Class 8

8th Standard English Guide Term 3 Unit 2 Poem My Computer Needs A Break Book Back Answers

Students can Download English Poem 2 My Computer Needs A Break Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 8th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

My Computer Needs A Break 8th Standard English 2nd Lesson Question and Answer

Read And Understand

A. Answer the following.

Question 1.
How does the poet describe her computer?
Answer:
The computer of the poet is very brainy and smart. It is a mountain of information. It gives the answers very quickly with a click of the mouse.

Question 2.
What happened to the computer?
Answer:
The computer was behaving badly, forgets to save and store anything. A virus has got into it and it became sick.

Question 3.
List four things that the computer could not do after it became absent-minded.
Answer:

1. The computer forgets to save the files.
2. It makes the files vanish away.
3. It does not check the spelling, whether they are right or wrong.
4. It hides the files so that the files vanish from sight.

Question 4.
What made the poet squirm?
Answer:
When the computer actually gobbled a worm, a virus, it behaved unsteadily. So the poet squirmed.

Question 5.
Why did the poet call the doctor?
Answer:
Once, the computer caught a virus and fell very sick. So, she had to call in a doctor.

B. Fill in the blanks.

1. Computers are ________ and ________
2. We get answers for questions by a ________
3. The computer forgot to ________ the poet’s work.
4. The computer actually gobbled a ________
5. The poet feels that his computer needs ________

Answer:

1. brainy, smart
2. click
3. save
4. worm
5. a holiday

C. Pick out the rhyming words from the poem.

1. Smart – ______
2. click – ______
3. right – ______
4. sick – ______

Answer:

1. heart
2. quick
3. sight
4. quick

D. Match the poetic lines with Figures of speech

1. So brainy – (a) personification
2. Mountains – (b) personification
3. It’s so absent-minded – (c) hyperbole
4. Computer gobbled a worm – (d) metaphor
5. Very sick – (e) metaphor

Answer:

1. (d)
2. (c)
3. (a)
4. (b)
5. (e)

E. Find the alliterating words from the poem

1. Save – ______
2. Doctor – ______
3. Virus – ______
4. makes – ______

Answer:

1. store
2. double
3. very
4. most

Parallel Reading

Additional Questions:

Question 1.
Who bought us closer and then made us more distant?
Answer:
Technology brought us closer and then made us more distant.

Question 2.
What did technology do?
Answer:
It introduced us to more friends and invited enemies. It gave us more publicity and exploited us. It saved us more and it is spent to be busier.

Question 3.
What does the poet say about Technology?
Answer:
The technology is an entrapping blessing in disguise.

Question 4.
Which is our new addiction?
Answer:
Technology is our new addiction.

Question 5.
Can we live without Technology?
Answer:
No, we cannot live without technology.

My Computer Needs A Break Additional Questions

I. Poem Comprehension:

Question 1.
My computer has always been so brainy and smart – It seems to know mountains of information by heart.

(a) What has the computer always been to the poet?
Answer:
The computer had always been so brainy and smart.

(b) Why does the poet use the word ‘mountains’?
Answer:
She had used this word to insist that the computer knows a lot of information.

(c) What do you mean by ‘brainy’.
Answer:
‘Brainy’ means “intelligent”.

Question 2.
It forgets to ‘save’ my work, and store it away,
And instead, makes it vanish in the most dreadful way.

(a) What does the computer forget?
Answer:
The computer forgets to save the poet’s work.

(b) What vanishes in the most dreadful way?
OH Her works vanish in the most dreadful way.

(c) What do you mean by the word “dreadful”?
Answer:
“Dreadful” means “unpleasant”.

II. Poem Appreciation:

Question 1.
If I type in a question and give my mouse a click,
My computer always gives me the answer really quick!

(a) Pick out the rhyming words.
Answer:
click – quick are the rhyming words.

(b) Name the alliterated words.
Answer:
my – mouse is the alliterated words.

Question 2.
And one day, my naughty computer actually gobbled a worm,
And behaved so erratically that it made me squirm.

(a) What is the figure of speech used here?
Answer:
Personification is used here. The computer is given human traits. It gobbled a worm.

(b) Pick out the alliterated words.
Answer:
made – me are the alliterated words.

III. Very Short Questions and Answers.

Question 1.
What do you mean by the phrase “mountains of information”?
Answer:
It means a lot of information.

Question 2.
Who is absent-minded?
Answer:
The poet’s computer is absent-minded.

Question 3.
What do you mean by the word “erratically”?
Answer:
“Erratically” means “unsteadily or unpredictably”.

Question 4.
Why did the computer fall very sick according to the poet?
Answer:
The computer fell very sick because it caught a virus.

Question 5.
What vanished from the poet’s sight?
Answer:
Files vanished from the poet’s sight.

IV. Short Questions and Answers :

Question 1.
What happened one day?
Answer:
One day, the poet’s computer actually gobbled a worm that was a virus and behaved so unsteadily. This made the poet discomfort.

Question 2.
Why did the poet call a doctor?
Answer:
The poet called in a doctor to examine her computer which caught a virus and fell
very sick.

Question 3.
Why did the poet say that she didn’t know what to do?
Answer:
Her computer behaved badly and became absent-minded. So the poet said that she didn’t know what to do.

V. Paragraph Question with Answer.

Question 1.
What are the problems faced by the poet with her computer?
Answer:
The poet’s computer had always been so intelligent and fashionable. It gave a lot of information quickly. But after some days, it gave her a lot of problems. She says that it was absent-minded. She did not know what to do about it. It forgot to save her work and store it away. It made her work vanish in the most unpleasant way. It didn’t check her spelling. It hid her files and vanished them from her sight. One day it gobbled a virus and fell very sick. So she had to call a doctor (a technician) to set it right.

Warm-Up

Question 1.
In pairs, tell each other how the computer plays a vital role in all fields.
Answer:

1. A computer helps students to learn new things.
2. It is a boon to modern society.
3. It is widely used in all fields such as Economics, Science, Medical, Law, Engineering, Designing, Graphics and Film making, etc.
4. Computer literacy is vital to success in today’s world.
5. Computers have helped in the design of safety equipment in sports.

My Computer Needs A Break Summary

The poet’s computer had always been very intelligent and fashionable. It seemed to know a lot of information by heart. If she typed a question and gave her mouse a click, it gave her the answer quickly. But recently, her computer had been behaving badly. It was so absent-minded and forgot to save her work and store it. Instead, her work vanished in the most dreadful way. It also didn’t check her spellings and hid her files, vanishing them from sight.

One day, the naughty computer actually swallowed a worm, that is a virus, and behaved so unsteadily. The poet was in discomfort. As it caught the virus, it became sick. So she had to call the doctor (technician) to examine her computer. She just asked him whether his tired computer needed some relaxation (break) because she had been using it for many days.

Samacheer Kalvi 8th English Book Back Answers

• Special Hero Book Back Answers
• My Hobby: Reading Book Back Answers
• Making Life Worth While Book Back Answers
• A Thing of Beauty Book Back Answers
• Fire Work Night Book Back Answers
• Lessons in Life Book Back Answers
• My Computer Needs A Break Book Back Answers

8th Standard English Guide Term 1 Unit 1 Prose Hobby – Turns A Successful Career Book Back Answers

Students can Download English lesson 1 Hobby – Turns A Successful Career Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 8th English Book Solutions Guide Pdf  helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Hobby – Turns A Successful Career 8th Standard English 1st Lesson Question and Answer

Section – I

Read And Understand

A. Say true or false.

1. Mani’s hobby is playing cricket.
2. Hobbies are unique to people.
3. People have hobbies only to pass time.

Answer:

1. False
2. True
3. False

B. Choose the correct answer.

Question 1.
Mani imagined a world with_______
(a) music and musicians
(b) Building and sculptures
(c) magic and magicians
Answer:
(c) magic and magicians

Question 2.
When Mani started writing things, he was unable to_______
(a) describe them accurately
(b) describe them fluently
(c) describe them imaginatively
Answer:
(a) describe them accurately

Question 3.
Mani feels writing is like_______
(a) painting the voice
(b) engraving the voice
(c) designing the voice
Answer:
(a) painting the voice

Section – II
Read And Understand

A. Say true or false.

1. Ajay Garg is an artist.
2. Asha Devi taught Ajay the miniature painting.
3. Ajay displayed 150 paintings.
4. Ajay was awarded by the Indian president Dr. A. P. J. Abdul Kalam.

Answer:

1. True
2. True
3. True
4. True

B. Answer the following questions.

Question 1.
What was Ajay gifted with?
Answer:
Ajay was gifted with painting.

Question 2.
How was he honoured?
Answer:
He was honoured by the Indian President Dr. A.P.J. Abdul Kalam with a national award of accomplishment.

Question 3.
What did Ajay’s current goal?
Answer:
His current goal is to revitalize the dying art of traditional miniature. Indian paintings.

Question 4.
Is Ajay’s hobby become a successful career? How?
Answer:
Yes, Ajay’s hobby has become a successful career because his work became popular all throughout India, the United States and the United Kingdom. He has also become the executive member of the ‘Deaf and Dumb’ Association in Rajasthan

Section – III

Read And Understand

A. Fill in the blanks.

1. Mani feels writing is like
2. Ajay mastered the techniques of
3. Ilavazhagi won her first match against

Answer:

1. Paintaing the voice
2. Prepare colours and brushes
3. Her father

B. Choose correct synonyms for the italic word.

Question 1.
Writing is a unique hobby.
(a) common
(b) beneficial
(c) uncommon
Answer:
(c) uncommon

Question 2.
Ajay started to paint happily.
(a) depressed
(b) joyfully
(c) unhappily
Answer:
(b) joy fully

Question 3.
Miniature paintings depict Indian culture.
(a) warp
(b) portray
(c) twist
Answer:
(b) portray

C. Choose correct antonyms for the italic word.

Question 1.
Ajay suffered an injury.
(a) sorrow
(b) endure
(c) hurt
Answer:
(b) endure

Question 2.
Ajay mastered the techniques of painting.
(a) skilled
(b) proficient
(c) unskilled
Answer:
(c) unskilled

Question 3.
Ilavazhagi won the world championship in
(a) win
(b) succeed
(c) defeat
Answer:
(c) defeat

D. Answer the following questions in one or two sentences.

Question 1.
What happened to Ajay at the age of three?
Answer:
Ajay suffered an injury in his ear at the age of three. He had taken treatments but the treatment left him permanently deaf.

Question 2.
What did Asha Devi teach Ajay?
Answer:
Asha Devi taught Ajay the old and dying technique of ‘Traditional Indian Miniature Painting’.

Question 3.
Why writing is beneficial according to Mani?
Answer:
According to Mani, writing is very beneficial, because it opens his mind to think beyond the little things with a broader mind.

