Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

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Tamilnadu Samacheer Kalvi 10th Social Science Solutions Geography Chapter 5 India: Population, Transport, Communication, and Trade

Samacheer Kalvi 10th Social Science India: Population, Transport, Communication, and Trade Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The scientific study of different aspects of population is called:
(a) Photography
(b) Demography
(c) Choreography
(d) Population density
Answer:
(b) Demography

Question 2.
The state with highest literacy rate as per 2011 census is ……..
(a) Tamil nadu
(b) Karnataka
(c) Kerala
(d) Uttarpradesh
Answer:
(c) Kerala

Question 3.
Human Development is measured in terms of:
(a) Human Resource Index
(b) Per capita index
(c) Human Development Index
(d) UNDP
Answer:
(c) Human Development Index

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 4.
……… Transport provides door to door services.
(a) Railways
(b) Roadways
(c) Airways
(d) Waterways
Answer:
(b) Roadways

Question 5.
The length of Golden Quadrilateral superhighways in India is:
(a) 5846 km
(b) 5847 km
(c) 5849 km
(d) 5800 km
Answer:
(a) 5846 km

Question 6.
The length of navigable Inland waterways in India is ……..
(a) 17,500 km
(b) 5000 km
(c) 14,500 km
(d) 1000 km
Answer:
(c) 14,500 km

Question 7.
The National Remote sensing Centre(NRSC) is located at:
(a) Bengaluru
(b) Chennai
(c) Delhi
(d) Hyderabad
Answer:
(d) Hyderabad

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 8.
The transport useful in the inaccessible areas is ……..
(a) Roadways
(b) Railways
(c) Airways
(d) Waterways
Answer:
(c) Airways

Question 9.
Which of the following is associated with helicopter service?
(a) Air India
(b) Indian Airlines
(c) Vayudoot
(d) Pavan Hans
Answer:
(d) Pavan Hans

Question 10.
The major import item of India is
(a) Cement
(b) Jewells
(c) Tea
(d) Petroleum
Answer:
(d) Petroleum

II. Match the following

Question 1.
Match the Column I with Column II.
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 1
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 2
Answer:
A. (v)
B. (i)
C. (iv)
D. (ii)
E. (iii)

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

III. Answer the following questions briefly

Question 1.
What is Human Development?
Answer:
It is a process of enlarging the range of people’s choice, increasing their opportunities for education, health care, income and empowerment. It covers a full range of human choices from a sound physical environment to economic, social and political freedom.

Question 2.
What is migration? State its types.
Answer:
Migration is the movement of people across regions and territories. It can be internal (within a country) or international (between the countries). Migration depends on

Push factor: Unemployment and underemployment in rural areas.

Pull factor: higher wages, employment opportunity, industrial development.

Question 3.
Write any four advantages of railways.
Answer:

  1. Railways are the price mode of transport for goods and passengers in India.
  2. They make it possible to conduct varied activities like business, sightseeing and pilgrimage along with transportation of goods over longer distances.
  3. They are suitable for long-distance travel and play an important role in national integration.
  4. They bind the economic life of the country as well as the accelerate the development of the industry and agriculture.

Question 4.
Write a note on Pipeline network transport in India.
Answer:
Pipelines are the very convenient transport to connect oil and natural gas fields, refineries and to the markets.
It can be laid through difficult Terrain as well as under water.
It ensures steady supply of goods and reduces the transhipment loss and delays.

Three important network of pipeline:

  1. Oil fields in upper Assam and Kanpur.
  2. From Salaya in Gujarat to Jalandhar in Punjab.
  3. From Hazira in Gujarat to Jagadispur in Uttar Pradesh

Question 5.
State the major Inland waterways of India.
Answer:
The major inland waters ways of India are :

  1. National Waterway I: It extends between Haldia and Allahabad, measures 1620 km and includes the stretches of the Ganga – Bhagirathi – Hooghly river system.
  2. National Waterway 2: This waterway includes the stretch of the Brahmaputra river between Dhubri and Sadiya a distance of 891 km.
  3. National Waterway 3: This waterway extends between Kollam and Kottappuram in the state of Kerala. It is the first national waterway in the country with 24 hour navigation facilities along its entire stretch of 205 km.

Question 6.
What is communication? What are its types?
Answer:
Communication is a process that involves exchange of information, thoughts , and ideas. lt is categorized into two types.

Personal Communication: Post and telegraph, telephone, mobile phone, short message services, fax, internet, e-mail etc.
Mass Communication: Electronic media-Radio, T.V, internet and print media-News papers.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 7.
Define “International trade”.
Answer:
When trade takes place between two countries it is known as international trade.

Question 8.
State the merits of Roadways.
Answer:

  1. Roadways can provide door to door services.
  2. Easy and cheap to construct and maintain. Indian roads are cost-efficient.
  3. The most universal mode of transport. It is used by all sections of people in society.
  4. Can establish easy contact between farms, fields, factories and markets.

IV. Distinguish between

Question 1.
Density of population and Growth of population.
Answer:
Density of Population:

  1. Density of population is the number of persons living per square kilometre.
  2. It is calculated per 1000.
  3. As per 2011 census the average density of population in India is 382 persons living in per sq.km.

Growth of Population:

  1. Growth of population refers to the change in the number of in habitants of a country/territory in a specific period of time.
  2. It is expressed in percentage.
  3. The year 1921 is called the year of Great Demographic Divide.

Question 2.
Persona! communication and Mass communication.
Answer:
Personal communication:

  1. Exchange of communication between the individuals.
  2. Enables the user to establish direct contact.
  3. It includes post and telegraph, telephones, mobile phones, short message service (SMS) fax, Internet, e-mail etc.

Mass communication:

  1. Millions of people get the information at the same time.
  2. Provide information through print media and electronic media.
  3. It includes Radio, Television and Internet. Electronic media, News paper, Magazines, books, journals etc.

Question 3.
Print Media and Electronic Media.
Answer:
Print Media:

  1. Most powerful means of communication.
  2. Many news papers carry information on local national and international events to the people.
  3. Knowledge of reading is essential.

Electronic Media:

  1. Communication to millions of people through electronic gadgets at the same time.
  2. Radio, Television, Internet provide and create awareness on various national policies and programme.
  3. A powerful audio-visual medium.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 4.
Roadways and Railways.
Answer:
Roadways:

  1. Highly suitable for short distance services.
  2. Most common mode of transport, easy, cheap to construct and maintain.
  3. Based on the construction and maintenance divided into National Highways, State Highways, District Roads, Village roads, Border roads, International highways, Express ways.
  4. Second longest network in the world.
  5. Door to door services possible

Railways:

  1. Ideal for long distance services.
  2. Construction depend upon the climatic and physical factors like terrain. Costly when compared with roadways.
  3. On the basis of width of the track falls under 4 types, Broad gauge, meter gauge, narrow gauge and light gauge.
  4. Largest in Asia and second largest in the World.
  5. Cannot provide door to door services.

Question 5.
Waterways and Airways.
Answer:
Waterways:

  1. Oldest and Cheapest means of transport.
  2. Most suitable for carrying heavy and bulky materials.
  3. Two types, inland waterways-river and backwaters and canals and oceanic routes connect coastal areas.

Airways:

  1. Modem and costliest means of transport.
  2. Only limited weight can be carried. Cannot carry heavy and bulky materials.
  3. Domestic air sendee within the country and International Airways connect major cities of the World.

Question 6.
internal trade and International trade.
Answer:
Internal trade:

  1. Trade is carried on within the domestic territory.
  2. Land transport plays a major role (Roadways and Railways).
  3. Local currency is used.
  4. Helps to promote a balanced regional growth.

International trade:

  1. Trade is carried on between two or more countries.
  2. Waterways and Airways play a vital role.
  3. Foreign currency is involved.
  4. Helps to promote country’s economy and raise the standard of living

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

V. Answer the following in a paragraph

Question 1.
What is urbanization? Explain its impacts.
Answer:
The process of society’s transformation from rural to urban is known as urbanization. The level of urbanization of a place is assessed based on the size of population of the towns and cities and the proportion of population engaged in non agricultural sectors. These two are closely linked to the process of industrialization and expansion of the secondary and tertiary sectors of economy.
Impacts of urbanization:

  • Urbanization and population concentration go hand – in – hand and are closely related to each other. A rapid rate of urbanization in a society is taken as an indicator of its economic development.
  • Urbanization is increasing rapidly in the developing countries including India.
  • Rural to urban migration leads to population explosion in urban areas. Metropolitan cities like Mumbai, Kolkatta and Delhi have more population than that can accommodate.
    The following are the major problem of urbanization in India.
  • It creates urban sprawl.
  • It makes overcrowding in urban centres.
  • It leads to shortage of houses in urban areas.
  • It leads to the formation of slums.
  • It increases traffic congestion in cities.
  • It creates water scarcity in cities.
  • It creates drainage problems.
  • It poses the problem of solid waste management.
  • It increases the rate of crime.

Question 2.
Explain the importances of satellite communication in India.
Answer:

  1. The communication through satellites emerged as a new era in the system of communication in our country.
  2. The use of Satellite in getting a continuous and synoptic view of larger area has made this communication system very vital for the country.
  3. Satellite images are used for weather forecasting.
  4. Monitoring of natural calamities.
  5. Surveillance of border areas.
  6. The INSAT (series) is a multipurpose system for telecommunication, meterological observation and for various other programs.
  7. They are used for relaying signals to television, telephone, radio and mobile phone.
  8. And also useful in weather detection, internet and military applications.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 3.
Bring out the distribution and density of population in India.
Answer:
The term ‘Population Distribution’ refers to the way the people are spaced over the earth’s surface. The distribution of population in India is quite uneven because of the vast variation ki the availability of resources. The population is mostly concentrated in the regions of industrial centres and the good agricultural lands. On the other hand, the areas such as high mountains and lands thickly forested areas and some remote comers are very thinly populated and some areas are even uninhabited.

