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Haryana State Board HBSE 6th Class Social Science Solutions Civics Chapter 8 Rural Livelihoods Textbook Exercise Questions and Answers.

Haryana Board 6th Class Social Science Solutions Civics Chapter 8 Rural Livelihoods

HBSE 6th Class Civics Rural Livelihoods Textbook Questions and Answers

Class 6 Civics Chapter 8 HBSE Rural Livelihoods Question 1.
Based on the above diagram would you say the Thulasi earns money throughout the year?
Answer:
No. Thulasi works in seasonal job which is agriculture she earns only during those seasons when she has work on her fields. According to the given data, it can be important that from June to December, she earns.

Class 6 Civics Chapter 8 Question Answer HBSE  Question 2.
Describe the work that Thulasi does. How is it different from the work that Raman does?
Answer:
Thulasi transplants paddy in the fields. Thulasi’s husband, is also a labourer. They don’t own any land. While Thulsi transplants, Raman sprays pesticides.

Rural Livelihoods Class 6 Civics Chapter 8 Question Answer Question 3.
Thulasi gets paid very little money for the work. She does, why do you think agricultural labourers like her are forced to accept low wages.
Answer:
1. Agricultural labourers are unskilled and do not have the knowledge or access to higher paid jobs.
2. They are under the debt of landlords or moneylenders who employ them on low wages.

Question Answer Class 6 Civics Chapter 8 HBSE Rural Livelihoods Question 3.
In what ways would her way of earning a living have been different if Thulasi own some farm land? Discuss.
Answer:
1. If Thulasi would have owned a farm land she would have been the possessor of all the process.
2. She would have an increased income by selling all her-crop to the market.
3. This would have an impact on her livelihood.

Question 4.
What are the crops grown in your region or nearby rural are? What kinds of works do agricultural labourers do?
Answer:
Maize, Wheat and Rice is planted in the neighbourhood village.
Agricultural labourers do the following works:
(a) Ploughing fields
(b) Preparing saplings
(c) Transplanting saplings
(d) Irrigating crops
(e) Weeding
(f) Harvesting
(g) Storage of crops.

Question 5.
What work does Sekar’s family do. Why do you think Sekar does not usually employ labourers for doing farm work?
Answer:
Sekar’s family works in the field, harvests crop and does all sort of agricultural work.
Sekar might not be employing labourers for farm work to cut cost on manual labour. Money has to be given to labour for doing work.

Question 6.
Why does Sekar not go to the town market to get a better price for his paddy?
Answer:
Sekar earns very less to satisfy his needs. He cannot earn a higher income. There is very little produce lift. Some is used in loans, some so satisfy their home needs.

Question 7.
Sekar’s sister Mina had also taken a loan from the trader. She does not loan to sell her paddy to him but will pay back her loan. Write an imaginary conversation between Mina and the trader’s agent and the arguments given by each person.
Answer:
AGENT: Mina/when will you repay me loan? How long can I excuse you?
MINA : Just fifteen days more, Sir/ I will repay it as soon as I sell my paddy.
AGENT : But I have already offered you excuse for one month.
MINA : Sorry, Sir.
AGENT : Next time, I’ll not take off.

Question 8.
What are the similarities between Sekar’s and Thulasi lives. Your answer could be based on the land they have, the need to work on the land that belong to others or the loans that they need and their earning.
Answer:
Both Sekar and Thulasi are agricultural labourers who accept loans (even at higher interest to meet then daily needs, they have low wage difference.
Thulasi owns no land while Sekar owns two acres of field on which he works.
Thulasi has a definite source of income paid by Ramalingam while Sekar’s income depends upon his crop produce.

Read again Sekar’s and Thulasi’s accounts. What do they say about Ramalingam, the large farmer? Together with what you have read fill in the details below.

Question i.
How much land does he have?
Answer:
20 acres.

Question ii.
What does Ramalingam do with the paddy grown in his field?
Answer:
The paddy is sold to traders in nearby towns.

Question iii.
A part from farming how else does he earn?
Answer:
His family owns a rice mill and shop selling seeds, pesticides. This given them extra substantial income.

Question 9.
Why do both Sekar’s and Aruna’s family have to borrow? What similarities and differences do you find?
Answer:
Sekar’s family had to borrow loan to as keep their produce while Aruna buys loan for the maintenance of their requirement. Sekar buys loan from an informal source of credit while Aruna busy loans from formal source of credit.

Question 10.
Have you heard of Tsunami? What is this and what damage do you think it might have done to the life of fishing families like Aruna’s.
Answer:
Do yourself.

EXERCISE

Short Answer Type Questions

Question 1.
You have probably noticed that people in Kalpattu are engaged in a variety of non-farm work. List five of these.
Answer:
(i) Making Baskets,
(ii) Making pots
(iii) Making bullocks carts
(iv) Making utensil
(v) Making bricks. .

Question 2.
List the different types of people you read about in Kalpattu who depend on farming. Who is the poorest among them and why?
Answer:
In Kalpattu, there are people who provides service such as blacksmiths, nurses, washerman, weavers, barbers, cycle repair mechanics, people whose work depends on farming are shopkeepers, traders, peasants etc.
Labour is the poorest of all sense it works only when in demand.

Question 3.
Imagine you are a member of a fishing family and you are discussing whether to take a loan from the bank for an engine. What would you say?
Answer:
If I were a member of the fishing family, I would say we must take loan from a bank or cooperative to improve the condition of our occupation because:
(a) Catamaran engine is useful because with their help, fishes can be caught in deep waters.
(b) It will help increase income, reduce time and help in overall development.

Question 4.
Poor rural labourers like Thulasi often do not have access to good medical facilities, good schools, and other resources. You have read about inequality in the first unit of this text. The difference between her and Ramalingam is one of inequality. Do you think this is a fair situation? What do you think can be done?
Discuss in class.
Answer:
Thulasi is a labourer while Ramalingam is a rich landlord.
1. Thulasi has a poor living standard, Ramalingam has a higher living standard.
2. Thulasi leads a sad life while Ramalingam leads happy life.

Question 5.
What do you think the government can do to help farmers like Sekar when they get into debt? Discuss.
Answer:
Debts-trap is a common problem for farmers in India.
(i) Decrease the rate of interest at which loan is given.
(ii) Give compensation in case of a natural calamity.
(iii) Additional time can be given.
(iv) Formal sources of credit should be increase.

Question 6.
Compare the situation of Sekar and Ramalingam by filling in the following table:

Answer:

HBSE 6th Class Civics Rural Livelihoods Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is the main work of the people in the village of Kalpattu?
Answer:
Paddy is the main crop that is grown in irrigated lands. Most of the families earn a living through agriculture.

Question 2.
What is the type of cultivation in Nagaland?
Answer:
Terrace farming.

Question 3.
What are catamarans?
Answer:
A catamarans is a boat with turn hull, side by side or a raft made of logs of wood or pieces of timber lashed together.

Question 4.
Who are seasonal workers?
Answer:
Most small farmers, agricultural labourers, fishing families, crafts persons in the villages are seasonal

Question 5.
What operations do farming activities involve?
Answer:
Working on farms involve operations such as preparing the lands, sowing, weeding, harvesting of crops.

Question 6.
Name the main crop grown in irrigated lands.
Answer:
Paddy.

Short Answer Questions

Question 1.
Why is agriculture a seasonal job?
Answer:
Process of agricultures take place only during some seasons. Not being able to earn money forces people in many rural areas to travel long distances in search of work. This travel or migration takes place during some seasons.

Question 2.
List the different sources of income of rural areas?
Answer:
(a) Many people in rural areas depend upon forest, animal husbandry, dairy produce, fishing, etc.
(b) Collecting mahua, tendu leave, honey to be sold to traders is an important source of additional income.
(c) Selling .milk, to cooperative is another source.

Question 3.
What is the involvement coasts in crop production?
Answer:
Equipments that are related to agriculture, seeds, pesticides, fertilizers, minerals water, irrigational facilities.

Question 4.
Name the three category of farmers.
Answer:
The three categories of farmers are:
(i) Big land owners, who employ other farmers to work on their fields.
(ii) Farmers who have very small plots of land. They have to do other work, when there is a lean season on the farm. The produce is not sufficient to meet the needs of their family.
(iii) Landless peasants, who work on the fields of big land owners and are paid for the work done by them.

Question 5.
How do money-lenders and traders exploit the poor farmers and others? Give examples.
Answer:
The traders and money-lenders exploit the poor farmers and others by taking a part of their produce at a lower price, e.g., Sekar has to; sell his paddy at a lower price than the market rate. Aruna has to sell her fish catch to the trader and cannot auction it.

Long Answer Questions

Question 1.
Write about the everyday work of the village fisherman.
Answer:
Most of the fisherman live close to the sea. Their work starts at 7 o’clock in the morning, when there is lot of activity on the beach. This is the time when catamarans return with their catch and women gather to buy and sell fish.

The fisherman go in groups to the sea in catamaran. Some have engines fitted in the catamaran, so that they can go far into the sea. The money is equally divided in the group, keeping one part for the equipment. The women folk sell the fish till afternoon. In the evening the fishermen untangle and repair the nets. Next day at 2 A.M. they will set out to sea again.

Rural Livelihoods Class 6 HBSE Notes

• Blacksmith: A man engaged in making iron goods, tools, instruments etc.
• Villager: A citizen of India living in a village and occupied with agriculture.
• Chakhesang Community: People of Chizami Village in Nagaland.
• Community: The people living at one place, district or country are combined known with this term.
• Family: The smallest unit of society consisting of parents and their children.
• Heterogeneous: Consisting of many different kinds of people.
• Catamaran: Catamaran is a small boat used for catching fish.
• Migration: Movement of people from one place to another to find jobs is termed as migration.
• Fisherman: A person engaged in the work of catching fish.

Haryana State Board HBSE 6th Class Social Science Solutions Civics Chapter 7 Urban Administration Textbook Exercise Questions and Answers.

Haryana Board 6th Class Social Science Solutions Civics Chapter 7 Urban Administration

HBSE 6th Class Civics Urban Administration Textbook Questions and Answers

Urban Administration Class 6 Question Answers HBSE Question 1.
List at least four different tasks that the Municipal does.
Answer:
1. It manages to collect garbage and disposes at proper place.
2. It takes care of the situation of the town.
3. It builds roads and repair them whenever required.
4. It takes care of water supply and keeps the streets and roads clean.

Class 6 Civics Chapter 7 Fill In The Blanks HBSE Question 2.
Fill in the blanks in the sentences below.
1. In a panchayat the elected members are called ———— .
2. The city is divided into several ———— .
3. In a municipal corporation the elected members are called ———— .
4. Group of councillors deal with issues that affect the ———— .
5. Elections are held once every ———— years for the panchayat as well as for the municipality.
6. While the councillors make decisions the administrative staff led by the Commissioner ———— these.
Answer:
(1) Panchs
(2) Wards
(3) Councillors
(4) entire city
(5) five
(6) implement

Urban Administration Question Answer HBSE Question 3.
What was Gangabai protesting about?
Answer:
Gangabai was protesting about the filthy environment of her locality.