E. Answer the following questions in 100 words.

Question 1.
How did Ajay’s father find his son’s talent?
Answer:
Ajay Kumar Garg’s childhood was going well, until he suffered an injury at the age of three. Ajay was treated, but that treatment left him permanently deaf. His parents tried many hospitals to cure his disability but it was useless. One day, his father bought him a paint set to play. Ajay started painting happily on the walls and floors. He looked happy, while using the paint set. Soon Ajay father realised that Ajay was talented in painting. When Ajay was five years, a court artist of Dholpur, Shri Sua Lai was impressed with Ajay’s work. He took Ajay with him and started training him.

Question 2.
Write a note on Ilavazhagi.
Answer:
Ilavalazhagi is an accomplished carrom player. She was born into an economically backward family in Vyasarpadi, Chennai. Her father was a fish – cart driver. He was a district-level champion in carrom, but he could not become a champion, because of his poverty. He encouraged his daughter to become a champion in carom. He used to take ger to the club to play the game. One day her father called her for a match, and she won the match, which gave her confidence. The confidence made her gain the world championship. She has played for the state for almost 14 years. Illavalgi is a member of the Thiruvalluvar District Carrom Association. Her Victory included the Asia Cup, and SAARC Cup, and the World Championship. In 2008, she gained the Indian National Carrom Championship. In the same year, she won Carrom World Championships and became a World Champion in the women’s Singles.

Vocabulary

Homophone

a. Pick out the correct homophone.

1. I am not ____ to drink soda, (aloud, allowed)
2. The wind _____ the leaves, (blue, blew)
3. I will _____ my friend, (meat, meet)
4. He will play the _____ in the second play, (role, roll)
5. I have_______movie before.(scene,seen)

Answer:

1. allowed
2. blew
3. meet
4. role
5. seen

b. Choose the correct homophone.

Question 1.
Every morning my father likes to look at the ______ on the grass
(a) due
(b) dew
Answer:
(b) dew

Question 2.
Chennai is the____ of Tamilnadu
(a) capitol
(b) capital
Answer:
(b) capital

Question 3.
Their words had a negative
(a) affect
(b) effect
Answer:
(b) effect

Question 4.
I was very last weekend because my friends made plans that
(a) bored
(b) board
Answer:
(a) bored

Question 5.
The cat caught the
(a) scent
(b) cent
Answer:
(a) scent

Abbreviation And Acronyms

(a) & (b) Expand the following:
Answer:

1.	GST	Goods and Services Tax.
2.	ECG	Electrocardiography
3.	ILO	International Labour Organization
4.	SCERT	State Council of Educational Research and Training
5.	IIT	Indian Institute of Technology
6.	ISRO	Indian Space Research Organisation
7.	FIR	First Information Report
8.	UNESCO	United Nations Educational, Scientific and Cultural Organization
9.	UNICEF	United Nations International Children’s Emergency Fund
10	NASA	National Aeronautics and Space Administration

Listening

Questions on passage. (*Listening passage is given at the end of the third unit.)

Question 1.
What is philately?
Answer:
Stamp collection is known as philately.

Question 2.
What is numismatics?
Answer:
The hobby of collecting coins from different countries is known as numismatics.

Question 3.
Do you know where the rarest and the most beautiful shells are found?
Answer:
The rarest and the most beautiful shells are found in the Indo Pacific, Caribbean, and Mediterranean regions.

Question 4.
Painting can help unleash your creative side. How?
Answer:
Painting allows you to tap into the thoughts, desires and feelings in your head and translate them into something beautiful. It relaxes our mind and helps us to focus better.

Question 5.
What is your friend’s hobby? How does it differ from yours? Discuss and write.
Answer:
My hobby is cooking. It is a life skill. My friend’s bobby is gardening.

Speaking

Role-play the conversation in the class.
Continue this conversation with your friends atleast five to ten dialogues each.
Answer:
Naveen: Hi Ajay! How are you?
Ajay: I am Fine, and what about you?
Naveen: I am fine too. What are you doing in this early morning?
Ajay: I am taking care of my garden. The garden is running to weeds.
Naveen: oh! What a beautiful garden it is !
Ajay: It is beautiful as I work every day in it.
Naveen: Who helps you in your work?
Ajay: I have done this. It is my and who I have cultivated vegetable hobby.
Naveen: Oh! thats nice. How long have you been doing this?
Ajay: I have been doing this for three years.
Naveen: What vegetables are grown here?
Ajay: Tomatoes, Chillies, Brinjals, Lady’s fingers, and some greens.
Naveen: Wonderful! You have a mini vegetable market here.
Ajay: Of course, My mother uses these vegetables for cooking
Naveen: It is a very good idea.
Ajay: Yes it is! You benefit a lot by this.
Naveen: It is good for our health because we don’t use any chemicals here.
Ajay: Exactly! Gardening also develops a green atmosphere and the environment.
Naveen: It’s interesting. Let me also do this.

Reading (Tongue Twisters)

• I Scream. I scream you scream, let’s all scream, for ice cream!
• No need to light a night-light on a light night like tonight.
• Green glass globes glow greenly.
• She sells seashells on the seashore.
• He would chuck, he would, as much as he could.

Writing

Using the given letter as a model, write a complaining letter on any one of the topics given below.

a) To the Police commissioner about the noise caused by the loudspeaker in a music shop near your school.
Answer:

Chennai,
3rd March 2019.

From
Adiya
VIII Std ‘A’,
Government Higher Secondary School,
Chennai – 600 Oil.

To
The Police Commissioner,
Perambur, Chennai – 600 Oil.

Sir,
Sub: Complaint about the noise caused by the loudspeaker regarding.
We, the students of Government Higher Secondary School, are getting distracted by the noise caused by the loudspeaker in a music shop near our school. We are unable to listen to the lessons taught and concentrate on our studies. The noise from the loudspeaker is unbearable.

We request you to look into the matter immediately and take the necessary action to stop this problem.

Thank you,
Yours faithfully,

Aditya
VIII Std ‘A’ Section.
Address on the Cover:

To
The Police Commissioner, Perambur,
Chennai – 600011.

b) To the Postmaster General on non – receipt of book parcel.
Answer:
Tiruchirappalli,
10th May 2019.

From
G. Alex
School Pupil Leader,
SSR Higher Secondary School,
Tiruchirappalli – 620 001

To
The Postmaster General
Sangiliyandapuram,
Tiruchirappalli – 620 001.

Sir,
Sub: Complaint on the non-receipt of a book parcel – reg.

I had ordered some books from a publishing company, two weeks back. When I contacted the publishing company, they told us that they had sent us the books in a registered parcel two days back. But so far, we have not received the parcel.

You are requested to enquire into the matter and do the needful.

Thank you,
Yours faithfully,
G. Alex.

Address on the Cover:

To
The Postmaster General
Sangiliyandapuram,
Tiruchirappalli – 620 001,

c) To the Transport Manager, TNSTC on the non- stopping of buses near your school bus stop.
Answer:
Chennai,
4th June 2019.

From
Manisha. K VIII Std ‘B’,
Anitha Hr. Sec. School,
Vepery, Chennai – 600 007.

To
The Transport Manager,
TNSTC,
Chennai – 600 002.

Sir,
Sub: Non-stopping of buses near the school bus stop – reg.

This is to bring to your notice that TNSTC buses do not stop near our school bus stop. The students are finding it difficult to commute to school. They had to walk a long distance to come to school and then go home from school. It takes a lot of time for all of us.

We request you to kindly look ‘ into this matter and take necessary action for the benefit of our school students.

Thank you,

Yours faithfully,
Manisha. K

Address on the Cover:

To
The Transport Manager, TNSTC,
Chennai – 600 002.

Grammar

A. NOUN
a. Pick out the words from the list and put them appropriately on the table.

David, Madurai, parrot, happiness, book, thought, green, tiger, Trichy, computer, Narayanan, fan, sister, Kalam, woman, pencil, Bharthi, problem, brother, Mumbai, swan, elephant, Vijay, solution, success, school, convent.
Answer:

Person	Place	Animals	Things	Ideas/feelings
David	Madurai	parrot	book	happiness
Narayanan	Trichy	tiger	computer	thought
sister	Mumbai	swan	fan	green
Kalam	school	elephant	pencil	problem
woman	convent			solution
Bharthi				success
brother
Vijay

b. Write the common noun for the following proper nouns.
Answer:

No.	Proper Noun	Common Noun
1.	Priya	girl
2.	Saran	boy
3.	Tiger	animal
4.	India	country
5.	Peacock	Bird
6.	Coimbatore	place
7.	Lion	animal
8.	Dove	bird
9.	Kabilan	boy
10.	Saleema	girl

c. Read the following paragraph and pick out the different types of nouns and put them in the table.
A hobby is an activity we enjoy doing in our free time. It keeps us busy in our leisure time. People choose their hobbies on the basis of their interests and personality. Do you know what Dr. A. P. J. Abdul Kalam’s hobby was?
Answer:

Proper Noun	Common Noun	Collective Noun	Abstract Noun
Dr. A.P.J. Abdul Kalam	Hobby	activity	personality
	people	leisure	interests
			time
			busy

d. Fill in the blanks.
( flock, swarm, bunch, herd)

1. a______ of cows.
2. a______ of birds.
3. a______ of ants.
4. a______ of grapes.

Answer:

1. herd
2. flock
3. swarm
4. bunch

B.pronoun

a. Complete with the subject personal pronoun.

1. My name is Preethi. (Preethi)______ am a student and this is my family.
2. My mom’s name is princy a housewife.
3. Albert is my dad. (My dad)
4. Benjamin is my brother. (Benjamin)
5. (Preethi and Benjamin) are twins.

Answer:

1. I
2. She
3. He
4. He
5. They

b. Change the underlined words with the correct pronoun.

(they, he, him, it, she )
Question 1.
I saw Mr. Balu this morning and gave Mr. Balu my homework.
Answer:
I saw Mr. Balu this morning and gave him my homework.

Question 2.
Muthu likes computer games but he doesn’t play computer games very often.
Answer:
Muthu likes computer games but he doesn’t play it very often.

Question 3.
My aunt lives in Trichy but, my aunt often comes to visit my family.
Answer:
My aunt lives in Trichy but, she often comes to visit my family.

Question 4.
Neil Armstrong was born in 1930. Neil Armstrong landed on the moon in 1969.
Answer:
Neil Armstrong was born in 1930. He landed on the moon in 1969.

C. Change the possessive adjective into a possessive pronoun.

1. This is my pen.
2. She missed her purse.
3. Her speech is nice.
4. My book is new.
5. My bike is costly.

Answer:

1. This pen is mine.
2. The missed purse is hers.
3. The nice speech is hers.
4. The new book is mine.
5. The costly bike is mine.

