Bhagya

Punjab State Board PSEB 8th Class English Book Solutions English Reading Comprehension Picture / Poster Based Exercise Questions and Answers, Notes.

PSEB 8th Class English Reading Comprehension Picture / Poster Based

Look at the pictures carefully and answer the questions that follow:

Question 1.
What is the purpose of this advertisement ?
(a) to prevent people from using motor vehicle.
(b) to spread awareness about traffic rules.
(c) to stop people from walking on the road.
(d) to secure people of road ancient.
Answer:
(b) to spread awareness about traffic rules.

Question 2.
While on scooter or bike, which thing can help to save our lives:
(a) scarf
(b) cap
(c) helmet
(d) seat belt.
Answer:
(c) helmet

Question 3.
Zebra crossing is meant for:
(a) four wheelers
(b) bikers
(c) cyclists
(d) pedestrians
Answer:
(d) pedestrians

Question 4.
One should stop the vehicle when it is a:
(a) red light
(b) yellow light
(c) green light
(d) none of these
Answer:
(a) red light

Question 5.
Road accidents can be prevented by:
(a) driving within a speed limit
(b) not driving while drinking
(c) obeying the traffic rules
(d) all of the above.
Answer:
(d) all of the above.

Working Together to Keep Our Children Safe.

Question 1.
What is the purpose of this advertisement ?
(a) To make children happy
(b) Teaching children how to drive a bike or a car
(c) Taking chidren to the park
(d) Promoting road safety awareness among children
Answer:
(d) Promoting road safety awareness.

Question 2.
Children should be aware of:
(a) speed limit while driving
(b) road safety rules
(c) parking their vehicles at a safe place
(d) all these.
Answer:
(d) all these.

Question 3.
Parking of vechiles on the roadside can result in:
(a) an accident
(b) theft of the vehicle
(c) traffic jam
(d) all the above.
Answer:
(d) all the above.

Question 4.
For safe driving the driver should have the knowledge of:
(a) signboards on the roadside
(b) his R.C.
(c) the vehicles coming behind him
(d) the condition of his vehicle.
Answer:
(a) signboards on the roadside.

Question 5.
While driving we should:
(a) not drink
(b) not use our mobile
(c) not cross green light
(d) not drive below speed limit.
Answer:
(a) and (b)

Question 1.
What is the theme of the picture ?
(a) The Values and Advantages of Games and Sports
(b) Good Manners
(c) The Value of Reading Books
(d) The hazards of Pollution.
Answer:
(b) Good Manners.

Question 2.
Which of the following is not a good habit ?
(a) helping old people
(b) planting trees
(c) getting up early in the morning
(d) keeping your classroom dirty.
Answer:
(d) keeping your classroom dirty.

Question 3.
Which kind of words ‘please’ and thankyou’ are ?
(a) bad words
(b) harsh words
(c) polite words
(d) difficult words.
Answer:
(c) polite words.

Question 4.
We should wait for our turn by standing in the line.
(a) quietly
(b) uneasily
(c) impatiently
(d) angrily.
Answer:
(a) quietly.

Question 5.
‘Early to bed, early to rise makes a man healthy, and wise.’
(a) dull
(b) poor
(c) wealthy
(d) foolish.
Answer:
(c) wealthy.

Question 1.
The poster tells us that:
(a) India is a land of festivals.
(.b) we celebrate many festivals in India.
(c) festivals of all religions are celebrated in India.
(d) all these.
Answer:
(d) all these.

Question 2.
What is the importance of festivals in our life ?
(a) They give us new energy.
(b) They keep our culture alive.
(c) They entertain us.
(d) All these.
Ans, (d) All these.

Question 3.
Pushkar fair is celebrated:
(a) all over India
(b) in Rajasthan.
(c) in South India
(d) None of these
Answer:

Question 4.
Holi is a festival of:
(a) lights
(b) colours.
(c) praying in mosques
(d) cleaning our houses and shops
Answer:
(b) colours.

Question 5.
Which of the following festivals, in particular, would promote Hindu Muslim unity ?
(a) Diwali and Christmas
(b) Eid and christmas
(c) Pushkar Fair and Christmas
(d) Diwali and Eid.
Answer:
(d) Diwali and Eid.

Question 1.
What is the purpose of this poster/advertisement about ?
(a) Women Backwardness
(b) Women Education
(c) Women Empowerment
(d) Sources of Entertainment for Women.
Answer:
(c) Women Empowerment.

Question 2.
Daughter’s Day gives the message of:
(a) loving daughters only
(b) having daughters only
(c) Beti Bachao Beti Padhao
(d) marry your daughters in their chile
Answer:
(c) Beti Bachao Beti Padhao

Question 3.
Women’s Day is observed on:
(a) 5th September
(b) First sunday of May
(c) 15th September
(d) 8th March.
Answer:
(d) 8th March.

Question 4.
Women feel empowered when they :
(a) use their power to empower others
(b) use their power to belittle others
(c) win elections to rule the country
(d) all these.
Answer:
(a) use their power to empower others

Question 5.
Mother’s Day is celebrated to:
(a) inspire women to become mother soon after their marriage
(b) to honour mothers of the world
(c) to teach uneducated mothers
(d) none of these.
Answer:
(b) to honour mothers of the world

Polio Drops and Healthy Life

Question 1.
The most suitable title for this advertisement is:
(a) Healthy Life
(b) Medication Vs Yoga
(c) Old Age and Yoga
(d) Eating is Better than Yoga.
Answer:
(a) Heatlhy Life.

Question 2.
We should avoid eating:
(a) fruits and vegetables
(b) balanced food
(c) junk food
(d) cooked food.
Answer:
(c) junk food.

Question 3.
Yoga is kind of:
(a) exercise to please Swami Ramdev
(b) diet to grow tall
(c) excercise to keep us fit and healthy
(d) prayer to please god.
Answer:
(c) exercise to keep us fit and healthy.

Question 4.
Which of the following activity is included in a trip to healthy life ?
(a) walking and laughing loudly
(b) crying and yelling
(c) eating food three times a day
(d) taking medicine now and then.
Answer:
(a) walking and laughing loudly.

Question 5.
Polio drops are given to the children of:
(a) two years
(b) Three years
(c) four years
(d) five years.
Answer:
(d) five years.

