Bhagya

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TN State Board 12th English Model Question Paper 1

Time: 2 1/2 Hours
Maximum Marks: 90

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding option code.
• Part II has got two sections. The questions are of two marks each. Question numbers 21 to 26 in Section I and Question numbers 27 to 30 in Section II are to be answered in about one or two sentences each.
• Question numbers 31 to 40 in Part III are of three marks each and have been divided in three sections. These are to be answered as directed.
• Question numbers 41 and 47 in Part IV are of five marks each. These are to be answered as directed.

Part -1

I. Answer all the questions. [20 x 1 = 20]
Choose the correct synonyms for the underlined words from the options given:

Question 1.
We may choose to be wise or ridiculous.
(a) funny (b) earnest (c) critical (d) serious
Answer:
(a) funny

Question 2.
Tenzing kicked steps in a long traverse.
(a) vertical movement (b) diving (c) jumping (d) crisscross
Answer:
(d) crisscross

Question 3.
Suffering ennobles you.
(a) destroys (b) dignifies (c) distresses (d) encourages
Answer:
(b) dignifies

Choose the correct antonyms for the underlined words from the options given:

Question 4.
Liberties of everybody must be curtailed.
(a) restricted (b) reduced (c) constrained (d) increased
Answer:
(d) increased

Question 5.
With effort I could muster my arms and shoulder.
(a) collect (b) give (c) lose (d) gather
Answer:
(c) lose

Question 6.
He was a walking horror.
(a) delight (b) shock (c) disgust (d) fear
Answer:
(a) delight

Question 7.
Choose the correct combination for the compound word ‘merciless’.
(a) Noun + Verb (b) Noun + Adjective (c) Gerund + Noun (d) Preposition + Noun
Answer:
(b) Noun + Adjective

Question 8.
Choose the correct expansion of VAT.
(a) Value Added Task (b) Value Added Time, (c) Value Addictive Task (d) Value Added Tax
Answer:
(d) Value Added Tax

Question 9.
Choose the meaning of the foreign word in the sentence:
The guerrilla fighters took control of the capital of the country, which gave them control of the government.
(a) civil war (b) monkey war (c) little war (d) giant war
Answer:
(c) little war

Question 10.
Choose the right combination for the blended word adflation.
(a) ad + flation (b) advertisement + inflation (c) advt + elation (d) advt + flation
Answer:
(a) ad + flation

Question 11.
Choose the clipped word for influenza.
(a) flu (b) influ (c) inf (d) fluenz
Answer:
(a) flu

Question 12.
The craze for collecting postage stamps is known as ………………………
(a) Melomania (b) parricide (c) pragmatist (d) stampomania
Answer:
(d) stampomania

Question 13.
Form a derivative by adding the right suffix to the word ‘reconcile’.
(a) -ment (b) -ion (c) -able (d) -ly
Answer:
(a) -ment

Question 14.
Fill in the blanks with a suitable relative pronoun.
The management is the one …………………….. promised to block it off.
(a) which (b) where (c) who (d) whose
Answer:
(a) which

Question 15.
Fill in the blanks with a suitable preposition.
I got ………………………. the horse.
(a) above (b) by (c) onto (d) from
Answer:
(c) onto

Question 16.
Choose the correct question tag for the following statement.
You met Seth yesterday, ………………………….?
(a) won’t you (b) didn’t you (c) will you (d) don’t you
Answer:
(b) didn’t you

Question 17.
Choose the suitable meaning for the idiom found in the following sentence.
Sasidharan was accusing us of stealing his phone until he found it, and now he’s trying to sweep it under the rug.
(a) hiding under the rug (b) pretend that the incident never happened (c) sweeping the floor (d) make up for his mistake
Answer:
(b) pretend that the incident never happened

Question 18.
Replace the underlined word with a euphemistic expression.
The Government bans ethnic cleansing of the disabled.
(a) infanticide (b) matricide (c) regicide (d) genocide
Answer:
(d) genocide

Question 19.
Choose the correct sentence pattern for the following sentence.
Their selfless action gave a promise for human society.
(a) SVIODO (b) SVOA (c) SVOC (d) SVO
Answer:
(b) SVOA

Question 20.
Fill in the blank with a suitable phrasal verb.
The burglars had to ……………………… all the jewels and money that had gone missing.
(a) account to (b) account in (c) account for (d) account towards
Answer:
(c) account for

Part II
Section – 1

Read the following sets of poetic lines and answer any four from it. [4 x 2 = 8]

Question 21.
“All through the summer at ease we lay,
And daily from the turret wall
We watched the mowers in the hay’”
(a) Who does ‘we’ refer to?
(b) How did the soldiers spend the summer days?
Answer:
(a) “We” refers to the brave and loyal soldiers in the castle.
(b) They spent the summer days gazing out of the castle. They were ready to shoot the enemy at sight who were at half-a kilometer distance. But none came near. So, they were relaxed.

Question 22.
“ Thro ’ scudding drifts the rainy Hyades Vext the dim sea: I am become a name;”
(a) What is the figure of speech employed in the above lines?
(b) What is Hyades?
Answer:
(a) Personification is the figure of speech employed in the above lines.
(b) Hyades is a group of stars in the constellation Taurus often associated with rain.

Question 23.
“Full-galloping: nor bridle drew Until he reached the mound.”
(a) Why was the rider in a hurry?
(b) Where did the soldier stop? Why?
Answer:
(a) The rider was carrying an urgent message to Napoleon Bonaparte.
(b) The soldier stopped at the mound. Napoleon Bonaparte was expecting news about the outcome of the battle at Ratisbon.

Question 24.
“Mine inner vision rose a form sublime,
Thy form, O Tree, as in my happy prime
I saw thee, in my own loved native clime.”
(a) What rose in the mind of the poet?
(b) What was found in the sublime inner vision?
Answer:
(a) A glorious inner vision rose in the mind of the poet.
(b) The tree’s youthful and handsome figure was found in the inner vision of the poet.

Question 25.
“Brutes have been gentled where lashes failed.”
(a) When does the Government or law use lashes?
(b) What can transform brutes?
Answer:
(a) When a brute has committed a crime, the law uses lashes to punish him.
(b) Gentleness can transform brutes.

Question 26.
“Then a soldier.
Full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,”
(a) Describe the two traits of a soldier.
(b) What is the poetic device used in : ‘bearded like the pard’?
Answer:
(a) A soldier is always ready to swear and is full of oaths. He is ever ready to compete for honour and glory.
(b) The poet uses a simile for comparison.

Section – 2

Answer any three of the following questions. [3 x 2 = 6]

Question 27.
Report the following dialogue:
Answer:
Annette : I want to order a big pineapple cake for my husband’s birthday.
Confectioner : When is his birthday and at what time do you want it to be delivered?
Annette told the Confectioner that she wanted to order a big pineapple cake for her husband’s birthday. The confectioner asked Annette when his birthday was and at what time she wanted it to be delivered.

Question 28.
You must read the clause carefully. Otherwise you will not understand it. (combine using if)
Answer:
If you do not read the clause carefully, you will not understand it.

Question 29.
Rewrite the sentence making an inversion in the conditional clause.
If they had arrived sooner, they could have prevented this tragedy!
Answer:
Had they arrived sooner, they could have prevented this tragedy!

Question 30.
Even though he suffered from arthritis, he studied hard because he wanted to go to medical school. (Change the following into a simple sentence)
Answer:
In spite of him suffering from arthritis, he studied hard because he wanted to go to medical school.
Or
Despite suffering from arthritis, he studied hard because he wanted to go to medical school.

Part-III
Section -1

III. Answer any two of the following with reference to the context. [2 x 3 = 6]

Question 31.
Sung darkling from our tree, while men repose.
Answer:
Reference: These lines are from the poem ‘Our Casuarina Tree” written by Torn Dutt.

Context: The poet says this while describing Casuarina and its inhabitants at night.

Explanation: The Casuarina Tree allows the birds and animals to abide in it. At night, the garden overflows with endless melodious song sung by the dark king from the Casuarina Tree. It is a kind of a lullaby to sleeping humans.

Question 32.
They have their exits and their entrances;
And one man in his time plays many parts,
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.

Context: The poet says this while hinting at the beginning and the end of life.

Explanation: The poet divides man’s life into seven stages. The first stage symbolises birth and the last stage death. So, he uses the words “entrances and exits”.

Question 33.
To see your flag-birdflap his vans Where I, to heart’s desire, Perched him!
Answer:
Reference: These lines are from the poem ‘Incident of the French Camp” written by Robert Browning.

Context: The narrator says these through the boy-soldier while explaining his role in the final stages of storming of Ratisbon.

Explanation: The boy soldier flung himself in the midst of battle and risked his life. He did not bother about his death. He doggedly carried out the mission of hoisting French national flag. After announcing that the battle was won he proudly says that he himself hoisted French Flag to his heart’s content at the Market place in Ratisbon.

Section – 2

Answer any two of the following questions in about 30 words. [2 x 3 = 6]

Question 34.
How does George Orwell explode the myth about the harmfulness of tea leaves?
Answer:
In some countries, tea pots are fitted with little dangling baskets under the spout to catch the stray leaves which are supposed to be harmful. The author asserts that one can swallow tea leaves in considerable quantities without ill-effect. He claims that if the tea is not loose in the pot it never infuses properly with the hot beverage (i.e.) tea.

Question 35.
What did Tenzing and Edmund Hillary gift to the God of lofty Summit? How did they do it?
Answer:
A hole was made in the ice where Tenzing placed reverentially a bar of chocolate, a packet of biscuits, and a handfull of lollies. As Hillary remembered that his team head Col. Hunt had requested to place a crucifix after reaching the peak. So, he also made a hole in the snow and placed the crucifix beside Tenzing’s gift to the Gods. Devout Buddhists believed that at least a small token of gift should be left with Gods who have their homes in that lofty Everest.

Question 36.
What is ‘liberty’ according to the old lady?
Answer:
According to the Russian old lady, liberty is the right to walk in the middle of the road. She is liberty-drunk. She wasn’t aware that if the pedestrian chooses to walk down the middle of the road, then the end of such liberty would be a universal chaos.

Section – 3

Answer any three of the following questions in about 30 words. [3 x 3 = 9]

Question 37.
Study the pie chart given and answer the questions that follow.

(a) What is the mode of transport preferred by most of the students?
(b) How many students prefer bus?
(c) What does the remaining students prefer?
Answer:
(a) Bicycle is preferred by most of the students.
(b) Thirty percent of students prefer to come by bus.
(c) The remaining students prefer to come by car or by walk to school.

Write a dialogue of minimum 3 exchanges between a fresher and a company manager.
Answer:
Manager : Good Morning Vikram. please take your seat.
Vikram : Good morning Sir. Thank you.
Manager : I understand that you are a First class Engineering graduate. Tell us about yourself.
Vikram : Well, I hail from a village in Iddayangudi and my parents are fanners.
Manager : If given an opportunity, can you handle project presentations on your own?
Vikram : Yes certainly Sir as I am well versed in Computer and have done various presentations during my course of study.

Question 39.
Describe the process of how to knot a bow tie.
Answer:
Place the tie around the neck, with one end slightly longer than the other.
Then place the longer end over the other and pass it upwards and behind the point where the two ends cross.
Next, take the other end of the tie and bend it twice to form an ‘S’ shape.
Bring the longer end down and in front, so that it holds the ‘S’ curve in place.
Now comes the trickiest part of the process when you have to take the long end of the tie and form a similar ‘S’ shape before passing it through the narrow gap behind the other end.
This creates a knot and the bow should now be held securely in place.
Finally, adjust both sides of the bow to make it symmetrical.

Question 40.
Complete the proverbs using the word given below.
(a) A bad ……………………….. makes a bad ending, (plan, beginning, plot)
(b) All roads lead to ……………………….. (Paris, London, Rome)
(c) ……………………….. have ears. (Walls, Floors, Windows)
Answer:
(a) beginning (b) Rome (c) Walls

Part – IV

Answer the following questions: [7 x 5 = 35]
Answer in a paragraph in about 150 words.

Question 41.
Summarise George Orwell’s distinctive ideas in “A Nice Cup of Tea”.
Answer:
Tea is one of the mainstays of civilizations of UK, New Zealand, Australia and Eire. But, it is not found in the recipes of cookeiy books. The nicest manner in which tea must be prepared has become a subject matter of violent disputes among tea loving persons even within a family. Indian tea fits in with the author’s description of “nicest cup of tea” as it rejuvenates the drinker with wisdom, stimulation, bravery and optimism. One cup of strong tea is equal to twenty weak ones.

Best tea is prepared by boiling tea leaves first and then adding milk to taste. Milk first schools and tea first schools still fight. Tea prepared in earthen pot or china ware is good. The real flavour of tea should never be imprisoned by strainers. Consuming tea leaves will not harm one’s health. Adding sugar to tea will naturally ruin its real flavour and taste.

The author advises misguided people who drink sweet tea to desist from the practice for a fortnight. He guarantees them that they would like the natural taste of tea and would never again add sugar to tea. Tea should be taken in a cylindrical or breakfast cup to enjoy it. Shallow and flat cups don’t retain wannth and taste for a long time.

Or

Life is unjust and cruel to certain people. Do they all resign themselves to their fate? Can you think of some who have fought their disabilities heroically and remained a stellar example for others? (for e.g. the astrophysicist Stephen Hawking, a paraplegic). Give an account of one such person and his/her struggle to live a fruitful life.

Alexis Leon lives in Kakkanad, Kerala. He passed B.Tech from the university of Kerala with first rank. Then he did his M.Tech. He met with an accident in 1993 which left him paralysed from chest down and confined him to a wheel chair for the rest of his life. After a brief spell of dejection owing to a suspended marriage, he made up his mind to write books. He has written 50 books for Engineering graduates.