C- VERB
List out the transitive and intransitive verbs in the following words.
(come, buy, teach, learn, arrive, sing, run, draw, jump, write )
Answer:

Transitive	Intransitive
buy	come
teach	arrive
learn	run
sing	jump
draw
write

Hobby – Turns A Successful Career Additional Questions

I. Choose the correct Synonyms for the italic word.

Question 1.
We do it in our spare time.
(a) busy time
(b) down time
(c) free time
(d) work time
Answer:
(c) free time

Question 2.
Writing is just a hobby for me, yet it is very beneficial.
(a) useful
(b) harmful
(c) wonderful
(d) frightful
Answer:
(a) useful

Question 3.
He was very impressed with Ajay’s work.
(a) disappointed
(b) bored
(c) lazy
(d) amazed
Answer:
(d) amazed

Question 4.
I was unable to describe them accurately.
(a) exactly
(b) definitely
(c) commonly
(d) suddenly
Answer:
(a) exactly

Question 5.
Soon he became an accomplished artist.
(a) finished
(b) unskilled
(c) fulfilled
(d) worked
Answer:
(a) finished

Question 6.
Ajay exhibited his work and sold 144 paintings.
(a) closed
(b) showed
(c) hid
(d) conceded
Answer:
(b) showed

Question 7.
Her passion for carom took her to the level of world champion.
(a) desire
(b) dread
(c) enmity
(d) honour
Answer:
(a) desire

Question 8.
Her father’s dream shattered.
(a) combined
(b) built
(c) destroyed
(d) fulfilled
Answer:
(c) destroyed

Question 9.
The confidence led her to win the Asia Cup.
(a) peace
(b) love
(c) hatred
(d) hope
Answer:
(d) hope

Question 10.
His parents tried several hospitals to cure his impairment.
(a) ability
(b) disability
(c) failure
(d) destruction
Answer:
(b) disability

II. Choose the correct Antonyms for the italic word.

Question 1.
Mani found it difficult to read and write.
(a) hard
(b) easy
(c) sharp
(d) bad
Answer:
(b) easy

Question 2.
Same is the case with me.
(a) fair
(b) Identical
(c) different
(d) ordinary
Answer:
(c) different

Question 3.
I sit with small children and entertain them.
(a) amuse
(b) cheer
(c) fascinate
(d) Annoy
Answer:
(d) Annoy

Question 4.
It opens my mind to think with a broader mind,
(a) narrower
(b) outer
(c) wider
(d) smother
Answer:
(a) narrower

Question 5.
But that treatment left Ajay permanently deaf.
(a) temporarily
(b) abruptly
(c) suddenly
(d) accurately
Answer:
(a) temporarily

Question 6.
Asha Devi taught Ajay the old and dying technique of Traditional Indian Miniature Painting’
(a) conventional
(b) modern
(c) standard
(d) normal
Answer:
(b) modern

Question 7.
His family did not encourage and support him.
(a) support
(b) motivate
(c) entertain
(d) discourage
Answer:
(d) discourage

Question 8.
Hard working people always chase their dream.
(a) follow
(b) run
(c) walk
(d) retreat
Answer:
(d) retreat

Question 9.
Her victories included the Asia Cup and SAARC Cup.
(a) added
(b) excluded
(c) existed
(d) outcast
Answer:
(b) excluded

Question 10.
She won the Championship beating a former world champion Rashmi kumari in the final.
(a) previous
(b) first
(c) future
(d) beginner
Answer:
(c) future

III. Choose the Correct Answer (MCQ)

Question 1.
Mani found it difficult to read and ______
(a) learn
(b) write
(c) speak
(d) talk
Answer:
(b) write

Question 2.
Hobbies are the_____that we do in our spare time.
(a) projects
(b) tasks
(c) activities
(d) assignments
Answer:
(c) activities

Question 3.
When I write thines. my_____is at its best.
(a) handwriting
(b) creations
(c) work
(d) imagination
Answer:
(d) imagination

Question 4.
It makes me more_____about mv life.
(a) passionate
(b) pleasant
(c) inquisitive
(d) accurate
Answer:
(a) passionate

Question 5.
Aiay started to paint_____
(a) sorrowful
(b) vaguely
(c) happily
(d) regretful
Answer:
(C) happily

Question 6.
Ajay turned his hobby into a _____career.
(a) medical
(b) successful
(c) vocational
(d)good
Answer:
(b) successful

Question 7.
Irudavarai was determined to achieve his_____
(a) preference
(b) request
(c) dream
(d) fancy
Answer:
(c) dream

Question 8.
The confidence led Ilavazhaki to win her_____
(a) victories
(b) demands
(c) claims
(d) conditions
Answer:
(a) victories

Question 9.
Ilavazhaki’s father was a_____level champion in carom
(a) world
(b) state
(c) International
(d) district
Answer:
(d) district

Question 10.
Her passion for_____took her to the level of world champion.
(a) carom
(b) cricket
(c) hockey
(d) kabaddi
Answer:
(a) carom

IV. Very Short Questions with Answers.

Question 1.
What is a hobby?
Answer:
The hobby is an activity we enjoy doing in our free time.

Question 2.
What is your hobby?
Answer:
My hobbies are Reading and Singing.

Question 3.
What does the term ‘pastime’ mean?
Answer:
The term ‘pastime’ means, an activity regularly done for enjoying rather than work. To pass the leisure time relaxingly.

Question 4.
Who is the best friend for a reader while travelling on a train?
Answer:
Books are the best friends for a reader while travelling alone on a train.

Question 5.
Name any two socially useful hobbies.
Answer:
Cooking and Painting.

V. Short Questions with Answers.

Question 1.
What was Mani’s problem?
Answer:
Mani had difficulty in reading and writing.

Question 2.
What did his Grandfather do?
Answer:
His grandfather helped him by reading bedtime stories.

Question 3.
Why does Mani say that writing is a unique hobby?
Answer:
Mani says that writing is unique because most people don’t want to exhaust themselves in their spare time.

Question 4.
What did Ajay’s father buy him?
Answer:
Ajay’s father bought him a paint set to play with.

Question 5.
What was Ajay gifted with?
Answer:
Ajay was gifted with the practice of creating images on a grain of rice.

Question 6.
What was Irudayaraj?
Answer:
Irudayaraj was a fish-cart driver, who used to transport materials like pipes to make a living.

Question 7.
Who did Ilavazhagi beat in the women’s singles?
Answer:
Ilavazhagi beat P. Nirmala in the finals of the Carom World Championship with 25-11, 25-11 victory.

VI. Paragraph Question with Answer.

Question 1.
How beneficial was the hobby of writing to Mani?
Answer:
Mani felt the hobby of writing very beneficial to him. He felt writing is like painting his voice’. When he got free time, he picked a pen and a notebook. He started writing stories, poems, and sometimes about his feelings. His mind used to relax when he poured his thoughts as words. His imagination is at its best when he writes things. Writing opened his mind to think beyond the little things with a broader mind. It made him more passionate about his life and learn new things. It also helped him in school too and made him a good writer.

Hobby – Turns A Successful Career Grammar Additional

I. Guide proper names for the common nouns :

Eg: Butter – Amul

1. Pencil – ________
2. Ball Pen – ________
3. Hospital – ________
4. Soft Drink – ________
5. Car – ________

Answer:

1. Apsara
2. Reynolds
3. Apollo
4. Pepsi
5. Toyota

II. Collective nouns:

1. ______ of bees.
2. ______ of keys.
3. ______ of cards.
4. ______ of players
5. ______ of books.

Answer:

1. Swarm
2. Bunch
3. Pack
4. Team
5. Libary

III.Form abstract nouns from these adjectives:

Eg: Soft – Softness

1. Cruel – ________
2. Polite – ________
3. Honest – ________
4. Beautiful – ________
5. Lazy – ________

Answer:

1. Cruelty
2. Politeness
3. Honesty
4. Beauty
5. Laziness

IV. Use singular pronouns and fill in the blanks :

1. _____gave_____a gift.
2. _____went out when_____rained.
3. _____do not want_____to steal things.
4. When he met_____I gave_____an advice.

Answer:

1. He, her
2. She, it
3. I, you
4. me, him

V. Use Plural pronouns and fill in the blanks:

1. Raj and Rahul are brothers and live in the same house.
2. They are my friends, I meet in the park.
3. My father and I went to the mall, but it was closed.
4. We won the match. Everybody congratulated

Answer:

1. They
2. Them
3. we
4. us

VI. Fill in the blanks with possessive pronouns:

1. The travellers brought a lot of goods. The goods were _______
2. A man was riding a motor bike. The bike was _______
3. She stitched the torn dresses from her cupboard. The dresses were _______
4. The certificates belong to me. They are _______
5. This classroom belongs to us. It is certainly _______
6. The photos look so good. They are _______
7. The dog has a puppy. It is _______

Answer:

1. theirs
2. his
3. hers
4. mine
5. ours
6. yours
7. its

VII. Use the simple present tense of the verb in the bracket and fill the blanks:

1. Look! He (fly)______the kite high
2. Yesterday he went by car, but today he (go)on a cycle.
3. If he (pass)the exam, he will get a good job.
4. Mohan will jump in joy if he (see) you coming.
5. If Vijay (run)fast, he will be able to catch the train.

Answer:

1. Flies
2. goes
3. passes
4. sees
5. runs

VIII. Use simple past tense of the verb in the bracket and fill the blanks:

1. I (buy) this coat from the boutique.
2. Dad (leave)for office an hour ago.
3. My sister (write) an essay that was published in the magazine.
4. We (come)home, (take) some rest, and (spend)time playing cards.

Answer:

1. bought
2. left
3. wrote
4. came / took /spent

IX. Use simple future tense of the verb in the bracket and fill the blanks:

1. I (leave)_______for Paris next week.
2. We (party)_______at the Taj on my birthday.
3. You (move)_______to Delhi next month on promotion.
4. Ten students failed the exam. They (write)_______their exams again from Monday.
5. The sun rise was at 6 a.m. It (set)_______at 5.45 p.m this evening.

Answer:

1. will leave
2. will party
3. will move
4. will write
5. will set

X. Insert the correct form of infinitive using the given verbs in the brackets:

1. It would be nice to______(meet) Dr. Abdul Kalam again.
2. To______(cook) this recipe, you will need 500 grams of minced meet
3. Ravi wants to______(take) raji to the amusement park
4. I need to______(wash) my hands before I have my food.
5. We discussed whether to______(invite) Rajiv or not.

Answer:

1. meet
2. cook
3. take
4. wash
5. invite

XI. Fill in the blanks with suitable Gerund (verb+ing):

1. ______(meet) is a periodic event.
2. She loves______(eat) ice creams.
3. I like______(celebrate) festivals.
4. ______(wait) is torturing.
5. He was awarded for______(run) fast
6. I achieve by______(work)
7. I don’t like______(play) cards
8. ______(cook) is my hobby.
9. Do you enjoy______(swim)
10. He loves______(sing) songs.