Question 1.
The best title for this poster is:
(a) Growing and cutting down the trees
(b) Resting and playing under trees
(c) Planting trees in rainy season
(d) Benefits of growing and protecting trees.
Answer:
(d) Benefits of growing and protecting trees.

Question 2.
Trees give us:
(a) fruits
(b) medicines
(c) firewood
(d) all the above.
Answer:
(d) all the above.

Question 3.
Trees serve us by:
(a) giving out oxygen
(b) taking in carbon dioxide
(c) giving us cool shade in summer
(d) all the above.
Answer:
(d) all the above.

Question 4.
Without trees climate would be:
(a) drier and cooler
(b) drier and hotter.
(c) warmer and cooler
(d) drier and hotter wetter and hotter.
Answer:
(b) drier and hotter.

Question 5.
What is our duty towards trees ?
(a) growing more trees and taking proper care of them
(b) cutting down trees only in winter
(c) planting only fruit trees
(d) not to let birds sit in trees.
Answer:
(a) growing more trees and taking proper care of them.

Effects of Noise Pollution

Question 1.
What is purpose of this poster ?
(a) to create awareness against noise pollution.
(b) to use loudspeakers to check noise pollution.
(c) to put hands on ears on hearing a noise
(d) to prevent people from making noise during the day.
Answer:
(a) to create awareness against noise pollution.

Question 2.
Which of the following activity is responsible for noise pollution ?
(a) high volume of loudspeakers
(b) running factories
(c) vehicles running fast on roads
(d) all these.
Answer:
(d) all these.

Question 3.
Too much noise may:
(a) make us deaf
(b) increase the speed of our vehicles
(c) incresae our hearing power
(d) increase our energy to work.
Answer:
(a) make us deaf.

Question 4.
We should not blow horns or ring bells near a hospital because—
(a) it may spoil the medicines
(b) it may disturb the resting patients
(c) the doctors may go on strike
(d) none of these.
Answer:
(b) it may disturb the resting patients.

Question 5.
To avoid noise pollution we should
(a) not blow horns unnecessarily
(b) avoid the use of loudspeakers
(c) not use old vehicles that produce screeching sound.
(d) all these.
Answer:
(d) all these.

Punjab State Board PSEB 8th Class English Book Solutions English Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 8th Class English Reading Comprehension Unseen Passages

I. Read the given passages and answer the questions that follow:

(1) Trees are as beautiful as they are useful. Wherever they are, they make that place look nice and green. They give us fruits, shade and wood. Birds build nests in their branches. Trees make the whole place like a garden. They are indeed nature’s precious gift to us.

Every tree is a living and breathing creature, like us. But unlike us, it prepares its own food from raw materials such as carbon dioxide, water and sunlight. Also, unlike us, it lacks a well-developed nervous system although it responds to many external stimuli. The, tree breathes through its leaves.

How does the environment affect the growth of the tree ? If there is a lack of water, the roots go down deeper and spread out far and wide, backward and forward, in search of food material. If there are too many trees in one place, they grow higher and higher to reach the sunshine. If there is a strong wind all the time, the tree takes firmer hold of the ground with its roots.

The tree is a strong fighter. It may bend before the wind but it does not always break. It protects itself very well against snow, frost and hail. It can defeat most of its enemies. But human beings defeat the tree every time by cutting it down. Litde do they know that by destroying trees at such a large scale, they are actually destroying themselves.

Question 1.
Trees are natures precious gift to us because:
(a) they provide us food, shade and wood
(b) they provide shelter to the birds
(c) they turn the earth into a beautiful place
(d) all of the above.
Answer:
(d) all of the above.

Question 2.
When there is lack of water, the tree:
(a) grow taller in order to get rain
(b) takes firm hold on the ground
(c) starts breathing through its leaves
(d) sends its roots deep, far and wide.
Answer:
(d) sends its roots deep, far and wide.

Question 3.
Which of the following statement is true for both humans and trees ?
(a) both breathe and grow
(b) both can move and run
(c) both buy their own food
(d) both have a nervous system.
Answer:
(a) both breathe and grow.

Question 4.
Trees fight many enemies but they are not able to defeat :
(a) snow
(b) wind
(c) water scarcity
(d) human beings
Answer:
(d) human beings.

Question 5.
Trees are strong fighters because :
(a) they can adapt themselves to all circumstances
(b) they can kill other trees for their growth
(c) they can defeat all their enemies
(d) they have strong roots and trunk.
Answer:
(a) they can adapt themselves to all circumstances.

(2) Schools all over India celebrate Childrens Day’ on 14th November every year. On this day, our great Prime Minister who had a great love for children was born. His ancestors came down from Kashmir to the rich plains below. Kaul had been his family name; this changed to Kaul-Nehru: and in later years. Kaul was dropped and they became simply Nehrus. Jawahar Lai Nehru was the only son of his prosperous parents. His two sisters were much younger to Jawahar Lai Nehru, And so, he grew up and spent his early years as a lonely child with no companion of his own age. Private tutors were in charge of his education. Then, he went to England and was educated at Harrow and at Trinity College, Cambridge.

Question 1.
Childrens Day is celebrated on:
(a) 15th August
(b) 26th January
(c) 14th November
(d) 30th January.
Answer:
(c) 14th November.

Question 2.
Nehrus ancestors came from:
(a) Delhi
(b) Allahabad
(c) Kashmir
(d) Raibareli.
Answer:
(c) Kashmir.

Question 3.
Jawahar Lai Nehru was educated at:
(a) Wilson College, Mumbai
(b) Harrow and Trinity College, Cambridge
(c) Presidency University, Kalkata
(d) Jesus and Mary College, Delhi.
Answer:
(b) Harrow and Trinity College, Cambridge.

Question 4.
Why is 14th November celebrated as Childrens Day ?
Or
What is the importance of 14th November ?
(a) Pt. Nehru was born on this day.
(b) Mahatma Gandhi was born on this day.
(c) Indira Gandhi was born on this day.
(d) None of the above.
Answer:
(a) Pt. Nehru was born on this day.

Question 5.
Nehru ji belonged to:
(a) a poor family
(b) a family of farmers
(c) a rich / prosperous family
(d) none of these.
Answer:
(c) a rich / prosperous family.