His notable work is ‘Internet for Everyone’ and ‘A Guide to Software Configuration Management’. He is also a mentor at International Mentoring Network Association. He offers software consultancy to international IT companies. He travels across the world and delivers lectures to graduates and Corporates. His indomitable will has made him strong. He has become a light house for many aspirants in the software industry.

Question 42.
Shakespeare has skilfully brought out the parallels between the life of man and actors on stage. Elaborate this statement with reference to the poem.
Answer:
Shakespeare has beautifully compared the growth of humans by stages with his emergent role during that stage. In the first stage man plays the role of an infant. As an infant, he does represent characterisation of mewling and puking. In the second Act, he does the role of a school boy with the characteristics of unwillingness to go to schools and innocence shining in his face. In the third Act, he performs the role of a lover head over heels in love with a beautiful lady. He composes woeful romantic ballads and sings serenades to impress his love. In the fourth act, he plays the impressive role of a short-tempered, honour pursuing soldier.

He is ready to put his mouth in the Cannon’s mouth for conquering the bubble like honour in order to defend the territory of his countiy. In the fifth Act, he performs the role of a mature and fair judge criticising the ways of the world often spicing up his conversations with wise remarks and wit. His pot belly and well-cut beard shows the social status he enjoys in life. In the sixth act, he is old. He performs the role of a thin old man wearing ill-fitting loose garments with a changed treble in his voice.

He is bespectacled and slow in walking. In the final act, he becomes a total invalid losing all senses of hearing, taste and sight. Then the performer leaves the stage (i.e.) the lonely planet.

[OR]

How according to the poet is it possible for his son to bring changes into a world that resents change?
The poet advises his son to introspect often and not feel ashamed of being called a fool especially when he pursues knowledge. He can examine himself and remove his follies. He must enjoy the advantages of solitude. Solitude would help him to be creative. He would invariably learn that final decisions are always taken in silent rooms. Being alone, he can identify his innate potential and talents. His free imagination will bring about changes even if the world resents them. The zest to bring about changes will elevate him to the level of Shakespeare, Pasteur, Wright Brothers, Pavlov and Michael Faraday. Thus he will be remembered as one of the great men who changed the world.

Question 43.
Develop the hints into a paragraph.
Aksionov, meek – fellow prisoners respect – prison authorities – old man unjustly punished – speaks truth – prison shelf dug – newcomer Semyonich – could be the culprit – suspicion – all prisoners summoned – denied – some quiet – betray – Governor – enormous trust in Aksionov – stayed quiet.
Answer:

Aksionov was a merchant who was condemned for a murder he had not committed. He waits for God’s justice. Meanwhile in prison he learnt to make boots and earned a little money. Aksionov was meek. Fellow prisoners, having realized that he was innocent, respected him and called him grandfather. The prison authorities also believed that the religious old man must have been unjustly punished. They knew Aksionov never told lies. The prison authorities found out that someone had been digging the prison shelf to escape. They doubted that newcomer Semyonich could be the culprit. But they could not nail him on the grounds of suspicion alone. All the prisoners were summoned and enquired including Semyonich. All denied any knowledge of it. Those who knew kept quiet because they knew Semyonich will get flogged if someone betrayed him. As no one was ready to tell the truth, the Governor, who had enormous trust in the nobility and honesty of Aksionov asked him to tell the truth. But Aksionov preferred to stay quiet.

[OR]

Richard Parker calmed Pi – irony of life – peace, purpose and wholeness – same boat – Pi’s personality – Parker’s presence – realized danger and peace – innate strength – tiger – less tormenting – loneliness and despair – chances of survival – middle of Pacific Ocean – optimistic, balanced and stoical – thanks to the presence of Parker.
Answer:
Pi was alone and orphaned. He was drifting in the middle of pacific ocean hanging on to the oar. The elements of nature which threatened his survival the previous night had calmed down. The life boat did not sink. It was Richard Parker who calmed down Pi. The irony of life is that the animal that scared him witless to start with was the same who brought him peace, purpose and even wholeness. They were literally and figuratively in the same boat. They had to live together.

A part of Pi’s personality was glad about Parker’s presence, who gave him the will to hang on to life under the toughest conditions. He realized danger and peace can co-exist when one is aware of one’s innate strength. He realizes the fact that staying with a tiger is less tormenting than living alone with despair. Loneliness and despair can easily kill a person.

He decided to focus on the chances of survival and shut off the tragic circumstances he was placed in a lonely boat in the middle of Pacific Ocean. In short, he became optimistic, balanced and stoical in his attitude to life. Thanks to the presence of Richard Parker.

Question 44.
Write a summary or make notes of the following passage.
Answer:
There is an enemy beneath our feet – an enemy more deadly for his complete neutrality. He recognizes no national frontiers, no political parties. Everyone in the world is endangered by him. The enemy is the earth itself. When an earthquake strikes, the world trembles. The power of a quake is greater than anything man himself can produce.

But today scientists are directing a great deal of their effort into finding some way of battling earthquakes, and it is possible that at some time in the near future mankind will discover a means of protecting itself from earthquakes. An earthquake strikes without warning.

When it does, its power is immense. If it strikes a modem city, the damage is very immense. When gas cylinders burst, explosions are caused and fires start. Underground railways are wrecked. Buildings collapse, bridges fall, dams burst, and gaping crevices appear in busy streets. If the quake strikes at sea, huge tidal waves sweep inland. If it strikes in mountain regions, avalanches roar down into the valley.

Consider the terrifying statistics from the past 1755: Lisbon, capital of Portugal – the city destroyed entirely and 450 were killed. 1970: Peru: 50,000 were killed. In 1968 an earthquake struck Alaska. As this is a relatively depopulated part, only a few people were killed. But it is likely that this was one of the most powerful quakes ever to have hit the world.

Geologists estimate that during the tremors, the whole of the state moved over 80 feet farther west into the Pacific Ocean. Imagine the power of something that can move an entire subcontinent!

This is the problem that the scientists face. They are dealing with forces so immense that man cannot hope to resist them. All that can be done is to try to pinpoint just where the earthquake will strike and work from there. At least some protective measures can then be taken to save lives and some of the property.

Summary

No. of words given in the original passage: 326
No. of words to be written in the summary: 326/3 = 108 ± 5

Rough Draft
The enemy, Earth more deadly for neutrality, recognizes no national frontiers, no political parties Everyone is endangered by him. When an earthquake strikes, the world trembles and is powerful. Today-scientists are finding ways of battling earthquakes which strikes without warning. In a modem city the damage is immense.

When gas cylinders burst, explosions trigger fires and damages occur. If the quake strikes at sea, huge tidal waves sweep inland. Iin mountain regions, avalanches roar down the valley. Once Alaska escaped because it is thinly populated unlike other countries. Geologists estimate-fiiat the state moved over 80 feet farther west into the Pacific Ocean.

This is the problem that the scientists face. They are dealing with forces soAmnjense. We should try to pinpoint just where the earthquake will strike for some protective measured

Fair draft:
The enemy, Earth more deadly for neutrality, recognizes none and endangers all. When an earthquake strikes, the world trembles. Scientists find ways to battle powerful earthquakes which strikes without warning. In modem cities, the damage is immense. When gas cylinders burst, explosions trigger fire. If the quake strikes at sea, huge tidal waves sweep inland.

In mountain regions, avalanches roar down the valley. Once Alaska escaped because it is depopulated unlike others. Geologists estimate that the state moved over 80 feet farther west into the Pacific Ocean.

Scientists face such problems with forces so immense. We should try to pinpoint just where the earthquake will strike for some protective measures.

No. of words in the summary: 110
[OR]

Notes
Title : Earthquake – the deadly enemy of mankind.
Deadly Enemy of Mankind

• strikes all without a distn. – national boundary/political affiliation
• Power greater than man-made weap.
• Scientists finding means to combat eq.; they will find ways to protect themselves from eq.

Damage Cause by an Earthquake in General

• Strikes without warning.
• Modem city when struck red. to a primitive village.

Damage Caused by an Earthquake in Particular

• strikes plains, seas, mnts. – all round destm.
• In 1755, Lisbon destyd, 450 killed.
• In 1970, Peru struck, 50,000 killed.
• In 1968, Alaska hit, subcontinent moved 80 feet into the Pacific Ocean.

Problems Scientists Face

• cannot resist powerful eq.
• can predict the place of origin
• precaution can be taken to save man & property.

Abbreviations used: eq. – earthquakes; distn. – distinction; weap. – weapon; red. – reduced; mnts. -mountains; destm. -destruction; destyd.-destroyed;

Question 45.
Write a letter to the editor of the newspaper on the mindless felling of trees on the road side.
Answer:
12.11.2020
From
XXX
YYY

To
The Editor
The Hindu Anna Salai
Chennai-600 002.

Sir,
Sub: Frequent felling of Trees

I am the Secretary of the YYY Colony Association. Our locality was surely a haven to us in the past few years. The fruits of labour that we had sown about ten years ago is well nurtured and utilised.

Most of the roads in our locality was filled with greenery because of the Flame of the Forest trees that we had planted and taken care of. The roads had enough shade and was filled with beautiful flowers.

In the recent past, we have encountered the mindless felling of trees by the Corporation staff for various reasons. We kindly request you to publish the same in the newspapers to bring to notice the mindless felling of trees to the public and the authorities concerned.

We hope that such a scenario doesn’t take place in our locality again and looking forward to a positive change in this situation.

Thank you,

Yours sincerely,
XXX

Address on the envelope
To
The Editor The Hindu Anna Salai
Chennai-600 002

[OR]

Write a paragraph of 150 words on “India-A Tourist’s Paradise”.
Answer:

India-A Tourist’s Paradise

India is a huge country with varied cultures and races. It is a land with primordial traditional heritage and rich history. India is the only country which has all the four seasons viz summer, autumn, winter and spring. It has beauty in all its places whether they are mountains, deserts, plains, seas or rivers. All these and the modem cities like Delhi, Mumbai, Chennai, Hyderabad and Bangalore as well as the places of pilgrimage make it one of the most sought-after tourist destinations in the world.

India is blessed with places that have natural beauty like Kashmir, Darjeeling, Matheran, Goa, Puri, Kerala, Valparai Ooty, Kodaikanal etc. All such beautiful places are often visited by people who try to combine spirituality with pleasure. India has a variety of beautiful locations to visit which makes it a tourist’s paradise.

Question 46.
Spot the errors and rewrite the sentences correctly.
(a) He spoke the English perfectly.
(b) I and she have booked tickets for the latest Harry Potter movie.
(c) My little sister is afraid for cockroaches.
(d) You can type quickly, won’t you?
(e) Though I had a cup of coffee yet I feel tired.
Answer:
(a) He spoke English perfectly.
(b) She and I have booked tickets for the latest Harry Potter movie.
(c) My little sister is afraid of cockroaches.
(d) You can type quickly, can’t you?
(e) Though I had a cup of coffee, I feel tired.

[OR]

Fill in the blanks correctly.
(a) When I was walking ……………………. foot outside my farm house, I saw a ……………………. under the grape vine, (bear/bare)
(b) I ……………………. arrive home late this evening. (Use a modal in the given blank.)
(c) He ……………………. buy a new car. (Use a semi-modal)
(d) ……………………. you ……………………. (see) my wallet and keys recently? (use proper tense)
Answer:
(a) bare/bear
(b) shall
(c) ought to
(d) Have, seen

Question 47.
Identify each of the following sentences with the fields given below:
(a) The deliberate suppression of texts by organisations shrouds the subject.
(b) The study of stars and stellar evolution is fundamental to our understanding of the universe.
(c) This shift to farming may have occurred because of climate change.
(d) Sovereign power may be vested on an individual or it may be vested on a group.
(e) Some medicines may not be safe during pregnancy.
[Astronomy; Medicine; Politics; Literature; Agriculture]
Answer:
(a) Literature
(b) Astronomy
(c) Agriculture
(d) Politics
(e) Health

[OR]

Read the following passage and answer in your own words.
Answer:
The earth is losing its forests. Presently, trees cover about 30 percent of the earth’s surface, but they are being destroyed at an alarming rate, especially in the tropics. Timber harvesting is a major reason for the destruction of the forests. Trees are used for building houses, making furniture, and providing pulp for paper products, such as newspapers and magazines. At least 40 hectares of rainforest are being felled every minute, mostly in order to extract the valuable timber.

Another way that man is destroying the world’s forests is by burning them down. In the Amazon, for example, rainforests are being burnt down at a rate of 20 hectares a minute. The main reason of burning down the rainforests is to clear the land for farming. Farmers in rainforest countries are often poor and cannot afford to buy land. Instead, these farmers clear rainforest land to raise their animals or grow their crops.

Because tropical rainforest soil is so poor in nutrients, farmers cannot reuse the same land year after year. In the following years, farmers just clear more land, destroying the forest piece by piece. Already more than 30 tropical countries have reached a critical level of forest destruction and one-time exporters of timber such as Nigeria and Thailand now have to import timber for their domestic needs.

Questions:
a. What is the major reason for the destruction of forests?
b. Mention any two uses of trees.
c. Why do some people burn down the rainforests?
d. How many tropical countries have reached a critical level of forest destruction?
e. From where do Nigeria and Thailand get timber for their domestic needs?
Answers:
(a) Timber harvest is the major reason for the destruction of forests.
(b) Trees are cut and their logs are used for making furniture and building houses.
(c) Some people bum down rain forests to clear the land for farming.
(d) More than 30 tropical countries have reached a critical level of forest destruction.
(e) Nigeria and Thailand get their timber from foreign countries.

