)
| = ` (8000+500+500+500 + 500)
| | (for the 5th year) | (for the 5th year)
= ` 10000 Observe that we are getting a list of numbers 8000, 8500, 9000, 9500, 10000,.
..
the 6th year?
The 15th year?
And, assuming that she will still be working in the job, what about the monthly salary for the 25th year?
You would calculate this by adding ` 500 each time to the salary of the previous year to give the answer.
Can we make this process shorter?
Let us see.
You may have already got some idea from the way we have obtained the salaries above.
Salary for the 15th year = Salary for the 14th year + ` 500 =
= ` [8000 + 14 × 500] = ` [8000 + (15 – 1) × 500] = ` 15000
| | | 2 = | a + d = a + (2 – 1) d
| --- | --- | --- | ---
| i.e., | First salary + (15 – 1) × Annual increment.
| In the same way, her monthly salary for the 25th year would be | ` [8000 + (25 – 1) × 500] = ` 20000
| = First salary + (25 – 1) × Annual increment
| This example would have given you some idea about how to write the 15th term, or the 25th term, and more generally, the nth term of the AP.
| Let a 1, a 2, a 3, be an AP whose first term a 1 is a and the common difference is d.
| Then,
| the second term a
| the third term | a | 3 = | a 2 + d = (a + d) + d = a + 2d = a + (3 – 1) d
| the fourth term | a | 4 = | a 3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d
Looking at the pattern, we can say that the nth term a
n
= a + (n – 1) d. So, the nth term an of the AP with first term a and common difference d is given by an =
a
+ (n – 1) d.
an is also called the general term of the AP.
If there are m terms in the AP, then am represents the last term which is sometimes also denoted by l.
Let us consider some examples.
Example 3 : Find the 10th term of the AP : 2, 7, 12,.
..
| Solution : Here, a = 2, | d = 7 – 2 = 5 | and | n = 10.
| --- | --- | --- | ---
| We have a n = a + (n – 1) d | | |
| So, a 10 = 2 + (10 – 1) × 5 = 2 + 45 = 47 | | |
Therefore, the 10th term of the given AP is 47.
| Example 4 : Which term of the AP : 21, 18, 15, . . . is | | 81? Also, is any term 0? Give
| --- | --- | ---
| reason for your answer.
| Solution : Here, a = 21, | d = 18 – 21 = – 3 and a | n = – 81, and we have to find n.
| As a n = a + ( n – 1) d, | |
| we have 81 = 21 + (n – 1)(– 3) | |
| 81 = 24 – 3n
| 105 = 3n | |
| So, | n = | 35
| Therefore, the 35th term of the given AP is – 81.
| Next, we want to know if there is any n for which a | | n = 0. If such an n is there, then
| 21 + (n – 1) (–3) = 0,
| i.e., | 3(n – 1) = 21 |
| i.e., | n = 8 |
So, the eighth term is 0.
Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution : We have a
| | 3 = a + (3 – 1) d = | a + 2d = 5 | (1)
| --- | --- | --- | ---
| and | a 7 = a + (7 – 1) d = | a + 6d = 9 | (2)
Solving the pair of linear equations (1) and (2), we get a = 3, d
=
1 Hence, the required AP is 3, 4, 5, 6, 7,.
..
Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23,.
..
Solution : We have : a
2
–
a
1 = 11 –
5
= 6, a
3
– a 2 = 17 – 11 = 6, a
4
– a 3 = 23 – 17 = 6
| As a | k + 1 – a | k is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
| --- | --- | ---
| Now, | a = 5 | and | d = 6.
| --- | --- | --- | ---
Let 301 be a term, say, the nth term of this AP.
| We know that | a | n = | a + (n – 1) d
| --- | --- | --- | ---
| So, | 301 = 5 + (n – 1) × 6 | |
| i.e., | 301 = 6n – 1 | |
| So, | n = | 302 | 151
| --- | --- | --- | ---
| Is this an AP? Yes it is. Here, | a = 12, | d = 3, a | n = 99.
| --- | --- | --- | ---
| 6 3 But n should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.
| Example 7 : How many two-digit numbers are divisible by 3? Solution : The list of two-digit numbers divisible by 3 is :
| 12, 15, 18, . . . , 99
| As | a | n = | a + (n – 1) d,
| we have | 99 = 12 + (n – 1) × 3 | |
| i.e., | 87 = (n – 1) × 3 | |
| i.e., | n – 1 = | | 87 3 = 29
| i.e., | n = 29 + 1 = 30 | |
So, there are 30 two-digit numbers divisible by 3.
Example 8 : Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4,.
.., – 62.
Solution : Here, a = 10, d
=
7 – 10 = – 3, l
=
– 62, where l
=
a + (n – 1) d To find the 11th term from the last term, we will find the total number of terms in the AP.
| So, | 62 = 10 + (n – 1)(–3)
| i.e., | 72 = (n – 1)(–3)
| i.e., | n – 1 = 24
| or | n = 25
| So, there are 25 terms in the given AP.
| The 11th term from the last term will be the 15th term. (Note that it will not be the 14th term. Why?)
| So, | a 15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32
i.e., the 11th term from the last term is – 32.
Alternative Solution :
If we write the given AP in the reverse order, then a
=
– 62 and d
=
3 (Why?) So, the question now becomes finding the 11th term with these a and d.
So, a 11 = 62 + (11 – 1) ×
3
= – 62 + 30 = – 32 So, the 11th term, which is now the required term, is – 32.
Example 9
:
A sum of ̀ 1000 is invested at 8% simple interest per year.
Calculate the interest at the end of each year.
Do these interests form an AP?
If so, find the interest at the end of 30 years making use of this fact.
Solution : We know that the formula to calculate simple interest is given by
Simple Interest = P× R ×T
100
So, the interest at the end of the 1st year = 1000 × 8×1 100 `
=
` 80
The interest at the end of the 2nd year = 1000 × 8× 2 100 `
=
` 160
| The interest at the end of the | 3rd year = | | 1000 × 8× 3 |
| --- | --- | --- | --- | ---
| Now, | | | | | a 30 = | a + (30 – 1) d = 80 + 29 × 80 = 2400 | | | | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| 100 ` = ` 240 Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
| So, the interest (in `) at the end of the 1st, 2nd, 3rd, . . . years, respectively are 80, 160, 240, . . .
| It is an AP as the difference between the consecutive terms in the list is 80, i.e., d = 80. Also, a = 80.
| So, to find the interest at the end of 30 years, we shall find a.
| 30
| So, the interest at the end of 30 years will be ` 2400.
| Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
| Solution : The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
| 23, 21, 19, . . ., 5
| It forms an AP (Why?). Let the number of rows in the flower bed be n.
| Then | a = 23, | | | | d = 21 – 23 = – 2, a | | | | n = 5 | | | | | | | |
| As, | | | | | a n = | a + (n – 1) d | | | | | | | | | | |
| We have, | | | | | 5 = 23 + (n – 1)(– 2) | | | | | | | | | | | |
| i.e., | 18 = (n – 1)(– 2)
| --- | ---
| i.e., | n = 10
So, there are 10 rows in the flower bed. | jemh105.pdf |
5 | CBSE | Class10 | Mathematics | EXERCISE 5.2
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a
n the nth term of the AP:
| | a | d | n | a n
| --- | --- | --- | --- | ---
| (i) | 7 | 3 | 8 | . . .
| (ii) 18. . . 10 0
| (iii) | . . . | – 3 | 18 | – 5
| (iv) 18.9 2.5. . . 3.6
| (v) | 3.5 | 0 | 105 | . . .
2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4,.
.. , is
| (A) 97 | (B) 77 | (C) –77 | (D) 87
| --- | --- | --- | ---
| (ii) 11th term of the AP: – 3, | 1 | |
| 2
| , 2, . . ., is
| (A) 28 | (B) 22 | (C) –38 | (D) 48 1
2
3. In the following APs, find the missing terms in the boxes :
| (i) 2, | , 26
| --- | ---
| (ii) | , | 13, | , 3
| --- | --- | --- | ---
| (v) | , | 38, | , | , | , | 22
| --- | --- | --- | --- | --- | --- | ---
| (iii) 5,,, 1 9 2
| (iv) 4,,,,, 6
| 4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
| 5. Find the number of terms in each of the following APs :
| (i) 7, 13, 19, . . . , 205 | | | | | | (ii) 18,
1 15 2
13,.
.. , – 47 6
. Check whether – 150 is a term of the AP : 11, 8, 5,
.
7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106.
Find the 29th term.
9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
10.
The 17th term of an AP exceeds its 10th term by 7.
Find the common difference.
11.
Which term of the AP : 3, 15, 27, 39,.
.. will be 132 more than its 54th term?
12.
Two APs have the same common difference.
The difference between their 100th terms is 100, what is the difference between their 1000th terms?
13.
How many three-digit numbers are divisible by 7?
14.
How many multiples of 4 lie between 10 and 250?
15.
For what value of n, are the nth terms of two APs: 63, 65, 67,.
.. and 3, 10, 17,.
.. equal?
16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
17.
Find the 20th term from the last term of the AP : 3, 8, 13,.
.., 253.
18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is
44.
Find the first three terms of the AP.
19.
Subba Rao started work in 1995 at an annual salary of ̀ 5000 and received an increment of ` 200 each year.
In which year did his income reach ̀ 7000?
20.
Ramkali saved ` 5 in the first week of a year and then increased her weekly savings by ` 1.75.
If in the nth week, her weekly savings become ̀ 20.75, find n.
5.4 Sum of First n Terms of an AP
Let us consider the situation again given in Section 5.1 in which Shakila put ̀ 100 into her daughter’s money box when she was one year old, ` 150 on her second birthday, ` 200 on her third birthday and will continue in the same way.
How much money will be collected in the money box by the time her daughter is 21 years old?
Here, the amount of money (in ̀) put in the money box on her first, second, third, fourth.
.. birthday were respectively 100, 150, 200, 250,.
.. till her 21st birthday.
To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up.
Don’t you think it would be a tedious and time consuming process?
Can we make the process shorter?
This would be possible if we can find a method for getting this sum.
Let us see.
We consider the problem given to Gauss (about whom you read in Chapter 1), to solve when he was just 10 years old.
He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050.
Can you guess how did he do?
He wrote :
S
=
1
+ 2 + 3 +.
.. + 99 + 100 And then, reversed the numbers to write S = 100 + 99 +.
.. +
3
+ 2 + 1 Adding these two, he got 2S = (100 + 1) + (99 + 2) +.
.. + (3 + 98) + (2 + 99) + (1 + 100) = 101 + 101 +.
.. + 101 + 101 (100 times) , i.e., the sum = 5050.
So, S = 100 101 5050 We will now use the same technique to find the sum of the first n terms of an AP : a, a + d, a + 2d,.
..
The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms of the AP.
We have
| S = a + (a + d | ) + (a + 2d) + . . . + [a + (n – 1) d] | (1)
| --- | --- | ---
| S = [a + (n – 1) d | ] + [a + (n – 2) d | ] + . . . + (a + d | ) + a | (2)
| --- | --- | --- | --- | ---
| 2S = | [2 | ( | 1)] [2 | ( | 1)] | [2 | ( | 1)] [2 | ( | 1)]
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Rewriting the terms in reverse order, we have
| On adding (1) and (2), term-wise. we get
Now, if there are only n terms in an AP, then a
a n
d
a n d
a
n d a n d
n
[2a + (n – 1) d]
2 2
| | n | times
| --- | --- | ---
| or, | 2S = | n [2a + (n – 1) d] | (Since, there are n terms)
| --- | --- | --- | ---
| or, | S = | |
So, the sum of the first n terms of an AP is given by
S = 2
S
=
2
n (a + l) (4) n [2a + (n – 1) d]
n
[a +
a
+ (n – 1) d]
n
(a + a
2
| n) | (3)
| n = l, the last term.
| We can also write this as | S =
| i.e., | S =
From (3), we see that
This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Now we return to the question that was posed to us in the beginning.
The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,..
., were 100, 150, 200, 250,.
.., respectively.
This is an AP.
We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21.
Using the formula : 2 (1) we have S
=
= 12600
S =
n
a
n d , 2
21 2 100 (21 1) 50 2 21 1200 2
= 21 200 1000 2
= So, the amount of money collected on her 21st birthday is ` 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
in place of S to denote the sum of first n terms of the AP.
We write S to denote the sum of the first 20 terms of an AP.
The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.
We also use S
n 20
Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a
n
= S n
–
S n – 1.
Let us consider some examples.
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2,.
..
Solution : Here, a = 8, d
=
3 – 8 = –5, n = 22.
We know that
S =
2 (1) 2
n
a
n d
Therefore, S =
22 16 21(5) 2
= 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution : Here, S
14
= 1050, n = 14, a = 10.
| As | S
| --- | ---
| | n = | 2 | | 2 ( a | 2 ( 1) n n | 2 ( d | 2 ( | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| so, | 1050 = | | | 20 | 13 14 | | |
2
d
= 140 + 91d
| i.e., | 910 = 91d
| --- | ---
| or, | d = 10
| Therefore, | a 20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18,.
.. must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, S
| | n = 78. We need to find n.
| --- | ---
| We know that | S
| --- | ---
| n = | | 2 ( | 1) | 2 | | n | a | n | d
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, 78 = | | | | | 1)( | | | 3) 48 ( |
| 2 n n = 51 3 2 n n
| or 3n2 – 51n + 156 = 0 | | | | | | | | |
| or n2 – 17n + 52 = 0 | | | | | | | | |
| or (n – 4)(n – 13) = 0 | | | | | | | | |
| or n = 4 or 13 | | | | | | | | |
Both values of n are admissible.
So, the number of terms is either 4 or 13.
Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be zero.
This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution :
(i) Let S
=
1 + 2
+
3 +.
.. + 1000
Using the formula S
n = () 2 n a l for the sum of the first n terms of an AP, we have
S
1000 = 1000 (1 1000) 2
= 500 × 1001 = 500500 So, the sum of the first 1000 positive integers is 500500.
(ii) Let S
n
=
1
+ 2 + 3 +.
.. + n Here a
=
1 and the last term l is n.
Therefore,
So, the sum of first n positive integers is given by
S
e have n = 24, a = 5, d = 2.
24
n n n n
or S
2
n
= (1) 2
n
= (1)
n n
S
n = (+ 1) 2
Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by a
Solution :
| As | a
| so, | a
n
=
3
+ 2n
n
=
3
+ 2n, 1
=
3 + 2
=
5 a 2
=
3 + 2
×
2 = 7
a
3 = 3
+
2 × 3
=
9 List of numbers becomes 5, 7, 9, 11,.
..
| Here, | 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
| --- | ---
| Therefore, | S 24 = | 24 2 | 24 2 | 24 2 5 (24 1) 2 2 | 24 2 | = | 12 10 46 = 672
| --- | --- | --- | --- | --- | --- | --- | ---
| So, it forms an AP with common difference d = 2.
| To find S
| So, sum of first 24 terms of the list of numbers is 672.
| Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
| (i) the production in the 1st year (ii) the production in the 10th year
| (iii) the total production in first 7 years
| Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
| Let us denote the number of TV sets manufactured in the nth year by a | | | | | | | . n
| 7 = 700
| Then, a | | | | | | |
| 3 = 600 and a
| or, | | a + 2d = 600 | | | | |
| and | | a + 6d = | | 700 | | |
| Solving these equations, we get | | d = 25 | | and a = 550. | | |
Therefore, production of TV sets in the first year is 550.
(ii) Now a 10 =
a
+ 9d = 550 +
9
× 25 = 775
So, production of TV sets in the 10th year is 775.
| (iii) Also, | S
| --- | ---
| 7 = | | 7 2 | 550 | (7 | 1) | 25 | 2 | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = | | 1100 150 | | | | 7 | | | | |
| | 2 | | | | | | = 4375 | | | |
Thus, the total production of TV sets in first 7 years is 4375. | jemh105.pdf |
6 | CBSE | Class10 | Mathematics | 5.4 Sum of First n Terms of an AP
Let us consider the situation again given in Section 5.1 in which Shakila put ̀ 100 into her daughter’s money box when she was one year old, ` 150 on her second birthday, ` 200 on her third birthday and will continue in the same way.
How much money will be collected in the money box by the time her daughter is 21 years old?
Here, the amount of money (in ̀) put in the money box on her first, second, third, fourth.
.. birthday were respectively 100, 150, 200, 250,.
.. till her 21st birthday.
To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up.
Don’t you think it would be a tedious and time consuming process?
Can we make the process shorter?
This would be possible if we can find a method for getting this sum.
Let us see.
We consider the problem given to Gauss (about whom you read in Chapter 1), to solve when he was just 10 years old.
He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050.
Can you guess how did he do?
He wrote :
S
=
1
+ 2 + 3 +.
.. + 99 + 100 And then, reversed the numbers to write S = 100 + 99 +.
.. +
3
+ 2 + 1 Adding these two, he got 2S = (100 + 1) + (99 + 2) +.
.. + (3 + 98) + (2 + 99) + (1 + 100) = 101 + 101 +.
.. + 101 + 101 (100 times) , i.e., the sum = 5050.
So, S = 100 101 5050 We will now use the same technique to find the sum of the first n terms of an AP : a, a + d, a + 2d,.
..
The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms of the AP.
We have
| S = a + (a + d | ) + (a + 2d) + . . . + [a + (n – 1) d] | (1)
| --- | --- | ---
| S = [a + (n – 1) d | ] + [a + (n – 2) d | ] + . . . + (a + d | ) + a | (2)
| --- | --- | --- | --- | ---
| 2S = | [2 | ( | 1)] [2 | ( | 1)] | [2 | ( | 1)] [2 | ( | 1)]
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Rewriting the terms in reverse order, we have
| On adding (1) and (2), term-wise. we get
Now, if there are only n terms in an AP, then a
a n
d
a n d
a
n d a n d
n
[2a + (n – 1) d]
2 2
| | n | times
| --- | --- | ---
| or, | 2S = | n [2a + (n – 1) d] | (Since, there are n terms)
| --- | --- | --- | ---
| or, | S = | |
So, the sum of the first n terms of an AP is given by
S = 2
S
=
2
n (a + l) (4) n [2a + (n – 1) d]
n
[a +
a
+ (n – 1) d]
n
(a + a
2
| n) | (3)
| n = l, the last term.
| We can also write this as | S =
| i.e., | S =
From (3), we see that
This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Now we return to the question that was posed to us in the beginning.
The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,..
., were 100, 150, 200, 250,.
.., respectively.
This is an AP.
We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21.
Using the formula : 2 (1) we have S
=
= 12600
S =
n
a
n d , 2
21 2 100 (21 1) 50 2 21 1200 2
= 21 200 1000 2
= So, the amount of money collected on her 21st birthday is ` 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
in place of S to denote the sum of first n terms of the AP.
We write S to denote the sum of the first 20 terms of an AP.
The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.
We also use S
n 20
Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a
n
= S n
–
S n – 1.
Let us consider some examples.
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2,.
..
Solution : Here, a = 8, d
=
3 – 8 = –5, n = 22.
We know that
S =
2 (1) 2
n
a
n d
Therefore, S =
22 16 21(5) 2
= 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution : Here, S
14
= 1050, n = 14, a = 10.
| As | S
| --- | ---
| | n = | 2 | | 2 ( a | 2 ( 1) n n | 2 ( d | 2 ( | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| so, | 1050 = | | | 20 | 13 14 | | |
2
d
= 140 + 91d
| i.e., | 910 = 91d
| --- | ---
| or, | d = 10
| Therefore, | a 20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18,.
.. must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, S
| | n = 78. We need to find n.
| --- | ---
| We know that | S
| --- | ---
| n = | | 2 ( | 1) | 2 | | n | a | n | d
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, 78 = | | | | | 1)( | | | 3) 48 ( |
| 2 n n = 51 3 2 n n
| or 3n2 – 51n + 156 = 0 | | | | | | | | |
| or n2 – 17n + 52 = 0 | | | | | | | | |
| or (n – 4)(n – 13) = 0 | | | | | | | | |
| or n = 4 or 13 | | | | | | | | |
Both values of n are admissible.
So, the number of terms is either 4 or 13.
Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be zero.
This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution :
(i) Let S
=
1 + 2
+
3 +.
.. + 1000
Using the formula S
n = () 2 n a l for the sum of the first n terms of an AP, we have
S
1000 = 1000 (1 1000) 2
= 500 × 1001 = 500500 So, the sum of the first 1000 positive integers is 500500.
(ii) Let S
n
=
1
+ 2 + 3 +.
.. + n Here a
=
1 and the last term l is n.
Therefore,
So, the sum of first n positive integers is given by
S
e have n = 24, a = 5, d = 2.
24
n n n n
or S
2
n
= (1) 2
n
= (1)
n n
S
n = (+ 1) 2
Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by a
Solution :
| As | a
| so, | a
n
=
3
+ 2n
n
=
3
+ 2n, 1
=
3 + 2
=
5 a 2
=
3 + 2
×
2 = 7
a
3 = 3
+
2 × 3
=
9 List of numbers becomes 5, 7, 9, 11,.
..
| Here, | 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
| --- | ---
| Therefore, | S 24 = | 24 2 | 24 2 | 24 2 5 (24 1) 2 2 | 24 2 | = | 12 10 46 = 672
| --- | --- | --- | --- | --- | --- | --- | ---
| So, it forms an AP with common difference d = 2.
| To find S
| So, sum of first 24 terms of the list of numbers is 672.
| Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
| (i) the production in the 1st year (ii) the production in the 10th year
| (iii) the total production in first 7 years
| Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
| Let us denote the number of TV sets manufactured in the nth year by a | | | | | | | . n
| 7 = 700
| Then, a | | | | | | |
| 3 = 600 and a
| or, | | a + 2d = 600 | | | | |
| and | | a + 6d = | | 700 | | |
| Solving these equations, we get | | d = 25 | | and a = 550. | | |
Therefore, production of TV sets in the first year is 550.
(ii) Now a 10 =
a
+ 9d = 550 +
9
× 25 = 775
So, production of TV sets in the 10th year is 775.
| (iii) Also, | S
| --- | ---
| 7 = | | 7 2 | 550 | (7 | 1) | 25 | 2 | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = | | 1100 150 | | | | 7 | | | | |
| | 2 | | | | | | = 4375 | | | |
Thus, the total production of TV sets in first 7 years is 4375. | jemh105.pdf |
7 | CBSE | Class10 | Mathematics | n
[2a + (n – 1) d]
2 2
| | n | times
| --- | --- | ---
| or, | 2S = | n [2a + (n – 1) d] | (Since, there are n terms)
| --- | --- | --- | ---
| or, | S = | |
So, the sum of the first n terms of an AP is given by
S = 2
S
=
2
n (a + l) (4) n [2a + (n – 1) d]
n
[a +
a
+ (n – 1) d]
n
(a + a
2
| n) | (3)
| n = l, the last term.
| We can also write this as | S =
| i.e., | S =
From (3), we see that
This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Now we return to the question that was posed to us in the beginning.