(3) Once a bee felt thirsty. It flew to a pond to drink water. While drinking water, the bee fell into the pond. A dove was sitting on the branch of a tree. It saw all and decided to save the bee’s life. The dove threw a leaf. The bee climbed over the leaf, dried its wings and flew away.

After a few days a hunter came to the forest. He aimed at the dove. Luckily the bee saw the hunter. It flew to the hunter and stung him hard on the hand. The hunter missed his aim. The dove heard the gunshot and flew away. The dove thanked the bee for this timely help.

Question 1.
Where did the bee fly to drink water ?
(a) a canal
(b) a pond
(c) a river
(d) a stream.
Answer:
(b) a pond

Question 2.
What happened to the bee while drinking water ?
(a) It fell from the tree
(b) It fell into the pond
(c) It was shot by the hunter
(d) None of the above.
Answer:
(b) It fell into the pond.

Question 3.
Who saved the bee’s life?
(a) a dove
(b) a hunter
(c) a fish
(d) a tortoise.
Answer:
(a) a dove.

Question 4.
What did the bee do to save the dove’s life?
(a) It killed the hunter
(b) It stung the hunter on the hand
(c) It shouted hard
(d) It did nothing.
Answer:
(b) It stung the hunter on the hand.

Question 5.
What is the moral of the story ?
(a) Do good, have good
(b) Revenge is the best policy
(c) Pride hath a fall
(d) Union is strength.
Answer:
(a) Do good, have good.

(4) Garbage is a great environmental hazard. It comes from various sources-used paper, tiffin packings, plastic bags, ice-cream wrappers, bottle caps, fallen leaves from trees and many more. Garbage makes the premises ugly, unkempt and breeds diseases.

A lot of trash that is thrown away contains material that can be recycled and reused such as paper, metals and glass which can be sent to the nearest recycling centre or disposed of to the junkdealer. It also contains organic matter such as leaves which can enrich land fertility.

A compost pit can be made at a convenient location where the refuse can be placed with layers of soil and and occasional sprinkling of water. This would help decomposition to make valuable manure (fertilizer). This would also prevent pollution that is usually caused by burning such organic waste.

Question 1.
Garbage is a great environmental hazard because it makes the premises:
(a) ugly
(b) unkempt
(c) breed diseases
(d) all these.
Answer:
(d) all these.

Question 2.
What happens to the disposed material at the recycling centre ?
(a) sent back to homes
(b) takes a new shape.
(c) thrown into rivers.
(d) all the above.
Answer:
(b) takes a new shape.

Question 3.
How can we make use of waste organic matter ?
(a) send it to junkdealer
(b) burn it
(c) change it into valuable manure (fertilizer)
(d) all these
Answer:
(c) change it into valuable manure (fertilizer.)

Question 4.
Proper disposal of garbage
(a) spreads pollution
(b) prevents pollution
(c) spoils mineral wealth
(d) none of these
Answer:
(b) prevents pollution.

Question 5.
The organic waste that can be recycled and reused is
(a) paper
(b) glass
(c) metals
(d) all these.
Answer:
(d) all these.

(5) Yoga is the ancient Indian system to keep a person fit in body and mind. It is basically a system of self-treatment. According to the yogic view, diseases, disorders and ailments are the result of some faulty ways of living, bad habits, lack of proper knowledge and unsuitable food. The diseases are thus the resultant state of a short or prolonged malfunctioning of the body system. The root cause of a disease lies in not correcting the mistakes by the same individual. The yogic practice of treatment comprises three steps, namely proper diet, proper yogic practice and proper knowledge of things concerning the self.

Question 1.
The benefit of the system of yoga is:
(a) It keeps a person fit in body and mind.
(b) It is a modern Indian system.
(c) It makes a person religious.
(d) Comprises three steps.
Answer:
(a) It keeps a person fit in body and mind.

Question 2.
What type of system is this basically?
(a) It is a costly treatment
(b) It is a self-treatment.
(c) It avoids bad habits.
(d) All of the above.
Answer:
(b) It is a self-treatment.
Or
Diseases, disorders and ailments are the results of
(a) Some faulty ways of living.
(b) Some normal ways of living
(c) Some cosdy ways of living.
(d) All of the above.
Answer:
(a) Some faulty ways of living.

Question 3.
What is the root cause of diseases ?
(a) Mistakes of the doctors.
(b) Mistakes of the parents.
(c) Mistakes of the governments.
(d) Mistakes of the individual.
Answer:
(d) Mistakes of the individual.

Question 4.
How many steps does yoga practice keep?
(a) Only one step.
(b) Only three steps.
(c) Only two steps.
(d) Only four steps.
Answer:
(b) Only three steps.
Or
Which is the first step of yoga?
(a) Proper yogic practice.
(b) Proper knowledge of things.
(c) Proper diet.
(d) Proper exercise of body.
Answer:
(c) Proper diet.

Question 5.
Which is the third step of yoga?
(a) Proper knowledge.
(b) Proper counselling.
(c) Proper thinking.
(d) None of the above.
Answer:
(a) Proper knowledge.
Or
Whose efforts cure the person ?
(a) The doctor’s efforts.
(b) The yoga experts efforts.
(c) The efforts of the society.
(d) The patients efforts.
Answer:
(d) The patients efforts.

(6) There is an incident which occurred at the examination during my first year at the high school. Mr. Giles, the Education Inspector, had come on a visit of inspection. He had set us five words, to write as a spelling exercise. One of the words was ‘ketde’. I had mis-spelt it. The teacher tried to prompt me with the point of his boot, but I would not be prompted. It was beyond me to see that he wanted me to copy the spelling from my neighbour’s slate for I had thought that the teacher was there to supervise us against copying.

Question 1.
When did the incident occur ?
(a) In the second year.
(b) In the first year.
(c) In the third year.
(d) None of the above.
Answer:
(b) In the first year.
Or
Who was the Education Inspector?
(a) Mr. Gordon
(b) Mr. Graham
(c) Mr. Giles
(d) Mr. George.
Answer:
(c) Mr. Giles

Question 2.
Which exercise was given to write?
(a) Dictation exercise.
(b) Handwriting exercise.
(c) Yoga exercise.
(d) Spelling exercise.
Answer:
(d) Spelling exercise
Or
How many words were given to us ?
(a) Three words.
(b) Five words.
(c) No word was given.
(d) Four words.
Answer:
(b) Five words.