Students can Download Tamil Nadu 12th English Model Question Paper 2 Pdf, Tamil Nadu 12th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th English Model Question Paper 2

Time: 2 1/2 Hours
Maximum Marks: 90

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding option code.
• Part II has got two sections. The questions are of two marks each. Question numbers 21 to 26 in Section I and Question numbers 27 to 30 in Section II are to be answered in about one or two sentences each.
• Question numbers 31 to 40 in Part III are of three marks each and have been divided in three sections. These are to be answered as directed.
• Question numbers 41 and 47 in Part IV are of five marks each. These are to be answered as directed.

Part -1

I. Answer all the questions. [20 x 1 = 20]

Choose the correct synonyms for the underlined words from the options given:

Question 1.
I scraped the ice off the gauges.
(a) applied (b) saw (c) scratched (d) melted
Answer:
(c) scratched

Question 2.
We hoisted our oxygen gear on to our backs.
(a) dropped (b) swayed (c) hauled (d) hooked up
Answer:
(c) hauled

Question 3.
Liberty is a social contract.
(a) contact (b) confront (c) encounter (d) agreement
Answer:
(d) agreement

Choose the correct antonyms for the underlined words from the options given:

Question 4.
There was a curious smothering noise from my friend.
(a) expressed (b) smoothened (c) smoothened (d) suffocated
Answer:
(a) expressed

Question 5.
The serenity was much admired.
(a) calmness (b) tranquillity (c) surrender (d) bustle
Answer:
(d) bustle

Question 6.
One can swallow considerable quantities without ill-effects.
(a) significant (b) sizeable (c) partly (d) large
Answer:
(c) partly

Question 7.
Choose the correct combination for the compound word ‘Tax payer’.
(a) Noun + Verb (b) Noun + Noun (c) Gerund + Noun (d) Preposition + Noun
Answer:
(b) Noun + Noun

Question 8.
Choose the correct expansion of IMP.
(a) Indian Maritime Department (b) India Meteorological Device (c) Indian Meteorological Department (d) India Meteorological Department
Answer:
(d) India Meteorological Department

Question 9.
Choose the meaning of the foreign word in the sentence:
At the meeting, the professors were schmoozing with the president of the club.
(a) talking in a friendly manner (b) brainstorming (c) gossiping (d) fighting
Answer:
(a) talking in a friendly manner

Question 10.
Choose the right combination for the blended word Interpol.
(a) Inter + Pol (b) India + Police (c) International + Police (d) Inter + Pal
Answer:
(c) International + Police

Question 11.
Choose die clipped word for gymnasium.
(a) gymum (b) gymna (c) gin (d) gym
Answer:
(d) gym

Question 12.
A strong desire to pen poetry is known as ……………….. .
(a) calligraphy (b) metromania (c) melodrama (d) decathlon
Answer:
(b) metromania

Question 13.
Form a derivative by adding the right prefix to the word ‘suspecting’.
(a) re- (b) im- (c) un- (d) mis-
Answer:
(c) un-

Question 14.
Fill in the blanks with a suitable relative pronoun.
It was Ausable …………………….. closed the door behind him.
(a) who (b) whose (c) whom (d) which
Answer:
(a) who

Question 15.
Fill in the blanks with a suitable preposition.
There was a delay before we went ………………………. the tunnel.
(a) across (b) above (c) into (d) through
Answer:
(d) through

Question 16.
Choose the correct question tag for the following statement. It’s not hot today ………………………..?
(a) is it (b) was it (c) will it (d) won’t it
Answer:
(a) is it

Question 17.
Choose the suitable meaning or idiom found in the following sentence.
Hari had a hard time comparing the iPhone to the Samsung phone because to him they were apples and oranges.
(a) similar (b) cannot be compared (c) comparing with fruits (d) was hungry
Answer:
(b) cannot be compared

Question 18.
Substitute the underlined word with the appropriate polite alternative. Surya was lazy.
(a) motivational (b) emotional (c) motivationally deficient (d) sensitive
Answer:
(c) motivationally deficient

Question 19.
Choose the correct sentence pattern for the following sentence. He smiled uncomfortably.
(a) SVO (b) SVIODO (c) SVCA (d) SVA
Answer:
(d) SVA

Question 20.
Fill in the blank with a suitable phrasal verb. The lorry was going slowly, blocking our way, but we managed to ……………………….
(a) pull ahead (b) pull through (c) pull towards (d) pull behind
Answer:
(a) pull ahead

Part II
Section – 1

Read the following sets of poetic lines and answer any four from it. [4 x 2 = 8]

Question 21.
“Oh then our maze of tunneled stone ‘
Grew thin and treacherous as air”
(a) What is the figure of speech in the first line?
(b) Mention the poetic device employed in the second line.
Answer:
(a) The figure of speech in the first line is metaphor.
(b) Simile is the poetic device employed in the second line.

Question 22.
“One equal temper of heroic hearts,
Made weak by time and fate, but strong in will To strive, to seek, to find, and not to yield”
(а) Though made weak by time and fate, the hearts are heroic. Explain.
(b) Pick out the words in alliteration in the above lines.
Answer:
(a) Ulysses and his compatriots might have become old and may not have the same strength they had in youth. But they still share the thirst for travel and pursuit of knowledge in the unexplored world. Their bravery and spiritual strength are intact. .
(b) Strive, seek, heroic, hearts are the words that alliterate.

Question 23.
“Atfirst the infant,
Mewling and puking in the nurse’s arms.”
(а) What does man do in the first stage of life?
(b) Explain, ‘Mewling and pucking’.
Answer:
(a) In the first stage of life man plays the role of an infant. He is always crying and vomiting in the nurse’s arms.
(b) It means crying and vomiting.

Question 24.
“ What is that dirge-like murmur that I hear Like the sea breaking on a shingle-beach?
It is the tree’s lament, an eerie speech,”
(а) What do you mean by dirge?
(b) Why does the tree lament?
Answer:
(a) ‘Dirge’ means a lament for the dead.
(b) The tree laments the absence of the poet.

Question 25.
“The Chief’s eye flashed; his plans Soared up again like fire.”
(а) Who is described as the chief?
(b) Identify the figure of speech employed.
Answer:
(a) Napoleon Bonaparte is described as the chief.
(b) Simile.

Question 26.
“and guide him among sudden betrayals and tighten him for slack moments
(a) What could guide the son among unexpected betrayals?
(b) What could happen to the boy during slack moments?
Answer:
(a) Rock/steel-like would guide the son among betrayals.
(b) During slack moments, the boy may be betrayed by his trusted friends.

Section – 2

Answer any three of the following questions. [3 x 2 = 6]

Question 27.
Report the following dialogue:
Answer:
Padma : Did you see the dentist last week?
Devi : No, I will be seeing the dentist next week.
Padma asked Devi if she had seen the dentist the previous week. Devi said that she had not and that she would be seeing the dentist the following week.

Question 28.
She must buy a ticket. Otherwise she will not be allowed to watch the programme, (combine using if)
Answer:
If she does not buy a ticket, she will not be allowed to watch the programme.

Question 29.
Rewrite the sentence making an inversion in the conditional clause.
If I had been there, this problem wouldn’t have happened.
Answer:
Had I been there, this problem wouldn’t have happened.

Question 30.
Though Andrew has an unconventional swing, he manages to drive the ball farther than some professional golfers. (Change the following into a compound sentence)
Answer:
Andrew has an unconventional swing but he manages to drive the ball farther than some professional golfers.

Part-III
Section -1

Explain any two of the following with reference to the context. [2 x 3 = 6]

Question 31.
Our only enemy was gold
Answer:
Reference: This line is from Edwin Muir’s poem “The Castle”.

Context: The helpless soldier says this while being surprised by the enemies who bribed the wicked wicket gate keeper.

Explanation: The soldier was initially proud of their fortified castle, brave and loyal soldiers and pile of arms and the well-stocked granary to last a siege. There was vigil behind the turret wall. But he was not aware of the enemy within the human soul: the greed. Gold was their enemy for which they had no weapon. The wizened warder had let the enemy in through the wicket gate by taking “gold”. The weakness of the gatekeeper for gold made the strong castle weak.

Question 32.
How dull it is to pause, to make an end,
To rust unburnished, not to shine in use!
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.

Context: The poet says these words while discussing the mental agony of Ulysses who is unable to settle down with his ageing wife Penelope and son Telemachus.

Explanation: Ulysses finds doling out justice to a savage people as ‘boring’. He does not want to settle down and die in Ithaca. He compares himself to a sword which may rust if left unused. He wants to lead an active and adventurous life till his death.

Question 33.
“‘I’m killed, Sire! ’And, his Chief beside,
Smiling, the boy fell dead.
Answer:
Reference: These lines are from the poem ‘Incident of the French Camp” written by Robert Browning.

Contex: The poet says these words while explaining the hurt reaction of the boy-soldier.

Explanation: Instead of being carried away for first aid, he hurried on horse back to communicate the news of conquest of Ratisbon despite his chest being split into two. When emperor Napoleon expressed his grief on his wounded status, the boy soldier said, “Nay I’m killed sire.”

Section – 2

Answer any two of the following questions in about 30 words. [2 x 3 = 6]

Question 34.
What made the boys join the resistance movement against the Germans?
Answer:
The boys hated the Germans who had destroyed their town and killed their father. So, they joined the resistance movement against the Germans.

Question 35.
Elucidate the author’s ideas about teapots.
Answer:
The teapot should be made of China or earthen ware. Silver or Britannia ware teapot produce inferior quality tea. Enamel pots are worse. Though curiously enough a pewter teapot, a rarity nowadays, is not so bad.

Question 36.
How were the achievers welcomed by fellow climbers?
Answer:
Two figures came towards them a couple of hundred feet about their camp. They were George Lowe and Wilfrid Noyce, and it was thoughtful of them because Hillary’s oxygen ran out just then.

Section – 3

Answer any three of the following questions in about 30 words. [3 x 3 = 9]

Question 37.
Study the pie chart given and answer the questions that follow. Expenditure in book Publishing
(a) For what does the publisher spend the maximum while publishing a book?
(b) How much expenditure is incurred on printing?
(c) How much percent royalty goes to the author?
Answer:
(a) A publisher spends the maximum in binding the books.
(b) Twenty five percent of the expenditure is incurred on printing.
(c) The royalty for the author is 15%.

Question 38.
Write a dialogue of minimum 3 exchanges between a Teacher and a student.
Answer:
Teacher : Why are you late today?
Student: I went to hospital with my mother.
Teacher : Who is sick?
Student: My mother is down with a viral infection.
Teacher : Couldn’t your father accompany her to the hospital?
Student: I am afraid that there wasn’t any alternative as my father was out of station.

Question 39.
Describe the process of making bricks.
Answer:

• In the first stage, the raw material clay is dug from the soil by a digger and lumps of clay are placed on a metal grid and broken into smaller pieces, which fall through a roller.
• They are later mixed with sand and water to make a uniform mixture.
• Brick-shaped pieces are subsequently formed either by using a mould or by cutting the clay mix with a wire cutter.
• The raw bricks are then hardened, stacked in a drying oven and left for one or two days. The dried bricks are finally heated in a kiln, first to a moderate temperature and then to a high temperature.
• They are finally transferred to a chamber to cool and harden slowly over two to three days before being transported to customers.

Question 40.
Complete the proverbs using the word given below.
(a) When in Rome as Romans do. (do, act, behave)
(b) A leopard can’t change its (tail, nature, spots)
(c) A burnt child fire, (fears, dreads, hates)
Answer:
(a) do (b) spots (c) dreads

Part – IV

Answer the following questions: [7 x 5 = 35]
Answer in a paragraph in about 150 words.

Question 41.
Adversity brings out the best as well as the worst in people. Elucidate this statement with reference to the story, ‘Two Gentlemen of Verona’.
Answer:
Whenever confronted by an adversity, there is a psychological reaction among most of us. Either fight or flee from it. Even adults run away from crisis and seek asylum or support of others. We find hordes of people, when living becomes difficult, migrate to other places to survive. In the face of adversity, even educated people turn to evil ways. They worked hard from dawn to midnight. They did any job that came their way. They shined shoes, hawked newspapers, sold wild fruits and took the people round the city.

They spent very little on their food and clothes. Every week they cycled to Poleta and paid Lucia’s medical bill systematically. In this story, the nurse of the private hospital reveals the bitter truth about the economy of Verona. Jobs are scarce. There is inflation. It is difficult to buy food with limited money. When the hospital laid the condition that the little boys Nicola and Jacopo should pay the weekly medical bill for their sister Lucia’s treatment for tuberculosis, they did not back off or give up. Their best qualities came out during adversity. Like tea leaves, they gave their best while in hot waters.

[OR]

‘There is no height, no depth that the spirit of man, guided by higher Spirit cannot attain’. Discuss the above statement in the context of the achievement of Edmund Hillary and Tenzing.
Answer:
Man is naturally endowed with an indomitable spirit. Guided by powers above, man reaches great height. The grit and persistence of both Hillary and Tenzing stand testimony to the above maxim. The conditions were really overwhelming. Hillary slipped many times. Once he even sought the advisability of continuing the climb under such conditions. But Tenzing and Hillary resolved to persist and conquer the peak. As a reward to 400 feet climb near the south summit, they got two bottles of oxygen which in fact kept them alive almost up to their base camp.

Both pick up the pieces of hopes only when they come into contact with firmer rock-like ice as they moved up. Though they had to inch their way up clearing snow with the ice-axe and making a path to haul themselves up ridge after ridge in the elusive terrain, they did not give up. As Hillary’s ice-axe bit into the first steep slop of the ridge, his hopes were realized. The snow was crystalline and firm. With just two or three blows, Hillary could make a step large enough for their oversized high altitude boots.