The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,..
., were 100, 150, 200, 250,.
.., respectively.
This is an AP.
We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21.
Using the formula : 2 (1) we have S
=
= 12600
S =
n
a
n d , 2
21 2 100 (21 1) 50 2 21 1200 2
= 21 200 1000 2
= So, the amount of money collected on her 21st birthday is ` 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
in place of S to denote the sum of first n terms of the AP.
We write S to denote the sum of the first 20 terms of an AP.
The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.
We also use S
n 20
Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a
n
= S n
–
S n – 1.
Let us consider some examples.
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2,.
..
Solution : Here, a = 8, d
=
3 – 8 = –5, n = 22.
We know that
S =
2 (1) 2
n
a
n d
Therefore, S =
22 16 21(5) 2
= 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution : Here, S
14
= 1050, n = 14, a = 10.
| As | S
| --- | ---
| | n = | 2 | | 2 ( a | 2 ( 1) n n | 2 ( d | 2 ( | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| so, | 1050 = | | | 20 | 13 14 | | |
2
d
= 140 + 91d
| i.e., | 910 = 91d
| --- | ---
| or, | d = 10
| Therefore, | a 20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18,.
.. must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, S
| | n = 78. We need to find n.
| --- | ---
| We know that | S
| --- | ---
| n = | | 2 ( | 1) | 2 | | n | a | n | d
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, 78 = | | | | | 1)( | | | 3) 48 ( |
| 2 n n = 51 3 2 n n
| or 3n2 – 51n + 156 = 0 | | | | | | | | |
| or n2 – 17n + 52 = 0 | | | | | | | | |
| or (n – 4)(n – 13) = 0 | | | | | | | | |
| or n = 4 or 13 | | | | | | | | |
Both values of n are admissible.
So, the number of terms is either 4 or 13.
Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be zero.
This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution :
(i) Let S
=
1 + 2
+
3 +.
.. + 1000
Using the formula S
n = () 2 n a l for the sum of the first n terms of an AP, we have
S
1000 = 1000 (1 1000) 2
= 500 × 1001 = 500500 So, the sum of the first 1000 positive integers is 500500.
(ii) Let S
n
=
1
+ 2 + 3 +.
.. + n Here a
=
1 and the last term l is n.
Therefore,
So, the sum of first n positive integers is given by
S
e have n = 24, a = 5, d = 2.
24
n n n n
or S
2
n
= (1) 2
n
= (1)
n n
S
n = (+ 1) 2
Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by a
Solution :
| As | a
| so, | a
n
=
3
+ 2n
n
=
3
+ 2n, 1
=
3 + 2
=
5 a 2
=
3 + 2
×
2 = 7
a
3 = 3
+
2 × 3
=
9 List of numbers becomes 5, 7, 9, 11,.
..
| Here, | 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
| --- | ---
| Therefore, | S 24 = | 24 2 | 24 2 | 24 2 5 (24 1) 2 2 | 24 2 | = | 12 10 46 = 672
| --- | --- | --- | --- | --- | --- | --- | ---
| So, it forms an AP with common difference d = 2.
| To find S
| So, sum of first 24 terms of the list of numbers is 672.
| Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
| (i) the production in the 1st year (ii) the production in the 10th year
| (iii) the total production in first 7 years
| Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
| Let us denote the number of TV sets manufactured in the nth year by a | | | | | | | . n
| 7 = 700
| Then, a | | | | | | |
| 3 = 600 and a
| or, | | a + 2d = 600 | | | | |
| and | | a + 6d = | | 700 | | |
| Solving these equations, we get | | d = 25 | | and a = 550. | | |
Therefore, production of TV sets in the first year is 550.
(ii) Now a 10 =
a
+ 9d = 550 +
9
× 25 = 775
So, production of TV sets in the 10th year is 775.
| (iii) Also, | S
| --- | ---
| 7 = | | 7 2 | 550 | (7 | 1) | 25 | 2 | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = | | 1100 150 | | | | 7 | | | | |
| | 2 | | | | | | = 4375 | | | |
Thus, the total production of TV sets in first 7 years is 4375. | jemh105.pdf |
8 | CBSE | Class10 | Mathematics | S = 2
S
=
2
n (a + l) (4) n [2a + (n – 1) d]
n
[a +
a
+ (n – 1) d]n
[a +
a
+ (n – 1) d]n
(a + a
2
| n) | (3)
| n = l, the last term.
| We can also write this as | S =
| i.e., | S =
From (3), we see that
This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Now we return to the question that was posed to us in the beginning.
The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,..
., were 100, 150, 200, 250,.
.., respectively.
This is an AP.
We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21.
Using the formula : 2 (1) we have S
=
= 12600
S =
n
a
n d , 2
21 2 100 (21 1) 50 2 21 1200 2
= 21 200 1000 2
= So, the amount of money collected on her 21st birthday is ` 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
in place of S to denote the sum of first n terms of the AP.
We write S to denote the sum of the first 20 terms of an AP.
The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.
We also use S
n 20
Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a
n
= S n
–
S n – 1.
Let us consider some examples.
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2,.
..
Solution : Here, a = 8, d
=
3 – 8 = –5, n = 22.
We know that
S =
2 (1) 2
n
a
n d
Therefore, S =
22 16 21(5) 2
= 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution : Here, S
14
= 1050, n = 14, a = 10.
| As | S
| --- | ---
| | n = | 2 | | 2 ( a | 2 ( 1) n n | 2 ( d | 2 ( | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| so, | 1050 = | | | 20 | 13 14 | | |
2
d
= 140 + 91d
| i.e., | 910 = 91d
| --- | ---
| or, | d = 10
| Therefore, | a 20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18,.
.. must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, S
| | n = 78. We need to find n.
| --- | ---
| We know that | S
| --- | ---
| n = | | 2 ( | 1) | 2 | | n | a | n | d
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, 78 = | | | | | 1)( | | | 3) 48 ( |
| 2 n n = 51 3 2 n n
| or 3n2 – 51n + 156 = 0 | | | | | | | | |
| or n2 – 17n + 52 = 0 | | | | | | | | |
| or (n – 4)(n – 13) = 0 | | | | | | | | |
| or n = 4 or 13 | | | | | | | | |
Both values of n are admissible.
So, the number of terms is either 4 or 13.
Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be zero.
This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution :
(i) Let S
=
1 + 2
+
3 +.
.. + 1000
Using the formula S
n = () 2 n a l for the sum of the first n terms of an AP, we have
S
1000 = 1000 (1 1000) 2
= 500 × 1001 = 500500 So, the sum of the first 1000 positive integers is 500500.
(ii) Let S
n
=
1
+ 2 + 3 +.
.. + n Here a
=
1 and the last term l is n.
Therefore,
So, the sum of first n positive integers is given byS =
n
a
n d , 2
21 2 100 (21 1) 50 2 21 1200 2
= 21 200 1000 2
= So, the amount of money collected on her 21st birthday is ` 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
in place of S to denote the sum of first n terms of the AP.
We write S to denote the sum of the first 20 terms of an AP.
The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.
We also use S
n 20
Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a
n
= S n
–
S n – 1.
Let us consider some examples.
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2,.
..
Solution : Here, a = 8, d
=
3 – 8 = –5, n = 22.
We know thatWe also use S
n 20
Remark : The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., a
n
= S n
–
S n – 1.
Let us consider some examples.
Example 11 : Find the sum of the first 22 terms of the AP : 8, 3, –2,.
..
Solution : Here, a = 8, d
=
3 – 8 = –5, n = 22.
We know that | jemh105.pdf |
9 | CBSE | Class10 | Mathematics | Therefore, S =
22 16 21(5) 2
= 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution : Here, S
14
= 1050, n = 14, a = 10.
| As | S
| --- | ---
| | n = | 2 | | 2 ( a | 2 ( 1) n n | 2 ( d | 2 ( | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| so, | 1050 = | | | 20 | 13 14 | | |
2
d
= 140 + 91d
| i.e., | 910 = 91d
| --- | ---
| or, | d = 10
| Therefore, | a 20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18,.
.. must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, S
| | n = 78. We need to find n.
| --- | ---
| We know that | S
| --- | ---
| n = | | 2 ( | 1) | 2 | | n | a | n | d
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, 78 = | | | | | 1)( | | | 3) 48 ( |
| 2 n n = 51 3 2 n n
| or 3n2 – 51n + 156 = 0 | | | | | | | | |
| or n2 – 17n + 52 = 0 | | | | | | | | |
| or (n – 4)(n – 13) = 0 | | | | | | | | |
| or n = 4 or 13 | | | | | | | | |
Both values of n are admissible.
So, the number of terms is either 4 or 13.
Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be zero.
This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution :
(i) Let S
=
1 + 2
+
3 +.
.. + 1000Solution : Here, S
14
= 1050, n = 14, a = 10.
| As | S
| --- | ---
| | n = | 2 | | 2 ( a | 2 ( 1) n n | 2 ( d | 2 ( | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| so, | 1050 = | | | 20 | 13 14 | | |
2
d
= 140 + 91d
| i.e., | 910 = 91d
| --- | ---
| or, | d = 10
| Therefore, | a 20 = 10 + (20 – 1) × 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP : 24, 21, 18,.
.. must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, S
| | n = 78. We need to find n.
| --- | ---
| We know that | S
| --- | ---
| n = | | 2 ( | 1) | 2 | | n | a | n | d
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, 78 = | | | | | 1)( | | | 3) 48 ( |
| 2 n n = 51 3 2 n n
| or 3n2 – 51n + 156 = 0 | | | | | | | | |
| or n2 – 17n + 52 = 0 | | | | | | | | |
| or (n – 4)(n – 13) = 0 | | | | | | | | |
| or n = 4 or 13 | | | | | | | | |
Both values of n are admissible.
So, the number of terms is either 4 or 13.Remarks:
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th to 13th will be zero.
This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.
Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integersSolution :
(i) Let S
=
1 + 2
+
3 +.
.. + 1000Using the formula S
n = () 2 n a l for the sum of the first n terms of an AP, we haveS
1000 = 1000 (1 1000) 2
= 500 × 1001 = 500500 So, the sum of the first 1000 positive integers is 500500.
(ii) Let S
n
=
1
+ 2 + 3 +.
.. + n Here a
=
1 and the last term l is n.(ii) Let S
n
=
1
+ 2 + 3 +.
.. + n Here a
=
1 and the last term l is n. Therefore,
So, the sum of first n positive integers is given by | jemh105.pdf |
10 | CBSE | Class10 | Mathematics | n n
S
n = (+ 1) 2
Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by a
Solution :
| As | a
| so, | a
n
=
3
+ 2n
n
=
3
+ 2n, 1
=
3 + 2
=
5 a 2
=
3 + 2
×
2 = 7
a
3 = 3
+
2 × 3
=
9 List of numbers becomes 5, 7, 9, 11,.
..
| Here, | 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
| --- | ---
| Therefore, | S 24 = | 24 2 | 24 2 | 24 2 5 (24 1) 2 2 | 24 2 | = | 12 10 46 = 672
| --- | --- | --- | --- | --- | --- | --- | ---
| So, it forms an AP with common difference d = 2.
| To find S
| So, sum of first 24 terms of the list of numbers is 672.
| Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
| (i) the production in the 1st year (ii) the production in the 10th year
| (iii) the total production in first 7 years
| Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
| Let us denote the number of TV sets manufactured in the nth year by a | | | | | | | . n
| 7 = 700
| Then, a | | | | | | |
| 3 = 600 and a
| or, | | a + 2d = 600 | | | | |
| and | | a + 6d = | | 700 | | |
| Solving these equations, we get | | d = 25 | | and a = 550. | | |
Therefore, production of TV sets in the first year is 550.
(ii) Now a 10 =
a
+ 9d = 550 +
9
× 25 = 775
So, production of TV sets in the 10th year is 775.
| (iii) Also, | S
| --- | ---
| 7 = | | 7 2 | 550 | (7 | 1) | 25 | 2 | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = | | 1100 150 | | | | 7 | | | | |
| | 2 | | | | | | = 4375 | | | |
Thus, the total production of TV sets in first 7 years is 4375.S
n = (+ 1) 2
Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is given by a
Solution :
| As | a
| so, | a
n
=
3
+ 2n
n
=
3
+ 2n, 1
=
3 + 2
=
5 a 2
=
3 + 2
×
2 = 7
a
3 = 3
+
2 × 3
=
9 List of numbers becomes 5, 7, 9, 11,.
..
| Here, | 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
| --- | ---
| Therefore, | S 24 = | 24 2 | 24 2 | 24 2 5 (24 1) 2 2 | 24 2 | = | 12 10 46 = 672
| --- | --- | --- | --- | --- | --- | --- | ---
| So, it forms an AP with common difference d = 2.
| To find S
| So, sum of first 24 terms of the list of numbers is 672.
| Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
| (i) the production in the 1st year (ii) the production in the 10th year
| (iii) the total production in first 7 years
| Solution : (i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
| Let us denote the number of TV sets manufactured in the nth year by a | | | | | | | . n
| 7 = 700
| Then, a | | | | | | |
| 3 = 600 and a
| or, | | a + 2d = 600 | | | | |
| and | | a + 6d = | | 700 | | |
| Solving these equations, we get | | d = 25 | | and a = 550. | | |
Therefore, production of TV sets in the first year is 550.
(ii) Now a 10 =
a
+ 9d = 550 +
9
× 25 = 775
So, production of TV sets in the 10th year is 775.
| (iii) Also, | S
| --- | ---
| 7 = | | 7 2 | 550 | (7 | 1) | 25 | 2 | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| = | | 1100 150 | | | | 7 | | | | |
| | 2 | | | | | | = 4375 | | | |
Thus, the total production of TV sets in first 7 years is 4375. | jemh105.pdf |
11 | CBSE | Class10 | Mathematics | EXERCISE 5.3
1. Find the sum of the following APs:
(i) 2, 7, 12,.
.., to 10 terms.
(ii) 37, –33, –29,.
.., to 12 terms.
(iii) 0.6, 1.7, 2.8,.
.., to 100 terms.
(iv) 1 1 1,,
15 12 10
,. .., to 11 terms.
2. Find the sums given below :
(i) 7
+
1 10 2 + 14 +.
.. + 84 (ii) 34 + 32 + 30 +.
.. + 10
(iii) –5 + (–8) + (–11) +.
.. + (–230)
3. In an AP:
(i) given a = 5, d = 3, a
n
= 50, find n and S n.
(ii) given a = 7, a 13 = 35, find d and S 13.
(iii) given a 12 = 37, d = 3, find a and S 12.
(iv) given a
3
= 15, S 10 = 125, find d and a 10.
(v) given d = 5, S
9 = 75, find a and a 9
.
(vi) given a = 2, d = 8, S
n
= 90, find n and a n.
(vii) given a = 8, a
n
= 62, S
n
= 210, find n and d.
(viii) given a
n
= 4, d = 2, S
n
= –14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms.
Find a.
4. How many terms of the AP : 9, 17, 25,.
.. must be taken to give a sum of 636?
5. The first term of an AP is 5, the last term is 45 and the sum is 400.
Find the number of terms and the common difference.
6. The first and the last terms of an AP are 17 and 350 respectively.
If the common difference is 9, how many terms are there and what is their sum?
7. Find the sum of first 22 terms of an AP in which d
=
7 and 22nd term is 149.
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
10. Show that a
1 2
a,.
.., a n,
. .. form an AP where a n is defined as below
(i) a
n
= 3 + 4n (ii) a
n
= 9 – 5n
Also find the sum of the first 15 terms in each case.
11.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S
1)?
What is the sum of first two terms?
What is the second term?
Similarly, find the 3rd, the 10th and the nth terms.
12.
Find the sum of the first 40 positive integers divisible by 6.
13.
Find the sum of the first 15 multiples of 8.
14.
Find the sum of the odd numbers between 0 and 50.
15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ̀ 200 for the first day, ̀ 250 for the second day, ̀ 300 for the third day, etc., the penalty for each succeeding day being ̀ 50 more than for the preceding day.
How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
16.
A sum of ` 700 is to be used to give seven cash prizes to students of a school for their overall academic performance.
If each prize is ̀ 20 less than its preceding prize, find the value of each of the prizes.
17.
In a school, students thought of planting trees in and around the school to reduce air pollution.
It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII.
There are three sections of each class.
How many trees will be planted by the students?
18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,.
.. as shown in Fig. 5.4.
What is the total length of such a spiral made up of thirteen consecutive
semicircles?
(Take = [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2
×
5 + 2 × (5 + 3)]
22 7
)
Fig. 5.4
[Hint : Length of successive semicircles is l, l, l, l,.
.. with centres at A, B, A, B,.
.., respectively.]
1
2
3 4
19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5).
In how many rows are the 200 logs placed and how many logs are in the top row?
Fig. 5.5
20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line.
There are ten potatoes in the line (see Fig. 5.6).
Fig. 5.6
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket.
What is the total distance the competitor has to run?
EXERCISE 5.4 (Optional)*
1. Which term of the AP : 121, 117, 113,.
.., is its first negative term?
[Hint : Find n for a
n < 0] 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8.
Find the sum of first sixteen terms of the AP.
3. A ladder has rungs 25 cm apart.
(see Fig. 5.7).
The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top.
If the top and the bottom rungs are m apart, what is the length of the wood required for the rungs?
1
2
2
1 4 1 2
Fig. 5.7
[Hint : Number of rungs = 250 1 25
]
4. The houses of a row are numbered consecutively from 1 to 49.
Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
Find this value of x.
[Hint :
S
– 1 = S 49 – S x] 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.
Each step has a rise of m and a tread of m. (see Fig. 5.8).
Calculate the total volume
of concrete required to build the terrace.
| [Hint : Volume of concrete required to build the first step = | 31 | 1 50 m
| --- | --- | ---
| 4 | 2 | | ]
| --- | --- | --- | ---
Fig. 5.8
* These exercises are not from the examination point of view.
5.5 Summary
In this chapter, you have studied the following points :
1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term.
The fixed number d is called the common difference.
The general form of an AP is a, a + d, a + 2d, a + 3d,.
..
2. A given list of numbers a
1, a 2, a 3, is an AP, if the differences a
2
– a 1, a
3
– a 2 ,
a
4
–
a 3,.
.., give the same value, i.e., if a
k
+ 1 –
a
k is the same for different values of k.
3. In an AP with first term a and common difference d, the nth term (or the general term) is given by a
n
=
a
+ (n – 1) d.
4. The sum of the first n terms of an AP is given by :
| S = 2 | 1) | 2 (
| --- | --- | ---
| n a | n | d
5. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by :
S = ()
2
n a l | jemh105.pdf |
12 | CBSE | Class10 | Mathematics | 3. In an AP:
(i) given a = 5, d = 3, a
n
= 50, find n and S n.
(ii) given a = 7, a 13 = 35, find d and S 13.
(iii) given a 12 = 37, d = 3, find a and S 12.
(iv) given a
3
= 15, S 10 = 125, find d and a 10.
(v) given d = 5, S
9 = 75, find a and a 9
.
(vi) given a = 2, d = 8, S
n
= 90, find n and a n.
(vii) given a = 8, a
n
= 62, S
n
= 210, find n and d.
(viii) given a
n
= 4, d = 2, S
n
= –14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms.
Find a.
4. How many terms of the AP : 9, 17, 25,.
.. must be taken to give a sum of 636?
5. The first term of an AP is 5, the last term is 45 and the sum is 400.
Find the number of terms and the common difference.
6. The first and the last terms of an AP are 17 and 350 respectively.
If the common difference is 9, how many terms are there and what is their sum?
7. Find the sum of first 22 terms of an AP in which d
=
7 and 22nd term is 149.
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.10. Show that a
1 2
a,.
.., a n,
. .. form an AP where a n is defined as below
(i) a
n
= 3 + 4n (ii) a
n
= 9 – 5n
Also find the sum of the first 15 terms in each case.
11.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S
1)?
What is the sum of first two terms?
What is the second term?
Similarly, find the 3rd, the 10th and the nth terms.
12.
Find the sum of the first 40 positive integers divisible by 6.
13.
Find the sum of the first 15 multiples of 8.
14.
Find the sum of the odd numbers between 0 and 50.
15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ̀ 200 for the first day, ̀ 250 for the second day, ̀ 300 for the third day, etc., the penalty for each succeeding day being ̀ 50 more than for the preceding day.
How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
16.
A sum of ` 700 is to be used to give seven cash prizes to students of a school for their overall academic performance.
If each prize is ̀ 20 less than its preceding prize, find the value of each of the prizes.
17.
In a school, students thought of planting trees in and around the school to reduce air pollution.
It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII.
There are three sections of each class.
How many trees will be planted by the students?
18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,.
.. as shown in Fig. 5.4.
What is the total length of such a spiral made up of thirteen consecutive
semicircles?
(Take = [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2
×
5 + 2 × (5 + 3)]
22 7
)Fig. 5.4
[Hint : Length of successive semicircles is l, l, l, l,.
.. with centres at A, B, A, B,.
.., respectively.]
1
2
3 4
19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5).
In how many rows are the 200 logs placed and how many logs are in the top row?Fig. 5.5
20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line.
There are ten potatoes in the line (see Fig. 5.6).Fig. 5.6
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket.
What is the total distance the competitor has to run? | jemh105.pdf |
0 | CBSE | Class10 | Mathematics | TRIANGLES
6.1 Introduction
You are familiar with triangles and many of their properties from your earlier classes.
In Class IX, you have studied congruence of triangles in detail.
Recall that two figures are said to be congruent, if they have the same shape and the same size.
In this chapter, we shall study about those figures which have the same shape but not necessarily the same size.
Two figures having the same shape (and not necessarily the same size) are called similar figures.
In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
Can you guess how heights of mountains (say Mount Everest) or distances of some long distant objects (say moon) have been found out?
Do you think these have been measured directly with the help of a measuring tape?
In fact, all these heights and distances have been found out using the idea of indirect measurements, which is based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3 and also Chapters 8 and 9 of this book).
6.2 Similar Figures
In Class IX, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent.
Now consider any two (or more) circles [see Fig. 6.1 (i)].
Are they congruent?
Since all of them do not have the same radius, they are not congruent to each other.
Note that some are congruent and some are not, but all of them have the same shape.
So they all are, what we call, similar.
Two similar figures have the same shape but not necessarily the same size.