Class 6 Civics Chapter 7 HBSE Urban Administration Question 4.
Why do you think Gangabai decided to approach the Ward Councillor?
Answer:
Gangabai decided to approach the Ward Councillor because it was Ward Councillor’s duty to look after the sanitation in his locality.

Class 6 Urban Administration HBSE Question 5.
What did Gangabai say when the Commissioner told that there were not enough trucks in the city?
Answer:
When the Commissioner said that there were not enough trucks in the city, Gangabai said. But you seem to have enough trucks to, collect garbage from the rich localities.

EXERCISE

Question 1.
Why did the children go to Yasmin Khala’s house?
Answer:
The children had broken street lights while playing cricket. They had no knowledge how the light was to be replaced and to whom they should pay the compensation. They first went to Rehana’s mother who gave them knowledge that it was the duty of municipal corporation to replace the lights. She directed them to Yasmin Khala’s house as she had just retired from the Municipal Corporation. So, children went to Yasmin Khala’s house to get their problems solved.

Question 2.
List four ways in which the work of the Municipal Corporation affects the life of city dweller.
Answer:
The four ways in which the Municipal Corporation affects the life of a city dweller are:
(i) It takes care of street lights, garbage collection, water supply, keeping the streets and the market clean.
(ii) It ensures that diseases do not break out in the city.
(iii) It runs school, hospitals and dispensaries.
(iv) It makes garbage around the city and maintains them.

Question 3.
Who is a Municipal Councillor?
Answer:
Municipal Councillor is an elected representative of a ward. He is also known as the ward councilor.

Question 4.
What did Gangabai do and why?
Answer:
Gangabai and women in the locality were upset over the garbage lying in the streets and unhygienic condition of the society. Gangabai proposed that it was the Ward Councillor who they should go to. Gangabai was confident that they had the right to do so as they were ones who elected him.

Question 5.
How does the Municipal Corporation earn the money to do its work?
Answer:
The Municipal Corporation is responsible for all round development of the society. It provides many services which requires a lot of money. The Municipal Corporation earn the money by collecting taxes from the people. It levied taxes on property, water and other services. Taxes are levied on the measurement and consumption of services. Taxes are also levied on education and other amenities like owning a hotel, entertainment etc.

Question 6.
Discuss:


In the two photograph you see different ways of collecting disposing garbage.
1. Which way do you think safety to person disposing garbage?
2. What are the dangers of collecting in the manner shown in the first photograph?
3. Why do you think that the proper ways of disposing garbage are not available to those who work in municipalities?
Answer:
1. In photograph-2, safer means for disposal of garbage are being adopted.

2. Dangers of collecting garbage in the manner shown in the photograph-1 are:
(i) The animals roaring around garbage dump may cause injury/harm/ hurt to them personally.
(ii) Always a fear of being get seriously ill or fall in the grip of serious diseases.

3. Municipalities have less facilities and transportation available to dispose garbage as compared to the corporations. In other words corporations are meant for big cities or Metropolitian cities. Due to growing population and slums a number of steps were taken by corporations to over come these problems. So, ways of disposing garbage are less available to those who work in municipalities.

Question 7.
Several poor people in the work as domestic servants as well as work for the Corporation, keeping the city clean. Yet the slums in which they live are quite filthy. This is because these slums seldom have any water and sanitation facilities. The reason often given by the Municipal Corporation is that the land in which the poor have set up their homes does not belong to them and that slum-dwellers do not pay taxes. However, people living in middle class neighbourhood, pay very little in taxes compared to the amount of money the corporation spends on them in setting up parks, street lighting facilities, regular garbage collection etc. Also as you read in this chapter, the property taxes collected by the Municipality makes up only 25-30 per cent of its money. Why do you think it is important that the Corporation should spend more money on slum localities? Why is it important that the Municipal Corporation provide the poor in the city with the same facilities that the rich get?
Answer:
The Corporation should spend of money on slum localities because:
(i) The share of general public in corporations earning stand between 70-75 per cent in terms of collection of various axes excluding only property tax.
(ii) Sanitation and light fall in the category of basic needs mandatory to provide by corporation like authority in a city.
(iii) These effects part of corporation would in long terms, bring buoyancy to its earning through taxes when parks, schools, water and electricity etc. facilities are provided with the poor in slums. Again it would inspire them to prepare for better in future.

HBSE 6th Class Civics Urban Administration Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is a Ward?
Answer:
Ward is a division of city for the purpose of municipal or corporation election.

Question 2.
What is a property tax?
Answer:
People who own homes pay proper tax. The larger the house the more the tax.

Question 3.
What is a Metropolitan city?
Answer:
A city in which people from different, religion, state and caste live together is called a metropolitan city.

Question 4.
Whose job is to replace the street light?
Answer:
It is the job of the Municipal Corpo-ration of the city to replace the street lights.

Short Answer Type Questions

Question 1.
What is sub-contracting? What did Yasmin Khala tell about this in case of garbage collection?
Answer:
Subcontracting means that the work that was earlier being done by government workers is how being done by a private company. Yasmin Khala told that in order to save money the Commissioners of several munici-palities across the country had hired private contractors to collect and process garbage.

Question 2.
What are hazards and dis-advantages associated with the job of collecting garbage?
Answer:
(i) The contract workers get paid less.
(ii) Their jobs are temporary.
(iii) It is quite a dangerous job.
(iv) The contract’ workers do not have any access to safety measures.
(v) They are not taken care of if they are injured while working.

Question 3.
How does the state government help the Municipalities?
Answer:
The sources of income of the Municipalities are very limited while the expenditures are more then the income. The state government helps the Municipalities by giving grants to them.

Question 4.
How are complicated decisions taken?
Answer:
Complicated decisions usually affect the entire city and therefore these decision are taken by groups of councillor who form committees to decide and debts issue. For example, if bus stands need to be improved or a crowded market place needs to have its garbage cleared more regularly.

Question 5.
What are major features and facilities available in a city?
Answer:
Major features and facilities available in a city are:
(a) Crowded markets
(b) Bigger in size then villages
(c) Many buses and cars
(d) Water facilities
(e) Electricity facilities
(f) Hospitals and schools

Long Answer Type Questions

Question 1.
What are the three main functions of the district administration?
Answer:
The three functions of the district administration are:
(i) To maintain law and order: To perform this task, all officers work under the supervision of the Collector.

(ii) To maintain land record and realise land revenue: For this purpose. Tehsildar, Naib Tehsildar, Kanungo and Lekhpal (Patwari) assist the Collector. He also provides relief when unforeseen situations as epidemics, famine are befallen.

(iii) To provide civic amenities: The district administration provide civic amenities as health facilities, education, means of transport, repair of roads etc.

Question 2.
What are the main functions performed by the Municipality or Municipal Corporation?
Answer:
The functions of the Municipality or a Municipal Corporation are generally similar. These are.
1. Functions for maintaining public health:
(i) They arrange for the disposal wastes causes numerous diseases.
(ii) They construct sewage, drain out dirty water from our cities and villages.
(iii) They protect the people from epidemics and other diseases like small-pox, cholera. They make suffice arrangements for vaccination and inoculation of the people.
(iv) They construct public latrines and urinals.
(v) They maintain dispensaries and hospitals to treat the sick.

2. Functions of Public Convenience:
(i) They arrange for water and electricity.
(ii) They maintain and repair roads, bridges and streets.
(iii) They plant trees along the roads for the convenience of travellers.
(iv) They make arrangements for the primary education of the children. For this, schools are maintained with proper basic facilities.
(v) They also manage libraries, museums and zoological parks etc.
(vi) It makes arrangements for playgrounds, parks and gardens.

3. Maintaining optional or miscellaneous departments:
(i) It maintains fire engines for extinguishing fire.
(ii) It checks food adulteration.
(iii) They remove obstacles from the public roads.
(iv) They maintain records of births and deaths.
(iv) It approves the plans of houses to be built in its areas.
(vi) They make arrangement for cremation and burial grounds.
(vii) Corporation issue permission to set-up petrol pumps, cinema houses etc.
(viii) Corporation also set-up homes for helpless women and children.

4. Tax Collection:
It impose and collect taxes like house tax, vehicles tax etc.

Urban Administration Class 6 HBSE Notes

• Municipal Corporation: Civic body in big cities is called Municipal Corporation.
• Municipal Council (Municipality): The civic body in small towns and cities is termed as municipal Council.
• Ward Councillor: Elected representative of a ward in a corporation is called Ward Councillor.
• Municipal Commissioner: The chief executive and administrator of the Municipal Corporation is called Municipal Commissioner.
• Public Defence: The maintenance of fire engines for extinguishing fire, checking food adulteration etc. .
• Mayor: Presiding officer or Chairman of a municipal corporation.
• Aldermen: Members of municipal bodies choosen for their experience and distinguished services.
• Democracy i A government of the people, for the people and by the people.
• Collector: The highest officer of the district.
• Tehsildar: He is in-charge of everything relating to land and land disputes of district.
• The Sessions Judge: He conducts/administers the highest criminal court in the district

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution :
We have, ∆ABC ~ ∆DEF

Hence, BC = 11.2 cm.

Question 2.
Diagonals of trapezium ABCD with AB || CD intersects each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:
We have, AB || CD and AB = 2CD
In triangles AOB and COD,
∠ABO = ∠CDO (alternate interior ∠s)
∠AOB = ∠COD (vertically opposite ∠s)
∆AOB ~ ∆COD (By AA similarity criterion)
⇒ \(\frac{\ {ar}(\Delta \mathrm{AOB})}{\ {ar}(\Delta \mathrm{COD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2}\)
⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{(2 \mathrm{CD})^2}{\mathrm{CD}^2}\)
⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{4 \mathrm{CD}^2}{\mathrm{CD}^2}\)
⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{4}{1}\)
Hence, ar(∆AOB) : ar(∆COD) = 4 : 1.

Question 3.
In given figure ABC and DBC are two triangles on the same base BC.. If AD intersects BC at O, show that \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}\).

Solution :
Given : ∆ABC and ∆DBC are two triangles on the same base BC
To Prove: \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}\)
Construction : Draw AL ⊥ BC and DM ⊥ BC
Proof : In ∆ALO and ∆DMO

∠ALO = ∠DMO = 90° (By construction)
∠AOL = ∠DOM (vertically opposite ∠s)
∆ALO ~ ∆DMO (By AA similarity criterion)
⇒ \(\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\)
[corresponding sides of similar triangles are proportional]

Hence proved.