Answera:

1. Meeting
2. eating
3. celebrating
4. Waiting
5. running
6. working
7. playing
8. Cooking
9. swimming
10. singing

Section – I

Hobby – Turns A Successful Career Summary

A boy named Mani had difficulty in reading and writing. His grandfather noticed this and helped  him by reading bedtime stories. Gradually, Mani started reading on his own. After some days, he also  started writing his own stories. In this section, he wrote an essay on hobbies.

Hobbies are the activities done during our free time. They are special to people. Some people  like gardening, a few like photography and many have book reading as their hobby. Hobbies help people to learn new things, relax after a tiring day, and energies our body and mind. Mani’s hobby was writing. Though many people feel writing is tough, he felt ‘writing’ was like ‘painting our voice’.

When he wrote things, he imagined a world with magic and magicians. Writing opened his mind to think beyond the little things with a broader mind. It made him learn new things and enjoy every moment of his life. His hobby of writing helped him in school too. He could describe things better in  his stories. He had no problem in writing English essays or stories, as it was his hobby. Thus it is important to always learn and improve Our hobbies.

Section – II

Ajay Kumar Garg, a very talented young artist lived in Jaipur, India. He suffered an injury at the age of three and became deaf. His parents tried many hospitals to cure his disability. But it was useless. Ajay used to paint on the walls and floors. Realizing his interest in painting, a court artist of Dholpur, Shri Sua Lai started educating him in painting. After completing his education, Ajay became an apprentice of Asha Devi, who taught him the ‘Miniature Painting’. Soon, Ajay became an ‘accomplished artist’ in the miniature style. He was gifted in the practice of creating images on a grain of rice.

Ajay exhibited his work and sold 144 out of 150 paintings displayed in the exhibition. His work became famous throughout India, the United States, and the United Kingdom. In 2004, he was awarded a national award of accomplishment from the Indian President, Dr. A.P.J. Abdul Kalam.

Ajay also involved in the welfare of people with hearing disabilities and gave them free training. He has been the executive member of’Deaf and Dumb’ Association in Rajasthan. One of Ajay’s current goals is to re-energize the dying art of traditional miniature Indian paintings. Thus, Ajay’s life shows us that we can turn our hobby into a successful career if we are ready to learn and develop our skills.

Section – III

Ilavazhagi achieved her dream amidst all odds. Her father, A. Irudayaraj, was a district level champion in Carom. His dream of achieving success in it, shattered, as he was not encouraged and supported by his family members. He determined to achieve his dream through his daughter. So he began coaching her and carried her to the local clubs frequently. She won the match against her father and that gave her the confidence in achieving success. Later she won the Asia Cup, the SAARC cup, and the world championship.

She lived with her family in a small one-room apartment in Vysarpadi, Chennai. This was not enough to accommodate her family and to preserve her trophies. She has been playing for the state for almost 14 years yet, she found it difficult to finance her trips for the matches.

Ilavazhagi is a member of the Thiruvallur District Carom Association. She also represented India at the Carom World Championships. She won the 2008 Indian National Carom Championship and became a world champion in the women’s singles after beating P. Nirmala in the finals with 25-11, 25-11 victory. Thus her passion for Carom took her to the level of world champion.

Samacheer Kalvi 8th English Prose

• The Nose-Jewel Book Back Answers
• Hobby – Turns A Successful Career Book Back Answers
• Sir Isaac Newton – The Ingenious Scientist Book Back Answers
• My Reminiscence Book Back Answers
• Being Safe Book Back Answers
• Friendship Book Back Answers
• Cyber Safety Book Back Answers

Students can Download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Question 1.
Factorise the following by taking out the common factor
(i) 18xy – 12yz
(ii) 9x⁵y³ + 6x³y² – 18x²y
(iii) x(b – 2c) + y(b – 2c)
(iv) (ax + ay) + (bx + by)
(v) 2x²(4x – 1) – 4x + 1
(vi) 3y(x – 2)² – 2(2 – x)
(vii) 6xy – 4y² + 12xy – 2yzx
(viii) a³ – 3a² + a – 3
(ix) 3y³ – 48y
(x) ab² – bc² – ab + c²
Solution:
(i) 18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)
Taking out the common factors 2, 3, y, we get
= 2 × 3 × y (3x – 2z) = 6y (3x- 2z)

i) 9x⁵y³ + 6x³y² – 18x²y = (3 × 3 × x² × x³ × y × y²) + (2 × 3 × x² × x × y × y)
Taking out the common factors 3, x², y, we get
= 3 × x² × y (3x³ y² + 2xy – 6)
= 3x²y (3x³ y² + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)
Taking out the binomial factor (b – 2c) from each term, we have
= (b – 2c)(x + y)

(iv) (ax + ay) + (bx + by)
Taking at ‘a’ from the first term and ‘b’ from the second term we have
(ax + ay) + (bx + by) = a (x + y) + b (x + y)
Now taking out the binomial factor (x + y) from each term
= (x + y)(a + b)

(v) 2x²(4x – 1) – 4x + 1
Taking out -1 from last two terms
2x² (4x – 1) – 4x + 1 = 2x² (4x – 1) – 1 (4x- 1)
Taking out the binomial factor 4x – 1, we get
= (4x – 1)(2x² – 1)

(vi) 3y(x – 2)² – 2(2 – x)
3y(x – 2)² – 2(2 – x) = 3y(x – 2)(x – 2)-2(-1) (x – 2) [∵ Taking out -1 from 2 – x]
= 3y (x – 2) (x – 2) + 2 (x – 2)
Taking out the binomial factor x – 2 from each term, we get
= (x – 2) [3y (x – 2) + 2]

(vii) 6xy – 4y² + 12xy – 2yzx
= 6xy + 12xy – 4y² – 2yzx [∵ Addition is commutative]
= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
Taking out 6 × x × y from first two terms and (-1) × 2 × y from last two terms we get
= 6 × x × y(1 + 2) + (-1)(2)y[2y + zx]
= 6xy (3) -2y(2y + zx)
= (2 × 3 × 3 × x × y) – 2xy (2y + zx)
Taking out 2y from two terms
= 2y (9x – (2y + zx)) = 2y (9x – 2y – xz)

(viii) a² – 3a² + a – 3 = a² (a – 3) + 1 (a – 3) [∵ Grouping the terms suitably]
= (a – 3) (a² + 1)

(ix) 3y² – 48y = 3 × y × y² – 3 × 16 × y
Taking out 3 × y = 3y (y² – 16) = 3y (y² – 4²)
Comparing y² – 4² with a² – b²
a = y, b = 4
a² – b² = (a + b) (a – b)
y² – 4² = (y + 4) (y – 4)
∴ 3y (y² – 16) = 3y (y + 4) (y – 4)

(x) ab² – bc² – ab + c²
Grouping suitably
ab² – bc² – ab + c² = b ((ab – c²) – 1(ab – c²)
Taking out the binomial factor ab – c² = (ab – c²) (b – 1)

Question 2.
Factorise the following expressions
(i) x² + 14x + 49
(ii) y² – 10y + 25
(iii) c² – 4c – 12
(iv) m² + m – 72
(v) 4x² – 8x + 3
Solution:
x² + 14x + 49 = x² + 14x + 72
Comparing with a² + 2ab + b² = (a + b)² we have a = x and b = 7
⇒ x² + 2(x) (7) + 7² = (x + 7)²
∴ x² + 14x + 49 = (x + 7)²

(ii) y² – 10y + 25 = y² – 10y + 5²
Comparing with a² – 2ab + b² = (a – b)² we get a = y ; b = 5
⇒ y² – 2(y) (5) + 5² = (y – 5)²
∴ y² – 10y + 25 = (y – 5)²

(iii) c² – 4c – 12
This is of the form ax² + bx + c
Where a = 1,b = – 4 c = – 12, x = c
Now the product ac = 1 × – 12 = – 12 and the sum b = -4

(iv) m² + m – 72

This is of the form ax² + bx + c
where a = 1, b = 1, c = -12
Product a × c = 1 × -72 = -72
Sum b = 1
The middle term m can be written as 9m – 8m
m² + m – 72 = m² + 9m – 8m – 12
= m (m + 9) – 8 (m + 9)
Taking out (m + 9)
= (m + 9) (m – 8)
∴ m² + m – 72 = (m + 9) (m – 8)

(v) 4x² – 8x + 3
This is of the form ax² + bx + c with a = 4 b = -8 c = 3
Product ac = 4 × 3 = 12
Sum b = -8

Question 3.
Factorize the following expressions using a³ + b³ = (a + b)(a² – ab + b²) identity
(i) h³ + k²
(ii) 2a³ + 16
(iii) x³y³ + 27
(iv) 64m³ + n³
(v) r⁴ + 27p³r
Solution:
(i) h³ + k³
Comparing h³ + k³ with a³ + b³ = (a + b)(a² – ab + b²) we have a = h, b = k
∴ h³ + k³ = (h + k)(h² – hk + k²)

(ii) 2a² + 16
(2 × a³) + (2 × 8) = 2(a³ + 8) = 2 (a³ + 2³)
∴ 2a³ + 16 = 2 (a³ + 2³)
Comparing with a³ + b³ we have a = a and b = 2
a³ + b³ = (a + b)(a² – ab + b²)
2(a³ + 2³) = 2[(a + 2) (a² – (a) (2) + 2²)] = 2[(a + 2) (a² – 2a + 4)]
2a³ +16 = 2 (a + 2) (a² – 2a + 4)

(iii) x³ y³ + 27 = (xy)³ + 3³
Comparing with a³ + b³ we have a = xy ;b = 3
a³ + b³ = (a + b) (a² – ab + b²)
(xy)³ + 3³ = (xy + 3) ((xy)² – (xy) (3) + 3²) = (xy + 3) (x²y² – 3xy + 9)
∴ x³y³ + 27 = (xy + 3)(x²y² – 3xy + 9)

(iv) 64m³ + n³ = (4³m³) + n³
= (4m)³ + n³
Comparing this with a³ + b³ we have a = 4m; b = n
a³ + b³ = (a + b) (a² – ab + b²)
(4m)³ + n³ = (4m + n) [(4m)2 – (4m) (n) + n2]
= (4m + n) [4²m² – 4mn + n²]
= (4m + n) [ 16m² – 4mn + n²]
64m³ + n³ = (4m + n) (16m² – 4mn + n²)

(v) r³ + 27p³r = r (r³ + 27p³) = r [r³ + 3³p³]
Comparing r³ + (3p)³ with a³ + b³ we have a = r; b = 3p
a³ + b³ = (a + b)(a² – ab + b²)
r[r³ + (3p)³] = r[(r + 3p)(r² – r(3p) + (3p)²)]
= r[(r+ 3p) (r² – 3rp + 3²p²]
= r(r + 3p) (r² – 3rp + 9p²)
r⁴ + 27p³r = r(r + 3p) (r² – 3rp + 9p²)

