AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry InText Questions

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry InText Questions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry InText Questions

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry InText Questions

Try This

Can you give any two axioms from your daily life. [Page No. 63]
Solution:
Example -1 : If we flip coin then there are only two chances that is head or tail.
Example – 2 : Things which are equal to the same thing are also equal to one another.
Example – 3 : The whole is greater than the part.

AP Board 9th Class English Solutions Chapter 1C Little Bobby

AP Board 9th Class English Solutions Chapter 1C Little Bobby

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 1C Little Bobby Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 1C Little Bobby

9th Class English Chapter 1C Little Bobby Textbook Questions and Answers

I. Answer the following questions.

Question 1.
Which is the most humorous action? Give reasons in support of your answer.
Answer:
Stealing Mother Mary’s statue and writing to God as kidnapping is the most humorous action. The reasons are many.

  1. Persons are kidnapped; not statues.
  2. Mother Mary’s statues are aplenty in so many other churches.
  3. God can get back that statue even without giving Bobby a bike.
  4. Bobby’s creativity – though used for a wrong cause – is both admirable and humorous.

Question 2.
If you wanted a bike, what would you write to God?
Answer:
If I wanted a bike, I wouldn’t think of writing a letter to God. If I were to write one, would very humbly request God for a bike. I would also explain the need in a detailed way.

AP Board 9th Class English Solutions Chapter 1C Little Bobby

Question 3.
Would you like the behaviour of Little Bobby? Why or why not? Give a reason.
Answer:
No, I would not like Little Bobby’s trouble making behaviour. I do not like stealing Mother Mary’s statue. I do not like the action of kidnapping God’s mother and demanding a bike. But I do like Bobby’s honesty in admitting his misbehaviour. I do admire his creativity.

Project Work

Collect data from 30 families (including men, women and children) on the following programmes they watch on TV.
Family Information Sheet
AP Board 9th Class English Solutions Chapter 1C Little Bobby 1

Follow the process given below.
Before conducting survey: Form into three groups.
During the survey: Collect the information to know how frequently people watch different programmes on TV depending on their gender and age.

Each group should collect data from ten families by using the above format.

After the survey:
On the basis of the 10 individual sheets, prepare one consolidated sheet giving the profile of the families visited.
Analyse the results in groups and write a brief report on the result of your survey.

You may use the following phrases:
Very large number of men/women/children (above 75%) ……………………..
Very few number of men/women/children (below 15%) ……………………..
A good number of men/women/children (15%-75%) ……………………..

Consolidated Sheet of Group ONE
Giving the Profile of the 10 Families Visited.
AP Board 9th Class English Solutions Chapter 1C Little Bobby 2

Report of Group ONE :

This report is prepared on the basis of the data collected from ten families on the above mentioned programmes they watch on TV. It reflects how frequently people watch different programmes on TV depending on their gender and age. According to the analysis very large number of women watch movies and cookery programmes. A good number of women, men and children watch movies. Very large number of men watch movies, discussions and film reviews. Only a few number of children and women watch discussions. A large number of children watch games and sports. People of ail age groups and genders spend their time to watch TV.

AP Board 9th Class English Solutions Chapter 1C Little Bobby

Consolidated Sheet of Group TWO
Giving the Profile of the 10 Families Visited.
AP Board 9th Class English Solutions Chapter 1C Little Bobby 3

Report of Group TWO :

This report is prepared on the basis of the data collected from ten families on the above mentioned programmes they watch on TV. It reflects how frequently people watch different programmes on TV depending on their gender and age. Very few number of women watch news, games and sports and cartoon programmes. Very few number of children spend their time to watch discussions and cookery programmes. Most of the children watch games and sports. Children watch movies more than women and men. Daily serials and film reviews are great attractions for all. All most all people sit before their TV sets every day.

AP Board 9th Class English Solutions Chapter 1C Little Bobby

Consolidated Sheet of Group THREE
Giving the Profile of the 10 Families Visited.
AP Board 9th Class English Solutions Chapter 1C Little Bobby 4

Report of Group THREE :

This report is prepared on the basis of the data collected from ten families on the above mentioned programmes they watch on TV. It reflects how frequently people watch different programmes on TV depending on their gender and age. Games and sports are major attractions for children. Reality shows, cartoon programmes and movies are other important attractions for them. Most of women spend their time to watch daily serials, cookery programmes and movies. Very few women watch news, discussions and games and sports. Men are interested in watching movies, news, games and sports and reality shows. A good number of men like to watch daily serials. All the members of every family spend their time before their TV sets daily.

Little Bobby Summary in English

“Little Bobby” is the story of a naughty boy by Ritcha Rao. He was a troublemaker both at home and school. He one day asked his mother a bike for his coming birthday. Mother wanted Bobby to know how he had been behaving. So she asked Bobby to write a letter to God. She wanted him to write the reasons that made him eligible for a bike. Bobby started writing a letter. But he was not good to deserve a bike. So, he wrote four letters and tore them up as they were not true or helpful. Then he went to the nearby church. He stole the statue of Mother Mary. He came back and wrote to God “I’VE KIDNAPPED YOUR MOM. IF YOU WANT TO SEE HER AGAIN, SEND THE BIKE!”

Little Bobby Glossary

troublemaker (n) : one who causes troubles

deserved (v-past tense) : was eligible to have something

reflect (v) : think about seriously

stumped (v-past tense) : walked angrily and noisily

altar (n) : (The word ‘alter’ (v) is a different word.) place in a prayer hall where statues are placed

kidnapped (v-past tense) : took someone by force with demand for money or things

AP Board 9th Class English Solutions Chapter 1B The Duck and the Kangaroo

AP Board 9th Class English Solutions Chapter 1B The Duck and the Kangaroo

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 1B The Duck and the Kangaroo Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 1B The Duck and the Kangaroo

9th Class English Chapter 1B The Duck and the Kangaroo Textbook Questions and Answers

I. Answer the following questions.

Question 1.
Why did the Duck get bored in life? What did it want to do?
Answer:
The Duck lived in a pond. It cannot hop like a Kangaroo and go round the world. When the Duck saw the Kangaroo going round the places, it felt bored. It wanted to tour places. Limiting to the pond made the Duck unhappy.

Question 2.
What problem did the Kangaroo find with the Duck? How was it solved?
Answer:
The Kangaroo was willing to take the Duck to new places. But it objected to the wet cold feet of the Duck. The Kangaroo said that would make him ill. But the Duck had a solution. It got socks, a coat, and cigars ready to save them from cold.

AP Board 9th Class English Solutions Chapter 1B The Duck and the Kangaroo

Question 3.
Who was happy in the end? Why?
Answer:
Both the Kangaroo and the Duck were happy in the end. The Duck was happier as its desire to see places was fulfilled. The Kangaroo too was happy as it could help his friend.

Question 4.
What is the underlying message in the poem?
Answer:
The poem is basically humorous. But it has valuable messages too ! Desire to see new places is a natural way to better oneself. Seeking others’ help promotes cooperative living. Foreseeing the problems and providing solutions is a good lesson.

II. Read the above poem once again and choreograph it as described here under.

Choreography

Choreography is the art of designing sequences of movements or actions for a script or a poem with a voice or a song in background.
Do you know how to process choreography?
Here is the process.
Think of the following.

1. How do we begin the poem?
(Sing the poem three or four times to register the tune and rhythm.)

2. What are the characters in the poem?
(The Duck and the Kangaroo)

3. What is the theme of the poem?
(Wish of the Duck to go round the world)

4. What is the theme of each stanza? (sub themes)

Stanza Theme
1. First Wish of the Duck to hop and go round the world
2. Second Requesting the Kangaroo to take her for a ride
3. Third Conditional acceptance of the Kangaroo
4. Fourth Assurance of comfort from the Duck’s side
5. Fifth Fulfilment of desire

5. Convert the themes into incidents and performable actions.

6. What are the actions involved in the poem?

  • Boredom with present life (feelings of boredom, pain)
  • Desire to go round the world (miming)
  • Wearing socks and smoking cigarettes (joy)
  • Displeasure due to wet and cold feet (gestures)
  • Sitting on Kangaroo at the end of the tail (miming)
  • Joy (facial expression and gestures)

AP Board 9th Class English Solutions Chapter 1B The Duck and the Kangaroo

7. Who are the characters involved?
(The Duck, the Kangaroo and other creatures in the pond like frogs, fish, etc.)

8. Where does the incident take place (location)?
(By the side of the pond)

9. Presentation of the choreography on the stage:

  • The chorus team sings the poem
  • The characters perform their actions
  • The action of the Duck and the Kangaroo- expressing boredom, wish, discomfort, joy, etc.
  • The actions related to miming, facial expressions and gestures
  • Actions related to hopping, sitting, walking, jumping, etc.