Therefore, all circles are similar.
What about two (or more) squares or two (or more) equilateral triangles [see Fig. 6.1 (ii) and (iii)]?
As observed in the case of circles, here also all squares are similar and all equilateral triangles are similar.
From the above, we can say that all congruent figures are similar but the similar figures need not be congruent.
Can a circle and a square be similar?
Can a triangle and a square be similar?
These questions can be answered by just looking at the figures (see Fig. 6.1).
Evidently these figures are not similar.
(Why?) Fig. 6.1
Fig. 6.2
What can you say about the two quadrilaterals ABCD and PQRS (see Fig 6.2)?Are they similar?
These figures appear to be similar but we cannot be certain about it.Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not.
For this, let us look at the photographs given in Fig. 6.3:
Fig. 6.3
You will at once say that they are the photographs of the same monument (Taj Mahal) but are in different sizes.
Would you say that the three photographs are similar?
Yes,they are.
What can you say about the two photographs of the same size of the same person one at the age of 10 years and the other at the age of 40 years?
Are these photographs similar?
These photographs are of the same size but certainly they are not of the same shape.
So, they are not similar.
What does the photographer do when she prints photographs of different sizes from the same negative?
You must have heard about the stamp size, passport size and postcard size photographs.
She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm).
Thus, if we consider any line segment in the smaller photograph (figure), its corresponding line
segment in the bigger photograph (figure) will be of that of the line segment.
45 35
55 or 35
This really means that every line segment of the smaller photograph is enlarged (increased) in the ratio 35:45 (or 35:55).
It can also be said that every line segment of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35).
Further, if you consider inclinations (or angles) between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations(or angles) are always equal.
This is the essence of the similarity of two figures and in particular of two polygons.
We say that: Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
Note that the same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons.
You must have heard that world maps (i.e., global maps) and blue prints for the construction of a building are prepared using a suitable scale factor and observing certain conventions.
In order to understand similarity of figures more clearly, let us perform the following activity:
Activity 1 : Place a lighted bulb at a point O on the ceiling and directly below it a table in your classroom.
Let us cut a polygon, say a quadrilateral ABCD, from a plane cardboard and place this cardboard parallel to the ground between the lighted bulb and the table.
Then a shadow of ABCD is cast on the table.
Mark the outline of this shadow as ABCD (see Fig.6.4).
Note that the quadrilateral ABCD is an enlargement (or magnification) of the quadrilateral ABCD.
This is because of the property of light that light propogates in a straight line.
You may also note that A lies on ray OA, B lies on ray OB, C lies on OC and D lies on OD.
Thus, quadrilaterals ABCD and ABCD are of the same shape but of different sizes.
Fig. 6.4
So, quadrilateral ABCD is similiar to quadrilateral ABCD.
We can also say that quadrilateral ABCD is similar to the quadrilateral ABCD.
Here, you can also note that vertex A corresponds to vertex A, vertex B corresponds to vertex B, vertex C corresponds to vertex C and vertex D corresponds to vertex D. Symbolically, these correspondences are represented as A A, B B, C C and D D. By actually measuring the angles and the sides of the two quadrilaterals, you may verify that
(i)
A
= A,
B
= B,
C
= C,
D
= D and
(ii) AB BC CD DA A B B C C D D A .
This again emphasises that two polygons of the same number of sides are similar, if (i) all the corresponding angles are equal and (ii) all the corresponding sides are in the same ratio (or proportion).
From the above, you can easily say that quadrilaterals ABCD and PQRS of Fig. 6.5 are similar.
Fig. 6.5 Remark : You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon.
You may note that in the two quadrilaterals (a square and a rectangle) of Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the same ratio.
Fig. 6.6
So, the two quadrilaterals are not similar.
Similarly, you may note that in the two quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the same ratio, but their corresponding angles are not equal.
Again, the two polygons (quadrilaterals) are not similar.
Fig. 6.7
Thus, either of the above two conditions (i) and (ii) of similarity of two polygons is not sufficient for them to be similar. | jemh106.pdf |
1 | CBSE | Class10 | Mathematics | 6.1 Introduction
You are familiar with triangles and many of their properties from your earlier classes.
In Class IX, you have studied congruence of triangles in detail.
Recall that two figures are said to be congruent, if they have the same shape and the same size.
In this chapter, we shall study about those figures which have the same shape but not necessarily the same size.
Two figures having the same shape (and not necessarily the same size) are called similar figures.
In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
Can you guess how heights of mountains (say Mount Everest) or distances of some long distant objects (say moon) have been found out?
Do you think these have been measured directly with the help of a measuring tape?
In fact, all these heights and distances have been found out using the idea of indirect measurements, which is based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3 and also Chapters 8 and 9 of this book).6.2 Similar Figures
In Class IX, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent.
Now consider any two (or more) circles [see Fig. 6.1 (i)].
Are they congruent?
Since all of them do not have the same radius, they are not congruent to each other.
Note that some are congruent and some are not, but all of them have the same shape.
So they all are, what we call, similar.
Two similar figures have the same shape but not necessarily the same size.
Therefore, all circles are similar.
What about two (or more) squares or two (or more) equilateral triangles [see Fig. 6.1 (ii) and (iii)]?
As observed in the case of circles, here also all squares are similar and all equilateral triangles are similar.
From the above, we can say that all congruent figures are similar but the similar figures need not be congruent.
Can a circle and a square be similar?
Can a triangle and a square be similar?
These questions can be answered by just looking at the figures (see Fig. 6.1).
Evidently these figures are not similar.
(Why?) Fig. 6.1
Fig. 6.2
What can you say about the two quadrilaterals ABCD and PQRS (see Fig 6.2)?Are they similar?
These figures appear to be similar but we cannot be certain about it.Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not.
For this, let us look at the photographs given in Fig. 6.3:
Fig. 6.3
You will at once say that they are the photographs of the same monument (Taj Mahal) but are in different sizes.
Would you say that the three photographs are similar?
Yes,they are.
What can you say about the two photographs of the same size of the same person one at the age of 10 years and the other at the age of 40 years?
Are these photographs similar?
These photographs are of the same size but certainly they are not of the same shape.
So, they are not similar.
What does the photographer do when she prints photographs of different sizes from the same negative?
You must have heard about the stamp size, passport size and postcard size photographs.
She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm).
Thus, if we consider any line segment in the smaller photograph (figure), its corresponding line
segment in the bigger photograph (figure) will be of that of the line segment.
45 35Fig. 6.2
What can you say about the two quadrilaterals ABCD and PQRS (see Fig 6.2)?Are they similar?
These figures appear to be similar but we cannot be certain about it.Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not.
For this, let us look at the photographs given in Fig. 6.3:
Fig. 6.3
You will at once say that they are the photographs of the same monument (Taj Mahal) but are in different sizes.
Would you say that the three photographs are similar?
Yes,they are.
What can you say about the two photographs of the same size of the same person one at the age of 10 years and the other at the age of 40 years?
Are these photographs similar?
These photographs are of the same size but certainly they are not of the same shape.
So, they are not similar.
What does the photographer do when she prints photographs of different sizes from the same negative?
You must have heard about the stamp size, passport size and postcard size photographs.
She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm).
Thus, if we consider any line segment in the smaller photograph (figure), its corresponding line
segment in the bigger photograph (figure) will be of that of the line segment.
45 35 | jemh106.pdf |
2 | CBSE | Class10 | Mathematics | 55 or 35
This really means that every line segment of the smaller photograph is enlarged (increased) in the ratio 35:45 (or 35:55).
It can also be said that every line segment of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35).
Further, if you consider inclinations (or angles) between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations(or angles) are always equal.
This is the essence of the similarity of two figures and in particular of two polygons.
We say that: Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
Note that the same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons.
You must have heard that world maps (i.e., global maps) and blue prints for the construction of a building are prepared using a suitable scale factor and observing certain conventions.
In order to understand similarity of figures more clearly, let us perform the following activity:
Activity 1 : Place a lighted bulb at a point O on the ceiling and directly below it a table in your classroom.
Let us cut a polygon, say a quadrilateral ABCD, from a plane cardboard and place this cardboard parallel to the ground between the lighted bulb and the table.
Then a shadow of ABCD is cast on the table.
Mark the outline of this shadow as ABCD (see Fig.6.4).
Note that the quadrilateral ABCD is an enlargement (or magnification) of the quadrilateral ABCD.
This is because of the property of light that light propogates in a straight line.
You may also note that A lies on ray OA, B lies on ray OB, C lies on OC and D lies on OD.
Thus, quadrilaterals ABCD and ABCD are of the same shape but of different sizes.
Fig. 6.4
So, quadrilateral ABCD is similiar to quadrilateral ABCD.
We can also say that quadrilateral ABCD is similar to the quadrilateral ABCD.
Here, you can also note that vertex A corresponds to vertex A, vertex B corresponds to vertex B, vertex C corresponds to vertex C and vertex D corresponds to vertex D. Symbolically, these correspondences are represented as A A, B B, C C and D D. By actually measuring the angles and the sides of the two quadrilaterals, you may verify that
(i)
A
= A,
B
= B,
C
= C,
D
= D and
(ii) AB BC CD DA A B B C C D D A .
This again emphasises that two polygons of the same number of sides are similar, if (i) all the corresponding angles are equal and (ii) all the corresponding sides are in the same ratio (or proportion).
From the above, you can easily say that quadrilaterals ABCD and PQRS of Fig. 6.5 are similar.
Fig. 6.5 Remark : You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon.
You may note that in the two quadrilaterals (a square and a rectangle) of Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the same ratio.
Fig. 6.6
So, the two quadrilaterals are not similar.
Similarly, you may note that in the two quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the same ratio, but their corresponding angles are not equal.
Again, the two polygons (quadrilaterals) are not similar.
Fig. 6.7
Thus, either of the above two conditions (i) and (ii) of similarity of two polygons is not sufficient for them to be similar.Fig. 6.4
So, quadrilateral ABCD is similiar to quadrilateral ABCD.
We can also say that quadrilateral ABCD is similar to the quadrilateral ABCD.
Here, you can also note that vertex A corresponds to vertex A, vertex B corresponds to vertex B, vertex C corresponds to vertex C and vertex D corresponds to vertex D. Symbolically, these correspondences are represented as A A, B B, C C and D D. By actually measuring the angles and the sides of the two quadrilaterals, you may verify that
(i)
A
= A,
B
= B,
C
= C,
D
= D and
(ii) AB BC CD DA A B B C C D D A .
This again emphasises that two polygons of the same number of sides are similar, if (i) all the corresponding angles are equal and (ii) all the corresponding sides are in the same ratio (or proportion).
From the above, you can easily say that quadrilaterals ABCD and PQRS of Fig. 6.5 are similar.
Fig. 6.5 Remark : You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon.
You may note that in the two quadrilaterals (a square and a rectangle) of Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the same ratio. | jemh106.pdf |
3 | CBSE | Class10 | Mathematics | EXERCISE 6.1
1. Fill in the blanks using the correct word given in brackets :
(i) All circles are. (congruent, similar)
(ii) All squares are. (similar, congruent)
(iii) All triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are.
(equal, proportional)
2. Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
3. State whether the following quadrilaterals are similar or not:
Fig. 6.8
6.3 Similarity of Triangles
What can you say about the similarity of two triangles?
You may recall that triangle is also a polygon.
So, we can state the same conditions for the similarity of two triangles.
That is: Two triangles are similiar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion).
Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles.
A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same.
It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.
Thales (640 – 546 B.C.)
To understand the Basic Proportionality
Theorem, let us perform the following activity:
Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB.
Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9).
Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from Fig. 6.9 Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
your constructions that
AD 3
DB 2
?
Measure AE and AE 3 EC.
What about AE EC ?
Observe that EC is also equal to 2.
Thus, you can see that
AD AE DB EC
. Is it a coincidence?
No, it is due to the following in ABC, DE || BC and theorem (known as the Basic Proportionality Theorem):
Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10).
We need to prove that
AD AE DB EC
. Let us join BE and CD and then draw DM AC and EN AB.
Fig. 6.10
| Now, area of ADE (= | 1 2 base × height) = | 1 2 AD × EN.
| --- | --- | ---
| Recall from Class IX, that area of ADE is denoted as ar(ADE).
| So, ar(ADE) = 1 | |
| 2 AD × EN
| Similarly, ar(BDE) = 1 | |
2 DB × EN,
| | ar(ADE) = | 1 2 AE × DM and ar(DEC) = | 1 2 EC × DM.
| --- | --- | --- | ---
| Therefore, | ar(ADE) ar(BDE) = | |
| | 1 AD × EN | | | | AD2 |
| --- | --- | --- | --- | --- | --- | ---
| | 1 | | | | DBDB |
| × EN
| | 2 | | | | | (1)
| and | ar(ADE) ar(DEC) = | | | | |
| | 1 AE × DM | | | | AE2 |
| | 1 | | | | ECEC |
| × DM
| | 2 | | | | | (2)
Note that BDE and DEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC) (3)
Therefore, from (1), (2) and (3), we have :
AD DB = AE EC
Is the converse of this theorem also true (For the meaning of converse, see Appendix 1)?
To examine this, let us perform the following activity:
Activity 3 : Draw an angle XAY on your notebook and on ray AX, mark points B, B, B 4 and B such that AB 1
=
B 1 B
2
= B 2 B
1 2, B
3
3 = B
| 3 B 4 = B 4 B. | | | | |
| --- | --- | --- | --- | --- | ---
| 1 | C | 1 and BC are parallel to each other, i.e.,
| --- | --- | ---
| 1 C 1 | || | BC | (1)
| --- | --- | --- | ---
| Similarly, on ray AY, mark points
| C 1 , C 2
| , C 3 , C 4 and C such that AC 1 = C 1 | | | | | C 2 =
| C 2 C 3 = C 3 C 4 = C 4 C. Then join B
| 1 C 1 and BC (see Fig. 6.11). Fig. 6.11 Fig. 6.13 Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
| Note that
| 1 1
| AB B B =
| 1 1
| AC C C
| (Each equal to
| 1 4) You can also see that lines B
| B
Similarly, by joining B
2
C
2
, B 3
C
3 and B
4
C 4 , you can see that: 2 2 AB B
B
= 2 2 AC C
C
2 3
and B
2
C 2 || BC (2)
3 3
AB B
B
= 3 3 AC C
C
3 2
and B
3
C 3 || BC (3)
4 4
AB B B = 4 4 AC C C 4 1
and B
4
C
4 || BC (4)
From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY.
Each time, you will arrive at the same result.
Thus, we obtain the following theorem, which is the converse of Theorem 6.1: that
This theorem can be proved by taking a line DE such
AD AE DB EC
and assuming that DE is not parallel to BC (see Fig. 6.12).
Fig. 6.12 Example 2 : ABCD is a trapezium with AB || DC.
E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB If DE is not parallel to BC, draw a line DE parallel to BC.
So, AD
DB =
| | AE | E C | (Why ?)
| --- | --- | --- | ---
| Therefore, | AE EC = | AE E C | (Why ?)
| --- | --- | --- | ---
| Adding 1 to both sides of above, you can see that E and E must coincide. (Why ?)
| Let us take some examples to illustrate the use of the above theorems.
| Example 1 : If a line intersects sides AB and AC of a ABC at D and E respectively
| and is parallel to BC, prove that AD
| AB =
| AE
| AC (see Fig. 6.13).
| Solution : | DE || BC | (Given)
| --- | --- | ---
| So, | AD DB = AE EC | (Theorem 6.1)
| or, | DB |
AD = EC AE
or,
DB 1
= EC 1 AE
| or, | AB |
| --- | --- | ---
| AD = AC AE
| So, | AD AB = | AE AC
(see Fig. 6.14).
Show that AE BF
ED FC .
Solution : Let us join AC to intersect EF at G (see Fig. 6.15).
Fig. 6.14
AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other) Now, in ADC, EG || DC (As EF || DC)
So, AE ED = AG GC (Theorem 6.1) (1) Similarly, from CAB,
Fig. 6.15
CG AG = CF BF
i.e., AG GC = BF FC (2) Therefore, from (1) and (2), AE ED = BF FC
Example 3 : In Fig. 6.16,
PS SQ =
PT TR and PST = PRQ.
Prove that PQR is an isosceles triangle.
Fig. 6.16
Solution : It is given that
PS PT SQ TR
| So, | ST || QR | (Theorem 6.2)
| --- | --- | ---
| Therefore, | PST = | PQR | (Corresponding angles) | (1)
| --- | --- | --- | --- | ---
| Also, it is given that | PST = | PRQ | (2)
| --- | --- | --- | ---
| So, | PRQ = PQR [From (1) and (2)] | |
| Therefore, | PQ = PR | (Sides opposite the equal angles) |
i.e., PQR is an isosceles triangle. | jemh106.pdf |
4 | CBSE | Class10 | Mathematics | 1. Fill in the blanks using the correct word given in brackets :
(i) All circles are. (congruent, similar)
(ii) All squares are. (similar, congruent)
(iii) All triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are.
(equal, proportional)
2. Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
3. State whether the following quadrilaterals are similar or not:(i) All circles are. (congruent, similar)(ii) All squares are. (similar, congruent)(iii) All triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are.
(equal, proportional)
2. Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
3. State whether the following quadrilaterals are similar or not:Fig. 6.86.3 Similarity of Triangles
What can you say about the similarity of two triangles?
You may recall that triangle is also a polygon.
So, we can state the same conditions for the similarity of two triangles.
That is: Two triangles are similiar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion).
Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles.
A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same.
It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.
Thales (640 – 546 B.C.)
To understand the Basic Proportionality
Theorem, let us perform the following activity:
Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB.
Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9).
Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from Fig. 6.9 Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
your constructions that
AD 3
DB 2
?
Measure AE and AE 3 EC.
What about AE EC ?
Observe that EC is also equal to 2.
Thus, you can see that
AD AE DB EC
. Is it a coincidence?
No, it is due to the following in ABC, DE || BC and theorem (known as the Basic Proportionality Theorem):
Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10).
We need to prove thatThales (640 – 546 B.C.)
To understand the Basic Proportionality
Theorem, let us perform the following activity:
Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB.
Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9).
Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from Fig. 6.9 Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
your constructions that
AD 3
DB 2
?
Measure AE and AE 3 EC.
What about AE EC ?
Observe that EC is also equal to 2.
Thus, you can see that
AD AE DB EC
. Is it a coincidence?
No, it is due to the following in ABC, DE || BC and theorem (known as the Basic Proportionality Theorem):
Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10).
We need to prove thatTo understand the Basic Proportionality
Theorem, let us perform the following activity:
Activity 2 : Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB.
Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9).
Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from Fig. 6.9 Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
your constructions that
AD 3
DB 2
?
Measure AE and AE 3 EC.
What about AE EC ?
Observe that EC is also equal to 2.
Thus, you can see that
AD AE DB EC
. Is it a coincidence?
No, it is due to the following in ABC, DE || BC and theorem (known as the Basic Proportionality Theorem):
Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10).
We need to prove thatAD 3 | jemh106.pdf |
5 | CBSE | Class10 | Mathematics | AD AE DB EC
. Is it a coincidence?
No, it is due to the following in ABC, DE || BC and theorem (known as the Basic Proportionality Theorem):
Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10).
We need to prove thatAD AE DB EC
. Let us join BE and CD and then draw DM AC and EN AB.
Fig. 6.10
| Now, area of ADE (= | 1 2 base × height) = | 1 2 AD × EN.
| --- | --- | ---
| Recall from Class IX, that area of ADE is denoted as ar(ADE).
| So, ar(ADE) = 1 | |
| 2 AD × EN
| Similarly, ar(BDE) = 1 | |
2 DB × EN,
| | ar(ADE) = | 1 2 AE × DM and ar(DEC) = | 1 2 EC × DM.
| --- | --- | --- | ---
| Therefore, | ar(ADE) ar(BDE) = | |
| | 1 AD × EN | | | | AD2 |
| --- | --- | --- | --- | --- | --- | ---
| | 1 | | | | DBDB |
| × EN
| | 2 | | | | | (1)
| and | ar(ADE) ar(DEC) = | | | | |
| | 1 AE × DM | | | | AE2 |
| | 1 | | | | ECEC |
| × DM
| | 2 | | | | | (2)
Note that BDE and DEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC) (3)
Therefore, from (1), (2) and (3), we have :Fig. 6.10
| Now, area of ADE (= | 1 2 base × height) = | 1 2 AD × EN.
| --- | --- | ---
| Recall from Class IX, that area of ADE is denoted as ar(ADE).
| So, ar(ADE) = 1 | |
| 2 AD × EN
| Similarly, ar(BDE) = 1 | |
2 DB × EN,
| | ar(ADE) = | 1 2 AE × DM and ar(DEC) = | 1 2 EC × DM.
| --- | --- | --- | ---
| Therefore, | ar(ADE) ar(BDE) = | |
| | 1 AD × EN | | | | AD2 |
| --- | --- | --- | --- | --- | --- | ---
| | 1 | | | | DBDB |
| × EN
| | 2 | | | | | (1)
| and | ar(ADE) ar(DEC) = | | | | |
| | 1 AE × DM | | | | AE2 |
| | 1 | | | | ECEC |
| × DM
| | 2 | | | | | (2)
Note that BDE and DEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC) (3)
Therefore, from (1), (2) and (3), we have :So, ar(BDE) = ar(DEC) (3)
Therefore, from (1), (2) and (3), we have :AD DB = AE EC
Is the converse of this theorem also true (For the meaning of converse, see Appendix 1)?
To examine this, let us perform the following activity:
Activity 3 : Draw an angle XAY on your notebook and on ray AX, mark points B, B, B 4 and B such that AB 1
=
B 1 B
2
= B 2 B
1 2, B
3
3 = B
| 3 B 4 = B 4 B. | | | | |
| --- | --- | --- | --- | --- | ---
| 1 | C | 1 and BC are parallel to each other, i.e.,
| --- | --- | ---
| 1 C 1 | || | BC | (1)
| --- | --- | --- | ---
| Similarly, on ray AY, mark points
| C 1 , C 2
| , C 3 , C 4 and C such that AC 1 = C 1 | | | | | C 2 =
| C 2 C 3 = C 3 C 4 = C 4 C. Then join B
| 1 C 1 and BC (see Fig. 6.11). Fig. 6.11 Fig. 6.13 Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
| Note that
| 1 1
| AB B B =
| 1 1
| AC C C
| (Each equal to
| 1 4) You can also see that lines B
| B
Similarly, by joining B
2
C
2
, B 3
C
3 and B
4
C 4 , you can see that: 2 2 AB B
B
= 2 2 AC C
C
2 3
and B
2
C 2 || BC (2)
3 3
AB B
B
= 3 3 AC C
C
3 2
and B
3
C 3 || BC (3)
4 4
AB B B = 4 4 AC C C 4 1
and B
4
C
4 || BC (4)
From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY.