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given : Let two triangles ABC and PQR such that ar(ABC) = ar(APQR)
To Prove: ∆ABC ∆PQR
Proof: ar(∆ABC) ≅ ar(∆PQR)
⇒ ∆ABC ~ ∆PQR
⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
and \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)

⇒ AB = PQ, BC = QR, CA = RP
∴ ∆ABC ≅ ∆PQR (By SSS congruence criteriah)
Hence Proved.

Question 5.
D, E and F are respectively the mid points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:
We have,
∵ D, E and F, are the mid points of sides AB, BC and CA respectively.
∴ DF = \(\frac{1}{2}\)BC, DE = \(\frac{1}{2}\)AC and EF = \(\frac{1}{2}\)AB
[Mid point Theorem]

Hence, ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 6.
Prove that ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians.
Solution:
Given: Two triangles ABC and PQR such that AABC – APQR and AD, PM are their medians respectively.

Hence proved.

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution :
Given : A square ABCD and two equilateral ∆s BCE and ACF have been described on side BC and diagonal AC respectively.
To Prove: Area (∆BCE) = \(\frac{1}{2}\) Area (∆ACF)
Proof: We have ∆BCE and ∆ACF are equilateral triangles.

∠B = ∠A (each = 60°)
∠C = ∠C (each = 60°)
∠E = ∠F (each = 60°)
∆ BCE ~ ∆ ACF (By AAA similarity criterion)
⇒ \(\frac{\ {ar}(\triangle \mathrm{BCE})}{\ {ar}(\triangle \mathrm{ACF})}=\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\)
[By theorem 6.6] ………………..(1)
But AB = BC (sides squares)
and AC² = AB² + BC² [By Pythagoras theorem]
⇒ AC² = 2BC² ……………..(2)
From (1) and (2), we get
\(\frac{\ {ar}(\Delta B C E)}{\ {ar}(\Delta A C F)}=\frac{B^2}{2 B C^2}\)
⇒ \(\frac{\ {ar}(\triangle B C E)}{\ {ar}(\triangle A C F)}=\frac{1}{2}\)
⇒ ar(∆BCE) = \(\frac{1}{2}\) ar(∆ACF)
Hence Proved.

In question 8 and 9, tick the correct answer and justify:
Question 8.
ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of ∆ABC and ∆BDE is :
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution :
Correct answer is (C).
Justification : ∆ABC and ∆BDE are two equilateral triangles.
Let AB = BC = CA = a unit
Then BD = \(\frac{a}{2}\) unit
[∵ D is the mid point of BC]
∵ ∠CAB = ∠DBE, ∠ABC = ∠BDE, ∠ACB = ∠BED [each is 60°]
∆ABC ~ ∆BDE (By AAA similarity criterion)
⇒ \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\Delta \mathrm{BDE})}=\frac{\mathrm{AB}^2}{\mathrm{BD}^2}\) [By theorem (6.6)]

⇒ ar(∆ABC) : ar(∆BDE) = 4 : 1.

Question 9.
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution :
Correct answer is (D).
Justification : We have,
\(\frac{\text { Side of } 1^{\text {st }} \text { triangle }}{\text { Side of } 2^{\text {nd }} \text { triangle }}=\frac{4}{9}\)
We know that areas of two similar triangles are in the ratio of the squares of their corresponding sides.
∴ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{(\text { side of } 1 \text { st } \Delta)^2}{(\text { side of } 2 \text { nd } \Delta)^2}\)
⇒ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{4^2}{9^2}\)
⇒ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{16}{81}\)
⇒ Area of 1st triangle : Area of IInd triangle = 16 : 81.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangles in below figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

Solution:
(i) In ∆ABC and ∆PQR
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
By AAA similarity criterion.
∆ABC ~ ∆PQR.

(ii) In ∆ABC and ∆QRP
\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}\)
\(\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{2 \cdot 5}{5}=\frac{1}{2}\)
\(\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\)
∵ \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{AC}}{\mathrm{PQ}}\)
∴ By SSS similarity criterion
∆ABC ~ ∆QRP

(iii) In ∆LMP and ∆DEF
\(\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2 \cdot 7}{4}\) = 0.675
\(\frac{\mathrm{MP}}{\mathrm{EF}}=\frac{2}{5}\) = 0.4
\(\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{3}{6}\) = 0.5
∵ \(\frac{\mathrm{LM}}{\mathrm{DF}} \neq \frac{\mathrm{MP}}{\mathrm{EF}} \neq \frac{\mathrm{LP}}{\mathrm{DF}}\)
∴ These two triangles are not similar because they do not satisfy the SSS similarity criterion.

(iv) In ∆MNL and ∆QPR
∠M = ∠Q = 70°
\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{2 \cdot 5}{6}\) (approx)
\(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}\) = 0.5
∵ \(\frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}\)
∴ These two triangles are not similar because they do not satisfy the SAS similarity criterion.

(v) In ∆ABC and ∆FDE
∠A = ∠F = 80°
\(\frac{\mathrm{AB}}{\mathrm{DF}} \neq \frac{\mathrm{AC}}{\mathrm{EF}}\) [∵ AC is not given]
∴ These two triangles are not similar because they do not satisfy the SAS similarity criterion.

(vi) In ∆DEF and ∆PQR
∠P = 180° (80° + 30°)
∠P = 180° – 110°
∠P = 70°
Now, ∠D = ∠P = 70°
∠E = ∠Q = 80°
By AA similarity criterion ∆DEF ~ ∆PQR.

Question 2.
In given figure ∆ODC – ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
We have, ∠CDO = 70° and ∠COB = 125°
∵ BD is a line and CO ray on it
∠DOC + ∠COB = 180°
∠DOC + 125° = 1800
∠DOC = 180° – 125°
∠DOC = 55°
In ∆COD, we have
∠DOC + ∠DCO + ∠CDO = 180° [Sum of ∠s of a triangle is 180°]
55° + ∠DCO + 70° = 180°
125° + ∠DCO = 180°
∠DCO = 180° – 125°
∠DCO = 55°
Now ∆ODC ~ ∆OBA (given)
∴ ∠OAB = ∠DCO
(Corresponding ∠s of similar ∆s)
∠OAB = 55°
Hence, ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB DC ntersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\).
Solution:
Given : ABCD is a trapezium in which AB || DC. Its diagonals AC and BD meet at O.
To Prove: \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
∠OAB = ∠OCD [∵ AB || DC]
∴ Alternate ∠s are equal.
∠AOB = ∠COD (vertically opposite ∠s)

∴ ∆OAB ~ ∆OCD (By AA similarity criterion)
∴ \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
Hence Proved.

Question 4.
In given figure \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:
Given: \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2.
To Prove: ∆PQS ~ ∆TQR
Proof: We have

∴ ∆PQS ~ ∆TQR [By SAS similarity criterion]
Hence Proved.

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ – ∆RTS.
Solution :
Given : S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS
To Prove: ∆RPQ ~ ∆RTS.

Proof: In ∆RPQ and ∆RTS.
∠P = ∠RTS (given)
∠R = ∠R (common)
∴ ∆RPQ ~ ∆RTS (By AA similarity criterion)
Hence Proved.

Question 6.
In given figure, If ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.

Solution:
Given: ∆ABE ≅ ∆ACD
ToProve: ∆ADE ~ ∆ABC
Proof: ∆ABE ~ ∆ACD
∴ AB = AC
and AD = AE [CPCT]
⇒ \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = 1 and \(\frac{\mathrm{AD}}{\mathrm{AE}}\) = 1
⇒ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
Now, In ∆ADE and ∆ABC, we have
⇒ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)
and ∠DAE = ∠BAC (each = ∠A)
∆ADE ~∆ABC. [By AA similarity criterion]
Hence Proved .

Question 7.
In given figure altitudes AD and CE of AABC intersect each other at the point P. Show that :
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC.

Solution :
Given : CE ⊥ AB and AD ⊥ BC.
To Prove : (i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC.
Proof : (i) In ∆AEP and ∆CDP
∠AEP = ∠CDP (each = 90°)
∠APE = ∠CPD (vertically opposite ∠s)
∴ ∆AEP ~ ∆CDP
(By AA similarity criterion) Hence Proved.

(ii) In ∆ABD and ∆CBE
∠ADB =∠CEB (each = 90°)
∠B = ∠B (common)
∴ ∆ABD ~ ∆CBE
(By AA similarity criterion) Hence Proved.

(iii) In ∆AEP and ∆ADB
∠AEP = ∠ADB (each = 90°)
∠PAE = ∠DAB (same angle)
∴ ∆AEP ~ ∆ADB
(By AA similarity criterion) Hence Proved.

(iv) In ∆PDC and ∆BEC
∠PDC = ∠BEC (each = 90°)
∠PCD = ∠BCE (same angle)
∆PDC ~ ∆BEC
(By AA similarity criterion) Hence Proved.

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.

Solution :
Given : E is a point on side AD produced of a ||gm ABCD and BE intersects CD at F.
To Prove : ∆ABE ~ ∆CFB.
Proof : In ∆ABE and ∆CFB.
∠AEB = ∠CBF (alternate interior ∠s)
∠A = ∠C (opposite ∠s of a ||gm )
∴ ∆ABE ~ ∆CFB
(By AA similarity criterion) Hence Proved.

Question 9.
In given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
(i) ∆ABC ~ ∆AMP.
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\).

Solution:
Given: ∆ABC and ∆AMP are two right angled triangles right angles at B and M respectively.
To Prove :(i) ∆ABC ~ ∆AMP.
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\).
Proof : (i) In ∆ABC and ∆AMP.
∠ABC = ∠AMP (each = 90°)
∠A = ∠A (common)
∴ ∆ABC ~ ∆AMP (By AA similarity criterion)
Hence Proved.

(ii) ∵ ∆ABC ~ ∆AMP
∴ \(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
(corresponding sides of similar triangles)
Hence Proved.

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:
(i) \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
Solution:
Given : CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on the sides AB and EF of ∆ABC and ∆EFG and ∆ABC ~ ∆FEG.
To Prove :
(i) ∆DCB ~ ∆HGE
(ii) ∆DCA ~ ∆HGF
(iii) \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
Proof : (i) ∆ABC ~ ∆FEG (given)
∠ACB = ∠EGF (corresponding ∠s of similar ∆s)
\(\frac{1}{2}\) ∠ACB = \(\frac{1}{2}\) ∠EGF
[CD and GH are bisectors of ∠C and ∠G.]

∠DCB = ∠HGE ………………..(1)
and ∠B = ∠E
(corresponding ∠s of similar ∆s)
⇒ ∠DBC = ∠HEG ……………….(2)
From (1) and (2), we have
∆DCB ~ ∆HGE
Hence Proved.