They could create comfortable belays and trudge forward with confidence. As the humps were continuously seen, their original zest started declining. It was at this point Hillary saw a narrow ridge up to a snowy summit. With a few more whacks of the ice-axe in the form of snow, they reached the top.

Question 42.
Why did the narrator say that their “only enemy was gold”?
Answer:
The narrator was very proud of the awe-inspiring castle. The soldiers were least bothered about the enemy who was just half a mile away. They had a well -guarded, tall, smooth and insurmountable fort. They were led by a brave captain. The soldiers were loyal and ready to lay down their lives for protecting the castle and the trusting citizens of their country. There was constant vigil from the turret wall to shoot down the enemy at sight.

They were proud that only a bird could have the temerity to scale over the fortified castle. They were even prepared for a siege as they had well-stocked granaries. In the event of the outbreak of a war, allies were ready to rush with their assured forces in support of them. But the soldiers were unaware of the truth, “the strength of the strongest chain lies in its weakest link”. An aged warder of a wicked wicket gate had a cancerous growth of greed.

As the enemy was hidden within the soul of their own citizen, they had no weapon to fight it. Gold was ultimately their enemy. The wicked gate keeper betrayed his country for a bag full of gold. The great citadel fell without even a groan of protest and without shedding “a jot of blood.”

[OR]

Napoleon was a great source of inspiration to his army Justify.
Answer:
Napoleon was a powerful orator and was able to muster the support of young soldiers who could gladly throw away their lives for the glory of France and for fulfilling the ambitious plans of territorial expansion of Napoleon Bonaparte. He inspired unprecedented courage among the soldiers.

They never worried about the strength of the enemy army or their pile of armaments. They faced the battles with the single minded determination to ‘do or die’ or do and die. They kissed death for the glorification of France and for making Napoleon proud of their heroism, sacrifices and patriotism.

Question 43.
Write a paragraph of about 150 words by developing the following hints:
Martha and John – grudge – underpaid two decades – punishment – condemned the bribe – abusive language – one hundred thousand dollars – changed attitude – testify – three words – corrupt practices.
Answer:
Martha and John cherished a grudge against Gresham for having underpaid Baldwin for about two decades. They openly said that he deserved punishment. Initially Gresham condemned the bribe claimed as a difference in salary paid. Had he been made a similar offer he claimed he would have asked him to go to the devil. Baldwin said that having been a friend he couldn’t use such an abusive language. Casually John asked how much he offered.

Baldwin said that it was one hundred thousand dollars. Suddenly Martha and John changed their attitude towards Gresham. John viewed a shame if Gresham got indicted because he shared his name. Martha also desperately tried to convince him not to testify against Gresham. John even suggested that he could say those three words “I don’t remember” as the depositor would not lose a cent. Baldwin felt miserable because the family members who he wanted to feel proud of his uprightness wanted him to crossover to the side of corrupt practices because of the generous offer made by Gresham.

[OR]

Atisable, private detective-automatic gun – Berlin – nice story – balcony – displeasure – hotel management – Ausable’s room – times of danger – pre-ordered drinks – blatant – lie – Max jumped – suicidal jump – outwitted the spy.
Answer:
Ausable was a shrewd private detective. He did not become panicky on seeing Max with an automatic gun in his room. Even if he were, he did not show it off. Instead, he expressed surprise that he expected him to be in Berlin. He cooked up a nice story that it was the second time someone had broken into his room through the balcony of the neighbouring apartment which reaches down under his window.

He expressed his displeasure that he would raise hell with the hotel management for not blocking that balcony. This gave a strong suggestion to Max, that he could have used the balcony instead of the pass key to enter Ausable’s room. Being a criminal or spy, a person always looks for various routes of escape in times of danger. Very rarely he starts direct encounter risking his life. When Henry, the waiter who arrived with the pre-ordered drinks, knocked, Mr. Ausable simply smiled.

When Max asked who it was, he told a blatant lie that it was the police who had come for his extra protection and wouldn’t hesitate to fire as the door wasn’t locked but just closed. Max, assuming that there is a balcony extending below Ausable’s window, jumped. He never knew that it was a suicidal jump from the sixth floor. Thus, Ausable outwitted the spy, Max.

Question 44.
Write a summary or Make notes of the following passage.
Answer:
Residents of the Bhirung Raut Ki Gali, where Ustad Bishmillah Khan was bom on March 21,1916, were in a state of shock. His cousin, 94-year -old Mohd Idrish Khan had tears welled up in his eyes. Shubhan Khan, the care-taker of Bismillah’s land, recalled “Whenever in Dumaraon, he would give rupees two to the boys and rupees five to the girls of the locality”. He was very keen to play Shehnai again in the local Bihariji’s Temple where he had started playing Shehnai with his father, Bachai Khan, at the age of six.

His original name was Quamaruddin and became Bishmillah only after he became famous as a Shehnai player in Varanasi. His father Bachai Khan was the official Shehnai player of Keshav Prasad Singh, the Maharaja of the erstwhile Dumaraon estate, Bismillah used to accompany him. For Bismillah Khan, the connection to music began at a very early age. By his teens, he had already become a master of the Shehnai. On the day India got freedom, Bismillah Khan, then a sprightly young man of 31 years old, had the rare honour of playing from Red Fort.

But Bishmillah Khan won’t just be remembered for elevating the Shehnai from an installment heard only in weddings and Naubatkhanas to one that was appreciated in concert halls globally across the world. His life was a witness to the plurality that is India. A practicing Muslim, he would take a daily dip in River Ganges in his younger days after a bout of kusti in Benia Baga Akhada. Every morning, Bismillah Khan would do riyaaz at the Balaji temple on the banks of the river. Even during his final hours in a Varanasi hospital, music didn’t desert Bismillah Khan.

A few hours before he left this worldly home early on Monday, the Shehnai wizard hummed a thumri to show that he was feeling better. This was typical of a man for whom life revolved around music. Throughout his life, he abided by the principle that all religions are one and the same. What marked Bismillah Khan was his simplicity and disregard for the riches that come with musical fame. Till the very end, he used a cycle rickshaw to travel around Varanasi. But the pressure of providing for some 60 family members took its toll during his later years.

Summary

No. of words given in the original passage: 388
No. of words to be written in the summary: 388/3 = 129 ± 5

Rough Draft
Residents of the Bhirung Raut Ki Gali were shocked, where Ustad Bismillah Khan was born on March 21, 1916. Subhan Khan recalled that in Dumaraon, he would give Rs.2/- and 5/- to boys and girls respectively. He wanted to play Shehnai in Bihariji’s Temple to recall his childhood days withhls-Tather, Bachai Khan. His original name was Quamaruddin and became Bishmillah as a Shehnai player in varanasi. His father was the official Shehnai player of the Maharaja of Dumaraon estate. On the day India got freedom, 31 year old Bismillah, had the rare honour of playing from Red Fort.

He was appreciated globally. His life was a witness to the plurality that is India. A few hours before he died the Shehnai wizard hummed a thumri, typical of a man whose life revolved around music. Throughout his life, he abided by the principle that all religions are one and the same. His simplicity and disregard for the riches in spite of being famous was seen in his cycle rickshaw rides.

Fair Draft: Bismillah Khan – A Legend
Residents were shocked about Bishmillah Khan’s demise. Tears welled in an aged cousin. Shubhan recalled his generosity to children. Quamaruddin became Bishmillah as a Shehnai player and wanted to play in Bihariji’s Temple to recall his childhood days with his father, the official Shehnai player of Dumaraon. Bismillahmastered it during teenage. On the day of Independence, young Bismillah, played from Red Fort and appreciated globally.

He was a witness to the plurality in India. Being Muslim, he would dip in River Ganges and do riyaaz at the Balaji temple. During his final hours, the Shehnai wizard hummed a thumri, typical of a man whose life revolved around music. Throughout he believed that all religions were same. His disregard for riches was seen in his cycle rickshaw rides but providing for his family took its toll.

No. of words in the summary: 133

[OR]
Notes
Title: Bismillah Khan – A Legend
1. Shocked Death

• cousin Khan in tears
• caretaker of Dumaraon recalls
• Boys – Rs. 21- to boys ; Girls – Rs. 5/-

2. Early Years

• org. name Quamaruddin
• played shehnai with father at temp.
• got famous as Bismillah at Varanasi.

3. Music, a family heritage

• father: Dumarao’s court poet
• played Shehnai from 6 yrs.
• played Shehnai at Red Fort in 1947, age 31

4. Routine in Vans.

• dip in Ganges
• riyaz at Balaji Temp.
• Hummed thumri before his last breath.

5. Strong Belief in Secularism

• Bismillah bel. in secularism
• all religions are one and same
• a plurality of life, India
• Pract Islam

Abbreviations used: Rs – rupees; Org. – original; Temp. – temple; Yrs. – years; Fml. – family Mem. – member; Pract. – Practised; Bel. – believed.

Question 45.
You are Raja. You are upset about the bad influence of TY channels on the young children. Write a letter to the editor of a leading newspaper suggesting measures to upgrade the standard.
Answer:
20th Sept, 2020
From
XXX
YYY
zzz

To
The Mayor,
Municipal Corporation,
Chennai-600012.

Sir,
Sub: Provision for additional street lamps.
This is to bring to your notice that there is provision bf just one street light in every street on YYY Nagar in ZZZ locality. It is very scary to walk on the road as there are empty lands here and there with plants and bushes grown robustly. As a result, we have insects and snakes that come out of its shelter in search of food in the nights. There is also the problem of chain snatching. Additional lights will help us to walk safe in the locality.

We thereby request you to look into the matter, make provisions for additional lamp posts and also change the tube lights that aren’t working.

Thank you,

Yours faithfully,
XXX

Address on the envelope
To
The Mayor,
Municipal Corporation
Chennai- 600012

[OR]

Write a paragraph of 150 words on “Common Man’s Woes During Monsoon”.
Answer:
Common Man’s Woes During Monsoon

The much anticipated Monsoon brings relief from the sweltering heat. However Monsoon and madness visit us at the same time. Every year during the monsoons, pandemonium rules supreme. The roads are flooded; the sewage system falls; a huge loss of crops, fruits, life, and property is caused. Water logging and breeding of mosquitoes together becomes the reason for the rise of many diseases. Network connectivity gets weak.

Though we have benefitted from independence since 1947, the Indian Government has not been able to tackle the flood issues caused by monsoons. Recent floods have laid bare the emptiness of the claims of the civic authorities of their readiness. The poor had to bear the brunt of the problem.

One of the main causes behind this issue is the lack of stringent laws and accountability of the officials and the Ministries responsible for tackling the floods. The Government should formulate and pass strict laws regarding the projects launched to relieve people of their distress but there is hardly a sign of effective implementation of the scheme. There needs to be strict accountability.

When the Ministers will be accountable, they will ensure the officers, officials involved in the projects surely create better systems benefitting the public.

Question 46.
Spot the errors and rewrite the sentences correctly
(а) Mr. Johnson with his wife and children were present at the show.
(b) The quality of goods produced here are not upto the international standard.
(c) The prices quoted for this commodity at different shops is different.
(d) Either Sanjay or Raja are responsible for this crime.
(e) Bread and butter are his lunch.
Answer:
(a) Mr. Johnson along with his wife and children were present at the show.
(b) The quality of goods produced here is not up to the international standard.
(c) The price quoted for this commodity at different shops is different.
(d) Either Sanjay or Raja is responsible for this crime.
(e) Bread and butter is his lunch.

[OR]

Fill in the blanks correctly.
(a) Her baby is ………………. in December when the ………………. on the grass will be a rare sight to see. (dew/due)
(b) When I was young, I ………………. eat tons of ice cream! (use a modal verb)
(c) I ………………. drink coffee when I was young, (use a semi-modal)
(d) We ………………. (eat) the lasagna here, (use a proper tense)
Answer:
(a) due/dew
(b) could
(c) used to
(d) have eaten

Question 47.
Identify each of the following sentences with the field in the list given below, by understanding the word or words serving as the clue.
(a) Sometimes A-class facilities are extended to even those imprisoned.
(b) Did an asteroid hit south India millions of years ago?
(c) The investigations included megascopic and microscopic studies of rocks.
(d) Many foreigners come to watch Jallikatu at Madurai.
(e) The rapid depletion in indigenous aquatic plants would prove detrimental to wetlands.
[Astronomy; Law; Flora; Geology; Sports]
Answer:
(a) Law
(b) Astronomy
(c) Geology
(d) Sports
(e) Flora

[OR]

Read the following passage and answer the questions in your own words.
Answer:
The little boy of twelve will remember this day forever. It is not every day you get to see a world championship at Hyatt Regency on November 19, 20Question 13. The visit of the Children was the result of an initiative by the organizers of the tournament and the School education Department to get school Children by a part of the world championship. Many children from schools gathered to witness the ongoing world Championship and there was an overwhelming response.

Children got a chance to see Viswanathan Anand, Garry Kasparov, Magnus Carlsen and Vijayalakshmi. The entry being free, Children found it a rare and unforgettable opportunity. Student visitors were given a chance to spend an hour in the chess puzzles contest. No wonder the visitors would be hailed heroes in their schools after their visit to the spot. Children enthusiastically tried to move the big chess pieces at the hotel lobby, thereby founding that the city of Chennai would produce many masters of chess dazzling with intelligence and talent.