Question 3.
Who tried to prompt the speaker/writer ?
(a) The teacher.
(b) The Inspector.
(c) The students.
(d) All of the above.
Answer:
(a) The teacher.
Or
What mistake had the writer committed ?
(a) A word mistake.
(b) A meaning mistake.
(c) A spelling mistake.
(d) A speaking mistake.
Answer:
(c) A spelling mistake.

Question 4.
What was the point of indication used by the teacher ?
(a) The point of his hand finger.
(b) The point of his right foot finger.
(c) The point of his left hand thumb.
(d) The point of his boot.
Answer:
(d) The point of his boot.

Question 5.
Which word was mis-spelt ?
(a) Kettle.
(b) Catde.
(c) Settle.
(d) Metal.
Answer:
(a) Kettle.

(7) People often curse poverty as a great evil, and it seems to be an accepted belief that if people only had plenty of money, they would be happy and useful and get more out of life. But the reality is that while palaces give a comfortable life, peace and contentment dwell in cottages. I always pity the sons and daughters of rich parents who are attended by servants and governesses. It is because I know how sweet and happy and pure the home of honest poverty is and how loving and united the members of poor families are in common interests. It is for these reasons that so many strong, eminent and self-reliant men have always sprung from poor families.

Question 1.
What do the people often think about the poverty?
(a) It is a curse and great evil.
(b) It is a boon of God.
(c) It is a self-created act.
(d) It is a social evil.
Answer:
(a) It is a curse and great evil.
Or
Who are attended by servants and governesses?
(a) The kings of the world.
(b) The members of the poor families.
(c) The sons and daughters of rich parents.
(d) None of the above.
Answer:
(b) The sons and daughters of rich parents.

Question 2.
Who are happy according to accepted belief ?
(a) People who have no money.
(b) People who have plenty of money.
(c) People who have a higher education.
(d) People who have no higher education.
Answer:
(a) People who have no money.
Or
What is the reality of happy life.
(a) To live in luxurious palaces.
(b) To live in the forests.
(c) To live peaceful and contented life in a hut.
(d) To live in the king’s palaces.
Answer:
(c) To live peaceful and contented life in a hut.

Question 3.
What does the home of poverty provide us?
(a) A life of prosperity.
(b) A sweet, happy and pure home.
(c) A dirty, bad and disturbed life.
(d) A life of dissatisfaction.
Answer:
(b) A sweet, happy and pure home.

Question 4.
Whose members are loving and united ?
(a) Members of poor families.
(b) Members of rich families.
(c) Members of tribal families.
(d) Members of royal families.
Answer:
(a) Members of poor families.
Or
Who have sprung from poor families ?
(a) Weak, cowardly and religious persons.
(b) Educated, rich and royal persons.
(c) Prosperous honoured and noble persons.
(d) Strong, eminent and self-reliant persons.
Answer:
(d) Strong, eminent and self-reliant persons.

Question 5.
Whose plus points are highlighted in the passage ?
(a) The rich.
(b) The noble,
(c) The honoured.
(d) The poor.
Answer:
(d) The poor.

(8) Books have much value in our life. They are our lifeline and best companion. Everything comes to an end but they live for ever. They never deceive the readers. They help us in difficulties. We get much knowledge and entertainment from them. We get new meanings and beauties in books. By reading books, we just get confidence in life. This world would be quite dark without books. They tell us about people, their culture and profession.

Question 1.
Books have:
(a) no value.
(b) less value.
(c) much value.
(d) different value.
Answer:
(c) much value.
Or
Books are our:
(a) lifeline.
(b) best companion.
(c) companion and enemy
(d) both (a) and (b).
Answer:
(d) both (a) and
(b) lifeline and best companion.

Question 2.
A quality of books is:
(a) They come to an end.
(b) They don’t live for ever.
(c) They live for ever.
(d) All the above.
Answer:
(c) They live for ever.
Or
Books help us:
(a) in trouble.
(b) in difficulties.
(c) Both (a) and (b).
(d) in sorrows.
Answer:
(b) in difficulties.

Question 3.
We get from books:
(a) waste paper
(b) much knowledge.
(c) entertainment.
(d) Both (b) and (c).
Answer:
(d) both (b) and (c) much knowledge and entertainment.
Or
By reading books we get:
(a) confidence in life
(b) difficulties in life
(c) popularity and prosperity
(d) All the above.
Answer:
(a) confidence in life.

Question 4.
Without books the world would become:
(a) quite happy.
(b) quite bright.
(c) quite dark.
(d) Both (a) and (b).
Answer:
(c) quite dark.

Question 5.
Books tell us about:
(a) people.
(b) their culture.
(c) their profession.
(d) All the above.
Answer:
(d) All the above.

(9) Diwali is the greatest festival of Hindus. It is celebrated throughout the world. It is a festival of lights and candles. It comes in the month of October or November every year. On this day people worship Goddess Lakshmi. They put on new clothes and buy sweets. They also give presents to their friends and relatives. Some people gamble on this day which is an evil practice. Children play crackers and fireworks.

Question 1.
Diwali is the greatest festival of
(a) Muslims
(b) the poor
(c) the rich
(d) Hindus
Answer:
(d) Hindus

Question 2.
Diwali is celebrated:
(a) only in India.
(b) throughout Asia
(c) throughout the world.
(d) None of these.
Answer:
(c) throughout the world.
Or
Diwali is a festival of:
(a) lights.
(b) Candles
(c) swings
(d) both (a) and (b)
Answer:
(d) both (a) and (b) lights and candles.

Question 3.
Diwali fells in the month of:
(a) October.
(b) October or November.
(c) December
(d) None of these.
Answer:
(b) October or November.
Or
On this day people worship:
(a) Goddess Kali.
(b) Goddess Lakshmi.
(c) Both (a) and (b).
(d) None of these.
Answer:
(b) Goddess Lakshmi.

Question 4.
On this day people:
(a) buy houses
(b) buy sweets.
(c) give up bad habbits
(d) All the above.
Answer:
(b) buy sweets.
Or
An evil practice related to Diwali is:
(a) drinking
(b) fighting
(c) smoking
(d) gambling.
Answer:
(d) gambling

Question 5.
Who plays cracker and fire works on this day ?
(a) rich people
(b) poor people
(c) Children
(d) Goddess Laxmi
Answer:
(c) Children

(10) Mohan and Sohan were fast friends. Mohan was very selfish and cunning while Sohan was very loyal and dependable. One day they set out on a long journey. Both decided to help each other. While crossing the forest, they saw a bear coming towards them. Mohan at once climbed up a tree. But Sohan did not know how to climb up. He lay on the ground and held his breath. The bear came and took Sohan as dead. After the bear had gone, Mohan came down the tree and asked Sohan what the bear had said in his ear. Sohan replied that the bear had told him never to trust a false friend.