(ii) In ∆DCA and ∆HGF
∠C = ∠G
(corresponding ∠s of similar ∆s)
∠ACD = ∠HGF [CD and GH are bisectors of ∠C and ∠G] ………………..(1)
∠A = ∠F
(corresponding ∠s of similar ∆s) ……………….(2)
∠ADC = ∠FHG
(third ∠s of two triangles) …………………(3)
∴ ∆DCA ~ ∆HGF. (By AAA similarity criterion)
Hence Proved.

(iii) ∵ ∆DCA ~ ∆HGF.
∴ \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
(corresponding sides of similar ∆s)
Hence Proved.

Question 11.
In given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution :
Given : AB = AC, AD ⊥ BC and EF ⊥ AC.
To Prove : ∆ABD ~ ∆ECF
Proof : AB = AC (given)
∠B = ∠C [∵ ∠s opposite to equal side are equal]
Now, In ∆ABD and ∆ECF,
∠ABD = ∠ECF [∵ ∠B = ∠C]
∠ADB = ∠EFC = 90°
[∵ AD ⊥ BC and EF ⊥ AC]
∴ ∠ABD ~ ∆ECF
(By AA similarity criterion)
Hence Proved.

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see figure below). Show that ∆ABC ~ ∆PQR.

Solution:

(By SSS similarity criterion)
∠B = ∠Q
(Corresponding ∠s of Similar As)
Now, in ∆ABC and ∆PQR.
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) (given)
∠B = ∠Q (Proved above)
∴ ∆ABC ~ ∆PQR [By SAS similarity criterion]
Hence Proved.

Question 13.
D is a point on the side BC of a trianlge ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD
Solution:
Given : ∠ADC = ∠BAC
To prove : CA² = CB.CD
Proof : In ∆ABC and ∆DAC
∠ADC = ∠BAC (given)
∠C = ∠C (common)
∆ABC ~ ∆DAC (By AA similarity criterion)

⇒ CA² = CB × CD.
Hence Proved.

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution :
Given : In ∆ABC and ∆PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
To Prove : ∆ABC ~ ∆PQR
Construction : Produce AD to E such that AD = DE and PM to N such that PM = MN Join CE and NR.
Proof : In ∆ADB and ∆EDC
AD = ED (By construction)
BD = CD (AD is median)
∠ADB = ∠CDE (vertically opposite ∠s)
∆ADB ≅ ∆EDC (By SAS Congruence criterion)
⇒ AB = EC (By CPCT) …………….(1)

Similarly, ∆PMQ ≅ ∆NMR (By SAS congruence criterion)
⇒ PQ = NR (By CPCT) …………..(2)
Now, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)(given)
⇒ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
[From (1) and (2), AB = EC and PQ = NR]
⇒ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\)
⇒ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
[By construction AD = DE and PM = MN)
∴ ∆ACE ~ ∆PRN (By SSS similarity criterion)
∠1 = ∠3 ……………….(3)
(Corresponding ∠s of similar triangles)
Similarly ∠2 = ∠4 ………………(4)
Adding (3) and (4) we get
∠1 + ∠2 = ∠3 + ∠4.
⇒ ∠A = ∠P.
Now ∆ABC and ∆PQR.
∠A = ∠P (Proved above)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\) (given)
∴ ∆ABC ~ ∆PQR.

Question 15.
A vertical pole of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.

Solution:
Let AB be the vertical pole of length 6m casts a shadow BC and PQ be the tower casts a shadow QR. Let height of tower berm. Join AC and PR.
In ∆ABC and ∆PQR
∠B = ∠Q (each = 90°)
∠C = ∠R (Angle of elevation of the Sun)
∴ ∆ABC ~ ∆PQR (By AA similarity criterion)
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)
[Corresponding sides of similar triangles]
⇒ \(\frac{6}{x}=\frac{4}{28}\)
⇒ x = \(\frac{6 \times 28}{4}\)
x = 42.
Hence, height of tower = 42 m.

Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\).

Solution :
Given : AD and PM are medians of ∆ABC and ∆PQR respectively and ∆ABC ~ ∆PQR.
To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Proof : ∆ABC ~ ∆PQR
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) ………………(1)
(corresponding sides of similar triangles)
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\)
[∵ AD and PM are medians
∴ BD = CD and QM = MR]
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
and ∠B = ∠Q
[Corresponding ∠s of similar ∆s ABC and PQR]
∴ ∆ADB ~ ∆PMQ [By SAS similarity criterion]
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (Corresponding sides of similar ∆s.)
Hence Proved.

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.7

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

Question 1.
Find the volume of the right circular cone with :
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm.
Solution:
(i) We have, Radius of cone (r) = 6 cm
Height of the cone (h) = 7 cm
∴ Volume e the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7\)
= 264 cm³.
(ii) We have,
Radius of the cone (r) = 3.5 cm
Height of the cone (h) = 12 cm
∴ Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times 12\)
= 154 cm³
Hence, volume of the cone (i) 264 cm³, (ii) 154 cm³.

Question 2.
Find the capacity in litres of a conical vessel with :
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm.
Solution:
(i) We have, Radius of the conical vessel (r) = 7 cm
Slant height of the conical vessel (l) = 25 cm
Let the height of the conical vessel be h cm.
∴ l² = h² + r²
⇒ 25² = h² + 7²
⇒ h² = 25² – 7²
⇒ h² = (25 + 7) (25 – 7)
⇒ h² = 32 × 18
⇒ h² = 576
⇒ h= \(\sqrt{576}\)
⇒ h = 24 cm
∴ Volume of the conical vessel = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24\)
= 1232 cm³
= \(\frac{1232}{1000}\) litres
= 1.232 litres.

(ii) We have, Slant height of the conical vessel (l) = 13 cm
Height of the conical vessel (h) = 12 cm
Let the radius of the conical vessel be r cm.
∴ l² = h² + r²
⇒ 13² = 12² + r²
⇒ r² = 13² – 12²
⇒ r² = (13 + 12) (13 – 12)
⇒ r² = 25
⇒ r = \(\sqrt{25}\)
⇒ r = 5 cm
∴ Volume of the conical vessel = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12\)
= \(\frac{2200}{7} \mathrm{~cm}^3\)
= \(\frac{2200}{7 \times 1000} \text { litres }\)
= \(\frac{11}{35} \text { litres }\)
Hence, volume of the cone (i) 1.232 litres, (ii) \(\frac{11}{35} \text { litres }\).

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Solution:
Let the radius of base of a cone be r сm.
We have, Height of the cone (h) = 15 cm
volume of the cone = 1570 cm³
⇒ \(\frac{1}{3}\)πr²h = 1570
⇒ \(\frac{1}{3}\) × 3.14 × r² × 15 = 1570
⇒ 15.7 × r² = 1570
⇒ r² = \(\frac{1570}{157}\)
⇒ r² = 100
⇒ r = \(\sqrt{100}\)
⇒ r = 10 cm
Hence, radius of the base of the cone = 10 cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Let the radius of the basse of the cone be r сm.
We have, Height of the cone (h) = 9 cm
Volume of the cone = 48π cm³
⇒ \(\frac{1}{3}\)πr²h = 48π
⇒ \(\frac{1}{3}\)π × r² × 9 = 48π
⇒ 3πr² = 48π
⇒ r² = \(\frac{48 \pi}{3 \pi}\)
⇒ r² = 16
⇒ r = \(\sqrt{16}\)
⇒ r = 4 cm
∴ Diameter of the base of the cone = 2 × 4
= 8 cm
Hence,diameter of the base of cone = 8 cm

Question 5.
A conical pit of top diameter 3.5 m is 12m deep. What is its capacity in kilolitres ?
Solution:
We have,
Depth of conical pit (h) = 12 m
Diameter of conical pit (d) = 3.5 m
∴ Radius of conical pit (r) = \(\frac{3.5}{2}\) = 1.75 m
∴ Capacity of the conical pit = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(1.75)^2 \times 12\)
= 38.5 m³
= 38.5 kilolitres
(∵ 1 m³ = 1 kilolitres)
Hence,capacity of the conical pit = 38.5 kilolitres.

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find :
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
Solution:
We have,
The volume of the cone = 9856 cm³
The diameter of the base of the cone (d) = 28 cm
∴ Radius of the base of the cone (r) = \(\frac{28}{2}\) cm = 14 cm
(i) Let the height of the cone be h cm.
Volume of the cone = 9856 cm³
⇒ \(\frac{1}{3}\)πr²h = 9856
⇒ \(\frac{1}{3} \times \frac{22}{7} \times 14^2 \times h\) = 9856
⇒ \(\frac{616}{3}\) × h = 9856
⇒ h = \(\frac{9856 \times 3}{616}\)
⇒ h = 480cm

(ii) Let the slant height of the cone be 1 cm, then
⇒ l² = h² + r²
⇒ l² = 48² + 14²
⇒ l² = 2304 + 196
⇒ l2 = 2500
⇒ l = \(\sqrt{2500}\)
⇒ l = 50 cm.

(iii) Curved surface of the cone = πrl
= \(\frac{22}{7}\) × 14 × 50
= 2200 cm².
Hence, (i) Height of the cone = 48 cm, (ii) Slant height of the cone = 50 cm, (iii) Curved surface area of the cone = 2200 cm².

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
We have,
Sides of the triangle ABC are AB = 12 cm, BC = 5 cm and AC = 13 cm
Since ΔABC is revolved about the side AB(= 12 cm)
∴ We get a cone (AC’C) as shown in the figure.

Then,radius of the base of the cone (r) = 5 cm
and height of the cone (h) = 12 cm
∴ Volume of the cone (solid) = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 5² × 12
= 100π cm³
Hence, volume of the cone so obtained = 100π cm³.

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Since ΔABC is revolved about BC(= 5 cm).
∴ We get a cone (CAA’) as shown in the figure.

Then,radius of the base of the cone (r) = 12 cm
and height of the cone (h) = 5 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 12² × 5
= 240π cm³
Ratio of the volumes of the two solids obtained in Q. 7 and Q. 8
= 100π : 240π = \(\frac{100 \pi}{240 \pi}\)
= \(\frac{5}{2}\) = 5 : 12
Hence,volume of the cone (solid) = 240π cm³,
and required ratio of their volumes = 5 : 12.

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:
We have,
Height of the conical heap (h) = 3 m
Diameter of the base of the conical heap (d) = 10.5 m³

∴ Radius of the base of the conical heap (d) = \(\frac{10.5}{2}\) = 5.25 m
∴ Slant height of a conical heap (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{3^2+(5.25)^2}\)
= \(\sqrt{9+27.56}\)
= \(\sqrt{36.56}\)
= 6.05 m
∴ Volume of the conical heap = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(5.25)^2 \times 3\)
= \(\frac{606.375}{7}\)
= 86.625 m³
Required area of canvas = Curved surface
area of the conical heap = πrl
= \(\frac{22}{7}\) × 5.25 × 6.05
= 16.5 × 6.05
= 99.825 m²
Hence,volume of the heap = 86.625 m³ and required area of canvas = 99.825 m².