The Duck and the Kangaroo Summary in English

‘The Duck and the Kangaroo’ is a poem with a touch of humour. It was the creation of a British poet called Edward Lear. The Duck says that it is bored with its life in the pond.
It wants the Kangaroo to carry it around the world. The Kangaroo says that the Duck’s cold and wet feet will make him ill. The Duck says it has a solution to that problem. It will wear socks. It will also have a coat and smoke. Then the Kangaroo asks the Duck to sit on his tail. The Kangaroo hops around the world thrice with the Duck sitting on his tail. The
poet leaves the question – as to who is so happy? – to the reader!

The Duck and the Kangaroo Glossary

Good gracious! (phr) : an exclamation expressing surprise and emphasis

nasty (adj) : unpleasant

long (v) : have a strong desire

the Dee and the Jelly Bo Lee (n) : wonderful places imagined by the poet

reflection (n) : thinking, consideration

roo-matiz (n) : rheumatism (n) – a disease that makes joints and muscles painful; arthritis

worsted socks : woollen socks

cloak (n) : coat

pale (adj) : low, dull

steady (adj) : remaining in the same position

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 1A The Snake And The Mirror Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 1A The Snake And The Mirror

9th Class English Chapter 1A The Snake And The Mirror Textbook Questions and Answers

Look at the picture and answer the questions that follow.
AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror 1

Question 1.
What did the lion do on looking at its own reflection?
(a)enjoyed its beauty
(b) felt threatened
(c) felt angry
Give reasons for your option.
Answer:
(c) felt angry
The lion was the king of the forest. It became angry when it saw another lion in his kingdom.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

Question 2.
Have you ever heard or read a story of a pet or an animal looking into a mirror? If yes, how did the animal feel?
Answer:
Yes, I have heard of my friend’s dog looking at its reflection in a mirror. The dog barks at its own image till someone takes him away.

Comprehension

I. Answer the following questions.

Question 1.
The doctor heard a sound coming from above three times. What was it?
Answer:
The doctor heard a noise from above three times. It was made by the snake moving under the roof. But the doctor thought it was the movement of rats.

Question 2.
Why did the doctor sit like a stone?
Answer:
A full blooded cobra was coiling itself around the doctor’s left hand. The snake’s wide open hood was just four inches away from his face. Terrible fear made the doctor sit like a stone. His active mind told him any movement would make the snake bite him. So he remained like a statue.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

Question 3.
What made the doctor forget his fears and have a sense of relief?
Answer:
The snake looked at its image in the mirror. It seemed it was interested in its beauty. It slipped down from the doctor’s hand on to the table. It moved closer to the mirror, the doctor felt relieved. He quietly rose from his chair. He ran out of the room.

Question 4.
Did the snake admire its beauty? Yes or no? Pick the lines from the passage in support of your answer.
Answer:
The snake appeared to admire its beauty. It was certainly looking at its image in the mirror. It wanted to enjoy its reflection at closer quarters.

Question 5.
Why did the doctor want to marry a woman doctor? What kind of woman doctor would he like to marry?
Answer:
The doctor says he was poor. So he wanted to marry a woman doctor with good practice and who had plenty of money. He also wanted her to be fat. That would help him – when he made some silly mistake – run away from her easily. That way he wanted to save himself from her possible attacks.

Question 6.
What did the doctor think about the snake when it looked into the mirror?
Answer:
The doctor thought that the snake was admiring its beauty. He also thought that it was going to take an important decision about its beauty. It was probably thinking of growing moustache or applying eye shadow and mascara.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

Question 7.
Why did the doctor run to his friend’s house? Did the snake want to harm the doctor? Why or why not?
Answer:
The doctor ran to his friend’s house as he was afraid of the snake. But the snake had no interest in harming the doctor. Even when it was on the doctor’s hand, it did not harm him. Normally snakes harm persons when they feel threatened by those persons.

II. Here is a list of sentences from the story. Tick (✓) the sentences that tell you that the author was afraid of the snake.

Sentence Boxes for Ticking
1. I didn’t tremble. 1 didn’t cry out.
2. I sat there holding my breath.
3. I forgot my danger and smiled feebly at myself.
4. Holding my breath 1 got up from the chair.
5. I leapt into the yard and ran for all I was worth.
6. I don’t claim that it was the first snake that had ever looked into a mirror.

Answer:

Sentence Boxes for Ticking
1. I didn’t tremble. 1 didn’t cry out.
2. I sat there holding my breath.
3. I forgot my danger and smiled feebly at myself.
4. Holding my breath 1 got up from the chair.
5. I leapt into the yard and ran for all I was worth.
6. I don’t claim that it was the first snake that had ever looked into a mirror.

Vocabulary

I. Tick (✓) the correct meaning of the words underlined below.

1. The house was not electrified.
a) There was no electricity in the house.
b) The house was not away from the electricity board.
c) The house was not lit up.
Answer:
a) There was no electricity in the house.

2. The earnings were meagre.
a) very high
b) very small
c) nil
Answer:
b) very small

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

3. I also possessed one solitary black coat,
a) single
b) unique
c) expensive
Answer:
a) single

4. I made another earth-shaking decision,
a) valuable
b) dangerous
c) important
Answer:
c) important

5. The room was surrounded by darkness,
a) full of
b) covered
c) encircled
Answer:
a) full of

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

6. Some thief had removed most of my things.
a) damaged
b) destroyed
c) stolen
Answer:
c) stolen

II. Read the following sentence from the text and notice the underlined word. Suddenly there came a dull thud as if a rubber tube had fallen to the ground.
Thud’ means to fall with a low dull sound.

Here is a list of words related to sounds. Say what these words mean. Refer to a dictionary.

Word Meaning
bang
clang
buzz
ring

bang (n) : a sudden loud noise
clang (v) : make a loud ringing sound like that of metals being hit
buzz(n) : a continuous sound like the one a bee makes make
ring (v) : a sound like the telephone or a bell makes

III. Usethelinkersgivenintheboxtocompletethestory.Alinkermaybeusedmorethanonce.

A young doctor lived in a village. The doctor told his friends an interesting story about himself and a cobra. It was a hot summer night. ___(1)___ he opened the door, he heard a sound ___(2)___ was well known to him. He and rats shared the same room. There was no electricity in the house, ___(3)___ he took out a box of matches and lit a kerosene lamp. He took off his clothes and lay down ___(4)___ he could not sleep. He got up from the bed and started reading a story-book. Near the table there was a mirror, ___(5)___ tempted him to look at himself. He was unmarried ___(6)___ he was particular about his handsomeness. He wanted to marry a woman doctor ___(7)___ had plenty of money and should be fat. ___(8)___ he was having such thoughts, suddenly he felt ___(9)___ a rubber tube had fallen to the ground ___(10)___ a snake landed on his shoulder. At that time the only thing he could do was to pray to God. The snake would bite him at any moment ___(11)___ suddenly it turned its head and started admiring its beauty in the mirror. The doctor was thus saved.
AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror 3
Answer:

  1. When
  2. which
  3. so
  4. but
  5. which
  6. so
  7. who
  8. While
  9. as if
  10. but
  11. but

Grammar

I. The Past Perfect and the Simple Past
Read the following sentences from the story.

We found we had little to carry. Some thief had removed most of my things.

In the above two sentences, the first one is in the simple past (found, had) and the second is in the past perfect tense (had removed).

As you can see, the action which occurred earlier, is expressed in the past perfect tense, whereas the action which occurred later, is expressed in the simple past tense. Here are a few more examples.
1. Sujatha had arrived at the station before the train came.
2. The doctor arrived after the patient had died.
3. I bought the ticket as soon as I boarded the bus.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

When two actions occur nearly simultaneously in the past, both are expressed in the simple past tense.
Now combine the following pairs of sentences using the past perfect and the simple past.
(use after, before or when as linkers,)
The first one has been done for you.

1. I brushed my teeth. I took my breakfast.
Answer:
After I had brushed my teeth, I took my breakfast.
Or
I had brushed my teeth before I took my breakfast.

2. She started to go to market. It rained heavily.
Answer:
After she had started to go to market, it rained heavily.
Or
She had started to go to market before it rained heavily.

3. He completed his homework. He went to bed.
Answer:
After he had completed his homework, he went to bed.

4. I finished my dinner. Then my father came.
Answer:
When I had finished my dinner, my father came.

5. I fell seriously ill. I met my family doctor.
Answer:
When I had fallen seriously ill, I met my family doctor.

6. I locked the door. I went out to meet my friend at a restaurant.
Answer:
After I had locked the door, I went out to meet my friend at a restaurant.

II . Phrasal Verbs

Look at the following phrasal verbs and their meanings.

1. take out – to remove something from inside
2. set up – to arrange for something to happen / to start
3. take off – to remove
4. lie down – to take rest
5. take time off – to take some time for something
6. look into – to observe keenly
7. look at – to have a look at something
8. get up – to wake up
9. change into – to turn into
10. leave behind – to leave something as a token of memoray or gift

As you can see, a phrasal verb is a unit of a verb and one or more particles/a preposition. It cannot be understood on the basis of the meanings of the individual parts in isolation, but rather it must be taken as a whole.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

Use the above phrasal verbs in the blanks. Make necessary changes in the verb forms.