Each time, you will arrive at the same result.
Thus, we obtain the following theorem, which is the converse of Theorem 6.1: that
This theorem can be proved by taking a line DE such
AD AE DB EC
and assuming that DE is not parallel to BC (see Fig. 6.12).
Fig. 6.12 Example 2 : ABCD is a trapezium with AB || DC.
E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB If DE is not parallel to BC, draw a line DE parallel to BC.
So, AD
DB =
| | AE | E C | (Why ?)
| --- | --- | --- | ---
| Therefore, | AE EC = | AE E C | (Why ?)
| --- | --- | --- | ---
| Adding 1 to both sides of above, you can see that E and E must coincide. (Why ?)
| Let us take some examples to illustrate the use of the above theorems.
| Example 1 : If a line intersects sides AB and AC of a ABC at D and E respectively
| and is parallel to BC, prove that AD
| AB =
| AE
| AC (see Fig. 6.13).
| Solution : | DE || BC | (Given)
| --- | --- | ---
| So, | AD DB = AE EC | (Theorem 6.1)
| or, | DB |
| jemh106.pdf |
6 | CBSE | Class10 | Mathematics | EXERCISE 6.2
1. In Fig
6.17, (i) and (ii), DE || BC.
Find EC in (i) and AD in (ii).
Fig. 6.17
2. E and F are points on the sides PQ and PR
respectively of a PQR.
For each of the following cases, state whether EF || QR :
(i) PE = 3
9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Fig. 6.18
(iii) PQ = 1
28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
3. In Fig
6.18, if LM || CB and LN || CD, prove that
| AM | AN | |
| --- | --- | --- | ---
| AB AD | AD | AD |
4. In Fig
6.19, DE || AC and DF || AE.
Prove that
BF BE FE EC
Fig. 6.19
5. In Fig
6.20, DE || OQ and DF || OR.
Show that EF || QR.
6. In Fig
6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR.
7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
(Recall that you have proved it in Class IX).
8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
(Recall that you have done it in Class IX).
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
Fig. 6.21
that
AO CO BO DO
10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that
AO CO BO DO
Show that ABCD is a trapezium.
6.4 Criteria for Similarity of Triangles
In the previous section, we stated that two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
That is, in ABC and DEF, if
(i)
A
= D,
B
= E,
C
= F and
(ii) AB BC CA, DE EF FD then the two triangles are similar (see Fig. 6.22).
Fig. 6.22
Fig. 6.20
Here, you can see that A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ‘ ABC ~ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’.
The symbol ‘~’ stands for ‘is similar to’.
Recall that you have used the symbol ‘’ for ‘is congruent to’ in Class IX.
It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices.
For example, for the triangles ABC and DEF of Fig. 6.22, we cannot write ABC ~ EDF or ABC ~ FED.
However, we can write BAC ~ EDF.
Now a natural question arises : For checking the similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their corresponding angles ( A
=
D,
B
= E,
C
= F) and all the equality relations of the ratios
of their corresponding sides ?
Let us examine.
You may recall that
| | AB | BC | CA |
| --- | --- | --- | --- | ---
| | DE | EF | FD |
| | | | |
only three pairs of corresponding parts (or elements) of the two triangles.
Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts of the two triangles, instead of all the six pairs of corresponding parts.
For this, let us perform the following activity:
Activity 4 : Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively.
Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°.
Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).
Fig. 6.23
Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that
B
= E,
C
= F and
A
= D. That is, corresponding angles of these two triangles are equal.
What can you say about their corresponding sides ?
Note that
BC 3
0.6. EF 5
What about AB DE and CA FD ?
On measuring AB, DE, CA and FD, you
will find that and are also equal to 0.6 (or nearly equal to 0.6, if there is some
AB DE
CA FD
| error in the measurement). Thus, | AB | BC | CA
| --- | --- | --- | ---
| DE | EF | FD | | | You | can | repeat | this | activity | by
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
constructing several pairs of triangles having their corresponding angles equal.
Every time, you will find that their corresponding sides are in the same ratio (or proportion).
This activity leads us to the following criterion for similarity of two triangles.
Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
A
= D,
B
= E and
C
= F (see Fig. 6.24)
Fig. 6.24
Cut DP = AB and DQ = AC and join PQ.
| So, | ABC | DPQ | (Why ?)
| --- | --- | --- | ---
| This gives | B = P = E and PQ || EF (How?)
| --- | ---
| Therefore, | DP PE = | DQ QF | (Why?)
| --- | --- | --- | ---
| i.e., | AB | DE = | AC DF | (Why?)
| --- | --- | --- | --- | ---
| Similarly, | AB | BC | AB | BC | AC |
| --- | --- | --- | --- | --- | --- | ---
| | DE = | EF and so | DE EF DF | DF | DF . |
Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
Therefore, AAA similarity criterion can also be stated as follows:
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
This may be referred to as the AA similarity criterion for two triangles.
You have seen above that if the three angles of one triangle are respectively equal to the three angles of another triangle, then their corresponding sides are proportional (i.e., in the same ratio).
What about the converse of this statement?
Is the converse true?
In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal?
Let us examine it through an activity :
Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25).
Fig. 6.25
So, you have : AB BC CA
DE EF FD
(each equal to
2 3
) Now measure A, B, C, D, E and F. You will observe that
A
= D,
B
= E and
C
= F, i.e., the corresponding angles of the two triangles are equal.
You can repeat this activity by drawing several such triangles (having their sides in the same ratio).
Everytime you shall see that their corresponding angles are equal.
It is due to the following criterion of similarity of two triangles: Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that AB BC CA DE EF FD (< 1) (see Fig. 6.26):
Fig. 6.26
Cut DP = AB and DQ = AC and join PQ.
| It can be seen that | | DP PE = DQ QF | and PQ || EF (How?) | |
| --- | --- | --- | --- | --- | ---
| So, | | P = E | and | Q = F. |
| Therefore, | | DP | | |
| DE = DQ DF = PQ EF
| So, | | DP | | |
| DE = DQ DF = BC EF (Why?)
| So, | | BC = PQ | | (Why?) |
| Thus, | | ABC | DPQ | (Why ?) |
| So, | A = D, | B = E | and | C = F | (How ?)
Remark : You may recall that either of the two conditions namely, (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar.
However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other.
Let us now recall the various criteria for congruency of two triangles learnt in Class IX.
You may observe that SSS similarity criterion can be compared with the SSS congruency criterion.This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles.
For this, let us perform an activity.
Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm,
A
= 50°, AC = 4 cm, DE = 3 cm,
D
= 50° and DF = 6 cm (see Fig.6.27).
Fig. 6.27
| AC | Here, you may observe that AB | 2
| --- | --- | ---
| DE = | DF (each equal to | 3 ) and A (included
between the sides AB and AC) =
D (included between the sides DE and DF).
That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion).
Now let us measure B, C, E and F.
You will find that
B
= E and
C
= F. That is,
A
= D,
B
= E and
C
= F. So, by AAA similarity criterion, ABC ~ DEF.
You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional.
Everytime, you will find that the triangles are similar.
It is due to the following criterion of similarity of triangles: Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
This criterion is referred to as the SAS (Side–Angle–Side) s imi la r i ty c
r
i te r ion for two triangles.
As before, this theorem can be proved by taking two triangles ABC and DEF such that
AB AC DE DF ( 1) and A = D
(see Fig. 6.28).
Cut DP = AB, DQ = AC and join PQ.
Fig. 6.28
| Now, | PQ || EF and | ABC | DPQ | (How ?)
| --- | --- | --- | --- | ---
| So, | A = D, B = P and C = Q | | |
| Therefore, | ABC ~ DEF | | | (Why?)
We now take some examples to illustrate the use of these criteria.
Example 4 : In Fig. 6.29, if PQ || RS, prove that POQ ~ SOR.
| Solution : | Fig. 6.29 PQ || RS | (Given)
| --- | --- | ---
| So, | P = S | (Alternate angles)
| and | Q = | R
| Also, | POQ = SOR | (Vertically opposite angles)
| Therefore, | POQ ~ SOR | (AAA similarity criterion)
Example 5 : Observe Fig. 6.30 and then find P.
Fig. 6.30 Solution : In ABC and PQR,
AB 3.8 1, RQ 7.6 2
BC 6 1 QP 12 2
and CA 3
3
1 PR 26 3
That is,
AB BC CA RQ QP PR
| So, | ABC ~ | RQP | (SSS similarity)
| --- | --- | --- | ---
| Therefore, | C = | P | (Corresponding angles of similar triangles)
| But | C = 180° – A – B | | (Angle sum property)
| = 180° – 80° – 60° = 40°
| So, | P = 40° | |
Example 6 : In Fig. 6.31, OA.
OB = OC.
OD.
Show that
A
= C and
B
= D.
Solution : OA . OB = OC . OD (Given)
Fig. 6.31
| So, | OA OC = | OD OB | (1)
| --- | --- | --- | ---
| Also, we have | AOD = | COB | (Vertically opposite angles) (2)
Therefore, from (1) and (2), AOD ~ COB (SAS similarity criterion)
So, A = C and D = B (Corresponding angles of similar triangles)
Example 7
:
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s.
If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution : Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post (see Fig. 6.32).
Fig. 6.32
Fig. 6.33
From the figure, you can see that DE is the shadow of the girl.
Let DE be x metres.
Now, BD = 1.2 m
×
4 = 4.8 m.
Note that in ABE and CDE,
B
= D (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground) = 3.6
| and | E = | E | (Same angle)
| --- | --- | --- | ---
| So, | ABE ~ | CDE | (AA similarity criterion)
| Therefore, | BE | |
| DE = AB CD
| i.e., | 4.8 + x | |
x
0.9
| (90 cm = | 90 100
| i.e., | 4.8 + x = 4x
| i.e., | 3x = 4.8
| i.e., | x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
Example 8 : In Fig. 6.33, CM and RN are respectively the medians of ABC and PQR.
If ABC ~ PQR, prove that :
(i) AMC ~ PNR
(ii) CM AB RN PQ
(iii) CMB ~ RNQ Solution : (i) ABC ~ PQR (Given)
So,
AB PQ = BC CA
QR RP
| | (1)
| --- | ---
| and | A = P, | B = Q and C = R | (2)
| --- | --- | --- | ---
| But | AB = 2 AM and PQ = 2 PN | |
| (As CM and RN are medians)
| So, from (1), | 2AM 2PN = | CA RP |
| i.e., | AM PN = | CA RP | | (3)
| --- | --- | --- | --- | ---
| Also, | MAC = | NPR | [From (2)] | (4)
| So, from (3) and (4), | AMC ~ | PNR | (SAS similarity) (5) |
| (ii) From (5), | CM RN = | CA RP | | (6)
| But | CA RP = | AB PQ | [From (1)] (7) |
| Therefore, | CM RN = | AB PQ | [From (6) and (7)] (8) |
| (iii) Again, AB
| PQ =
| BC QR [From (1)]
| Therefore, | CM RN = | BC | QR | [From (8)] (9)
| Also, | CM RN = | AB 2 BM PQ 2 QN | |
| i.e., | CM | | |
| RN =
| BM QN (10)
| i.e., | CM RN = | BC BM QR QN | [From (9) and (10)] |
| Therefore, | CMB ~ | RNQ | (SSS similarity) |
[Note : You can also prove part (iii) by following the same method as used for proving part (i).] | jemh106.pdf |
7 | CBSE | Class10 | Mathematics | Fig. 6.22
Fig. 6.20
Here, you can see that A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ‘ ABC ~ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’.
The symbol ‘~’ stands for ‘is similar to’.
Recall that you have used the symbol ‘’ for ‘is congruent to’ in Class IX.
It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices.
For example, for the triangles ABC and DEF of Fig. 6.22, we cannot write ABC ~ EDF or ABC ~ FED.
However, we can write BAC ~ EDF.
Now a natural question arises : For checking the similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their corresponding angles ( A
=
D,
B
= E,
C
= F) and all the equality relations of the ratios
of their corresponding sides ?
Let us examine.
You may recall that
| | AB | BC | CA |
| --- | --- | --- | --- | ---
| | DE | EF | FD |
| | | | |
only three pairs of corresponding parts (or elements) of the two triangles.
Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts of the two triangles, instead of all the six pairs of corresponding parts.
For this, let us perform the following activity:
Activity 4 : Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively.
Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°.
Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).
Fig. 6.23
Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that
B
= E,
C
= F and
A
= D. That is, corresponding angles of these two triangles are equal.
What can you say about their corresponding sides ?
Note that
BC 3
0.6. EF 5
What about AB DE and CA FD ?
On measuring AB, DE, CA and FD, you
will find that and are also equal to 0.6 (or nearly equal to 0.6, if there is some
AB DE
CA FD
| error in the measurement). Thus, | AB | BC | CA
| --- | --- | --- | ---
| DE | EF | FD | | | You | can | repeat | this | activity | by
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
constructing several pairs of triangles having their corresponding angles equal.
Every time, you will find that their corresponding sides are in the same ratio (or proportion).
This activity leads us to the following criterion for similarity of two triangles.
Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
A
= D,
B
= E and
C
= F (see Fig. 6.24)
Fig. 6.24
Cut DP = AB and DQ = AC and join PQ.
| So, | ABC | DPQ | (Why ?)
| --- | --- | --- | ---
| This gives | B = P = E and PQ || EF (How?)
| --- | ---
| Therefore, | DP PE = | DQ QF | (Why?)
| --- | --- | --- | ---
| i.e., | AB | DE = | AC DF | (Why?)
| --- | --- | --- | --- | ---
| Similarly, | AB | BC | AB | BC | AC |
| --- | --- | --- | --- | --- | --- | ---
| | DE = | EF and so | DE EF DF | DF | DF . |
Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
Therefore, AAA similarity criterion can also be stated as follows:
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
This may be referred to as the AA similarity criterion for two triangles.
You have seen above that if the three angles of one triangle are respectively equal to the three angles of another triangle, then their corresponding sides are proportional (i.e., in the same ratio).
What about the converse of this statement?
Is the converse true?
In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal?
Let us examine it through an activity :
Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25).
Fig. 6.25
So, you have : AB BC CA
DE EF FD
(each equal to
2 3
) Now measure A, B, C, D, E and F. You will observe that
A
= D,
B
= E and
C
= F, i.e., the corresponding angles of the two triangles are equal.
You can repeat this activity by drawing several such triangles (having their sides in the same ratio).
Everytime you shall see that their corresponding angles are equal.
It is due to the following criterion of similarity of two triangles: Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that AB BC CA DE EF FD (< 1) (see Fig. 6.26):
Fig. 6.26
Cut DP = AB and DQ = AC and join PQ.
| It can be seen that | | DP PE = DQ QF | and PQ || EF (How?) | |
| --- | --- | --- | --- | --- | ---
| So, | | P = E | and | Q = F. |
| Therefore, | | DP | | |
| DE = DQ DF = PQ EF
| So, | | DP | | |
| DE = DQ DF = BC EF (Why?)
| So, | | BC = PQ | | (Why?) |
| Thus, | | ABC | DPQ | (Why ?) |
| So, | A = D, | B = E | and | C = F | (How ?)
Remark : You may recall that either of the two conditions namely, (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar.
However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other.
Let us now recall the various criteria for congruency of two triangles learnt in Class IX.
You may observe that SSS similarity criterion can be compared with the SSS congruency criterion.This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles.
For this, let us perform an activity.
Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm,
A
= 50°, AC = 4 cm, DE = 3 cm,
D
= 50° and DF = 6 cm (see Fig.6.27).
Fig. 6.27
| AC | Here, you may observe that AB | 2
| --- | --- | ---
| DE = | DF (each equal to | 3 ) and A (included
between the sides AB and AC) =
D (included between the sides DE and DF).
That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion).
Now let us measure B, C, E and F.
You will find that
B
= E and
C
= F. That is,
A
= D,
B
= E and
C
= F. So, by AAA similarity criterion, ABC ~ DEF.
You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional.
Everytime, you will find that the triangles are similar.
It is due to the following criterion of similarity of triangles: Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
This criterion is referred to as the SAS (Side–Angle–Side) s imi la r i ty c
r
i te r ion for two triangles.
As before, this theorem can be proved by taking two triangles ABC and DEF such that
AB AC DE DF ( 1) and A = D
(see Fig. 6.28).
Cut DP = AB, DQ = AC and join PQ.
Fig. 6.28
| Now, | PQ || EF and | ABC | DPQ | (How ?)
| --- | --- | --- | --- | ---
| So, | A = D, B = P and C = Q | | |
| Therefore, | ABC ~ DEF | | | (Why?)
We now take some examples to illustrate the use of these criteria.
Example 4 : In Fig. 6.29, if PQ || RS, prove that POQ ~ SOR.
| Solution : | Fig. 6.29 PQ || RS | (Given)
| --- | --- | ---
| So, | P = S | (Alternate angles)
| and | Q = | R
| Also, | POQ = SOR | (Vertically opposite angles)
| Therefore, | POQ ~ SOR | (AAA similarity criterion)
Example 5 : Observe Fig. 6.30 and then find P.
Fig. 6.30 Solution : In ABC and PQR,
AB 3.8 1, RQ 7.6 2
BC 6 1 QP 12 2
and CA 3
3
1 PR 26 3
That is,
AB BC CA RQ QP PR
| So, | ABC ~ | RQP | (SSS similarity)
| --- | --- | --- | ---
| Therefore, | C = | P | (Corresponding angles of similar triangles)
| But | C = 180° – A – B | | (Angle sum property)
| = 180° – 80° – 60° = 40°
| So, | P = 40° | |
Example 6 : In Fig. 6.31, OA.
OB = OC.
OD.
Show that
A
= C and
B
= D.
Solution : OA . OB = OC . OD (Given)
Fig. 6.31
| So, | OA OC = | OD OB | (1)
| --- | --- | --- | ---
| Also, we have | AOD = | COB | (Vertically opposite angles) (2)
Therefore, from (1) and (2), AOD ~ COB (SAS similarity criterion) | jemh106.pdf |
8 | CBSE | Class10 | Mathematics | Fig. 6.20
Here, you can see that A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ‘ ABC ~ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’.
The symbol ‘~’ stands for ‘is similar to’.
Recall that you have used the symbol ‘’ for ‘is congruent to’ in Class IX.
It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices.
For example, for the triangles ABC and DEF of Fig. 6.22, we cannot write ABC ~ EDF or ABC ~ FED.
However, we can write BAC ~ EDF.
Now a natural question arises : For checking the similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their corresponding angles ( A
=
D,
B
= E,
C
= F) and all the equality relations of the ratios
of their corresponding sides ?
Let us examine.
You may recall that
| | AB | BC | CA |
| --- | --- | --- | --- | ---
| | DE | EF | FD |
| | | | |
only three pairs of corresponding parts (or elements) of the two triangles.
Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts of the two triangles, instead of all the six pairs of corresponding parts.
For this, let us perform the following activity:
Activity 4 : Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively.
Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°.
Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).
Fig. 6.23
Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that
B
= E,
C
= F and
A
= D. That is, corresponding angles of these two triangles are equal.
What can you say about their corresponding sides ?
Note that
BC 3
0.6. EF 5
What about AB DE and CA FD ?
On measuring AB, DE, CA and FD, you
will find that and are also equal to 0.6 (or nearly equal to 0.6, if there is some
AB DE
CA FD
| error in the measurement). Thus, | AB | BC | CA
| --- | --- | --- | ---
| DE | EF | FD | | | You | can | repeat | this | activity | by
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
constructing several pairs of triangles having their corresponding angles equal.
Every time, you will find that their corresponding sides are in the same ratio (or proportion).
This activity leads us to the following criterion for similarity of two triangles.
Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
A
= D,
B
= E and
C
= F (see Fig. 6.24)
Fig. 6.24
Cut DP = AB and DQ = AC and join PQ.
| So, | ABC | DPQ | (Why ?)
| --- | --- | --- | ---
| This gives | B = P = E and PQ || EF (How?)
| --- | ---
| Therefore, | DP PE = | DQ QF | (Why?)
| --- | --- | --- | ---
| i.e., | AB | DE = | AC DF | (Why?)
| --- | --- | --- | --- | ---
| Similarly, | AB | BC | AB | BC | AC |
| --- | --- | --- | --- | --- | --- | ---
| | DE = | EF and so | DE EF DF | DF | DF . |
Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
Therefore, AAA similarity criterion can also be stated as follows:
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
This may be referred to as the AA similarity criterion for two triangles.
You have seen above that if the three angles of one triangle are respectively equal to the three angles of another triangle, then their corresponding sides are proportional (i.e., in the same ratio).
What about the converse of this statement?
Is the converse true?
In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal?
Let us examine it through an activity :
Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25). | jemh106.pdf |
9 | CBSE | Class10 | Mathematics | BC 3
0.6. EF 5
What about AB DE and CA FD ?
On measuring AB, DE, CA and FD, you
will find that and are also equal to 0.6 (or nearly equal to 0.6, if there is someAB DECA FD
| error in the measurement). Thus, | AB | BC | CA
| --- | --- | --- | ---
| DE | EF | FD | | | You | can | repeat | this | activity | by
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
constructing several pairs of triangles having their corresponding angles equal.
Every time, you will find that their corresponding sides are in the same ratio (or proportion).
This activity leads us to the following criterion for similarity of two triangles.
Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
A
= D,
B
= E and
C
= F (see Fig. 6.24)Fig. 6.24
Cut DP = AB and DQ = AC and join PQ.
| So, | ABC | DPQ | (Why ?)
| --- | --- | --- | ---
| This gives | B = P = E and PQ || EF (How?)
| --- | ---
| Therefore, | DP PE = | DQ QF | (Why?)
| --- | --- | --- | ---
| i.e., | AB | DE = | AC DF | (Why?)
| --- | --- | --- | --- | ---
| Similarly, | AB | BC | AB | BC | AC |
| --- | --- | --- | --- | --- | --- | ---
| | DE = | EF and so | DE EF DF | DF | DF . |
Remark : If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
Therefore, AAA similarity criterion can also be stated as follows:
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
This may be referred to as the AA similarity criterion for two triangles.
You have seen above that if the three angles of one triangle are respectively equal to the three angles of another triangle, then their corresponding sides are proportional (i.e., in the same ratio).
What about the converse of this statement?
Is the converse true?
In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal?
Let us examine it through an activity :
Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25).Fig. 6.25
So, you have : AB BC CA
DE EF FD
(each equal to
2 3
) Now measure A, B, C, D, E and F. You will observe that
A
= D,
B
= E and
C
= F, i.e., the corresponding angles of the two triangles are equal.