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 1.
In given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution :
(i) we have, AD = 1.5 cm, AE = 1 cm, DB = 3 cm
DE || BC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By theorem 6.1 (BPT)]
\(\frac{1 \cdot 5}{3 \cdot 0}=\frac{1}{\mathrm{EC}}\) = 2
EC = \(\frac{3.0}{1.5}\) = 2
Hence, EC = 2 cm.

(ii) We have, DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm and DE || BC
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) [By Theorem 6.1 (BPT)]
\(\frac{\mathrm{AD}}{7 \cdot 2}=\frac{1 \cdot 8}{5 \cdot 4}\)
AD = \(\frac{7.2 \times 1.8}{5.4}\) = 2.4
Hence, AD = 2.4 cm.

Question 2.
E and F are points on the sides PQ and PR respectively of a APQR. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 018 cm and PF = 0.36 cm.

Solution :
(i) we have PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3.0}\) = 1.3
and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3 \cdot 6}{2 \cdot 4}\) = 1.5
∵ \(\frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF is not parallel to QR.

(ii) We have,
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{4}{4 \cdot 5}=\frac{8}{9}\) and
\(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9}\)
∵ \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF || QR

(iii) We have,
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm,
EQ = PQ – PE = 1.28 – 0.18 = 11 cm,
PR = PF = 2.56 – 0.36 = 2.2 cm.
Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0 \cdot 18}{1 \cdot 1}=\frac{9}{55}\)
and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2 \cdot 2}=\frac{9}{55}\)
∵ \(\frac{P E}{E Q}=\frac{P F}{F R}\)
∴ EF || QR

Question 3.
In given figure, if LM || CB and LN || CD, prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\).

Solution:
Given: In □ABCD, LM || CB and LN || CD
To Prove: \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)
Solution: In ∆ABC LM || CB (given)

Question 4.
In given figure DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).

Solution:
Given : In ∆ABC DE || AC and DF || AE
To prove: \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).
Proof: In ∆BAE,
DF || AE (given)
∴ \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}\) ………………..(1)
[By Theorem 6.1 (BPT)]
In ∆ABC, DE || AC (given)
∴ \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}\) [By Theorem 6.1 (BPT)]
From(1) and (2) we get
\(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)
Hence Proved.

Question 5.
In given figure DE || OQ and DF || OR. Show that EF || QR.

Solution:
Given: In ∆POQ, DE || OQ and in POR, DF || OR
To prove: EF || QR
Proof: In ∆POQ DE || OQ (given)
∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PE}}{\mathbf{E Q}}\) ……………..(1)
[By Theorem 6.1 (BPT)]
In ∆POR DF || OR
∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}}\) ……………..(2)
[By Theorem 6.1 (BPT)]
From (1) and (2), we get
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF | | QR [By converse of BPT]

Question 6.
In given figure, A. B, and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:
Given: In figure A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
To prove: BC || QR
Proof: In ∆OPQ
AB || PQ (given)
\(\frac{O A}{A P}=\frac{O B}{B Q}\) …………………(1)
[By theorem 6.1 (BPT)]
In ∆OPR, AC || PR (given)
\(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}\) …………….(2)
[By theorem 6.1 (BPT)]
From (1) and (2) we get
\(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)
Thus B and C are respectively points on sides OQ and OR of ∆OQR, such that \(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\).
⇒ BC || QR [By converse of BPT]
Hence Proved.

Question 7.
Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in class IX).
Solution:
Given : A ∆ABC in which D is the mid-point of AB and DE || BC and meeting AC at E.

To Prove : AE = EC
Proof : Since, DE || BC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ……………..(1)
[By Theorem 6.1 (BPT)]
But AD = DB
[∵ D is the mid point of AB]
∴ \(\frac{\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ………………(2)
⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1
⇒ AE = EC.

Question 8.
Using theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in class IX).
Solution:
Given : A ∆ABC in which D is the mid-point of AB and E is the mid-point of AC
To prove : DE || BC
Proof : Since D and E are the mid-points of AB and AC respectively
AD = DB
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = 1
and AE = EC
⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
⇒ DE || BC
[By Theorem of 6.2 (converse of BPT)]
Hence Proved.

Question 9.
ABCD is a trapezium in which AB | DC and its diagonals intersect each other at the point O. Show that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\).
OR
PQRS is a trapezum in which PQ || RS and its diagonals intersect each other at the point O. Prove \(\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{RO}}{\mathrm{SO}}\)
Solution:
Given : A trapezium ABCD in which AB || CD and its diagonals AC and BD intersect at O.

To Prove: \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Construction: Draw OE || AB through O, which meets AD at E.
Proof : We have,
EO || AB (By construction) …………(1)
DC || AB (given) …………(2)
From (1) and (2), we get
EO || DC
∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}\) ………………..(3)
[By Theorem 6.1 (BPT)]
In ∆ DAB EO || AB
\(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{OB}}\)
∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}}\) ………………(4)
[By Theorem 6.1 (BPT)]
From (3), (4), we get
\(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}\)
⇒ \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Hence Proved.

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\). Show that ABCD is a trapezium.
OR
The diagonals of a quadrilateral PQRS intersect each other at the point O, such that \(\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{RO}}{\mathrm{SO}}\). Show that PQRS is a trapezium.
Solution:
Given : A quadrilateral ABCD whose diagonals AC and BD intersect at a point O such that

\(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
To Prove : ABCD is a trapezium i.e. AB || DC.
Construction : Draw OE || DC which meets AD at E.
Proof: In ∆ACD,
0E || DC (By construction)
∴ \(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{AE}}{\mathrm{ED}}\) ……………..(1)
[By Theorem 6.1 (BPT)]
But, \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\) (given)
⇒ \(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{DO}}\) ………………..(2)
From (1) and (2), we get
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)
⇒ EO || AB
[By Theorem 6.2 (converse of BPT)1
But, OE || DC (By construction)
From (3) and (4), we get
AB || CD
Hence, ABCD is a trapezium. Hence Proved.

Haryana State Board HBSE 6th Class English Solutions A Pact with the Sun Chapter 2 The Friendly Mongoose Textbook Exercise Questions and Answers.

Haryana Board 6th Class English Solutions A Pact with the Sun Chapter 2 The Friendly Mongoose

HBSE 6th Class English The Friendly Mongoose Textbook Questions and Answers

The Friendly Mongoose HBSE 6th Class English Question 1.
Why did the farmer bring a baby mongoose into the house?
Answer:
The farmer brought a baby mongoose into the house to provide a companion to his son.

The Friendly Mongoose Question Answer HBSE 6th Class English Question 2.
Why didn’t the farmer’s wife want to leave the baby alone with the mongoose?
Answer:
The farmer’s wife didn’t want to have the baby alone with the mongoose because she thought the mongoose could harm her son.

The Friendly Mongoose Class 6 Questions And Answers Question 3.
What was the farmer’s comment on his wife’s fears?
Answer:
The farmer commented that the mongoose was a friendly animal and that he was the best of his son’s friend.

The Friendly Mongoose Summary HBSE 6th Class English Question 4.
What did the farmer’s wife strike the mongoose with her basket?
Answer:
The farmer’s wife struck the mongoose with the basket because she thought the mongoose had killed her son.

The Friendly Mongoose Questions And Answers HBSE 6th Class English Question 5.
Did she repent her hasty action? How does she show her repentance?
Ans.
Yes, she repented her hasty actions as she shed tears later.

The Friendly Mongoose Question Answers 6th Class HBSE English  Question 6.
Do you have a pet a cat or a dog? If not, would you like one? How would you look after it? Are you for or against keeping birds in a cage as pets ?
Ans.
Do it yourself.

HBSE 6th Class English The Friendly Mongoose Important Questions and Answers

Friendly Mongoose Question Answer HBSE 6th Class English Question 1.
Who said this to whom and when? ‘You need n’t be a fraid.’
Answer:
The farmer said this to his wife. He said this when she was going to the market and she told that she did not want her baby to stay alone with mongoose.

Summary Of The Friendly Mongoose HBSE 6th Class English Question 2.
Why did the farmer not return for quite some time?
Answer:
The farmer did not return for quite some time because he met some friends on the way back from fields.

Question 3.
What was customary for the mongoose?
Answer:
It was customary for the mongoose to welcome the farmer’s wife when she returned home.

Question 4.
How did the mongoose prove himself to be a true friend of the baby?
Answer:
When the baby’s parents were not at home the mongoose saw a snake near the baby. He attacked the snake and tore it into pieces. Thus he saved the child. Hence, he proved himself to be a true friend of the baby.

Question 5.
What is the moral of the story?
Answer:
The moral of the story is ‘Don’t act hastily’, and ‘Think twice before you act’.

The Friendly Mongoose Word Meanings

village (विलेज) = गाँव, pet (पेट) = पालतू जानवर, companion (कम्पेनियन) = साथी, tiny (टाइनी) = सूक्ष्म, mongoose (मेन्गूस) = नेवला, fuilly grown (फूलीग्रोन) = पूर्ण विकसित, lovely (लवली) = सुन्दर, रम्य, shining (शाइनी) = चमकदार, bushy (बुशी)- झाड़ीदार, tail (टेल) = पूँछ। Cradle (क्रेडल) = पालना, alternately (आल्टरनेटली) = बारी-बारी से, rocked (रॉक्ड) = हिलाया, basket (बास्केट) टोकरी, friendly (फ्रेन्डली) = मित्रतापूर्ण तरीके से, fields (फिल्डस) = खेत, return (रिटन) = वापसी, finished (फिनिश्ड) = खत्म किया, shooping (शोपिंग) = खरीददारी, groceries (ग्रोसरीज) = घरेलू सामान, welcome (वेलकम) = स्वागत, customary (कास्टमरी) = रिवाज के अनुसार, screamed (स्क्रीम्ड) = चीखी, blood (ब्लड) = खून, paws (पॉज) = पंजे, smeared (स्मीअड) = लेप किया हुआ, सना हुआ, wicked (विकेड) = दुष्ट, screamed (स्क्रीम्ड) = चीखी, hysterically (हिस्ट्रीकली) = उन्मत्त (पागल) जैसा, blind (ब्लाइन्ड) = अंधी, rage (रेग) = नाराज, गुस्सा, strength (स्ट्रेन्थ) = ताकत, askeep (एसलीप) = सोया हुआ, torn (टान) = कटा-फटा होना, bleeding (ब्लीडिंग) = खून बहते हुए, saved (सेव्ड) = बचाया, unaware (अनवेयर) = अनजान था, sobbing (सॉबिंग) = सुबकते हुए, hastily (हेस्टली) = जल्दी से, rashly (रेशली) = लापरवाही से, stared (स्टेयड) = ताका, wiping (वाईपिंग) = पोंडते हुए, tears (टीअर्स) = आँसू, feed (फीड) = पिलाना।

The Friendly Mongoose Summary in English

There lived a farmer, his wife and their small son in a village. The farmer and his wife brought a mongoose to give their son a companion. One day the farmer and his wife had to go out of their house leaving their son alone with the mongoose. When the farmer’s wife returned home, she found the mongoose at the entrance with blood smeared on his face and paws. She thought her son had been killed by the mongoose and as such she killed the mongoose then and there. But when she went to her son, she learnt the truth. In fact the mongoose had guarded her son from the attack of a snake and that’s why her face and paws were smeared with blood. She went to the mongoose and repented a lot. But it was of no use.