Question 1.
Mohan was:
(a) loyal.
(b) very helpful.
(c) very selfish and cunning.
(d) All the above.
Answer:
(c) very selfish and cunning.

Question 2.
Sohan was very:
(a) selfish.
(b) cunning.
(c) loyal and dependable.
(d) a false friend
Answer:
(c) loyal and dependable.

Question 3.
One day Mohan and Sohan set out:
(a) on a tour.
(b) on a long journey.
(c) on an expedition.
(d) on a short trip.
Answer:
(b) on a long journey.

Question 4.
What did they see in the forest?
(a) An elephant.
(b) A lion.
(c) A tiger.
(d) A bear.
Answer:
(d) A bear.
Or
What did Sohan do?
(a) He ran away.
(b) He lay on the ground.
(c) He held his breath.
(d) Both (b) and (c).
Answer:
(d) both (b) and (c) He lay on the ground and held his breath.

Question 5.
What did the bear do ?
(a) It attacked Sohan.
(b) It killed Sohan.
(c) It took Sohan as dead.
(d) It took Mohan as dead
Answer:
(c) It took Sohan as dead.
Or
The bear had told Sohan not to
(a) trust anybody.
(b) climb up a tree.
(c) go on a journey.
(d) trust a false friend.
Answer:
(d) trust a false friend.

(11) His first ‘Satyagraha in India was in Champaran, in Bihar. The peasants of that district were being cruelly treated by the British indigo planters. Gandhiji left for Champaran to find out the truth. The news that a Mahatma had arrived to inquire into their sufferings attracted thousands of peasants who flocked to Champaran to have has darshan. The Government got alarmed and Gandhiji was asked to leave the district. He refused and was asked to appear before magistrate. Later, the case was withdrawn.Gandhiji lived with the peasants for some time in order to learn about their hard lot. But, he also taught them to be free and to stand on their feet. At last, he succeeded in securing justice for the poor peasants.

Question 1.
Gandhiji left for Champaran to find
(a) out the truth
(b) out the peasants
(c) out the magistrate
(d) none of the above.
Answer:
(a) out the truth.

Question 2.
Who were cruelly treated ?
(a) The British indigo planters
(b) Gandhiji and his followers
(c) The peasants of Champaran
(d) The peasants all over India
Answer:
(c) The peasants of Champaran.

Question 3.
Gandhiji was asked to leave because:
(a) his life was in danger.
(b) the government was alarmed.
(c) he had lost his energy to unite the peasants.
(d) all the above.
Answer:
(b) the government was alarmed.

Question 4.
When refused, Gandhiji was asked to appear before:
(a) the peasants
(b) the public meeting
(c) the indigo planters
(d) the magistrate.
Answer:
(d) the magistrate.

Question 5.
Gandhiji’s first ‘Satyagraha’ was a:
(a) success
(b) failure
(c) false show
(d) poor show
Answer:
(a) success.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions

Try These : (Textbook Page No.4)

Question 1.
Fill in the blanks in the following table :

Answer:

[Note : Rational numbers are not closed under division.]
e.g., \(\frac {2}{3}\) ÷ 0 = ? This is not defined. That’s why our answer in the table is ‘No’.

Try These : (Textbook Page No.6)

Question 1.
Complete the following table:

Answer:

Try These : (Textbook Page No.9)

Question 1.
Complete the following table:

Answer:

Think, Discuss and Write : (Textbook Page No.11)

1. If a property holds for rational numbers, will it also hold for integers ? For whole numbers ? Which will ? Which will not ?
Answer:
( i ) Any property which is true for rational numbers is also true for integers except for any integers ‘a’ and ‘b’ (a ÷ b) is not necessarily an integer.
(ii) All properties which are true for rational numbers are also true for whole numbers also except:

• For ‘a’ and ‘b’ being whole numbers (a – b) may not be a whole number.
• For ‘a’ and ‘b’ being whole numbers (b ≠ 0), a ÷ b may not be a whole number.

Try These : (Textbook Page No.13)

1. Find using distributivity :

Question (i).
\(\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}\)
Answer:
\(\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}\)
= \(=\frac{7}{5} \times\left[\frac{-3}{12}+\frac{5}{12}\right]\)
= \(\frac{7}{5} \times\left[\frac{-3+5}{12}\right]\)
= \(\frac{7}{5} \times \frac{2}{12}\)
= \(\frac{7}{5} \times \frac{1}{6}\)
= \(\frac {7}{30}\)

Question (ii).
\(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}\)
Answer:
\(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}\)
= \(\frac{9}{16} \times\left[\frac{4}{12}+\left(\frac{-3}{9}\right)\right]\)
= \(\frac{9}{16} \times\left[\frac{12+(-12)}{36}\right]\) …..(LCM = 36)
= \(\frac{9}{16} \times\left[\frac{0}{36}\right]\)
= \(\frac{9}{16} \times 0\)
= 0.

Try These : (Textbook Page No.17)

1. Write the rational number for each point labelled with a letter:

Question (i).

Answer:

Here, the rational number for-
the point A is \(\frac {1}{5}\)
the point B is \(\frac {4}{5}\)
the point C is \(\frac {5}{5}\) or 1.
the point D is \(\frac {8}{5}\)
the point E is \(\frac {9}{5}\)

Question (ii).

Answer:

Here, the rational number for-
the point F is \(\frac {-2}{6}\) or \(\frac {-1}{3}\).
the point G is \(\frac {-5}{6}\).
the point H is \(\frac {-7}{6}\).
the point I is \(\frac {-8}{6}\) or \(\frac {-4}{3}\).
the point J is \(\frac {-11}{6}\).