You can repeat this activity by drawing several such triangles (having their sides in the same ratio).
Everytime you shall see that their corresponding angles are equal.
It is due to the following criterion of similarity of two triangles: Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that AB BC CA DE EF FD (< 1) (see Fig. 6.26):So, you have : AB BC CA
DE EF FD
(each equal to
2 3
) Now measure A, B, C, D, E and F. You will observe that
A
= D,
B
= E and
C
= F, i.e., the corresponding angles of the two triangles are equal.
You can repeat this activity by drawing several such triangles (having their sides in the same ratio).
Everytime you shall see that their corresponding angles are equal.
It is due to the following criterion of similarity of two triangles: Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that AB BC CA DE EF FD (< 1) (see Fig. 6.26): | jemh106.pdf |
10 | CBSE | Class10 | Mathematics | Fig. 6.26
Cut DP = AB and DQ = AC and join PQ.
| It can be seen that | | DP PE = DQ QF | and PQ || EF (How?) | |
| --- | --- | --- | --- | --- | ---
| So, | | P = E | and | Q = F. |
| Therefore, | | DP | | |
| DE = DQ DF = PQ EF
| So, | | DP | | |
| DE = DQ DF = BC EF (Why?)
| So, | | BC = PQ | | (Why?) |
| Thus, | | ABC | DPQ | (Why ?) |
| So, | A = D, | B = E | and | C = F | (How ?)
Remark : You may recall that either of the two conditions namely, (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar.
However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other.
Let us now recall the various criteria for congruency of two triangles learnt in Class IX.
You may observe that SSS similarity criterion can be compared with the SSS congruency criterion.This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles.
For this, let us perform an activity.
Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm,
A
= 50°, AC = 4 cm, DE = 3 cm,
D
= 50° and DF = 6 cm (see Fig.6.27).
Fig. 6.27
| AC | Here, you may observe that AB | 2
| --- | --- | ---
| DE = | DF (each equal to | 3 ) and A (included
between the sides AB and AC) =
D (included between the sides DE and DF).
That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion).
Now let us measure B, C, E and F.
You will find that
B
= E and
C
= F. That is,
A
= D,
B
= E and
C
= F. So, by AAA similarity criterion, ABC ~ DEF.
You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional.
Everytime, you will find that the triangles are similar.
It is due to the following criterion of similarity of triangles: Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
This criterion is referred to as the SAS (Side–Angle–Side) s imi la r i ty c
r
i te r ion for two triangles.
As before, this theorem can be proved by taking two triangles ABC and DEF such thatFig. 6.27
| AC | Here, you may observe that AB | 2
| --- | --- | ---
| DE = | DF (each equal to | 3 ) and A (included
between the sides AB and AC) =
D (included between the sides DE and DF).
That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion).
Now let us measure B, C, E and F.
You will find that
B
= E and
C
= F. That is,
A
= D,
B
= E and
C
= F. So, by AAA similarity criterion, ABC ~ DEF.
You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional.
Everytime, you will find that the triangles are similar.
It is due to the following criterion of similarity of triangles: Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
This criterion is referred to as the SAS (Side–Angle–Side) s imi la r i ty c
r
i te r ion for two triangles.
As before, this theorem can be proved by taking two triangles ABC and DEF such thatAB AC DE DF ( 1) and A = D
(see Fig. 6.28).
Cut DP = AB, DQ = AC and join PQ.
Fig. 6.28
| Now, | PQ || EF and | ABC | DPQ | (How ?)
| --- | --- | --- | --- | ---
| So, | A = D, B = P and C = Q | | |
| Therefore, | ABC ~ DEF | | | (Why?)
We now take some examples to illustrate the use of these criteria.
Example 4 : In Fig. 6.29, if PQ || RS, prove that POQ ~ SOR.
| Solution : | Fig. 6.29 PQ || RS | (Given)
| --- | --- | ---
| So, | P = S | (Alternate angles)
| and | Q = | R
| Also, | POQ = SOR | (Vertically opposite angles)
| Therefore, | POQ ~ SOR | (AAA similarity criterion)
Example 5 : Observe Fig. 6.30 and then find P.
Fig. 6.30 Solution : In ABC and PQR,
AB 3.8 1, RQ 7.6 2
BC 6 1 QP 12 2
and CA 3
3
1 PR 26 3
That is,
AB BC CA RQ QP PR
| So, | ABC ~ | RQP | (SSS similarity)
| --- | --- | --- | ---
| Therefore, | C = | P | (Corresponding angles of similar triangles)
| But | C = 180° – A – B | | (Angle sum property)
| = 180° – 80° – 60° = 40°
| So, | P = 40° | |
Example 6 : In Fig. 6.31, OA.
OB = OC.
OD.
Show that
A
= C and
B
= D. | jemh106.pdf |
11 | CBSE | Class10 | Mathematics | Fig. 6.30 Solution : In ABC and PQR,
AB 3.8 1, RQ 7.6 2
BC 6 1 QP 12 2
and CA 3
3
1 PR 26 3
That is,AB BC CA RQ QP PR
| So, | ABC ~ | RQP | (SSS similarity)
| --- | --- | --- | ---
| Therefore, | C = | P | (Corresponding angles of similar triangles)
| But | C = 180° – A – B | | (Angle sum property)
| = 180° – 80° – 60° = 40°
| So, | P = 40° | |
Example 6 : In Fig. 6.31, OA.
OB = OC.
OD.
Show that
A
= C and
B
= D.Solution : OA . OB = OC . OD (Given)
Fig. 6.31
| So, | OA OC = | OD OB | (1)
| --- | --- | --- | ---
| Also, we have | AOD = | COB | (Vertically opposite angles) (2)
Therefore, from (1) and (2), AOD ~ COB (SAS similarity criterion)So, A = C and D = B (Corresponding angles of similar triangles)
Example 7
:
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s.
If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution : Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post (see Fig. 6.32).
Fig. 6.32
Fig. 6.33
From the figure, you can see that DE is the shadow of the girl.
Let DE be x metres.
Now, BD = 1.2 m
×
4 = 4.8 m.
Note that in ABE and CDE,
B
= D (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground) = 3.6
| and | E = | E | (Same angle)
| --- | --- | --- | ---
| So, | ABE ~ | CDE | (AA similarity criterion)
| Therefore, | BE | |
| DE = AB CD
| i.e., | 4.8 + x | |
x
0.9
| (90 cm = | 90 100
| i.e., | 4.8 + x = 4x
| i.e., | 3x = 4.8
| i.e., | x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
Example 8 : In Fig. 6.33, CM and RN are respectively the medians of ABC and PQR.
If ABC ~ PQR, prove that :
(i) AMC ~ PNR
(ii) CM AB RN PQ
(iii) CMB ~ RNQ Solution : (i) ABC ~ PQR (Given)
So,
AB PQ = BC CA
QR RP
| | (1)
| --- | ---
| and | A = P, | B = Q and C = R | (2)
| --- | --- | --- | ---
| But | AB = 2 AM and PQ = 2 PN | |
| (As CM and RN are medians)
| So, from (1), | 2AM 2PN = | CA RP |
| i.e., | AM PN = | CA RP | | (3)
| --- | --- | --- | --- | ---
| Also, | MAC = | NPR | [From (2)] | (4)
| So, from (3) and (4), | AMC ~ | PNR | (SAS similarity) (5) |
| (ii) From (5), | CM RN = | CA RP | | (6)
| But | CA RP = | AB PQ | [From (1)] (7) |
| Therefore, | CM RN = | AB PQ | [From (6) and (7)] (8) |
| (iii) Again, AB
| PQ =
| BC QR [From (1)]
| Therefore, | CM RN = | BC | QR | [From (8)] (9)
| Also, | CM RN = | AB 2 BM PQ 2 QN | |
| i.e., | CM | | |
| RN =
| BM QN (10)
| i.e., | CM RN = | BC BM QR QN | [From (9) and (10)] |
| Therefore, | CMB ~ | RNQ | (SSS similarity) |
[Note : You can also prove part (iii) by following the same method as used for proving part (i).]Fig. 6.32Fig. 6.33
From the figure, you can see that DE is the shadow of the girl.
Let DE be x metres.
Now, BD = 1.2 m
×
4 = 4.8 m.
Note that in ABE and CDE,
B
= D (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground) = 3.6
| and | E = | E | (Same angle)
| --- | --- | --- | ---
| So, | ABE ~ | CDE | (AA similarity criterion)
| Therefore, | BE | |
| DE = AB CD
| i.e., | 4.8 + x | |
x
0.9
| (90 cm = | 90 100
| i.e., | 4.8 + x = 4x
| i.e., | 3x = 4.8
| i.e., | x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
Example 8 : In Fig. 6.33, CM and RN are respectively the medians of ABC and PQR.
If ABC ~ PQR, prove that :
(i) AMC ~ PNR
(ii) CM AB RN PQ
(iii) CMB ~ RNQ Solution : (i) ABC ~ PQR (Given)(ii) CM AB RN PQ
(iii) CMB ~ RNQ Solution : (i) ABC ~ PQR (Given)So,
AB PQ = BC CA
QR RP
| | (1)
| --- | ---
| and | A = P, | B = Q and C = R | (2)
| --- | --- | --- | ---
| But | AB = 2 AM and PQ = 2 PN | |
| (As CM and RN are medians)
| So, from (1), | 2AM 2PN = | CA RP |
| i.e., | AM PN = | CA RP | | (3)
| --- | --- | --- | --- | ---
| Also, | MAC = | NPR | [From (2)] | (4)
| So, from (3) and (4), | AMC ~ | PNR | (SAS similarity) (5) |
| (ii) From (5), | CM RN = | CA RP | | (6)
| But | CA RP = | AB PQ | [From (1)] (7) |
| Therefore, | CM RN = | AB PQ | [From (6) and (7)] (8) |
| (iii) Again, AB
| PQ =
| BC QR [From (1)]
| Therefore, | CM RN = | BC | QR | [From (8)] (9)
| Also, | CM RN = | AB 2 BM PQ 2 QN | |
| i.e., | CM | | |
| RN =
| BM QN (10)
| i.e., | CM RN = | BC BM QR QN | [From (9) and (10)] |
| Therefore, | CMB ~ | RNQ | (SSS similarity) |
[Note : You can also prove part (iii) by following the same method as used for proving part (i).] | jemh106.pdf |
12 | CBSE | Class10 | Mathematics | QR RP
| | (1)
| --- | ---
| and | A = P, | B = Q and C = R | (2)
| --- | --- | --- | ---
| But | AB = 2 AM and PQ = 2 PN | |
| (As CM and RN are medians)
| So, from (1), | 2AM 2PN = | CA RP |
| i.e., | AM PN = | CA RP | | (3)
| --- | --- | --- | --- | ---
| Also, | MAC = | NPR | [From (2)] | (4)
| So, from (3) and (4), | AMC ~ | PNR | (SAS similarity) (5) |
| (ii) From (5), | CM RN = | CA RP | | (6)
| But | CA RP = | AB PQ | [From (1)] (7) |
| Therefore, | CM RN = | AB PQ | [From (6) and (7)] (8) |
| (iii) Again, AB
| PQ =
| BC QR [From (1)]
| Therefore, | CM RN = | BC | QR | [From (8)] (9)
| Also, | CM RN = | AB 2 BM PQ 2 QN | |
| i.e., | CM | | |
| RN =
| BM QN (10)
| i.e., | CM RN = | BC BM QR QN | [From (9) and (10)] |
| Therefore, | CMB ~ | RNQ | (SSS similarity) |
[Note : You can also prove part (iii) by following the same method as used for proving part (i).]EXERCISE 6.3
1. State which pairs of triangles in Fig. 6.34 are similar.
Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Fig. 6.34
2. In Fig
6.35, ODC ~ OBA, BOC = 125° and CDO = 70°.
Find DOC, DCO and OAB.
3. Diagonals AC and BD of a trapezium ABCD
with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA OB OC OD Fig. 6.35
4. In Fig
6.36, QR QT
QS PR
and
1
= 2.
Show that PQS ~ TQR.
5. S and T are points on sides PR and QR of PQR such that P = RTS
and QR of PQR such that
P
= RTS.
Show that RPQ ~ RTS.
Fig. 6.36
6. In Fig
6.37, if ABE ACD, show that ADE ~ ABC.
7. In Fig
6.38, altitudes AD and CE of ABC intersect each other at the point P. Show that:
(i) AEP ~ CDP
(ii) ABD ~ CBE
(iii) AEP ~ ADB
Fig. 6.37
(iv) PDC ~ BEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE ~ CFB.
9. In Fig
6.39, ABC and AMP are two right triangles, right angled at B and M respectively.
Prove that:
(i) ABC ~ AMP
Fig. 6.38
(ii) CA BC PA MP
10. CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively.
If ABC ~ FEG, show that:
(i) CD AC GH FG
(ii) DCB ~ HGE
Fig. 6.39
(iii) DCA ~ HGF
11. In Fig
6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC.
If AD BC and EF AC, prove that ABD ~ ECF.
12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of PQR (see Fig. 6.41).
Show that ABC ~ PQR.
Fig. 6.40
13.
D is a point on the side BC of a triangle ABC such that ADC = BAC.
Show that CA2 = CB.CD.
14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.
Show that ABC ~ PQR.
Fig. 6.41
15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long.
Find the height of the tower.
16.
If AD and PM are medians of triangles ABC and PQR, respectively where
| ABC ~ PQR, prove that | AB | AD
| --- | --- | ---
| PQ | PM |
6.5 Summary
In this chapter you have studied the following points :
1. Two figures having the same shape but not necessarily the same size are called similar figures.
2. All the congruent figures are similar but the converse is not true.
3. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).
4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
6. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion).
7. If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity criterion).
8. If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity criterion).
9. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar (SAS similarity criterion).Fig. 6.342. In Fig
6.35, ODC ~ OBA, BOC = 125° and CDO = 70°.
Find DOC, DCO and OAB.
3. Diagonals AC and BD of a trapezium ABCD
with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA OB OC OD Fig. 6.35 | jemh106.pdf |
13 | CBSE | Class10 | Mathematics | QS PR
and
1
= 2.
Show that PQS ~ TQR.5. S and T are points on sides PR and QR of PQR such that P = RTS
and QR of PQR such that
P
= RTS.
Show that RPQ ~ RTS.
Fig. 6.36Fig. 6.366. In Fig
6.37, if ABE ACD, show that ADE ~ ABC.7. In Fig
6.38, altitudes AD and CE of ABC intersect each other at the point P. Show that:
(i) AEP ~ CDP
(ii) ABD ~ CBE
(iii) AEP ~ ADB
Fig. 6.37
(iv) PDC ~ BEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE ~ CFB.Fig. 6.37
(iv) PDC ~ BEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE ~ CFB.9. In Fig
6.39, ABC and AMP are two right triangles, right angled at B and M respectively.
Prove that:
(i) ABC ~ AMP
Fig. 6.38
(ii) CA BC PA MP
10. CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively.
If ABC ~ FEG, show that:
(i) CD AC GH FG
(ii) DCB ~ HGE
Fig. 6.39
(iii) DCA ~ HGFFig. 6.38
(ii) CA BC PA MP
10. CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively.
If ABC ~ FEG, show that:
(i) CD AC GH FG
(ii) DCB ~ HGE(i) CD AC GH FG
(ii) DCB ~ HGEFig. 6.39
(iii) DCA ~ HGF11. In Fig
6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC.
If AD BC and EF AC, prove that ABD ~ ECF.
12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of PQR (see Fig. 6.41).
Show that ABC ~ PQR.
Fig. 6.40
13.
D is a point on the side BC of a triangle ABC such that ADC = BAC.
Show that CA2 = CB.CD.
14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.
Show that ABC ~ PQR.
Fig. 6.41
15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long.
Find the height of the tower.
16.
If AD and PM are medians of triangles ABC and PQR, respectively where
| ABC ~ PQR, prove that | AB | AD
| --- | --- | ---
| PQ | PM |
6.5 Summary
In this chapter you have studied the following points :
1. Two figures having the same shape but not necessarily the same size are called similar figures.
2. All the congruent figures are similar but the converse is not true.
3. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).
4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
6. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion).
7. If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity criterion).
8. If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity criterion).
9. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar (SAS similarity criterion). | jemh106.pdf |
0 | CBSE | Class10 | Mathematics | COORDINATE GEOMETRY
7.1 Introduction
In Class IX, you have studied that to locate the position of a point on a plane, we require a pair of coordinate axes.
The distance of a point from the y-axis is called its x-coordinate, or abscissa.
The distance of a point from the x-axis is called its y-coordinate, or ordinate.
The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y).
Here is a play for you.
Draw a set of a pair of perpendicular axes on a graph paper.
Now plot the following points and join them as directed: Join the point A(4, 8) to B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to form a triangle.
Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle.
Lastly join S to the points (0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6).
What picture have you got?
Also, you have seen that a linear equation in two variables of the form ax + by +
c
= 0, (a, b are not simultaneously zero), when represented graphically, gives a straight line.
Further, in Chapter 2, you have seen the graph of y = ax2 + bx + c (a ¹ 0), is a parabola.
In fact, coordinate geometry has been developed as an algebraic tool for studying geometry of figures.
It helps us to study geometry using algebra, and understand algebra with the help of geometry.
Because of this, coordinate geometry is widely applied in various fields such as physics, engineering, navigation, seismology and art! In this chapter, you will learn how to find the distance between the two points whose coordinates are given.
You will also study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio.
7.2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it.
Let us see.
This situation can be represented graphically as shown in Fig. 7.1.
You may use the Pythagoras Theorem to calculate this distance.
Fig. 7.1
Now, suppose two points lie on the x-axis.
Can we find the distance between them?
For instance, consider two points A(4, 0) and B(6, 0) in Fig. 7.2.
The points A and B lie on the x-axis.
From the figure you can see that OA = 4 units and OB = 6 units.
Therefore, the distance of B from A, i.e., AB = OB – OA =
6
– 4 = 2 units.
So, if two points lie on the x-axis, we can easily find the distance between them.
Now, suppose we take two points lying on the y-axis.
Can you find the distance between them.
If the points C(0, 3) and D(0, 8) lie on the y-axis, similarly we find that CD =
8
– 3 = 5 units (see Fig. 7.2).
Fig. 7.2
Next, can you find the distance of A from C (in Fig. 7.2)?
Since OA = 4 units and OC = 3 units, the distance of A from C, i.e., AC = 2 23 4 = 5 units.
Similarly, you can find the distance of B from D = BD = 10 units.
Now, if we consider two points not lying on coordinate axis, can we find the distance between them?
Yes!
We shall use Pythagoras theorem to do so.
Let us see an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant.
How do we use Pythagoras theorem to find the distance between them?
Let us draw PR and QS perpendicular to the x-axis from P and Q respectively.
Also, draw a perpendicular from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0), respectively.
So, RS = 2 units.
Also, QS = 8 units and TS = PR = 6 units.
Therefore, QT = 2 units and PT = RS = 2 units.
Now, using the Pythagoras theorem, we have
PQ2 = PT2 + QT2
= 22 + 22 = 8 So, PQ =
2
2 units How will we find the distance between two points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3) (see Fig. 7.4).
Draw QS perpendicular to the x-axis.
Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point R.
Fig. 7.3
Fig. 7.4
Then PT = 11 units and QT = 7 units.
(Why?) Using the Pythagoras Theorem to the right triangle PTQ, we get PQ = 2 211 7 = 170 units.
Let us now find the distance between any two points P(x 1
,
y 1 ) and Q(x 2
,
y 2 ). Draw PR and QS perpendicular to the x-axis.
A perpendicular from the point P on QS is drawn to meet it at the point T (see Fig. 7.5).
Then, OR = x
1 , OS = x 2.
So, RS =
x
2 – x
1
= PT.
Also, SQ = y
2 1 1
T = PR = y
1. So, QT =
y
2 – y 1.
Now, applying the Pythagoras theorem in PTQ, we get
Fig. 7.5
PQ2 = PT2 + QT2
| = (x | 2 – x 1 )2 + (y | 2 – y 1 )2
| --- | --- | ---
| Therefore, | PQ = | | 2 2
| --- | --- | --- | ---
| 2 | 1 | 2 | 1x x | y | y | |
| --- | --- | --- | --- | --- | --- | --- | ---
Note that since distance is always non-negative, we take only the positive square root.
So, the distance between the points P(x, y) and Q(x which is called the distance formula.
Remarks :
| | 2 , y 2
| --- | ---
| PQ = | 2 | 1 x | 1– + | 1– + y | 2 1– + | 1– + –x | 1– + y | 1– + | 2 2 | ,
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| ) is
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
| OP = | 2 | y. | 2x
| --- | --- | --- | ---
| 1 | 2 2x x | 2x x | 1 2x x y | 2x x | 2x x y | 2x x | . (Why?)
| --- | --- | --- | --- | --- | --- | --- | ---
| 2. We can also write, PQ = 2 2
Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle?
If so, name the type of triangle formed.
Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points.
We have
| PQ = | 2) 2 | (2 | | 3) 2 | | 5 | 2 | | 5 | 2(3 | | 50 | = 7.07 (approx.)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| QR = | 2) 2 2(–2 – | (–3 2(–2 – | – 2(–2 – | 3) 2 2(–2 – | 2(–2 – | (– 2(–2 – | 4) 2 2(–2 – | 2(–2 – | 2(–2 – | (– 2(–2 – | 6) 2(–2 – | 2(–2 – | | 52 | = 7.21 (approx.)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR = | 2) 2 | (2 | – | 3) 2 | | 1 | 2 | | ( | | 1) | 2(3 – | | 2 | = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle.
Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have
P
= 90°.
Therefore, PQR is a right triangle.
Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points.
One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal.
Now,
| AB = 2(1 – | 4) 2 2(1 – | (7 2(1 – | 2) 2(1 – | 2(1 – | | | 9 25 | | | 34 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BC = | 1) 2 | (2 34 | 1) 2(4 34 | 34 | 34 | 25 34 | 9 34 | 34 | 34 | |
| CD = | 4) 2 | (–1 – 4) 2(–1 34 | 34 | 34 | 34 | 34 | 9 25 34 | 34 | 34 | 34 |
| DA = | 4) 2 | (7 – 4) 2(1 34 | 34 | 34 | 34 | 34 | 25 9 34 | 34 | 34 | 34 |
| AC = | 1) 2 68 | (7 1) 2(1 68 | 68 | 68 | 68 | 4 68 | 64 68 | 68 | | |
| BD = | 4) 2 | (2 68 | 4) 2(4 68 | 68 | 68 | 68 | 64 4 68 | 68 | 68 | 68 |
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.