Question 4.
Factorize the following expressions using a³ – b³ = (a – b)(a² + ab + b²) identity
(i) y³ – 27
(ii) 3b³ – 192c³
(iii) -16y³ + 2x³
(iv) x³ y³ – 7³
(v) c³ – 27b³ a³
Solution:
(i) y³ – 27 = y³ – 3³
Comparing this with a³ – b³ , we have a = y and b = 3
a³ – b³ = (a + b) (a² + ab + b² )
y³ – 3³ = (y – 3)(y² + (y) (3) + 3² ) = (y – 3) (y² + 3y + 9)
y³ – 27 = (y – 3) (y² + 3y + 9)

(ii) 3b³ + 192c³ = (3 × b³) – (3 × 4 × 4 × 4 × c³) = 3(b³ – 4³ c³)
= 3(b³ – (4c)³ )
Comparing b³ – (4c)³ with a³ – b³ we have a = b and b = 4c
a³ – b³ = (a – b) (a² + ab + b²)
3(b³ – (4c)³) = 3[(b – 4c)(b² + (b)(4c) + (4c)²)]
= 3[(b – 4c)(b² + 4bc + 4² c²)]
3b³ – 192c³ = 3 [(b – 4c) (b² + 4bc + 16c²)]

(iii) -16y³ + 2x³ = 2x³ – 16y³ [∵ Addition is commatative]
= 2(x³ – 8y³) = 2(x³ – 23y³)
= 2(x³ – (2y)³)
Comparing x³ – (2y)³ with a³ – b³ we have a = x and b = 2y
a³ – b³ = (a – b)(a² + ab + b²)
2[x³ – (2y)³] = 2[(x – 2y) (x² + (x) (2y) + (2y)²)]
= 2[(x – 2y) (x² + 2xy + 2² y²)]
-16y³ + 2x³ = 2[(x – 2y)(x² + 2xy + 4y²)]

(iv) x³y³ – 7³ = (xy)³ – 7³
Comparing with a³ – b³ we have a = xy and b = 7
a³ – b³ = (a – b)(a² + ab + b²)
(xy)³ – 7³ = (xy – 7) ((xy)² + (xy) (7) + 7²)
x³y³ – 7³ = (xy – 7) (x²y² + 7xy + 49)

(v) c³ – 27 b³ a³ = c³ – 3³b³ a³ = c³ – (3ba)³
Comparing this with a³ – b³ we have a = x and b = 3ba
a³ – b³ = (a – b)(a² + ab + b²)
∴ c³ – (3ba)³ = (c – 3ba) (c² + (c) (3ba) + (3ba)²)
= (c – 3ba) (c² + 3bac + 3² b²a²)
c³ – 27b³a³ = (c – 3ab) (c² + 3bac + 9a²b²)

Students can Download Maths Chapter 4 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions

Additional Questions And Answers

Very Short Answers [2 Marks]

Question 1.
In the given figure if ∠A = ∠C then prove that ∆AOB ~ ∆COD.

Solution:
In triangles ∆AOB and ∆COD
∠A = ∠C (Y given)
∠AOB = ∠COD [∵ Vertically opposite angles]
∠ABO = ∠CDO [Remaining angles of ∆AOB and ∆COD]
∴ ∆AOB ~ ∆COD [∵ AAA similarity]
∵ ∆AOB ~ ∆COD [∵ AAA similarity]

Question 2.
In the figure AB ⊥ BC and DE ⊥ AC prove that ∆ABC ~ ∆AED.

Solution:
In triangles ∆ABC and ∆AED
∠ABC = ∠AED = 90°
∠BAC = ∠EAD [Each equal to A]
∠ADE = ∠ACB [∵ Remaining angles]
∴ By AAA criteria of similarity ∆ABC ~ ∆AED

Short Answers [3 Marks]

Question 1.
In the figure with respect to ABEP and CPD prove that BP × PD = EP × PC.

Solution:
Proof:
In ∆EPB and ∆DPC
∠PEB = ∠PDC = 90° [given]
∠EPB = ∠DPC [Vertically opposite angles]
∠EPB = ∠PCD [∵ Remaining angles]
Thus,

Long Answers [5 Marks]

Question 1.
P and Q are points on sides AB and AC respectively of ∆ABC. If AP = 3 cm PB = 6cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.
Solution:

AB = AP + PB
= 3 + 6 cm = 9 cm
AC = AQ + QC = 510 cm = 15

Students can Download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions

Construct the quadrilaterals with the following measurements and also find their area.

Question 1.
ABCD, AB = 5 cm, BC 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.
Solution:
Given AB = 5 cm,
BC = 4.5 cm,
CD = 3.8 cm,
DA = 4.4 cm,
AC = 6.2 cm

Steps:
1. Draw a line segment AB = 5 cm
2. With A and B as centers drawn arcs of radii 6.2 cm and 4.5cm respectively and let them cut at C.
3. Joined AC and BC.
4. With A and C as centrers drawn arcs of radii 4.4cm and 3.8 cm respectively and let them at D.
5. Joined AD and CD.
6. ABCD is the required quadrilateral.
Calculation of Area:

Question 2.
KITE, KI = 5.4 cm, IT = 4.6 cm, TE= 4.5 cm, KE = 4.8 cm and IE = 6 cm.
Solution:
Given, KI = 5.4 cm,
IT = 4.6 cm,
TE= 4.5 cm,
KE = 4.8 cm,
IE = 6 cm.

Steps:
1. Draw a line segment KI = 5.4 cm
2. With K and I as centers drawn arcs of radii 4.8 cm and 6 cm respectively and let them cut at E.
3. Joined KE and IE.
4. With E and I as centers, drawn arcs of radius 4.5cm and 4.6 cm respectively and let them cut at T.
5. Joined ET and IT.
6. KITE is the required quadrilateral.
Calculation of Area:

Question 3.
PLAY, PL = 7 cm, LA = 6 cm, AY= 6 cm, PA = 8 cm and LY = 7 cm.
Solution:
Given PL = 7 cm,
LA = 6 cm,
AY= 6 cm,
PA = 8 cm,
LY = 7 cm

Steps:
1. Draw a line segment PL = 7 cm
2. With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A.
3. Joined PA and LA.
4. With L and A as centers, drawn arcs of radii 7 cm and 6 cm respectively and let them cut at Y.
5. Joined LY, PY and AY.
6. PLAY is the required quadrilateral.
Calculation of Area:

Question 4.
LIKE, LI = 4.2 cm, IK = 7 cm, KE = 5 cm, LK = 6 cm and IE = 8 cm.
Solution:
LI = 4.2 cm,
IK = 7 cm,
KE = 5 cm,
LK = 6 cm,
IE = 8 cm

Steps:
1. Draw a line segment LI = 4.2 cm
2. With L and I as centers, drawn arcs of radii 6 cm and 7 cm respectively and let them cut at K.
3. Joined LK and IK.
4. With I and K as centers, drawn arcs of radius 8 cm and 5 cm respectively and let them cut at E.
5. Joined LE, IE and KE.
6. LIKE is the required quadrilateral.
Calculation of Area:

Question 5.
PQRS, PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q =120° .
Solution:
PQ = QR = 3.5 cm,
RS = 5.2 cm,
SP = 5.3 cm ,
∠Q =120°

Steps:
1. Draw a line segment PQ = 3.5 cm
2. Made ∠Q = 120°. Drawn the ray QX.
3. With Q as centre drawn an arc of radius 3.5 cm. Let it cut the ray QX at R.
4. With R and P as centres drawn arcs of radii 5.2cm and 5.5 cm respectively and let them cut at S.
5. Joined PS and RS.
6. PQRS is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral PQRS = 18 cm²

Question 6.
EASY, EA = 6 cm, AS = 4 cm, SY = 5 cm, EY = 4.5 cm and ∠E = 90°.
Solution:
EA = 6 cm,
AS = 4 cm,
SY = 5 cm,
EY = 4.5 cm,
∠E = 90°

1. Drawn a line segment EA = 6 cm
2. Made ∠E = 90°. From E drawn the ray EX.
3. With E as center drawn an arc of 4.5 cm radius. Let of cut the ray EX at Y.
4. With A and Y as centres drawn arcs of radii 4 cm and 5 cm respectively and let them cut at S.
5. Joined AS and YS.
6. EASY is the required quadrilateral.
Calculation of Area:

∴ Area of the quadrilateral = 22.87 cm²

Question 7.
MIND, MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°.
Solution:
MI = 3.6 cm,
ND = 4 cm,
MD= 4 cm,
∠M = 50°,
∠D = 100°

1. Draw a line segment MI = 3.6 cm
2. At M on MI made an angle ∠IMX = 50°
3. Drawn an arc with center M and radius 4 cm let it cut MX it D
4. At D on DM made an angle ∠MDY = 100°
5. With I as center drawn an arc of radius 4 cm, let it cut DY at N.
6. Joined DN and IN.
7. MIND is the required quadrilateral.
Calculation of Area:

Area of the quadrilateral = 9.6 cm²

Question 8.
WORK, WO = 9 cm, OR = 6 cm, RK = 5 cm, ∠O = 100° and ∠R = 60°.
Solution:
WO = 9 cm,
OR = 6 cm,
RK = 5 cm,
∠O = 100°,
∠R = 60°

Steps:
1. Drawn a line segment WO = 9 cm
2. At O on WO made an angle ∠WOR = 100° and drawn the ray OX.
3. Drawn an arc of radius 6 cm with center O. Let it intersect OX at R.
4. At R on OR, made ∠ORY = 60°, and drawn the ray RY.
5. With center R drawn an arc of radius 5 cm, let it intersect RY at K.
6. Joined WK.
7. WORK is the required quadrilateral.
Calculation of Area:

Area of the quadrilateral = 31.59 cm²

Question 9.
AGRI, AG = 4.5 cm, GR = 3.8 cm, ∠A = 90°, ∠G = 110° and ∠R = 90°.
Solution:
AG = 4.5 cm,
GR = 3.8 cm,
∠A = 90°,
∠G = 110°,
∠R = 90°

1. Draw a line segment AG = 4.5 cm
2. At G on AG made ∠AGX =110°
3. With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
4. At R on GR made ∠GRZ = 90°
5. At A on AG made ∠GAY = 90°
6. AY and RZ meet at I.
7. AGRI is the required quadrilateral.
Calculation of Area:

Question 10.
YOGA, YO = 6 cm, OG = 6 cm, ∠O = 55°, ∠G = 55° and ∠A = 55°.
Solution:
YO = 6 cm,
OG = 6 cm,
∠O = 55°,
∠G = 55°,
∠A = 55°

Steps:
1. Draw a line segment OG = 6 cm
2. At G on DG made an angle ∠OGY = 55°
3. AT G on GO made ∠GOX = 55°.
4. GY and OX meet cut A.
5. At A on OA made ∠OAZ = 55°
6. Drawn an arc of radius 6 cm with center O. It cut AZ at Y. Joined OY.
7. YOGA is the required quadrilateral.
Calculation of Area:

Area of the quadrilateral YOGA = 28.08 cm²

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions

Try this Page No. 35

Question 1.
\(\frac{22}{7}\) and 3.14 are rational numbers. Is ‘π’ a rational number? Why?
Solution:
\(\frac{22}{7}\) and 3.14 are rational numbers n has non-terminating and non -repeating decimal expansion. So it is not a rational number. It is an irrational number.