Venkata Rao was a rich man. He wanted to ___(1)___ some business in the town. So he ___(2)___ a lot of money from the bank. On an auspicious day he started his medical business, which prospered by leaps and bounds. He became so busy that he could not ___(3)___ for any rest. He used to ___(4)___ very early in the morning and go to his office. In the afternoon he would ___(5)___ his shirt and ___(6)___ on his couch in the office. After a nap he would put on his shirt. Then he would ___(7)___ the mirror and ___(8)___ himself admiringly. Anyhow he ___(9)___ some fame in the town even when he went abroad. ___(10)___ a new man. Everybody believed that he
Answer:

  1. set up
  2. took out
  3. take time off
  4. get up
  5. take off
  6. lie down
  7. look into
  8. look at
  9. left behind
  10. had changed into

Writing

I. Write a descriptive essay on any unusual or humorous incident you may have witnessed or experienced.
Answer:
Have you ever heard of or seen a dog walking on water ? Most of or rather all of you would say a firm ‘No’. Nay, you may even doubt if there could be a more foolish question on this earth than the one above. You have every right to doubt and you are right in doubting too I But the world is so full of surprising and unusual incidents and creatures that we can’t believe their existence unless we see them !

I am lucky and also proud to be a witness to one of such unbelievable and exciting incidents. I have a friend whose uncle from the USA gifted him on his last birthday a very strange looking young dog. My friend Rohit named his prized possession ‘Walkie’. Rohit loves to spend every second of his free time with Walkie. Even Walkie likes to spend more with Rohit than with any other member of the family. Walkie accompanies Rohit to every place except, of course, school.

On one hot summer day we all thought of going to the river bank to cool ourselves by swimming in the river. As usual, Walkie joined us along with Rohit. And as Rohit walked into the water in his swim clothes, Walkie too stepped into the river. And then came the stunning discovery that made us dumbstruck ! Walkie was walking on the surface of the water as if it was land ! For a few moments, none of us could believe our eyes ! To be doubly sure of truth of what we saw, we made Walkie go this way, that way in – no, on – the river. And as if unaware of its unusual ability, Walkie jumped„hopped, leapt and ran all around the river – on the surface – to our commands.

That evening everyone of our friends was telling Walkie’s story to every person we came across. And everyone of us got the same response – “Don’t be a fool and don’t try to make us a fool !

The news spread like a wildfire ! The following day the whole village was on the bank of the river. And Walkie was the focus of everyone’s attention. And to thrill every spectator, Walkie crossed the river – walking on the surface – twice or thrice ! Walkie became the hero of our village and Rohit enjoyed his pet’s popularity to his heart’s content.

And it has become our hobby now to frequent to the riverside along with Walkie !

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

II. Preparation of a Poster :

Teli the names of some important events/happenings.
If you want to publicize the matter, what would you do?
Look at the poster given below.
AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror 2
Now reflect on the following questions.

1. What are the elements of an effective poster?
Answer:
All the elements like colour, size, layout, font size, language are important.

2. Who issued the poster?
Answer:
(Here) THE KANDUKURU LAUGHING CLUB, issued the poster.

3. What is the purpose of the poster?
Answer:
The purpose of the poster is to invite entries for Raja Babu award.

4. What is the deadline?
Answer:
July 27 is the dead line.

5. What are the other details highlighted?
Answer:
The other details highlighted are
1) Award value
2) Who can participate
3) Contact address, etc.

6. What should be highlighted in a poster?
Answer:
All the necessary details are to be highlighted.

7. Who are the target audience of the poster?
Answer:
(Here) The print media journalists in Telugu, Urdu and English are the target audience.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

Now prepare a poster using the information given below by arranging it in a proper sequence.

Highlights Information
Theme Humour
Occasion The Laughter Day
Category Stories, Songs, Poems, Essays, Articles and Cartoons
Award Best Comic Writer Award
Prize Amount ₹ 10,000
Contact No. 98xxxxxx49
Deadline 9th August
Issued by ABC Laughing Club, Mummidivaram, East Godavari Dist. -533216
Target Audience Writers, Journalists, Poets, Essayists,Cartoonists and Lyricists

Answer:
AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror 4

Study Skills

Here is a short narration. Read it carefully and make a list of important points. Suggest a suitable title for it.

The young doctor married a rich fat foreign woman. He told her ‘the episode of the snake and the mirror’. She laughed heartily. He went on talking about cobras and some funny facts (associated) related to them. He said to her “Cobras are the most poisonous, active and dangerous snakes. But, surprisingly, most Hindus worship them. It is funny to know that some women drop eggs and pour milk into the ant hills where cobras usually reside.

It is funny to know that some people believe cobras dance to the tunes of charmers. In fact, they have no ears. It is equally humorous to be aware of the fact that most cobra victims go to the charmers instead of the doctors. The charmers enchant some so-called mantras taking some neem leaves into their hands. The cobra victim dies and they say the victim came late to them. It is funnier to know that some farmers cremate cobras and perform funeral rites as if one of the members of their family has died.” “Are there any other funny stories related to cobras?” The young wife enquired. “Yes”. There are so many stories to tell but! will tell them tomorrow. The patients may be waiting for me at hospital.” The doctor said and left for hospital in a hurry.
Answer:
Cobras : Some Beliefs Among Hindus

  • An Indian doctor tells his foreign wife about cobras and some beliefs about them.
  • Cobras are the most poisonous, active, and dangerous snakes.
  • Yet, most Hindu women worship cobras.
  • Women offer eggs and milk to cobras.
  • People believe cobras dance to the tunes of charmers. Cobras do not have ears.
  • Cobra victims go to charmers, instead of to doctors.
  • Some farmers cremate cobras and perform funerals.

Listening

Practise listening carefully. Then you will be able to speak.
Your teacher will tell you the story, ‘Master of the Game.’ Listen to it carefully and answer the following questions.

Master of the Game

An old man who lived in a small side street in the city of Mumbai had to put up with the nuisance of having boys play cricket outside his house, at night.

One evening when the boys were particularly noisy, he went out to talk to them. He explained that he was a pensioner, who was happiest when he could see or hear boys playing his favourite game, cricket. He said he would give them ? 25 each week to play in the street at night. The boys were thrilled. They were being paid to do something they enjoyed I At the end of the first week they knocked at the old man’s house and asked to be paid.

He did so.
The second week when they asked for payment he said he had run out of money and sent them away with only ₹15.

The third week the man said he had not yet received his pension and gave them only ₹10

The boys were very disappointed but there was not much they could do about it.

The fourth week the man said he could not afford to pay them ₹25 as he had promised, but would give them ₹ 5 each week without fail.

This was too much for the boys.
“You expect us to play seven days a week for a measly ₹5!” they yelied. “Go to blazes.” They stormed away and never played on the street again.

AP Board 9th Class English Solutions Chapter 1A The Snake And The Mirror

Now, answer the following questions.

1. What made the old man talk to the boys?
Answer:
The old man was living in a small side street in Mumbai. Some boys were playing cricke outside his house every night. They were very noisy. The old man Was unable to put up with that nuisance. So, he decided to talk to the boys.

2. Why did the boys stop playing cricket in the night outside the old man’s house?
Answer:
The old man promised to pay Rs. 25 a week to the boys for playing his favourite game near his house. After some time he paid Rs. 15 a week. Later he reduced it to Rs. 10. When he said he couldn’t pay more than Rs. 5 a week, the boys said it was too little for a week’s play. So they stopped playing cricket there.

The Snake And The Mirror Summary in English

Vaikom Muhammad Basheer is the writer of ‘The Snake and the Mirror”. He is a famous story teller from Kerala. The story is humorous. It tells us the experience of a young home doctor with a snake. The doctor was living alone in a small room with tiled roof. It was a hot summer night. He entered his room late at night. He tried to sleep but couldn’t. It was hot and he sat on a chair in front of a mirror and tried to read a book. There was a noise in the roof once or twice. But he did not bother about it because there were a lot many rats moving around. He started dreaming about marrying a fat, rich lady doctor.

Suddenly there was a ‘thud’ like sound. As he turned back to see what it was, there was a strong, long cobra coiling around his left hand with its hood wide open and just four inches away from his face. He at once became like a stone in fear. But his mind was active. As he was thinking of how to save himself, the snake saw its image in the mirror. It seemed that the snake was interested in its beauty. So it slipped down on to the table and moved towards the mirror. The doctor took this opportunity to run away from the room to a friend’s house. The following morning he came! back to his room with his friend. There was no snake. The room was almost empty. Some! thief had taken away all his belongings except a dirty vest. The doctor answers a listener’s question by saying he married a woman who was lean and an expert in running!