Thereore, ABCD is a square.
Alternative Solution : We find the four sides and one diagonal, say, AC as above.
Here AD2 + DC2 = 34 + 34 = 68 = AC2.
Therefore, by the converse of Pythagoras theorem,
D
= 90°.
A quadrilateral with all four sides equal and one angle 90° is a square.
So, ABCD is a square.
Example 3 : Fig. 7.6 shows the arrangement of desks in a classroom.
Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a line?
Give reasons for your answer.
Fig. 7.6
Solution : Using the distance formula, we have
| AB = | 3) 2 | (4 | | 1) 2(6 | | 9 | 9 | | 18 | | 3 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BC = | 6) 2 | (6 | – | 4) | 2(8 – | | 4 | | 4 | | 8 | | 2 | 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| AC = 2(8 – | 3) 2 2(8 – | (6 2(8 – | – 2(8 – | 1) 2(8 – | | | 25 | | 25 | | | 50 | | 5
Since, AB + BC =
3
2 2 2
5
2 AC, we can say that the points A, B and C are collinear.
Therefore, they are seated in a line.
Example 4 : Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).
Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
We are given that AP = BP. So, AP2 = BP2
| i.e., | (x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2
| --- | ---
| i.e., | x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25
| i.e., | x – y = 2
which is the required relation.
Remark : Note that the graph of the equation x
–
y = 2 is a line.
From your earlier studies, you know that a point which is equidistant from A and B lies on the perpendicular bisector of AB.
Therefore, the graph of x
–
y = 2 is the perpendicular bisector of AB (see Fig. 7.7).
Example 5 : Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Solution : We know that a point on the y-axis is of the form (0, y).
So, let the point P(0, y) be equidistant from A and B. Then
Fig. 7.7
Fig. 7.8
(6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2 So, the required point is (0, 9).
| i.e., | 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y
| i.e., | 4y = 36
| i.e., | y = 9
| Let us check our solution : | AP = | 0) 2 | (5 | – | 9) | 2(6 – | | 36 | | 16 | | 52
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BP = | – | 0) 2 | (3 | – | 9) | 2(– 4 | | 16 | | 36 | | 52
Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB. | jemh107.pdf |
1 | CBSE | Class10 | Mathematics | 7.2 Distance Formula
Let us consider the following situation:
A town B is located 36 km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it.
Let us see.
This situation can be represented graphically as shown in Fig. 7.1.
You may use the Pythagoras Theorem to calculate this distance.
Fig. 7.1
Now, suppose two points lie on the x-axis.
Can we find the distance between them?
For instance, consider two points A(4, 0) and B(6, 0) in Fig. 7.2.
The points A and B lie on the x-axis.
From the figure you can see that OA = 4 units and OB = 6 units.
Therefore, the distance of B from A, i.e., AB = OB – OA =
6
– 4 = 2 units.
So, if two points lie on the x-axis, we can easily find the distance between them.
Now, suppose we take two points lying on the y-axis.
Can you find the distance between them.
If the points C(0, 3) and D(0, 8) lie on the y-axis, similarly we find that CD =
8
– 3 = 5 units (see Fig. 7.2).
Fig. 7.2
Next, can you find the distance of A from C (in Fig. 7.2)?
Since OA = 4 units and OC = 3 units, the distance of A from C, i.e., AC = 2 23 4 = 5 units.
Similarly, you can find the distance of B from D = BD = 10 units.
Now, if we consider two points not lying on coordinate axis, can we find the distance between them?
Yes!
We shall use Pythagoras theorem to do so.
Let us see an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant.
How do we use Pythagoras theorem to find the distance between them?
Let us draw PR and QS perpendicular to the x-axis from P and Q respectively.
Also, draw a perpendicular from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0), respectively.
So, RS = 2 units.
Also, QS = 8 units and TS = PR = 6 units.
Therefore, QT = 2 units and PT = RS = 2 units.
Now, using the Pythagoras theorem, we have
PQ2 = PT2 + QT2
= 22 + 22 = 8 So, PQ =
2
2 units How will we find the distance between two points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3) (see Fig. 7.4).
Draw QS perpendicular to the x-axis.
Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point R.
Fig. 7.3
Fig. 7.4
Then PT = 11 units and QT = 7 units.
(Why?) Using the Pythagoras Theorem to the right triangle PTQ, we get PQ = 2 211 7 = 170 units.
Let us now find the distance between any two points P(x 1
,
y 1 ) and Q(x 2
,
y 2 ). Draw PR and QS perpendicular to the x-axis.
A perpendicular from the point P on QS is drawn to meet it at the point T (see Fig. 7.5).
Then, OR = x
1 , OS = x 2.
So, RS =
x
2 – x
1
= PT.
Also, SQ = y
2 1 1
T = PR = y
1. So, QT =
y
2 – y 1.
Now, applying the Pythagoras theorem in PTQ, we get
Fig. 7.5
PQ2 = PT2 + QT2
| = (x | 2 – x 1 )2 + (y | 2 – y 1 )2
| --- | --- | ---
| Therefore, | PQ = | | 2 2
| --- | --- | --- | ---
| 2 | 1 | 2 | 1x x | y | y | |
| --- | --- | --- | --- | --- | --- | --- | ---
Note that since distance is always non-negative, we take only the positive square root.
So, the distance between the points P(x, y) and Q(x which is called the distance formula.
Remarks :
| | 2 , y 2
| --- | ---
| PQ = | 2 | 1 x | 1– + | 1– + y | 2 1– + | 1– + –x | 1– + y | 1– + | 2 2 | ,
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| ) is
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
| OP = | 2 | y. | 2x
| --- | --- | --- | ---
| 1 | 2 2x x | 2x x | 1 2x x y | 2x x | 2x x y | 2x x | . (Why?)
| --- | --- | --- | --- | --- | --- | --- | ---
| 2. We can also write, PQ = 2 2
Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle?
If so, name the type of triangle formed.
Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points.
We have
| PQ = | 2) 2 | (2 | | 3) 2 | | 5 | 2 | | 5 | 2(3 | | 50 | = 7.07 (approx.)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| QR = | 2) 2 2(–2 – | (–3 2(–2 – | – 2(–2 – | 3) 2 2(–2 – | 2(–2 – | (– 2(–2 – | 4) 2 2(–2 – | 2(–2 – | 2(–2 – | (– 2(–2 – | 6) 2(–2 – | 2(–2 – | | 52 | = 7.21 (approx.)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR = | 2) 2 | (2 | – | 3) 2 | | 1 | 2 | | ( | | 1) | 2(3 – | | 2 | = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle.
Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have
P
= 90°.
Therefore, PQR is a right triangle.
Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points.
One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal.
Now,
| AB = 2(1 – | 4) 2 2(1 – | (7 2(1 – | 2) 2(1 – | 2(1 – | | | 9 25 | | | 34 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BC = | 1) 2 | (2 34 | 1) 2(4 34 | 34 | 34 | 25 34 | 9 34 | 34 | 34 | |
| CD = | 4) 2 | (–1 – 4) 2(–1 34 | 34 | 34 | 34 | 34 | 9 25 34 | 34 | 34 | 34 |
| DA = | 4) 2 | (7 – 4) 2(1 34 | 34 | 34 | 34 | 34 | 25 9 34 | 34 | 34 | 34 |
| AC = | 1) 2 68 | (7 1) 2(1 68 | 68 | 68 | 68 | 4 68 | 64 68 | 68 | | |
| BD = | 4) 2 | (2 68 | 4) 2(4 68 | 68 | 68 | 68 | 64 4 68 | 68 | 68 | 68 |
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.
Thereore, ABCD is a square.
Alternative Solution : We find the four sides and one diagonal, say, AC as above.
Here AD2 + DC2 = 34 + 34 = 68 = AC2.
Therefore, by the converse of Pythagoras theorem,
D
= 90°.
A quadrilateral with all four sides equal and one angle 90° is a square.
So, ABCD is a square.
Example 3 : Fig. 7.6 shows the arrangement of desks in a classroom.
Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a line?
Give reasons for your answer.
Fig. 7.6
Solution : Using the distance formula, we have
| AB = | 3) 2 | (4 | | 1) 2(6 | | 9 | 9 | | 18 | | 3 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BC = | 6) 2 | (6 | – | 4) | 2(8 – | | 4 | | 4 | | 8 | | 2 | 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| AC = 2(8 – | 3) 2 2(8 – | (6 2(8 – | – 2(8 – | 1) 2(8 – | | | 25 | | 25 | | | 50 | | 5
Since, AB + BC =
3
2 2 2
5
2 AC, we can say that the points A, B and C are collinear.
Therefore, they are seated in a line.
Example 4 : Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).
Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
We are given that AP = BP. So, AP2 = BP2
| i.e., | (x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2
| --- | ---
| i.e., | x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25
| i.e., | x – y = 2
which is the required relation.
Remark : Note that the graph of the equation x
–
y = 2 is a line.
From your earlier studies, you know that a point which is equidistant from A and B lies on the perpendicular bisector of AB.
Therefore, the graph of x
–
y = 2 is the perpendicular bisector of AB (see Fig. 7.7).
Example 5 : Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Solution : We know that a point on the y-axis is of the form (0, y).
So, let the point P(0, y) be equidistant from A and B. Then
Fig. 7.7
Fig. 7.8
(6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2 So, the required point is (0, 9).
| i.e., | 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y
| i.e., | 4y = 36
| i.e., | y = 9
| Let us check our solution : | AP = | 0) 2 | (5 | – | 9) | 2(6 – | | 36 | | 16 | | 52
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BP = | – | 0) 2 | (3 | – | 9) | 2(– 4 | | 16 | | 36 | | 52
Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB. | jemh107.pdf |
2 | CBSE | Class10 | Mathematics | Fig. 7.1
Now, suppose two points lie on the x-axis.
Can we find the distance between them?
For instance, consider two points A(4, 0) and B(6, 0) in Fig. 7.2.
The points A and B lie on the x-axis.
From the figure you can see that OA = 4 units and OB = 6 units.
Therefore, the distance of B from A, i.e., AB = OB – OA =
6
– 4 = 2 units.
So, if two points lie on the x-axis, we can easily find the distance between them.
Now, suppose we take two points lying on the y-axis.
Can you find the distance between them.
If the points C(0, 3) and D(0, 8) lie on the y-axis, similarly we find that CD =
8
– 3 = 5 units (see Fig. 7.2).Fig. 7.2
Next, can you find the distance of A from C (in Fig. 7.2)?
Since OA = 4 units and OC = 3 units, the distance of A from C, i.e., AC = 2 23 4 = 5 units.
Similarly, you can find the distance of B from D = BD = 10 units.
Now, if we consider two points not lying on coordinate axis, can we find the distance between them?
Yes!
We shall use Pythagoras theorem to do so.
Let us see an example.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant.
How do we use Pythagoras theorem to find the distance between them?
Let us draw PR and QS perpendicular to the x-axis from P and Q respectively.
Also, draw a perpendicular from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0), respectively.
So, RS = 2 units.
Also, QS = 8 units and TS = PR = 6 units.
Therefore, QT = 2 units and PT = RS = 2 units.
Now, using the Pythagoras theorem, we have
PQ2 = PT2 + QT2
= 22 + 22 = 8 So, PQ =
2
2 units How will we find the distance between two points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3) (see Fig. 7.4).
Draw QS perpendicular to the x-axis.
Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point R.PQ2 = PT2 + QT2
= 22 + 22 = 8 So, PQ =
2
2 units How will we find the distance between two points in two different quadrants?
Consider the points P(6, 4) and Q(–5, –3) (see Fig. 7.4).
Draw QS perpendicular to the x-axis.
Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point R.Fig. 7.3
Fig. 7.4
Then PT = 11 units and QT = 7 units.
(Why?) Using the Pythagoras Theorem to the right triangle PTQ, we get PQ = 2 211 7 = 170 units.
Let us now find the distance between any two points P(x 1
,
y 1 ) and Q(x 2
,
y 2 ). Draw PR and QS perpendicular to the x-axis.
A perpendicular from the point P on QS is drawn to meet it at the point T (see Fig. 7.5).
Then, OR = x
1 , OS = x 2.
So, RS =
x
2 – x
1
= PT.
Also, SQ = y
2 1 1
T = PR = y
1. So, QT =
y
2 – y 1.
Now, applying the Pythagoras theorem in PTQ, we getFig. 7.4
Then PT = 11 units and QT = 7 units.
(Why?) Using the Pythagoras Theorem to the right triangle PTQ, we get PQ = 2 211 7 = 170 units.
Let us now find the distance between any two points P(x 1
,
y 1 ) and Q(x 2
,
y 2 ). Draw PR and QS perpendicular to the x-axis.
A perpendicular from the point P on QS is drawn to meet it at the point T (see Fig. 7.5).
Then, OR = x
1 , OS = x 2.
So, RS =
x
2 – x
1
= PT.
Also, SQ = y
2 1 1
T = PR = y
1. So, QT =
y
2 – y 1.
Now, applying the Pythagoras theorem in PTQ, we getThen, OR = x
1 , OS = x 2.
So, RS =
x
2 – x
1
= PT.Also, SQ = y
2 1 1
T = PR = y
1. So, QT =
y
2 – y 1.
Now, applying the Pythagoras theorem in PTQ, we get | jemh107.pdf |
3 | CBSE | Class10 | Mathematics | PQ2 = PT2 + QT2
| = (x | 2 – x 1 )2 + (y | 2 – y 1 )2
| --- | --- | ---
| Therefore, | PQ = | | 2 2
| --- | --- | --- | ---
| 2 | 1 | 2 | 1x x | y | y | |
| --- | --- | --- | --- | --- | --- | --- | ---
Note that since distance is always non-negative, we take only the positive square root.
So, the distance between the points P(x, y) and Q(x which is called the distance formula.
Remarks :
| | 2 , y 2
| --- | ---
| PQ = | 2 | 1 x | 1– + | 1– + y | 2 1– + | 1– + –x | 1– + y | 1– + | 2 2 | ,
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| ) is
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
| OP = | 2 | y. | 2x
| --- | --- | --- | ---
| 1 | 2 2x x | 2x x | 1 2x x y | 2x x | 2x x y | 2x x | . (Why?)
| --- | --- | --- | --- | --- | --- | --- | ---
| 2. We can also write, PQ = 2 2
Example 1 : Do the points (3, 2), (–2, –3) and (2, 3) form a triangle?
If so, name the type of triangle formed.
Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points.
We have
| PQ = | 2) 2 | (2 | | 3) 2 | | 5 | 2 | | 5 | 2(3 | | 50 | = 7.07 (approx.)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| QR = | 2) 2 2(–2 – | (–3 2(–2 – | – 2(–2 – | 3) 2 2(–2 – | 2(–2 – | (– 2(–2 – | 4) 2 2(–2 – | 2(–2 – | 2(–2 – | (– 2(–2 – | 6) 2(–2 – | 2(–2 – | | 52 | = 7.21 (approx.)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR = | 2) 2 | (2 | – | 3) 2 | | 1 | 2 | | ( | | 1) | 2(3 – | | 2 | = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle.
Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have
P
= 90°.
Therefore, PQR is a right triangle.
Example 2 : Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points.
One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal.
Now,
| AB = 2(1 – | 4) 2 2(1 – | (7 2(1 – | 2) 2(1 – | 2(1 – | | | 9 25 | | | 34 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| BC = | 1) 2 | (2 34 | 1) 2(4 34 | 34 | 34 | 25 34 | 9 34 | 34 | 34 | |
| CD = | 4) 2 | (–1 – 4) 2(–1 34 | 34 | 34 | 34 | 34 | 9 25 34 | 34 | 34 | 34 |
| DA = | 4) 2 | (7 – 4) 2(1 34 | 34 | 34 | 34 | 34 | 25 9 34 | 34 | 34 | 34 |
| AC = | 1) 2 68 | (7 1) 2(1 68 | 68 | 68 | 68 | 4 68 | 64 68 | 68 | | |
| BD = | 4) 2 | (2 68 | 4) 2(4 68 | 68 | 68 | 68 | 64 4 68 | 68 | 68 | 68 |
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.
Thereore, ABCD is a square.
Alternative Solution : We find the four sides and one diagonal, say, AC as above.
Here AD2 + DC2 = 34 + 34 = 68 = AC2.
Therefore, by the converse of Pythagoras theorem,
D
= 90°.
A quadrilateral with all four sides equal and one angle 90° is a square.
So, ABCD is a square.
Example 3 : Fig. 7.6 shows the arrangement of desks in a classroom.
Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a line?
Give reasons for your answer. | jemh107.pdf |
4 | CBSE | Class10 | Mathematics | EXERCISE 7.1
1. Find the distance between the following pairs of points :
(i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b)
2. Find the distance between the points (0, 0) and (36, 15).
Can you now find the distance between the two towns A and B discussed in Section 7.2.
3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8.
Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?”
Chameli disagrees.
Using distance formula, find which of them is correct.
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
7.3 Section Formula
Let us recall the situation in Section 7.2.
Suppose a telephone company wants to position a relay tower at P between A and B is such a way that the distance of the tower from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1
:
2 (see Fig. 7.9).
If we take A as the origin O, and 1 km as one unit on both the axis, the coordinates of B will be (36, 15).
In order to know the position of the tower, we must know the coordinates of P. How do we find these coordinates?
Fig. 7.9
Let the coordinates of P be (x, y).
Draw perpendiculars from P and B to the x-axis, meeting it in D and E, respectively.
Draw PC perpendicular to BE.
Then, by the AA similarity criterion, studied in Chapter 6, POD and BPC are similar.
Therefore ,
OD OP 1 PC PB 2
, and
PD OP 1 BC PB 2
So, 1
36 2 x x and 1 15 2 y y
These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB =
1
: 2.
Now let us use the understanding that you may have developed through this example to obtain the general formula.
Consider any two points A(x, y
1 1
) and B(x, y
2
2) and assume that P (x, y) divides AB internally in the ratio m
1
:
m 2, i.e.,
Fig. 7.10
1 2
PA PB m m
(see Fig. 7.10).
Draw AR, PS and BT perpendicular to the x-axis.
Draw AQ and PC parallel to the x-axis.
Then, by the AA similarity criterion, PAQ ~ BPC
| , y | ) and B(x
| --- | ---
| 1 | 1
| --- | ---
| Therefore, | PA | AQ
| --- | --- | ---
| Now, | AQ = RS = OS – OR = x – x
| --- | ---
| | | BP | | PC | = | PQ BC | | | | | (1) | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| 1
| PC = ST = OT – OS = x
| 2 – x
| PQ = PS – QS = PS – AR = y – y
| 1
| BC = BT– CT = BT – PS = y
| 2 – y
| | m m | | x | x | y | y | | | | | | |
| | 1 2 | = | | 1 | | 1 | | | | | | |
| x x y y
| | | | 2 | | 2 | | | | | | | |
| Substituting these values in (1), we get
| Taking | 1 2 m m | = | 2 x x | 1 x x | , we get x = | | 1 2 1 2 m x m | 2 m x | 2 m x m m | 2 m x | 2 | | 1 x
| Similarly, taking | 1 2 m m | = | 2 y y | 1 y y | , we get y = | | 1 2 1 2 m y m | 2 m y | 2 m y m m | 2 m y | 2 2 m y | 2 m y | 1 y
So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x
| , y | ), internally, in the ratio m
| --- | ---
| 2 | 2 1 : m
| |
| | x | m x | m y | m y
| --- | --- | --- | --- | ---
| | 1 | 2 | 2 | 1 | 1 | 2 | 2 | 1 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | | | | | | | |
| | m | m | | | m | m | | |
| | | 1 | 2 | 2,m | 2,m | 1 2,m | | |
2 are
(2)
| 1 ) and B(x | | | | | 1
| --- | --- | --- | --- | --- | ---
| This is known as the section formula.
| This can also be derived by drawing perpendiculars from A, P and B on the y-axis and proceeding as above.
| If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be
| | | | | |
| kx x ky y k k | | | | |
| 2 1 2 1, | | | | |
| 1 1 | | | | |
Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1.
Therefore, the coordinates of the mid-point P of the join of the points A(x, y
| 2 | 2 ) is
| --- | ---
| x | | x | | y | | | y | | | | x | | x y | y |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| 1 1 | | 2 1 | | 1 | 1, | | 2 | | | | 1 | | 2 1 , | 21 | .
| 1 | 1 | | | 1 | | 1 | | | | | | 2 | | 2 |
Let us solve a few examples based on the section formula.
Example 6 : Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3
:
1 internally.
Solution : Let P(x, y) be the required point.
Using the section formula, we get
Therefore, (7, 3) is the required point.
Example 7 : In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?
Solution : Let (– 4, 6) divide AB internally in the ratio m formula, we get
, y = = 3(8) 1(4) 7 3 1 3(5) 1(–3) 3 3 1
Recall that if (x, y) = (a, b) then x
=
a and y = b.
So,
1
:
m 2
Using the section
| | |
| --- | --- | ---
| | 3 | | –8 m | | | 6 10,m m m
| --- | --- | --- | --- | --- | --- | ---
| (– 4, 6) = | 1 | 2 | | 1 | | 2
| | | | | | |
| | m | | m | | m | m
| | 1 | | 1 | | 2 | 2
(1)
| 3 | 6m m
| --- | ---
| 4 = | 1 | 2
| --- | --- | ---
| m | m | | and | 1 | 2
| --- | --- | --- | --- | --- | ---
| 1 | | 2 | | |
| 3 | | 6m m | | |
| 4 = 1 | | 2 | | |
| m | | m | gives us | |
| 1 | | 2 | | |
| 4m 4m 6m 1 2 = 3m 1 | | 2 | | |
| 1 = 2m 2 | | | | |
| 1 : m 2 = 2 : 7 | | | | |
m
m
m m
1
| 8 |
| 8 10m m |
| 2
| = | (Dividing throughout by m 2 )
| m | m
| 1 | 1
| 1 | 2
2
2 8 10 7
6
2 1 = 7
| 6 | m | m
| --- | --- | ---
| | 8 | 10
| m m
| | 1 2 |
| Now, | |
| --- | --- | ---
| i.e., | | 7m
| i.e., | m |
You should verify that the ratio satisfies the y-coordinate also.
Now, 1 2
Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Alternatively : The ratio m :1, m m or k : 1.
Let (– 4, 6) divide AB internally in the ratio k : 1.
Using the section formula, we get
| 1 : m | 2 can also be written as | 1 2
| --- | --- | ---
| (– 4, 6) = 3 1 k k | 6 1 1 k | 8 1 k k k | 1 k | 1 k | 10, | | (2)
| --- | --- | --- | --- | --- | --- | --- | ---
4
=
3 6 1
k
k
4k –
4
= 3k – 6
7k =
2
k : 1
=
2 : 7
So, i.e.,
i.e., i.e., You can check for the y-coordinate also.