The Friendly Mongoose Summary in Hindi

एक गाँव में एक किसान, उसकी पत्नी और उनका छोटा लड़का रहते थे। किसान और उसकी पत्नी एक नेवले को अपने पुत्र को एक साथी की संगति देने के लिए ले आये। एक दिन किसान और उसकी पत्नी को अपने बच्चे को अकेला छोड़कर घर से बाहर जाना पड़ा। जब किसान की पत्नी घर वापस लौटी, उसने नेवले को प्रवेश द्वार पर उसके चेहरे व पंजे को खून से सना हुआ पाया। उसने सोचा नेवले ने उसके पुत्र को मार दिया ऐसा सोचकर उसने भी उस नेवले को मार दिया। लेकन जब वह अपने पुत्र के पास गई तो उसे सच का पता चला। वास्तव में नेवले ने सांप के हमले से उसके लड़के की रक्षा की थी और इसी वजह से उसका चेहरा और पंजे खून से सन गये थे। वह नेवले के पास गई और काफी पश्चात्ताप किया। लेकिन इसका कोई फायदा नहीं था।

Haryana State Board HBSE 6th Class English Solutions A Pact with the Sun Chapter 3 The Shepherds Treasure Textbook Exercise Questions and Answers.

Haryana Board 6th Class English Solutions A Pact with the Sun Chapter 3 The Shepherds Treasure

HBSE 6th Class English The Shepherds Treasure Textbook Questions and Answers

The Shepherd’s Treasure Summary HBSE 6th Class Question 1.
The shepherd hadn’t been to school because :
(i) he was very poor.
(ii) there were very few schools in those days.
(iii) he wasn’t interested in studies.
Choose the right answer.
Answer:
(ii) There were very few schools in those days.

The Shepherd’s Treasure Question Answer HBSE 6th Class Question 2.
Who visited the shepherd one day, and why ?
Answer:
The king of the country visited him one day. He wanted to meet him and find out the truth.

The Shepherd’s Treasure Question Answers HBSE 6th Class Question 3.
Why did the other governors x grow jealous of the shepherd ?
Answer:
The other governors grew jealous of the shepherd because he grew very famous as a fair and wise governor of the country.

The Shepherd’s Treasure Class 6 Questions And Answers Question 4.
Why was the new governor called to the palace ?
Answer:
The new governor was called to the palace because the king wanted to know the secret of the iron chest.

The Shepherd’s Treasure Moral HBSE 6th Class Question 5.
Why was everyone delighted to see the iron chest on the camel’s back ?
Answer:
Everyone was delighted to see the iron chest on the camel’s back because they thought that this would prove beyond doubt that the charges levelled against the governor were true.

Summary Of The Shepherd’s Treasure HBSE 6th Class Question 6.
(i) What did the iron chest contain ?
(ii) Why did the shepherd always carry it ?
(iii) Is it an example of the shepherd’s humility or wisdom or both ?
Answer:
(i) The iron chest contained an old blanket.
(ii) The shepherd always carried it because he regarded it as his best friend. This blanket could protect him if the king took away new clothes.
(iii) It was an example of the shepherd’s humility as well as wisdom.

The Shepherd’s Treasure Questions And Answers HBSE 6th Class Question 7.
How did the king reward the new governor ?
Answer:
The king rewarded him by making him the governor of a much bigger district that very day.

HBSE 6th Class English The Shepherds Treasure Important Questions and Answers

Class 6 The Shepherd’s Treasure Question Answer HBSE Question 1.
How did the poor shepherd become famous ?
Answer:
The shepherd though poor, was very wise. He could understand people’s sorrows and troubles. He helped them to face them wisely with courage. Thus he became famous for his wise and friendly nature.

The Shepherd Treasure Question Answer HBSE 6th Class Question 2.
Why did the king disguise himself as a shepherd ?
Answer:
The king wanted to test the shepherd’s wisdom and friendliness so he disguised himself as a shepherd and came to the cave. He did not want to disclose his identity to the shepherd.

Class 6 The Shepherd’s Treasure HBSE Question 3.
How did the shepherd welcome the king who come to him as a poor traveller ?
Answer:
The shepherd welcomed the traveller. He served him water and a share of his own simple meal.

Question 4.
Why did the old governors of the kingdom talk against, the new governor ?
Answer:
As the new governor was humble and wise, he soon became very famous. The old governors became very jeolous so they talked against the new governor.

Question 5.
What is the moral of the story ?
Answer:
The moral of the story is that humility and wisdom always pay in the long run.

The Shepherds Treasure Word Meanings

shepherd (शेफड) = गड़रिया, cottage (कॉटेज) = झोंपड़ी, uneducated (अनएज्यूकेटेड) = अशिक्षित, wise (वाइज) = बुद्धिमान, sorrows (सोरोज) = दु:ख, troubles (ट्रब्लस) = परेशानियाँadvice (एडवाइस) = सलाह, famous (फेमस)प्रसिद्ध, wisdom (विजडम) = बुद्धिमानी, country (कन्ट्री)देश, meeting (मीटिंग) = मिलना, disguised (डिसगाईज्ड)छद्म वेश में, mule (म्यूल) = खच्चर, cave (केव)- गुफा, meagre (मीगर) = अपर्याप्त, greatly (ग्रेटली) = काफी ज्यादा, impressed (इम्प्रेस्ड) = प्रभावित, hospitality (हास्पेटिलिटी) = मेजबानी, अतिथि सत्कार, conversation (कन्वर्शन) = बातचीत, depart (डिपार्ट) = विदा होना, kindness (काइन्डनेस) = दया, permit (परमिट) = आज्ञा देना, leave (लीव) = छुट्टी, छोड़ना, guest (गेस्ट) = मेहमान, majesty (मेजेस्टी)- हुजूर, compliment (कम्पलीमेन्ट) = आदर-सत्कार, astonished (एस्टोनिस्ड) = आश्चर्यचकित indeed (इन्डीड) = वास्तव में, appointed (अपायन्टेड) = नियुक्त किया, humble (हम्बल) = विनम्र, governor (गवर्नर) = राज्यपाल, district (डिस्ट्रीक्ट) = जिला, dignity (डिगनिटी) = सम्मान, sympathy (सेम्पैथी) = सद्भावना, सहानुभूति, goodness (गुडनेस) = अच्छापन, just (जस्ट) = न्यायप्रिय, throughout(धू-आउट) = सारी जगह, provinces (प्रोविन्सेज)- प्रान्त, terribly (टेरिबली) = भयंकर रूप से। jealous (जेलयस) ईर्ष्यालु, dishonest (डिसओनस्ट) = बेईमान, collect (कलैक्ट) = इकट्ठा करता है, added (एडेड) कहा (यहाँ), iron chest (आयरनचेस्ट) = लोहे की पेटी (यहाँ) treasure (ट्रेजर) = खजाना, secretly (सिक्रेटली) = गुप्त रूप से, attention (अटेन्शन) = ध्यान, ignore (इग्नोर) = नजर अन्दाज करना, endless (एन्डलस) = अंतहीन, summoned (सम्मनअड) = बुलाया, palace (पैलेस) = महल, camel (कमल) ऊँट, delight (डिलाइट) खुशी, fastened (फास्टअनड) = बाँध ना, securely (सिक्योरअली)- सुरक्षित ढंग से, contain (कन्टेन) = रखता है, smiled (स्माइल्ड) = मुस्कराया, eagerly (इगअरली) = उत्सुकता से, astonishment (एस्टचोनिशमेन्ट) = आश्चर्य में, gold (गोल्ड) = सोना, silver (सिल्वर) = चाँदी, blanket (ब्लेन्कट) कम्बल, holding (होल्डिंग) = पकड़े हुए, proudly (प्राउडलि)- घमंड से, treasure (ट्रेजर)- खजाना, dignity (डिगनिटी) = सम्मान, take away (टेकअवे) = ले लेना, cloaks (क्लोक्स)- कपड़े, embarrassed (एम्ब्रेअसड) = परेशान, jealous (जेलस) = ईर्ष्यालु, wisest (वाइजेस्ट) = सबसे ज्यादा बुद्धिमान।

The Shepherds Treasure Summary in English

Once there lived a shepherd who was uneducated but very wise and helpful. He was very famous for his wisdom. Once the king went to meet him disguised as a shepherd on a mule. He behaved and greeted him very nicely. This poor shepherd could make out that his visitor was the king of his kingdom. The king was impressed with his wisdom. He made him the governor of a small district. Other governors grew jealous of this shepherd and hatched a conspiracy. The king was frequently complained of the shepherd’s dishonesty as a new governor. One day he was summoned to the palace. Then he was asked to explain why he always carried an iron chest with him. When the iron chest was opened, it was not found containing any treasures as his enemies had been alleging. It contained an old blanket which the shepherd regarded as his oldest friend.

The Shepherds Treasure Summary in Hindi

एक बार एक गडरिया जो कि अशिक्षित परन्तु बुद्धिमान और सहायता करने वाला रहता था। वह अपनी बुद्धिमता के लिए बहुत प्रसिद्ध था। एक बार राजा एक खच्चर पर बैठकर गडरिये के रूप में वेश बदलकर उससे मिलने गया। उसने (गड़रिये ने) अच्छी तरह से स्वागत-सत्कार तथा व्यवहार किया। यह गरीब गडरिया समझ गया था कि उसका दर्शक राज्य का राजा था। राजा उसकी बुद्धिमता से बहुत प्रभावित हुआ। राजा ने उसे छोटे से जिले का राज्यपाल बना दिया। अन्य राज्यपाल इस गड़रिये से बहुत ईर्ष्यालु हो गए और उन्होंने एक षड्यंत्र रचा। राजा को नये राज्यपाल अर्थात् गड़रिये की बेईमानी के बारे में बार-बार शिकायतें की जाती थी। एक दिन नये राज्यपाल (गड़रिया) को महल बुला लिया गया। वहाँ उससे अपने साथ हमेशा लोहे की पेटी रखने का कारण पूछा गया। जब लोहे की पेटी खोली गई तो जैसा कि उसके दुश्मन आरोप लगा रहे थे ऐसा कोई खजाना नहीं पाया गया। इसमें एक पुरानी कम्बल थी जिसे गड़रिया अपना सबसे पुराना मित्र मानता था।

Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (- 5, 7), (- 1, 3)
(iii) (a, 6), (- a, – b)
Solution :
(i) Let the given points be A(2, 3) and B(4, 1), then
AB = \(\sqrt{(4-2)^2+(1-3)^2}\)
⇒ AB = \(\sqrt{(2)^2+(-2)^2}\)
⇒ AB = \(\sqrt{4+4}\)
⇒ AB = √8
⇒ AB = \(\sqrt{2 \times 2 \times 2}\)
⇒ AB = 2√2
Hence, distance (AB) = 2√2.