Questions:
a. What made the day memorable for children?
b. Name any two celebrities mentioned in the passage.
c. Mention the purpose for which the visit of the children had been organized.
d. Which places were the student visitors permitted in to?
e. How did the children try to prove their powers and interest in the game?
Answers:
(a) World Championship at Hyatt made the day memorable for children.
(b) Children got a chance to see Viswanathan Anand and Garry Kasparov.
(c) The children from many schools were brought to witness the ongoing world Championship in Chess.
(d) The children were permitted into the Hyatt Regency lobby.
(e) Children enthusiastically tried to move the big chess pieces at the hotel lobby. Student visitors were also given a chance to spend an hour in the chess puzzles contest.

Students can download Maths Chapter 6 Trigonometry Ex 6.4 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 6 Trigonometry Ex 6.4

Question 1.
From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree, (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the second tree be “h”
ED = (h – 13) m
Let AB = x m
In the right ∆ ABC, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 13 }{ x } \)
x = 13 \(\sqrt { 3 }\) ……..(1)

In the right ∆ CED, tan 45° = \(\frac { DE }{ EC } \)
1 = \(\frac { h-13 }{ x } \)
x = h – 13 ……..(2)
From (1) and (2) we get
h – 13 = 13 \(\sqrt { 3 }\) ⇒ h = 13 \(\sqrt { 3 }\) + 13
= 13 × 1.732 + 13
= 22.52 + 13 = 35.52 m
∴ Height of the second tree = 35.52 m

Question 2.
A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30° . Calculate the distance of the hill from the ship and the height of the hill. (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the hill BE be “h” m and the distance of the hill from the ship be “x” m
In the right ∆ ABD
tan 30° = \(\frac { AD }{ DB } \)
\(\frac{1}{\sqrt{3}}=\frac{40}{x}\)
x = 40 \(\sqrt { 3 }\) ……..(1)
In the right ∆ CDE
tan 60° = \(\frac { CE }{ DC } \)
\(\sqrt { 3 }\) = \(\frac { h-40 }{ x } \)
x = \(\frac{h-40}{\sqrt{3}}\) ……..(2)

From (1) and (2) we get
\(\frac{h-40}{\sqrt{3}}\) = 40\(\sqrt { 3 }\)
h – 40 = 40 × 3
h = 120 + 40 = 160 m
Height of the hill = 160 m
Distance of the hill from the ship = 40 × \(\sqrt { 3 }\) = 40 × 1.732 = 69.28 m

Question 3.
If the angle of elevation of a cloud from a point ‘h’ metres above a lake is θ1 and the angle of depression of its reflection in the lake is θ2. Prove that the height that the cloud is located from the ground is \(\frac{h\left(\tan \theta_{1}+\tan \theta_{2}\right)}{\tan \theta_{2}-\tan \theta_{1}}\)
Answer:
Let P be the cloud and Q be its reflection.
Let A be the point of observation such that AB = h
Let the height of the cloud be x. (PS = x)
PR = x – h and QR = x + h

Let AR = y
In the right ∆ ARP, tan θ₁ = \(\frac { PR }{ AR } \)
tan θ₁ = \(\frac { x-h }{ y } \) ………(1)
In the ∆ AQR,
tan θ₂ = \(\frac { QR }{ AR } \)
tan θ₂ = \(\frac { x+h }{ y } \) ……….(2)
Add (1) and (2)

Question 4.
The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30° . If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.
Answer:
Let the height of the cell phone tower be “h” m
AD is the height of the apartment; AD = 50 m
Let AB be “x”
In the right triangle ABC
tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)

x = \(\frac{h}{\sqrt{3}}\) …….(1)
In the right triangle ABD, tan 30° = \(\frac { AD }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 50 }{ x } \)
x = 50 \(\sqrt { 3 }\) ……(2)
From (1) and (2) We get
\(\frac{h}{\sqrt{3}}\) = 50 \(\sqrt { 3 }\)
h = 50\(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3 = 150
Height of the cell phone tower is 150 m.
Yes, the cell phone tower meets the radiation norms.

Question 5.
The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find
(i) The height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. (\(\sqrt { 3 }\) = 1.732)
Answer:
(i) Let the height of the lamp post AE be “h” m
DE = h – 66
Let AB be “x”
In the right ∆ ABC, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}=\frac{66}{x}\)
x = 66 \(\sqrt { 3 }\) ……(1)

In the right ∆ CDE, tan 60° = \(\frac { DE }{ DC } \)
\(\sqrt { 3 }\) = \(\frac { h-66 }{ x } \) ⇒ \(\sqrt { 3 }\) x = h – 66
x = \(\frac{h-66}{\sqrt{3}}\) ………….(2)
From (1) and (2) we get
\(\frac{h-66}{\sqrt{3}}\) = 66 \(\sqrt { 3 }\)
h – 66 = 66 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 66 × 3
h – 66 = 198 ⇒ h = 198 + 66
h = 264 m
(i) the height of the lamp post = 264 m
(ii) Difference of the height of lamp post and apartment = 264 – 66
= 198 m
(ii) Distance between the lamp post and the apartment = 66 \(\sqrt { 3 }\) m
= 66 × 1.732 = 114.31 m

Question 6.
Three villagers A, B and C can see each other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate:
(i) the vertical height between A and B.
(ii) the vertical height between B and C. (tan 20° = 0 .3640, \(\sqrt { 3 }\) = 1. 732)
Answer:
Let AD is the vertical height between A and B
In the right ∆ ABD

tan 20° = \(\frac { AD }{ BD } \)
0.3640 = \(\frac { AD }{ 8 } \)
AD = 0.3640 × 8 = 2.912 km
∴ AD = 2.91 km
CE is the vertical height between C and B
In the right ∆ BCE, tan 30° = \(\frac { CE }{ BE } \)
\(\frac{1}{\sqrt{3}}=\frac{C E}{12} \Rightarrow \sqrt{3} C E=12\)
CE = \(\frac{12}{\sqrt{3}}=\frac{12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{12 \times \sqrt{3}}{3}\)
= 4 \(\sqrt { 3 }\) = 4 × 1.732 = 6.928
= 6.93 km
(i) The vertical height between A and B = 2.91 km
(ii) The vertical height between B and C = 6.93 km

Students can download Maths Chapter 6 Trigonometry Ex 6.3 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 6 Trigonometry Ex 6.3

Question 1.
From the top of a rock 50 \(\sqrt { 3 }\) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:
Let the distance of the car from the rock is “x” m

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)
\(\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}\)
x = 50 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3
= 150 m
∴ Distance of the car from the rock = 150 m

Question 2.
The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer:
Let the height of the first building AD be “x” m
∴ EC = 120 – x
In the right ∆ CDE,
tan 45° = \(\frac { CE }{ CD } \)

1 = \(\frac { 120-x }{ 70 } \) ⇒ 70 = 120 – x
x = 50 cm
∴ The height of the first building is 50 m

Question 3.
From the top of the tower 60 m high the anles of depression the top and bottom of a vertical lamp post are observed be 38° and 60° respectively
Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC,

tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 60 }{ x } \)
x = \(\frac{60}{\sqrt{3}}\) ……..(1)
In the right ∆ DEC, tan 38° = \(\frac { EC }{ DE } \)
0.7813 = \(\frac { 60-h }{ x } \)
x = \(\frac { 60-h }{ 0.7813 } \) …….(2)
From (1) and (2) we get
\(\frac{60}{\sqrt{3}}\) = \(\frac { 60-h }{ 0.7813 } \)
60 × 0.7813 = 60 \(\sqrt { 3 }\) – \(\sqrt { 3 }\) h
\(\sqrt { 3 }\) h = 60 \(\sqrt { 3 }\) – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac { 57.04 }{ 1.732 } \)
h = \(\frac { 570440 }{ 1732 } \) = 32.93 m
∴ Height of the lamp post = 32.93 m

Question 4.
An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are
60° and 30° respectively. Find the distance between the two boats. (\(\sqrt { 3 }\) = 1.732)
Answer:
C and D are the position of the two boats.
Let the distance between the two boats be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BD } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 1800 }{ x+y } \)
x + y = 1800 \(\sqrt { 3 }\)
y = 1800 \(\sqrt { 3 }\) – x ……(1)
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \)
\(\sqrt { 3 }\) = \(\frac { 1800 }{ y } \)
y = \(\frac{1800}{\sqrt{3}}\) ……….(2)
From (1) and (2) we get
\(\frac{1800}{\sqrt{3}}\) = 1800 \(\sqrt { 3 }\) – x
1800 = 1800 × 3 – \(\sqrt { 3 }\)x
\(\sqrt { 3 }\)x = 5400 – 1800
x = \(\frac{3600}{\sqrt{3}}=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \times \sqrt{3}}{3}\)
= 1200 × 1.732 = 2078.4 m
Distance between the two boats = 2078.4 m

Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac{4 h}{\sqrt{3}}\) m.
Answer:
A and C be the position of two ships.
Let AB be x and BC be y. Distance between the two ships is x + y

In the right ∆ ABD, tan 60° = \(\frac { BD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ……(1)
In the right ∆ BCD,
tan 30° = \(\frac { BD }{ BC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ y } \)
y = \(\sqrt { 3 }\) h
Distance between the two ships (x + y) = \(\frac{h}{\sqrt{3}}+\sqrt{3} h\)
= \(\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}}\)
Hence it is verified

Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30 \(\sqrt { 3 }\) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer:
Let the speed of the lift is “x” feet / minute
Distance AB = 2 x feet (speed × time)
BC = (90 – 2x)
In the right ∆ BCD,

tan 30° = \(\frac { BC }{ DC } \)
\(\frac{1}{\sqrt{3}}=\frac{90-2 x}{30 \sqrt{3}}\)
\(\sqrt { 3 }\) (90 – 2x) = 30\(\sqrt { 3 }\)
(90 – 2x) = \(\frac{30 \sqrt{3}}{\sqrt{3}}\) ⇒ (90 – 2x) = 30
2x = 60
x = \(\frac { 60 }{ 2 } \) = 30
x = 30 feet/minute
Speed of the lift = 30 feet / minute (or) [ \(\frac { 30 }{ 60 } \) second) 0.5 feet / second

Students can download Maths Chapter 6 Trigonometry Ex 6.2 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 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 \(\sqrt { 3 }\) m.
Answer:
Height of the tower (AC) = 10 \(\sqrt { 3 }\) m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = \(\frac { AC }{ AB } \)
= \(\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle of inclination is 30°

Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt { 3 }\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 4\(\sqrt { 3 }\) m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = \(\frac { AD }{ AP } \)

\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x}\)
x = 4 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m

In the right ∆ CDF, tan 45° = \(\frac { CE }{ CD } \)
1 = \(\frac { x }{ 5 } \) ⇒ x = 5
In the right ∆ CDE, tan 60° = \(\frac { CE }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { x+h }{ 5 } \) ⇒ x + h = 5\(\sqrt { 3 }\)
5 + h = 5 \(\sqrt { 3 }\) (substitute the value of x)
h = 5 \(\sqrt { 3 }\) – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, \(\sqrt { 3 }\) = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h+1.6 }{ x } \)
x = \(\frac{h+1.6}{\sqrt{3}}\) ……..(1)

In the right ∆ ABC, tan 40° = \(\frac { AC }{ AB } \)
0.8391 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.8391 } \)
Substitute the value of x in (1)
\(\frac{h}{0.8391}=\frac{h+1.6}{\sqrt{3}}\)
(h + 1.6) 0.8391 = \(\sqrt { 3 }\) h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = \(\frac { 1.34 }{ 0.89 } \) = \(\frac { 134 }{ 89 } \) = 1.5 m
Height of the pedestal = 1.5 m

Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\(\sqrt { 3 }\) = 1.732)

Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = \(\frac { AD }{ AB } \)

\(\frac{1}{\sqrt{3}}=\frac{r}{r+12}\)
\(\sqrt { 3 }\) r = r + 12
\(\sqrt { 3 }\) r – r = 12 ⇒ r (\(\sqrt { 3 }\) – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = \(\frac { 12 }{ 0.732 } \) ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = \(\frac { AE }{ AC } \)
1 + \(\frac { r+h }{ r+7 } \)
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”

In the right ∆ ABC, tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 15 }{ x } \)
x = \(=\frac{15}{\sqrt{3}}=\frac{15 \times \sqrt{3}}{3}\)
= 5\(\sqrt { 3 }\)
In the right ∆ CDE, tan 30° = \(\frac { EC }{ DE } \)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{x}\) ………….(1)
Substitute the value of x = 5 \(\sqrt { 3 }\) in (1)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{5 \sqrt{3}} \Rightarrow \sqrt{3}(15-h)=5 \sqrt{3}\)
(15 – h) = \(\frac{5 \sqrt{3}}{\sqrt{3}}\) ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem

\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
\(\frac { 9x }{ x } \) = \(\frac { AD }{ 25 } \) ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD² = AC² + CD²
(225)² = (10x)² + 25²
50625 = 100x² + 625
∴ 100x² = 50625 – 625 = 50000
x² = \(\frac { 50000 }{ 100 } \) = 500
x = \(\sqrt { 500 }\) = \(\sqrt{5 \times 100}=10 \sqrt{5}\)
∴ AC = 10 × 10\(\sqrt { 5 }\) = 100 \(\sqrt { 5 }\) (AC = 10x)
∴ Height of the pole = 100 \(\sqrt { 5 }\) m

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = \(\frac { AC }{ AC } \)

0.1405 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.1405 } \) ………..(1)
In ∆ ABC, tan 4° = \(\frac { AC }{ AB } \)
0.0699 = \(\frac { h }{ x+1 } \) ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = \(\frac { h-0.0699 }{ 0.0699 } \) ………(2)
Equation (1) and (2) we get,
\(\frac { h-0.0699 }{ 0.0699 } \) = \(\frac { h }{ 0.1405 } \)
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = \(\frac { 0.0098 }{ 0.0706 } \) = \(\frac { 98 }{ 706 } \) = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 1.
Solve the following system of linear equations by matrix inversion method.
(i) 2x + 5y = -2, x + 2y = -3
Solution:

(ii) 2x – y = 8, 3x + 2y = -2
Solution:

x = 2, y = -4

(iii) 2x + 3y – z = 9, x + y + z = 9, 3x – y – z = -1
Solution:

|A| = 2(-1+1)-3(-1-3)-1(-1-3)
= 0 + 12 + 4 =16 ≠ 0 A^-1 exists.