So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Note : You can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear.
Example 8 : Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).
Solution : Let P and Q be the points of trisection of AB i.e., AP = PQ = QB
Fig. 7.11
(see Fig. 7.11).
Therefore, P divides AB internally in the ratio 1 : 2.
Therefore, the coordinates of P, by applying the section formula, are
| | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| 1(7) 2(2) 1(4) 2(2),
| | | | | | | | |
1
2
1
2 , i.e., (–1, 0)
, i.e., (– 4, 2) Now, Q also divides AB internally in the ratio 2 : 1.
So, the coordinates of Q are Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).
Note : We could also have obtained Q by noting that it is the mid-point of PB.
So, we could have obtained its coordinates using the mid-point formula.
| | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| 2(7) 1(2) 2(4) 1(2),
| | | | | | |
| | | 2 | 1 | 2 | 1 |
Example 9 : Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4).
Also find the point of intersection.
Solution : Let the ratio be k : 1.
Then by the section formula, the coordinates of the
point which divides AB in the ratio k
:
1 are 5 4 6,
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
| Therefore, | 5
| So, | k = 5
That is, the ratio is 5 : 1.
Putting the value of k = 5, we get the point of intersection as = 0
| . | 13 0, 3
| --- | ---
| | | | |
| --- | --- | --- | --- | ---
| k | k | | |
| k 1 | k | 1 | |
k k
1
Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
Solution : We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
| i.e., | 6 | 9 | 1 | 4,
| --- | --- | --- | --- | ---
| |
| --- | ---
| 2 | 2
| --- | ---
| | | | = | | 8 | 2 | p | 5, 2 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = 8 | 2 | 3, | 2 | 2 p | | | | |
| 2 2
| i.e., | 15 5, | | | | | | | |
| so, | 15 2 = 8 2 p | | | | | | | |
| i.e., | | p = 7 | | | | | | |
| jemh107.pdf |
5 | CBSE | Class10 | Mathematics | 7.3 Section Formula
Let us recall the situation in Section 7.2.
Suppose a telephone company wants to position a relay tower at P between A and B is such a way that the distance of the tower from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1
:
2 (see Fig. 7.9).
If we take A as the origin O, and 1 km as one unit on both the axis, the coordinates of B will be (36, 15).
In order to know the position of the tower, we must know the coordinates of P. How do we find these coordinates?
Fig. 7.9
Let the coordinates of P be (x, y).
Draw perpendiculars from P and B to the x-axis, meeting it in D and E, respectively.
Draw PC perpendicular to BE.
Then, by the AA similarity criterion, studied in Chapter 6, POD and BPC are similar.
Therefore ,
OD OP 1 PC PB 2
, and
PD OP 1 BC PB 2
So, 1
36 2 x x and 1 15 2 y y
These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB =
1
: 2.
Now let us use the understanding that you may have developed through this example to obtain the general formula.
Consider any two points A(x, y
1 1
) and B(x, y
2
2) and assume that P (x, y) divides AB internally in the ratio m
1
:
m 2, i.e.,
Fig. 7.10
1 2Fig. 7.9
Let the coordinates of P be (x, y).
Draw perpendiculars from P and B to the x-axis, meeting it in D and E, respectively.
Draw PC perpendicular to BE.
Then, by the AA similarity criterion, studied in Chapter 6, POD and BPC are similar.
Therefore ,
OD OP 1 PC PB 2
, and
PD OP 1 BC PB 2
So, 1
36 2 x x and 1 15 2 y y
These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB =
1
: 2.
Now let us use the understanding that you may have developed through this example to obtain the general formula.
Consider any two points A(x, y
1 1
) and B(x, y
2
2) and assume that P (x, y) divides AB internally in the ratio m
1
:
m 2, i.e.,Therefore ,
OD OP 1 PC PB 2
, and
PD OP 1 BC PB 2
OD OP 1 PC PB 2
, andPD OP 1 BC PB 2
So, 1
36 2 x x and 1 15 2 y y
These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB =
1
: 2.
Now let us use the understanding that you may have developed through this example to obtain the general formula.
Consider any two points A(x, y
1 1
) and B(x, y
2
2) and assume that P (x, y) divides AB internally in the ratio m
1
:
m 2, i.e.,36 2 x x and 1 15 2 y y
These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB =
1
: 2.
Now let us use the understanding that you may have developed through this example to obtain the general formula.
Consider any two points A(x, y
1 1
) and B(x, y
2
2) and assume that P (x, y) divides AB internally in the ratio m
1
:
m 2, i.e.,Fig. 7.10
1 2 | jemh107.pdf |
6 | CBSE | Class10 | Mathematics | So,
1
:
m 2
Using the section
| | |
| --- | --- | ---
| | 3 | | –8 m | | | 6 10,m m m
| --- | --- | --- | --- | --- | --- | ---
| (– 4, 6) = | 1 | 2 | | 1 | | 2
| | | | | | |
| | m | | m | | m | m
| | 1 | | 1 | | 2 | 2
(1)
| 3 | 6m m
| --- | ---
| 4 = | 1 | 2
| --- | --- | ---
| m | m | | and | 1 | 2
| --- | --- | --- | --- | --- | ---
| 1 | | 2 | | |
| 3 | | 6m m | | |
| 4 = 1 | | 2 | | |
| m | | m | gives us | |
| 1 | | 2 | | |
| 4m 4m 6m 1 2 = 3m 1 | | 2 | | |
| 1 = 2m 2 | | | | |
| 1 : m 2 = 2 : 7 | | | | |
m
m
m m
1
| 8 |
| 8 10m m |
| 2
| = | (Dividing throughout by m 2 )
| m | m
| 1 | 1
| 1 | 2
2
2 8 10 7
6
2 1 = 7
| 6 | m | m
| --- | --- | ---
| | 8 | 10
| m m
| | 1 2 |
| Now, | |
| --- | --- | ---
| i.e., | | 7m
| i.e., | m |
You should verify that the ratio satisfies the y-coordinate also.
Now, 1 2
Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Alternatively : The ratio m :1, m m or k : 1.
Let (– 4, 6) divide AB internally in the ratio k : 1.
Using the section formula, we get
| 1 : m | 2 can also be written as | 1 2
| --- | --- | ---
| (– 4, 6) = 3 1 k k | 6 1 1 k | 8 1 k k k | 1 k | 1 k | 10, | | (2)
| --- | --- | --- | --- | --- | --- | --- | ---
4
=
3 6 1
k
k
4k –
4
= 3k – 6
7k =
2
k : 1
=
2 : 7
So, i.e.,
i.e., i.e., You can check for the y-coordinate also.
So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Note : You can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear.
Example 8 : Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).
Solution : Let P and Q be the points of trisection of AB i.e., AP = PQ = QB
Fig. 7.11
(see Fig. 7.11).
Therefore, P divides AB internally in the ratio 1 : 2.
Therefore, the coordinates of P, by applying the section formula, are
| | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| 1(7) 2(2) 1(4) 2(2),
| | | | | | | | |
1
2
1
2 , i.e., (–1, 0)
, i.e., (– 4, 2) Now, Q also divides AB internally in the ratio 2 : 1.
So, the coordinates of Q are Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).
Note : We could also have obtained Q by noting that it is the mid-point of PB.
So, we could have obtained its coordinates using the mid-point formula.
| | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| 2(7) 1(2) 2(4) 1(2),
| | | | | | |
| | | 2 | 1 | 2 | 1 |
Example 9 : Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4).
Also find the point of intersection.
Solution : Let the ratio be k : 1.
Then by the section formula, the coordinates of the
point which divides AB in the ratio k
:
1 are 5 4 6,
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
| Therefore, | 5
| So, | k = 5
That is, the ratio is 5 : 1.
Putting the value of k = 5, we get the point of intersection as = 0
| . | 13 0, 3
| --- | ---
| | | | |
| --- | --- | --- | --- | ---
| k | k | | |
| k 1 | k | 1 | |
| jemh107.pdf |
7 | CBSE | Class10 | Mathematics | k k
1
Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
Solution : We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
| i.e., | 6 | 9 | 1 | 4,
| --- | --- | --- | --- | ---
| |
| --- | ---
| 2 | 2
| --- | ---
| | | | = | | 8 | 2 | p | 5, 2 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = 8 | 2 | 3, | 2 | 2 p | | | | |
| 2 2
| i.e., | 15 5, | | | | | | | |
| so, | 15 2 = 8 2 p | | | | | | | |
| i.e., | | p = 7 | | | | | | |
EXERCISE 7.2
1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each.
100 flower pots have been placed at a distance of 1m from each other along AD, as shown
in Fig. 7.12.
Niharika runs th the distance AD on the 2nd line and posts a green flag.
Preet runs th the distance AD on the eighth line and posts a red flag.
What is the distance between both the flags?
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
and P lies on the line segment AB.
1 4
1 5
Fig. 7.12
4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis.
Also find the coordinates of the point of division.
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y.
7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that
AP =
3 AB 7
9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in
order.
[Hint : Area of a rhombus =
1 2
(product of its diagonals)]
7.4 Summary
In this chapter, you have studied the following points :
1. The distance between P(x
2. The distance of a point P(x, y) from the origin is 2 2.x y
3. The coordinates of the point P(x, y) which divides the line segment joining the points A(x
| | |
| --- | --- | ---
| x | m x | m y | m y
| --- | --- | --- | ---
| 1 | 2 | 2 | 1 | 1 | 2 | 2 | 1
| --- | --- | --- | --- | --- | --- | --- | ---
| m | m | | | m | m | |
| | 1 | 2 | 2,m | 2,m | 1 2,m | |
4. The mid-point of the line segment joining the points P(x
x
x
y y
.
| 1 | 2 | 1 | 2,
| --- | --- | --- | ---
| 2 | | 2 |
| ) and Q(x | , y
| --- | ---
| 1 , y 1 | 2 | 2) is | 1 1() .x x | 2 2 2 2 1() ( ) y y |
| --- | --- | --- | --- | --- | ---
1
,
y
1 ) and B(x 2
,
y 2 ) internally in the ratio m
| 1 : m 2 | are
| , y | ) and Q(x
| 1 | 1
, y 2 2) isFig. 7.12
4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis.
Also find the coordinates of the point of division.
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y.
7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that
AP =
3 AB 7
9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in
order.
[Hint : Area of a rhombus =
1 2
(product of its diagonals)]AP =
3 AB 7
9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in
order.
[Hint : Area of a rhombus =
1 2
(product of its diagonals)] | jemh107.pdf |
0 | CBSE | Class10 | Mathematics | INTRODUCTION TO TRIGONOMETRY
There is perhaps nothing which so occupies the middle position of mathematics as trigonometry.
– J.F.
Herbart (1890)
8.1 Introduction
You have already studied about triangles, and in particular, right triangles, in your earlier classes.
Let us take some examples from our surroundings where right triangles can be imagined to be formed.
For instance :
1. Suppose the students of a school are
visiting Qutub Minar.
Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made, as shown in Fig 8.1.
Can the student find out the height of the Minar, without actually measuring it?
river.
She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river.
A right triangle is imagined to be made in this situation as shown in Fig.8.2.
If you know the height at which the person is sitting, can you find the width of the river?
2.
Suppose a girl is sitting on the balcony of her house located on the bank of a
Fig. 8.1
Fig. 8.2
3. Suppose a hot air balloon is flying in the air.
A girl happens to spot the
balloon in the sky and runs to her mother to tell her about it.
Her mother rushes out of the house to look at the balloon.Now when the girl had spotted the balloon intially it was at point A. When both the mother and daughter came out to see it, it had already travelled to another point B. Can you find the altitude of B from the ground?
Fig. 8.3
In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’.
The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure).
In fact, trigonometry is the study of relationships between the sides and angles of a triangle.
The earliest known work on trigonometry was recorded in Egypt and Babylon.
Early astronomers used it to find out the distances of the stars and planets from the Earth.
Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.
In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle.
We will restrict our discussion to acute angles only.
However, these ratios can be extended to other angles also.
We will also define the trigonometric ratios for angles of measure 0° and 90°.
We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.
8.2 Trigonometric Ratios
In Section 8.1, you have seen some right triangles imagined to be formed in different situations.
Let us take a right triangle ABC as shown in Fig. 8.4.
Note that the position of sides change when you consider angle C in place of A (see Fig. 8.5).
| | Here, CAB (or, in brief, angle A) is an
| --- | ---
| acute angle. Note the position of the side BC with respect to angle A. It faces A. We call it the hypotenuse of the right triangle and the side AB is a part of adjacent | side opposite A. So, we call it the to angle A. | to angle A. AC is the side | Fig. 8.4
| --- | --- | --- | ---
You have studied the concept of ‘ratio’ in your earlier classes.
We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios.
The trigonometric ratios of the angle A in right triangle ABC (see Fig. 8.4) are defined as follows :
Fig. 8.5
| sine of A = side opposite to angle A | hypotenuse AC | BC |
| --- | --- | --- | ---
| cosine of A = | side adjacent to angle A hypotenuse AC | AC | AB
tangent of A =
side opposite to angle A BC side adjacent to angle A AB
| cosecant of A = | 1 | hypotenuse | AC
| --- | --- | --- | ---
| sine of | | A | | side opposite to angle A | | | BC |
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| secant of A = 1 | | | | hypotenuse | | | AC |
| cosine of | | A | | side adjacent to angle A | | | AB |
| cotangent of A = | 1 | | | side adjacent to angle A | | | |
| tangent of | | | | A side opposite to angle A BC | BC | | BC |
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively.
Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A.
Also, observe that tan A = BC BC sin AAC
ABAB cos A AC
and cot A =.
cosA sin A
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right triangle?
(See Fig. 8.5)
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500.
Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course.
When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is.
The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin.
Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe.
An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
C.E.
476 – 550
The origin of the terms ‘cosine’ and ‘tangent’ was much later.
The cosine function arose from the need to compute the sine of the complementary angle.
Aryabhatta called it kotijya.
The name cosinus originated with Edmund Gunter.
In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’.
sin A is not the product of ‘sin’ and A.
‘sin’ separated from A has no meaning.
Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.
Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of A in PAM differ from those of A in CAB or from those of A in
Fig. 8.6
QAN?
To answer this, first look at these triangles.
Is PAM similar to CAB?
From Chapter 6, recall the AA similarity criterion.
Using the criterion, you will see that the triangles PAM and CAB are similar.
Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.
So, we have AM AB = AP MP AC BC
From this, we find
| MP AP = | BC sin A AC .
AM AB AP AC
= cos A,
MP BC tan A AM AB
and so on.
This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of (sin A)2, (cos A)2, etc., respectively.
But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A).
sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle.
If we know any one of the ratios, can we obtain the other ratios?
Let us see.
If in a right triangle ABC, sin A = then this means that
1, 3
BC 1
AC 3
, i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1
:
3 (see Fig. 8.7).
So if BC is equal to k, then AC will be 3k, where k is any positive number.
To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB.
Do you remember the Pythagoras theorem?
Let us use it to determine the required length AB.
Fig. 8.7 Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
| | AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 | 2 k)2
| --- | --- | ---
| Therefore, | AB = | 2 | 2 k
| --- | --- | --- | ---
| So, we get | AB = 2 2 k | (Why is AB not – 2 2 k | ?)
| --- | --- | --- | ---
| Now, | cos A = | AB 2 2 | | 2 | | 2
| --- | --- | --- | --- | --- | --- | ---
| | | AC 3 3 k k | 3 k | 3 k | 3 k |
Similarly, you can obtain the other trigonometric ratios of the angle A.
Let us consider some examples.
Example 1 : Given tan A
=
4 3, find the other trigonometric ratios of the angle A.
Solution : Let us first draw a right ABC (see Fig 8.8).
Now, we know that tan A = BC 4
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
Then AC
AB =
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have
AC PR = AB BC PQ QR
Then, by using Theorem 6.4, ACB ~ PRQ and therefore,
B
= Q.
Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10).
Determine the values of
(i) cos2 + sin2 ,
(ii) cos2 – sin2
Solution : In ACB, we have
Fig. 8.10 Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that.
| | AC = | 2 | BC | 2AB | = | 2 | (21) | 2(29)
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = (29 | 21)(29 | 21) 20units | 20units | (8) 20units | (50) 20units | 20units | 400 20units | 20units | 20units |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, | AC 20 | BC 21, cos = | sin = | | | | | | | |
| | AB 29 | | AB 29 | | | | | | | |
| | | 2 | 2 | 2 | | 2 | | | | |
| Now, (i) cos2 + sin2 =
| | 20 | 21 | | 20 | 21 | | 400 | | 441 | |
|
| | | | | | | | | | | |
| | | | | | | | | | | |
| 2
| 1,
| | 29 | 29 | | | | | | 84129 | | |
| | 21 | 20 | | (21 | 20)(21 | | | 20) | | |
| | | 2 | 2 | | | | | | | |
| and (ii) cos2 – sin2 =
|
| | 29 | 29 | | | 84129 | 84129 | 84129 | 84129 | 84129 | 84129 |
2
2 sin A cos A = 1.
Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11)
Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
| jemh108.pdf |
1 | CBSE | Class10 | Mathematics | 8.2 Trigonometric Ratios
In Section 8.1, you have seen some right triangles imagined to be formed in different situations.
Let us take a right triangle ABC as shown in Fig. 8.4.
Note that the position of sides change when you consider angle C in place of A (see Fig. 8.5).
| | Here, CAB (or, in brief, angle A) is an
| --- | ---
| acute angle. Note the position of the side BC with respect to angle A. It faces A. We call it the hypotenuse of the right triangle and the side AB is a part of adjacent | side opposite A. So, we call it the to angle A. | to angle A. AC is the side | Fig. 8.4
| --- | --- | --- | ---
You have studied the concept of ‘ratio’ in your earlier classes.
We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios.
The trigonometric ratios of the angle A in right triangle ABC (see Fig. 8.4) are defined as follows :
Fig. 8.5
| sine of A = side opposite to angle A | hypotenuse AC | BC |
| --- | --- | --- | ---
| cosine of A = | side adjacent to angle A hypotenuse AC | AC | AB
tangent of A =
side opposite to angle A BC side adjacent to angle A AB
| cosecant of A = | 1 | hypotenuse | AC
| --- | --- | --- | ---
| sine of | | A | | side opposite to angle A | | | BC |
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| secant of A = 1 | | | | hypotenuse | | | AC |
| cosine of | | A | | side adjacent to angle A | | | AB |
| cotangent of A = | 1 | | | side adjacent to angle A | | | |
| tangent of | | | | A side opposite to angle A BC | BC | | BC |
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively.
Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A.
Also, observe that tan A = BC BC sin AAC
ABAB cos A AC
and cot A =.
cosA sin A
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right triangle?
(See Fig. 8.5)
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500.
Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course.
When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is.
The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin.
Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe.
An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
C.E.
476 – 550
The origin of the terms ‘cosine’ and ‘tangent’ was much later.
The cosine function arose from the need to compute the sine of the complementary angle.
Aryabhatta called it kotijya.
The name cosinus originated with Edmund Gunter.
In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’.
sin A is not the product of ‘sin’ and A.
‘sin’ separated from A has no meaning.
Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.
Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of A in PAM differ from those of A in CAB or from those of A in
Fig. 8.6
QAN?
To answer this, first look at these triangles.
Is PAM similar to CAB?
From Chapter 6, recall the AA similarity criterion.
Using the criterion, you will see that the triangles PAM and CAB are similar.
Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.
So, we have AM AB = AP MP AC BC
From this, we find
| MP AP = | BC sin A AC .
AM AB AP AC
= cos A,
MP BC tan A AM AB
and so on.
This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of (sin A)2, (cos A)2, etc., respectively.
But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A).
sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle.
If we know any one of the ratios, can we obtain the other ratios?
Let us see.
If in a right triangle ABC, sin A = then this means that
1, 3
BC 1
AC 3
, i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1
:
3 (see Fig. 8.7).
So if BC is equal to k, then AC will be 3k, where k is any positive number.
To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB.
Do you remember the Pythagoras theorem?
Let us use it to determine the required length AB.
Fig. 8.7 Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
| | AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 | 2 k)2
| --- | --- | ---
| Therefore, | AB = | 2 | 2 k
| --- | --- | --- | ---
| So, we get | AB = 2 2 k | (Why is AB not – 2 2 k | ?)
| --- | --- | --- | ---
| Now, | cos A = | AB 2 2 | | 2 | | 2
| --- | --- | --- | --- | --- | --- | ---
| | | AC 3 3 k k | 3 k | 3 k | 3 k |
Similarly, you can obtain the other trigonometric ratios of the angle A.
Let us consider some examples.
Example 1 : Given tan A
=
4 3, find the other trigonometric ratios of the angle A.
Solution : Let us first draw a right ABC (see Fig 8.8).
Now, we know that tan A = BC 4
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
Then AC
AB =
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have
AC PR = AB BC PQ QR
Then, by using Theorem 6.4, ACB ~ PRQ and therefore,
B
= Q.
Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10).
Determine the values of
(i) cos2 + sin2 ,
(ii) cos2 – sin2
Solution : In ACB, we have
Fig. 8.10 Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that.
| | AC = | 2 | BC | 2AB | = | 2 | (21) | 2(29)
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = (29 | 21)(29 | 21) 20units | 20units | (8) 20units | (50) 20units | 20units | 400 20units | 20units | 20units |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, | AC 20 | BC 21, cos = | sin = | | | | | | | |
| | AB 29 | | AB 29 | | | | | | | |
| | | 2 | 2 | 2 | | 2 | | | | |
| Now, (i) cos2 + sin2 =
| | 20 | 21 | | 20 | 21 | | 400 | | 441 | |
|
| | | | | | | | | | | |
| | | | | | | | | | | |
| 2
| 1,
| | 29 | 29 | | | | | | 84129 | | |
| | 21 | 20 | | (21 | 20)(21 | | | 20) | | |
| | | 2 | 2 | | | | | | | |
| and (ii) cos2 – sin2 =
|
| | 29 | 29 | | | 84129 | 84129 | 84129 | 84129 | 84129 | 84129 |
2
2 sin A cos A = 1.
Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11)
Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
| jemh108.pdf |
2 | CBSE | Class10 | Mathematics | side opposite to angle A BC side adjacent to angle A AB
| cosecant of A = | 1 | hypotenuse | AC
| --- | --- | --- | ---
| sine of | | A | | side opposite to angle A | | | BC |
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| secant of A = 1 | | | | hypotenuse | | | AC |
| cosine of | | A | | side adjacent to angle A | | | AB |
| cotangent of A = | 1 | | | side adjacent to angle A | | | |
| tangent of | | | | A side opposite to angle A BC | BC | | BC |
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively.
Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A.
Also, observe that tan A = BC BC sin AAC
ABAB cos A AC
and cot A =.
cosA sin A
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right triangle?
(See Fig. 8.5)
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500.
Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course.
When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is.
The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin.
Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe.
An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
C.E.
476 – 550
The origin of the terms ‘cosine’ and ‘tangent’ was much later.
The cosine function arose from the need to compute the sine of the complementary angle.
Aryabhatta called it kotijya.
The name cosinus originated with Edmund Gunter.
In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’.
sin A is not the product of ‘sin’ and A.
‘sin’ separated from A has no meaning.
Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.
Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of A in PAM differ from those of A in CAB or from those of A in
Fig. 8.6
QAN?
To answer this, first look at these triangles.
Is PAM similar to CAB?
From Chapter 6, recall the AA similarity criterion.
Using the criterion, you will see that the triangles PAM and CAB are similar.
Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.
So, we have AM AB = AP MP AC BC
From this, we find
| MP AP = | BC sin A AC .
AM AB AP AC
= cos A,
MP BC tan A AM AB
and so on.
This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of (sin A)2, (cos A)2, etc., respectively.
But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A).
sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle.
If we know any one of the ratios, can we obtain the other ratios?
Let us see.
If in a right triangle ABC, sin A = then this means that
1, 3
BC 1
AC 3
, i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1
:
3 (see Fig. 8.7).
So if BC is equal to k, then AC will be 3k, where k is any positive number.
To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB.
Do you remember the Pythagoras theorem?
Let us use it to determine the required length AB.
Fig. 8.7 Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
| | AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 | 2 k)2
| --- | --- | ---
| Therefore, | AB = | 2 | 2 k
| --- | --- | --- | ---
| So, we get | AB = 2 2 k | (Why is AB not – 2 2 k | ?)
| --- | --- | --- | ---
| Now, | cos A = | AB 2 2 | | 2 | | 2
| --- | --- | --- | --- | --- | --- | ---
| | | AC 3 3 k k | 3 k | 3 k | 3 k |
Similarly, you can obtain the other trigonometric ratios of the angle A.
Let us consider some examples.
Example 1 : Given tan A
=
4 3, find the other trigonometric ratios of the angle A.
Solution : Let us first draw a right ABC (see Fig 8.8).
Now, we know that tan A = BC 4
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
Then AC
AB =
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have
AC PR = AB BC PQ QR
Then, by using Theorem 6.4, ACB ~ PRQ and therefore,
B
= Q.
Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10).
Determine the values of
(i) cos2 + sin2 ,
(ii) cos2 – sin2
Solution : In ACB, we have
Fig. 8.10 Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that.
| | AC = | 2 | BC | 2AB | = | 2 | (21) | 2(29)
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = (29 | 21)(29 | 21) 20units | 20units | (8) 20units | (50) 20units | 20units | 400 20units | 20units | 20units |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, | AC 20 | BC 21, cos = | sin = | | | | | | | |
| | AB 29 | | AB 29 | | | | | | | |
| | | 2 | 2 | 2 | | 2 | | | | |
| Now, (i) cos2 + sin2 =
| | 20 | 21 | | 20 | 21 | | 400 | | 441 | |
|
| | | | | | | | | | | |
| | | | | | | | | | | |
| 2
| 1,
| | 29 | 29 | | | | | | 84129 | | |
| | 21 | 20 | | (21 | 20)(21 | | | 20) | | |
| | | 2 | 2 | | | | | | | |
| and (ii) cos2 – sin2 =
|
| | 29 | 29 | | | 84129 | 84129 | 84129 | 84129 | 84129 | 84129 |
2
2 sin A cos A = 1.
Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11)
Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
| jemh108.pdf |
3 | CBSE | Class10 | Mathematics | ABAB cos A AC
and cot A =.
cosA sin A
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right triangle?
(See Fig. 8.5)
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500.
Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course.
When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is.
The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin.
Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe.
An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
C.E.
476 – 550
The origin of the terms ‘cosine’ and ‘tangent’ was much later.
The cosine function arose from the need to compute the sine of the complementary angle.
Aryabhatta called it kotijya.
The name cosinus originated with Edmund Gunter.
In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’.
sin A is not the product of ‘sin’ and A.
‘sin’ separated from A has no meaning.
Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.
Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of A in PAM differ from those of A in CAB or from those of A in
Fig. 8.6
QAN?
To answer this, first look at these triangles.
Is PAM similar to CAB?
From Chapter 6, recall the AA similarity criterion.
Using the criterion, you will see that the triangles PAM and CAB are similar.
Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.
So, we have AM AB = AP MP AC BC
From this, we find
| MP AP = | BC sin A AC .
AM AB AP AC
= cos A,
MP BC tan A AM AB
and so on.
This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of (sin A)2, (cos A)2, etc., respectively.
But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A).
sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle.
If we know any one of the ratios, can we obtain the other ratios?
Let us see.
If in a right triangle ABC, sin A = then this means that
1, 3
BC 1
AC 3
, i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1
:
3 (see Fig. 8.7).
So if BC is equal to k, then AC will be 3k, where k is any positive number.
To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB.
Do you remember the Pythagoras theorem?
Let us use it to determine the required length AB.
Fig. 8.7 Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
| | AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 | 2 k)2
| --- | --- | ---
| Therefore, | AB = | 2 | 2 k
| --- | --- | --- | ---
| So, we get | AB = 2 2 k | (Why is AB not – 2 2 k | ?)
| --- | --- | --- | ---
| Now, | cos A = | AB 2 2 | | 2 | | 2
| --- | --- | --- | --- | --- | --- | ---
| | | AC 3 3 k k | 3 k | 3 k | 3 k |
Similarly, you can obtain the other trigonometric ratios of the angle A.
Let us consider some examples.
Example 1 : Given tan A
=
4 3, find the other trigonometric ratios of the angle A.
Solution : Let us first draw a right ABC (see Fig 8.8).
Now, we know that tan A = BC 4
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
Then AC
AB =
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have
AC PR = AB BC PQ QR
Then, by using Theorem 6.4, ACB ~ PRQ and therefore,
B
= Q.
Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10).
Determine the values of
(i) cos2 + sin2 ,
(ii) cos2 – sin2
Solution : In ACB, we have
Fig. 8.10 Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that.
| | AC = | 2 | BC | 2AB | = | 2 | (21) | 2(29)
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = (29 | 21)(29 | 21) 20units | 20units | (8) 20units | (50) 20units | 20units | 400 20units | 20units | 20units |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, | AC 20 | BC 21, cos = | sin = | | | | | | | |
| | AB 29 | | AB 29 | | | | | | | |
| | | 2 | 2 | 2 | | 2 | | | | |
| Now, (i) cos2 + sin2 =
| | 20 | 21 | | 20 | 21 | | 400 | | 441 | |
|
| | | | | | | | | | | |
| | | | | | | | | | | |
| 2
| 1,
| | 29 | 29 | | | | | | 84129 | | |
| | 21 | 20 | | (21 | 20)(21 | | | 20) | | |
| | | 2 | 2 | | | | | | | |
| and (ii) cos2 – sin2 =
|
| | 29 | 29 | | | 84129 | 84129 | 84129 | 84129 | 84129 | 84129 |
2
2 sin A cos A = 1.
Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11)
Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
| jemh108.pdf |
4 | CBSE | Class10 | Mathematics | cosA sin A
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right triangle?
(See Fig. 8.5)
The first use of the idea of ‘sine’ in the way we use it today was in the work Aryabhatiyam by Aryabhata, in A.D. 500.
Aryabhata used the word ardha-jya for the half-chord, which was shortened to jya or jiva in due course.
When the Aryabhatiyam was translated into Arabic, the word jiva was retained as it is.
The word jiva was translated into sinus, which means curve, when the Arabic version was translated into Latin.
Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe.
An English Professor of astronomy Edmund Gunter (1581–1626), first used the abbreviated notation ‘sin’.
C.E.
476 – 550
The origin of the terms ‘cosine’ and ‘tangent’ was much later.
The cosine function arose from the need to compute the sine of the complementary angle.
Aryabhatta called it kotijya.
The name cosinus originated with Edmund Gunter.
In 1674, the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’.
sin A is not the product of ‘sin’ and A.
‘sin’ separated from A has no meaning.
Similarly, cos A is not the product of ‘cos’ and A. Similar interpretations follow for other trigonometric ratios also.
Now, if we take a point P on the hypotenuse AC or a point Q on AC extended, of the right triangle ABC and draw PM perpendicular to AB and QN perpendicular to AB extended (see Fig. 8.6), how will the trigonometric ratios of A in PAM differ from those of A in CAB or from those of A in
Fig. 8.6
QAN?
To answer this, first look at these triangles.
Is PAM similar to CAB?
From Chapter 6, recall the AA similarity criterion.
Using the criterion, you will see that the triangles PAM and CAB are similar.
Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.
So, we have AM AB = AP MP AC BC
From this, we find
| MP AP = | BC sin A AC .
AM AB AP AC
= cos A,
MP BC tan A AM AB
and so on.
This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of (sin A)2, (cos A)2, etc., respectively.
But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A).
sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle.
If we know any one of the ratios, can we obtain the other ratios?
Let us see.
If in a right triangle ABC, sin A = then this means that
1, 3Fig. 8.6
QAN?
To answer this, first look at these triangles.
Is PAM similar to CAB?
From Chapter 6, recall the AA similarity criterion.
Using the criterion, you will see that the triangles PAM and CAB are similar.
Therefore, by the property of similar triangles, the corresponding sides of the triangles are proportional.
So, we have AM AB = AP MP AC BC
From this, we find
| MP AP = | BC sin A AC .
AM AB AP AC
= cos A,
MP BC tan A AM AB
and so on.
This shows that the trigonometric ratios of angle A in PAM not differ from those of angle A in CAB.
In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in QAN also.
From our observations, it is now clear that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of (sin A)2, (cos A)2, etc., respectively.
But cosec A = (sin A)–1 sin–1 A (it is called sine inverse A).
sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well.
Sometimes, the Greek letter (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle.
If we know any one of the ratios, can we obtain the other ratios?
Let us see.
If in a right triangle ABC, sin A = then this means that
1, 3So, we have AM AB = AP MP AC BC | jemh108.pdf |
5 | CBSE | Class10 | Mathematics | AC 3
, i.e., the lengths of the sides BC and AC of the triangle ABC are in the ratio 1
:
3 (see Fig. 8.7).
So if BC is equal to k, then AC will be 3k, where k is any positive number.
To determine other trigonometric ratios for the angle A, we need to find the length of the third side AB.
Do you remember the Pythagoras theorem?
Let us use it to determine the required length AB.
Fig. 8.7 Remark : Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or, in particular, equal to 1).
| | AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 | 2 k)2
| --- | --- | ---
| Therefore, | AB = | 2 | 2 k
| --- | --- | --- | ---
| So, we get | AB = 2 2 k | (Why is AB not – 2 2 k | ?)
| --- | --- | --- | ---
| Now, | cos A = | AB 2 2 | | 2 | | 2
| --- | --- | --- | --- | --- | --- | ---
| | | AC 3 3 k k | 3 k | 3 k | 3 k |
Similarly, you can obtain the other trigonometric ratios of the angle A.
Let us consider some examples.
Example 1 : Given tan A
=
4 3, find the other trigonometric ratios of the angle A.
Solution : Let us first draw a right ABC (see Fig 8.8).
Now, we know that tan A = BC 4
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
Then AC
AB =
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have
AC PR = AB BC PQ QR
Then, by using Theorem 6.4, ACB ~ PRQ and therefore,
B
= Q.
Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10).
Determine the values of
(i) cos2 + sin2 ,
(ii) cos2 – sin2
Solution : In ACB, we have
Fig. 8.10 Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that.
| | AC = | 2 | BC | 2AB | = | 2 | (21) | 2(29)
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = (29 | 21)(29 | 21) 20units | 20units | (8) 20units | (50) 20units | 20units | 400 20units | 20units | 20units |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, | AC 20 | BC 21, cos = | sin = | | | | | | | |
| | AB 29 | | AB 29 | | | | | | | |
| | | 2 | 2 | 2 | | 2 | | | | |
| Now, (i) cos2 + sin2 =
| | 20 | 21 | | 20 | 21 | | 400 | | 441 | |
|
| | | | | | | | | | | |
| | | | | | | | | | | |
| 2
| 1,
| | 29 | 29 | | | | | | 84129 | | |
| | 21 | 20 | | (21 | 20)(21 | | | 20) | | |
| | | 2 | 2 | | | | | | | |
| and (ii) cos2 – sin2 =
|
| | 29 | 29 | | | 84129 | 84129 | 84129 | 84129 | 84129 | 84129 |
2
2 sin A cos A = 1.
Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11)
Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
| jemh108.pdf |
6 | CBSE | Class10 | Mathematics | Now, we know that tan A = BC 4
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
AB 3 .
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).
Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
| AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2 | | | | | |
| --- | --- | --- | --- | --- | --- | ---
| So, | | AC = 5k | | | |
| Now, we can write all the trigonometric ratios using their definitions.
| sin A = BC 4 4 AC 5 5
| k k
| cos A = AB 3 3 AC 5 5
| k k
| Therefore, cot A = 1 | | 3 1 5, cosec A = | | | |
| tan A | | 4 sin A 4 and sec A = | 1 cos A | | 5 3 |
Example 2 : If B and Q are acute angles such that sin B = sin Q, then prove that
B
= Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. 8.9).Fig. 8.9
| We have | sin B = | AC AB
| --- | --- | ---
| and | sin Q = | PR
| jemh108.pdf |
7 | CBSE | Class10 | Mathematics | AB =
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have
AC PR = AB BC PQ QR
Then, by using Theorem 6.4, ACB ~ PRQ and therefore,
B
= Q.
Example 3 : Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10).
Determine the values of
(i) cos2 + sin2 ,
(ii) cos2 – sin2
Solution : In ACB, we have
Fig. 8.10 Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that.
| | AC = | 2 | BC | 2AB | = | 2 | (21) | 2(29)
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| | = (29 | 21)(29 | 21) 20units | 20units | (8) 20units | (50) 20units | 20units | 400 20units | 20units | 20units |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| So, | AC 20 | BC 21, cos = | sin = | | | | | | | |
| | AB 29 | | AB 29 | | | | | | | |
| | | 2 | 2 | 2 | | 2 | | | | |
| Now, (i) cos2 + sin2 =
| | 20 | 21 | | 20 | 21 | | 400 | | 441 | |
|
| | | | | | | | | | | |
| | | | | | | | | | | |
| 2
| 1,
| | 29 | 29 | | | | | | 84129 | | |
| | 21 | 20 | | (21 | 20)(21 | | | 20) | | |
| | | 2 | 2 | | | | | | | |
| and (ii) cos2 – sin2 =
|
| | 29 | 29 | | | 84129 | 84129 | 84129 | 84129 | 84129 | 84129 |
2
2 sin A cos A = 1.
Solution : In ABC, tan A = BC AB = 1 (see Fig 8.11)
Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
PR PQ
| Therefore, | AC
| --- | ---
| | | AB | | | | k | , say PQ | | | | | | (1)
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| PR =
| Now, using Pythagoras theorem,
| | BC = | | | | 2 | | AC | 2AB | | | | |
| and | QR = | | | | 2 | | PR | | | 2PQ – | | |
| So,
| BC QR =
| 2 2 | 2 | | 2 2 | | 2 | | | | 2 | 2 | | |
| 2 2 | 2 | | | 2 | | | | | | 2 2 | | |
| AB AC | PQ | | PR | | | | | PQ | | PR | | |
| PQ PR | PQ | PR k | | | | | PQ k | | | PR k | | k | (2)
From (1) and (2), we have | jemh108.pdf |
8 | CBSE | Class10 | Mathematics | Fig. 8.11
| i.e., | BC = AB
| --- | ---
| Now, | AC = | 2 | BC | 2AB
| --- | --- | --- | --- | ---
| | = | 2 | ( | k | ) | 2() | | k | 2k
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Therefore, | | | | | sin A = | BC AC | | 1 | and | cos A = | 2 AB AC | | 1 2
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| Let AB = BC = k, where k is a positive number.
| So, | 2 sin A cos A = | | | | 2 | | 1, | | | | | |
| 1 1
| | | | | | 2 | 2 | | | | | | |
| which is the required value.
| Example 5 : In OPQ, right-angled at P,
| OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
| Determine the values of sin Q and cos Q.
| Solution : In OPQ, we have
| OQ2 = OP2 + PQ2
| i.e., | (1 + PQ)2 = OP2 + PQ2 | | | | | | (Why?) | | | | | |
| i.e., | 1 + PQ2 + 2PQ = OP2 + PQ2 | | | | | | | | | | | |
| i.e., | 1 + 2PQ = 72 | | | | | | (Why?) | | | | | |
| i.e., | | PQ = 24 cm and OQ = 1 + PQ = 25 cm | | | | | | | | | | |
| So, | sin Q = 7 25 | | | | and cos Q = | | 24 25 | | | | Fig. | 8.12 |
| jemh108.pdf |
9 | CBSE | Class10 | Mathematics | 2. In Fig
8.13, find tan P – cot R.3. If sin A = 3,
4 calculate cos A and tan A.
4. Given 15 cot A = 8, find sin A and sec A.Fig. 8.13 Trigonometric Ratios of 45°
5. Given sec = 13,
12 calculate all other trigonometric ratios.
6. If A and B are acute angles such that cos A = cos B, then show that
A
= B.
7. If cot
=
7 , 8 evaluate : (i) (1 sin) (1 sin) , (1 cos) (1 cos)
(ii) cot2
8. If 3 cot A = 4, check whether
2 2
1 tan A
1
+ tan A
= cos2 A – sin2A or not.
9. In triangle ABC, right-angled at B, if tan A = 1,
3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
10.
In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm.
Determine the values of sin P, cos P and tan P.
11.
State whether the following are true or false.
Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12 5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin = 4
3 for some angle .8.3 Trigonometric Ratios of Some Specific Angles
From geometry, you are already familiar with the construction of angles of 30°, 45°, 60° and 90°.
In this section, we will find the values of the trigonometric ratios for these angles and, of course, for 0°.
In ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e.,
A
= C = 45° (see Fig. 8.14).
| So, BC = AB (Why?) | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | ---
| Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = AB2 + BC2 = a2 + a2 = 2a2,
| and, therefore, AC = 2a Using the definitions of the trigonometric ratios, we have :
| sin 45° = side opposite to angle 45° | | | BC | | | | | 1
| hypotenuse | | | AC | | a | 2 | | 2 a
| cos 45° = side adjacent toangle 45° | | | AB | | | | | 1
| hypotenuse | | | AC | | a | 2 | | 2 a
tan 45° =, sec 45° =, cot 45° =.
side opposite to angle 45° BC 1 side adjacent to angle 45° AB a a
Also, cosec 45° =
1 2 sin 45 1 2 cos 45
side opposite to angle 45° BC 1 side adjacent to angle 45° AB a a
Also, cosec 45° =
1 2 sin 45 1 2 cos 45
Also, cosec 45° =
1 2 sin 45 1 2 cos 45
Fig. 8.15Fig. 8.14
1 1 tan 45
Trigonometric Ratios of 30° and 60°
Let us now calculate the trigonometric ratios of 30° and 60°.
Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,
A
= B
=
C = 60°.
Draw the perpendicular AD from A to the side BC (see Fig. 8.15).
| Now | ABD ACD | (Why?)
| --- | --- | ---
| Therefore, | BD = DC |
| and | BAD = | CAD (CPCT)
Now observe that: ABD is a right triangle, right-angled at D with BAD = 30° and ABD = 60° (see Fig. 8.15).
As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle.
So, let us suppose that AB = 2a.
| Then, | | BD = 1 BC = | | | | | | | |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
| | | | 2 | a | | | | | |
| and AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2, | | | | | | | | | |
| Therefore, | | AD = | 3a | | | | | | |
| Now, we have :
| sin 30° = BD | | 1 | | | | | | | |
| AB 2 a | | , cos 30° = | 2 a | AD AB | | 2 2 a | 3 2 a a | 2 a | 2 a | 3
| | tan 30° = | BD | 1
| --- | --- | --- | ---
| | AD | 3 | 3 a | a | | | .
| --- | --- | --- | --- | --- | --- | --- | ---
| Also, | cosec 30° = | 1 2, sin 30 | 2, sin 30 | 2, sin 30 | 2, sin 30 | sec 30° = | 1 cos 30 | | 2 3
| --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
cot 30° =.
1 3 tan 30
Similarly,
| sin 60° = | AD | 3 | 3
| --- | --- | --- | ---
| AB | 2 | 2 a a | | | , cos 60° = | 1 2, tan 60° = | 3 | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
1 2, 3 sec 60° = 2 and cot 60° = 3
cosec 60° = | jemh108.pdf |
10 | CBSE | Class10 | Mathematics | 1 3 tan 30
Similarly,
| sin 60° = | AD | 3 | 3
| --- | --- | --- | ---
| AB | 2 | 2 a a | | | , cos 60° = | 1 2, tan 60° = | 3 | ,
| --- | --- | --- | --- | --- | --- | --- | --- | ---
1 2, 3 sec 60° = 2 and cot 60° = 3
cosec 60° =Trigonometric Ratios of 0° and 90°
Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC (see Fig. 8.16), till it becomes zero.
As A gets smaller and smaller, the length of the side BC decreases.The point C gets closer to point B, and finally when A becomes very close to 0°, AC becomes almost the same as AB (see Fig. 8.17).Fig. 8.16
Fig. 8.17
When A is very close to 0°, BC gets very close to 0 and so the value of
| sin A = | BC AC | is very close to 0. Also, when A is very close to 0°, AC is nearly the
| --- | --- | ---
| same as AB and so the value of cos A = | AB AC | is very close to 1.
This helps us to see how we can define the values of sin A and cos A when A = 0°.
We define : sin 0° = 0 and cos 0° = 1.
Using these, we have :
tan 0° = = 0, cot 0° = which is not defined.
(Why?)Fig. 8.17
When A is very close to 0°, BC gets very close to 0 and so the value of
| sin A = | BC AC | is very close to 0. Also, when A is very close to 0°, AC is nearly the
| --- | --- | ---
| same as AB and so the value of cos A = | AB AC | is very close to 1.
This helps us to see how we can define the values of sin A and cos A when A = 0°.
We define : sin 0° = 0 and cos 0° = 1.
Using these, we have :
tan 0° = = 0, cot 0° = which is not defined.
(Why?)Using these, we have :
tan 0° = = 0, cot 0° = which is not defined.
(Why?) | jemh108.pdf |