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution.
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= \(\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}\)
= \(\frac{\sin 18^{\circ}}{\sin 18^{\circ}}\) = 1
[∵ cos (90° – θ) = sin θ]

(ii) \(\frac{\tan 26^{\circ}}{\cos 64^{\circ}}=\frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)}\)
= \(\frac{\tan 26^{\circ}}{\tan 26^{\circ}}\) = 1
[∵ cot (90°- θ) = tan θ]

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
[∵ cos (90° – 0) = sin O]
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59°
=cosec 31° – sec (90° – 31°)
= cosec 31° – cosec 31°
[∵ sec (90° – θ) = cosec θ].

Question 2.
Show that:
(i) tan 4 tan 230 tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) L.H.S.
= tan 48° tan 23° tan 42° tan 67°
= tan 48° × tan 23° × tan (90° – 48°) × tan (90° – 23°)
= tan48° × tan 23° × cot48° × cot 23°
= tan 48C × tan 23° × \(\frac{1}{\tan 48^{\circ}}\) × \(\frac{1}{\tan 23^{\circ}}\) = 1
∴ L.H.S. = R.H.S.

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°
= cos 38° × cos (90 – 38°) – sin 38° × sin (90° – 38°)
= cos 38° × sin 38° – sin 38° × cos 38
= 0.
∴ L.H.S. = RH.S.

Question 3.
If tan 2A = cot (A – 18°) where 2A is an acute angle, find the value of A.
Solution:
Given: tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°)
[cot (90° – θ) = tan θ]
⇒ 90°- 2A = A – 18°
⇒ 3A = 108°
⇒A = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
Given that: tan A = cot B
⇒ tan A = tan(90° – B)
[∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B.
⇒ A + B = 90°..

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Given that: sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A — 20°)
[∵ cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20°
⇒ 5A = 110°
⇒ A = 22°.

Question 6.
If A, B and C interior angles of a triangle ABC, then show that: \(\sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)\)
Solution:
Since, A, B and C are interior angles of a triangle
∴ A + B + C = 180°
[Sum of three angles of a triangle is 180°]
⇒ B + C = 180° – A
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}\)
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
Taking sin on both sides, we get
⇒ \(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
[∵ sin (90° – θ) = cos θ].

Question 7.
Express sin 67° + cos 75° in terms of Trigonometric ratios of angles between 0° and 45°.
Solution:
Given that: sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
[∵ sin(90° – θ) = cos θ and cos (90° – θ) = sin θ].

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.6

1. Draw a line XY and point P not lying on XY. Draw a line parallel to XY passing through P with the help of ruler and compasses.
Solution:
Steps of Construction:
1. Draw a line XY and point P not lying on it.

2. Take any point Q, anywhere on line XY.
3. Join PQ.

4. Now take Q as centre, draw arc AB of any radius on XY. Similarly, draw an arc CD of same radius on line segment PQ from point P.

5. Measure arc AB with compasses.
6. Draw an arc equal to radius AB from point C witch intersect CD on E.

7. Join PE and produce it. So, the line l is the required line parallel to XY.

2. Draw a line p parallel to line m passing through a point A which is not lying on line m with the help of set squares.
Solution:
Steps of Construction:
1. Given a line m with point A not lying on it.
2. Place one of the edge of a ruler along the line m and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Hold the ruler firmly, slide the set square along the line m till its vertical side reaches the point A.

5. Firmly hold the set square in this position, take another set square and place it in such a way that one of its edges containing right angle concides with previous set square as shown.
6. Now draw a line p along side of second set square passing through A.

Thus p \(\text { ॥ } \) m passing through A.

3. Given a line AB and the point X is not lying on it. Draw a line parallel to AB passing through X.

Question (i)
By a ruler and compasses
Solution:
By a ruler and compasses:
Let us consider a line AB and point X not lying on it.

Steps of Construction:
1. Take any point, say Y anywhere on line AB.
2. Join XY.

3. Now take Y as centre, draw an arc PQ of any radius on AB. Similarly draw an arc RS of same radius on line segment XY from point X.

4. Measure arc PQ with compasses.
5. Draw an arc equal to radius PQ from point R which intersect RS on T.
6. Join XT and produce it.

So the line m is the required line parallel to AB.

Question (ii)
By set squares.
Solution:
By set squares.

Steps of Construction:
1. Given a line AB with point X not lying on it.
2. Place one of the edge of a ruler along the line AB and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Hold the ruler firmly, slide the set square along the line AB till its vertical side reaches the point X.

5. Firmly hold the set square in this position, take another set square and place it in such a way that one of its edges containing right angle concides with previous set square as shown.
6. Now draw a line l along side of second set square passing through X.

7. Thus l \(\text { ॥ } \) AB passing through X.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.5

1. Draw the following angles in both directions (Left and right) by protractor:

Question (i)
75°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on ray OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 75° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 75°.

If ray OA lies to the left of the centre (midpoint) of the baseline, start reading the angle on the outer scale from 0° and mark 75°. Join OB, then \(\angle AOB\) = 75°.

Question (ii)
110°
Solution:
Steps of Construction:
1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 110° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 110°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 110°. Join OB, then \(\angle AOB\) = 110°.

Question (iii)
62°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-62° base line along OA.

3. Mark a point B on the paper against the mark of 62° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 62°.

If the ray OA lies to the left of the centre (midpoint) of the baseline, start reading the angle on the outer scale from 0° and mark 62°. Join OB, then \(\angle AOB\) = 62°.

Question (iv)
165°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° baseline along OA.

3. Mark a point B on the paper against the mark of 165° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle
\(\angle AOB\) = 165°

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 165°. Join OB, then \(\angle AOB\) = 165°.

Question (v)
170°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on ray OA such that its centre lies on the initial point O and 0-480° base line along OA.

3. Mark a point B on the paper against the mark of 170° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 170°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 170°. Join OB, then \(\angle AOB\) = 170°.

Question (vi)
32°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 32° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 32°.
If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 32°. Join OB, then \(\angle AOB\) = 32°.

Question (vii)
128°
Solution:
Steps of Construction:

1. Draw a ray OA.

Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 128° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle
\(\angle AOB\) = 128°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 128°. Join OB, then \(\angle AOB\) = 128°.

Question (viii)
25°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 25° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 25°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 25°. Join OB, then \(\angle AOB\) = 25°.

Question (ix)
80°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 80° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 80°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 80°. Join OB, then \(\angle AOB\) = 80°.

Question (x)
135°.
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 135° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 135°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 135°. Join OB, then \(\angle AOB\) = 135°.