Try this Page No. 38

Question 1.
The given circular figure is divided into six equal parts. Can we call the parts as sectors? Why?

Solution:
No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle.
Here the boundaries are not radii.

Try these Page No. 38

Question 1.
Fill the central angle of the shaded sector (each circle is divided into equal sectors)

Try this Page No. 44

Question 1.
If the radius of a circle is doubled, what will the area of the new circle so formed?
Solution:
If r = 2r₁ ⇒ Area of the circle = πr² = π(2r₁)² = π4r₁² = 4πr₁²
Area = 4 × old area.

Try this Page No. 49

Question 1.
All the sides of a rhombus are equal. Is it a regular polygon?
Solution:
For a regular polygon all sides and all the angles must be equal. But in a rhombus all the
sides are equal. But all the angles are not equal
∴ It is not a regular polygon.

Try this Page No. 53

Question 1.
In the above example split the given mat as into two trapeziums and verify your answer.
Solution:
Area of the mat = Area of I trapezium + Area of II trapezium

∴ Cost per sq.feet = ₹ 20
Cost for 28 sq. feet = ₹ 20 × 28 = ₹ 560
∴ Total cost for the entire mat = ₹ 560
Both the answers are the same.

Try these Page No. 54

Question 1.
Show that the area of the unshaded regions in each of the squares of side ‘a’ units are the same in all the cases given below.

Solution:

Question 2.
If π = \(\frac{22}{7}\), show that the area of the unshaded part of a square of side ‘a’ units is approximately \(\frac{3}{7}\) a² sq. units and that of the shaded part is approximately \(\frac{4}{7}\) a² sq. units for the given figure.

Solution:

Try this Page No. 57

Question 1.
List out atleast three objects in each category which are in the shape of cube, cuboid,
cylinder, cone and sphere.
Solution:
(i) Cube – dice, building blocks, jewel box.
(ii) Cuboid – books, bricks, containers.
(iii) Cylinder – candles, electric tube, water pipe.
(iv) Cone – Funnel, cap, ice cream cone
(v) Sphere – ball, beads, lemon.

Try this Page No. 58

Question 1.
Tabulate the number of faces(F), vertices(V) and edges(E) for the following polyhedron. Also find F + V – E
Solution:

From the table F + V – E = 2 for all the solid shapes.

Try this Page No. 58

Question 1.
Find the area of the given nets.

Solution:
Area = 6 × Area of a square of side 6 cm
= 6 × (6 × 6) cm²
= 216 cm²
(ii) Area = Area of 2 rectangles of side (8 × 6) cm² + Area of 2 rectangles of side (8 × 4) cm² + Area of 2 rectangles of side (6 × 4) cm²
= (8 × 6) + (8 × 4) + (6 × 4)cm²
= 48 + 32 + 24 cm²
= 104 cm²

Students can Download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

Question 1.
Expand
(i) (3m + 5)²
(ii) (5p – 1)²
(iii) (2n – 1)(2n + 3)
(iv) 4p² – 25q²
Solution:
(i) (3m + 5)²
Comparing (3m + 5)² with (a + b)² we have a = 3m and b = 5
(a + b)² = a² + 2 ab + b²
(3m + 5)² = (3m)² + 2 (3m) (5) + 5²
= 3²m² + 30m + 25 = 9m² + 30m +25

(ii) (5p – 1)²
Comparing (5p – 1)² with (a – b)² we have a = 5p and b = 1
(a – b)² = a² – 2ab + b²
(5p – 1)² = (5p)² – 2 (5p) (1) + 1²
= 5²p² – 10p + 1 = 25p² – 10p + 1

(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3
(x + a) (x + b) = x² + (a + b)x + ab
(2n +(- 1)) (2n + 3) = (2n)² + (-1 + 3)2n + (-1) (3)
= 2²n² + 2 (2n) – 3 = 4n² + 4n – 3

(iv) 4p² – 25q² = (2p)² – (5q)²
Comparing (2p)² – (5q)² with a² – b² we have a = 2p and b = 5q
(a² – b²) = (a + b)(a – b) = (2p + 5q) (2p – 5q)

Question 2.
Expand
(i) (3 + m)³
(ii) (2a + 5)³
(iii) (3p + 4q)³
(iv) (52)³
(v) (104)³
Solution:
(i) (3 + m)³
Comparing (3 + m)³ with (a + b)³ we have a = 3; b = m
(a + b)³ = a² + 3a²b + 3 ab² + b³
(3 + m)³ = 3³ + 3(3)² (m) + 3 (3) m² + m³
= 27 + 27m + 9m² + m³ = m³ + 9 m² + 27m + 27

(ii) (2a + 5)³
Comparing (2a + 5)³ with (a + b)³ we have a = 2a, b = 5
(a + b)³ = a³ + 3a²b + 3ab² + b³ = (2a)³ + 3(2a)² 5 + 3 (2a) 5² + 5³
= 2³a³ + 3(2²a²) 5 + 6a (25) + 125
= 8a³+ 60a² + 150a + 125

(iii) (3p + 4q)³
Comparing (3p + 4q)³ with (a + b)³ we have a = 3p and b = 4q
(a + b) ³ = a³ + 3a²b + 3ab² + b³
(3p + 4q)³ = (3p)³ + 3(3p)² (4q) + 3(3p)(4q)² + (4q)³
= 3³p³ +3 (9p²) (4q) + 9p (16q²) + 43q³
= 27p³ + 108p²q + 144pq² + 64q³

(iv) (52)³ = (50 + 2)³
Comparing (50 + 2)³ with (a + b)³ we have a = 50 and b = 2
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(50 + 2)³ = 50³ + 3 (50)²2 + 3 (50)(2)² + 2³
52³ = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
52³ = 1,40,608

(v) (104)³ = (100 + 4)³
Comparing (100 + 4)³ with (a + b)³ we have a = 100 and b = 4
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(100 + 4)³ = (100)³ + 3 (100)² (4) + 3 (100) (4)² + (4)³
= 10,00,000 + 3(10000) 4 + 300 (16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

Question 3.
Expand
(i) (5 – x)³
(ii) (2x – 4y)³
(iii) (ab – c)³
(iv) (48)³
(v) (97xy)³
Solution:
(i) (5 – x)³
Comparing (5 – x)³ with (a – b)³ we have a = 5 and b = x
(a – b)³ = a³ – 3a²b + 3ab² – b³
(5 – x)³ = 5³ – 3 (5)² (x) + 3(5)(x²) – x³
= 125 – 3(25)(x) + 15x² – x³ = 125

(ii) (2x – 4y)³
Comparing (2x – 4y)³ with (a – b)³ we have a = 2x and b = 4y
(a – b)³ = a³ – 3a²b + 3ab³ – b³
(2x – 4y)³ = (2x)³ – 3(2x)² (4y) + 3(2x) (4y)² – (4y)³
= 2³x³ – 3(2²x²) (4y) + 3(2x) (4²y²) – (4³y³)
= 8x³ – 48x²y + 96xy² – 64y³

(iii) (ab – c)³
Comparing (ab – c)³ with (a – b)³ we have a = ab and b = c
(a – b)³ = a³ – 3a²b + 3ab² – b³
(ab – c)³ = (ab)³ – 3 (ab)² c + 3 ab (c)² – c³
= a³b³ – 3(a²b²) c + 3abc² – c³
= a³b³ – 3a²b² c + 3abc² – c³

(iv) (48)³ = (50 – 2)³
Comparing (50 – 2)³ with (a – b)³ we have a = 50 and b = 2
(a – b)³ = a³ – 3a²b + 3ab² – b³
(50 – 2)³ = (50)³ – 3(50)²(2) + 3 (50)(2)² – 2³
= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592
= 1,10,592

(v) (97xy)³
= 97³ x³ y³ = (100 – 3)³ x³y³
Comparing (100 – 3)³ with (a – b)³ we have a = 100, b = 3
(a – b)³ = a³ – 3a²b + 3ab² – b³
(100 – 3)³ = (100)³ – 3(100)² (3) + 3 (100)(3)² – 3³
97³ = 10,00,000 – 90000 + 2700 – 27
97³ = 910000 + 2673
97³ = 912673
97x³y³ = 912673x³y³

Question 4.
Simplify (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² (ab + bc + ca) x + abc
= (5y)³ + (1 + 2 + 3) (5y)² + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 5³y³ + 6(5²y²) + (2 + 6 + 3)5y + 6
= 125³ + 150y² + 55y + 6

(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + be + ca) x + abc
= p³ + (-2 + 1 + (-4))p² + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)
= p³ + (-5 )p² + (-2 + (-4) + 8)p + 8
= p³ – 5p² + 2p + 8

Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
Given side of the cube = (x + 1) cm
Volume of the cube = (side)³ cubic units = (x + 1)³ cm³
We have (a + b)³ = (a³ + 3a²b + 3ab² + b³) cm³
(x + 1)³ = (x³ + 3x² (1) + 3x (1)² + 1³) cm³
Volume = (x³ + 3x² + 3x + 1) cm³

Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).
Solution:
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units³
= (x + 2) (x – 1) (x – 3) units³
We have (x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + bc+ ca)x + abc
∴ (x + 2)(x – 1) (x – 3) = x³ + (2 – 1 – 3)x² + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)
x³ – 2x² + (-2 + 3 – 6)x + 6
Volume = x³ – 2x² – 5x + 6 units³

Students can Download Maths Chapter 2 Measurements Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is _________.
(ii) A line segment which joins any two points on a circle is a ______.
(iii) The longest chord of a circle is _______.
(iv) The radius of a circle of diameter 24 cm is ______.
(v) A part of circumference of a circle is called as _____.
Solution:
(i) π
(ii) chord
(iii) diameter
(iv) 12 cm
(v) an arc

Question 2.
Match the following

Solution:
(i) 3
(ii) 4
(iii) 5
(iv) 2
(v) 1

Question 3.
Find the central angle of the shaded sectors (each circle is divided into equal sectors)

Solution:

Question 4.
For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d = 12.6 cm
(iii) central angle 60°, r = 36 cm
(iv) central angle 72°, d = 10 cm
Solution:
(i) Central angle 45°, r = 16 cm
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
l = \(\frac{45^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 16 cm
l = \(\frac{1}{8}\) × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
A = \(\frac{45^{\circ}}{360^{\circ}}\) × 3.14 × 16 × 16
A = 100.48 cm²
Perimeter of the sector
P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm

(ii) Central angle 120°, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3 cm
Length of the arc


Area of the sector missing
Perimeter of the sector
P = l + 2r units
P = 6.28 + 2(5) cm
P = 6.28 + 10 cm
P = 16.28 cm

Question 5.
From the measures given below, find the area of the sectors.

Solution:
(i) Area of the sector
A = \(\frac{l r}{2}\) sq. units
l = 48 m
r = 10 m
= \(\frac{48 \times 10}{2}\) m²
= 24 × 10 m²
= 240 m²
Area of the sector = 240 m²

(ii) length of the arc l = 12.5 cm
Radius r = 6 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{12.5 \times 6}{2}\)
= 12.5 × 3 cm² cm²
= 37.5 cm²
Area of the sector = 37.5 cm²

(iii) length of the arc l = 50 cm
Radius r = 13.5 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{50 \times 13.5}{2}\)
= 25 × 13.5 cm² cm²
= 337.5 cm²
Area of the sector = 337.5 cm²

Question 6.
Find the central angle of each of the sectors whose measures are given below (π = \(\frac{22}{7}\))

Solution:
(i) Radius of the sector = 21 cm
Area of the sector = 462 cm²
\(\frac{l r}{2}\) = 462
\(\frac{l \times 21}{2}\) = 462

∴ Central angle of the sector = 120°

(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm²

(iii) Radius of the sector = 35 m
Length of the arc l = 44 m

Question 7.
Answer the following questions:
(i) A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
(ii) A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution:
(i) Radius of the circle r = 120 m
Number of equal sectors = 8

(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5

Note: We can solve this problem using A = \(\frac{1}{n}\) πr² sq. units also.

Question 8.
Find the area of a sector whose length of the arc is 50 mm and radius is 14 mm.
Solution:
Length of the arc of the sector l = 50 mm
Radius r = 14 mm
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{50 \times 14}{2}\) mm² = 50 × 7 mm² = 350 mm²
Area of the sector = 350 mm²

Question 9.
Find the area of a sector whose perimeter is 64 cm and length of the arc is 44 cm.
Solution:
Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64 cm
l + 2r = 64 cm
44 + 2 r = 64 .
2 r = 64 – 44
2 r = 20
r = \(\frac{20}{2}\) = 10 cm²
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{44 \times 10}{2}\) cm² = 22 × 10 cm² = 220 cm²
Area of the sector = 220 cm²

Question 10.
A sector of radius 4.2 cm has an area 9.24 cm². Find its perimeter
Solution:
Radius of the sector r = 4.2 cm
‘ Area of the sector = 9.24 cm²
\(\frac{l r}{2}\) = 9.24
\(\frac{l \times 4.2}{2}\) = 9.24
l × 2.1 = 9.24
l = \(\frac{9.24}{2.1}\)
l = 4.4 cm
Perimeter of the sector = 1 + 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 cm

Question 11.
Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the pot in which the plants grow. (π = 3.14)

Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet

Area of the quadrant = 3.14 sq. feet (approximately)

Question 12.
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector, (π = 3.14).

Solution:
Side of the square = 30 cm
∴ Radius of the sector design = 30 cm
Given design in the design of a circular quadrant.
Area of the quadrant = \(\frac{1}{4}\) πr² sq. units
= \(\frac{1}{4}\) × 3.14 × 30 × 30 cm²
= 3.14 × 15 × 15 cm²
∴ Area of the sector design = 706.5 cm² (approximately)

Question 13.
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite. (π = \(\frac{22}{7}\))

Solution:
Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of the each sector = \(\frac{1}{n}\) πr² sq. units
= \(\frac{1}{8} \times \frac{22}{7}\) × 56 × 56 cm² = 1232 cm²
Area of each sector = 1232 cm² (approximately)

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Recap Page No. 66 and 67

Question 1.
Write the numbers of terms in the following expressions.
(i) x + y + z – xyz
Solution:
4 terms

(ii) m²n²c
Solution:
1 term

(iii) a²b²c – ab²c² + a²bc² + 3abc
Solution:
4 terms

(iv) 8x² – 4xy + 7xy²
Solution:
3 terms

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
Question 1.
2x² – 5xy + 6y² + 7x – 10y + 9
Solution:
Numerical co efficient in 2x² is 2
Numerical co efficient in -5xy is -5
Numerical co efficient in 6y² is 6
Numerical co efficient in 7x is 7
Numerical co efficient in -10y is – 10
Numerical co-efficient in 9 is 9

Question 2.
\(\frac{x}{3}+\frac{2 y}{5}-x y+7\)
Solution:

Numerical co efficient in -xy is -1
Numerical co efficient in 7 is 7

Question 3.
Pick out the like terms from the following.

Solution:

Question 4.
Add : 2x, 6y, 9x – 2y
Solution:
2x + 6y + 9x – 2y = 2x + 9x + 6y – 2y = (2 + 9)x + (6 – 2)y = 11x + 4y

Question 5.
Simplify : (5x³ y³ – 3x² y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
Solution:
(5x³y³ – 3x²y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
= 5x³y³ + x³y³ – 3x²y² + 2x²y² + xy + 2xy + 7 – 5
= (5 + 1)x³y³ + (-3 + 2)x²y² +(1 +2)xy + 2
= 6x³y³ – x²y² + 3xy + 2

Question 6.
The sides of a triangle are 2x – 5y + 9, 3y + 6x – 7 and -4x + y +10 . Find perimeter of the triangle.
Solution:
Perimeter of the triangle = Sum of three sides
= (2x – 5y + 9) + (3y + 6x – 7) + (-4x + y + 10)
= 2x – 5y + 9 + 3y + 6x – 7 – 4x + y + 10
= 2x + 6x – 4x – 5y + 3y + y + 9 – 7 + 10
= (2 + 6 – 4)x + (-5 + 3 + 1)y + (9 – 7 + 10)
= 4x – y + 12
∴ Perimeter of the triangle = 4x – y + 12 units.

Question 7.
Subtract -2mn from 6mn.
Solution:
6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn

Question 8.
Subtract 6a² – 5ab + 3b² from 4a² – 3ab + b².
Solution:
(4a² – 3ab+ b²) – (6a²– 5ab + 3b²)
= (4a² – 6a²) + (- 3ab -(-5 ab)] + (b²– 3b²)
= (4 – 6) a² + [-3ab + (+ 5ab)] + (1 – 3) b²
= [4 + (- 6)] a² + (-3 + 5) ab + [1+ (-3)]b²
= -2a² + 2ab – 2b²

Question 9.
The length of a log is 3a + 4b – 2 and a piece (2a – b) is remove from it. What is the length of the remaining log?
Solution:
Length of the log = 3a + 4b – 2

Length of the piece removed = 2a – b
Remaining length of the log = (3a + 4b – 2) – (2a – b)
= (3a – 2a) + [4b – (-b)] – 2
= (3 – 2)a + (4 + 1)b – 2
= a + 5b – 2

Question 10.
A tin had ‘x’ litre oil. Another tin had (3x² + 6x – 5) litre of oil. The shopkeeper added (x + 7) litre more to the second tin. Later he sold (x² + 6) litres of oil from the second tin. How much oil was left In the second tin?
Solution:
Quantity of oil in the second tin = 3x² + 6x – 5 litres.
Quantity of oil added = x + 7 litres
∴ Total quantity of oil in the second tin
= (3x² + 6x – 5) + (x + 7) litres
= 3x² + (6x + x) + (-5 + 7)
= 3x² + (6 + 1)x + 2
= 3x² + 7x + 2 litres
Quantity of oil sold = x + 6 litres
∴ Quantity of oil left in the second tin = (3x² + 7x + 2) – (x² + 6)(3x² – x² ) + 7x + (2 – 6)
= (3 – 1)x² + 7x + (-4) = 2x² + 7x – 4
Quantity of oil left = 2x² + 7x – 4 litres

Try this Page No. 70

Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Solution:
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
Eg. 2y² + 5y^-1 – 3 is a an algebraic expression. But not a polynomial.

Try this Page No. 71

Question 2.
-(5y² + 2y – 6) Is this correct? If not, correct the mistake.
Solution:
Taking -(5y² + 2y – 6) = 5y² + [(-)(+) 2y] + [(-) × (-)6]
= -5y² – 2y + 6
≠ -5y² – 2y + 6
∴ Correct answer is -5y² + 2y – 6 = -(5y² + 2y + 6)

Try this Page No 71

(i) 3ab², -2a²b³
(ii) 4xy, 5y²x, (-x²)
(iii) 2m, -5n, -3p
Solution:
(i) (3ab²) × (-2a²b²) = (+) × (-) × (3 × 2) × (a × a²) × (b² × b³) = -6a³ b⁵

(ii) (4xy) × (5y²x) × (-x²)
= (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x²) × (y × y²)
= -20x⁴y³

(iii) (2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30mnp = 30 mnp

Try this Page No. 71

Question 1.
Why 3 + (4x – 7y) ≠ 12x – 21y?
Solution:
Addition and multiplication are different
3 + (4x – 7y) = 3 + 4x – 7y
We can add only like terms.

Try this Page No. 72

Question 1.
Which is corrcet? (3a)² is equal to
(i) 3a²
(ii) 3²a
(iii) 6a²
(iv) 9a²
Solution:
(3a) =3²a² = 9a²
(iv) 9a² is the correct answer

Try These Page No.72

Question 1.
Multiply
(i) (5x² + 7x – 3) by 4x²
Solution:
(5x² + 7x – 3) × 4x²
= 4x²(5x² + 7x – 3) Multiplication is commutative
= 4x² (5x² + 4x² (7x) + 4x² (-3)
= (4 × 5)(x² × x²) + (4 × 7)(x² × x) + (4 × -3)(x²)
= 20x⁴ + 28x³ – 12x²

(ii) (10x – 7y + 5z) by 6xyz
Solution:
(10x – 7y + 5z) by 6xyz
(10x – 7y + 5z) × 6xyz = 6xyz (10x – 7y + 5z) [∵ Multiplication is commutative]
= 6xy (10x) + 6xyz (-7y) + 6xyz (5z)
= (6 × 10)(x × x × y × z) + (6 × -7) + (x × y × y × z) + (6 × 5)(x × y × z × z)
= 60x²yz + (-42xy²z) + 30xyz²
= 60x²yz – 42x²z + 30xyz²

(iii) (ab + 3bc – 5ca) by – 3abc
Solution:
(ab + 3bc – 5ca) × (- 3abc) = (-3abc) (ab + 3bc – 5ca)
[∵ Multiplication is commutativel
= (-3abc) (ab) + (-3abc) (3bc) + (-3abc) (5ca)
= (-3)(a × a × b × b × c) + (- 3 × 3) + (a × b × b × c × c)
= -3a²b²c – 9ab²c² – 30a²bc²

Try these Page No. 74

Question 1.
Multiply
(i) (a – 5) and (a + 4)
Solution:
(a – 5) (a + 4) = a(a + 4) – 5 (a + 4)
= (a × a) + (a × 4) + (-5 × a) + (-5 × 4)
= a² + 4a – 5a – 20 = a² – a – 20

(ii) (a + b) and (a – b)
Solution:
(a + b) (a – b) = a(a – b) + b (a – b)
= (a × a) + (a × -b)+(b × a) + b(-b)
= a² – ab + ab – b² = a² – b²

(iii) (m⁴ + n⁴) and (m – n)
Solution:
(m⁴ + n⁴)(m – n) = m⁴(m – n) + n⁴(m – n)
= (m⁴ × m) + (m⁴ × (-n)) + (n⁴ × m (n⁴ × (-n))
= m⁵ – m⁴n + mn⁴ – n⁵

(iv) (2x + 3)(x – 4)
Solution:
(2x + 3)(x – 4) = 2x(x – 4) + 3(x – 4)
= (2x² × x) – (2x × 4) + (3 × x) – (3 × 4)
= 2x² – 8x + 3x – 12 = 2x² – 5x – 12

(v) (x – 5)(3x + 7)
Solution:
(x – 5)(3x + 7) = x(3x + 7) – 5(3x + 7)
= (x × 3x) + (x × 7) + (-5 × 3x) + (-5 × 7)
= 3x² + 7x – 15x – 35
= 3x² – 8x – 35

(vi) (x – 2)(6x – 3)
Solution:
(x – 2)(6x – 3) × (6x – 3) – 2(6x – 3)
= (x × 6x)+(x × (-3) × (2 × 6x) – (2 × 3)
= 6x² – 3x – 12x + 6
= 6x² – 15x + 6

Try this Page No. 74

Question 2.
3x² (x⁴ – 7x³ + 2), what is the highest power in the expression.
Solution:
3x²(x⁴ – 7x³ + 2) = (3x²) (x⁴) + 3x² (-7x³)+ (3x²)²
= 3x⁶ – 21x⁵ + 6x²
Highest power is 6 in x⁶.

Exercise 3.2

Try this Page No. 77

Question 1.
Are the following correct?
(i) \(\frac{x^{3}}{x^{8}}=x^{8-3}=x^{5}\)
(ii) \(\frac{10 m^{4}}{10 m^{4}}=0\)
(iii) When a monomial is divided by itself, we will get I?
Solution:

Try this Page No. 77

Question 1.
Divide

Solution:

Try this Page No. 78

Question 1.
Are the following divisions correct ?

Solution:

Try this Page No. 78

Question 1.

Solution:

Exercise 3.3

Try these Page No. 81

Question 1.
1. (p + 2)² = …….
2. (3 – a)² = …….
3. (6² – x²) = ………
4. (a + b)² – (a – b)² = …….
= a² + 2ab + b² – a² – 2ab – b²
= (1 – 1)a² + (2 + 2)ab + (+1 – 1 )b² = 4ab
5. (a + b)² = (a + b) × (a + b)
6. (m + n)( m – n) = m² – n²
7. (m + 7)² = m² + 14m + 49
8. (k² – 36) ≡ k² – 6² = (k + 6)(k – 6)
9. m² – 6m + 9 = (m – 3)²
10. (m – 10)(m + 5) = m² + (-10 + 5)m + (-10)(5) = m² – 5m – 50
Solution:

Try these page No. 83

Question 1.
Expand using appropriate identities
Question 1.
(3p + 2q)²
Solution:
(3p + 2q)²
Comparing (3p + 2q)² with (a + b)², we get a = 3p and b = 2q.
(a + b)² = a² + 2ab + b²
(3p + 2q)² = (3p)²+ 2(3p) (2q) + (2q)²
= 9p² + 12pq + 4q²

Question 2.
(105)²
Solution:
(105)² = (100 + 5)²
Comparing (100 + 5)² with (a + b)², we get a = 100 and b = 5.
(a + b)² = a² + 2ab + b²
(100 + 5)² = (100)² + 2(100)(5) + 5² = 1oooo + 1000 + 25
105² = 11,025

Question 3.
( 2x – 5d)²
Solution:
(2x – 5d)²
Comparing with (a – b)², we get a = 2x b = 5d.
(a – b)² = a² – 2ab + b²
(2x – 5d)² = (2x)² – 2(2x)(5d) + (5d)²
= 2x² – 20 xd + 5²d² = 4x² – 20 xd + 25d²

Question 4.
(98)²
Solution:
(98)² = (100 – 2)²
Comparing (100 – 2)² with (a – b)² we get
a = 100, b = 2
(a – b)² = a² – 2ab + b²
(100 – 2)² = 100² – 2(100)(2) + 2²
= 10000 – 400 + 4 = 9600 + 4 = 9604

Question 5.
(y – 5)(y + 5)
Solution:
(y – 5)(y + 5)
Comparing (y – 5) (y + 5) with (a – b) (a + b) we get
a = y; b = 5
(a – b)(a + b) = a² – b²
(y – 5)(y + 5) = y² – 5² = y² – 25

Question 6.
(3x)² – 5²
Solution:
(3x)² – 5²
Comparing (3x)² – 5² with a² – b² we have
a = 3x; b = 5
(a² – b²) = (a + b)(a – b)
(3x)² – 5² = (3x + 5)(3x – 5) = 3x(3x – 5) + 5(3x – 5)
= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)
= 9x² – 15x + 15x – 25 = 9x² – 25

Question 7.
(2m + n)(2m +p)
Solution:
(2m + n) (2m + p)
Comparing (2m + n) (2m + p) with (x + a) (x + b) we have
x = 2n; a = n ;b = p
(x – a)(x + b) = x² + (a + b)x + ab
(2m +n) (2m +p) = (2m²) + (n +p)(2m) + (n) (p)
= 2²m² + n(2m) + p(2m) + np
= 4m² + 2mn + 2mp + np

Question 8.
203 × 197
Solution:
203 × 197 = (200 + 3)(200 – 3)
Comparing (a + b) (a – b) we have
a = 200, b = 3
(a + b)(a – b) = a² – b²
(200 + 3)(200 – 3) = 200² – 3²
203 × 197 = 40000 – 9
203 × 197 = 39991

Question 9.
Find the area of the square whose side is (x – 2)
Solution:
Side of a square = x – 2
∴ Area = Side × Side
= (x – 2) (x – 2) = x(x – 2) – 2(x – 2)
= x(x) + (x)(-2) + (-2)(x) + (-2)(-2)
= x – 2x – 2x + 4x² – 4x + 4

Question 10.
Find the area of the rectangle whose length and breadth are (y + 4) and (y – 3).
Solution:
Length of the rectangle = y+ 4
breadth of the rectangle = y – 3
Area of the rectangle = length × breadth
= (y + 4)(y – 3) = y² + (4 +(-3))y + (4)(-3)
= y² + y – 12

Try these Page No. 88

Question 1.
Expand :
Question 1.
(x + 4)³
Solution:
Comparing (x + 4)³ with (a + b)³, we have a = x and b = 4.
(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 4)³ = x³ + 3x²(4) + 3(x)(4)² + 4³
= x³ + 12x² + 48x + 64

Question 2.
( y – 2)²
Solution:
Comparing (y – 2) with (a – b)³ we have a = y b = z
(a – b)³ = a³ – 3a²b + 3ab² – b³
(y – 2)² = y³ – 3y(2) + 3y(2)² + 2³
= y³ – 6y² + 12y + 8

Question 3.
(x + 1)(x + 3)(x + 5)
Solution:
Comparing (x + 1) (x + 3) (x + 5) with (x + a) (x + b) (x + c) we have
a = 1
b = 3
and c = 5

Exercise 3.4

Try These Page No.92

Question 1.
Factorize the following:
Question 1.
3y + 6
Solution:
3y + 6
3y + 6 = 3 × y + 2 × 3
Taking out the common factor 3 from each term we get 3 (y + 2)
∴ 3y + 6 = 3(y + 2)

Question 2.
10x² + 15y
Solution:
10x² + 15y²
10x² + 15y² = (2 × 5 × x × x) + (3 × 5 × y × y)
Taking out the common factor 5 we have
10x² + 15y² = 5(2x² + 3y²)

Question 3.
7m(m – 5) + 1(5 – m)
Solution:
7m(m – 5) + 1(5 – m)
7m(m – 5) + 1(5 – m) = 7m(m – 5) + (-1)(-5 + m)
= 7m(m – 5) – 1 (m – 5)
Taking out the common binomial factor (m – 5) = (m – 5)(7m – 1)

Question 4.
64 – x²
Solution:
64 – x²
64 – x² = 8² – x²
This is of the form a² – b²
Comparing with a² – b² we have a = 8, b = x
a² – b² = (a + b)(a – b)
64 – x² = (8 + x)(8 – x)

Students can Download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 1.
Fill in the blanks:

Solution:
(i) \(\frac{18 m^{4}\left(n^{8}\right)}{2 m^{(3)} n^{3}}\) = 9 mn⁵
(ii) \(\frac{l^{4} m^{5} n^{(7)}}{2 l m^{(3)} n^{6}}=\frac{l^{3} m^{2} n}{2}\)
(iii) \(\frac{42 a^{4} b^{5}\left(c^{2}\right)}{6(a)^{4}(b)^{2}}\) = (7)b³c²

Question 2.
Say True or False:
(i) 5x³y ÷ 4x² = 2xy
(ii) 7ab² ÷ 14ab = 2b²
Solution:
(i) True
(ii) False

Question 3.
(i) 27y³ ÷ 3y
(ii) x³y² ÷ x²y
(iii) 45x³y²z⁴ ÷ (-15xyz)
(iv) (3xy)² ÷ 9xy
Solution:

Question 4.

Solution:

Question 5.
Divide
(i) 32y² – 8yz by 2y
(ii) (4m² n³ + 16m⁴ n² – mn) by 2 mn
(iii) 10 (4x – 8y) by 5 (x – 2y)
(iv) 81 (9⁴q²r³ + 2p³q³r² – 5p²q²r²) by (3pqr)²
Solution:

Question 6.
Find Adirai’s percentage of marks who scored 25m³n²p out of 100m²np
Solution:
Total marks = 100 m²np
Adirai’s score = 25 m³n²p

Question 7.
Identify the error and correct them.
(i) 7y² – y² + 3y² = 10y²
(ii) 6xy + 3xy = 9x²y²
(iii) m (4m – 3) = 4m² – 3
(iv) (4n)² – 2n + 3 = 4n² – 2n + 3
(v) (x – 2) (x + 3) = x² – 6
Solution:
(i) 7y² – y² + 3y² = (7 – 1 + 3)y² = (6 + 3)y² = 9y²
(ii) 6xy + 3xy = (6 + 3)xy = 9xy
(iii) m (4m – 3) = m (4m) + m (-3) = 4m² – 3m
(iv) (4n)² – 2n + 3 = 16n² – 2n + 3
(v) (x – 2) (x + 3) = x (x + 3) – 2 (x + 3) = x (x) + (x) × 3 + (-2) (x) + (-2) (3)
= x² + 3x – 2x – 6 = x² + x – 6