The Snake And The Mirror Glossary

coiled (v-past tense) : turned round and round

full-blooded (adj) : strong

homeopath (n) : a doctor practising home system

discuss (v) : talk about (Note : The word ‘discuss’ should not be followed by ‘about’. Its noun form ‘discussion’ can take ‘about’. The spelling of ‘discuss’ has two ‘s’ letters at the end.)

restaurant (n) : a place where food items are served (Make a note of the spelling !)

returned (v-past tense) : came back (Note : The word ‘back’ should NOT be used after ‘return’)

meagre (adj) : too small

solitary (adj) : single; only one

vest (n) : a sleeveless undershirt

gables (n-plural) : (here) long wooden bars supporting the roof

take time off (phrasal verb) : be free from work; take rest

parting (n) : the separating area

moustache (n) : hair grown on a man’s upper lip

earthshaking (adj) : very surprising; having great importance

resumed (v-past tense) : took back; occupied again

wriggled (v-past tense) : twisted and turned the body to move

Simultaneous (adj) : happening at the same time as another act

tremble (v) : shiver/shake in fear

slithered (v-past) : moved smoothly, glided

at closer quarters (phrasal verb) : from a short distance away

hood (n) : an expanded part of snake’s head

leaden (adj) : made of lead

lurked (v-past tense) : was around secretly

feebly (adv) : weakly

reflection (n) : image

mascara (n) : something applied on eye lashes to make them look dark and thick

vermilion (n) : bright red coloured powder applied on the forehead as a round spot

unwound (v-past tense) (wind-wound x unwound) : freed itself

crept (v-past tense of creep) : moved on all limbs

closer (adj-comparative degree) : nearer

no mere image cut in granite : no more afraid and so not like a statue

a man of flesh and blood : a normal man

leapt (v- past tense of leap) : jumped

heaved a sigh of relief : felt relieved

companion ife : wife

reedy (adj) : very thin like a reed

sprinter (n) : one who participates in running competitions

smeared (v-past tense) : applied; rubbed

taken with (phrasal verb) : attracted by

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation InText Questions

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Think, Discuss and Write

Question
Which of the following expressions are polynomials ? Which are not ? Give reasons. [Page No. 28]
Solution:
i) 4x2 + 5x – 2 is a polynomial.
ii) y2 – 8 is a polynomial.
iii) 5 is a constant polynomial.
iv) \(2 x^{2}+\frac{3}{x}-5\) is not a polynomial as x is in denominator.
v) √3x2 + 5y is a polynomial.
vi) \(\frac{1}{x+1}\) is not a polynomial as the variable x is in denominator.
vii) √x is not a polynomial as its exponent is not an integer.
viii) 3xyz is a polynomial.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Do These

Question
Write two polynomials with variable ‘x’. [Page No. 29]
Solution:
5x2 + 2x – 8 and 3x2 – 2x + 6.

Question
Write three polynomials with variable ‘y’.
Solution:
y3 – y2 + y ; 2y2 + 7y – 9 + 3y3; y4 – y + 6 + 2y2.

Question
Is the polynomial 2x2 + 3xy + 5y2 in one variable ?
Solution:
No. It is in two variables x and y.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question
Write the formulae of area and volume of different solid shapes. Find out the variables and constants in them. [Page No. 29]
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions 1

Question 1.
Write the degree of each of the following polynomials. [Page No. 30]
Solution:
i) 7x3 + 5x2 + 2x – 6 – degree 3
ii) 7 – x + 3x2 – degree 2
iii) 5p – √3 – degree 1
iv) 2 – degree 0
v) – 5 xy2 – degree 3

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question 2.
Write the co-efficient of x2 in each of the following. [Page No. 30]
Solution:
i) 15 – 3x + 2x2 : co-efficient of x2 is 2
ii) 1 -x2 : co-efficient of x2 is -1
iii) πx2 – 3x + 5 : co-efficient of x2 is π
iv) √2x2 + 5x – 1 : co-efficient of x2 is √2

Think, Discuss and Write

Question
How many terms a cubic (degree 3) polynomial with one variable can have? Give examples. [Page No. 31]
Solution:
A cubic polynomial can have atmost 4 terms.
E.g.: 5x3 + 3x2 – 8x + 4; x3 – 8

Try These

Question 1.
Write a polynomial with 2 terms in variable x. [Page No. 31]
Solution:
2x + 3x2

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question 2.
How can you write a polynomial with 15 terms in variable ‘x’. [Page No. 31]
Solution:
a14p14 + a13p13 + a12p12+ …………….+ a1p + a0

Do This

Question
Find the value of each of the follow ing polynomials for the indicated value of variables. [Page No. 33]
(i) p(x) = 4x2 – 3x + 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
4(1)2 – 3(1) + 7 = 8

ii) q(y) = 2y3 – 4y + √11 at y = 1.
Solution:
The value of q(y) at y = 1 is
2(1)3 – 4(1) + √11 = -2 + √11

iii) r(t) = 4t4 + 3t3 – t2 + 6 at t = p, t ∈ R.
Solution:
The value of r(t) at t = p is
4p4 + 3p3 – p2 + 6

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

iv) s(z) = z3 – 1 at z – 1.
Solution:
The value of s(z) at z = 1 is 13 – 1 = 0

v) p(x) = 3x2 + 5x – 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
3(1)2 + 5(1) — 7 = 1.

vi) q(z) = 5z3 – 4z + √2 at 7. = 2.
Solution:
The value of q(z) at z = 2 is
5(2)3 – 4(2) + √2 = 40 – 8 + √2
= 32 + √2

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Try These

Question
Find zeroes of the following polyno¬mials. [Page No. 34]

1. 2x-3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)
∴ x = \(\frac{3}{2}\) is the zero of 2x – 3.

2. x2 – 5x + 6
Solution:
x2 – 5x + 6 = 0
⇒ x2 – 3x – 2x + 6 = 0
⇒ x (x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x – 2 = 0 or x – 3 = 0
⇒ x = 2 or x = 3
∴ x = 2 or 3 are the zeroes of x2 – 5x + 6.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

3. x + 5
Solution:
x + 5 = 0
x = – 5
∴ x = – 5

Do This

Fill in the bianks : [Page No. 35]

Linear polynomial Zero of the polynomial
x + a – a
x – a a
ax + b \( \frac{-b}{a} \)
ax – b \( \frac{b}{a} \)

Solution:

Linear polynomial Zero of the polynomial
x + a – a
x – a a
ax + b \( \frac{-b}{a} \)
ax – b \( \frac{b}{a} \)

Think, and Discuss

Question 1.
x2 + 1 has no zeroes. Why ? [Page No. 36]
Solution:
x2 + 1 = 0 ⇒ x2 = -1
No real number exists such that whose root is – 1.
∴ x2 + 1 has no zeroes.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question 2.
Can you tell the number of zeroes of a polynomial of degree ‘n’ will have? [Page No. 36]
Solution:
A polynomial of degree n will have n- zeroes.

Do These

Question 1.
Divide 3y3 + 2y2 + y by ‘y’ and write division fact. [Page No. 38]
Solution:
(3y3 + 2y2 + y) ÷ y = \(\frac{3 y^{3}}{y}+\frac{2 y^{2}}{y}+\frac{y}{y}\)
= 3y2 + 2y + 1
Division fact = (3y2 + 2y + 1) y
= 3y3 + 2y2 + y

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question 2.
Divide 4p2 + 2p + 2 by ‘2p’ and write division fact.
Solution:
4p2 + 2p ÷ 2 = \(\frac{4 p^{2}}{2 p}+\frac{2 p}{2 p}+\frac{2}{2 p}\)
= 2p + 1 + \(\frac{1}{\mathrm{P}}\)
Division fact:
(2p + 1 + \(\frac{1}{\mathrm{P}}\)).2p = 4p2 + 2p + 2

Try These

Show that (x – 1) is a factor of xn – 1. [Page No. 45]
Solution:
Let p(x) = xn – 1
Then p(1) = 1n – 1 = 1 – 1 = 0
As p(1) = 0, (x – 1) is a factor of p(x).

Do These

Question
Factorise the following. [Page No. 46]

1. 6x2 + 19x + 15
Solution:
6x2 + 19x + 15 = 6x2 + 10x + 9x + 15
= 2x (3x + 5) + 3 (3x + 5)
= (3x + 5) (2x + 3)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

2. 10m2 – 31m – 132
Solution:
10m2 – 31m – 132
= 10m2 – 55m + 24m – 132
= 5m (2m- 11) + 12 (2m- 11)
= (2m – 11) (5m + 12)

3. 12x2 + 11x + 2
Solution:
12x2 + 11x + 2
= 12x2 + 8x + 3x + 2
= 4x (3x + 2) + 1 (3x + 2)
= (3x + 2) (4x + 1)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Try This

Question
Try to draw the geometrical figures for other identities. [Page No. 49]
i) (x + y)2 ≡ x2 + 2xy + y2
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions 2
Step – 1 : Area of fig. I = x x = x2
Step – 2 : Area of fig. II = x y = xy
Step – 3 : Area of fig. III = x y = xy
Step – 4 : Area of fig. IV = y y = y2

Area of big square = sum of the areas of figures I, II, III and IV
∴ (x + y) (x + y) = x2 + xy + xy + y2
(x + y)2 = x2 + 2xy + y2

ii) (x + y) (x – y) ≡ x2 – y2
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions 3
Step -1: Area of fig! I = x (x – y) = x2 – xy
Step – 2: Area of fig. II = (x – y) y = xy – y2
Area of big rectangle = sum of areas of figures I & II
(x + y) (x – y) = x2 – xy + xy – y2
= x2 – y2
∴ (x + y) (x-y) = x2-y2

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

iii) (x + a) (x + b) ≡ x2 + (a + b) x + ab
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions 4
Step – 1 : Area of fig. I = x2
Step – 2 : Area of fig. II = ax
Step – 3 : Area of fig. Ill = bx
Step – 4 : Area of fig. IV = ab
∴ Area of big rectangle = Sum of areas of four small figures.
∴ (x + a) (x + b) = x2 + ax + bx + ab
(x + a) (x + b) = x2 + (a + b) x + ab

Do These

Question
Find the following product using appropriate identities. [Page No. 49]
i) (x + 5) (x + 5)
Solution:
(x + 5) (X + 5) = (x + 5)2
= x2 + 2(x) (5) + 52
= x2 + 10x + 25

ii) (p – 3) (p + 3)
Solution:
(p – 3) (p + 3)
= p2 – 32
= p2 – 9

iii) (y – 1) (y – 1)
Solution:
(y – 1) (y – 1)
= (y – 1)2
= y2 – 2y + 1

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

iv) (t + 2) (t + 4)
Solution:
(t + 2) (t + 4)
= t2 + t(2 + 4) + 2 x 4
= t2 + 6t + 8

v) 102 x 98
Solution:
102 x 98 = (100 + 2) (100 -2)
= 1002 – 22
= 10000 – 4
= 9996

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Do These

Question
Factorise the following using appro-priate identities. [Page No. 50]
i) 49a2 + 70ab + 25b2
Solution:
49a2 + 70ab + 25b2
= (7a)2 + 2 (7a) (5b) + (5b)2
= (7a + 5b)2
= (7a + 5b)(7a + 5b)

ii) \(\frac{9}{16} x^{2}-\frac{y^{2}}{9}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions 5

iii) t2 – 2t + 1
= (t)2 – 2(t) (1) + (1)2
= (t – 1)2 = (t – 1) (t – 1)

iv) x2 + 3x + 2
Solution:
x2 + 3x + 2 = x2 + (2 + 1) x + (2 x 1)
(x + 2) (x + 1)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Do These

Question i)
Write (p + 2q + r)2 in expanded form. [Page No. 52]
Solution:
(p + 2q + r)2 = (p)2 + (2q)2 + (r)2
+ 2 (P) (2q) + 2 (2q) (r) + 2(r) (p)
= p2 + 4q2 + r2 + 4pq + 4qr + 2rp

Question ii)
Expand (4x – 2y – 3z)2 using identity. [Page No. 52]
Solution:
(4x – 2y – 3z)2 = (4x)2 + (- 2y)2 + (- 3z)2 + 2 (4x) (- 2y) + 2 (- 2y) (- 3z) + 2 (- 3z) (4x)
= 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx.

Question iii)
Factorise 4a2 + b2 + c2 – 4ab + 2bc – 4ca
using identity. [Page No. 52]
Solution:
4a2 + b2 + c2 – 4ab + 2bc – 4ca
= (2a)2 + (- b)2 + (- c)2 + 2(2a) (- b) + 2 (- b) (- c) + 2(- c) (2a)
= (2a – b – c)2 = (2a – b – c) (2a – b – c)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Try These

Question
How can you find (x – y)3 without actual multiplication ? Verify with actual multiplication. [Page No. 52]
Solution:
(x – y)3 = x3 – 3x2y + 3xy2 – 3y3 from identity.
By actual multiplication
(x – y)3 = (x – y)2 (x – y)
= (x2 – 2xy + y2) (x – y)
= x3 – 2x2y + xy2 – x2y + 2xy2 – y3
= x3 – 3x2 y + 3xy2 – y3
Both are equal.

Do These

Question 1.
Expand (x + 1)3 using an identity. [Page No. 54]
Solution:
(x + 1)3 = (x)3 + (1)3 + 3 (x) (1) (x + 1)
= x3 + 1 + 3x (x + 1)
= x3 + 1 + 3x2 + 3x = x3 + 3x2 + 3x + 1

Question 2.
Compute (3m – 2n)3. [Page No. 54]
Solution:
(3m-2n)3
=(3m)3 – 3 (3m)2 (2n) + 3 (3m) (2n)2 – (2n)3
= 27m3 – 54m2n + 36mn2 – 8n3

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question 3.
Factorise a3 – 3a2b + 3ab2 – b3. [Page No. 54]
Solution:
a3 – 3a2b + 3ab2 – b3
= (a)3 – 3 (a)2 (b) + 3 (a) (b)2 – (b)3
= (a – b)3
= (a – b) (a – b) (a – b)

Do These

Question 1.
Find the product (a – b – c) (a2 + b2 + c2 – ab + be – ca) without actual multi-plication. [Page No. 55]
Solution:
The problem is incorrect.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

Question 2.
Factorise 27a3 + b3 + 8c3 – 18abc using identity. [Page No. 55]
Solution:
27a3 + b3 + 8c3 – 18abc
= (3a)3 + (b)3 + (2c)3 – 3(3a) (b) (2c)
= (3a + b + 2c) (9a2 + b2 + 4c2 – 3ab – 2be – 6ca)

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers InText Questions

Do This

Question 1.
Represent \(\frac{-3}{4}\) on a number line. (Page No.3)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 1
Step – 1: Divide each unit into four equal parts to the right and left side of zero on the number line.
Step – 2 : Take 3 parts after zero on its left side.
Step -3 : It represents \(\frac{-3}{4}\)

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Question 2.
Write 0, 7, 10, -4 in p/q form (Page No.2)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 2

Question 3.
Guess my number : Your friend chooses an integer between 0 and 100. You have to find out that number by asking questions, but your friend can only answer ‘Yes’ or ‘No’. What strategy would you use ? (Page No.3)
Solution:
Let my friend choosen 73; then my questions may be like this.

Q : Does it lie in the first 50 numbers ?
A: No .

Q : Does it lie between 50 and 60 ?
A: No

Q : Does it lie between 60 and 70 ?
A: No

Q : Does it lie between 70 and 80 ?
A: Yes
[then my guess would be “it is a number from 70 to 80]

Q : Is it an even number ?
A: No
[my guess : it should be one of the numbers 71, 73, 75, 77 and 79]

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Q : Is it a prime number ?
A: Yes
[my guess : it may be 71, 73 or 79]

Q : Is it less than 75 ?
A: Yes
[my guess : it may be either 71 or 73]

Q : Is it less than 72 ?
A: No

Then the number is 73.
Therefore the strategy is
→ Use number properties such as even, odd, composite or prime to determine the number.

Do This

Question (i)
Find five rational numbers between 2 and 3 by mean method. (Page No.4)
Solution:
We know that \(\frac{a+b}{2}\) is a rational
between any two numbers a and b.
Let a = 2 and b = 3
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 3

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Question (ii)
Find 10 rational numbers between \(\frac{-3}{11}\) and \(\frac{8}{11}\) (Page No.4)
Solution:
\(\frac{-3}{11}<\left[\frac{-2}{11}, \frac{-1}{11}, \frac{0}{11}, \frac{1}{11}, \frac{2}{11}, \frac{3}{11}, \frac{4}{11}, \frac{5}{11}, \frac{6}{11}, \frac{7}{11}\right]<\frac{8}{11}\)

Do This

Question
Find the decimal form of
(i) \(\frac{1}{17}\) (Page No.5)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 4
\(\frac{1}{19}\) = 0.052631

(ii) \(\frac{1}{19}\)(Page No.5)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 5

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Try These

Question 1.
Find the decimal values of the following. (Page No.6)
Solution:
i) \(\frac{1}{2}\) = 0.5
ii) \(\frac{1}{2^{2}}=\frac{1}{4}\) = 0.25
iii) \(\frac{1}{5}\) = 0.2
iv) \(\frac{1}{5 \times 2}=\frac{1}{10}\) = 0.1
v) \(\frac{3}{10}\) = 0.3
vi) \(\frac{27}{25}\)
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 6

vii) \(\frac{1}{3}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 7

viii) \(\frac{7}{6}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 8

ix) \(\frac{5}{12}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 9

x) \(\frac{1}{7}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 10

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Think, Discuss and Write

Question 1.
Kruthi said √2 can be written as \(\frac{\sqrt{2}}{1}\) which is in form. So √2 is a rational number. Do you agree with her argument ? (Page No.10)
Solution:
No.
Writing √2 = \(\frac{\sqrt{2}}{1}\) is not in the \(\frac { p }{ q }\) form.
Since p and q are integers and √2 is not an integer.

Try These

Question 1.
Find the value of √3 upto six decimals.(Page No.10)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 11

Step 1 :
Write 3 as 3.00 00 00 00 00 00 00
Step – 2 :
Group the zeros in pairs (i.e.) make periods.
Step – 3 :
Find the square root using long division method.
∴ √3 = 1.732050

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Try These

Question 1.
Locate √5 and -√5 on the number line [Hint 5 = 22 + 12](Page No.12)
Solution:
Step – 1 :
At zero draw a rectangle of length 2 units and breadth 1 unit.

Step – 2 :
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 12

Step – 3 :
Using compass draw an arc of radius OB with centre ‘O’ which cuts the num-ber line at D and D1.

Step – 4 :
D represents √5 and D1 represents – √5 on the number line.
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 13

Activity

Question
Constructing the ‘Square root spiral’. (Page No. 15)
Solution:
Take a large sheet of paper and construct the ‘Square root spiral’ in the following manner.
Step – 1 : Start with point ‘O’ and draw a line segment \(\overline{\mathrm{OP}}\) of 1 unit length.
Step – 2 : Draw a line segment \(\overline{\mathrm{PQ}}\) perpendicular to \(\overline{\mathrm{OP}}\) of unit length (where OP = PQ = 1) (see Fig.)
Step – 3 : Join 0, Q. (OQ = √2 )
Step – 4 : Draw a line segment OR of unit length perpendicular to \(\overline{\mathrm{OQ}}\)
Step – 5 : Join O, R. (OR = √3 )
Step – 6 : Draw a line segment RS of unit length perpendicular to \(\overline{\mathrm{OR}}\).
Step – 7 : Continue in this manner for some more number of steps, you will create a
beautiful spiral made of line segment \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}, \overline{\mathrm{RS}}, \overline{\mathrm{ST}}, \overline{\mathrm{TU}}\) …………etc. Note that the line segments \(\overline{\mathrm{OQ}}, \overline{\mathrm{OR}}, \overrightarrow{\mathrm{OS}}, \overline{\mathrm{OT}}, \overline{\mathrm{OU}}\) …… etc., denote the lengths √2,√3, √4, √5, √6 respectively.
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 14

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Do This

Question 1.
Find rationalising factors of the denominators of (Page No. 20)
i) \(\frac{1}{2 \sqrt{3}}\)
Solution:
\(\frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{2 \times 3}=\frac{\sqrt{3}}{6}\)
∴ The R.F is √3

ii) \(\frac{1}{\sqrt{5}}\)
Solution:
\(\frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}\)
∴ The R.F is √5

iii) \(\frac{1}{\sqrt{8}}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions 15 (2)

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Do This;

Question
Simplify : (Page No. 23)

i) (16) 1/2
Solution:
(16)1/2 = (4 x 4)1/2 = (42)1/2 = 42/2 = 4

ii) (128)1/7
Solution:
(128)1/7 =(2 X 2 X 2 X 2 X 2 X 2 X 2)1/7 = (27)1/7 = 2

iii) (343)1/5
Solution:
(343)1/5= (3 x 3 x 3 x 3 x 3)1/5 = (35)1/5= 3

Question 1.
Write the following surds in exponen-tial form (Page No. 24)
i) √2
Solution:
√2 = 21/2

ii) 3√9
Solution:
\(\sqrt[3]{9}=\sqrt[3]{3 \times 3}=\sqrt[3]{3^{2}}=3^{\frac{2}{3}}\)

AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions

iii) \(\sqrt[5]{20}\)
Solution:
\(\begin{aligned}
\sqrt[5]{20}=\sqrt[5]{2 \times 2 \times 5} &=\sqrt[5]{2^{2} \times 5} \\
&=2^{\frac{2}{5}} \times 5^{\frac{1}{5}}
\end{aligned}\)

iv) 17√19
Solution:
\(\sqrt[17]{19}=19^{\frac{1}{17}}\)

Question 2.
Write the surds in radical form. (Page No. 24)
i) 51/7 = \(\sqrt[7]{5}\)
ii) 171/6 = \(\sqrt[6]{17}\)
iii) 52/5 = \(\sqrt[5]{5^{2}}=\sqrt[5]{5 \times 5}=\sqrt[5]{25}\)
iv) 1421/2 = \(\sqrt{142}\)

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.4

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

Question 1.
State which of the following are mathematical statements and which are not ? Give reason.
i) She has blue eyes.
Solution:
This is not a mathematical statement. no mathematics is involved in it.

ii) x + 7 = 18
Solution:
This is not a statement, as its truthness cant be determined.

iii) Today is not Sunday.
Solution:
This is not a statement. This is an am biguous open sentence.

iv) For each counting number x, x + 0 = x.
Solution:
This is a mathematical statement.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

v) What tune is it?
Solution:
This is not a riathematical statement.

Question 2.
Find counter examples to disprove the following statements.
i) Every rectangle is a square.
Solution:
AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4 1
A rectangle and square are equiangular i.e., all the four angles are right angles. This doesn’t mean that they have equal sides.

ii) For any integers x and y,
\(\sqrt{x^{2}+y^{2}}\) = x + y
Solution:
Let x = 3; y = 8
\(\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+8^{2}}\)
= \(\sqrt{9+64}=\sqrt{73}\)

x + y = 3 + 8 = 11
Here, √73 ≠ 11
i.e., \(\sqrt{x^{2}+y^{2}}\) ≠ x + y

iii) If n is a whole number then 2n2 +11 is a prime.
Solution:
If n = 11 then 2n2+ 11 = 2 (11)2 + 11
= 11 (2 × 11 + 1) = 11 × (22 + 1)
= 11 × 23 is not a prime.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

iv) Two triangles are congruent if all their corresponding angles are equal.
Solution:
AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4 2
If the corresponding angles are equal then the triangles are only similar.

v) A quadrilateral with all sides are equal is a square.
Solution:
A rhombus is not a square, but all its sides are equal.

Question 3.
Prove that the sum of two odd numbers is even.
Solution:

Steps Reasons
1) (2m + 1); (2n + 1) be the two odd numbers General form of an odd number.
2) (2m + 1) + (2n + 1) = (2m + 2n + 2)
= 2 (m + n + 1) = 2K
Hence proved.
Adding the two numbers
General form of an even number.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

Question 4.
Prove that the product of two even numbers is an even number.
Solution:

Steps Reasons
1) Let 2m and 2n be two even numbers. General form of an even number.
2) 2m.2n = 4mn = 2(2mn) = 2K Taking the product Rearranging the numbers.
3) 2K where K = 2mn K=2mn
4) Even number
Hence proved.
General form of an even number.

Question 5.
Prove that if x is odd, then x2 is also odd.
Solution:
Let x be an odd number.
Then x = 2m + 1
(general form of ah odd number) Squaring on both sides,
x2 = (2m + 1)2
= 4m2 + 4m +1
= 2 (2m2 + 2m) + 1
= 2K + 1 where K = 2m2+ 2
Hence x2 is also odd.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

Question 6.
Examine why they work ?
Choose a number. Double it. Add nine. Add your original number. Divide by three. Add four. Subtract your original number. Your result is seven.
Solution:
Choose a number = x say
Double it = 2x
Add nine = 2x + 9
Add your original number
= 2x + 9 + x = 3x + 9
Divide by 3 = (3x + 9) ÷ 3
= \(\frac{3 x}{3}+\frac{9}{3}\) = x + 3
Add 4 ⇒ x + 3 + 4 = x + 7
Subtract your original number =
x + 7 – x = 7
Your result is 7 – True.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4

ii) Write down any three digit number (for example, 425). Make a six digit number by repeating these digits in the same order 425425. Your new number is divisible by 7, 11 and 13.
Solution:
Let a three digit number be xyz.
Repeat the digit = xyzxyz
= xyz × (1001)
= xyz × (7 × 11 × 13)
Hence the given conjecture is true.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.3

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

Question 1.
Take any three consecutive odd numbers and find their product, for example 1 × 3 × 5 = 15;
3 × 5 × 7 = 105: 5 × 7 × 9 = ……………
ii) Take any three consecutive even numbers and add them, say,
2 + 4 + 6 = 12; 4 + 6 + 8 = 18:
6 + 8 + 10 = 24; 8 + 10 + 12 = 30 ….
so on. Is there any pattern you can guess in these sums ? What can von conjecture about them ?
Solution:
i) 1 × 3 × 5 = 15
3 × 5× 7 = 105
5 × 7 × 9 = 315
7 × 9 × 11 = 693