2. Bisect the following angles by compasses:

Question (i)
48°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 48°.
4. Join OB. Then \(\angle AOB\) = 48°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point
E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 48°.
Measure \(\angle AOB\) and \(\angle AOB\)
\(\angle AOB\) = \(\angle AOB\) = 24°.

Question (ii)
140°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 140°.
4. Join OB. Then \(\angle AOB\) = 140°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 140°.
On measurement \(\angle AOE\) = \(\angle BOE\) = 70°.

Question (iii)
75°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 75°.
4. Join OB. Then \(\angle AOB\) = 75°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 75°.
On measurement
\(\angle AOE\) = \(\angle BOE\) = 37.5°.

Question (iv)
64°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 64°.
4. Join OB. Then \(\angle AOB\) = 64°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 64°.
On measurement
\(\angle AOE\) = \(\angle BOE\) = 32°.

Question (v)
124°.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 124°.
4. Join OB. Then \(\angle AOB\) = 124°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 124°.
On measurement \(\angle AOE\) = \(\angle BOE\) = 62°.

3. Draw an angle of 80° and bisect it in to four equal parts by compasses.
Solution:
1. Draw a line OY of any length.
2. Place the centre of the protractor at O.
3. Starting with 0 mark a point X at 80°.
4. Join OX. Then \(\angle XOY\) = 80°.
5. With O as centre and using compass, draw an arc that cuts both rays of \(\angle XOY\). Name the point of intersection as X’ and Y’.
6. With Y’ as centre, draw an arc whose radius is more than half the length X’Y’.
7. With the same radius and with X’ as a centre, draw another arc which cut the first arc at point C.
8. With O as centre and using compass, draw an arc that cuts both rays of \(\angle COY\). Name the points of intersection as B and A.
9. With A as centre, draw an arc whose radius is more than half the length AB.
10. With the same radius and with B as centre, draw another arc which cuts the first arc at point S.
11. With O as centre and using compass, draw an arc that cuts both rays of \(\angle XOC\) . Name the points of intersection as D and E.
12. With E as centre, draw an arc whose radius is more than half the length DE.
13. With the same radius and with E as centre, draw another arc which bisects the first arc at T. Then OT is the bisector of \(\angle XOC\).
Thus \(\overline{\mathrm{OS}}, \overline{\mathrm{OC}} \text { and } \overline{\mathrm{OT}}\) divide, \(\angle AOB\) = 80° into four equal parts.

4. Draw a right angle and bisect it.
Solution:
1. Draw a ray OB.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point A at 90°.

4. Join OA. Then \(\angle AOB\) = 90°
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\). Name the points of intersection as A’ and B’.
6. With B’ as centre, draw an arc whose radius is more than half of the length B’A’.
7. With the same radius and with A’ as a centre, draw another arc which cuts the first arc at point C. Join OC bisects \(\angle AOB\).

5. Draw the following angles by ruler and compasses:

Question (i)
30°
Solution:
To Construct angle of 30°

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.

3. With C as centre and same radius as before draw another arc cutting the previous arc at E.
4. Join OE and produce it to B. \(\angle AOB\) = 60°.
5. Bisect \(\angle AOB\).
Thus \(\angle AOM\) = \(\angle MOB\) = 30°.

Question (ii)
45°
Solution:
To Construct Angle of 45°:

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.

3. With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.
4. With P and Q as centres and any suitable radius (more than half of PQ) or even the same radius draw arc cutting each other at R.
5. Join OR and produce it to B. Then \(\angle AOB\) = 90°.
6. Bisect \(\angle AOB\).
7. OD is the bisector of \(\angle AOB\).
\(\angle BOD\) = \(\angle DOA\) = 45°.

Question (iii)
135°
Solution:
To Construct Angle of 135°.

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.
3. With C as centre and the same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q and then with Q as centre the same radius cut off the arc again at R.
4. With Q and R as centres and radius more than half of RQ draw arcs cutting each other at L.
5. Join OL and produce it to B. Then \(\angle AOB\) = 150°.
6. Take a point M on the arc where OL intersects the arc.
7. With M and Q as centres and radius more than half of MQ draw arcs cutting each other at N.
8. Join ON and produce it to E. \(\angle AOE\) = 135°.

Question (iv)
180°
Solution:
To Construct Angle of 180°.

Steps of Construction:
1. Draw a line AB and mark a point C on it.
2. Taking C as centre and with any suitable radius, draw an are PQ cutting AB at P and Q.

3. Here \(\angle ACB\) = 180° (It is a straight line).

Question (v)
120°
Solution:
To Construct Angle of 120°:

Steps of Construction:
1. Draw a line segment OA.

2. With O as centre and any suitable radius draw an arc cutting OA at point M.
3. With M as centre and same radius draw an arc which cuts the arc at N and then with N as centre and the same radius cut off the arc again at Q.
4. Join OQ and produce it to B.
Then \(\angle AOB\) = 120°.

Question (vi)
75°.
Solution:
To Construct Angle of 75°.

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at C.
3. With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.
4. With P and Q as centres and radius more than half of PQ draw arcs cutting each other at R.
5. Join OR and produce it to B. \(\angle AOB\) = 90°.
6. Bisect \(\angle AOB\).
7. OD is the bisector of \(\angle AOB\).
\(\angle BOD\) = \(\angle DOA\) = 45°.
8. Again draw OE bisector of \(\angle DOB\).
Thus angle \(\angle EOA\) = 75°.

6. Draw an angle of 30° by protractor and bisect it by a ruler and compasses.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 30°.
4. Join OB. Then \(\angle AOB\) = 30°.
5. With O as centre and using compasses draw an arc that cuts both rays of \(\angle AOB\), with the point intersection as C and D.
6. With C as centre, draw an arc whose radius is more than half of the length of CD.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, it bisect \(\angle AOE\).