(ii) Let the given points be P(- 5, 7) and Q(- 1, 3), then
PQ = \(\sqrt{(-1+5)^2+(3-7)^2}\)
PQ = \(\sqrt{(4)^2+(-4)^2}\)
⇒ PQ = \(\sqrt{16+16}\)
⇒ PQ = √32
⇒ PQ = \(\sqrt{4 \times 4 \times 2}\)
⇒ PQ = 4√2
Hence, distance (PQ) = 4√2

(iii) Let the given points be P(a, b) and Q(- a, – 6), then
PQ = \(\sqrt{(-a-a)^2+(-b-b)^2}\)
⇒ PQ = \(\sqrt{(-2 a)^2+(-2 b)^2}\)
⇒ PQ = \(\sqrt{4 a^2+4 b^2}\)
⇒ PQ = \(\sqrt{4\left(a^2+b^2\right)}\)
⇒ PQ = 2\(\sqrt{a^2+b^2}\)
Hence, the distance (PQ) = 2\(\sqrt{a^2+b^2}\).

Question 2.
Find the distance between the points (0,0) and (36,15). Can you now find the distance between the two towns A and B.

Solution:
Let the given points be A(0,0) and B(36, 15) then
AB = \(\sqrt{(36-0)^2+(15-0)^2}\)
⇒ AB = \(\sqrt{(36)^2+(15)^2}\)
⇒ AB = \(\sqrt{1296+225}\)
⇒ AB = \(\sqrt{1521}\)
⇒ AB = 39
Hence, distance between two towns = 39 km.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Let the given points be A(1, 5), B(2, 3) and C(- 2, – 11). Then,

⇒ AC = √256 = 16.28 (approx)
Now,AB + BC = 2.24 + 14.56 = 16.8
AC = 16.28
∴ AB + BC ≠ AC
Hence, the given points (1, 5), (2, 3) and (- 2, – 11) are not collinear.

Question 4.
Check whether (5, – 2), (6,4) and (7, – 2) are the vertices of an isosceles triangle.
Solution :
Let the given points be A(5, – 2), B(6, 4) and C(7, – 2), then
AB = \(\sqrt{(6-5)^2+(4+2)^2}\)
= \(\sqrt{(1)^2+(6)^2}\)
= \(\sqrt{37}\)

BC = \(\sqrt{(7-6)^2+(-2-4)^2}\)
= \(\sqrt{(1)^2+(-6)^2}\)
= \(\sqrt{37}\)
Since, AB = BC
Therefore, the given joints are the vertices of an isosceles triangle.

Question 5.
In a class room, 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square ?” Chameli disagrees. Using distance formula, find which of them is correct ?

Solution:
From the figure, coordinates of A, B, C, D, are respectively (3, 4), (6, 7), (9, 4), (6, 1)

⇒ BD = √36
⇒ BD = 6.
Since, AB = BC = CD = DA and diagonal AC = diagonal BD.
Therefore, ABCD is a square.
Hence, Champa is correct.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer :
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (-1, – 4)
(iii) (4, 6), (7, 6), (4, 3), (1, 2)
Solution :
(i) Let the given points of the quadrilateral be A(- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0).
Join AC and BD

Since, AB = BC = CA = DA and diagonal AC = diagonal BD
Therefore, given points form a square.

(ii) Let the given points of the quadrilateral be A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, – 4).
Join AC and BD

Now,
AC + BC = √13 + √13 = 2√13
AB = 2√13
Since, AC + BC = AB.
Therefore, A, B, C are collinear. So, the given points do not form any quadrilateral.

(iii) The given points of a quadrilateral be A(4, 5), B(7, 6), C(4, 3) and D(l, 2). Join AC and

Diagonal AC ≠ diagonal BD.
Since, opposite sides are equal but diagonals are not equal.
Therefore, the given points form a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution :
Let the given points be A(2, – 5) and B(- 2, 9) and let the required point be POc, 0), then
PA = PB
⇒ PA² = PB²
⇒ (x – 2)² + (0 + 5)² = (x + 2)² + (0 – 9)²
⇒ x² + 4 – 4x + 25 = x² + 4 + 4x + 81
⇒ x² – 4x + 29 = x² + 4x + 85
⇒ x² – 4x – x² – 4x = 85 – 29
⇒ – 8x = 56.
⇒ x = \(\frac{56}{-8}\)
x = – 7
Hence, the required point = (- 7, 0)

Question 8.
Find the values ofy for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution :
The given points are P(2, – 3) and Q(dfgvbtyh10, y)
and PQ = 10 units
∴ PQ = \(\sqrt{(10-2)^2+(y+3)^2}\)
10 = \(\sqrt{(8)^2+y^2+9+6 y}\)
10 = \(\sqrt{64+y^2+9+6 y}\)
10 = \(\sqrt{73+y^2+6 y}\)
100 = 73 + y² + 6y (squaring both sides)
y² + 6y + 73 – 100 = 0
y² + 6y – 27 = 0
y² + (9 – 3)y – 27 = 0
y² + 9y – 3y – 27 = 0
[∵ 1 × – 27 = – 27
9 × – 3 = – 27
9 – 3 = 6]
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9) (y – 3) = 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = – 9 or y = 3.
Hence, y = – 9 or 3.

Question 9.
If Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution:
The given points are P(5, -3), Q(0, 1) and R(x, 6)
∵ Q is the equidistant from P and R.
∴ PQ = RQ
⇒ PQ² = RQ²
⇒ (0 – 5)² + (1 + 3)² = (0 – x)² + (1 – 6)²
⇒ (- 5)² + (4)² = (- x)² + (- 5)²
⇒ 25 + 16 = x² + 25
⇒ 25 – 25 + 16 = x²
⇒ x² = 16
⇒ x = ± 4

Case II: x = 4

PR = \(\sqrt{(-1)^2+81}\)
PR = \(\sqrt{1+81}=\sqrt{82}\)

Case II: If x = – 4

Hence, x = ± 4, QR = √41 and PR = √82 or 9√2.

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-. 3, 4).
Solution:
Let P(x, y) be the equidistant from the points M(3, 6) and B(- 3, 4), then
PA = PB
PA² = PB² .
⇒ (3 – x)² + (6 – y)²
⇒ (- 3 – x)² + (4 – y)²
⇒ 9 + x² – 6x + 36 + y² – 12y = 9 + x² + 6x + 16 + y² – 8y
⇒ x² + y² + 6x – 8y + 6x + 12y – x² -y² = 45 – 25
⇒ 12x + 4y = 20
⇒ 3x + y = 5
⇒ 3x + y – 5 = 0
Hence, the required relation is 3x + y – 5 = 0

Haryana State Board HBSE 9th Class Maths Solutions Chapter 12 Heron’s Formula Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 12 Heron’s Formula Exercise 12.1

Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’ is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
Let ABC be the signal board is shape of an equilateral triangle with side a.
s = \(\frac{a+a+a}{2}\)
⇒ s = \(\frac{3 a}{2}\)
By Heron’s formula, we have
Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}\)
= \(\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}\)
= \(\frac{a}{2} \times \frac{a}{2} \times \sqrt{3}\)
= \(\frac{a^2}{4} \times \sqrt{3}\)
= \(\frac{\sqrt{3}}{4}\) a² square units

Perimeter = 180 cm
⇒ 3a = 180
⇒ a = \(\frac{180}{3}\) = 60
∴ Area of triangle = \(\frac{\sqrt{3}}{4}\) × (60)²
= 900\(\sqrt{3}\) cm²
Hence, area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) a² square units and 900\(\sqrt{3}\) cm².

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:
Sides of a triangular wall of a flyover are 122 m, 22 m and 120 m. Here,
a = 122 m, b = 22 m and c = 120 m
s = \(\frac{122+22+120}{2}\)
⇒ s = \(\frac{264}{2}\) = 132 m
By Heron’s formula, we have
Area of side walls of flyover
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{132(132-122)(132-22)(132-120)}\)
= \(\sqrt{132 \times 10 \times 110 \times 12}\)
= \(\sqrt{11 \times 12 \times 10 \times 10 \times 11 \times 12}\)
= 10 × 11 × 12
= 1320 m²
∵ Rent paid by a company in 12 months for 1 m² = Rs. 5000
∴ Rent paid by a company in 1 month for 1 m² = Rs. \(\frac{5000}{12}\)
∴ Rent paid by company in 3 months for 1320 m² = Rs. \(\frac{5000 \times 3 \times 1320}{12}\)
= Rs. 1250 × 1320
= Rs. 1650000
Hence, rent paid by company = Rs. 1650000.

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “keep the park green and clean” (see Fig. 12.10). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Solution:
Sides of the triangular park are 15 m, 11 m and 6 m.
Here, a = 15 m, b = 11 m and c = 6 m
s = \(\frac{15+11+6}{2}\)
s = \(\frac{32}{2}\)
s = 16 m
By Heron’s formula, we have
Area of triangular park = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{16(16-15)(16-11)(16-6)}\)
= \(\sqrt{16 \times 1 \times 5 \times 10}\)
= \(\sqrt{4 \times 4 \times 1 \times 5 \times 5 \times 2}\)
= 4 × 5\(\sqrt{2}\)
= 20\(\sqrt{2}\) m²
Hence, area painted in colour of triangular park = 20\(\sqrt{2}\) m².

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
We have,
Two sides of a triangle are a = 18 cm and b = 10 cm
and perimeter of triangle (2s) = 42 cm
We know that,
Perimeter (2s) = a + b + c
⇒ 42 = 18 + 10 + c
⇒ 42 = 28 + c
⇒ c = 42 – 28 = 14 cm
Now 2s = 42
s = \(\frac{42}{2}\) = 21 cm
By Heron’s formula, we have
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{21(21-18)(21-10)(21-14)}\)
= \(\sqrt{21 \times 3 \times 11 \times 7}\)
= \(\sqrt{3 \times 7 \times 3 \times 11 \times 7}\)
= 3 × 7\(\sqrt{11}\)
= 21\(\sqrt{11}\) cm²
Hence,area of a triangle = 21\(\sqrt{11}\) cm².

Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
Ratio of sides of a triangle = 12 : 17 : 25 and perimeter is 540 cm.
Let sides of a triangle be 12x cm, 17x cm and 25x cm.
Perimeter = Sum of the sides of a triangle
⇒ 540 = 12x + 17x + 25x
⇒ 540 = 54x
x = \(\frac{540}{51}\) = 10
So,sides of the triangle = 12 × 10, 17 × 10, 25 × 10 = 120 cm, 170 cm, 250 cm
Here, a = 120 cm, b = 170 cm and c = 250 cm
s = \(\frac{120+170+250}{2}\)
⇒ s = \(\frac{540}{2}\)
⇒ s = 270 cm
By Heron’s formula, we have
Area of a triangler = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{270(270-120)(270-170)(270-250)}\)
= \(\sqrt{270 \times 150 \times 100 \times 20}\)
= \(\sqrt{3 \times 3 \times 30 \times 30 \times 5 \times 5 \times 20 \times 20}\)
= 3 × 5 × 20 × 30
= 9000 cm²
Hence,area of a triangle = 9000 cm².

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Two equal sides of an isosceles triangle are 12 cm and 12 cm and perimeter = 30 cm
Perimeter of a triangle = Sum of sides of a triangle
⇒ 30 = 12 + 12 + IIIrd side
⇒ 30 = 24 + IIIrd side
⇒ IIIrd side = 30 – 24 = 6 cm
Here, a = 12 cm, b = 12 cm and c = 6 cm
s = \(\frac{12+12+6}{2}\)
⇒ s = \(\frac{30}{2}\)
= 15 cm
By Heron’s formula, we have
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15(15-12)(15-12)(15-6)}\)
= \(\sqrt{15 \times 3 \times 3 \times 9}\)
= \(\sqrt{15 \times 3 \times 3 \times 3 \times 3}\)
= 3 × 3\(\sqrt{15}\)
= 9\(\sqrt{15}\) cm²
Hence, area of an isosceles triangle = 9\(\sqrt{15}\) cm².

Haryana State Board HBSE 6th Class Social Science Solutions Civics Chapter 1 Understanding Diversity Textbook Exercise Questions and Answers.

Haryana Board 6th Class Social Science Solutions Civics Chapter 1 Understanding Diversity

HBSE 6th Class Civics Understanding Diversity Textbook Questions and Answers

Understanding Diversity Class 6 Questions And Answers HBSE Question 1.
Draw up a list of different festivals celebrated in your locality. Which of these celebrations are shared by members of different regional and religious communities ?
Answer:
Festivals celebrated in our locality are:

1. Id
2. Gurupurab
3. Ram Navami
4. Christmas
5. Holi
6. Dussehra
7. Diwali
8. Buddha Purnima
9. Makar Sakranti
10. Ramzan
11. Lohri
12. Baishakhi
13. Teej
14. Rakshabandhan
15. Mahavir Jayanti.

We are all the children of one god. All the religions (Hinduism, Sikhism, Jainism, Buddhism, Islam and Christianity) teach us tolerance, equality and humanity. So the above mentioned festivals are shared by number of people in our community (locality). The festivals including national holidays are celebrated equally by people of different regional and religious communities.

HBSE 6th Class Social Science Understanding Diversity Question 2.
What do you think living in India with its rich heritage of diversity adds to your life?
Answer:
The people of India belong to different races and castes, and em brace different religions. The inhabitants of the India are greatly different from one another in dress, eating habits and customs. Foreign scholars have termed India as a museum of diverse religions, races, communities and cultures.

This is the only country where unity in diversity and vice-versa breathes comfortably. So we think that living in India with its rich heritage of diversity adds great experience in our life.

Understanding Diversity HBSE Social Science 6th Class Question 3.
Do you think the term ‘Unity in Diversity’ is an appropriate term to describe India? What do you think Nehru is trying to say about Indian unity in the sentence quoted above from his book ‘The Discovery of India’?
Answer:
Yes, this is an appropriate term to describe India. There is diversity in religion, languages, food, clothing, etc. in India. Everybody in India feels proud to call himself. Therself an Indian. They celebrate many festivals together and people of different religions visit each other’s places of worship. They live in the same community and share each other’s joys and sorrows.

They together fought to free India from British rule. Jawaharlal Nehru, the former Prime Minister and the writer of the ‘Discovery of India’ says that the Unity of India is deeper within its fold. The Indians are known for their tolerance, acknowledgment and encouragement for diverse religions and customs.

Question 4.
Underline the line in the poem sung after Jallianwalla Bagh Massacre, which according to you reflects India’s essential Unity.
Don’t forget the days of blood, O friend In the midst of your happiness remember to shed tear for us
The hunter has torn away every single flower Do plant a flower in the desert garden dear friend
Having fallen to bullets we slept in Jallianwala Bagh
Do light a lamp on this lonely grave O friend The blood of Hindus and Muslims flows together today
Do soak your robe in this river of blood dear friend
Some rot in jails while others lie in their graves
Do shed a few tears for them O friend.
Answer:
It is “The blood of Hindus and Muslims flows together today”.

Question 5.
Choose another region in India and do a similar study of the historical and geographical factors that influences the diversity found there. Are these historical and geographical factors connected to each other? How?
Answer:
We choose ‘Punjab’ region in India and do a similar study of the historical and geopgraphical factors there. Punjab is one of the most fertile plain in India. It is drained five rivers. The climate of Punjab veries from time to time. Like the whole part of the country, it undergoes four climatic changes in a year. In cold season, temperature goes down while in hot season, the temperature becomes very high. This plain is covered on north side by Himalayan ranges.

People of different religions live here. Hinduism, Islam and Sikhism are the religions mostly found here. Punjab produces a huge amount of wheat, rice and sugarcane. From industrial point of view, Punjab is also very rich. Like above mentioned geographical factors, history also influence this region.

From the beginning Punjab has been a number of attack from the foreign invaders, that makes the punjabi people very hard worker and brave. From the earliest time, it remained the very important part of India. All invaders who conquered Punjab were able to conquered India and those who were defeated at frontier state could never see India Punjab is known as the gateway of India. So, we can conclude that both geographical and historical factors are connected with each other. One has deep impact on other and vice-versa.

HBSE 6th Class Civics Understanding Diversity Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What are communal riots?
Answer:
When arson, looting, fire and killing break out between two communities due to different reasons it is called communal riots.

Question 2.
Name the European sailor who came in search of a sea route to India.
Answer:
The Portuguese discovered the sea route to India, from Europe, when Vasco da Gama landed with his ships here.

Question 3.
Why was Ladakh considered a good trade route?
Answer:
Ladakh was considered a good trade route as it has many passes through which caravans travelled to what is today called Tibet.

Question 4.
By what name is national epic of Tibet known as?
Answer:
Kesar Saga.

Question 5.
How did the caste system create inequality?
Answer:
In the caste system a person could not change his profession, it was not considered necessary for him to know anything more than what was needed in the profession. This created a situation of inequality.

Short Answer Type Questions

Question 1.
What is inequality ? Give one example of inequality.
Answer:
Inequality comes about when a person does not have the resources and opportunities that are available to other persons. Samir did not have the opportunity to go to school, because his parents are poor and do not have enough to eat and wear.

Question 2.
Why is only very little agriculture possible in Ladakh?
Answer:
Very little agriculture is possible in Ladakh because:
(i) the region does not receive any rain.
(ii) it is covered in snow for large part of the year-1

Question 3.
Who was Ibn Batuta ? What did he write about Kerala?
Answer:
Ibn Batuta was an Arab trader who travelled to Kerala a little less than seven hundred years ago. In a travelogue, he described the lives of Muslims. He wrote that they were a highly respected community.

Question 4.
How did people of India prove that though they had diverse culture, they were united?
Answer:
Indians fought unitedly for the freedom of the country though they were from different backgrounds. They went to jail together. They found different ways to oppose the British.

Long Answer Type Questions

Question 1.
How do the different cultures affect our lives?
Answer:
We are influenced by the different cultures because:
(a) We move from place to place for work and with each move our lives which are affected by different cultures that exist around us.
(b) In our own neighburhoods we live close to people from several communities.
(c) We hear stories about each other’s lives, customs and traditions and that too has an impact upon our lives.

Question 2.
Describe the Jallianwala Bagh Massacre. How does it reflect unity in diversity?
Answer:
(i) Men and women, Hindus, Sikhs and Muslims, rich and poor had gathered to peaceful protest against the British at Jallianwala Bagh on 13th April 1919.
(ii) A British General Dyer ordered his troops to open fire on a large group of these unarmed, peaceful people.
(iii) Thousands of people were killed. This shows how Indians proved that though they were different in race, culture, religion, languages and tradition; they were all united , they would do anything for India.

Question 3.
How can we explain diverity?
Answer:
Diversity can be explained in many ways, e.g.,
(i) Two hundred years before people travelled from one part of the world to another, in ships, on horses or camels or on foot. But after the advent of aeroplanes and trains, buses and cars people want to travel by them as journey is comfortable and takes less time.

(ii) Often people went in search of new lands or for trade. Because it took long to reach the destination, people stayed there for a long time. Sometimes people left their homes because of natural calamities, some went in search of work.

(iii) People come from villages to cities to find work. In cities it is often easy to forget how then- lives revolved around the surroundings. In villages they grow their own vegetables and grain. In the cities they depend on the market to buy things.

(iv) Sometimes when people make their homes in new places, they begin to change. Their languages, food, music and religion become a mix of old and new.

(v) Similarly people have to adapt themselves to geographical surroundings, e.g., living near the sea is very different rather living in a mountainous region.

Question 4.
Write how historical and geographical factors have influenced the life of the people in Kerala.
Answer:
Kerala is a state in south-western part of India, near the Arabian sea. It has sea on one side and hills on the other. A number of spices are grown on the hills. Spice trade was a flourishing trade. Pepper, cloves and cardamons were grown. It attracted lots of traders. Jewish and Arab traders were the first to come.

The Apostle of Christ, St. Thomas introduced Christianity to India. The Portuguese discovered the sea-route to India. These historical influences led people in Kerala to practise Judaism, Islam and Christianity together with Hinduism. The fertile land and climate are suitable for growing rice and majority of the people eat rice, fish and vegetables.

The fishing nets are called cheena-vala and utensils used for frying are called cheen-chatti and it is believed that word cheen must have come from China.

Understanding Diversity Class 6 HBSE Notes

• Diversity: The variety or differences in the traits, looks, behaviour, culture, religion, language, abilities, resources and opportunities related to different people.
• Habitat: The geographical area in which people live and adapt their lives.
• Resources: Anything that can be of use to the people.
• Unity in Diversity: Oneness among people despite diversity in traits and emotions.