Terrain, climate, soil, water bodies, mineral resources, industries, transport and urbanization are the major factors which affect the distribution of population in our country. Population density is a better measure of understanding the variation in distribution of population. It is expressed as number of person per unit area usually per sq.km. According to 2011, the average density of population of India is 382 person per sq.km. India is one of the most thickly populated ten countries of the world. The most densely populated state of India is Bihar and the state with lest population density is Arunachal Pradesh. Among union territories, Delhi is the densely populated one with 11,297 per sq.km, while Andaman and Nicobar Islands have the lowest density of population.

Question 4.
Explain the process of measuring Human Development.
Answer:
Dr. Mahabub-ul-haq defined Human Development as “It is a process of enlarging the range of people’s choice, increasing their opportunities for education, health care, income and empowerment. It covers the full range of human choices from a. sound physical environment to economic, social and political freedom”.

Measuring of Human Development (HDI): Human Development Index is a composite index focusing on three basic dimensions of human development.

  1. Health: Life expectancy at birth.
  2. Education: Expected years of schooling for school age children and average years of schooling for the adult population.
  3. Income: Measured by gross national income and per capita income.

Question 5.
Classify and explain the roadways in India.
Answer:

  1. The Indian roads are cost-efficient and the most popular dominant mode of transport
    linking different parts of our country.
  2. Roads stretch across the length of people in society.
  3. It is used by all sections of people in society.
  4. Road network in India is the second-longest in the world accounting for 3,314 millions of km.

Types of Roadways:

1. Village Roads:

  • Village roads link different villages with towns.
  • They are maintained by Village Panchayath.
  • In India village roads run to a length of 26,50,000 kms.

2. District Roads:

  • District roads link the towns with the district headquarters.
  • They are maintained by the Corporations and Municipalities.
  • It is used by all sections of people in the society.
  • In India district roads run to a total length of 4,67,763 kms.

3. State Highways:

  • State Highways links the State capitals with the different district headquarters.
  • These roads are maintained by the State Public Works Department.
  • The State Highways run to a length of 1,31,899 kms. (Ex: Cudalore – Chittor Road).

4. National Highways:

  • National Highways link the State Capitals with the National Capital.
  • They are the primary road system of our country and are maintained by the Central Public Works Department.
  • It runs to a length of 70,548 kms. (Ex: NH47 is a National Highway which connects Tamil Nadu and Kerala)

5. Golden Quadrilateral Super Highways:

  • It is a major road development project launched by the Government of India.
  • It runs to a length of 14,846 kms connecting the major cities of India.
  • The major objective of these roads is providing ‘connectivity’, ‘speed’ and ‘safety’.

6. Expressways:

  • Expressways are the technologically improved high class roads in the Indian Road network.
  • They are six lanes roads. They run to a length of more than 200 kms.
  • New Mumbai – Pune Road is an example for Expressway.

7. Border Road:

  • Border Roads are the roads constructed along the northern and north eastern borders of our country.
  • These roads are constructed and maintained by Border Roads Organization.
  • The Organization has constructed 46,780 km of roads in difficult terrain.

8. International Highways:
International Highways are the roads that link India with neighbouring countries for promoting harmonious relationship with them.

VI. On the outline map of India mark the following

Question 1.
National Highway NH-7.
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 3

Question 2.
Major seaports in India.
Answer:

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 4

Question 3.
Major International Airports in India.
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 5

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 4.
Densely populated state of India – Uttarpradesh
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 4

Question 5.
State of highest literacy in India – Kerala
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 4

Question 6.
Railways zones of India.
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 6

TB Page No. 151
Hots:

Question 1.
What could be the reasons for uneven distribution of population in India?
Answer:

  • Topography, favourable climate, fertility of soils, availability of fresh water, minerals are major geographical factors affecting population density of a region.
  • People prefer to live on plains more than mountains or plateaus and they live more in moderate climates than extreme hot or cold. From the agriculture point of view, fertile lands are preferred.
  • Areas with mineral deposits are more populated.
  • Some social factors that boost the density of population in a region are better housing, education and health facilities.
  • Places with cultural or historical significances are usually populated.
  • Employment opportunities is another attraction for large chunks of the population.

Question 2.
What are the reasons for the rapid growth of population in India?
Answer:

  1. The reasons for the rapid growth of population in India is mainly the , high birth rate and low death rate.
  2. Birth rate is high, death rate declined due to advanced medical facilities and immunization to dreadful diseases.
  3. And also migration of population due to employment, education and industrial development.

TB Page No. 154

Question 3.
The sex ratio in our country is always unfavourable to females. Give reasons.
Answer:
The number of females per thousand men is called the sex ratio.
Reasons:

  • Lesser care of female children.
  • Greater risk to womens’ life especially at the time of child brith.
  • Women are also killed or forced to die by the dowry seekers.
  • Due to illiteracy.
  • Lack of medical facilities for women, etc.

TB Page No. 158

Question 4.
Find out what are the functions of NHAI?
Answer:

  1. National Highway Authority of India was established in 1995.
  2. It is an autonomous body under the ministry of Surface Transport.
  3. Management of all major National Highways and the highways entrusted to it.
  4. Maintenance, development and operation of the National Highways is undertaken by NHAI.

Question 5.
What are the highlights and benefits of the Golden Quadrilateral Highways? Highlights:
Answer:

  • It is the largest Highway project completed in India.
  • It is the fifth largest highway project in the world.
  • The overall length of the Golden Quadrilateral is 5,846 km.
  • The Golden Quadrilateral passes through 13 states of India.
  • It connects four major metropolitan cities of the country in four directions.

Benefits:

  • Provides faster transport networks between major cities and ports.
  • Provides connectivity to major agricultural industrial and cultural centres of India.
  • Provides smoother movement of goods and people within the country.
  • Enables industrial development and job creation in smaller town through access to varied markets.
  • Farmers are able to transport their produce to major cities and towns for sale and export and there is less wastage and spoils.
  • More economic growth through construction and indirect demand for steel, cement, and other construction materials.
  • Giving an impetus to truck transports.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

TB Page No. 162

Question 6.
Why is air travel preferred in the North Eastern State?
Answer:
The North Eastern States are mountainous and forested. So construction of roadways or railways is very difficult due to terrains. Air transport has made accessibility easier. So air travel is preferred in the North eastern states.

TB Page No. 165

Question 7.
Find out the major trade blocs which are useful for multilateral trade.
Answer:
Trading blocs are usually groups of countries in specific regions that manage and promote trade activities. Trading blocs lead to trade liberalisation and trade creation between members, since they are treated favourably in comparison to non-members. The World Trade Organisation (WTO) permits the existence of trading bloc provided that they result in lower protection against outside countries than existed before the creation trading bloc.
The most significant trading blocs currently are:

  1. European Union (EU): A customs union, a single market and now with a single currency.
  2. European Free Trade Area (EFTA)
  3. North American Free Trade Agreement (NFATA) between the USA, Canada and Mexico.
  4. Mercosur: A customs union between Argentina, Brazil, Paraguay, Uruguay, and Venezuela.
  5. Association of Southeast Asian Nations (ASEAN)
  6. Association of Free Trade Area (AFTA)
  7. Common market of Eastern and Southern Africa (COMESA)
  8. South Asian Free Trade Area (SAFTA) created in 2006 with countries such as India and Pakistan.
  9. Pacific Alliance: 2013 – a regional trade agreement between Chile, Colombia, Mexico and Peru.

TB Page No. 159
Activity:

Question 1.
Prepare a seminar topic about “Role of Railways in Indian Economy”Key Points:
Answer:

  1. Large scale movement
  2. National integration promotion
  3. Commercialisation of agriculture
  4. Movement of perishable goods
  5. Avoids traffic congestion how?
  6. Engineering marvel
  7. Quiz regarding railways

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

TB Page No. 165

Question 2.
Collect the countries names and make it as a table of Bilateral trade and multilateral trade countries.
Answer:
India has made bilateral trade agreement with these countries.

Bilateral Trade Countries: U.S.A . China, Hongkong, Singapore, United Kingdom, Germany, Bangladesh.

Multilateral Trade Countries: The members of the SAARC (South Asian Association for Regional Cooperation) Korea, Japan, Bangladesh, Bhutan, Maldives, Nepal, Pakistan, Afghanistan, Srilanka, U.S.A and United Kingdom.

Samacheer Kalvi 10th Social Science India: Population, Transport, Communication, and Trade Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
…………………. is the second-most populous country next to China.
(a) India
(b) Pakistan
(c) Bangladesh
(d) Nepal
Answer:
(a) India

Question 2.
The fast movement of traffic are established by ……
(a) national highways
(b) Express highways
(c) International highways
Answer:
(a) national highways

Question 3.
As per 2011 census the average density of population of India is …………………. persons per Sq.Km.
(a) 302
(b) 382
(c) 100
(d) 365
Answer:
(b) 382

Question 4.
The ………. have more railways than the Himalayan Mountains.
(a) Northern Plains
(b) Coastal Plains
(c) Deccan Plateau
Answer:
(a) Northern Plains

Question 5.
The Grand Trunk Road extends from …………………. to ………………….
(a) Delhi to Mumbai
(b) Amritsar to Kolkatta
(c) Mumbai to Thane
(d) Srinagar to Amritsar.
Answer:
(b) Amritsar to Kolkatta

Question 6.
The cheapest mode of transport is …….
(a) roadways
(b) railways
(c) waterways
Answer:
(c) waterways

Question 7.
…………………. are multi-lane good quality highways for high speed traffic.
(a) National Highways
(b) State highways
(c) Border roads
(d) Express highways.
Answer:
(d) Express highways.

Question 8.
Trade carried on within the domestic territory of a country is known as …… trade.
(a) External
(b) Foreign
(c) Internal
Answer:
(c) Internal

Question 9.
The …………………. railway accounts for the longest route length.
(a) Northern Railways
(b) Central Railway
(c) Eastern railway
(d) Southern Railway
Answer:
(a) Northern Railways

Question 10.
A cost-efficient and most popular mode of transport in our country is ……..
(a) Airways
(b) Roadways
(c) Waterways
Answer:
(b) Roadways

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 11.
India is the …………………. largest ship owning country in Asia.
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(b) Second

Question 12.
The costliest and most modem means of transport is ……..
(a) Air transport
(b) Road transport
(c) Rail transport
Answer:
(a) Air transport

Question 13.
The is the first communication Satellite in INSAT Series.
(a) G.SAT
(b) EDUSAT
(c) INSAT-IB
(d) Kalpana-1
Answer:
(c) INSAT-IB

Question 14.
The major ports are managed and controlled by …….
(a) National Ports Corporation
(b) Port Trust of India
(c) Indian Airlines
Answer:
(b) Port Trust of India

Question 15.
Indian space Research organisation was established in the year:
(a) 1959
(b) 1969
(c) 1979
(d) 1996
Answer:
(b) 1969

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

II. Match the following

Question 1.
Match the Column I with Column II.
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 8
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 9
Answer:
A. (v)
B. (iv)
C. (i)
D. (iii)
E. (ii)

Question 2.
Match the Column I with Column II.
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 10
Answer:
A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

III. Answer the following questions briefly

Question 1.
Describe the three population density zones of India.
Answer:
The three population density zones of India are :

  1. High-density zone: The Northern plains above 500 people per sq.km. Northern plains and Kerala in the. South.
  2. Moderate or Medium density zone: Mountain region 250-500 people per sq.km. Ex. Assam and Peninsular states.
  3. Low-density zone: Plateau region below 250 people per sq.km. Ex. Jammu and Kashmir, Sikkim, Arunachal Pradesh.

Question 2.
What are the reasons for the uneven distribution of population in India?
Answer:
The uneven distribution of population in the country is the result of several factors such as physical, socio-economic and historical factors.

Physical factors: Relief, climate, water, natural vegetation, minerals and energy resources.

Socio-economic factors: Religion, culture, political issues, economy, human settlements, transport network, industrialization, urbanization, employment opportunity etc.

Question 3.
What is the major objective to develop the Super Highways?
Answer:
The major objective to develop the Super Highways is to reduce the time and distance between
the megacities of India facilitating the fast movement of traffic.

Question 4.
Which phase period in population growth of India is often referred as period of population explosion?
Answer:
During the third phase (1951-1981) the population of India grew from 361 million in 1951 to 683 million in 1981. Growth rate in this period is almost doubled. So this period (1951-1981) is often referred as the period of “population explosion”.

Question 5.
State some problems of road transport in our country.
Answer:

  1. Distribution of road is not uniform in the country.
  2. Keeping in view the volume of traffic and passengers, the road network is inadequate.
  3. About half of the roads are unmetalled and this restricts their usage during the rainy season.
  4. The roadways are highly congested in cities.
  5. Poor maintenance is also a big problem.

Question 6.
What are the four major shipyards in India?
Answer:
India as four major shipyards:

  1. Hindustan shipyard – Vishakapatnam
  2. Garden Reach workshop – Kolkata
  3. Mazagaon Dock – Mumbai
  4. Kochi shipyard – Kochi

Question 7.
State the highlights of India’s foreign trade policy since 2004.
Answer:

  1. Merchandise trade has been doubled.
  2. Thrust is given for employment generation; especially in semi-urban and rural areas.
  3. Trade procedure is simplified and transaction cost is reduced.
  4. Special focus is given to make India a global hub.
  5. A new scheme called Vishesh Krishi Upay Yojna has been introduced to boost exports of fruits, vegetables, flowers and minor forest products.

Question 8.
Mention the services provided by the Indian postal system in India.
Answer:
Pcstai system is one among the personal communication system. The postal service was open to the public in the country in 1837.

  1. Collecting and delivering mail is the primary function of the postal department.
  2. It introduced Quick Mail Service in 1975 on the basis ofPIN code (1972)
  3. The premium products include the Money order. E-money order, Speed post. Express parcel post, Business post, Media post, Retail post, Data post, Satellite post, Greeting post, Speed net and Speed passport services,

Question 9.
What are the advantages of communication network?
Answer:

  1. It has enhanced the efficiency of communication. Because it enables quick exchange of information with people any where in the world.
  2. It leads to enormous growth of trade.
  3. It helps the government to tackle various socio – economic problems in the society.
  4. It improves the quality of human life.
  5. It opens the door to the information age.
  6. It promotes Edusat programs.
  7. It plays a significant role in the economic and social growth of our country.

Question 10.
Write the full form of STD, ISD, PCO, Internet.
Answer:

  • STD – Subscriber Trunk Dialing
  • ISD – International Subscriber Dialing
  • PCO – Public Call Office
  • Internet – Inter connected network.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

Question 11.
What are major ports along the West Coast and East Coast?
Answer:
West Coast Ports :

  1. Kandla,
  2. Mumbai,
  3. Jawahar Lal Nehru,
  4. Marmagoa,
  5. New Mangalore and
  6. Cochin East Cost Ports:
    • Tuticorin,
    • Chennai,
    • Ennore,
    • Vishakhapatnam,
    • Paradip,
    • Haldia and
    • Kolkota.

Question 12.
Name the major ports on the East coast and on the West Coast of India.
Answer:
There are 13 major ports in India. They are administered by the Central Government.

  1. The major ports on the East coast are Kolkata, Haldia, Paradip,Visakhapatnam, Chennai, Ennore and Tuticorin.
  2. The major ports on the West Coast are Kandla, Mumbai, Nhava Seva (Jawaharial Nehru Port), New Mangalore, Marmagao and Kochi.

Question 13.
Write a short note on Internet.
Answer:

  1. Internet is a vast network of computers, Internet means interconnected network of net , works.
  2. It connects many of the world’s business institutions and individuals.
  3. It enables computer users throughout the world to send and receive messages and information in a variety of forms.
  4. It is fully a multimedia based system with capacity to deliver pictures, images, video and audio.
  5. The basic services of internet are e-mail. The World Wide Web and Internet Phone.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

IV. Distinguish between

Question 1.
Exports and Imports.
Exports:

  1. Goods and services sold for foreign currency.
  2. Major exports of India are tea, ores and minerals, marine products, textiles etc.
  3. Value of exports are more than the value of imports favourable balance of trade.
  4. India exports goods to nearly 190 countries of the world.

Imports:

  1. Goods and services bought from overseas producers.
  2. Major imports are petroleum products, gold, telecom instruments.
  3. Value of imports exceeds value of exports unfavourable balance of trade.
  4. Imports we get from nearly 140 countries.

Question 2.
Low Density and High Density of population.
Answer:
Low Density of Population:

  1. Areas that have 150 to 300 persons living per sq.km.
  2. Extreme climate, high mountains, remote areas, forested regions have low density of population.
  3. In India Arunachal Pradesh, Sikkim, Mizoram, Andaman and Nicobar islands are some states with low population density.

High Density of Population:

  1. Areas that have 500 to 1000 persons living per sq.km.
  2. Favourable climate, plains, employment opportunities, industrial centres area have high density of population.
  3. Punjab, Tamil Nadu, Uttar Pradesh and Kerala are the states with high population density.

Question 3.
National and State Highways.
Answer:
National Highways:

  1. Connects Capitals of States, major ports, rail junctions, industrial and tourist centres.
  2. Ministry of Road transport and Highways of India is responsible development and maintenance.
  3. Runs to a length of 1,01,011 kms as of 2016.

State Highways:

  1. Link important cities, towns and district head quarters within the State and connect them with national highways or highways of neighbouring states.
  2. Administered and financed by state Governments.
  3. Runs to the length of 1,76,166 kms as of 2016.

Question 4.
Domestic and International Airways.
Answer:
Domestic Airways:

  1. Fly within the boundaries of the country.
  2. Indian Airlines provides the domestic air services. NACIL (I)
  3. There are about 80 domestic airports and about 25 civil enclaves at defence air fields.

International Airways:

  1. Fly across the world and connect major cities of the world.
  2. Air India provides international air services. NACIL (A)
  3. There are 19 designated international airports in the country.

Question 5.
Harbour and Port.
Answer:
Harbour:

  1. Extensive stretch of deep water near the seashore.
  2. Vessels can anchor securely.

Port:

  1. Commercial part of a harbour.
  2. Facility of loading and unloading of goods and space for the storage of cargo
  3. Cuddalore, Ennore.

Question 6.
Birth and Death.
Answer:
Birth Rate:

  1. No. of live Births per 1000 people in a year.
  2. Also known as Nationality rate.

Death Rate:

  1. No. of Deaths per 1000 people in a year.
  2. Also known as Mortality rate.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

V. Answer the following in a paragraph

Question 1.
What is the importance of railway transport?
Answer:

  1. The Indian Railways have been a great integrating force for more than 150 years.
  2. They have a network of about 7,112 stations spread over a route length of 66,687 km with a fleet of 11,122 locomotives, 70,241 passenger coaches and 2,54,000 wagons.
  3. Railways are the prime mode of transport for goods and passengers in India.
  4. They make it possible to conduct varied activities like business, sightseeing and pilgrimage along with transportation of goods over longer distances.
  5. They are suitable for long distance travel and play an important role in national integration.
  6. They bind the economic life of the country as well as accelerate the development of the industry and agriculture.
  7. The Indian Railways is the largest public sector undertaking in the country. The first train streamed of from Mumbai to Thane in 1853, covering a distance of 34 km.

Question 2.
What do you mean by Census? How it is useful?
Answer:

  1. Population census is the total process of collecting, compiling, analyzing or otherwise disseminating demographic, economic and social data pertaining at a specific time, of all persons in a country or a well defined part of a country.
  2. Census is being taken in an interval of ten years.
  3. The data collected through the census are used for administration, planning, policy making, management and evaluation of various programmes by the government.
  4. In India 1st census was carried out in the year 1872.
  5. But the first complete and synchronous census was conducted in 1881 The 2011 census represents the fifteenth census of India.

Question 3.
Describe the importance of pipelines in India. Name three important networks of pipelines transportation in the country.
Answer:
Pipeline transport network is a new arrival on the transportation of India.

  1. Transport of crude oil, petroleum products and natural gas from oil and natural gas fields to refineries, fertiliser factories and big thermal power plants. ‘
  2. Even solids can be transported through pipelines when converted into slurry.
  3. The far inland locations of refineries and gas based fertiliser plants could be transported.
  4. Initial cost of lying pipelines is high but subsequent running costs are tninimal.
  5. It rules out trans-shipment losses or delays.

Three important networks of pipeline transportation in the country:

  • From oil field in the upper Assam to Kanpur (UP), via Guwahati, Baraumi and Allahabad.
  • From Salaya in Gujarat to Jalandhar in Punjab. Via Viramgam, Mathura, Delhi and Sonipat.
  • Gas pipeline from Hazira in Gujarat connects Jagdishpur in UP, via Vijaipur in MP.

Question 4.
What are the problems created by over population in India?
Answer:
In India, growing pressure of Population on resource base, created many socio economic, cultural, political, ecological and environmental problems. The population problem varies in space and time and differ from region to region.

Major issues created by the over population are – overcrowding, unemployment and under employment, low standard of living mal nutrition, mismanagement of natural and agricultural resources, unhealthy environment etc.

Question 5.
Write about the significance of Indian Railways.
Answer:

  1. Indian Railway system is the main artery of the country’s inland transport.
  2. Indian Railway network is the largest in Asia and second largest in the world.
  3. Railways are considered as the back bone of the surface transport system of India.
  4. Significances: It caters to the needs of large scale movement of traffic both for freight and passengers, there by contributing to economic growth.
  5. It promotes trade, tourism, education etc.
  6. It promotes national integration by bringing people together.
  7. Railways help in the commercialization of the agricultural sector by facilitating the quick movements of perishable goods.
  8. It provides invaluable service by transporting raw materials to industries and finished goods to market.

Question 6.
What is trade? What are the two types of trade? State its components:
Answer:

  1. Trade is an act (or) process of buying selling or exchanging of goods and services. Trade in general is of two types. They are Internal trade and International trade.
  2. Internal Trade: Trade carried on within the domestic territory of a country.
  3. International trade: Trade carried on between two or more countries.
  4. Exports and imports are the two components of international trade.
  5. The difference between value of exports and imports is called balance of trade.
  6. If the value of exports exceeds the value of imports trade is said to be favourable balance of trade.
  7. If the value of Imports exceeds value of exports it is said to be unfavourable balance of trade.
  8. The value of currency of a country depends upon the balance of trade of that country.

Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India: Population, Transport, Communication, and Trade

VI. On the outline map of India mark the following

Question 1.
North-South corridor
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 7

Question 2.
East-West corridor
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 7

Question 3.
Head quarters of Indian Railway – Delhi.
Answer:
Samacheer Kalvi 10th Social Science Guide Geography Chapter 5 India Population, Transport, Communication, and Trade 5

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Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
Check whether the which triangles are similar and find the value of x.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 1
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 2
Solution:
(i) In ∆ABC and ∆AED
\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 8 }{ 3 } \) = \(\frac{11}{\frac{2}{2}}\)
\(\frac { 8 }{ 3 } \) = \(\frac { 11 }{ 4 } \) ⇒ 32 ≠ 33
∴ The two triangles are not similar.

(ii) In ∆ABC and ∆PQC
∠PQC = 70°
∠ABC = ∠PQC = 70°
∠ACB = ∠PCQ (common)
∆ABC ~ ∆PQC
\(\frac { 5 }{ X } \) = \(\frac { 6 }{ 3 } \)
6x = 15
x = \(\frac { 15 }{ 6 } \) = \(\frac { 5 }{ 2 } \)
∴ x = 2.5
∆ ABC and ∆PQC are similar. The value of x = 2.5

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 2.
A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.
Solution:
Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.
∠ABC = ∠CED = 90° (vertical Pole)
∠ACB = ∠ECD (Laws of reflection)
∆ ABC ~ ∆DEC
\(\frac { AB }{ DE } \) = \(\frac { BC }{ EC } \)
\(\frac { 1.5 }{ x } \) = \(\frac { 0.4 }{ 87.6 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 3
x = \(\frac{1.5 \times 87.6}{0.4}\) = \(\frac{1.5 \times 876}{4}\)
= 1.5 × 219 = 328.5
The height of the Lamp Post = 328.5 m

Question 3.
A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
In ∆ABC and ∆PQR,
∠ABC = ∠PQR = 90° (Vertical Stick)
∠ACB = ∠PRQ (Same time casts shadow)
∆BCA ~ ∆QRP
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 4
\(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \)
\(\frac { 6 }{ x } \) = \(\frac { 4 }{ 28 } \)
4x = 6 × 28 ⇒ x = \(\frac{6 \times 28}{4}\) = 42
Length of the lamp post = 42m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 4.
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Solution:
In ∆PQT and ∆STR we have
∠P = ∠S = 90° (Given)
∠PTQ = ∠STR (Vertically opposite angle)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 5
By AA similarity
∆PTQ ~ ∆STR we get
\(\frac { PT }{ ST } \) = \(\frac { TQ }{ TR } \)
PT × TR = ST × TQ
Hence it is proved.

Question 5.
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 6
Solution:
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE (By AA similarity)
\(\frac { BC }{ DE } \) = \(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
In ∆ABC, AB2 = BC2 + AC2
= 122 + 52 = 144 + 25 = 169
AB = \(\sqrt { 169 }\) = 13
Consider, \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
∴ AE = \(\frac{5 \times 3}{13}\) = \(\frac { 15 }{ 13 } \)
AE = \(\frac { 15 }{ 13 } \) and DE = \(\frac { 36 }{ 13 } \)
Consider, \(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \)
DE = \(\frac{12 \times 3}{13}\) = \(\frac { 36 }{ 13 } \)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 6.
In the adjacent figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 7
Solution:
Given ∆ACB ~ ∆APQ
\(\frac { AC }{ AP } \) = \(\frac { BC }{ PQ } \) = \(\frac { AB }{ AQ } \)
\(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
Consider \(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \)
4 AC = 8 × 2.8
AC = \(\frac{8 \times 2.8}{4}\) = 5.6 cm
Consider \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
8 AQ = 4 × 6.5
AQ = \(\frac{4 \times 6.5}{8}\) = 3.25 cm
Length of AC = 5.6 cm; Length of AQ = 3.25 cm

Question 7.
If figure OPRQ is a square and ∠MLN = 90°. Prove that
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 8
(i) ∆LOP ~ ∆QMO
(ii) ∆LOP ~ ∆RPN
(iii) ∆QMO ~ ∆RPN
(iv) QR2 = MQ × RN.
Solution:
(i) In ∆LOP and ∆QMO
∠OLP = ∠OQM = 90°
∠LOP = ∠OMQ (Since OQRP is a square OP || MN)
∴ ∆LOP~ ∆QMO (By AA similarity)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

(ii) In ∆LOP and ∆RPN
∠OLP = ∠PRN = 90°
∠LPO = ∠PNR (OP || MN) .
∴ ∆LOP ~ ∆RPN (By AA similarity)

(iii) In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM (OP || MN)
∴ ∆QMO ~ ∆RPN (By AA similarity)

(iv) We have ∆QMO ~ ∆RPN
\(\frac { MQ }{ PR } \) = \(\frac { QO }{ RN } \)
\(\frac { MQ }{ QR } \) = \(\frac { QR }{ RN } \)
QR2 = MQ × RN
Hence it is proved.

Question 8.
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm2 and the area of ∆DEF is 16 cm2 and BC = 2.1 cm. Find the length of EF.
Solution:
Given ∆ABC ~ ∆DEF
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 9
\(\frac { 9 }{ 16 } \) = \(\frac{(2.1)^{2}}{\mathrm{E} \mathrm{F}^{2}}\)
(\(\frac { 3 }{ 4 } \))2 = (\(\frac { 2.1 }{ EF } \))2
\(\frac { 3 }{ 4 } \) = \(\frac { 2.1 }{ EF } \)
EF = \(\frac{4 \times 2.1}{3}\) = 2.8 cm
Legth of EF = 2.8 cm

Question 9.
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 10
Solution:
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
\(\frac { AP }{ BQ } \) = \(\frac { AC }{ BC } \)
\(\frac { 6 }{ y } \) = \(\frac { AC }{ BC } \)
∴ \(\frac { BC }{ AC } \) = \(\frac { y }{ 6 } \) …..(1)
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
\(\frac { RC }{ QB } \) = \(\frac { AC }{ AB } \)
\(\frac { 3 }{ y } \) = \(\frac { AC }{ AB } \)
\(\frac { AB }{ AC } \) = \(\frac { y }{ 3 } \) ……..(2)
By adding (1) and (2)
\(\frac { BC }{ AC } \) + \(\frac { AB }{ AC } \) = \(\frac { y }{ 6 } \) + \(\frac { y }{ 3 } \)
1 = \(\frac{3 y+6 y}{18}\)
9y = 18 ⇒ y = \(\frac { 18 }{ 9 } \) = 2
The Value of y = 2m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 10.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 2 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 2 }{ 3 } \) ).
Solution:
Given ∆PQR, we are required to construct another triangle whose sides are \(\frac { 2 }{ 3 } \) of the corresponding sides of the ∆PQR

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 11
Steps of construction:
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q1, Q2 and Q3 on QX.
So that QQ1 = Q1Q2 = Q2Q3
(iv) Join Q3 R and draw a line through Q2 parallel to Q3 R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 11.
Construct a triangle similar to a given triangle LMN with its sides equal to \(\frac { 4 }{ 5 } \) of the corresponding sides of the triangle LMN (scale factor \(\frac { 4 }{ 5 } \) ).
Solution:
Given a triangle LMN, we are required to construct another ∆ whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the ∆LMN.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 12

Steps of Construction:

  1. Construct a ∆LMN with any measurement.
  2. Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
  3. Locate 5 Points Q1, Q2, Q3, Q4, Q5 on MX.
    So that MQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q4Q5
  4. Join Q5 N and draw a line through Q4. Parallel to Q5N to intersect MN at N’.
  5. Draw a line through N’ parallel to the line LN to intersect ML at L’.
    ∴ ∆L’ MN’ is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 12.
Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac { 6 }{ 5 } \) of the corresponding sides of the triangle ABC (scale factor \(\frac { 6 }{ 4 } \)).
Solution:
Given triangle ∆ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct an ∆ABC with any measurement.
(ii) Draw a ray BX making an acute angle with BC.
(iii) Locate 6 points Q1, Q2, Q3, Q4, Q5, Q6 on BX such that
BQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q5Q6
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 13
(iv) Join Q5 to C and draw a line through Q6 parallel to Q5 C intersecting the extended line BC at C’.
(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.
∆A’BC’ is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1

Question 13.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 7 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 7 }{ 3 } \)).
Solution:
Given triangle ABC, we are required to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points Q1, Q2, Q3, Q4, Q5, Q6, Q7 on QX.
So that
QQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q5Q6 = Q6Q7

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.1 14
(iv) Join Q3 to R and draw a line through Q7 parallel to Q3R intersecting the extended line segment QR at R’.
(v) Draw a line through parallel to RP.
Intersecting the extended line segment QP at P’.
∴ ∆P’QR’ is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Solution:
Let the initial position of the man be “O” and his final
position be “B”.
By Pythagoras theorem
In the right ∆ OAB,
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 1
OB2 = OA2 + AB2
= 182 + 242
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 2.
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 2
Solution:
Distance between Sarah House and James House using “C street”.
AC2 = AB2 + BC2
= 22 + 1.52
= 4 + 2.25 = 6.25
AC = \(\sqrt { 6.25 }\)
AC = 2.5 miles
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 3
Distance covered by using “A Street” and “B Street”
= (2 + 1.5) miles = 3.5 miles
Difference in distance = 3.5 miles – 2.5 miles = 1 mile

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 3.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
In the right ∆ABC,
By Pythagoras theorem
AC2= AB2 + BC2 = 342 + 412
= 1156 + 1681 = 2837
AC = \(\sqrt { 2837 }\)
= 53.26 m
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 4
Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 5
Solution:
Let the length of the rectangle be “a” and the breadth of the rectangle be “b”.
XY + YZ = 17 cm
b + a = 17 …….. (1)
In the right ∆ WXZ,
XZ2 = WX2 + WZ2
(XZ)2 = a2 + b2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 6
XZ = \(\sqrt{a^{2}+b^{2}}\)
Similarly WY = \(\sqrt{a^{2}+b^{2}}\) ⇒ XZ + WY = 26
2 \(\sqrt{a^{2}+b^{2}}\) = 26 ⇒ \(\sqrt{a^{2}+b^{2}}\) = 13
Squaring on both sides
a2 + b2 = 169
(a + b)2 – 2ab = 169
172 – 2ab = 169 ⇒ 289 – 169 = 2 ab
120 = 2 ab ⇒ ∴ ab = 60
a = \(\frac { 60 }{ b } \) ….. (2)
Substituting the value of a = \(\frac { 60 }{ b } \) in (1)
\(\frac { 60 }{ b } \) + b = 17
b2 – 17b + 60 = 0
(b – 2) (b – 5) = 0
b = 12 or b = 5
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 7
If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 5.
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side of the right ∆ be x.
∴ Hypotenuse = 6 + 2x
Third side = 2x + 6 – 2
= 2x + 4
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 8
In the right triangle ABC,
AC2 = AB2 + BC2
(2x + 6)2 = x2 + (2x + 4)2
4x2 + 36 + 24x = x2 + 4x2 + 16 + 16x
0 = x2 – 24x + 16x – 36 + 16
∴ x2 – 8x – 20 = 0
(x – 10) (x + 2) = 0
x – 10 = 0 or x + 2 = 0
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 9
x = 10 or x = -2 (Negative value will be omitted)
The side AB = 10 m
The side BC = 2 (10) + 4 = 24 m
Hypotenuse AC = 2(10) + 6 = 26 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 6.
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
In the right ∆ABC,
BC2 = AC2 – AB2 = 52 – 42
= 25 – 16 = 9
BC = \(\sqrt { 9 }\) = 3m.
When the foot of the ladder moved 1.6 m toward the wall.
The distance between the foot of the ladder to the ground is
BE = 3 – 1.6 m
= 1.4 m
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 10
Let the distance moved upward on the wall be “h” m
The ladder touch the wall at (4 + h) M
In the right triangle BED,
ED2 = AB2 + BE2
52 = (4 + h)2 + (1.4)2
25 – 1.96= (4 + h)2
∴ 4 + h = \(\sqrt { 23.04 }\)
4 + h = 4. 8 m
h = 4.8 – 4
= 0.8 m
Distance moved upward on the wall = 0.8 m

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 7.
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2.
Solution:
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = \(\frac { 1 }{ 4 } \) QR …..(1)
QS = 3SR
SR = \(\frac { QS }{ 3 } \) ……..(2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 11
From (1) and (2) we get
\(\frac { 1 }{ 4 } \) QR = \(\frac { QS }{ 3 } \)
∴ QS = \(\frac { 3 }{ 4 } \) QR ………(3)
In the right ∆ PQS,
PQ2 = PS2 + QS2 ……….(4)
Similarly in ∆ PSR
PR2 = PS2 + SR2 ………..(5)
Subtract (4) and (5)
PQ2 – PR2 = PS2 + QS2 – PS2 – SR2
= QS2 – SR2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 12
PQ2 – PR2 = \(\frac { 1 }{ 2 } \) QR2
2PQ2 – 2PR2 = QR2
2PQ2 = 2PR2 + QR2
Hence the proved.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3

Question 8.
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2.
Solution:
Since the Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3 13
In the right ∆ABD,
AD2 = AB2 + BD2
AD2 = AB2 + x2 ……….(1)
In the right ∆ABE,
AE2 = AB2 + 2BE2
AE2 = AB2 + 4X2 ………..(2) (BE = 2x)
In the right ∆ABC
AC2 = AB2 + BC2
AC2 = AB2 + 9x2 …………… (3) (BC = 3x)
R.H.S = 3AC2 + 5AD2
= 3[AB2 + 9x2] + 5 [AB2 + x2] [From (1) and (3)]
= 3AB2 + 27x2 + 5AB2 + 5x2
= 8AB2 + 32x2
= 8 (AB2 + 4 x2)
= 8AE2 [From (2)]
= R.H.S.
∴ 8AE2 = 3AC2 + 5AD2

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Students can download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.2

Question 1.
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
(i) If \(\frac { AD }{ DB } \) = \(\frac { 3 }{ 4 } \) and AC = 15 cm find AE.
(ii) If AD = 8x – 7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.
Solution:
(i) Let AE be x
∴ EC = 15 – x
In ∆ABC we have DE || BC
By Basic proportionality theorem, we have
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 1
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 3 }{ 4 } \) = \(\frac { x }{ 15-x } \)
4x = 3 (15 – x)
4x = 45 – 3x
7x = 45 ⇒ x = \(\frac { 45 }{ 7 } \) = 6.43
The value of x = 6.43

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

(ii) Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 8x-7 }{ 5x-3 } \) = \(\frac { 4x-3 }{ 3x-1 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 2
(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x2 – 8x – 21x + 7 = 20x2 – 12x – 15x + 9
24x2 – 20x2 – 29x + 27x + 7 – 9 = 0
4x2 – 2x – 2 = 0
2x2 – x – 1 = 0 (Divided by 2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 3
2x2 – 2x + x – 1 = 0
2x(x -1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = -1 ⇒ x = – \(\frac { 1 }{ 2 } \) (Negative value will be omitted)
The value of x = 1

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 2.
ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ
Solution:
Join AC intersecting PQ at S.
Let AP be x
∴ AD = x + 18
In the ∆ABC, QS || AB
By basic proportionality theorem.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 4
\(\frac { AS }{ SC } \) = \(\frac { BQ }{ QC } \)
\(\frac { AS }{ SC } \) = \(\frac { 35 }{ 15 } \) ………(1)
In the ∆ACD; PS || DC
By basic proportionality theorem.
\(\frac { AS }{ SC } \) = \(\frac { AP }{ PD } \)
\(\frac { AS }{ SC } \) = \(\frac { x }{ 18 } \) ………..(2)
From (1) and (2) we get
\(\frac { 35 }{ 15 } \) = \(\frac { x }{ 18 } \)
15x = 35 × 18 ⇒ x = \(\frac{35 \times 18}{15}\) = 42
AD = AP + PD
= 42 + 18 = 60
The value of AD = 60 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 3.
In ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC.
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
Solution:
(i) Here AB = 12 cm; BD =12 – 8 = 4 cm; AE =12 cm; EC = 18 – 12 = 6 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 5
∴ \(\frac { AD }{ DB } \) = \(\frac { 8 }{ 4 } \) = 2
\(\frac { AE }{ EC } \) = \(\frac { 12 }{ 6 } \) = 2
\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
By converse of basic proportionality theorem DE || BC

(ii) Here AB = 5.6 cm; AD = 1.4 cm;
BD = AB – AD
= 5.6 – 1.4 = 4.2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 6
AC = 7.2 cm; AE = 1.8 cm
EC = AC – AE
= 7.2 – 1.8
EC = 5.4 cm
\(\frac { AD }{ DB } \) = \(\frac { 1.4 }{ 4.2 } \) = \(\frac { 1 }{ 3 } \)
\(\frac { AE }{ EC } \) = \(\frac { 1.8 }{ 5.4 } \) = \(\frac { 1 }{ 3 } \)
\(\frac { AE }{ EC } \) = \(\frac { AD }{ DB } \)
By converse of basic proportionality theorem DE || BC

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 4.
In fig. if PQ || BC and BC and PR || CD prove that
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 7
(i) \(\frac { AR }{ AD } \) = \(\frac { AQ }{ AB } \)
(ii) \(\frac { QB }{ AQ } \) = \(\frac { DR }{ AR } \)
Solution:
(i) In ∆ABC, We have PQ || BC
By basic proportionality theorem
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 8
\(\frac { AQ }{ AB } \) = \(\frac { AP }{ AC } \) ……(1)
In ∆ACD, We have PR || CD
basic proportionality theorem
\(\frac { AP }{ AC } \) = \(\frac { AR }{ AD } \) ………..(2)
From (1) and (2) we get
\(\frac { AQ }{ AB } \) = \(\frac { AR }{ AD } \) (or) \(\frac { AR }{ AD } \) = \(\frac { AQ }{ AB } \)

(ii) In ∆ABC, PQ || BC (Given)
By basic proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { AQ }{ QB } \) ………..(1)
In ∆ADC, PR || CD (Given)
By basic proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { AR }{ RD } \) ………(2)
From (1) and (2) we get
\(\frac { AQ }{ QB } \) = \(\frac { AP }{ RD } \) (or) \(\frac { QB }{ AQ } \) = \(\frac { RD }{ AR } \)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 5.
Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
Solution:
Let the side of the rhombus be “x”. Since PQRB is a Rhombus PQ || BC
By basic proportionality theorem
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 9
\(\frac { AP }{ AB } \) = \(\frac { PQ }{ BC } \) ⇒ \(\frac { 12-x }{ BC } \) = \(\frac { x }{ 6 } \)
12x = 6 (12 – x)
12x = 72 – 6x
12x + 6x = 72
18x = 72 ⇒ x = \(\frac { 72 }{ 18 } \) = 4
Side of a rhombus = 4 cm
PQ = RB = 4 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 6.
In trapezium ABCD, AB || DC , E and F are points on non-parallel sides AD and BC respectively, such that EF || AB.
Show that = \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)
Solution:
Given: ABCD is a trapezium AB || DC
E and F are the points on the side AD and BC
EF || AB
To Prove: \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 10
Construction: Join AC intersecting AC at P
Proof:
In ∆ABC, PF || AB (Given)
By basic proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { BF }{ FC } \) ………..(1)
In the ∆ACD, PE || CD (Given)
By basic Proportionality theorem
\(\frac { AP }{ PC } \) = \(\frac { AE }{ ED } \) …………..(2)
From (1) and (2) we get
\(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 7.
In figure DE || BC and CD || EE Prove that AD2 = AB × AF.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 11
Solution:
Given: In ∆ABC, DE || BC and CD || EF
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 12
To Prove: AD2 = AB × AF
Proof: In ∆ABC, DE || BC (Given)
By basic proportionality theorem
\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \) ……….. (1)
In ∆ADC; FE || DC (Given)
By basic Proportionality theorem
\(\frac { AD }{ AF } \) = \(\frac { AC }{ AE } \) ……..(2)
From (1) and (2) we get
\(\frac { AB }{ AD } \) = \(\frac { AD }{ AF } \)
AD2 = AB × AF
Hence it is proved

Question 8.
In ∆ABC, AD is the bisector of ∠A meeting side BC at D, if AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
Solution:
In ∆AABC AD is the internal bisector of ∠A
Given BC = 6 cm
Let BD = x ∴ DC = 6 – x cm
By Angle bisector theorem
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
\(\frac { x }{ 6-x } \) = \(\frac { 10 }{ 14 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 13
14x = 60 – 10x
24x = 60
x = \(\frac { 60 }{ 24 } \) = \(\frac { 10 }{ 4 } \) = 2.5
BD = 2.5 cm;
DC = 6 – x ⇒ 2.5 = 3.5 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 9.
Check whether AD is bisector of ∠A of ∆ABC in each of the following,
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm.
(ii) AB = 4 cm, AC 6 cm, BD = 1.6 cm and CD = 2.4 cm.
Solution:
(i) In ∆ABC, AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm
\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3.5 } \) = \(\frac { 15 }{ 35 } \) = \(\frac { 3 }{ 7 } \)
\(\frac { AB }{ AC } \) = \(\frac { 5 }{ 10 } \) = \(\frac { 1 }{ 2 } \)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 14
\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)
∴ AD is not a bisector of ∠A.

(ii) In ∆ABC, AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 15
\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \)
∴ \(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
By angle bisector theorem; AD is the internal bisector of ∠A

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 10.
In figure ∠QPR = 90°, PS is its bisector.
If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR.
Solution:
Given: ∠QPR = 90°; PS is the bisector of ∠P. ST ⊥ ∠PR
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 16
To prove: ST × (PQ + PR) = PQ × PR
Proof: In ∆ PQR, PS is the bisector of ∠P.
∴ \(\frac { PQ }{ QR } \) = \(\frac { QS }{ SR } \)
Adding (1) on both side
1 + \(\frac { PQ }{ QR } \) = 1 + \(\frac { QS }{ SR } \)
\(\frac { PR+PQ }{ PR } \) = \(\frac { SR+QS }{ SR } \)
\(\frac { PQ+PR }{ PR } \) = \(\frac { QR }{ SR } \) ……….(1)
In ∆ RST And ∆ RQP
∠SRT = ∠QRP = ∠R (Common)
∴ ∠QRP = ∠STR = 90°
(By AA similarity) ∆ RST ~ RQP
\(\frac { SR }{ QR } \) = \(\frac { ST }{ PQ } \)
\(\frac { QR }{ SR } \) = \(\frac { PQ }{ ST } \) ……..(2)
From (1) and (2) we get
\(\frac { PQ+PR }{ PR } \) = \(\frac { PQ }{ ST } \)
ST (PQ + PR) = PQ × PR

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 11.
ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.
Solution:
ABCD is a quadrilateral. AB = AD.
AE and AF are the internal bisector of ∠BAC and ∠DAC.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 17
To prove: EF || BD.
Construction: Join EF and BD
Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.
By Angle bisector theorem, we have,
∴ \(\frac { AB }{ AC } \) = \(\frac { BE }{ EC } \) ………(1)
In ∆ ADC, AF is the internal bisector of ∠DAC
By Angle bisector theorem, we have,
\(\frac { AD }{ AC } \) = \(\frac { DF }{ FC } \)
∴ \(\frac { AB }{ AC } \) = \(\frac { DF }{ FC } \) (AB = AD given) ………(2)
From (1) and (2), we get,
\(\frac { BE }{ EC } \) = \(\frac { DF }{ FC } \)
Hence in ∆ BCD,
BD || EF (by converse of BPT)

Question 12.
Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 18

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 19
Steps of construction

  1. Draw a line segment PQ = 4.5 cm
  2. At P, draw PE such that ∠QPE = 60°
  3. At P, draw PF such that ∠EPF = 90°
  4. Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
  5. With O as centre and OP as radius draw a circle.
  6. From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
  7. Join PR and RQ. PQR is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 13.
Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 20
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 21
Steps of construction

  1. Draw a line segment RQ = 5 cm.
  2. At R draw RE such that ∠QRE = 40°
  3. At R, draw RF such that ∠ERF = 90°
  4. Draw the perpendicular bisector to RQ, which intersects RF at O and RQ at G.
  5. With O as centre and OP as radius draw a circle.
  6. From G mark arcs of radius 4.4 cm on the circle. Mark them as P and S.
  7. Join PR and PQ. Then ∆PQR is the required triangle.
  8. From P draw a line PN which is perpendicular to RQ it meets at N.
  9. Measure the altitude PN.
    PN = 2.2 cm.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 14.
Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 22
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 23
Steps of construction

  1. Draw a line segment QR = 6.5 cm.
  2. At Q draw QE such that ∠RQE = 60°.
  3. At Q, draw QF such that ∠EQF = 90°.
  4. Draw the perpendicular of QR which intersects QF at O and QR at G.
  5. With O as centre and OQ as radius draw a circle.
  6. X Y intersects QR at G. On X Y, from G mark an arc at M. Such that GM = 4.5 cm.
  7. Draw AB through M which is parallel to QR.
  8. AB Meets the circle at P and S.
  9. join QP and RP.
    PQR is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 15.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 24
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 25
Steps of construction

  1. Draw a line segment AB = 5.5 cm.
  2. At A draw AE such that ∠BAE = 25°.
  3. At A draw AF such that ∠EAF = 90°.
  4. Draw the perpendicular bisector of AB which intersects AF at O and AB at G.
  5. With O as centre and OB as radius draw a circle.
  6. X Y intersects AB at G. On X Y, from G mark an arc at M. Such that GM = 4 cm.
  7. Through M draw a line parallel to AB intersect the circle at C and D.
  8. Join AC and BC.
    ABC is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 16.
Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 26
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 27
Steps of construction

  1. Draw a line segment BC = 5.6 cm.
  2. At B draw BE such that ∠CBE = 40°.
  3. At B draw BF such that ∠EBF = 90°.
  4. Draw the perpendicular bisector to BC which intersects BF at O and BC at G.
  5. With O as centre and OB as radius draw a circle.
  6. From C mark an arc of 4 cm on CB at D.
  7. The perpendicular bisector intersects the circle at I. Joint ID.
  8. ID produced meets the circle at A. Now Join AB and AC.
    This ABC is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2

Question 17.
Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 28
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.2 29
Steps of construction

  1. Draw a line segment PQ = 6.8 cm.
  2. At P draw PE such that ∠QPE = 50°.
  3. At P draw PF such that ∠EPF = 90°.
  4. Draw the perpendicular bisector to PQ which intersects PF at O and PQ at G.
  5. With O as centre and OP as radius draw a circle.
  6. From P mark an arc of 5.2 cm on PQ at D.
  7. The perpendicular bisector intersects the circle at I. Join ID.
  8. ID produced meets the circle at A. Now Joint PR and QR. This PQR is the required triangle.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.9

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Determine the quadratic equations, whose sum and product of roots are
(i) -9, 20
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (-9) x + 20 = 0 ⇒ x2 + 9x + 20 = 0

(ii) \(\frac { 5 }{ 3 } \), 4
Answer:
Sum of the roots = \(\frac { 5 }{ 3 } \); Product of the roots = 4
The Quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (\(\frac { 5 }{ 3 } \)) x + 4 = 0 ⇒ x2 – \(\frac { 5 }{ 3 } \) x + 4 = 0
3x2 – 5x + 12 = 0

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.9

(iii) \(\frac { -3 }{ 2 } \), -1
Answer:
Sum of the roots = \(\frac { -3 }{ 2 } \); Product of the roots = -1
The Quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (-\(\frac { 3 }{ 2 } \)) x + (-1) = 0 ⇒ x2 + \(\frac { 3 }{ 2 } \) x – 1 = 0
2x2 + 3x – 2 = 0

(iv) – (2 – a)2, (a + 5)2
Answer:
Sum of the roots = – (2 – a)2; Product of the roots = (a + 5)2
x2 – (sum of the roots) x + product of the roots = 0
x2 – [-(2 – a)2] x + (a + 5)2 = 0
x2 + (2 – a)2 x + (a + 5)2 = 0

Question 2.
Find the sum and product of the roots for each of the following quadratic equations
(i) x2 + 3x – 28 = 0
(ii) x2 + 3x = 0
(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(iv) 3y2 – y – 4 = 0
Solution:
(i) x2 – (-3)x + (-28) = 0.
Comparing this with x2 – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x2 + 3x = 0 = x2 – (-3)x + 0 = 0
x2 – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + \(\frac { 1 }{ a } \) = \(\frac{10}{a^{2}}\)
Answer:
Multiply by a2
3a2 + a = 10
3a2 + a – 10 = 0
Sum of the roots (α + β) = \(\frac { -1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -10 }{ 3 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.9

(iv) 3y2 – y – 4 = 0
Answer:
Sum of the roots (α + β) = \(\frac { -(-1) }{ 3 } \) = \(\frac { 1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -4 }{ 3 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x4 – 12x3 + 42x2 – 36x + 9
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 1
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8

(ii) 31 x2 – 28x3 + 4x4 + 42x + 9
Answer:
Rearrange the order we get
4x4 – 28x3 + 37x2 + 42x + 9
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8

(iii) 16x4 + 8x2 + 1
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 5
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 6

(iv) 121 x4 – 198x3 – 183x2 + 216x + 144
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 7
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 8

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 11
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x4 – 12x3 + 37x2 + bx + a
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 112
Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax4 + bx3 + 361x2 + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x2 + bx3 + ax4
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 12
Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 13
Since it is a perfect square
\(\frac { 1 }{ x } \) (m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8

(ii) x4 – 8x3 + mx2 + nx + 16
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.8 14
Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Find the square root of the following.
(i) \(\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7

(ii) \(\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 2

(iii) \(\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 3

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7

Question 2.
Find the square root of the following
(i) 4x2 + 20x + 25
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 4

(ii) 9x2 – 24xy + 30xz – 40yz + 25z2 + 16y2
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 5

(iii) \(1+\frac{1}{x^{6}}+\frac{2}{x^{3}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 6

(iv) (4x2 – 9x + 2)(7x2 – 13x – 2)(28x2 – 3x – 1)
Answer:
4x2 – 9x +2 = 4x2 – 8x – x + 2
= 4x(x – 2)-1 (x – 2)
= (x – 2)(4x – 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 7
7x2 – 13x – 2 = 7x2 – 14x + x – 2
= 7x (x – 2) + 1 (x – 2)
= (x – 2) (7x + 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 8
28x2 – 3x – 1 = 28x2 – 7x + 4x – 1
= 7x (4x – 1) + 1 (4x – 1)
= (4x – 1) (7x + 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 9

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7

(v) (2x2 + \(\frac { 17 }{ 6 } \)x + 1) (\(\frac { 3 }{ 2 } \) x2 + 4x + 2) (\(\frac { 4 }{ 3 } \) x2 + \(\frac { 11 }{ 3 } \) x + 2)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 10
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 12
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 13

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 134
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.7 14

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

Students can download Maths Chapter 3 Algebra Ex 3.6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.6

Question 1.
Simplify
(i) \(\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}\)
Answer:
\(\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}\) = \(\frac{x(x+1)+x(1-x)}{x-2}\)
= \(\frac{x^{2}+x+x-x^{2}}{x-2}\)
= \(\frac { 2x }{ x-2 } \)

(ii) \(\frac { x+2 }{ x+3 } \) + \(\frac { x-1 }{ x-2 } \)
Answer:
\(\frac { x+2 }{ x+3 } \) + \(\frac { x-1 }{ x-2 } \) = \(\frac{(x+2)(x-2)+(x-1)(x+3)}{(x+3)(x-2)}\)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

(iii) \(\frac{x^{3}}{x-y}+\frac{y^{3}}{y-x}\) = \(\frac{x^{3}}{x-y}+\frac{y^{3}}{-1(x-y)}\)
Answer:
= \(\frac{x^{3}}{x-y}-\frac{y^{3}}{x-y}\)
= \(\frac{x^{3}-y^{3}}{x-y}\) (using a3 – b3 = (a – b) (a2 + ab + b2))
= \(\frac{(x-y)\left(x^{2}+x y+y^{2}\right)}{x-y}\)
= x2 + xy + y2

Question 2.
Simplify
(i) \(\frac{(2 x+1)(x-2)}{x-4}\) – \(\frac{\left(2 x^{2}-5 x+2\right)}{x-4}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 2

(ii) \(\frac{4 x}{x^{2}-1}-\frac{x+1}{x-1}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 5
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 4

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

Question 3.
Subtract \(\frac{1}{x^{2}+2}\) from \(\frac{2 x^{3}+x^{2}+3}{\left(x^{2}+2\right)^{2}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 6

Question 4.
Which rational expression should be subtracted from \(\frac{x^{2}+6 x+8}{x^{3}+8}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 7

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

Question 5.
If A = \(\frac{2x+1}{2 x-1}\), B = \(\frac{2x-1}{2x+1}\) find \(\frac{1}{A-B}\) – \(\frac{2 \mathbf{B}}{\mathbf{A}^{2}-\mathbf{B}^{2}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 8

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

Question 6.
If A = \(\frac { x }{ x+1 } \) B = \(\frac { 1 }{ x+1 } \) prove that \(\frac{(\mathbf{A}+\mathbf{B})^{2}+(\mathbf{A}-\mathbf{B})^{2}}{\mathbf{A}+\mathbf{B}}=\frac{2\left(x^{2}+1\right)}{x(x+1)^{2}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 9
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6 10

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

Question 7.
Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?
Solution:
Pari: time required to complete the work = 4 hrs.
∴ In 1 hr. he will complete = \(\frac{1}{4}\) of the work.
= \(\frac{1}{4}\) w.
Yuvan: Time required to complete the work = 6 hrs.
∴ In 1 hr. he will complete the \(\frac{1}{6}\) of the work
= \(\frac{1}{6}\) w
Working together, in 1 hr. they will complete \(\frac{w}{4}+\frac{w}{6}\) of the work.
= \(\frac{6 w+4 w}{24}=\frac{5}{12} w\)
∴ To complete the total work time taken
= \(\frac{w}{\frac{5}{12} w}=\frac{12}{5}\) = 2.4 hrs. [∵ (4) hrs = 4 × 60 = 24 min]
= 2 hrs 24 minutes.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.6

Question 8.
Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought ? 1800 worth of apples and ₹ 600 worth bananas, then how many kgs of each fruit did she buy?
Let the weight of applies be a kg.
Let the weight of bananas be b kg.
a + b = 50
ax = ₹ 1800 ………….. (1)
by = ₹ 600 ………… (2)
x = 2y …………… (3)
Use (3) in (1) ⇒ a(2y) = 1800
y = \(\frac{900}{a}\) ………….. (4)
Use (4) in (2) ⇒ \(b\frac{900}{a}\) = 600
∵ 3b = 2a …………….. (5)
∵ a + b = 50
a + \(\frac{2 a}{3}\) = 50 ⇒ \(\frac{5 a}{3}\) = 50
⇒ a = 50 × \(\frac{3}{5}\)
= 30
∴ b = 20
∴ Iniya bought 30 kg of applies and 20 kg of bananas
= 30
.’. b = 20.
.’. Iniya bought 30 kg of applies and 20 kg of bananas.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

Students can download Maths Chapter 3 Algebra Ex 3.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 1.
Simplify
(i) \(\frac{4 x^{2} y}{2 z^{2}} \times \frac{6 x z^{3}}{20 y^{4}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

(ii) \(\frac{p^{2}-10 p+21}{p-7} \times \frac{p^{2}+p-12}{(p-3)^{2}}\)
Answer:
P2 – 10p + 21 = (p – 7) (p – 3)
p2 + p – 12 = (p + 4) (p – 3)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 2
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 4

(iii) \(\frac{5 t^{3}}{4 t-8} \times \frac{6 t-12}{10 t}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 5

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

Question 2.
Simplify
(i) \(\frac{x+4}{3 x+4 y} \times \frac{9 x^{2}-16 y^{2}}{2 x^{2}+3 x-20}\)
Answer:
9x2 – 16y2 = (3x)2 – (4y)2
= (3x + 4y) (3x – 4y)
2x2 + 3x – 20 = 2x2 + 8x – 5x – 20
= 2x (x + 4) – 5 (x + 4)
= (x + 4) (2x – 5)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 6

(ii) \(\frac{x^{3}-y^{3}}{3 x^{2}+9 x y+6 y^{2}} \times \frac{x^{2}+2 x y+y^{2}}{x^{2}-y^{2}}\)
Answer:
x3 – y3 = (x – y) (x2 + xy + y2)
x2 + 2xy + y2 = (x + y) (x + y)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 7
3x2 + 9xy + 6y2 = 3(x2 + 3xy + 2y2)
= 3 (x + 2y) (x + y)
(x2 – y2) = (x + y) (x – y)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 8

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

Question 3.
Simplify
(i) \(\frac{2 a^{2}+5 a+3}{2 a^{2}+7 a+6} \div \frac{a^{2}+6 a+5}{-5 a^{2}-35 a-50}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 9
2 a2 + 5a + 3a + 3 = 2a2 + 2a + 3a + 3
= 2a(a + 1) + 3 (a + 1)
= (a + 1) (2a + 3)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 10
2a2 + 7a + 6 = 2a2 + 3a + 4a + 6
= a(2a + 3) + 2 (2a + 3)
= (2a + 3) (a + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 11
a2 + 6a + 5 = (a + 5) + (a + 1)
-5a2 – 35a – 50 = -5(a2 + 7a + 10)
= -5(a + 5)(a + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 12
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 13

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

(ii) \(\frac{b^{2}+3 b-28}{b^{2}+4 b+4}+\frac{b^{2}-49}{b^{2}-5 b-14}\)
Solution:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 14
b2 + 3b – 28 = (b + 7) (b – 4)
b2 + 4b + 4 = (b + 2) (b + 2)
b2 – 49 = b2 – 72
= (b + 7) (b – 7)
b2 – 5b – 14 = (b – 7) (b + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 15
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 16
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 17

(iii) \(\frac{x+2}{4 y}+\frac{x^{2}-x-6}{12 y^{2}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 18
x2 – x – 6 = (x – 3) (x + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 19

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

(iv) \(\frac{12 t^{2}-22 t+8}{3 t} \div \frac{3 t^{2}+2 t-8}{2 t^{2}+4 t}\)
Answer:
12t2 – 22t + 8 = 2(6t2 – 11t + 4)
= 2[6t2 – 8t – 3t + 4]
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 20
= 2[2t (3t – 4) – 1 (3t – 4)]
= 2(3t – 4) (2t – 1)
3t2 + 2t – 8 = 3t2 + 6t – 4t – 8
= 3t(t + 2) – 4 (t + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 21
= (t + 2) (3t – 4)
2t2 + 4t = 2t(t + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 22

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5
Question 4.
If x = \(\frac{a^{2}+3 a-4}{3 a^{2}-3}\) and y = \(\frac{a^{2}+2 a-8}{2 a^{2}-2 a-4}\) find the value of x2y-2
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 25
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 23
The value of x2 y-2 = \(\frac{x^{2}}{y^{2}}\) = (\(\frac { x }{ y } \))2
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 27

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5

Question 5.
If a polynomial p(x) = x2 – 5x – 14 when divided by another polynomial q(x) gets reduced to \(\frac { x-7 }{ x+2 } \) find q(x).
Answer:
p(x) = x2 – 5x – 14
= (x – 7) (x + 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.5 28
By the given data
\(\frac { p(x) }{ q(x) } \) = \(\frac { (x-7) }{ x+2 } \)
\(\frac{(x-7)(x+2)}{q(x)}\) = \(\frac { (x-7) }{ x+2 } \)
q(x) × (x – 7) = (x – 7) (x + 2) (x + 2)
q(x) = \(\frac{(x-7)(x+2)(x+2)}{(x-7)}\)
= (x + 2)2
q(x) = x2 + 4x + 4

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.2

Question 1.
For what values of natural number n, 4th can end with the digit 6?
Answer:
4n = (22)n = 22n
= 2n × 2n
2 is a factor of 4n
∴ 4n is always even.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 2.
If m, n are natural numbers, for what values of m, does 2n × 5n ends in 5?
Solution:
2n × 5m
2n is always even for all values of n.
5m is always odd and ends with 5 for all values of m.
But 2n × 5m is always even and ends in 0.
∴ 2n × 5m cannot end with the digit 5 for any values of m. No value of m will satisfy 2n × 5m ends in 5.

Question 3.
Find the H.C.F. of 252525 and 363636.
Answer:
To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0
By division of Euclid’s algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0
Again by division of Euclid’s algorithm
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0
Again by division of Euclid’s algorithm.
30303 = 20202 + 10101
The remainder 10101 ≠ 0
Again by division of Euclid’s algorithm.
20202 = 10101 × 2 + 0
The remainder is 0
∴ The H.C.F. is 10101

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 4.
If 13824 = 2a × 3b then find a and b?
Answer:
Using factor tree method factorise 13824
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 2
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 1
13824 = 29 × 33
Given 13824 = 2a × 3b
Compare we get a = 9 and b = 3

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Aliter:
13824 = 29 × 33
Compare with
13824 = 2a × 3b
The value of a = 9 b = 3

Question 5.
If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1 p2, p3, p4 are primes in ascending order and x1, x2, x3, x4, are integers, find the value of p1,p2,p3,p4 and x1,x2,x3,x4.
Answer:
Given 113400 = p1x1 × p2x2 × p3x3 × p4x4
Using tree method factorize 113400
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 3
113400 = 23 × 34 × 52 × 7
compare with
113400 = p1x1 × p2x2 × p3x3 × p4x4
P1 = 2, x1 = 3
P2 = 3, x2 = 4
P3 = 5, x3 = 2
P4 = 7, x4 = 1

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Answer:
Factorise 408 and 170 by factor tree method
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 4
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 5
408 = 23 × 3 × 17
170 = 2 × 5 × 17
To find L.C.M. list all prime factors of 408 and 170 of their greatest exponents.
L.C.M. = 23 × 3 × 5 × 17
= 2040
To find the H.C.F. list all common factors of 408 and 170.
H.C.F. = 2 × 17 = 34
L.C.M. = 2040 ; HCF = 34

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
Answer:
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 6
24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
L.C.M = 23 × 32 × 5
= 360
To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360
The greatest number in 6 digits = 999999 – 279
= 999720
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 7

Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Solution:
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640
∴ Required number = 3647 which leaves remainder 7 in each case.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 9.
Find the least number that is divisible by the first ten natural numbers?
Answer:
Find the L.C.M of first 10 natural numbers
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 8
The least number is 2520

Modular Arithmetic
Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).

Euclid’s Division Lemma and Modular Arithmetic

Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.
This n = r (mod m)