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 15 Probability MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
When a balanced die is thrown, the probability of getting 3 is …………….. .
A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{6}\)
Answer:
D. \(\frac{1}{6}\)

Question 2.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a king is …………………. .
A. \(\frac{1}{52}\)
B. \(\frac{1}{26}\)
C. \(\frac{1}{13}\)
D. 1
Answer:
C. \(\frac{1}{13}\)

Question 3.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a card other than picture cards is ……………….. .
A. \(\frac{4}{13}\)
B. \(\frac{10}{13}\)
C. \(\frac{3}{13}\)
D. \(\frac{1}{13}\)
Answer:
B. \(\frac{10}{13}\)

Question 4.
When an unbiased coin is tossed thrice, the probability of receiving three heads is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{8}\)
Answer:
A. \(\frac{1}{8}\)

Question 5.
When three unbiased coins are tossed simultaneously, the probability of receiving exactly one tail is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
Answer:
D. \(\frac{3}{8}\)

Question 6.
When a balanced die is thrown, the probability of receiving an even number is ………………… .
A. \(\frac{1}{6}\)
B. \(\frac{5}{6}\)
C. \(\frac{1}{2}\)
D. \(\frac{1}{4}\)
Answer:
C. \(\frac{1}{2}\)

Question 7.
When a balanced die is thrown, the probability of receiving a prime number is ……………….. .
A. \(\frac{2}{3}\)
B. \(\frac{3}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
Answer:
D. \(\frac{1}{2}\)

Question 8.
When two balanced dice are thrown simultaneously, the probability of getting the total of numbers on dice as 9 is ………………. .
A. \(\frac{1}{9}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{12}\)
Answer:
A. \(\frac{1}{9}\)

Question 9.
Out of 100 days, the forecast predicted by the wheather department proved to be true on 20 days. Chosen any one day from these 100 days, the probability that the forecast proved to be false is ………………… .
A. \(\frac{1}{3}\)
B. \(\frac{1}{4}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
Answer:
D. \(\frac{4}{5}\)

Question 10.
The probability of a month of January having 5 Sundays is ………………….. .
A. \(\frac{2}{7}\)
B. \(\frac{3}{7}\)
C. \(\frac{5}{7}\)
D. \(\frac{1}{7}\)
Answer:
B. \(\frac{3}{7}\)

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.4

1. Draw a circle of the following radius:

Question (i)
3.5 cm
Solution:
Steps of construction.

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measure OA = 3.5 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (ii)
4 cm
Solution:
Steps of construction.

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measure OA = 4 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob.

Question (iii)
2.8 cm
Solution:
Steps of construction

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measures OA = 2.8 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (iv)
4.7 cm
Solution:
Steps of construction

1. Mark a point O on the page of your note book.
2. Take compasses fixed with sharp pencil and measures OA = 4.7 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (v)
5.2 cm.
Solution:
Steps of construction

1. Mark a point O on the page of your note book.
2. Take compasses fixed with sharp pencil and measures OA = 5.2 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

2. Draw a circle of diameter 6 cm.
Solution:
Steps of construction

1. Draw a line segment PQ = 6 cm.
2. Draw the perpendicular bisector of PQ intersecting PQ at O.
3. With O as centre and radius = OQ = 3 cm (= OP), draw a circle.
The circle thus drawn is the required circle.

3. With the same centre O, draw two concentric circles of radii 3.2 cm and 4.5 cm.
Solution:
Steps of Construction

1. Mark a point O on the page of your note book, where a circle is drawn.
2. Take compasses fixed with sharp pencil measuring OA = 4.5 cm using scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw complete arc by holding the compasses from its knob.
After completing one round, we get circle I.
4. Again with the same centre O and new radius = 3.2 cm draw another circle II following the same step 3.

4. Draw a circle of radius 4.2 cm with centre at O. Mark three points A, B and C such that point A is on the circle, B is in the interior and C is in the exterior of the circle.
Solution:
Steps of Construction

1. Mark a point O on the page of your note book, where a circle is to be drawn
2. Take compasses fixed with sharp pencil and measure OA = 4.2 cm using scale (∴ A is on the circle).
3. Without changing the opening of the compasses, keep the needle at point O and draw complete arc by rotating the compasses from the knob. After completing one round, we get required circle.
4. Mark point B in the interior of the circle and point C in the exterior of the circle.

5. Draw a circle of radius 3 cm and draw any chord. Draw the perpendicular bisector of the chord. Does the perpendicular bisector passes through the centre?
Solution:
Steps of Construction.

1. Draw a circle with C as centre and radius 3 cm.
2. Draw AB the chord of the circle.
3. Draw PQ the perpendicular bisector of chord AB.
4. We see that the perpendicular bisector of chord AB passes through the centre C of the circle.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.3

1. Draw a line r and mark a point P on it. Construct a line perpendicular to r at point P.

Question (i)
Using a ruler and compasses.
Solution:
Using ruler and compasses

Steps of Construction.

1. Draw a line r and mark a point P on it.

2. Draw an arc from P to the line r of any suitable radius which intersects line r at A and B.

3. Draw arcs of any radius which is more than half of arc made in step (2) from A and B which intersect at Q.

4. Join PQ.

Thus PQ is perpendicular to AB or line l or PQ ⊥ A.
Here P is called foot of perpendicular.

Question (ii)
Using a ruler and a set square.
Solution:
Using a ruler and a set square

Steps of Construction

1. Draw a line r and a point P on it.
2. Place one of the edges of a ruler along the line l and hold if firmly.

3. Place the set square in such a way that one of its edges contaning the right angle coincides with the ruler.
4. Holding the ruler, slide the set square along the line l till the vertical side reaches the point P.

5. Firmly hold the set square in this position. Draw PQ along its vertical edge. Now PQ is the required perpendicular to l ie. PQ ⊥ r.

2. Draw a line p and mark a point z above it. Construct a line perpendicular to p, from the point z.

Question (i)
Using a ruler and compasses.
Solution:
1. Draw a line p and mark a point z not lying on it.
2. From point z draw an arc which intersects line p at two points P and Q.


3. Using any radius and taking P and Q as centre, draw two arcs that intersect at point say B. On the other side (a shown in figure).
4. Join AB to obtain altitude to the line p.

Thus xz is altitude to line p.
i.e. xz ⊥ p.

Question (ii)
Using a ruler and set square
Solution:
Steps of constructions:
1. Draw a line p and mark a point z which is not lying on it.
2. Place one of the edge of a ruler along the line p and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Holding the ruler firmly, slide the set square along the line p till its vertical side reaches the point z.

5. Firmly hold the set square in this position, Draw xz along its vertical edge. Now xz is the required altitude to p i.e. xz ⊥ p.

3. Draw a line AB and mark two points P and Q on either side of line AB, