Class 9

You can Download The Race Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard English Solutions Unit 1 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard English Solutions Unit 4 Chapter 1 The Race

Std 9 English Textbook The Race Questions and Answers

The Race Question Answers Kerala Syllabus 9th Question 1.
Who do you like more, Tarun or his brother? Why?
Answer:
I like Tarun more because he knew his defects and he accepted them.

Kerala Syllabus 9th Standard English Notes Question 2.
Why did Tarun think that he was the black sheep of the family?
Answer:
Tarun thought he was the black sheep in the family because he was not good at studies. He was not a good singer, dancer, painter or even an actor.

The Race Question Answer Kerala Syllabus 9th Question 3.
What were the obstacles before Tarun in pursuing his dream? Give some suggestions to solve these
Answer:
He had two obstacles. One was the opposition from his parents who wanted him to concentrate on his studies. The other was his family had no money to send him for intensive and expert training. He could have asked his teachers to help him to solve his problems.

9th Class English Chapter The Race Question Answer Question 4.
“It was a day he wanted to wipe out from his memory….” Why?
Answer:
He wanted to wipe out that day from his memory because he failed in the terminal examinations. His father was angry and his friends made fun of him.

The Race 9th Standard Question Answer Kerala Syllabus Question 5.
What made Ram Narayan say “Life is full of ups and downs”?
Answer:
Ram Narayan said ‘Life is full of ups and downs’ because as an Olympian Gold Medalist he has seen that life Is not always smooth. He had seen Tarun running around the park for a long time and then lying on the bench breathing heavily.

The Race Lesson 9th Class Summary Kerala Syllabus Question 6.
Do you think Tarun’s meeting with Ram Narayan would change his life? Explain
Answer:
I do think Ram Narayan’s meeting would change Tarun’s life because Ram Narayan told him that he was one of the best runners he had ever seen.

Hss Live Guru 9th English Kerala Syllabus Question 7.
Complete the conversation.
a) Who is your coach?
Answer:
Tarun said to Ram Narayan, “I have no professional training.”

b) Oh! no professional training yet, I will train you if ….(b)…..
Sir, I will do it.
Answer:
Ram Narayan replied, “Oh! no professional training yet. I will train you if you win the race on Children’s Day at the Nehru Stadium.”

9th Standard English Question And Answer Kerala Syllabus Question 8.
What made Ram Narayan say that it was his job to get Tarun entry in the race?
Answer:
Ram Narayan said that it was his job to get Tarun entry in the race because he had seen how well Tarun was running. He told Tarun that he was one of the best runners he had ever seen.

9th Class English Chapter The Race Summary Kerala Syllabus Question 9.
How did Tarun prepare for the race?
Answer:
Tarun prepared for the race by training hard. Every day he would get up at four in the morning and run up to 10 miles. In the evening he would time himself according to the 1000-meter distance prescribed by the competition.

The Race Chapter Questions And Answers Kerala Syllabus Question 10.
“He also wanted someone to back him up and so he told everything to his mother.” Do you think this kind of sharing will help him face his problems better? Why?
Answer:
I certainly think sharing his problems will help him to face his problems better. When we share our problems with someone who loves us he/she will support us and help us in finding solutions.

9th Class English Chapter The Race Kerala Syllabus Question 11.
Why did Tarun really want to win the race?
Answer:
Tarun wanted to show his father that he was not really. the black sheep of the family and that he could be good at something.

The Race Story 9th Class Kerala Syllabus Question 12.
“Tarun felt very proud of himself.” What made him proud?
Answer:
Tarun felt very proud of himself because he saw that he was ahead of everybody else in the race.

9th Class English Chapter The Race Conversation Kerala Syllabus Question 13.
If you were Tarun, would you finish the race or give it up? Why?
Answer:
If I were Tarun, I would not give it up. Ups and downs are part of life. We learned to walk after falling many times and then got up and walked again.

9th English Notes Kerala Syllabus Question 14.
Do you think the presence of his mother and Ram Narayan helped Tarun in finishing the race? Give reasons.
Answer:
Yes, it did. The presence of his mother and Ram Narayan did help Tarun in finishing the race. When we do something in the presence of those we love, we want to do our best.

9th Standard English Textbook Activities Kerala Syllabus Question 15.
“You have won the toughest race, the race of your life.” What did Ram Narayan mean by this?
Answer;
Ram Narayan meant that Tarun had been trying hard to do his best in the race. He fell down three times. But each time he got and continued running. Ram Narayan knew that when facing problems later in his life, Tarun will behave in the same way with determination and will-power. Such persons will succeed in life. Tarun was a real-life hero.

Let’s revisit and Reflect

The Race Story Questions And Answers Kerala Syllabus Question 1.
The story ‘The Race’ gives us valuable lessons about life. Do you agree? Give reasons.
Answer:
The story certainly gives us valuable lessons about life! It teaches us that self-confidence, dedication, passion, and perseverance make one attain success in life.

The Race Story In Malayalam Kerala Syllabus Question 2.
Does the support of family and friends play an important role in one’s success? Justify your answer with reference to the story.
Answer:
The support of the family and friends certainly plays an important role in one’s success. The support from his mother in the gallery and the positive attitude given by Ram Narayan made Tarun gain self-confidence and determination.

Hsslive Guru 9th English Kerala Syllabus Question 3.
Tarun received a louder applause than the winner though he lost the race. If you were in the stands watching the race, would you applaud him? Substantiate.
Answer:
Surely I would applaud him. He didn’t give up the race even though he fell down three times. His determination and dedication made him run again. Failure is not the end of life. Some people expand the word ‘FAIL’ as “First Attempt In Learning”.

Activity-1 (Page 14)

Character sketch:

Question 1.
What is a character sketch?
Answer:
A character sketch gives the details about a character. It describes the appearance and character of a person.

• It describes the character’s physical appearance and personality
• It includes examples of how the character is developed throughout the story
• It gives your overall impression about the character.

Question 2.
How do we write a character sketch?
Answer:
We should be familiar with the words that describe. the appearance and character of the character. We may make use of the words given below.

Question 3.
To describe the personality of a person
Answer:
Intelligent, helpful, honest, kind, sympathetic, brave, solemn, smart, short-tempered, patient, gentle, cruel, lazy, wise, foolish, industrious, naughty, shy, humble, friendly, unfriendly, cheerful, responsible, absent-minded, determined, calm, as cunning as a fox, as lively as a squirrel, as stubborn as a mule, etc.

Question 4.
To describe the appearance of a person
Answer:
Young, old, tall, short, strong, weak, agile, slim, thin, fair complexioned, dark, green, or blue eyes, red-streaked eyes, huge moustache, thick black eyebrows, chubby, sharp face, bristly black lashes, strangled hair, hard of muscles, middle-aged, deep auburn hair, fresh, pale, grim, beautiful, handsome, etc.

When you write a character sketch, you are trying to give a good idea about that person. You want the reader to have a strong mental image of the person. He would like to know:
a) how the person talks
b) the person’s characteristic ways of doing things
c) something about the person’s value system.
Character sketches only give snapshots bf people

Read the story “The Race” and write a few words which best describe Tarun :

Answer:

Add a few more points to describe Tarun’s appearance, behavior, outlook, etc.
Answer:

• He had an athletic figure.
• His behavior was quite good and he wanted to prove that he too can do something.
• His outlook was one of optimism, he knew hard work brings success.

Activity – 2 (Page 15)

‘Tarun started practicing zealously. Every day, he would get up at four in the morning and run up to ten miles.’
This made him mentally and physically strong. But, there are so many factors, both mental and physical, that affect health. List such factors
1. fast food
2. …………..
3. …………..
4. …………..
5. …………..
6. …………..
Answer:
2. Lack of exercise
3. Lack of sleep
4. Unhealthy habits like smoking and drinking
5. Use of drugs
6. Spending too much time watching TV or playing videogames.

Discuss how these factors are related to lifestyle diseases and what the ways to prevent them are. Based on the discussion, prepare a write-up. In what way are they related to lifestyle diseases? Prepare a write-up.

Points to remember:

• A suitable title
• Discuss related ideas in small paragraphs
• Organize ideas
• Simple language
• Sentence varieties

Lifestyle Diseases and their Prevention

The decisions people make about their diet, exercise, smoking, and alcohol have an immediate impact on their health. Many people think that they are strong and they will not get diseases like cancer, diabetes, chronic lung diseases, or other lifestyle diseases. But, in fact, the choices people make can damage their health now and in the future. People with overweight, high cholesterol, high blood pressure, and such other problems are victims of lifestyle diseases.

People who smoke are at greater risk for cancer and chronic lung disease and often suffer breathing problems impacting daily life. People who drink too much don’t realize how alcoholism affects their physical and emotional well-being. Fast foods and aerated drinks make people fat and they fall easy victims to all kind of diseases. To prevent these lifestyle diseases people should: stop smoking, avoid alcohol, eat a balanced diet, avoid stress and get enough sleep. They should keep their bodies and minds fit by exercises and yoga.

Activity – 3 (Pagel 15)

The race Tarun participated in was quite exciting. Several factors made it lively. The announcement was one of them.

Read the script of an announcement about Tarun’s race.

Ladies and gentlemen,
Welcome to the Nehru Stadium for watching the race in connection with the Children’s Day Celebrations. The race is going to start soon. Young athletes from various schools are participating and it will be flagged off by Sri. Ram Narayan, the famous athletic coach.
Thank you.

Let’s discuss

Question 1.
What is the announcement about?
Answer:
The announcement is about a race to be held in connection with children’s day celebrations at the Nehru stadium.

Question 2.
Who is addressed here?
Answer:
The spectators.

Question 3.
When is the event held?
Answer:
The event is held on Children’s Day, 14^th November

Question 4.
Where does the event take place?
Answer:
The event takes place at the Nehru Stadium.

Question 5.
What is the intention of the announcement?
Answer:
The announcement is intended to invite the audience and inform them about the event that is going to take place.

Question 6.
What are the other details given?
Answer:
Details about the participants and the person who is flagging off the event.

Question 7.
What is special about the language?
Answer:
The announcement is short and to the point. It is in formal language.

The Sports Club of your school is organizing a meeting to congratulate Tarun. You are asked to make an announcement about the programme. Prepare a script of your announcement and present it.

Features of an announcement

• It should be direct, plain, complete and concise.
• It should be friendly and arousing the interest of the audience/spectators.
• It should be factual.

Respected teachers and dear friends,
We have assembled here to congratulate Tarun, the gem of our school, Tarun of Std IX. should be a role model for all of us. The confidence, perseverance, and determination shown by him to overcome obstacles have no parallel in the history of our school. The Sports Club of the school has decided to convene a meeting to honor him at the school auditorium today at 1.30 pm. All students and teachers are requested to gather in the school auditorium in time.
Thank you.

Activity 4 (Page 16)

Here is a news report about Tarun’s rise as an athletic champion.

A Twice-born Athlete:
Lucknow: The race conducted by the Children’s day celebration Committee on the Children’s Day of 2004 witnessed the rise of a new athletic star at the Nehru Stadium,
Lucknow. On the track, a boy fell down thrice. Undeterred, he continued running and finished the race, though in the last position. He is Tarun, son of Mr. Vimal Kapoor and Mrs. Rani Devi. Though he finished last in the race, the coach Ram Narayan recognized the fire within him and agreed to train him. It was a turning point in his life. “Tarun was born twice”, comments his parents.

His actual birthday was on 2^nd February 1992, and his birth as an athlete was on 14^th November 2004, the day on which he won recognition as an athlete. He was studying in the Gandhi Memorial High School then. Next year, he became the champion in the 100 meters race in the State School Athletic Meet. On the advice of his coach, he joined the State Sports Council School in Lucknow and completed his BP. Ed. in 2011. He became the fastest runner in the state in 2008, and a national champion in 2010. He is a self-motivated, confident and hard-working person. His residence, Varun Villa, is located at North Avenue near the Lucknow International Stadium where he practices vigorously to become an Olympian.

Question 1.
Athletes of today have many good sponsorships and job opportunities. Many institutions provide facilities for continuing their education and training. Tarun applies for such a post. Based on the above news report, complete the template of Tarun’s Curriculum Vitae.
Answer:
Curriculum Vitae gives details about a person and his qualifications, experiences, and special abilities. It is a brief and factual document giving information about one’s education, work experience, skills, and accomplishments. The key elements of curriculum vitae are the heading, career objective, educational profile, personal profile, professional experience and references. There is also a covering letter stating the most important facts and requesting for an interview. The term ‘Curriculum Vitae’ (shortened to CV) is also called Bio-Data. In America, a CV is called ‘Resume’.

Curriculum Vitae Of Tarun Kapoor

Tarun Kapoor
……………………..
……………………..
Contact number: …………………….. (Res.), …………………….. (Mob.)
Career objective: To become an Olympian
Qualifications :
……………………………
……………………………
Achievements :
……………………………
……………………………
Personal Profile
Father’s name:……………………….
Mother’s name:………………………
Date of birth:…………………………
Permanent address : ………………..
……………..………..
……………………….
Languages known : English, Hindi
Nationality:……………..
Gender:……………
Marital status: Single
Strengths :
Answer:
Tarun Kapoor
North Avenue
Near Lucknow International Stadium
Lucknow – 450 321
Contact number : 999555444; (Res)999555444
Career Objective: To become an Olympian
Qualifications: i) SSLC from St. John’s H.S., 2005, 84% marks
ii) +2 from St. John’s H.S., 2007,85% marks
iii) B.P.Ed, From State Sports Council School, Lucknow, 2011,87% marks’.
Achievements
Champion in 100m race in State School Athletic meet
Fastest runner in the State, 2008
National Champion, 2010

Personal Profile:
Father’s name: Vimal Kapoor
Mother’s name: Rani Devi
Date of birth: 2 Feb 1992
Permanent address: Varun villa,
North Avenue
Near Lucknow International
Stadium, Lucknow-450321
Languages known : English, Hindi
Nationality: Indian
Gender: Male
Marital status: Single
Strengths: Confidence, Perseverance, Hard work, Sincerity,

Declaration:

I hereby declare that the details given above are true to the best of my knowledge.
Place: Lucknow

Sd/

Date: 2 July 2011

Tarun

Activity 5 (page 18)

a. A part of the commentary of the race in which Tarun participated is given below.

Welcome to the Nehru Stadium for watching a wonderful race by young athletes. This is Raj Kumar in the commentary box. Today is 14th November, the birthday of Chacha Nehru. All the athletes are lined up at the starting point. It’s a 100 meters race. The family and friends of the participants are there in the stands to cheer them up. Athletes are not supposed to go away the track. If they do so, they will be disqualified. Oh! the whistle is blown.

The race has started. All are running along their own tracks. Wow! Tarun on the fifth track is running ahead of everybody. Just behind him, Pawan is on the fourth track. Oh, God! Tarun ……………………….. My congratulations to Pawan, the winner of the race and all the other athletes for your participation! This is Raj Kumar signing off. Thank you.

a) Do you think that

Question 1.
the commentator know the rules and background of the event?
i. Yes
ii. No
Answer:
i. Yes

Question 2.
the commentary will help the athletes to know about the status of the race?
i. Yes
ii. No
Answer:
i. Yes

Question 3.
the audience is informed about what is happening
i. Yes
ii. No
Answer:

Ques 4.
the commentary is meant to create excitement among the audience?
i. Yes
ii. No
Answer:
i. Yes

Ques 5.
the commentator’s language is simple and clear?
i. Yes
ii. No
Answer:
i. Yes

Question 6.
the commentary confuse the listeners?
i. Yes
ii. No
Answer:
ii. No

Question 7.
the commentator favor any of the participants?
i. Yes
ii. No
Answer:
i. Yes

Question 8.
the commentary has a suitable conclusion and leave-taking?
i. Yes
ii. No
Answer:
i. Yes

Now, complete the commentary given above.
Oh, God Tarun has fallen down. People are worried. But Tarun gets up and continues running. Yes, he is overtaking some others. But alas! he falls down again. He seems a little worried. But once again he gets up and runs. He is again with the lead runners. But, Oh my God! he falls again. Somebody is shouting, “Get up and run, Tarun!” Tarun gets up and runs but he can’t catch up with the others. Pawan comes first. There is great applause from the crowd. Tarun somehow finishes the line and the applause is greater for him! Yes, in spite of the falls he has crossed the finishing line!

b) Look at a few screenshots of the last over of the final of the Tri-series Cup cricket match between India and Sri Lanka, held at Queen’s Park Oval, Port of Spain, Trinidad, on July 11, 2013.

Commentary: India needs 15 runs. Only one over is left. The situation is electric. There is pin-drop silence. There is anxiety on the face of the Indian viewers. Eranga is the bowler and Dhoni is at the crease. Eranga bowls. Dhoni hits the ball hard. But alas! Dhoni has missed it. There is utter silence among the Indian crowd. 15 runs needed! Only 5 balls left. Eranga bowls. It is a six! The crowd applauds loudly. Fielding arrangements are changed by the Sri Lankan Captain. 9 runs needed. 4 balls left. Eranga bowls again. Dhoni hits a 4. The crowd is ecstatic. 5 runs needed. The 4th ball is bowled by Eranga. Dhoni hits it hard with his famous helicopter shot. It is a sixer! India has won with 2 balls to spare. What a victory!

You can Download The Trio Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard English Solutions Unit 5 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard English Solutions Unit 5 Chapter 3 The Trio (Story)

Std 9 English Textbook The Trio Questions and Answers

The story ‘Happiness Machine’ conveys that happiness cannot be materialistic. It is a state of mind. It comes from our relationships with other people, rather than from material things. Won’t we feel happy if our relationships are smooth and mind is at peace? Let’s read a story.

The Trio Question Answer Kerala Syllabus 9th Question 1.
What was the major occupation of people in Elliman Street?
Answer:
The major occupation of people in Elliman Street was oil-mongering (trading in oil).

The Trio Story Summary Kerala Syllabus 9th Question 2.
How was river Sarayu in the moonlight?
Rajam and Mani Questions and Answer:
It glistened like a silver belt across the North.

The Trio Summary In Malayalam 9th Question 3.
How is the evening at the river described here?
Answer:
There used to be crowds on the sand of the river bank. Birds filled the air with their cries. Downstream one could see cattle crossing the river. Country carts drawn by bullocks
could be seen, the cart-men humming low tunes. Soon after sunset, there would be a soft red in the west for some time.

Kerala Syllabus 9th Standard English Notes Question 4.
Rajam had a heroic entry into the class. Discuss.
Answer:
Rajam was a fresh arrival in first A. He had sauntered into the class on the opening day of the second term, walked up to the last bench and sat beside Mani.

Hss Live Guru 9th English Kerala Syllabus Question 5.
How was the dress of Rajam different from that of the others?
Swami and friends Questions Answer:
He was dressed very well. He was the only boy in the class who wore socks and shoes, fur cap and tie, and a wonderful coat and knickers.

9th Class English 5th Unit Kerala Syllabus Question 6.
Why were the boys not confident to speak to Rajam in English?
Answer:
There were rumors that he had come from some English boys’ school somewhere in Madras. He spoke very good English, exactly like a European. Many found it difficult to understand his English and they could not speak to him because of their broken English.

9th English Notes Kerala Syllabus Question 7.
How did Rajam turn out to be a power in the class and a rival to Mani?
Answer:
Rajam turned out to be a power in the class because of his English. He was the 2nd in the class. He became a rival to Mani because of his nonchalant attitude towards him. If Mani jabbed, Rajam jabbed. If Mani clouted, Rajam clouted. If Mani kicked, Rajam kicked. If Mani was the overlord of the class, Rajam seemed to be nothing less.

Kerala Syllabus 9th Standard English Guide Pdf Question 8.
What was Swaminathan’s fear when he was told about Mani’s intentions?
Answer:
Swaminathan’s fear was that if Mani went on troubling Rajam, he (Mani) Would have troubles because Rajam’s father was a police superintendent. Swaminathan thought the police people were an awful lot.

Hss Live English 9th Kerala Syllabus Question 9.
What did Mani call Swaminathan for being afraid of policemen?
Answer:
A milk-toothed coward.

Kerala Syllabus 9th English Notes  Question 10.
“How splendid it would have been!” According to Mani, what is missing in that fine evening?
Answer:
Lime pickles.

9th Std English Notes Kerala Syllabus Question 11.
Why did Rajam and Mani need a cord of communication between them? Who was it?
Answer:
Rajam and Mani needed a cord of communication between them because they were not on talking terms. The cord was Swaminathan.

Hsslive Guru 9th English Kerala Syllabus Question 12.
What made the teacher call Swaminathan?
Answer:
The teacher called Swaminathan because he noticed that Swaminathan was not attentive in the class. He was busy collecting the note from Mani and giving it to Rajam.

Hsslive Guru English 9th Kerala Syllabus Question 13.
What was the punishment Swami got from the teacher?
Answer:
Swami was asked to stand on the bench.

Hss Live Guru 9th English Notes Kerala Syllabus Question 14.
What was Swaminathan doing after he got the punishment?
Answer:
Standing on the bench, Swaminathan stood well over the whole class. He could see so many heads and he classified them according to the caps. There were 4 red caps, 25 Gandhi caps, 10 fur caps and so on.

Question 15.
The services of the mediator turned needless in the later part of the conversation. How did they communicate then?
Answer:
They communicated by shouting, Rajam shouting in one ear of Swaminathan and Mani shouting in the other ear.

Question 16.
What was the final settlement of their argument?
Answer:
The final settlement was that neither of them would come with guards. If anything happened to Rajam he would keep it out of his father’s knowledge.

Question 17.
What were the doubts arising in Mani’s mind when he was sitting on the river step with the club under his hand?
Answer:
He thought he would break Rajam’s head and throw his body into the river. But there was the possibility of the body being recovered. But then nobody would know that he had done it. Then there was the possibility of Rajam coming and troubling him at night as a spirit. Rajam could come and pull his hair at night. It would be better not to kill him. Just break his limbs and leave him. There was also the possibility of Swaminathan betraying him.

Question 18.
How did Rajam appear before them?
Answer:
He appeared before them dressed in khaki, carrying an air gun that was given to him on his birthday.

Question 19.
What did Rajam do to frighten Mani?
Answer:
He fired a shot in the air.

Question 20.
What made Mani hang down his head?
Answer:
Mani hanged down his head because he had not told Rajam he would come with a club. That was a breach of the promise.

Question 21.
According to Mani, what was the reason for his hatred towards Rajam?
Answer:
Rajam had called Mani a sneak before someone.

Question 22.
Now this was the happiest conclusion to all the unwanted trouble. How did they settle the issue?
Answer:
Rajam said he never called Mani a sneak. If this was the only reason for Mani’s anger, he should forget it and they should become friends. Mani agreed. Rajam lowered his gun and Mani dropped his club. The issue was thus settled.

Question 23.
How did Rajam show his goodwill?
Answer:
He showed his goodwill, by pulling out of his pocket half a dozen biscuits. The three friends shared them glowing with their new friendship.

Let’s Revisit and Reflect

Question 1.
“River Sarayu was the pride of Maigudi.”Why did R.K. Narayan say so?
Answer:
R.K. Narayan said so because its sandbanks were the evening resort of all the people of the town. The residents of the town took any distinguished visitor to the top of the Town Hall and proudly pointed to him Sarayu in moonlight, glistening like a silver belt across the North.

Question 2.
Why did the teacher punish Swaminathan? What was the punishment? How did he feel it?
Answer:
The teacher punished Swaminathan because he was not attentive in the class. The punishment was standing on the bench. He was glad that his punishment was standing on a bench and not getting beaten with the cane.

Question 3.
Why did Mani think that Rajam was his rival?
Answer:
Mani thought that Rajam was his rival because in his manner to Mani, Rajam assumed a nonchalant attitude to which Mani was not accustomed. If Mani jabbed, Rajam jabbed. If Mani clouted, Rajam clouted. If Mani kicked, Rajam kicked. If Mani was the overlord of the class, Rajam seemed to be nothing less.

Question 4.
Mani was planning to kill Rajam with his wooden club but what was the thought that stopped him?
Answer:
There were three things that stopped him from his planned killing of Rajam. Swaminathan could betray him to the police. Rajam’s father was a police superintendent. If Rajam is killed, his ghost may come and trouble Mani in the night.

Question 5.
The story ends with the sentence “Swaminathan felt at perfect peace with the world.” Why did he feel at perfect peace with the world?
Answer:
Swaminathan was at perfect peace with the world because he was the one between the two great rivals – Rajam and Mani. He had admired Rajam intensely and longed to be his friend. But if Mani came to know about this he would kill Swaminathan. So he was always in fear of both Rajam and Mani. He was virtually between the devil and the deep sea. Now since Mani and Rajam became friends, he felt peaceful.

Activity -1

Mani is a friend and guide for Swaminathan. He enjoys Mani’s companionship and does everything possible to keep him happy. He goes into the kitchen to get some pickle for Mani as promised and sees his mother. He pleads with his mother to give some lime pickle but she denies. Complete the conversation.
Swaminathan: Are you busy with your work amma?
Mother: No dear. Tell me what brought you to the kitchen
now.
Swaminathan : ………………………………………………
………………………………………………………………………
Mother:……………………………………………………….
………………………………………………………………………
Swaminathan : ……………………………………………..
……………………………………………………………………..
Mother:……………………………………………………….
……………………………………………………………………….
Answer:
Swaminathan: Amma, I want to take some lime pickle to give to somebody.
Mother: To whom do you want to give lime pickle?
Swaminathan: To my best friend, Mani. He is so good and I like him most. He also likes me.
Mother:: He may be your best friend. But that does not mean that you should give him lime pickle which I have made with great difficulty. I can’t give you any lime pickle.

Activity – 2

a) Rajam is the newcomer in-class First A. He impressed the whole class on the first day itself. Complete the boxes given below

Answer:
Personality: stylish and confident
Evidence: He saunters into the class, walks up to the last bench and sits beside Mani.
Appearance: Neat and Clean. Well-dressed
Evidence: He was the only boy in the class who wore socks and shoes, fur cap and tie, and a wonderful coat and knickers.
Actions: He is brave and his policy is an eye for an eye and a tooth for a tooth. He is not afraid of Mani, the bully.
Evidence: If Mani jabbed, Rajam jabbed. If Mani clouted, Rajam clouted. If Mani kicked, Rajam kicked. If Mani was the overlord of the class, Rajam seemed to be nothing less.

b) Now attempt a character sketch of Rajam using the points listed above and hints from the story.
Answer:
Character sketch:
Rajam is a fresh arrival in First A. He is stylish and confident and he saunters into the class, walks up to the last bench and sits beside Mani. When Mani gives him a jab in the ribs, he returns it. He impressed the class on the very first day. He was neat and clean and well-dressed. He was the only boy in the class who wore socks and shoes, fur cap and tie, and a wonderful coat and knickers. Rajam also was the best English speaker in the class. There were rumors that he had come from some English boys’ school somewhere in Madras. He spoke exactly like a European.

Many found it difficult to understand his English and they could not speak to him because of their broken English. His grades were also good as he was second in the class. Soon Mani realizes that Rajam was his rival. If Mani jabbed, Rajam jabbed. If Mani clouted, Rajam clouted. If Mani kicked, Rajam kicked. If Mani was the overlord of the class, Rajam seemed to be nothing less. When Mani challenges him fora duel Rajam accepts it. He goes well prepared for the fight. But soon Mani and Rajam decide to be friends and they, along with Swaminathan, share the biscuits Rajam had brought with him. Rajam is a boy who can be a good role model for many students of his age.

Activity – 3

Swaminathan was punished by the teacher for being inattentive in the class. Still, he paid no attention to the lessons and his mind began to wander. Infuriated by Swami’s behavior the teacher writes a letter to his father. What would the content of the letter be?
Answer:

Malgudi Primary School

10 December 2019

Mr. Laxminarayan
XV/146 Ayyangar Road
Malgudi
Dear Sir,
I am sorry to write such a letter to you regarding your son Swaminathan of First A. Initially he was a good boy and he was very attentive in the class. He was also getting reasonably good marks. But of late I have come to notice that he is least attentive in the class. He likes to sit on the backbench between two boys named Mani and Rajam. I find him talking to both these boys very often. He also passes some kind of written notes between them.

When I ask questions he stands and blabbers, not knowing the correct answers. You should ask him to be attentive in the class. We punish him here whenever he is caught inattentive. We have advised him, warned him, caned him and made him stand on the bench. But he is not improving. This letter is to tell you that if he continues to be inattentive in the class, he will surely fail in the final examination. So do whatever you can to bring some sense into him. Let him devote more time to his books and less time to his friends. I will be glad if you can come to the school one of these days so that we can have some further talks about the studies and behavior of your son.
Yours faithfully,
Abdulkarim Ibrahim
Class Teacher

Activity 4

Lena Auffmann experiences bouts of bliss watching sunset in the happiness machine. The story ‘The Trio’ begins with a description of the banks of river Sarayu at sunset. Sunset is the most magical and delightful moment of a day.
Let us now pen down our thoughts on:-
Sunset- An enthralling painting by mother nature.
(Hints- amazing moment in the western horizon- mixture of warm colors trees drenched in golden glow- beautified landscape-reflection on water- birds flying past and singing lullaby- sun fades into a long deep sleep.)
Answer:
Sunset – An Enthralling Painting by Mother Nature:
If you ask me what the most breathtaking sight from my home is, my answer is the sunset. It is an amazing moment in the western horizon. There is a mixture of warm colors that thrill your heart. Colors are mixed in such a way that only God can do it. It far excels the paintings of Michael Angelo, Leonardo da Vinci, Raphael, M.F. Hussein or our own Ravi Varma. The trees are drenched in golden glow.

The landscape is beautiful and the entire picture is reflected on water. Birds are flying back home singing lullabies. The butterflies and bees return to their abodes after having filled their bellies on the blooms. Slowly we see the sun disappearing beyond the hill as it is going away for having a deep sleep. Sunset is really an enthralling painting by Mother Nature to please her children. Each day we are given a different picture with different shades of colors.

Language Activity:

Read the following passage carefully.
Hi. My name is Mani. I often go to the banks of River Sarayu where people watch the sunset. Today, I invited Swami who is my best friend to watch the sunset. Swami, whose mother makes delicious pickle, has promised to bring me some pickle.

Add the missing relative pronouns.
1. Elliman street ………. was ten minutes walk from river Sarayu was always crowded. ( which/where)
2. The municipal resident …… was proud of the beauty of river Sarayu showed it to all the distinguished visitors. (who/whom)
3. Swami was a friend ………. Mani could trust. (whom/whose)
4. Mani, …….. anger towards Rajam knew no bounds wanted to harm him. (whom/whose)
5. The banks of the river ………….. people could enjoy the evening were very beautiful. (when/where)
Answer:
1. which
2. who
3. whom
4. whose
5. where

The villagers occupied the last street of the town.
Does the sentence make a complete sense without the bold portion?
Yes.

Defining and Non-Defining.
Rewriting the sentence:

River Sarayu, which glistened like a silver belt in moonlight, was the pride of Malgudi. (Non defining)
River Sarayu was the pride of Malgudi.

Is there a change in meaning when the relative clause is omitted?
No.

List down the features of defining and non defining relative clauses.

Answer:

Classify the sentences given below depending on the type of relative clause.
1. Rajam, who was a newcomer in class 1A, was admired by his classmates.
2. Children who love pickles are common.
3. The teacher, who was teaching History, punished Swami for being inattentive in the class.
4. Students who pay attention in the class score good marks.
5. The friends met near Nallappa’s grove, which was deserted in the evening, as promised
Answer:
1. non-defining
2. defining
3. non-defining
4. defining
5. non-defining

Vocabulary Activity (Page 195)

a. Find the word.
Question 1.
P r e l I g n p e x ………… a word that starts with the letter ‘p’.
i. This word gives the meaning ‘confusing’.
ii. A number of puzzling questions formed in his mind. (One of the words in this sentence can be replaced by the hidden word.
iii. ‘If you find something confusing, please inform me. (One of the words in this sentence can be replaced by the hidden word.)
iv. …………………………………
Answer:
1. perplexing

Question 2.
N E S A K –
i. This word gives the sense ‘secret’.
ii. This word can be formed by just interchanging the position of two letters of the word ‘snake’.
iii. The burglar tried to creep into the house. (The hidden word can replace the word given in bold letters in the sentence.)
iv. To escape from his father he ……………………… into the room.
Answer:
sneak

Question 3.
Q v u e r i-
i. It is a container for holding arrows, bolts, or darts. (noun)
ii. The word also means to shake rapidly. (verb)
iii. If the washing machine is overloaded, it will shake with rapid motion. (The hidden word can replace the word given in bold letters in the sentence.)
iv. Arjuna took out the arrows from the ……………………………
Answer:
quiver

Question 4.
S r o a
i. Kites do this in the sky. (fly high)
ii. The cost of living continued to ………………………………… (increase)
Answer:
soar

Question 5.
N l o v e
i. It is a form of literature. (noun) It also means something new.
ii. Even though the making of the happiness machine failed, it was a …………… idea.
Answer:
novel

b) Soul narrated his experience with the happiness machine to one of his friends. Some words are missing. Fill up the narration using the appropriate form of the words you identified in the above activity

Answer:
a) novel
b) perplexing
c) sneaked
d) soaring
e) quivered

Let’s Edit:

Read the diary entry of Swaminathan on the day he got a newcomer in his class. He has made some mistakes while writing the diary. The mistakes are given in bold letters. Correct them.

Answer:
a) who
b) striking
c) he has (remove also)
d) well-dressed
e) was impressed
f) whom

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Earlier Attempts for Classification of Elements

Kerala Syllabus 9th Standard Chemistry Notes Chapter 4 Question 1.
Explain the earlier attempt of classification by Lavoiser?
Answer:
Antoine Lavoisier classified the known elements into metals and nonmetals. But he was not able to duly classify metalloids.

9th Class Chemistry Periodic Table Kerala Syllabus Question 2.
Explain Newland’s law of octaves?
Answer:
Newlands arranged elements in the increasing order of atomic mass. He noticed that every eighth element has properties similar to those of the first elements. But this peculiarity could be noticed in elements upto calcium only.

Octaves of Newlands

9th Class Chemistry 4th Chapter Kerala Syllabus Question 3.
Define Mendeleev’s periodic law?
Answer:
In 1869 Mendeleev arranged the known 63 elements in horizontal and vertical columns and gave shape to the periodic table. He found that the chemical and physical properties of elements repeat at a regular intervals when they were arranged in the increasing order of atomic masses. Based on this Mendeleev proposed the periodic law of elements. The law states that physical and chemical properties of elements are periodic function of their atomic masses.

Chemistry Notes For Class 9 Periodic Table Kerala Syllabus Question 4.
What is meant by groups and periods in the periodic table?
Answer:
The vertical columns in the periodic table are known as groups and the horizontal rows are called periods.

Kerala Syllabus 9th Standard Chemistry Notes Question 5.
Evaluate the Mendeleev’s periodic table and find the following.
a) Total number of periods
b) Total number of Groups
c) Are the same elements showing similar properties arranged in the same group or same period?
Answer:
a) 6
b) 8
c) Same group

Class 9 Chemistry Notes Kerala Syllabus Question 6.
Advantages of Mendeleev’s periodic table?
Answer:
1. For the first time elements were comprehensively classified in such a way that elements of similar properties were placed in the same group. This has made the study of chemistry easy.
2. When the classification was made in such a way that the elements of similar properties came in the same group. It was noticed that certain their proper group. The reason for this was wrongly determined atomic masses and consequently, those wrong atomic masses were corrected.
Eg. The atomic mass of beryllium was known to be 14. Mendeleev reassessed it as a and assigned beryllium a proper place.
3. Columns were left vacant for elements which were not known at the time and their properties were predicted also. This gives an impetus to experiments in chemistry.

Ex Mendeleev give names Eka aluminum and Eka silicon to those elements which were to come below aluminum and silicon respectively in the periodic table and predicted their properties. Later when these elements gallium and germanium were discovered the prediction of Mendeleev turned out to be true.

Hsslive Guru 9th Chemistry Kerala Syllabus Question 7.
Explain the limitation of Mendeleev’s Periodic table?
Answer:
1. Elements with large difference in properties were included in the same group.
eg. Hard metal like copper [Cu], Silver [Ag] were included along with soft metals like sodium [Na] potassium [K],
2. No proper position could be given to element hy¬drogen. Non-metallic hydrogen was placed along with metals like sodium [Na] and potassium [K]
3. The increasing order of atomic mass was not strictly followed throughout.
eg. Co and Ni, Te and I
4. As isotopes are atoms of same element having different atomic masses, they should have been given different position while arranging them in the order of atomic mass. But this was not done.

Counting Atomic Calculator is a free online tool that displays the atomic mass for the given chemical formula.

Modern Period Table

Periodic Table 9th Class Pdf Kerala Syllabus Question 8.
State and explain modern periodic table and mod-ern periodic law?
Answer:
In 1913 Mosely through his x-ray diffraction experiments proved that the properties of elements depended on the atomic number not on the atomic mass.

According to this the periodic law of Mendeleev and the periodic table were modified consequently the modern periodic table was prepared by arranging elements in the increasing order of atomic number. The modern periodic law states that the physical and chemical properties of elements are periodic function of their atomic number.

Periodic Table Chapter Class 9 Kerala Syllabus Question 9.
How many periods in the modern periodic table?
Answer:
7

Labour India Class 9 Chemistry Kerala Syllabus Question 10.
Which is the shortest period?
Answer:
I period

9th Class Chemistry Chapter 4 Kerala Syllabus Question 11.
Number of elements in the third period?
Answer:
8

Class 9 Chemistry Periodic Table Kerala Syllabus Question 12.
Total number of groups?
Answer:
18

Periodic Table In 9th Class Kerala Syllabus Question 13.
Explain representative elements?
Answer:
Elements of group 1 and 2 also those in groups of 13 – 18 are called representative elements it belongs to metals, nonmetals, and metalloids.

Periodic Table Chemistry Class 9 Kerala Syllabus Question 14.
Do in representative elements do they include metalloids [eg. Si, Ge, As, Sb…) exhibiting the characteristics of metals and non-metals?
Answer:
Yes

Periodic Table Notes Pdf Class 9 Kerala Syllabus Question 15.
As there elements existing in solid, liquid and gaseous state find examples?
Answer:

• In solid-state- sodium, aluminum, carbon
• In liquid state – Bromine
• In gaseous state – oxygen, neon, argon

9th Class Chemistry Chapter 4 Notes Kerala Syllabus Question 16.
Write the electronic configuration of elements with atomic number 1-10
Answer:

The atom of the elements of these group show the periodically in electron filling they contain 1 -8 electron in their outermost shell. The elements of these groups are called representative elements.

Noble Gases

Kerala Syllabus 9th Standard Notes Chemistry Question 17.
List the elements in group 18
Answer:
Helium, Neon, Argon, Krypton, Xenon, Radon

9 Class Chemistry Chapter 4 Kerala Syllabus Question 18.
Now try to write their electronic configuration
Answer:
₂He – 2
₁₀Ne – 2, 8
₁₈Ar – 2, 8, 8
₃₆Kr – 2, 8, 18, 8
₅₄Xe – 2, 8, 18, 18, 8
₈₆Rn – 2, 8, 18, 32, 18, 8

Question 19.
How many electrons are there in the outermost shell of each element?
Answer:
8

Question 20.
The elements do not normally take part in chemical reactions. Find the reason?
Answer:
They have a stable configuration in the outermost shell.

Transition Elements

Question 21.
Which group of elements belong to transition elements?
Answer:
Elements of group 3-12 in the periodic table are transition elements.

Question 22.
Find out whether elements familiar to you are present in these groups?
Answer:
Copper, silver, gold, iron

Question 23.
Aren’t transition elements metals?
Answer:
Yes

Question 24.
What are the characteristics of transition elements?
1. They from coloured compound,
2. They show similarity in properties as well as in a period.
3. In compounds, they exhibit different oxidation state
eg. Fe²⁺ and Fe³⁺

Lanthanides and Actinoids

Question 25.
Which element is next to lanthanum with atomic number 57 of group 6 in the periodic table?
Answer:
Cerium with atomic No. 58

Question 26.
Find out the position allotted to the elements with atomic number 58-71?
Answer;
Separate position at the bottom of the periodic table.

Question 27.
Is the same way aren’t the elements with atomic number 90 to 103 of period 7 give separate positions at the bottom of the periodic table?
Answer:
Yes. These elements are called inner transition elements.

Question 28.
What is meant by inner transition elements?
Answer:
Inner transition elements from Cerium [Ce] to Lutecium [Lu] of period 6 are called lanthanides. Inner transition elements from Thorium (Th] to Lewrencium [Lr] of period 7 are called actinoids. Lanthanoids are also called rare earth. Actinoids are man-made artificial elements (except thorium and uranium).

Periodic trends in the periodic table

Question 29.
Electronic configuration of group I elements of the periodic table are given
Answer:

Question 30.
What is the peculiarity seen in the electronic configuration of the outer most shell of these elements?
Answer:
All these elements we can see one electron in the outermost shell.
Hence elements of group I exhibit similarity in chemical properties.

Question 31.
Which are the electrons shows the chemical properties of elements?
Answer:
Outermost electrons.

Question 32.
Is there any relationship between the group number and the number of electrons present in the outermost shell? What is it?
Answer:
Same, group number equal tot he number of election in the outermost shell for the elements in groups 1 and 2.

Question 33.
Observe figure the electronic configuration of the second-period elements of the group from 13-18 given below.

(i) Won’t we get the group number of these elements by adding 10 to the number of elements by adding 10 to the number of electrons in the outermost shell?
Answer:
Yes

(ii) Analyze table 3.1 and find whether there is any relation between the number of shells in an atom and the number of periods?
Answer:
Number of shells in an atom and the period number is same.

Size of an Atom in Group

Question 34.
Are you familiar with the Bohr model of an atom? See the Bohr model of atoms of certain elements, in group I.

(i) Which among them is the biggest?
Answer:
Potassium (K)

(ii) Which one is the smallest?
Answer:
Hydrogen [H]

(iii) What happens to the size of an atom when we move down the group?
Answer:
Increases

(iv) What is the reason for this?
Answer:
Number of shells increases.
As we move from top to bottom of a group in the periodic table the size of the atom increases as there is an increase in the number of shells.

Atomic Size in Period

See (Fig 3.3) the representation of Bohr model of elements with atomic numbers 3 to 9 in the second period of the periodic table.

Question 35.
Is there are in the number of shells with the increase in atomic number?
Answer:
No.

Question 36.
What happens to the nuclear charge with increase in atomic number?
Answer:
On moving from left to right in a period, as nuclear charge increases, the force of attraction on the outer-most electrons increases and consequently the size of atom decreases.

Ionisation Energy

Question 37.
You have understood how sodium chloride is formed by combining sodium and chlorine atoms. The Bohr model of sodium and chlorine are given below
Answer:

(i) Which among these atoms lose electrons?
Answer:
Sodium atom

(ii) Which one gains electrons
Answer:
Chlorine atom

Question 38.
How the ions are formed?
Answer:
Atom becomes charged when there is transfer of electrons [Lose or gain electrons] they are called ions.

Question 39.
Define ionization energy?
Answer:
The amount of energy required to liberate the most loosely bound electrons from the outermost shell of an isolated gaseous atom of an element is called ionization energy.

Question 40.
What are the factors affecting the ionization energy? Nuclear charge
Answer:
Size of the atom

Question 41.
When the size of an atom increases, does the attraction of the nucleus on the outermost electron increase or decrease?
Answer:
Decrease

Question 42.
Then what is the change in ionization energy?
Answer:
As the size of atom increases ionization energy decreases.

Question 43.
Can you find out how ionization energy changes as we move from top to bottom in a group?
Answer:
Ionization energy decreases.

Question 44.
What is the general trend in the variation of ionization energy on moving across a period from left to right?
Answer:
Ionization energy increases

Question 45.
Find how ionization energy changes with increase in nuclear charge?
Answer:
On moving from left to right in a period, as nuclear charge increases, the size of the atom decrease hence ionization energy increase.

Question 46.
Define electronegativity?
Answer:
In the case of two atoms joined by a covalent bond, electronegativity is the ability of each atom to attract the bonded electrons.

Question 47.
How size of an atom influence the electronegativity?
Answer:
As the size of an atom increases the distance between the nucleus and the outermost electron increases, hence the electronegativity decreases. As we move in the same period form left to right size of atom decrease hence electronegativity increases.

Question 48.
What is the basis for the chemical properties of metals and non-metals?
Answer:
Metals are the elements which give away the electrons and those that accept electrons are generally non-metals. Metals are electropositive elements because they lose electrons to form positive ions. Non-metals are called electronegative elements because they gain electrons in chemical reactions to form negative ions.

Question 49.
What is relationship between metallic character and the size of an atom?
Answer:
As the size of the atom increases metallic character also increases.

Question 50.
How do the metallic character and nonmetallic character vary while moving from left to right in a period? Arriving at a conclusion by assessing the size of atom?
Answer:
In the periodic table, while moving from to top to bottom in groups metallic character generally in-creases while non-metallic character decreases.
In a period as we move form left to right metallic character generally decreases while non-metallic character increases.

Question 51.
Don’t you think that there is a relationship between ionization energy and metallic -non-metallic character? Is the element with highest ionization energy metallic or non-metallic?
Answer;
Non-metallic

Question 52.
Then what about those having the low ionization energy?
Answer:
Metals

Question 53.
Isn’t there a relationship between electronegativity and metallic, non-metallic character? Explain the relationship?
Answer:
Non-metals are more electronegative.

Metalloids

Question 54.
Explain metalloids?
Answer:
Elements exhibiting the properties of both metal as well as nonmetal are called metalloids, eg. Silicon [Si], germanium [Ge] Arsenic [As], Antimony [Sb] and Tellurium [Te] belongs to this category.

Question 55.
You must understand certain periodic trends in the periodic table? Based on these (✓)the correct option given below in table 3.7.

Answer:

Let Us Assess

Question 1.
The table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Fill in the blanks.

Answer:

Question 2.
Complete the table
Answer:

Question 3.
Symbols of certain elements are given. Write their electronic configuration and find the period and group in which they are included.

Answer:
a) \(_{6}^{12} C\)
Electronic configuration 2, 4
Period – 2
Group – 14
b) \(_{12}^{24} \mathrm{Mg}\)
Electronic configuration 2,8,2
Period-3
Group – 2
c) \(_{17}^{35} \mathrm{Cl}\)
Electronic configuration 2,8,3
Period – 3
Group – 17
d) \(_{13}^{27} \mathrm{Al}\)
Electronic configuration 2,8,3
Period – 3
Group – 13
e) \(\begin{array}{l}{20} \\ {10}\end{array} \mathrm{Ne}\)
Electronic configuration 2,8
Period – 2
Group -18

Question 4.
There are three shells in the atom of element ‘X’, 6 electrons are present in its outermost shell.
a) Write the electronic configuration of the element.
b) What is its atomic number?
c) In which period does this element belong?
d) In which group is this element included?
e) Write the name and symbol of this element.
f) To which family of element does is this element belong to?
g) Draw and illustrate the Bohr atom model of this element.
Answer:
a) 2, 8, 6
b) 16
c) 3
d) 16
e) Sulphur, ‘S’
f) Oxygen family

Question 5.
Electronic configurations of elements P, Q, R, and S are given below. (These are not actual symbols).
P – 2, 2
Q -2, 8, 2
R – 2, 8, 5
S – 2, 8
a) Which among these elements are included in the same period?
b) Which are those included in the same group?
c) Which among them is a noble gas?
d) To which group and period does the element R belong?
Answer;
a) P and S, Q and R – belongs to same period
b) P and Q, belongs to same group
c) S
d) R belongs to 3rd period and 15th group

Question 6.
An incomplete form of the periodic table is given below. Write answers to the questions connecting the position of elements in it.

a) Which is the element with the biggest atom in group 1?
b) Which is the element having very lowest ionization energy in group 1?
c) Which element has the smallest atom in period 2?
d) Which among them are transition elements?
e) Which of the elements L and M has the lowest electronegativity?
f) Among B and I which has higher metallic character?
g) Which among these are included in the halogen family?
h) Which is the element that resembles E the most in its properties?
Answer:
a) D
b) D
c) M
d) G, H
e) L
f) B
g) M, N
h) F

Periodic Table Model Questions and Answers

Question 1.
Symbols of certain elements are given write down the electronic configuration and find the period and group in which they are included.

Answer:
a) \(\begin{array}{l}{23} \\ {11}\end{array} \mathrm{Na}\)
Electron configuration 2, 8, 1
Period – 3
Group – 1
b) \(_{17}^{35} \mathrm{Cl}\)
Electron configuration -2, 8, 7
Period – 3
Group-17
c) \(_{9}^{19} \mathrm{F}\)
Electron configuration -2, 7
Period – 3
Group – 17

Question 2.
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real)
A – 2, 2
B – 2, 8, 5
C – 2, 7
D – 2, 8, 2
a) Find the elements belongs to same period?
b) Find the elements belongs to same group.
c) ‘C’ belongs to which period and group?
Answer:
a) B, D, and A, C because the number of shells are same.
b) A, D because the number of electrons in the outermost shell is same.
c) ‘C’ belongs to second period and 17th group

Question 3.
Table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Make them in the correct order.

Answer:

Question 4.
An incomplete form of the periodic table is given below.
write answers in the question connecting the position of elements in it.

1. Which element has the largest atomic size in group I?
2. Write the transition elements?
3. Which element has the lowest ionization energy in the 2nd period?
4. Which element belongs to Noble gas elements?
5. Compare L, M which element has the lowest electronegativity?
6. Write the element belongs to Halogen family?
Answer:
1. D
2. F, G, H
3. C
4. N
5. L
6. M

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Textbook Page No. 100

Similar Triangles Class 9 Kerala Syllabus Question 1.
The perpendicular from the square corner of a right triangle cuts the opposite side into two parts of 2 and 3 centimetres length.

i. Prove that the two small right triangles cut by the perpendicular have the same angles.
ii. Taking the length of the perpendicular as h, prove that \(\frac{h}{2} = \frac{3}{h}\)
iii. Calculate the perpendicular sides of the large triangle.
iv. Prove that if the perpendicular from the square corner of a right triangle divides the opposite side into parts of lengths a and b and if the length of the perpendicular is h, then h² = ab.
Answer:
i. ∠ADB = 90°
Let ∠BAD = x°
∠DAC = ( 90 – x)°

∴ ∠B = 180 – (x + 90) = 90 – x
and ∠C = 180 – (90 – x + 90) = x
Angles of ΔABC = x°, (90 – x)°, 90°
Angles of ΔACD = x°, (90 – x)°, 90°
The two small triangles have the same angles.

ii. Triangle with the same angles, sides opposite equal angles are scaled by the factor.
2 : h = h : 3 ⇒ \(\frac{2}{h} = \frac{h}{3}\)
i.e \(\frac{h}{2} = \frac{3}{h}\)

iii. \(\frac{h}{2} = \frac{3}{h}\)
h² = 6 ⇒, h = \(\sqrt 6\)
In ΔABD
AB² = BD² + AD² = 2² + (\(\sqrt 6\))²
= 4 + 6 = 10
AB = \(\sqrt 10\)
In ΔADC
AC² = DC² + AD² = 3² + (\(\sqrt 6\))²
= 9 + 6 = 15
AC = \(\sqrt 15\)
The perpendicular sides are \(\sqrt 10\) and \(\sqrt 15\)

iv. Triangle with the same angles, sides opposite equal angles are scaled by the same factor.
BD : AD = AD : DC

\(\frac{h}{a} = \frac{b}{h}\)
h² = ab

Similar Triangles Class 9 Question 2.
At two ends of a horizontal line, angles of equal size are drawn and two points on the slanted lines are

i. Prove that the parts of the horizontal line and parts of the slanted line are in the same ratio.
ii. Prove that the two slanted lines at the ends of the horizontal line are also in the same ratio.
iii. Explain how a line of length 6 cm can be divided in the ratio 3 : 4
Answer:

i. ∠A = ∠B
∠AMC = ∠BMD
∴ ∠C = ∠D
(The triangles AMC and DMB are similar).
∴ \(\frac{MC}{MD} = \frac{AC}{BD} = \frac{AM}{MB}\)
\(\frac{AM}{MB} = \frac{MC}{MD}\)

ii. \(\frac{MC}{MD} = \frac{AC}{BD} = \frac{AM}{MB}\)
(From the similar triangles AMC and DMB)

iii. Draw a line segment of length 6 cm. At one end of this line AB, 3cm long and at the other end draw CD, 4cm long in the opposite direction.

∠CAB = ∠ACD
Draw BD to cut A at O.
Since AB : CD = OA : OC
∴ OA : OC = 3 : 4

Scert Class 9 Maths Similar Triangles  Question 3.
The mid point of the bottom side of a square is joined to the ends of the top side and extended by the same length. The ends of these lines are joined and perpendiculars are drawn from these points to the bottom side of the square extended.

i. Prove that the quadrilateral obtained thus is also a square.
ii. Explain how we can draw a square with two corners on a semicircle and the other two corners on its diameter as in the figure

Answer:

i. ΔDCM and ΔSRM are similar.
The angles of the ΔMBC and ΔMAD are equal.
ΔPMS, ΔQMR are also similar triangles.
∠D = ∠S; ∠C = ∠R; ∠A = ∠B; ∠B = ∠Q
Let the ratio of the equal sides are be k.
\(\frac{QR}{BC}\) = k ⇒ \(\frac{QR}{2a}\) = k
∴ QR = 2ak
Similarly, PS = PQ = SR = 2ka
∴ PQRS is a square.

ii.

Draw a line of AB, 8cm long. Find midpoint O on the line. Mark P and Q such that OP = OQ = 1cm. Draw the square PQRS whose sides are 2cm long.

Draw a semicircle with O as centre and OA as radius. Extend OS and OR to meet the semicircle at G and F. From G and F draw GD and FE perpendicular AB. Draw GF. Quadrilateral DEFG is the required square.

Kerala Syllabus 9th Standard Maths Chapter 7 Question 4.
The picture shows a square drawn sharing one corner with a right triangle and the other three corners on the sides of this triangle.

i. Calculate the length of a side of the square.
ii. What is the length of a side of the square drawn like this within a triangle of sides 3, 4 and 5 centmetres?
Answer:
i.

ΔABC and ΔAPQ are similar.
\(\frac{x + 2}{2} = \frac{x + 1}{x}\) = x(x + 2) = 2(x +1)
x² + 2x = 2x + 2
x² = 2; x = \(\sqrt 2\)
Side of square = \(\sqrt 2\)cm.

ii. AB = 4, BC = 3, AC = 5
ΔAPQ and ΔABC are similar triangles
∴ \(\frac{AP}{PB} = \frac{PQ}{BC} ⇒ \frac{4 – x}{4} = \frac{x}{3}\)
3(4 – x) = 4x
12 – 3 = 4x
12 – 7x; x = \(\frac {12}{7}\)

The length of the side of the square = \(1\frac {5}{7}\)

Chapter 7 Similar Triangles Kerala Syllabus Question 5.
Two poles of heights 3 m and 2 m are erected upright on the ground and ropes are stretched from the top of each to the foot of the other.

i. At what height above the ground do the ropes cross each other?
ii. Taking the heights of the poles as a and b and height above the ground of the point where the ropes cross each other as h, And the relation between a, b and h.
iii. Prove that this height would be the same, whatever be the distance between the poles.
Answer:
i. Consider ΔABC and ΔAFE. They are similar.
\(\frac{b}{h} = \frac{x + y}{x}\) ………..(1)

Consider ΔADB and ΔFEB. They are also similar
\( \frac{a}{h} = \frac{x + y}{y}\) ………..(1)
\( \frac{h}{a} = \frac{y}{x + y}, \frac{h}{b} = \frac{x}{x + y}\)
(1) + (2) ⇒ \( \frac{h}{a} + \frac{h}{b} = \frac{y}{x + y} + \frac{x}{x + y}\)
h\((\frac{1}{a} + \frac{1}{b}) = \frac{x + y}{x + y}\)
h\((\frac{1}{a} + \frac{1}{b})\) = 1
\(\frac{1}{a} + \frac{1}{b} = \frac{1}{h}\) ……..(2)
a = 3 and b = 2
\(\frac{1}{h} = \frac{1}{3} + \frac{1}{2} + \frac{5}{6}\)cm
h = \(\frac{6}{5}\) = 1.2

ii. The heights of the poles are a and b and the height above the ground of the point where the ropes cross each other is h, then the relation between a, b and h is \(\frac{h}{a} + \frac{h}{b}\) = 1 or \(\frac{a + b}{ab} = \frac{1}{h}\) from equation (3).

iii. Only change its height according to the height of the poles not the distance.

Textbook Page No. 107

9th Class Maths Notes Kerala Syllabus Question 1.
Draw a triangle of angles the same as those of the triangle shown and sides scaled by \(1\frac{1}{4}\)

Answer:
One side is 6 cm. Its \(1\frac{1}{4}\) part is 7.5 cm
Other sides are 4 × \(1\frac{1}{4}\) = 5 cm
8 × \(1\frac{1}{4}\) = 10 cm
Draw one side in 10 cm. Complete the triangle with given measures.

Kerala Syllabus Class 9 Maths Solutions Question 2.
See this picture of a quadrilateral.

i. Draw a quadrilateral with angles the same as those of this one and sides scaled by \(1\frac{1}{2}\)
ii. Draw a quadrilateral with angles different from those of this and sides scaled by \(1\frac{1}{2}\).
Answer:
i. Draw a quadrilateral by increasing all the sides including the diagonal by \(1\frac{1}{2}\)
4 × \(1\frac{1}{2}\) = 4 × 1.5 = 6
2 × \(1\frac{1}{2}\) = 2 × 1.5 = 3
3 × \(1\frac{1}{2}\) = 3 × 1.5 = 4.5
5 × \(1\frac{1}{2}\) = 5 × 1.5 = 7.5
6 × \(1\frac{1}{2}\) = 6 × 1.5 = 9

ii. Draw a quadrilateral by increasing all the sides increased by \(1\frac{1}{2}\) except the diagonal.

Textbook Page No. 111

Kerala Syllabus 9th Standard Maths Notes Question 1.
The picture shows two circles with the same centre and two triangles formed by joining the centre to the points of intersection of the circles with two radii of the larger circle. Prove that these triangles are similar

Answer:

AP = AQ (radius)
∠APQ = ∠AQP
(Base angles of an isosceles triangle)
∠A = ∠A (Common angle)
AC = AB (radius)
∴ ∠ACB = ∠ABC
∴ \(\frac{AP}{AC} = \frac{AQ}{AB}\)
That is two sides of these triangles are scaled by the same factor.
Since two sides of these triangles are scaled by the same factor and the angle between them the same ΔAPQ and ΔACB are similar. So the triangles are similar

Kerala Syllabus 9th Standard Maths Guide In Malayalam Question 2.
The lines joining the circumcentre of a triangle to the vertices are extended to meet another circle with the same centre and these points are joined to make another triangle.

i. Prove that the two triangles are similar.
ii. Prove that the scale factor of the sides of the triangle is the scale factor of the radii of the circles.
Answer:
∴ ΔAOB and ΔPOQ are similar triangles

∴ \(\frac{AB}{PQ} = \frac{OB}{OQ}\) ……..(1)
ΔBOC, ∠QOR are similar
∴ \(\frac{OB}{OQ} = \frac{BC}{QR} = \frac{OC}{OR}\) ……..(2)
Similarly
\(\frac{OC}{OR} = \frac{AC}{PR}\) ……..(3)
(1), (2), (3) consider
∴ \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}\)
ΔABC and ΔPQR are similar triangles.

ii. From equation (1), (2) and (3)
The scale factor of the sides of the triangle is the scale factor of the radii of the circles.

9th Class Maths Kerala Syllabus Question 3.
A point inside a quadrilateral is joined to its vertices and the lines are extended by the same scale factor. Their ends are joined to make another quadrilateral.

i. Prove that the sides of the two quadrilateral are scaled by the same factor.
ii. Prove that the angles of the two quadrilaterals are the same.
Answer:
i.

ΔAOB, ΔPQO are similar
\(\frac{OA}{OP} = \frac{AB}{PQ} = \frac{OB}{OQ}\) …….(1)
ΔOBC, ΔOQR are similar
\(\frac{OB}{OQ} = \frac{BC}{QR} = \frac{OC}{OR}\) …….(2)
ΔOCD, ΔORS are similar
\(\frac{OC}{OR} = \frac{CD}{RS} = \frac{OD}{OS}\) …….(3)
ΔODA, ΔOSP are similar
\(\frac{OD}{OS} = \frac{AD}{PS} = \frac{OA}{OP}\) …….(4)
Consider (1), (2), (3), (4)
\(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{DC}{SR} = \frac{AD}{PS}\)
Since the sides of quadrilateral EFGH and quadrilateral ABCD are scaled by the same factor.
∴ The angles of the quadrilateral are also same (Angles of similar triangles).

ii. Since the sides of quadrilateral PQRS and quadrilateral ABCD are scaled by the same factor, the angles of the large quadrilateral will be the same as the angles of the small quadrilateral.

Kerala Syllabus 9th Standard Maths Similar Triangles Exam Oriented Text Book Questions and Answers

Class 9 Maths Notes Kerala Syllabus Question 1.

ΔABC, ΔXYZ are similar triangles in the picture. Then
\(\frac{AB}{XY} = \frac{BC}{….} = \frac{….}{XZ}\)
Answer:
\(\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}\)

Hss Live Class 9 Maths Kerala Syllabus Question 2.
In ΔXYZ PQ is parallel to XY.

a. \(\frac{YP}{PZ} = \frac{XQ}{….}\)
b. \(\frac{x}{a} = \frac{…..}{b}\)
Answer:
a. \(\frac{YP}{PZ} = \frac{XQ}{QZ}\)
b. \(\frac{x}{a} = \frac{y}{b}\)

Kerala Syllabus 9th Standard Maths Solutions Question 3.


Fill in the blanks according to the picture.
∠K = ∠A; ∠B = …….
∠C = ……; KL = ……cm
KM = ….. cm
Answer:
∠B = ∠L
∠C = ∠M
KL = 8cm
KM = 10cm

9th Standard Maths Notes Kerala Syllabus Question 4.
In the diagram AP and BQ are perpendicular to AB. AP = 4cm, BQ = 2cm Then show that AC : CB = 2 : 1

Answer:
∠PAC = 90° ∠CBQ = 90°
∠ACP = ∠BCQ (opposite angles)
∴ ΔPAC ≅ ΔBCQ
\(\frac{PA}{BQ} = \frac{AC}{CB} = \frac{PC}{CQ}\)
\(\frac{AC}{CB} = \frac{PA}{BQ} = \frac{AC}{CB} = \frac{2}{1}\)
AC : CB = 2 : 1

Class 9 Maths Chapter 7 Kerala Syllabus Question 5.
In ΔABC and ΔPQR we have \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}\). In triangle ABC, the altitude through P meets BC at D, and in triangle PQR. The altitude through P meets QR at ‘S’. Prove that \(\frac{AB}{PQ} = \frac{AD}{PS}\)

Answer:
ΔABC, ΔPQR are similar.
∠A = ∠P; ∠B = ∠Q;
∠C = ∠R;
Consider ΔADB ΔPSQ
∠B = ∠Q, ∠ADB = ∠PSQ = 90°
(AD ⊥ BC & PS ⊥ QR)
[each equal to 90°]
∴ ΔADB, ΔPSQ are similar.
[Since the corresponding sides of similar triangles are proportional]
∴ \(\frac{AD}{PS} = \frac{DB}{SQ} = \frac{AB}{PQ} ⇒ \frac{AB}{PQ} = \frac{AD}{PS}\)

Chapter 7 Maths Class 9 Kerala Syllabus Question 6.
Prove that if the area of two similar triangles are proportional to the squares on their corresponding sides.

Answer:
ΔABC ~ ΔPQR.
AM, PN are perpendicular, \(\frac{AM}{PN} = \frac{BC}{QR}\)
Area of ΔABC = \(\frac{1}{2}\) BC × AM
Area of ΔPQR = \(\frac{1}{2}\) QR × PN
\(\frac{Area of ΔABC}{Area of ΔPQR}\)

Std 9 Maths Kerala Syllabus Kerala Syllabus Question 7.
In the given figure, PC ⊥ QR and QD ⊥ PR. Prove that ΔPCR and ΔQDR are similar.

Answer:
In ΔPCR, ∠PCR = 90°, and in ΔQDR = 90°
∠R is common to both the triangles
ΔPCR, ΔQDR. ∠PCR = ∠QDR = 90°
∠PRC = ∠QRD (common)
(Two triangles having two pairs of corresponding angles equal. The triangles are similar.)

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 8.
In ΔABC, P is a point on BC. Where D, E, F are the mid-points on BP, AP and CP respectively, then prove that ΔABC ~ ΔDEF.

Answer:
Consider ΔABP, ΔDPE.
ΔABP ~ ΔDPE (∠A is ∠BAP = ∠DEP and ∠ABP = ∠EDP)
∴ \(\frac{AB}{ED} = \frac{AP}{EP}\) …..(1)
similarly ΔAPC ≅ ΔEPF.
∴ \(\frac{AC}{EF} = \frac{AP}{EP}\) …..(2)
(1) = (2) \(\frac{AB}{ED} = \frac{AC}{EF} = \frac{AP}{EP}\)
ΔABC ~ ΔDEF

Kerala Syllabus 9th Standard Maths Guide Pdf Question 9.
In figure AD || BC, AB || DE prove that ΔABC – ΔEDA .

Answer:
∠DAE = ∠C
(AD || BC corresponding angles)
∠BAC = ∠DEA
(AB || DE corresponding angles)
ΔABC ~ ΔEDA

Question 10.
ΔABC is a right angled triangle. ∠B = 90° a perpendicular line from B to AC intersect AC at D. Prove that ΔABC, ΔADB and ΔBCD are similar to each other.
Answer:

ΔABC = ΔABD
[∠ABC = ∠ADB = 90°, ∠ACB = ∠ABD = 90 – X]
∠A (common)
similarly ΔABD ~ ΔBDC
In ΔABC, consider
∠A = X°
∠B = 90°
∠A + C = 90°, ∠C = 90 – X
In ΔABD, ∠ADB = 90°
Therefore ∠A = X°, ∠ABC = 90°
∠A + ∠ABD = 90°
∠ABD = 90 – ∠A = 90 – X;
In ΔBCD, ∠BDC = 90°
∠C = (90 – X); ∠C + ∠DBC = 90°;
∠DBC = 90 – (90 – X) = X°
Hence angles of three triangles are 90°, X° and 90° – X°.
Hence three triangles are similar to each other.

You can Download Chemical Bonding Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding

Chemical Bonding Textual Questions and Answers

Kerala Syllabus 9th Standard Chemistry Notes Chapter 2 Question 1.
What peculiarity do you see in the electronic configuration of noble elements except Helium? Except Helium all other elements have 8 electrons in the outermost shell, hence shall be considered to be chemically stable.

The arrangement of eight electrons in the outermost shell of atom is called octet electron configuration. In Helium atom there is only one shell. The maximum number of electrons in the first shell is 2. Hence two-electron pattern system of Helium also stable.

Kerala Syllabus 9th Standard Chemistry Chapter 2 Question 2.
The electronic configuration of some elements are given below.

Is the number of electrons in the outermost shell of these elements the same as that of the elements in Table (2.1).
Answer:
No

(i) You are familiar with the compounds of these elements. Write the names of some compounds?
Answer:
Magnesium chloride, Sodium oxide, Sodium chloride.

(ii) How are atoms in these compounds held together?
Answer:
Strong attractive force

(iii) What is meant by Chemical Bonding?
Answer:
The attractive force that holds the atoms together in the formation of a molecule is called chemical bonding.

Ionic Bonding

Class 9 Chemistry Chapter 2 Notes Kerala Syllabus Question 3.
In the formation of sodium chloride which atoms are combing.
Answer:
Sodium, chlorine

Chemical Bonding Notes Class 9 Kerala Syllabus Question 4.
How many electrons are there in the outermost shell of sodium atom?
Answer:
1

Chemical Bonding Questions And Answers Class 9 Kerala Syllabus Question 5.
How many electrons are there in outermost shell of chlorine?
Answer:
7

Chemical Bonding Questions And Answers Pdf Class 9 Kerala Syllabus Question 6.
How do chlorine and sodium attain stability?
Answer:
Sodium donates one electron to chlorine to become sodium ion [Na+] and chlorine become chloride ion [Cl]

Chemical Bonding Class 9 Notes Pdf Kerala Syllabus Question 7.
Analyze the electron transfer in each atom during the formation of sodium chloride.
Answer:

Chemical Bonding Class 9 Pdf Kerala Syllabus Question 8.
Draw the electron dot diagram of the transference of electron of sodium atom and chlorine atom. The diagram represents only electrons in the outermost shell because they are the only electrons participating- in chemical bonding.
Answer:

Questions On Chemical Bonding Class 9 Kerala Syllabus Question 9.
Complete Table 2.3 by examining the arrangement of electrons before and after the chemical reaction during the formation of sodium chloride.

a) Which atom donates electron? How many electrons?
b) Which atom accepts electron? How many electrons?
Answer:
a) Sodium, one electron
b) Chlorine, one electron

Kerala Syllabus 9th Standard Chemistry Notes English Medium Question 10.
Electron transfer during the formation of sodium chloride can be written in the form of an equation
Na → Na⁺+1e^–
Cl + 1e^– → Cl^–
Answer:
During the formation of sodium chloride sodium atom donates electron and gets converted to sodium ion (Na⁺) chlorine accepts an electron to form chloride ion (Cl^– ). Through this sodium and chlorine atoms complete an octet in their outermost shell to attain stability.

The oppositely charged ions thus formed are held together by electrostatic force of attraction. This attractive force is called Ionic Bond. Sodium chloride contains ionic bond.

Kerala Syllabus 9th Standard Chemistry Chapter 1 Question 11.
Define Ionic Bond?
Answer:
Ionic bond is a chemical bond formed by electron transfer in an ionic bond, the ions are held together by the electrostatic force of attraction between the oppositely charged ions.

Kerala Syllabus 9th Standard Chemistry Solutions Question 12.
Explain the formation of magnesium oxide from magnesium and oxygen?
Analyze the electron dot diagram and complete the table.
Answer:

To attain stability magnesium donates 2 electrons to become magnesium ion (Mg₂⁺) and – oxygen become [O₂^– ] ion. This type of bonding is ionic bonding.

Chemical Bonding Notes Class 9 Pdf Kerala Syllabus Question 13.
How the ionic bond formation of sodium oxide is represented?
[Hint: Atomic No. of sodium 11, oxygen 8]
Answer:

Question 14.
Draw the electron dot diagram of following compounds. [Hint: Atomic No. Na=11, F=9, Mg=12]
Answer:
1. Sodium Flouride [NaF]

Question 15.
Define ionic compounds?
Answer:
Compounds formed by ionic bonding are called ionic compound.

Covalent Bonding

Fluorine [F₂], Chlorine [Cl₂] oxygen [O₂] Nitrogen [N₂] etc. are diatomic molecules. Let us examine the formation of these molecules.
The Bohr atom model of fluorine is given in figure.

Question 16.
Write the atomic number of fluorine?
Answer:
9

Question 17.
The electronic configuration of Fluorine
Answer:
2, -7 ‘

Question 18.
How many electrons are required for one fluorine atom to attain the octet?
Answer:
1

Question 19.
Is there a possibility of transferring electrons from one fluorine atom to another fluorine atom?
Answer:
No.

Question 20.
How can the two fluorine atoms attain an octet arrangement?
Answer:
By sharing of electrons

Question 21.
The manner in which the two fluorine atoms in a fluorine molecule undergo chemical bonding is illustrated in fig. 2.6.
Answer:

Question 22.
What happens during the formation of fluorine molecule electron transfer or electron sharing?
Answer:
Electron sharing

Question 23.
How many pairs of electrons are shared?
Answer:
One pair

Question 24.
How covalent bonds are formed?
Answer:
The chemical bond formed as a result of the sharing of electrons between the combining atoms is called a covalent bond.

Question 25.
How single bonds are formed?
Answer:
Single bonds are formed by sharing one pair of electrons. It is represented by a small line between the symbols of the combining element, eg. Fluorine molecules can be represented as F – F. The atomic number of chlorine is 17.

Question 26.
Write down the electronic configuration?
Answer:
2, 8, 7

Question 27.
Draw the electron dot diagram of the formation of chlorine molecule by combining two chlorine atoms?
Answer:

Here one pair of electrons – sharing hence single bond is formed.

Question 28.
Examine the diagram illustrating the chemical bonding in the molecule of oxygen and nitrogen.
Answer:

In oxygen molecules, two pairs of electrons are shared hence this type of covalent bond is double bond.
In nitrogen molecule, three pairs of electrons are shared hence this type of covalent bond is triple bond.

Question 29.
Complete table 2.5 given below related to covalent bonding.

Answer:

Question 30.
Draw the chemical bond formation of hydrogen chloride [HCI]

a) How many electron pairs are shared?
b) Represent chemical bond by using symbols?
Answer:
a) One pair of electrons
b) H – Cl

Question 31.
Examples of some covalent compounds are given draw the chemical bonds of the compound by using electron dot diagram.
a)CH₄
b) HF
c)H₂O
Answer:
a)

Electronegativity

Question 32.
Define electronegativity?
Answer:
In a covalent bond the relative ability of each atom to attract the bonded pair of electrons towards itself is called electronegativity.

Question 33.
Who proposed the electronegativity scale?
Answer:
Linus Pauling

Question 34.
Some compounds and their nature are shown in table (2.6) complete the table by finding out the electron negativity difference between the constituent elements.

Answer:

Generally, the electronegativity difference of the component elements in a compound is 1.7 or more it shows ionic character. If it is less than 1.7 it shows covalent character.

Polar Nature

Question 35.
Consider the case of hydrogen molecule [HCl]
a) What is the electronegativity of hydrogen?
b) What is the electronegativity of chlorine?
c) The atomic nucleus of which of these elements has a greater tendency to attract the shared pair of electrons?
d) The chlorine atom with a higher electronegativity attracts the shared pair of electrons towards its nucleus. As a result, the chlorine atom in hydrogen chloride develops a partial negative charge g: (delta negative) and hydrogen atom develops a partial positive charge δ+ (delta positive) it can be represented below
Answer:
a) 2.2
b) 3.16
c) Chlorine
d) \(\begin{array}{l}{\delta^{+} \quad \quad \delta^{-}} \\ {H-C^{\prime}}\end{array}\) Compounds having partial electron charge separation with the molecule are called polar compound. HF, HBr, H20 are example of polar compounds.

Question 36.
Explain the properties of Ionic compounds and covalent compounds.
Answer:

Valency:

Question 37.
What is meant by valency?
Answer:
Valency is the combining capacity of the atoms of an element. It can be treated as the number of electrons lost gained or shared by an atom during chemical combination.

Question 38.
In the formation of sodium chloride- sodium donates one electron, chlorine accepts one electron write the valencies of each element?
Answer:
1

Question 39.
In the formation of magnesium oxide- How many electrons are donated by magnesium?
Answer:
2

Question 40.
How many electrons are accepted by oxygen?
Answer:
2

Question 41.
How is valency and electron transfer related in this case?
Answer:
Same

Question 42.
In the formation of hydrogen chloride, how many electron pairs are shared?
Answer:
One pair

Question 43.
What will be the valency of each atom?
Answer:
1

Question 44.
Complete the table given below analyze the change in the electronic arrangement of elements during the formation of each compound. Find how they are related to valency.

Answer:

From Valency to Chemical Formula

The chemical formula of some compounds are given
Sodium chloride – NaCl
Magnesium chloride – MgCl₂
Aluminium chloride – AICl₃
Carbon tetrachloride – CCl₄

Question 45.
The symbol of some elements and there valencies are given. Write the chemical formula of the compounds formed by them.

Answer:

Question 46.
Why does the number of chlorine atoms differ in these compounds? Try to find out by analyzing the valency of the elements Na, Mg, Al, Cl and C. Analyse Table 2.9
Answer:

Examine the above table and identify how to write the chemical formula from valency. Compare your findings with the following.

• First write the element with lower electronegativity.
• Exchange the valency of each element and write as suffix.
• Divide the suffix with the common factor.
• If the suffix is 1, it need not be written.

Let Us Assess

Question 1.
Complete the table given below and answer the following questions (symbols used are not true)

a) Which element in the table is the most stable one? Justify your answer.
b) Which element donates electrons in chemical reaction?
c) Write the chemical formula of the compound formed by combining element S with P.
Answer:

a) R – it contains an octet configuration in the outermost shell.
b) S
c) The valency of S = 2 and P is 1, hence the chemical formula

[The element in which the electron lose can be written first]

Question 2.
Electronegativity values of some elements are given. Using these values, find whether the following compounds are ionic or covalent.
(Electronegativity of Ca = 1, O = 3.5, C = 2.5, S = 2.58, H = 2.2, F = 3.98)
i) Sulphur dioxide (S02)
ii) Water (H₂O)
iii) Calcium fluoride (CaF₂)
iv) Carbon dioxide (CO₂)
Answer:
i) SO₂
Electronegativity difference = 3.5 – 2.58 = 0.92
If the electronegativity difference is less than 1.7, it shows covalent character.
ii) Water (H₂O)
Electronegativity difference = 3.5 – 2.2 = 1.3
If the electronegativity difference is less than 1.7 it forms covalent compounds.
iii) CaF₂
Electronegativity difference = 3.98 -1.0 = 2.98
If the electronegativity difference is greater than 1.7 if forms ionic compounds.
iv) CO₂
Electronegativity difference = 3.5 – 2. 5 = 1.0
If the electronegativity difference is less than 1.7 it shows covalent compounds.

Question 3.
Some elements and their valencies are given

a) Write the chemical formula of barium chloride
b) Write the chemical formula of zinc oxide
c) The chemical formula of calcium oxide is CaO. What is the valency of calcium?
Answer:

c) The valency of calcium is 2.

Question 4.
Examine the following chemical equations and answer the questions.
(Hint: Atomic Number Mg = 12 Cl = 17)
Mg + Cl₂ → MgCl₂
Mg → Mg²⁺ + ……..
Cl + 1e^– → ………..
a) Complete the chemical equations.
b) Which is the cation? Which is the anion?
c) Which type of chemical bond is present in MgCl₂?
Answer:
a) Mg → Mg^2- + 2e
Cl + 1e^– → Cl^–
b) The cation or positively charged ion is Mg²⁺ and the anion or negatively charged ion is Cl^–
c) Ionic Bonding

Extended Activities

Question 1.
Draw the electron dot diagram of chemical bonds in methane (CH₄) and ethane (C₂H₆).
Answer:

Question 2.
P, Q, R, S are four elements. Their atomic numbers are 8, 17, 12 and 16 respectively. Find the type of chemical bond in these compounds formed by combining the following pairs of elements. Construct and exhibit the type of bonds using different. substances (eg. pearls) (Electronegativity values:
P = 3.44, Q = 3.16, R = 1.31, S = 2.5)
1) P, R
2) P, S
3) Q, R
Answer:
1. PR
Electronegativity difference = 3.44 – 1.31 = 2.13
The electronegativity difference is greater 1.7 it shows ionic compound.
2. PS
Electronegativity difference = 3.44 – 2.58 = 0.86
The electronegativity difference is less than 1.7 it shows covalent compound.
3. QR
The electronegativity difference = 3.16 – 1.31 = 1.85
The electronegativity difference is greater than 1.7 it shows ionic compound.

Question 3.
Perform the experiments arranging the apparatus as shown in figure.

Record your observations and identify what type of compounds sodium chloride and sugar are
Answer:
When electricity is passed through sodium chloride solution in which carbon rod is immersed hydrogen and chlorine gas are produced. It is an ionic compound. In the second when electricity is passed through sugar solution there is no charge. Hence it belongs to covalent compound.

You can Download Parallel Lines Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines

Kerala Syllabus 9th Standard Maths Parallel Lines Text Book Questions and Answers

Textbook Page No. 88

Parallel Lines Class 9 Kerala Syllabus Question 1.
Draw an 8 cm long line and divide it in the ratio 2 : 3.
Answer:
Draw AB whose length is 8 cm. Draw AC with length 5 cm. Join BC. Let the point D divides AC in the ratio 2 : 3. Draw a line through D parallel to BC. This line divides AB in the ratio 2 : 3.

Kerala Syllabus 9th Standard Maths Chapter 6 Question 2.
Draw a rectangle of perimeter 15 cm and sides in the ratio 3 : 4
Answer:
Perimeter is 15 cm.
So length + breadth = 7.5 cm
Draw AB, a line segment of length 7.5 cm.
Draw AC from A, 4 + 3 = 7cm long.
Mark D on this line such that AD = 4 cm. Draw BC.
Draw a line parallel to BC and passing through D, which meets AB at E. Draw rectangle AEGF whose length as AE and breadth as EB.

Class 9 Maths Parallel Lines Kerala Syllabus Question 3.
Draw rectangles specified below each of perimeter 10 cm.
i. Equilateral triangle.
ii. Sides in the ratio 3 : 4 : 5
iii. Sides in the ratio 2 : 3 : 4
Answer:
i. Draw a line AB of length 10 cm.
Draw AC with length 9 cm. Join BC.
AD = DE = EC = 3
AF : FG : GB = 3 : 3 : 3 = 1 : 1 : 1
Divide it in the ratio 1 : 1 : 1 and draw a triangle.

Draw ΔFHG with FG as its side.

ii. Draw a line segment AB of length 10 cm. Draw AC = 12 cm, join BC. Divide it in the ratio 3 : 4 : 5.
AD : DE : EC = 3 : 4 : 5
AF : FG : GB = 3 : 4 : 5
Draw ΔGHB with sides AF, FG and GB.

iii. Draw a line segment AB of length 10 cm. Draw AC with length 9 cm, join BC. Divide it in the ratio 2 : 3 : 4.

AD : DE : EC = 2 : 3 : 4
AF : FG : GB = 2 : 3 : 4
Draw ΔGHB with sides AF, FG and GB.

Parallel Lines 9th Standard Kerala Syllabus Question 4.
In the picture below, the diagonals of the trapezium ABCD intersect at P. Prove that PA × PD = PB × PC.

Answer:

Sides AB, CD are parallel.
Draw EF parallel to AB and passing through P.
Now lines AB, EF and DC are parallel.
These lines cut the lines AC and BD in the same ratio.
So, PA : PC = PB : PD
i.e., \(\frac {PA}{PC} = \frac {PB}{PD}\)
From this by cross multiplication we get PA × PD = PC × PB

Textbook Page No. 93

Parallel Lines Class 9 Kerala Syllabus Question 1.
In the picture, the perpendicular is drawn from the midpoint of the hypotenuse of a right triangle to the base. Calculate the length of the third side of the large right triangle and the lengths of all three sides of small right triangle.

Answer:
AC = 10 cm
∴ AM = 5 cm

MN, CB are perpendicular to AB
MN, CB are parallel lines.
∴ MN divides AB and AC in the same ratio.
∴ AN = BN = 4cm
BC = \(\sqrt{10^{2} – 8^{2}} = \sqrt {100 – 64} = \sqrt {36}\) = 6 cm
M is the midpoint of the AC. N is the midpoint of AB.
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side. So MN is half of BC.
∴ MN = 3cm

Parallel Lines Chapter Class 9 Kerala Syllabus Question 2.
Draw a right triangle and the perpendicular from the midpoint of the hypotenuse to the base.

i. Prove that this perpendicular is half the perpendicular side of the large triangle.
ii. Prove that perpendicular bisects the bottom side of the larger triangle.
iii. Prove that in the large triangle the distances from the mid point of the hypotenuse to all the vertices are equal.
iv. Prove that the circumcentre of a right triangle is the mid point of its hypotenuse.
Answer:
MN is perpendicular to AB.
MN and CB are parallel lines. MN divides AC and AB in the same ratio.

i. AM = \(\frac{1}{2}\)AC, AN = \(\frac{1}{2}\)AB
∴ MN = \(\frac{1}{2}\)CB

ii. We get two right triangles of same base and perpendicular. So their hypotenuses are also equal.
∴ MA = MC = MB
AN = NB

iii. ∠ANM = ∠BNM = 90°
MN = MN
ΔANM ≅ ΔBNM
AM = MB …….(1)
M is the midpoint of AC
AM = MC …….(2)
(1) = (2) ⇒ AM = MB = MC

iv. The points A, B and C are at same distance from M. So a circle can be drawn with M as centre and pass through there three points. So this circle is the circum circle of ΔABC.

9th Standard Maths Notes Kerala Syllabus Question 3.
In the parallelogram ABCD, the line drawn through a point P on AB, parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R.

Prove that \(\frac {AP}{PB} = \frac {AR}{RD}\)
Answer:
In ΔABC, RQ the parallel line of BC, divides AB and BC in the same ratio When we consider the relation
\( \frac {AR}{RD} = \frac {AQ}{QC}, \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}\)

\( \frac {AR}{RD} = \frac {AQ}{QC}\)
\( \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}\)
∴ \( \frac {AP}{PB} = \frac {AR}{RD}\)

Hsslive Guru 9th Maths Kerala Syllabus Question 4.
In the picture below, two vertices of a parallelogram are joined to the mid points of two sides. Prove that these lines divide the diagonal in the picture into three equal parts.

Answer:

In the parallelogram ABCD, AB = CD and AB parallel CD.
P is the mid point of AB and Q is the midpoint of DC.
Since P and Q are the midpoints of AB and CD respectively.
PB = ½AB
DQ = ½CD = ½AB
∴ PB = DQ
i.e., PB = DQ and PB parallel DQ.
∴ PBQD is a parallelogram.
Since the parallel lines PD and BQ cut AB into equal parts, they cut AY also into equal parts.
AX = XY
The parallel lines PD and BQ cut CD into equal parts. So they cut CX also into equal parts.
YC = XY
∴ AX = XY = YC

9th Standard Maths Kerala Syllabus Question 5.
Prove that the quadrilateral formed by joining the mid points of a quadrilateral is a parallelogram. What if the original quadrilateral is a rectangle? What if it is a rhombus?
Answer:

P, Q, R and S are the midpoints of the sides of the quadrilateral ABCD. Draw the diagonals AC and BD.
The length of the line joining the mid-points of two sides of a trianglle is half the length of the third side.
In ΔABC, PQ = ½AC
In ΔADC, SR = ½AC
∴ PQ = SR
In ΔABD, PS = ½BD
In ΔBCD, QR = ½BD
∴ PS = QR
In quadrilateral PQRS, both pairs of opposite sides are equal. So it is a parallelogram.
PQ = ½AC, RS = ½AC
∴ RS = PQ = ½AC
PS = ½BD, QR = ½BD
∴ PS = QR = ½BD = ½AC
∴ PQ = RS = PS = QR
All sides of the quadrilateral PQRS are equal. So it is a rhombus.
Since ABCD is a rhombus. AC perpendicular to BD.
PQ and SP are perpendicular, i.e., ∠SPQ = 90°
One angle of the parallelogram PQRS is 90° and therefore it is a rectangle.

Kerala Syllabus 9th Standard Maths Parallel Lines Exam Oriented Text Book Questions and Answers

Class 9 Maths Chapter 6 Kerala Syllabus Question 1.
Draw ΔABC with AB = 5cm, BC = 4cm, AC = 6cm, ΔABC. In BC mark P such that BP = 1cm, PC = 3cm. What is the relation between the areas of ΔABP and ΔAPC?
Answer:
Area of ΔAPC is 3 times the area of ΔABP.

If two triangles have the common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.
⇒ \(\frac{Area of ΔABP}{Area of ΔAPC} = \frac{BP}{PC} = \frac{1}{3}\)

Chapter 6 Maths Class 9 Kerala Syllabus Question 2.
In ΔABC, the line parallel to BC meet AB and AC at D and E respectively. AE = 4.5cm, \(\frac{AD}{DB} = \frac{2}{5}\) = then find EC.
Answer:

\(\frac{AD}{DB} = \frac{AE}{EC}\)
\(\frac{2}{5} = \frac{4.5}{EC}\) = EC = \(\frac{4.5 \times 5}{2} = \frac{22.5}{2}\)
= 11.25 cm

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
In ΔPQR, PQ = 9cm, QR = 14 cm, PR = 12 cm The bisector of ∠PRS meets QR in S. Find QS and RS.
Answer:

\(\frac{PQ}{PR} = \frac{QS}{SR}\);
\(\frac{18}{24} = \frac{6}{RS}\)
= RS = \(\frac{24 \times 6}{18}\) = 8 cm

Hsslive Guru 9 Maths Kerala Syllabus Question 4.
In ΔABC, the point P is on AB such that the length of AP is double the length of PB. The line through P parallel to BC meets AC at Q and the length of AQ is 1 cm more than that of QC, what is the length of AC?
Answer:

AP = 2 × PB
AP : PB = 2 : 1
\(\frac{AP}{PB} = \frac{AQ}{QC} = \frac{2}{1}\) ⇒
AQ = 2QC
Consider QC = x cm
x + 1 = 2x ⇒ x = 1 ⇒ QC = 1
AQ = 2 × 1 = 2
AC = AQ + QC = 2 + 1 = 3 cm

Hss Live Guru Maths Kerala Syllabus Question 5.
In ΔABC, a line parallel to BC cuts a B and AC at X and Y

(a) If AB = 3.6cm, AC = 2.4cm, AX = 2.1 cm, what is the length of AY?
(b) If AB = 2cm, AC = 1.5cm, AY = 0.9cm
what is the length of BX?
Answer:
a. \(\frac{AB}{AX} = \frac{AC}{AY} ⇒ \frac{3.6}{2.1} = \frac{2.4}{AY}\);
⇒ 3.6 × AY = 2.4 × 2.1 ⇒ AY = 1.4 cm

b. Let length of AX = xcm
\(\frac{AX}{AB} = \frac{AY}{AC} ⇒ \frac{x}{2} = \frac{0.9}{1.5}\);
1.5 × x = 1.8 ⇒ x = 1.2;
BX = AB – AX; BX = 2 – 1.2 = 0.8 cm

Hss Live Guru Maths 9 Kerala Syllabus Question 6.
In the figure, ABCD is a trapezium, and AB || CD. P is the midpoint of AD. If PQ || AB, then prove that Q is the midpoint of BC.
Answer:

Given PQ || AB,
therefore
PQ || DC also. Line segments AD and BC are dividing 3 parallel lines EC, PQ and AB. Given P is the midpoint of AD
⇒ DP = PA ∴ \( \frac{DP}{PA}\) = 1, ∴ \( \frac{CQ}{QB}\) = 1
∴ CQ = QB ∴ Q is the midpoint of BC.

Ch 6 Maths Class 9 Kerala Syllabus Question 7.
In the figure AB || EH || DC, Answer the following questions given below.

a. If DE = 3cm, EA = 4cm and DG = 5cm, find GB?
b. If BH = 6cm,CB = 16cm and DB = 20cm, find DG?
c. If DA = 36cm, EA = 22cm and DG = 15cm find DB?
d. If DE = 20cm, EA = 25cm and HB = 22cm, find CH?
Answer:
a. In ΔDAB, AB || EG
∴ \(\frac{DE}{EA} = \frac{DG}{GB}; \frac{3}{4} = \frac{5}{GB} \Rightarrow GB = \frac{20}{3} = 6 \frac{2}{3}\)cm

b. In ΔBDC, DC || GH
∴ \(\frac{BH}{BC} = \frac{BG}{BD}; \frac{6}{16} = \frac{BG}{20}\);
BG = \(\frac{120}{16} = 7\frac{1}{2}\)cm
DG = BD – BG = 20 – \( 7\frac{1}{2}\) = \( 12\frac{1}{2}\)cm

c. In ΔABD, AB || EG
\(\frac{DA}{EA} = \frac{DB}{BG}; \frac{36}{22} = \frac{DB}{BG}\);
\(\frac{36}{22} \frac{DB}{DB – DG}\)
⇒ 36D(DB – DG) = DB × 22
14 DB = 540 DB = \(\frac{540}{14}\) = 38\(\frac{4}{7}\)cm

d. AB, EH and DC are 3 parallel lines.
\(\frac{DE}{EA} = \frac{CH}{HB}; \frac{20}{25} = \frac{CH}{22}\) ⇒ CH = \(\frac{88}{5}\) = 17\(\frac{3}{5}\)cm

Hss Live Guru Class 9 Maths Kerala Syllabus Question 8.
In ABC, a line parallel to BC cuts AB and AC at P and Q. Show that \(\frac{AP}{AB} = \frac{AQ}{AC}\)
Answer:

\(\frac{AB}{PB} = \frac{AQ}{QC}\) (BC || PQ)
\(\frac{AP}{PB}+1=\frac{AQ}{QC}+1; \frac{AP+PB}{PB} = \frac{AQ + QC}{QC}\)
\(\frac{AB}{PB} = \frac{AQ}{QC}\);
∴ \(\frac{AP}{AB} = \frac{AQ}{AC}\)

Question 9.
In the figure below, ABC is a right angled triangle. AB = 10cm, AC = 6cm, BC = 8cm, The midpoint of AB is M. Compute the lengths of the sides of ΔMBN
Answer:

(MN || AC)
∴ \(\frac{AM}{MB} = \frac{NC}{NB}\)
1 = \(\frac{NC}{NB}\) ∴ CN = NB = 4
∴ MB = 5, NB = 4, MN = 3

Question 10.
In the figure ABC is a right angled triangle and M is the midpoint of AB. Prove that MN = V2AC and also prove that MN = MA = MB.
Answer:

MN = \(\sqrt{x^{2} – y^{2}}\)
AC = \(\sqrt{4x^{2} – 4y^{2}} = 2\sqrt {x^{2} – y^{2}}\)
AC = 2 × MN or MN = \(\frac{1}{2}\)AC
MC = \(\sqrt{x^{2} – 4^{2} + 4^{2}}\); MC = X
∴ MC = MA = MB

You can Download Circle Measuress Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures

Circle Measures Textual Questions and Answers

Textbook Page No. 131

Circle Measures Class 9 Kerala Syllabus Question 1.
Prove that the circumcentre of an equilateral triangle is the same as its centroid.
i. Calculate the length of a side of an equilateral triangle with vertices on a circle of diameter 1 centimetre.
ii. Calculate the perimeter of such a triangle.
Answer:
Perpendicular bisector of sides of a triangle that can meet at a point is its circumcenter

Since the triangle is equilateral the perpendicular bisectors of the sides are also median. Since the triangle is equilateral the perpendicular bisectors of the sides are also centroid. That is circumcentre of an equilateral triangle is the same as its centroid.
i. Length of one side of an equilateral triangle is.

Circle Measures Class 9 Scert Chapter 9 Kerala Syllabus Question 2.
Calculate the perimeter of a square with vertices on a circle of diameter 1 centimetre.
Answer:
In the square ABCD
AO = 1/2 cm
In A AEO, ∠EAO = 45°
∠AEO = 90°
Since AO = 1/2 cm
AE = \(\frac{1}{2 \sqrt{2}} \mathrm{cm}\)
(Since Δ AEO is a isosceles right triangle, hypotenuse is √2 times of a perpendicular side.)

Perimeter = \(=4 \times \frac{\sqrt{2}}{2}=2 \sqrt{2} \mathrm{cm}\)

Circle Measures Class 9 Chapter 9 Kerala Syllabus Question 3.
Calculate the perimeter of a regu¬lar hexagon with vertices on a circle of diameter 1 centimetre.
Answer:
If we draw diagonals through centre of circle inside the regular hexagon it divides the regular hexagon into 6 equilateral triangles.
Let diameter be 1 cm
OA = 1/2 cm
OA = OB = AB.
therefore
One side = 1/2 cm
Perimeter of regular hexagon = 6 × 1/2 = 3 cm

Textbook Page No. 134

Hss Live Guru 9th Maths Chapter 9 Kerala Syllabus Question 1.
The perimeter of a regular hexagon with vertices on a circle is 24 centimetres.
i. What is the perimeter of a square with vertices on this circle?
ii What is the perimeter of a square with vertices on a circle of double the diameter?
iii. What is the perimeter of an equilateral triangle with vertices on a circle of diameter half that of the first circle?
Answer:
Perimeter of a regular hexagon = 24 cm
Length of one side of a regular hexagon = 24/6 = 4 cm
Length of one side of a regular hexagon is equal to the radius of the circle.
i. Diagonal of a square = 8 cm Diagonal of a square is equal to the diameter of the circle.
Let a be the side of the square
a² + a² = 8²
2a² = 64
a² = 64/2 = 32
a= 4√2
∴ One side of a square = 4√2 cm
Perimeter of a square = 4 × 4√2 cm = 16√2 cm

ii. The perimeter of a square with verti¬ces on a circle of double the diameter 2 × 16√2 = 32 √2 cm
(The perimeters of circles are scaled by the same factor as their diameters.)

iii. One side of an equilateral triangle with vertices on a circle of half the diameter of the first circle
\(=2 \sqrt{2^{2}-1^{2}}=2 \sqrt{3} \mathrm{cm}\)
Perimeter = 3 × 2√3 = 6√3 cm

Circles Class 9 State Syllabus Chapter 9 Kerala Syllabus Question 2.
A wire was bent into a circle of diameter 4 centimetres. What would be the diameter of a circle made by bending a wire of half the length?
Answer:
The ratio between the perimeters are equal to the ratio between their diameters. The perimeter of the first circle is twice the perimeter of the second circle.
Therefore diameter of the second circle is half of the diameter of the first circle. Diameter of the second circle.
= 4/2 = 2 cm

Kerala Syllabus 9th Standard Maths Solutions  Question 3.
The perimeter of a circle of diameter 2 metres was measured and found to be about 6.28 metres. How do we compute the perimeter of a circle of diameter 3 metres, without measuring?
Answer:
If diameter is 2 meters, perimeter is 6.28 meter.
If the diameter is 1 metre, perimeter is 6.28/2 meter.
If the diameter is 3 metres, perimeter = \(\frac{6.28}{2} \times 3=9.42 m\)

Textbook Page No. 137

Hss Live Guru Maths 9 Chapter 9 Kerala Syllabus Question 1.
In the pictures below, a regular hexagon, square and a rectangle are drawn with their vertices on a circle. Calculate the perimeter of each circle.

Answer:
a. AB = 2 cm
In the figure triangle are
equilateral triangles,
therefore
radius OA = 2 cm
Perimeter of circle = 2 πr
= 2 × π × 2 = 4π cm

b. ABCD is a square
AB = BC = 2 cm, ∠5 = 90°
AC = \(\sqrt{2^{2}+2^{2}}=\sqrt{8}=2 \sqrt{2}\)

Radius of circle = 1/2 × 2√2
= √2 cm
Perimeter of circle = 2π × √2 cm
= 2√2 π cm

c. PR = \(\sqrt{2^{2}+(1.5)^{2}}\)
= \(\sqrt{6.25}=2.5 \mathrm{cm}\)
Radius of circle = 1/2 × 2.5 = 1.25 cm
Perimeter of circle = 2 × π × 1.25
= 2.5 π cm

Kerala Syllabus 9th Standard Maths Notes Question 2.
An isosceles triangle with its vertices on a circle is shown in this picture.

What is the perimeter of the circle
Answer:
Consider the centre of circle as O and triangle as ABC
OC = Radius of circle = r
OD = 4 – r
AD = 2 cm
AO = r

Therefore, in triangle AOD
(AO)² = (AD)² + (OD)²
r²=2² + (4 – r)²
r²= 4 +16 – 8r + r²
8r=20;
r = 20/8 = 5/2 = 2.5 cm
∴ Perimeter = 2π × r = 2π × 2.5 cm
= 5π cm

Circles Class 9 Kerala Syllabus Chapter 9 Question 3.
In all the pictures below, the centres of the circles are on the same line. In the first two pictures, the small circles are of the same diameter.

Prove that in all pictures, the perimeters of the large circle is the sum of the perimeters of the small circles.
Answer:
a. Smaller circles have same diameters. Consider the diameter as d, perimeter of smaller circle
= π × diameter = π d
Perimeter of two small circles = 2πd

Diameter of the large circle = d + d = 2d
Perimeter of the large circle
= π × 2d = 2πd
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

b. Let d be the diameter of one small circle, then perimeter = π d
Sum of perimeter of three small circles =3 π d
Diameter of the large circle = d + d + d = 3d
Perimeter of the large circle = π × 3d = 3 π d
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

c. In figure diameter of three circless are different, let consider the diameters of small circles are p, q and r.

Perimeter of first small circle = πp
Perimeter of second small circle = πq
Perimeter of third small circle = πr
Sum of perimeters of three small circles
= πp + πq + πr = π (p + q + r)
Diameter of large circle = p + q + r
Perimeter of large circle = π (p + q + r)
Therefore also here the perimeters of the large circle is the sum of the perimeters of the small circles.

Hss Live Class 9 Maths Chapter 9 Kerala Syllabus Question 4.
In this picture, the circles have the same centre and the line drawn is a diameter of the large circle. How much more is the perimeter of the large circle than the perimeter of the small circle

Answer:
If ‘r’ is the radius of the small circle,
radius of the large circle = r + 1
Perimeter of the small circle = 2 π r
Perimeter of the large circle = 2 π (r + 1) = 2 π r + 2 π
i.e., perimeter of the large circle is 2 π units more than the perimeter of the small circle.

Textbook Page No. 141

Hss Live Guru 9 Maths Chapter 9 Kerala Syllabus Question 1.
In the pictures below, find the difference between the areas of the circle and the polygon, up to two decimal places.

Answer:
i. Radius of the small circle = 2cm
Area = π × 2² = 3.14 × 4
= 12.56 cm²
Daigonal of the square = 4 cm
One side of the square = 4/√2 cm
Area of the square = \(\frac{4}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=\frac{16}{2}=8 \mathrm{cm}\)
Differences between the areas = 12.56 – 8 = 4.56 cm

ii. Radius of the circle = 2 cm
Area= π × 2² = 3.14 × 4 = 12.56 cm²
One side of the regular hexagon=2 cm
Area of the regular hexagon = \(6 \times \frac{\sqrt{3} \times 2^{2}}{4}=6 \sqrt{3}\)
= 6 × 1.73 = 10.38 cm²
Differences between the areas = 12.56 – 10.38 = 2.18 cm²

9th Std Kerala Syllabus Maths Solutions Question 2.
The pictures below show circles through the vertices of a square and a rectangle

Calculate the areas of the circles
Answer:
i. One side of a square is 3 cm, therefore its diagonal is 3√2 cm
Diameter of the circle = 3√2 cm

ii. Rectangle inside the circle having length 4 cm and breadth 2 cm.
Diagonal = \(\sqrt{4^{2}+2^{2}}=\sqrt{16+4}\)

Diameter of the circle = √20 cm
Radius = \(\frac{\sqrt{20}}{2} \mathrm{cm}\)
Area = \(\pi \times\left(\frac{\sqrt{20}}{2}\right)^{2}=\pi \times \frac{20}{4}\)
= 5 π cm²

Hsslive Guru 9th Maths Chapter 9 Kerala Syllabus Question 3.
Draw a square and draw circles cen¬tered on the corners, of radius half the side of the square. Draw another square formed by four of the first square and a circle just fitting into it.

Prove that the area of the large circle is equal to the sum of the areas of the four small circles
Answer:
In the figure length of one side of the square is 2r
Radius of one small circle = r
Perimeter of one small circle = π r²

Radius of four small circles = 4 π r²
One side of a square in the second figure = 2r + 2r = 4r
Radius of the circle in the figure
= 4r/2 = 2r
perimeter of the circles in the figure = π × (2r)²
= π × 2r × 2r = 4πr²
The area of the large circle is equal to the sum of the areas of the four small circles

Hsslive 9th Maths Chapter 9 Kerala Syllabus Question 4.
In the two pictures below, the squares are of the same size.

Prove that the green regions are of the same area.
Answer:
Let ‘a’ be the side of the square in the picture.
Area of the square = a²
If the four sectors in the vertices are joined together, a circle is formed because the radius of each sector is a/2.
Area of the shaded part = Area of the square – area of the circle.

ii. In the second picture diameter of the
circle = a Radius = a/2
Area of the shaded part = Area of the square – area of the circle.

i.e., Areas of the shaded portions are equal.

Hsslive Guru 9 Maths Chapter 9 Kerala Syllabus Question 5.
Parts of circles are drawn inside a square as shown in the picture below. Prove that the area of the blue region is half the area of the square.

Answer:
Let ‘a’ be the side of the square
Area of the blue part in the half portion of the square is equal to half of the area of the circle having diameter ‘a’.
Area of the blue part in the half portion of the square \(=\frac{1}{2} \times \pi\left(\frac{a}{2}\right)^{2}=\frac{\pi a^{2}}{8}\)
We must subtract area of two circles
having diameter a/2 from the half the
area of the square to get the area of remaining blue shaded part.
Area of remaining blue shaded part.
= \(\frac{a^{2}}{2}-2 \times \frac{1}{4} \times \pi\left(\frac{a}{2}\right)^{2}\)
Area of blue shaded part.
= \(\frac{\pi a^{2}}{8}+\frac{a^{2}}{2}-\frac{\pi a^{2}}{8}=\frac{a^{2}}{2}\)
i.e., area of the blue part is equal to half the area of the square.

Hss Guru 9 Maths Chapter 9 Kerala Syllabus Question 6.
In the figure, semicircles are drawn with the sides of a right triangle as diameter.

Prove that the area of the largest semicircle is the sum of the areas of the smaller ones.
Answer:

In ΔABC, ∠5 = 90°
According to Pythagoras principle,
AB² + BC² = AC² ………. (1)
Radius of the semicircle with diameter
AB = AB/2
Area of the semicircle with diameter AB

= \(\frac{1}{2} \times \pi \times \frac{A C^{2}}{4}=\frac{\pi}{8} A C^{2}\)
Sum of the areas of the smaller semicircles = \(\frac{\pi}{8} A B^{2}+\frac{\pi}{8} B C^{2}=\frac{\pi}{8}\left(A B^{2}+B C^{2}\right)\)
= π/8 AC² = Area of the largest semicircle

Textbook Page No. 148

Hsslive Maths Class 9 Chapter 9 Kerala Syllabus Question 1.
In a circle, the length of an arc of central angle 40° is 3 π centimetres. What is the perimeter of the circle? What is its radius?
Answer:

Question 2.
In a circle, the length of an arc of central angle 25° is 4 centimetres.
i. In the same circle, what is the length of an arc of central angle 75°?
ii. In a circle of radius one and a half times the radius of this circle, what is the length of an arc of central angle 75°?
Answer:
i. Length of the arc having central angle 25° = 4 cm
Three times of 25 is 75.
Length of the arc having central angle 75° = 4 x 3 = 12 cm

ii. Length of the arc having central angle 75° and radius r = 12 cm
Length of the arc having central angle 75° and radius 1 1/2 r
= \(12 \times 1 \frac{1}{2}=18 \mathrm{cm}\)

Question 3.
From a bangle of radius 3 centimetres, a piece is to be cut out to make a ring of radius ^ centimetres.
i. What should be the central angle of the piece to be cut out?
ii. The remaining part of the bangle was bent to make a smaller bangle. What is its radius?
Answer:
Perimeter of the bangle having radius 3 cm = 6 π cm.
Perimeter of the bangle having radius 1/2 cm = π cm.
π is the 1/6 part of 6 π.
Therefore the central angle of the piece to be cut out = 360 × 1/6 = 60°
ii. Length of the remaining part of the bangle = 6π – π = 5π cm
Radius of the other small bangle = 5 π / 2π = 2.5 cm

Question 4.
The picture shows the parts of a circle centred at each vertex of an equilateral triangle and passing through the other two vertices.

What is the perimeter of this figure?
Answer:
Since the triangle is equilateral, each angle is 60°. There are in each side is in a circle of radius 4 cm and the central angle is 60°.

Question 5.
Parts of circles are drawn, centred at each vertex of a regular octagon and a figure is cut out as show below:

Calculate the perimeter of the fig¬ure.
Answer:
Sum of the angles in an octagon
= (n-2) × 18o° = 6x 180°= 1080° One angle of the regular octagon
= 1080 – 8 = 135°
One side of the regular octagon = 2 cm. Radius if the sectors having centre is each vertices of the circle= 1 cm The second picture shows the cut-down form of 8 sectors having centre angle 135° and radius 1 cm.
The perimeter is found by calculating the length of 8 arc having radius 1 cm and central angle 135°.

Textbook Page No. 151

Question 1.
What is the area of a sector of central angle 120° in a circle of radius 3 centimetres? What is the area of a sector of the same central angle in a circle of radius 6 centimetres?
Answer:
Area of the sector in the circle having radius 3 cm and angle of center 120°
π × 3² × \(\frac { 120 }{ 360 }\) = 3π cm²
Area of the sector in the circle having radius 6 cm and angle of center 120°
= π × 6² × \(\frac { 120 }{ 360 }\) = 12π cm²

Question 2.
Calculate the area of the green coloured part of this picture.

Answer:
In the picture area of the shaded part is the difference between the area of the two sectors.
Area of the large sector

Area of the shaded part = 9.42 – 4.19 = 5.23 cm²

Question 3.
Centred at each corner of a regular hexagon, a part of a circle is drawn and a figure is cut out as shown below:

What is the area of this figure?
Answer:
The area of the cut-down portion = Area of the regular hexagon – Area of 6 sectors Area of the regular hexagon

Area of the sector = 120°/360° part of area of the circle
(One angle of a regular hexagon is 120°)

Area of the 6 sectors = \(\frac{6 \times \pi}{3}=2 \pi \mathrm{cm}^{2}\)
Area of cut down portion = 6 √3 – 2 π cm²

Question 4.
The picture below shows two circles, each passing through the centre of the other:

Calculate the area of the region common to both.
Answer:
Consider the picture given below, we can divide the circle into two part by using the line AB, the area of the two portions are same.
That is we get the total area by multi¬plying area of the sector by 2.

We can find out the area of part above the line AB .
Given AB = 2cm
Circles having equal
radius. So,
AC = BC = 2 cm.
AABC is an equilateral triangle so angles are 60° each.
Add the area of sectors having centre A and B .
The area of ΔABC include twice, so we will subtract it once.
Area of sectors having centre A

Area of sectors having centre B

Area of ΔABC

Area of the part above the line AB = 2.09 + 2.09 – 1.7 = 2.45 cm²
The area of the region common to both = 2 × 2.45 = 4.90 cm²

Question 5.
The figure shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two ver¬tices;

Find the area of the region common to all three.
Answer:
The area of the common part = Area of three sectors – 2 × area of an equilateral triangle having side 2 cm

= 2 × 3.14 – 2 × 1.73
= 6.28 – 3.46
= 2.82 cm²
The area of the region common to all three = 2.82 cm²

Circle Measures Exam Oriented Questions And Answers

Question 1.
In the picture PQRS is a square of side 10 cm. A, B, C and D are midpoints of the sides of the square

Semi circles are drawn inside the square.
a. Compute the area of the square.
b. Compute the area of the semi-circle.
c. Compute the area of the shaded part.
Answer:
a. Since one side of the square is 10 cm,
Area = side x side = 10 × 10= 100 cm²

b. The diameter of one semicircle = half of the side of the square
Diameter = 5 cm
∴ Radius = 5/2 cm
Since the four semicircles are equal. Area of the 4 semicircles
= Area of 2 circles = 2 × πr²

c. Area of the shaded part = Area of the square – Area of four semicircles.

Question 2.
If a circular shaped dining table has an area of 31400 cm², find its radius. What will be its perimeter ?
Answer:
Area of the table = 31400 cm²

Question 3.
In the given picture shown two semicircular shaped iron bar can be cut down from a rectangular shaped iron bar. Calculate the area of the remaining shaded portion.

Answer:
Area of the rectangle = 24 × 14 = 336 cm²
Area of two semicircle = Area of a complete circle
Diameter of the circle = 14 cm
Radius = 7 cm
Area of the circle = π r²
= π × 72 =49
π = 153.86 cm2 Area of the remaining portion
= 336 – 153.86= 182.14 cm²

Question 4.
In the figure A, B, C and Dare the points on the square which touches the circle. If the radius of the circle is 6.
a. What is the length of one side of the square?
b. Find the area of the circle.
c. What will be the area of the shaded portion?

Answer:
a.Diameter of the circle = 6 × 2 = 12 cm
Side of the square = 12 cm
b. Area of the circle = n r²
= n × 6 × 6 = 36n =36 × 3.14 =113.04 cm²
c. Area of the square = 12 × 12 = 144 cm²
Area of the shaded portion = 144 – 113.04 = 30.96 cm²

Question 5.
In the ACB is the arc drawn by taking O as the centre and OA as the radius. Then find the area of the shaded region.

Answer:
The area of sector which has 7 cm radius and 90° central angle =

Area of the right angled triangle AOB
\(=\frac{7 \times 7}{2}=24.5\)
Area of the shaded part = 38.455 – 24.5 = 13.965 cm²

Question 6.
In the pictures given below, find the area of the shaded part?

Answer:
’The radius of the sector in the picture (1) is 5cm and its central angle is 40°.
Area of the shaded part \(=\pi \times 5^{2} \times \frac{40}{360}=\frac{25}{9} \pi \mathrm{cm}^{2}\)
The radius of the sector in the picture (2) is 6cm and its central angle is 300° (360 – 60).
Area = \(\pi \times 6^{2} \times \frac{300}{360}=36 \pi \times \frac{5}{6}\)
= 30 π cm²

Question 7.
What amount of reeper is needed to enclose a circular shaped dining table of area 6.28 cm²?
Answer:

Question 8.
A wheel which has 20 cm radius is rotating forward. After 10 rotations what distance will the wheel travel forward?
Answer:
Radius of the wheel = 20 cm
When the wheel rotates once it will travel the distance same as its area
Perimeter of the wheel = 2 × π × radius = 2 × 3.14 × 20 = 125.64cm
The distance travelled forward when the wheel rotates once = 125.64 cm
The distance travelled forward when the wheel rotates 10 times
= 125.64 × 10 = 1256.4 cm

Question 9.
In the picture the central angles of both the sectors are equal. Sum of the radii of the sectors is 18 cm.
Area of the shaded part is 18 π cm². Find the radii of the sector.

Answer:
‘Let ‘r’ be the radius of the small sector and ‘R’ be the radius of the large sector.
R + r = 18 …………. (1)
Area of the shaded portion = Area of the large sector – area of the small sector

Question 10.
In the figure O is the radius of the circle and OABC is a rectangle. OA = 8 cm, OC = 15 cm. Hence find the Area of the shaded portion

Answer:

Area of the shaded portion

Question 11.
The wheel of a vehicle has a diameter 60 cm. For this vehicle travels a distance of 200 m, how many times must this wheel rotates.
Answer:
‘The distance travelled when the wheel is rotated once =2 π r = 2 × π × 30
= 188.4 cm = 1.884 m.
The time required for the wheel to travel a distance of 200m = 200/1.884 = 106.16 = 106

Question 12.
In the below-given figures, there are two circles with the same centre. Then find the area of the second region.

Answer:
a. Area of the shaded portion =Area of the outer circle – Area of the inner circle
= π R² – π F= π × 10² – π × 8²
= 100π – 64π = 36π
=36 × 3.14 = 113.04 cm²

b. Outer radius = 7 + 2 = 9 cm Inner radius = 7 cm
Area of the shaded portion = π × 92- π × 72 = 81π – 49π = 32π = 32 × 3.14 =100.48 cm²

c. Outerradius = 10.5
Inner radius = 10.5 – 1 = 9.5
Area of the shaded portion = π × (10.5)² – π × (9.5)²
= π × (10.5² – 9.5)²
= π × (10.5² – 9.5²)
= π × (10.5 + 9.5) (10.5 – 9.5) = π × 20 × 1 = 62.8 cm²

d. Outer radius = \(\frac { 16π }{ 2π }\) = 8
Inner radius = \(\frac { 14π }{ 2π }\) = 7
Area of the shaded portion

Question 13.
Two semicircular pieces are cut out from a rectangular sheet. Find the area of the remaining portion.

Answer:
Area, of the rectangle = 50 × 20 = 1000 cm2 If the two semicircles cut out are joined it becomes a circle
Its radius = 20/2 = 10 cm
Area of the portion cut out
= πr² = π × 10 × 10 = 3.14 × 10 × 10 = 314 cm²
Area of the remaining portion
= 1000 – 314 = 686 cm²

Question 14.
If a square, equilateral triangle, regular hexagon and circle have the same perimeter. Which of these has the largest area ?
Answer:
If a square, equilateral triangle, regular hexagon and circle has a perimeter of 12 cm.
Equilateral triangle
One side = 12/3 = 4


Circle has the largest area.

Question 15.
In the figure, if ACB is the arc of circle having O as the centre and OA as the O, radius. Then find the area of the shaded portion

Answer:
Radius of the sector = 8cm Central angle = 90°

OBA is a right angled triangle
∴ Area of ΔOBA = 1/2 × 8 × 8 = 32 cm²
Area of shaded portion = 50.24 – 32 = 18.32 cm²

Question 16.
The area of an equilateral triangle is 17300 cm². Draw circles with radius half the length of one side of the triangle and vertices as the centre of the circle. Calculate the area of the shaded portion.
Answer:
Area of the equilateral triangle

One side = 200 cm,
Radius of the circle = 100 cm
Central angle of each sector = 60°
Area of one sector = \(\frac{\pi r^{2} \times 60^{\circ}}{360}\)
Area of the three sectors

Question 17.
In the figure, a side of the regular hexagon has length 20 cm. Find the area of the shaded portion.

Answer:
We have learnt that length of a side of the regular hexagon drawn inside a circle will be equal to the radius of the circle.
So the radius of the circle will be 20 cm.
Area of the circle = πr² = π × 20 × 20 = 400π = 1256 cm²
Area of the regular hexagon \(=\frac{6 \times \sqrt{3} \times a^{2}}{4}\)

Area of shaded portion = 1256 – 1039.2 = 216.8 cm²

Question 18.
If radius of a sector is 7 cm and its perimeter is 2.5 cm. Then find its area.
Answer:
Area of sector
= 2 × radius + length of the arc = 25
Length of the arc = 25 – 2 × radius = 25 – 14 = 11 cm
Let us check the ratio between the length of the arc and its area
The length of the arc: Area of the sector

11 : area of the sector = 2:7
Area of two sectors =7 × 11
Area of the. sector = \(\frac{7 \times 11}{2}\) = 38.5 cm²

Question 19.
In the figure if the length of one side of a square is 12 cm, then find the area of the shaded portion.

Answer:
Area of the square = 12² = 144
Diameter of the circle = 12; r = 6 Area of the circle
= πr² = π × 6 × 6 = 3.14 × 6 × 6 = 113.04 cm²
Area of shaded portion = 144 – 113.04 = 30.96 cm²

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