  1. The product of any three consecutive odd numbers is odd.
  2. The product of any three consecutive odd numbers is divisible by ’3′.
  3. 2 + 4 + 6 = 12; 4 + 6 + 8= 18;
    6 + 8 + 10 = 24; 8 + 10 + 12 = 30
  4. The sum of any three consecutive even numbers is even.
  5. The sum of any three consecutive even numbers is divisible by 6
  6. The sum of any three consecutive even numbers is a multiple of 6.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

Question 2.
Go back to Pascal’s triangle.
Line-1: 1=11°
Line-2: 11 = 111
Line-3 : 121 = 112
Make a conjecture about line – 4 and line – 5.
AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3 1
Does your conjecture hold ? Does your conjecture hold for line – 6 too ?
Solution:
Line-4 : 1331 = 113
Line-5 : 14641 = 114
Line – 6 : 115
∴ Line – n = 11n-1
Yes the conjecture holds good for line – 6 too.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

Question 3.
Look at the following pattern.
i) 28 = 22 × 71;
Total number of factors
(2+ 1)(1 + 1) = 3 × 2 = 6
28 is divisible by 6 factors i.e.,
1, 2, 4, 7, 14, 28.
ii) 30 = 21 × 31 × 51, Total number of .
factors (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 = 8
30 Is divisible by 8 factors i.e., 1, 2, 3, 5, 6, 10, 15 and 30
Find the pattern.
[Hint : Product of every prime base exponent +1)
Solution:
24 = 23 × 31
24 has (3+1) (1 + 1) = 4 × 2 = 8 factors
[1, 2, 3, 4, 6, 8, 12 and 24]
36 = 22 × 32
Number of factors = (2 + 1) (2 + 1)
3 × 3 = 9 [ 1, 2, 3. 4, 6, 9, 12, 18 and 36]
If N = ap. bq . cr…….. where N is a natural number.
a. b, c … are primes and p, q, r ……. are positive integers then, the number of factors of N =(p- 1)(q+ 1)(r + 1)

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

Question 4.
Look at the following pattern :
12 = 1
112 = 121
1112 = 12321;
11112 = 1234321
111112 = 123454321
Make a conjecture about each of the following
1111112 =
11111112 =
Check if your conjecture is true.
Solution:
1111112 = 12345654321
11111112 = 1234567654321
(111………. n-times)2 = (123 … (n- 1) n (n – 1) (n – 2) 1)
The conjecture is true.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

Question 5.
List five axioms (postulates) used in text book.
Solution:

  1. Things which are equal to the same things are equal to one another.
  2. If equals are added to equals, the sums are equal.
  3. If equals are subtracted from equals, the differences are equal.
  4. When a pair of parallel lines are in-tersected by a transversal, the pairs of corresponding angles are equal.
  5. There passes infinitely many lines through a given point.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3

Question 6.
In a polynomial p(x) = x2 + x + 41 put different values of x and find p(x). Can you conclude after putting different values of x that p(x) is prime for all. Is ‘x’ an element of N ? Put x = 41 in p(x). Now what do you find ?
Solution:
p(x) = x2 + x + 41
p(0) = 02 + 0 + 41 = 41 – is a prime
p(1) = 12 + 1 + 41 = 43 – is a prime
p(2) = 22 + 2 + 41 = 47 – is a prime
p(3) = 32 + 3 + 41 = 53 – is a prime
p(41) = 412 + 41 + 41
= 41(41 + 1 + 1)
= 41 x 43 is not a prime.
∴ p(x) = x2+ x + 41 is not a prime for all x.
∴ The conjecture “p(x) = x2 + x + 41 is a prime” is false.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.2

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2

Question 1.
Use deductive reasoning to answer the following.

i) Human beings are mortal. Jeevan is a human being. Based on these two statements, what can’ you conclude about Jeevan?
Solution:
From the above statements we can deduce that Jeevan is mortal as it is given that all humans are mortal and Jeevan is a human.

ii) All Telugu people are Indians. X is an Indian. Can you conclude that X belongs to Telugu people?
Solution:
No. X may belong to any other language like Tamil, Kannada, Malayali…. etc.

iii) Martians have red tongues. Gulag is a Martian. Based on these two statements, what can you conclude about Gulag?
Solution:
Gulag had red, tongue.

iv) What is the fallacy in the Raju’s reasoning in the cartoon below
AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2 1
Solution:
All smarts need not be a President. There could be some other persons who are smart too.

Question 2.
Once again you are given four cards. Each card has a number printed on one side and a letter on the other side. Which are the only two cards you need to turn .over to check whether the following rule holds ?
“If a card has a consonant on one side, then it has an odd number on the other side”.
AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2 2
Solution:
You need to turn over B and 8 only. If B has an even number then the rule has broken. Similarly if 8 has a consonant on the other side then also the rule has been broken.

Question 3.
Think of this puzzle. What do you need to find a chosen number from this square ? Four of the clues below are true but do nothing to help in finding the number. Four of the clues are necessary for finding it.
Here are eight clues to use:
a) The number is greater than 9.
b) The number is not a multiple of 10.
c) The number is a multiple of 7.
d) The number is odd.
e) The number is not a multiple of 11.
f) The number is less than 200.
g) Its ones digit is larger than its tens digit.
h) Its tens digit is odd.
What is the number ?
AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2 3
Solution:

Hint                 Conclusion
a The number may be from 10 to 99
(2nd clue implies, first clue is of no use) No use
b The number is not any of (10, 20, 30, …………90)
c The number may be any of (7, 14, 21,28, 35,42, …98)
d The number may be any of (7,21,35,49, 63, 77,91)
e The number may be any of (7, 21, 35, 49. 63. 91)
f No use
g The numbr may be 35, 49
h 35

 

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.1

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1

Question 1.
State whether the following sentences are always true, always false or ambiguous. Justify your answer.
i) There are 27 days in a month.
Solution:
Always false. Generally 30 days or 31 days make a month except February.

ii) Makarasankranthi fells on a Friday.
Solution:
Ambiguous. Makarasankranthi may fall on any day of the week.

iii) The temperature in Hyderabad is 2°C.
Solution:
Ambiguous. Sometimes the temperature may go down to 2°C in winter.

iv) The earth is the only planet where life exist.
Solution:
We can’t say always true. To the known fact, so far we can say this.

v) Dogs can fly.
Solution:
Always false,’as dogs can never fly.

vi) February has only 28 days.
Solution:
Ambiguous. A leap year has 29 days for February.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1

Question 2.
State whether the following statements are true or false. Give reasons for your answers.

i) The sum of the interior angles of a quadrilateral is 350°.
Solution:
False. Sum of the interior angles of a quadrilateral is 360°.

ii) For any real number x, x2 > 0
Solution:
True. This is true for all real numbers,

iii) A rhombus is a parallelogram.
Solution:
True. In a rhombus, both pairs of opposite sides are parallel and hence every rhombus is a parallelogram.

iv) The spm of two even numbers is even.
Solution:
True. This is true for any two even numbers.

v) Square numbers can be written as the sum of two odd numbers.
Solution:
Ambiguous. Since square of an odd number can’t be written as sum of two odd numbers.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1

Question 3.
Restate the following statements with appropriate conditions, so that they become true statements.

i) All numbers can be represented in prime factorization.
Solution:
Any natural number greater than 1 can be represented in prime factorization.

ii) Two times a real number is always even.
Solution:
Two times a natural number is always even.

iii) For any x, 3x + 1 > 4.
Solution:
For any x > 1; 3x + 1 > 4.

iv) For any x, x3 ≥ 0.
Solution:
For any x > 0; x3 ≥ 0.

v) In every triangle, a median is also an angle bisector.
Solution:
In an equilateral triangle, a median is also an angle bisector.

AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1

Question 4.
Disprove, by finding a suitable counter example, the statement
x2 > y2 for all x > y.
Solution:
If x = – 8 and y = – 10
Here x > y
x2 = (- 8)2 = 64 and y2 = (- 10)2 = 100
But x2 > y2 is false here. [ ∵ 64 < 100]
(This can be proved for any set of nega-tive numbers or a negative number and a positive number)

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 14th Lesson Probability Exercise 14.1

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

Question 1.
A die has six faces numbered from 1 to 6. It is rolled and number on the top face is noted. When this is treated as a random trial.
(Random trial: All possible outcomes are known before hand and the exact outcome can’t be predicted, then only the experiment is treated as a random experiment or a trial.)

a) What are the possible outcomes ?
Solution:
The possible outcomes are 1, 2, 3, 4, 5 and 6.

b) Are they equally likely ? Why ?
Solution:
Yes. All the outcomes are equally likely since every event has equal chance of occurrence or no event has priority to occur.

c) Find the probability of a composite number turning up on the top face.
Solution:
Even : Turning up of a composite number
Possible outcomes = 4, 6
No. of possible outcomes = 2
Total outcomes = 1, 2, 3, 4, 5 and 6
Number of total outcomes = 6
Probability = \( \frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }} \\\)
= \(\frac{2}{6}=\frac{1}{3}\)

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

Question 2.
A coin is tossed 100 times and the following outcomes are recorded. Head : 45 times; Tails : 55 times from the experiment.
a) Compute the probability of each outcome.
Solution:
Head = 45 times; Tails = 55 times
Total = 100
P(H),
Probability of getting Head = \(\frac{45}{100}\)
P(T),
Probability of getting Tail = \(\frac{55}{100}\)
\(\left[P=\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\right]\)

b) Find the sum of the probabilities of all outcomes.
Solution:
P(H) + P(T) = \(\frac{45}{100}+\frac{55}{100}=\frac{100}{100}\) = 1

Question 3.
A spinner has four colours as shown in the figure. When we spin it once, find
AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 1

a) At which colour, is the pointer more likely to stop ?
Solution:
Red = 5 sectors; Blue = 3 sectors;
Green = 3 sectors; Yellow = 1 sector
Total = 5 + 3 + 3+1 = 12 sectors
∴ Pointer is more likely to stop at Red.

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

b) At which colour, is the pointer less likely to stop ?
Solution:
Yellow; as only one sector is shaded in Yellow.

c) At which colours, is the pointer equally likely to stop ?
Solution:
Blue and Green have equal chances; as they are shaded in equal number of sectors.

d) What is the chance the pointer will stop on white ?
Solution:
No chance. Since no sector is shaded in white.

e) Is there any colour at which the pointer certainly stops ?
Solution:
No; as the experiment is a random ex-periment.

Question 4.
A bag contains five green marbles, three blue marbles, two red marbles and two yellow marbles. One marble is drawn out randomly. M)
a) Are the four different colour outcomes equally likely ? Explain.
Solution:
No. As they are not in equal number, they have different chances of occurrence.

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

b) Find the probability of drawing each colour marble, i.e., P (green), P (blue), P(red) and P (yellow).
Solution:
AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 2

c) Find the sum of their probabilities.
Solution:
P(Green) + P(Blue) + P(Red) +P(Yellow)
= \(\frac{5}{12}+\frac{3}{12}+\frac{2}{12}+\frac{2}{12}\)
= \(\frac{5+3+2+2}{12}=\frac{12}{12}\) = 1

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

Question 5.
A letter is chosen from English alphabet. Find the probability of the letters being
a) A vowel
b) A letter comes after P
c) A vowel or a consonant
d) Not a vowel
Solution:
Total letters ;= 26 [A, B, C ….. Z]
Probability = \(\frac{\text { No. of favourables }}{\text { Total no. of outcomes }}\)
a) Vowels = a, e, i, o, u [5]
∴ (vowels) = \(\frac{5}{26}\)

b) Letter after P = 10
[Q, R, S, T, U, V, W, X, Y, Z]
Probability of a letter that comes after
’P’ = \(\frac{10}{26}\) = \(\frac{5}{13}\)

c) A vowel or a consonant
Vowel or consonants = 26
[all letters from A to Z]
P(vowel or consonant) = \(\frac{26}{26}\) = 1

d) Not a vowel = 21
[other than A, E, I, O, U]
∴ P(not a vowel) = \(\frac{21}{26}\)

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

Question 6.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98,5.04,5.07,5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Total number of bags = 11
No. of bags with a weight more than 5 kg = 7
[5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07]
∴ Probability = \(\frac{\text { favourable outcomes }}{\text { total outcomes }}\)
P(E) = \(\frac{7}{11}\)

Question 7.
An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained is given in the following table.
AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 3
Find the probabilities of the following events for a driver chosen at random from the city:

i) The driver being in the age group 18-29 years and having exactly 3 accidents in one year.
Solution:
Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 +
360 + 45 + 35 + 15 + 9) = 2000
Event: The driver being in the age group (18 – 29) years and having exactly 3 accidents
= 61
P(E) = \(\frac{\text { No. of favourables }}{\text { Total outcomes }}=\frac{61}{2000}\)

ii) The driver being in the age group of 30 – 50 years and having one or more accidents in a year.
Solution:
Favourable outcomes = 125 + 60 + 22 + 18 = 225
Total outcomes = 2000
P(E) = \(\frac{225}{2000}=\frac{9}{80}\)

AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1

iii) Having no accidents in the year.
Solution:
Favourable outcomes = 440 + 505 + 360 = 1305
Total outcomes = 2000
P(E) = \(\frac{1305}{2000}=\frac{261}{400}\)

Question 8.
What is the probability that a randomly thrown dart hits the square board in shaded region ( Take π = \(\frac{22}{7}\) and express in percentage )
Solution:
AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 4
Radius of the circle r = 2 cm
Area of the circle A = πr2 = \(\frac{22}{7}\) × 2 × 2 = \(\frac{88}{7}\) cm2
Side of the square = 2 × radius
= 2 × 2 = 4cm
Area of the square = S2 = 4 × 4 = 16 cm2
Area of the shaded region = Area of the square – Area of the circle
AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 5

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions Exercise 13.2

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

Question 1.
Construct AABC in which BC = 7 cm,∠B = 75° and AB + AC =12 cm.
Solution:
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 1
Steps:

  • Draw a line segment BC = 7 cm.
  • Erect ∠B = 75°
  • MarkapointDon \(\overrightarrow{\mathrm{BX}}\) suchthat BD = AB + AC.
  • Join D, C and draw the perpendicular bisector of \(\overline{\mathrm{CD}}\) meeting BD at A.
  • Join A to C to form the ΔABC.

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

Question 2.
Construct ΔPQR in which QR = 8 cm, ∠B = 60° and AB – AC = 3.5 cm.
Read ∠Q = 60°and PQ – PR = 3.5 cm
Solution:
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 2
Steps: I

  • Draw QR = 8 cm.
  • Construct ∠RQX = 300 at Q.
  • Mark a point S on \(\overrightarrow{\mathrm{QX}}\) such that QS = PQ  –  PR = 3.5 cm.
  • Join S, R and draw the perpendicular bisector to \(\overline{\mathrm{QR}}\) meeting \(\overrightarrow{\mathrm{QX}}\) at P.
  • Join P, R to form the ΔPQR.

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

Question 3.
Construct ΔXYZ in which ∠Y = 30 °; ∠Z = 60 ° and XY + YZ + ZX = 10 cm.
Solution:
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 3
Steps:

  • Draw a line segment AB = XY + YZ + ZX = 10 cm.
  • Construct ∠BAP = \(\frac { 1 }{ 2 }\) ∠Y at A and ∠ABQ = \(\frac { 1 }{ 2 }\) ∠Z at B meeting at X.
  • Draw the perpendicular bisectors to XA and XB meeting \(\overline{\mathrm{AB}}\) at Y and Z respectively.
  • Join X to Y and Z to form the ΔXYZ.

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

Question 4.
Construct a right triangle whose base is 7.5 cm and sum of its hypotenuse and otherside is 15 cm.
Solution:
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 4
Steps:

  • Draw BC = 7.5 cm.
  • Construct ∠CBX = 90°
  • Mark a point D on \(\overrightarrow{\mathrm{BX}}\) such that BD = 15 cm.
  • Join C, D. ‘
  • Draw the perpendicular bisectors of \(\overline{\mathrm{CD}}\) meeting BD at A.
  • Join A, C to form the ΔABC.

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

Question 5.
Construct a segment of a circle on a chord of length 5 cm containing the following angles i) 90° ii) 45° iii) 120°
Solution:
i) 90°
A.
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 5
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 6

Steps:

  • Draw a rough sketch of ∠BAC = 90° and ∠BOC = 180°.
  • Draw a line segment BC = 5 cm.
  • Draw the perpendicular bisector of BC meeting \(\overline{\mathrm{BC}}\) at O
  • Draw an arc of radius OB or OC with centre O.
  • Mark any point A on the arc and join it with B and C.
  • ∠BAC = 90°

ii) 45°
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 7
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 8

Steps:

  • Draw a line segment BC = 5 cm.
  • Construct ABOC such that BC = 5 cm, ∠B = 45° = ∠C.
  • Draw a circle segment of radius OB or OC with centre ’O’.
  • Mark any point A on the segment and join it with B and C.
  • ∠BAC = 45°

AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2

iii) 120°
AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 9

Steps:

  • Draw a line segment AB = 5 cm. ,
  • Construct ΔAOB in which ∠A = 30°; ∠B = 30°; AB = 5 cm.
  • With ‘O’ as centre draw a circle segment.
  • On the opposite side make any point C and join it with B and C.
  • ∠ACB = 120°