∴ x = 2, y = 3, z = 4

(iv) x + y + z – 2 = 0, 6x – 4y + 5z – 31 = 0, 5x + 2y + 2z = 13
Solution:

AX = B
X = A^-1B
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
6 & -4 & 5 \\
5 & 2 & 2
\end{array}\right]\)
|A| = 1(-8-10)-1(12-25)+1(12+20)
= 18 + 13 +32 = 27
≠ 0
A^-1 Exists

∴ x = 3, y = -2, z = 1

Question 2.

Find the products AB and BA and hence solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2
Solution:

AB = BA = 4I₃

∴ x = 2, y = 1, z = -1

Question 3.
A man is appointed in a job with a monthly salary of a certain amount and a fixed amount of annual increment. If his salary was Rs 19,800 per month at the end of the first month after 3 years of service and Rs 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use the matrix inversion method to solve the problem.)
Solution:
Let the man starting the salary be Rs x and his annual increment be Rs y.
Given x + 3y = 19,800
x + 9y = 23,400
The equation can be written as
\(\left[\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
19800 \\
23400
\end{array}\right]\)
AX = B
X = A^-1B
A = \(\left[\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right]\)
To find A^-1
|A| = 9 – 3 = 6 ≠ 0 A^-1 Exists.

Monthly salary = Rs 18000
Annual increment = Rs 1800

Question 4.
4 men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the same work by Using the matrix inversion method.
Solution:
Let the time taken by one man alone be x days and one woman alone be y days.


∴ One man can do 18 days
One woman can do 36 days

Question 5.
The prices of three commodities A, B, and C are Rs x,y, and z per unit respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C. Person Q purchases 2 units of C and sells 3 units of A and one unit of B. Person R purchases one unit of A and sells 3 unit of B and one unit of C. In the process, P, Q and R earn Rs 15,000, Rs 1,000 and 14,000 respectively. Find the prices per unit of A, B, and C. (Use the matrix inversion method to solve the problem.)
Solution:
Let x, y, z are commodities of A, B, C
2x – 4y + 5z = 15,000 ………… (1)
3x + y – 2z = 1000 ………… (2)
-x + 3y + z = 4000 ………… (3)

|A| = 2(1 + 6)+ 4(3 – 2) + 5(9 + 1)
= 2(7) + 4(1) + 5(10)
= 14 + 4 + 50 = 68
≠ 0
A^-1 Exists.


x = Rs 2000, y = Rs 1000, z = Rs 3000
The prices per unit of A, B, and C are Rs 2000, Rs 1000, Rs 3000

Students can download Maths Chapter 6 Trigonometry Additional 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 6 Trigonometry Additional Questions

I. Multiple Choice Questions

Question 1.
(1 – sin² θ) sec² θ = …………
(1) 0
(2) 1
(3) tan² θ
(4) cos² θ
Answer:
(2) 1
Hint: (1 – sin² θ) sec² θ = cos² θ sec² θ = cos² θ \(\frac{1}{\cos ^{2} \theta}\) = 1

Question 2.
(1 + tan² θ) sin² θ = ………….
(1) sin² θ
(2) cos² θ
(3) tan² θ
(4) cot² θ
Answer:
(3) tan² θ
Hint: (1 + tan² θ) sin² θ = sec² θ sin² θ = \(\frac{1}{\cos ^{2} \theta}\) sin² θ = \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) = tan² θ

Question 3.
(1 – cos² θ) (1 + cot² θ) = ………..
(1) sin² θ
(2) 0
(3) 1
(4) tan² θ
Answer:
(3) 1
Hint: (1 – cos² θ) (1 + cot² θ) = sin² θ cosec² θ = sin² θ. \(\frac{1}{\sin ^{2} \theta}\)= 1

Question 4.
sin (90° – θ) cos θ + cos (90° – θ) sin θ = …………..
(1) 1
(2) 0
(3) 2
(4) -1
Answer:
(1) 1
Hint: sin (90° – θ) cos θ + cos (90° – θ) sin θ = cos θ cos θ + sin θ sin θ = cos² θ + sin² θ = 1

Question 5.
1 – \(\frac{\sin ^{2} \theta}{1+\cos \theta}\) = ……………….
(1) cos θ
(2) tan θ
(3) cot θ
(4) cosec θ
Answer:
(1) cos θ
Hint:

Question 6.
cos⁴ x – sin⁴ x = ……………
(1) 2 sin² x – 1
(2) 2 cos² x – 1
(3) 1 + 2 sin² x
(4) 1 – 2 cos² x
Answer:
(2) 2 cos² x – 1
Hint:
cos⁴ x – sin⁴ x (cos² x)² – (sin² x)²
= (cos² x + sin²x) (cos² x – sin²x)
= 1 (cos² x – sin² x)
= cos² x – (1 – cos² x)
cos² x – 1 + cos² x
= 2 cos² x – 1.

Question 7.
If tan θ = \(\frac { a }{ x } \) , then the value of \(\frac{x}{\sqrt{a^{2}+x^{2}}}\) = ………………
(1) cos θ
(2) sin θ
(3) cosec θ
(4) sec θ
Answer:
(1) cos θ
Hint:

Question 8.
If x = a sec θ, y = b tan θ, then the value of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = ………………..
(1) 1
(2) -1
(3) tan² θ
(4) cosec² θ
Answer:
(1) 1
Hint:

Question 9.
\(\frac{\sec \theta}{\cot \theta+\tan \theta}\) = ………..
(1) cot θ
(2) tan θ
(3) sin θ
(4) – cot θ
Answer:
(3) sin θ
Hint:

Question 10.

Answer:
(1) tan θ
(2) 1
(3) -1
(4) sin θ
Answer:
(2) 1
Hint:

Question 11.
In the adjoining figure, AC = ………….

(1) 25m
(2) 25 \(\sqrt { 3 }\) m
(3) \(\frac{25}{\sqrt{3}}\)
(4) 25 \(\sqrt { 2 }\) m
Answer:
(2) 25 \(\sqrt { 3 }\) m
Hint:
tan θ = \(\frac { AC }{ AB } \) ⇒ tan 60° = \(\frac { AC }{ 25 } \) ⇒ AC = 25 \(\sqrt { 3 }\) m

Question 12.
In the adjoining figure ∠ABC =

(1) 45°
(2) 30°
(3) 60°
(4) 50°
Answer:
(3) 60°
Hint:
tan B = \(\frac { AC }{ AC } \) = \(\frac{100 \sqrt{3}}{100}\) = \(\sqrt { 3 }\) ⇒ ∴ tan B = \(\sqrt { 3 }\) ⇒ ∠B = 60°

Question 13.
A man is 28.5 m away from a tower. His eye level above the ground is 1.5 m. The angle of elevation of the tower from his eyes is 45°. Then the height of the tower is …………..
(1) 30 m
(2) 27.5 m
(3) 28.5 m
(4) 27 m
Answer:
(1) 30 m
Hint:
tan 45° = \(\frac { AB }{ BC } \)

1 = \(\frac { x }{ 28.5 } \) ⇒ x = 28.5
Height of tower = 28.5 + 1.5 = 30 m

Question 14.
In the adjoining figure, sin θ = \(\frac { 15 }{ 17 } \) Then BC = ………….
(1) 85 m
(2) 65 m
(3) 95 m
(4) 75 m
Answer:
(4) 75 m
Hint:
sin θ = \(\frac { BC }{ AC } \) ⇒ \(\frac { 15 }{ 17 } \) = \(\frac { BC }{ 85 } \) ⇒ BC = \(\frac{85 \times 15}{17}\) = 75 m

Question 15.
(1 + tan² θ) (1 – sin θ) (1 + sin θ) = …………..
(1) cos² θ – sin² θ
(2) sin² θ – cos² θ
(3) sin² θ + cos² θ
(4) 0
Answer:
(3) sin² θ + cos² θ
Hint:
(1 + tan² θ) (1 – sin θ) (1 + sin θ) = (1 + tan² θ) (1 – sin² θ) = sec² θ × cos² θ = sec² θ × cos² θ = sec² θ × \(\frac{1}{\sec ^{2} \theta}\) = 1 = sin² θ + cos² θ

Question 16.
(1 + cot² θ) (1 – cos θ) (1 + cos θ) = ………………….
(1) tan² θ – sec² θ
(2) sin² θ – cos² θ
(3) sec² θ – tan² θ
(4) cos² θ – sin² θ
Answer:
(3) sec² θ – tan² θ
Hint:
(1 + cot² θ) (1 – cos θ) (1 + cos θ) = (1 + cot² θ) (1 – cos² θ) = cosec² θ. sin² θ
= \(\frac{1}{\sin ^{2} \theta}\) sin² θ = 1 = sec² θ – tan² θ.

Question 17.
(cos² θ – 1) (cot² θ + 1) + 1 = ……………….
(1) 1
(2) -1
(3) 2
(4) 0
Answer:
(4) 0
Hint:
(cos² θ – 1) (cot² θ + 1) + 1 = – sin² θ (cosec² θ) + 1 = – sin² θ \(\frac{1}{\sin ^{2} \theta}\) + 1 = -1 + 1 = 0

Question 18.
\(\frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}\) = …………….
(1) cos² θ
(2) tan² θ
(3) sin² θ
(4) cot² θ
Answer:
(2) tan² θ
Hint:

Question 19.
Sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) …………
(1) cosec² θ + cot² θ
(2) cosec² θ – cot² θ
(3) cot² θ – cosec² θ
(4) sin² θ – cos² θ
Answer:
(2) cosec² θ – cot² θ
Hint:
sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) = sin² θ + \(\frac{1}{\sec ^{2} \theta}\) = sin² θ + cos² θ = 1 cosec² θ – cot² θ

Question 20.
9 tan² θ – 9 sec² θ = ……..
(1) 1
(2) 0
(3) 9
(4) -9
Answer:
(4) -9
Hint:
9 tan² θ – 9 sec² θ = 9(tan² θ – sec² θ) = 9(-1) = -9

Question 21.
The length of shadow of a tower on the plane ground is \(\sqrt { 3 }\) times the height of the tower. The angle of elevation of sum is …………..
(1) 45°
(2) 30°
(3) 60°
(4) 90°
Ans.
(2) 30°
Hint: Let the height of the tower be “x”
Lenght of the shadow is \(\sqrt { 3 }\) x

In the right ∆ ABC, tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{x}{\sqrt{3} x}=\frac{1}{\sqrt{3}}\)
= tan 30°
θ = 30°

Question 22.
A ladder makes an angle of 60° with the ground, when placed against a wall. If the foot of the ladder is 2m away from the wall, then the length of the ladder (in metres) is …………
(1) \(\frac{4}{\sqrt{2}}\) m
(2) 4 \(\sqrt { 3 }\) m
(3) 2 \(\sqrt { 2 }\) m
(4) 4 m
Answer:
(4) 4 m
Hint:
Let the length of the ladder be x.

In ∆ ABC, cos 60° = \(\frac { BC }{ AC } \) = \(\frac { 2 }{ x } \)
\(\frac { 1 }{ 2 } \) = \(\frac { 2 }{ x } \) ⇒ x = 4m

Question 23.
The angle of depression of a car parked on the road from the top of a 150m high tower is 30°. The distance of the car from the tower (in metres) is ……….
(1) 150 \(\sqrt { 3 }\) m
(2) 150 \(\sqrt { 2 }\)
(3) 75 cm
(4) 50 \(\sqrt { 3 }\) m
Answer:
(4) 50 \(\sqrt { 3 }\) m
Hint:
Let the distance of the car from the tower is “x” m
In ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)

\(\frac{1}{\sqrt{3}}\) = \(\frac { 150 }{ x } \) ⇒ x = 150 \(\sqrt { 3 }\) m

Question 24.
The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45°. Then the height of the tower (in meters) is ……….
(1) 50 \(\sqrt { 3 }\) m
(2) 50 m
(3) \(\frac{50}{\sqrt{2}}\) m
(4) \(\frac{50}{\sqrt{3}}\) m
Answer:
(4) \(\frac{50}{\sqrt{3}}\) m
Hint:
Let the height of the AB be “x”
On ∆ ABC
tan 45° = \(\frac { AB }{ BC } \)

1 = \(\frac { x }{ 50 } \) ⇒ x = 50 m

Question 25.
If x = a cos θ and y = b sin θ, then b²x² + a²y² = ……………
(1) a²b²
(2) ab
(3) a4b4
(4) a² + b²
Answer:
(1) a²b²
Hint:
b²a² cos² θ + a² b² sin² θ = a²b² (cos² θ + sin² θ) = a²b² × 1 = a²b²

II. Answer The Following Questions.

Question 1.
Prove that sec² θ + cosec² θ = sec² θ cosec² θ
Answer:
L.H.S = sec² θ + cosec² θ

Question 2.
Prove that \(\frac{\sin \theta}{1-\cos \theta}\) = cosec θ + cot θ.
Answer:

Question 3.
Prove that \(\frac{\cos \theta}{\sec \theta-\tan \theta}\) = 1 + sin θ
Answer:

= 1 + sin θ = R.H.S
L.H.S = R.H.S

Question 4.
Prove that sec θ (1 – sin θ) (sec θ + tan θ) = 1
Answer:
L.H.S = sec θ (1 – sin θ) (sec θ + tan θ)

Question 5.
Prove that \(\frac{\sin \theta}{\csc \theta+\cot \theta}=1\) – cos θ
Answer:

Question 6.
Prove the identify \(\frac{\sin \theta}{\csc \theta}+\frac{\cos \theta}{\sec \theta}=1\)
Answer:

Question 7.
Prove the identify \(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\) = cosec θ – cot θ
Answer:

(∵ 1 – cos² θ = sin² θ)
L.H.S = R.H.S

Question 8.
Prove the identity [cosec (90° – θ) – sin (90° – θ)] [cosec θ – sin θ] [tan θ + cot θ] = 1
Answer:
Now, [cosec (90° – θ) – sin (90° – θ)] [cosec θ – sin θ] [tan θ + cot θ]

Question 9.
A kite is flying with a string of length 200 m. If the thread makes an angle 30° with the ground, find the distance of the kite from the ground level. (Here, assume that the string is along a straight line.)
Answer:
Let h denote the distance of the kite from the ground level.
In the figure, AC is the string
Given that ∠CAB = 30° and AC = 200 m.
In the right ∆ CAB,
sin 30 = \(\frac { BC }{ AC } \)

sin 30 = \(\frac { h }{ 200 } \)
⇒ h = 200 sin 30°
∴ h = 200 × \(\frac { 1 }{ 2 } \) = 100 m
Hence the distance of the kite from the ground level is 100 m.

Question 10.
Find the angular elevation (angle of elevation from the ground level) of the Sun when the length of the shadow of a 30 m long pole is 10 \(\sqrt { 3 }\) m.
Answer:
Let S be the position of the Sun and BC be the pole.
Let AB denote the length of the shadow of the pole.
Let the angular elevation of the Sun be θ.
Given that AB = 10 \(\sqrt { 3 }\) m and BC = 30 m
In the right ∆ CAB,tan θ = \(\frac { BC }{ AB } \) = \(\frac{30}{10 \sqrt{3}}=\frac{3}{\sqrt{3}}\)
⇒ tan θ = \(\sqrt { 3 }\)
∴ θ = 60°
Thus, the angular elevation of the Sun from the ground level is 60°

Question 11.
A ramp for unloading a moving truck, has an angle of elevation of 30°. If the top of the ramp is 0.9 m above the ground level, then find the length of the ramp.
Answer:
Let AC be the length of the ramp and AC = “x” metre
In the right angled ∆ ABC,
∠A = 30° and BC = 0.9 m
sin 30° = \(\frac { BC }{ AC } \)

\(\frac { 1 }{ 2 } \) = \(\frac { 0.9 }{ x } \)
x = 0.9 × 2 = 1.8 m
∴ Length of the ramp x = 1.8 m

Question 12.
A girl of height 150 cm stands in front of a lamp-post and casts a shadow of length 150 \(\sqrt { 3 }\) cm on the ground. Find the angle of elevation of the top of the lamp-post.
Answer:
The height of the girl (BC) = 150 cm
Length of the shadow = 150 \(\sqrt { 3 }\) cm
Let θ be the angle of elevation of the lamp post.

In θ = \(\frac{150}{150 \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)

tan 30° = \(\frac{1}{\sqrt{3}}\) ⇒ ∴ θ = 30°
∴ Angle of elevation of the lamp post = 30°

Question 13.
Prove that \(\sqrt{\cot ^{2} \theta-\cos ^{2} \theta}\) = cot θ . cos θ
Answer:

Question 14.
Prove that \(\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}\) = 2 sec θ
Answer:

Question 15.
A tower is 100 \(\sqrt { 3 }\) metres high. Find the angle of elevation of its top from a point 100 metres away from its foot.
Answer:
Let MN be the tower of height 100 \(\sqrt { 3 }\) m
“O” be the point of observation such that OM = 100 m

Let 0 be the angle of elevation
In the right ∆ OMN, we have
tan θ = \(\frac { MN }{ OM } \) = \(\frac{100 \sqrt{3}}{100}\)
tan θ = \(\sqrt { 3 }\) = tan 60°
∴ θ = 60°
Hence the angle of elevation is 60°

Question 16.
If sin θ = x and sec θ = y, then find the value of cot θ
Answer:
Given sin θ = x
y = sec θ = \(\frac{1}{\cos \theta}\)
∴ cos θ = \(\frac { 1 }{ y } \) ⇒ cot θ = \(\frac{\cos \theta}{\sin \theta}=\frac{1}{y} \div x\)
cot θ = \(\frac { 1 }{ xy } \)

Question 17.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° ( see figure).
Answer:
In the figure, let AC is the rope and AB is the pole. In right ∆ ABC, we have
\(\frac { AB }{ AC } \) = sin 30°
But, sin 30° = \(\frac { 1 }{ 2 } \)
\(\frac { AB }{ AC } \) = \(\frac { 1 }{ 2 } \) ⇒ \(\frac { AB }{ 20 } \) = \(\frac { 1 }{ 2 } \)
[∵ AC = 20 m]

AB = 20 × \(\frac { 1 }{ 2 } \) = 10 m
Thus, the required height of the pole is 10 m.

Question 18.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Answer:
Let in the right ∆ AOB,
OB = Length of the string
AB = 60 m = Height of the kite.
In the right ∆ OAB

sin 60° = \(\frac { AB }{ OB } \)
\(\frac{\sqrt{3}}{2}\) = \(\frac { 60 }{ OB } \)
\(\sqrt { 3 }\) × OB = 120
OB = \(\frac{120}{\sqrt{3}}=\frac{120 \times \sqrt{3}}{3}\)
Lenght of the string = 40 \(\sqrt { 3 }\) = 40 \(\sqrt { 3 }\) m

Question 19.
Prove that sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ = 1
Answer:
L. H. S = sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ
= (sin² θ)³ + (cos² θ)³ + 3 sin² θ cos² θ
= (sin² θ + cos² θ)³ – 3 × sin² θ cos² θ (sin² θ + cos² θ) + 3 × sin² θ cos² θ
[Using a³ + h³ = (a + h)³ – 3 ab (a + h)]
= 1³ – 3 sin² θ cos² θ (1) + 3 sin² θ cos² θ
= 1 – 3 sin² θ cos² θ + 3 sin² θ cos² θ
= 1
L.H. S = R. H. S

III. Answer The Following Questions.

Question 1.
Prove that \(\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=2\) cosec θ
Answer:

Question 2.
Prove that \(\frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A}\) = 2 cosec A.
Answer:

Question 3.
Prove that 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = θ
Answer:
consider, sin⁶ θ + cos⁶ θ
= (sin² θ)³ + (cos² θ)³
[Using a³ + b³ = (a + b)³ – 3 ab(a + b)]
= (sin² θ + cos² θ)³ – 3 sin² θ cos² θ (sin² θ + cos² θ)
= 1 – 3 sin² θ cos² θ
Now sin⁴ θ + cos⁴ θ = (sin² θ)² + (cos² θ)² (a² + b² = (a + b)² – 2 ab)
= (sin² θ + cos² θ)² – 2 (sin² θ cos² θ)
= 1 – 2 (sin² θ cos² θ)
L. H. S = 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1
= 2(1 – 3 sin² θ cos² θ) – 3 (1 – 2 sin² θ cos² θ) + 1
= 2 – 6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1
= 3 – 3 = 0
L.H.S. = R.H.S

Question 4.
Prove that \(\frac{\sin \left(90^{\circ}-\theta\right)}{1+\sin \theta}+\frac{\cos \theta}{1-\cos \left(90^{\circ}-\theta\right)}\) = 2 sec θ
Answer:

Question 5.
Prove that \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
Answer:

Question 6.
Prove that \(\frac{\sin \left(90^{\circ}-\theta\right)}{1-\tan \theta}+\frac{\cos \left(90^{\circ}-\theta\right)}{1-\cot \theta}\) = cos θ + sin θ
Answer:

L.H.S = R.H.S
Hence Proved

Question 7.
Prove that \(\frac{\tan \left(90^{\circ}-\theta\right)}{\csc \theta+1}+\frac{\csc \theta+1}{\cot \theta}\) = 2 sec θ
Answer:

Question 8.
Prove that

Answer:

Question 9.
Prove that (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2
Answer:
L.H.S = (1 + cot θ – cosec θ) (1 + tan θ + sec θ)

L. H. S = R. H. S
Hence proved

Question 10.
Prove that \(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}\)
Answer:

Question 11.
Prove that

Answer:

LHS = RHS
Hence proved

Question 12.
If tan θ = n tan α and sin θ = m sin α, then prove that cos² θ = \(\frac{m^{2}-1}{n^{2}-1}, n \neq \pm 1\)
Answer:
Given tan θ = n tan α and sin θ = m sin α
Let us eliminate using cosec² α – cot² α = 1

Question 13.
If sin θ, cos θ and tan θ are in G. P., then prove that cot⁶ θ – cot² θ = 1.
Answer:
Given, sin θ, cos θ, tan θ are in G. P., To prove cot⁶ θ – cot² θ = 1

Question 14.
A person in an helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30° and 45°. Find the width of the river. (\(\sqrt { 3 }\) = 1. 732)
Answer:
Let C be the point of observation.
The objects A and B lying opposite to each other on either bank of a river.
Width of the river AB = AD + BD
In the right ∆ ACD,

tan 30° = \(\frac { CD }{ AD } \) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 700 }{ AD } \)
∴ AD = 700 \(\sqrt { 3 }\)
In the right ∆ BCD tan 45° = \(\frac { CD }{ BD } \)
1 = \(\frac { 700 }{ BD } \) ⇒ BD = 700 m
∴ Width of the river = AD + BD
= 700 \(\sqrt { 3 }\) + 700 = 700 (\(\sqrt { 3 }\) + 1)
= 700 (1.732 + 1) = 700 × 2.732 m
= 1912.400 m
∴ Width of the river = 1912.4 m

Question 15.
A person X standing on a horizontal plane, observes a bird flying at a distance of 100 m from him at an angle of elevation of 30°. Another person Y standing on the roof of a 20 m high building, observes the bird at the same time at an angle of elevation of 45°. If X and Y are on the opposite sides of the bird, then find the distance of the bird from Y.
Answer:
Let the position of the bird be “B”
Given, AY = 20 m, CD = 20 m, BX = 100 m, ∠BXD = 30° and ∠BYC = 45°
In the right ∆ BXD, sin 30° = \(\frac { BD }{ BX } \)
\(\frac { 1 }{ 2 } \) = \(\frac { BD }{ 100 } \) ⇒ BD = \(\frac{100 \times 1}{2}\) = 50 m

BC = BD – DC
BC = 50 m – 20 m = 30 m
In the right ∆ YBC, sin 45° = \(\frac { BC }{ BY } \)
\(\frac{1}{\sqrt{2}}\) = \(\frac { 30 }{ BY } \) ⇒ BY = 30\(\sqrt { 2 }\)
∴ Distance of the bird from the person Y is 30\(\sqrt { 2 }\) m

Question 16.
A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Answer:
In the given figure A, B, C is a horizontal at the level of the boy
AD = A’D – AA’
AD = 30 – 1.5 = 28.5 m
Let the distance walked by the student CB is ‘x’ m.
Let AB be ‘y’ m
In the right ∆ ABD,tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 28.5}{ y } \)
y =\(\frac{28.5}{\sqrt{3}}\) …….(1)

In the right ∆ ACD,tan 30° = \(\frac { AD }{ AC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{x+y}\) ⇒ x + y = 28.5 \(\sqrt { 3 }\)
y = 28.5 \(\sqrt { 3 }\) – x ………..(2)
From (1) and (2) we get,
\(\frac{28.5}{\sqrt{3}}\) = 28.5 \(\sqrt { 3 }\) – x
28.5 = 28.5 × 3 – \(\sqrt { 3 }\) x
\(\sqrt { 3 }\) x = 28.5 × 3 – 28.5
\(\sqrt { 3 }\) x = 28.5 × (3 – 1) = 28.5 × 2
\(\sqrt { 3 }\) x = 57.0 m
x = \(\frac{57}{\sqrt{3}}=\frac{57 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) ⇒ x = \(\frac{57 \times \sqrt{3}}{3}\) = 19 \(\sqrt { 3 }\) m.
∴ The distance walked by the boy = 19 \(\sqrt { 3 }\) m.

Question 17.
A straight highway leads to the foot of a tower. A man standing on the top of the tower spots a van at an angle of depression of 30°. The van is approaching the tower with a uniform speed. After 6 minutes, the angle of depression of the van is found to be 60°. How many more minutes will it take for the van to reach the tower?
Answer:
Let A be the point of observation. B and C be the positions of the van. Let the height of the tower AD be x. Let the speed of the van be “s”. C is the position of van after 6 minutes.
Time taken by the van from B to C be minutes.
Distance BC = 6 × s = 6 s (speed × time)
Let distance between CD be t s (time × speed)

In the right ∆ ABD, tan 30° = \(\frac { AD }{ DB } \)


∴ 3 more minutes will be taken by the van to reach the tower.

Students can download Maths Chapter 5 Information Processing Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 6th 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 6th Maths Solutions Term 3 Chapter 5 Information Processing Ex 5.2

Miscellaneous Practice Questions 

Question 1.
Find HCF of 188 and 230 by Euclid’s game.
Solution:
By Euclid’s game HCF (a, b) = HCF (a, a – b) if a > b.
Here HCF (188, 230) = HCF (230, – 188) because 230 > 188
= HCF (188, 42) = HCF (146, 42)
= HCF (104, 42) = HCF (62, 42)
= HCF (42, 20) = HCF (22, 20)
= HCF (20,2) = HCF (18, 2) = 2
∴ HCF (230, 188) = 2

Question 2.
Write the numbers from 1 to 50. From that find the following.
i) The numbers which are neither divisible by 2 nor 7.
ii) The prime numbers between 25 and 40
iii) All square numbers upto 50.
Solution:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.
i) The numbers neither divisible by 2 nor 7 are 9, 11, 13, 15, 17, 19, 23, 25, 27, 29, 31, 33, 37, 39, 41, 43, 45, 47.
ii) The prime numbers between 25 and 40 are 29, 31, 37.
iii) Square numbers upto 50 are 1, 4, 9, 16, 25, 36, 49

Question 3.
Complete the following pattern.
(i) 1 + 2 + 3 + 4 = 10
2 + 3 + 4 + 5 = 14
___ + 4 + 5 + 6 = ___
4 + 5 + 6 + ___ = ___

(ii) 1 + 3 + 5 + 7 = 16
___ + 5 + 7 + 9 = 24
5 + 7 + 9 + ___ = ___
7 + 9 + ___ + 13 = ___

(iii) AB, DEF, HIJK, ___ , STUVWX
(vi) 20, 19, 17, ___ , 10, 5
Solution:
(i) 3, 18; 7, 22
(ii) 3; 11, 32; 11, 40
(iii) MNOPQ
(iv) 14

Question 4.
Complete the table by using the following instructions.

A : It is the 6th term in the Fibonacci sequence.
B : The predecessor of 2.
C : LCM of 2 and 3.
D : HCF of 6 and 20.
E : The reciprocal of 1/5.
F : The opposite number of -7.
G : The first composite number.
H : Area of a square of side 3 cm.
I : The number of lines of symmetry of an equilateral triangle.
After completing the table, what do you observe? Discuss.
Solution:
A – 8, B – 1, C – 6, D – 2, E – 5, F – 7, G – 4, H – 9, I – 3

Question 5.
Assign the number for English alphabets as 1 for A, 2 for B upto 26 for Z. Find the meaning of

Solution:
GOOD MORNING

Question 6.
Replace the letter with symbols as + for A, – for B, × for C, and ÷ for D. Find the answer for the pattern 4B3C5A30D2 by doing the given operations.
Solution:
Given the symbols + for A; – for B; × for C; + for D .
∴ 4B3C5A30D2 becomes
4 – 3 × 5 + 30 ÷ 2 Using BIDMAS rule
4 – 3 × 5 + 30 ÷ 2 = 4 – 3 × 5 + 15[× done first]
= 4 – 15 + 15 [+ done second]
= 4 – 0 [+ done third]
= 4 [- done last]

Question 7.
Observe the pattern and find the word by hiding the Numbers 1 H 2 0 3 W, 4 A 5 R 6 E, 7 Y 8 0 9 U.
Solution:
HOW ARE YOU

Question 8.
Arrange the following from the eldest to the youngest. What do you get?

Solution:
Arranging from eldest to the youngest we get
F – refers to grandparents
A – refers to parents
M – refers to an uncle
I – refers to elder sister
L – refers to me
Y – refers to the younger brother
So we get FAMILY

Challenge Problems

Question 9.
Prepare a daily time schedule for evening study at home.
Solution:
5.00 pm to 6.00 pm – Mathematics
6.0 pm to 7.00 pm – Science
7.0 pm to 8.00 pm – Social Science
8. pm to 9.00 pm – Dinner & Recreation
9. pm to 10.00 pm – Tamil and English

Question 10.
Observe the geometrical pattern and answer the following questions.

(i) Write down the number of sticks used in each iterative pattern,
(ii) Draw the next figure in the pattern also find the total number of sticks used in it.
Solution:
(i) 3, 9, 18

Question 11.
Find the HCF of 28, 35, 42 by Euclid’s game.
Solution:
HCF of 28, 35, 42
HCF of (28, 35 – 28, 42 – 28)
28 = 2 × 2 × 7
7 = 1 × 7
14 = 2 × 7
HCF of (28, 7, 14) = 7

Question 12.
Follow the given instructions to fill your name in the OMR sheet.
1. The name should be written in capital letters from left to right.
2. One alphabet is to be entered in each box.
3. If any empty boxes are there at the end they should be left blank.
4. Ballpoint pen is to be used for shading the bubbles for the corresponding alphabets.

Solution:
Do your self.

Question 13.
Consider the Postal index number (PIN) written on the letters as follows: 604506; 604516; 604560; 604506; 604516; 604516; 604560; 604516; 604505; 604470; 604515; 604520; 604303; 604509; 604470. How the letters can be sorted as per Postal Index Numbers?
Solution:
604 is common for all postal index numbers. Compare the remaining 3 digits, 303, 470, 505, 506 (two) 509, 510. 515, 516 (Four), 520, 560 (two).

Students can download Maths Chapter 4 Symmetry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 6th 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 6th Maths Solutions Term 3 Chapter 4 Symmetry Ex 4.1

Question 1.
Fill in the blanks
(i) The reflected image of the letter ‘q’ is …….
(ii) A rhombus has ………… lines of symmetry.
(iii) The order of rotational symmetry of the letter ‘Z’ is ……….
(iv) A figure is said to have rotational symmetry, if the order of rotation is atleast ……….
(v) ……… symmetry occurs when an object slides to new position.
Solution:
(i) P
(ii) two
(iii) 2
(iv) two
(v) Translation

Question 2.
Say True or False
(i) A rectangle has four lines of symmetry.
(ii) A shape has reflection symmetry if it has a line of symmetry.
(iii) The reflection of the name RANI is INAЯ.
(iv) Order of rotation of a circle is infinite.
(v) The number 191 has rotational symmetry.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
Match the following shapes with their number of lines of symmetry.

Solution:
(i) d
(ii) a
(iii) b
(iv) c

Question 4.
Draw the lines of symmetry of the following.

Solution:

Question 5.
Using the given horizontal line/ vertical line as a line of symmetry, complete each alphabet to discover the hidden word.

Solution:
(i) DECODE
(ii) KICK
(iii) BED
(iv) WAY
(v) MATY
(vi) TOMATO

Question 6.
Draw a line of symmetry of the given figures such that one hole coincides with the other hole(s) to make pairs.

Solution:

Question 7.
Complete the other half of the following figures such that the dotted line is the line of symmetry.

Solution:

Question 8.
Find the order of rotation for each of the following.

Solution:
(i) 2
(ii) 2
(iii) 4
(iv) 8
(v) 2

Question 9.
A standard die has six faces which are shown below. Find the order of rotational symmetry of each face of a die?

Solution:
(i) 4
(ii) 2
(iii) 2
(iv) 4
(v) 4
(vi) 2

Question 10.
What pattern is translated in the given border kolams?

Solution:

Objective Type Questions

Question 11.
Which of the following letter does not have a line of symmetry?
(a) A
(b) P
(c) T
(d) U
Hint: A, T, U have one line of symmetry
Solution:
(b) P

Question 12.
Which of the following is a symmetrical figure?

Solution:
(c)

Question 13.
Which word has a vertical line of symmetry?
(a) DAD
(b) NUN
(c) MAM
(d) EVE
Hint: D, N, E have no vertical line of symmetry
Solution:
D, N, E have no vertical line of symmetry

Question 14.
The order of rotational symmetry of 818 is ………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2

Question 15.
The order of rotational symmetry ★ is ___
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(a) 5

Students can download Maths Chapter 3 Perimeter and Area Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 6th 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 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Miscellaneous Practice Problems

Question 1.
A piece of wire is 36 cm long. What will be the length of each side if we form
(i) a square
(ii) an equilateral triangle
Solution:

Given the length of the wire = 36 cm
i) When a square is formed out of it
The perimeter of the square = 36 cm
4 × side = 36
side = \(\frac{36}{4}\) = 9 cm
Side of the square

ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm
i.e., side + side + side = 36 cm .
3 × side = 36 cm
side = \(\frac{36}{3}\) = 12 cm
One side of an equilateral triangle = 12 cm

Question 2.
From one vertex of an equilateral triangle with a side of 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion? The perimeter of the remaining portion
Solution:

= (40 + 34 + 6 + 34) cm
= 114 cm

Question 3.
Rahim and Peter go for a morning walk, Rahim walks around a. square path of side 50 m and Peter walks around a rectangular path with a length of 40 m and a breadth of 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Solution:
Distance covered by Rahim
= 50 × 4 m
= 200 m
If he walks 2 rounds, distance covered = 2 × 200 m
= 400 m
Distance covered by peter
= 2 (40 + 30) m
= 2(70)m
= 140 m
If he walks 2 rounds, distance covered = 2 × 140 m
= 280 m
∴ Rahim covers more distance by (400 – 280) = 120 m

Question 4.
The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park.
Solution:
Let the length be b + 14 m
breadth = b
perimeter = 200
2 (l + b) = 200
2 (b + 14 + b) = 200
2 (2b + 14) = 200
28 + 4b = 200
4b = 200 – 28
4b = 172 m
b = \(\frac{172}{4}\)
b = 43 m
Length = b + 14
= 43 + 14
Length l = 57 m
Area = l × b units
= 57 × 43 m²
= 2451 m²

Question 5.
Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at ₹ 10 per metre.
Solution:
a = 5 m
The perimeter of the garden
= 4 a units
= 4 × 5 m
= 20 m
For 1 row
Amount needed to fence l m= Rs 10
Amount needed to fence 20 m
= Rs 10 × 20
= Rs 200
For 2 rows
Total amount needed = 2 × Rs 200 = Rs 400

Challenge Problems

Question 6.
A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.
Solution:
Number of equal sides in the shape = 20
One of its side = 3 cm
Perimeter = length of one side × Number of equal sides
∴ Perimeter = (3 × 20) cm = 60 cm
∴ Perimeter = 60 cm

Question 7.
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.
Solution:
Area of rectangle = (length × breadth) units²
Length = 40 cm
Breadth = 20 cm
∴ Area = (40 × 20) cm² = 800 cm²
Area of rectangle = 800 cm²
Area of square = (side × side) units²
side = 10 cm
∴ Area of square = (10 × 10) cm² = 100 cm²
Required number of squares = \(\frac{\text { Area of Rectangle }}{\text { Area of } 1 \text { square }}=\frac{800 \mathrm{cm}^{2}}{100 \mathrm{cm}^{2}}\) = 8
8 squares can be formed.

Question 8.
The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.
Solution:
Given perimeter of a rectangle = 64 cm
Also given length is three times its breadth.
Let the breadth of the rectangle = b cm
∴ Length = 3 × b cm
Perimeter = 64 m
i.e., 2 × (l + b) = 64 m
2 × (3b + b) = 64 m
2 × 4b = 64m
4b = \(\frac{64}{2}\) = 32 m
b = \(\frac{32}{4}\) = 8 m
l = 3 × b = 3 × 8 = 24 m
∴ Breadth of the rectangle = 8 m
Length of the rectangle = 24 m

Question 9.
How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.
Solution:
Length of the string to be made into rectangle = 48 cm
∴ Perimeter of the rectangle = 48 cm
2 × (l + b) = 48 cm
l + b = \(\frac{48}{2}\)
l + b = 24 cm
Possible pairs of length and breadth are (1, 23), (2, 22) (3, 21), (4, 20), (5, 19),
(6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)
Number of different rectangles = 12.

Question 10.
Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Solution:

Perimeter of A = s + s + s + s units = 4 s units
Perimeter of B = (2s + 2s + 2s + 2s) units
= 8s units = 2 (4s) units.
∴ The perimeter of B is twice the perimeter of A

Question 11.
What will be the area of a new square formed if the side of a square is made one – fourth?
Solution:
Let the side of square is s units then area = (s × s) units²
If the side of the new square is made one fourth then side = \(\left(\frac{1 \times s}{4}\right)\) units
Then area = \(\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)\) units² = \(\frac{s \times s}{16}=\frac{1}{16}\) (s × s) units²
Area of the new square is reduced to \(\frac{1}{16}\) times to that of original area.

Question 12.
Two plots have the same perimeter. One . is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Solution:
a = 10 m, b = 8 m
Perimeter of the square plot
= 4 a units
= 4 × 10 m
= 40 m
Perimeter of the rectangular plot
40 = 2 (l + b) units
40 = 2 (l + 8) m
40 = 2 l + 16
2 l = 40 – 16
2 l = 24
l = \(\frac{24}{2}\)
l = 12 m
Area of the square plot
= a × a sq units
= 10 × 10 m²
= 100 m²
Area of the rectangular plot
= l × b sq units
= 8 × 12 m²
= 96 m²
Square plot has the greater area by 100 m² – 96 m² – 4 m²

Question 13.
Look at the picture of the house given and find the total area of the shaded portion.

Solution:
Total area of the shaded region = Area of a right triangle + Area of a rectangle
= (\(\frac{1}{2}\) × b × h) + (l × b) cm²
= [(\(\frac{1}{2}\) × 3 × 4) + (9 × 6)] cm²
= (6 + 54) cm² = 60 cm²

Question 14.
Find the approximate area of the flower in the given square grid.

Solution:
No of full squares = 11
No of half squares = 9
Area of 11 full squares
= 11 x 1 cm²
= 11 cm²
Area of 9 half squares
= 9 × \(\frac{1}{2}\) cm²
= 4.5 cm²
Area of the flower = (11 + 4.5) cm²
= 15.5 cm²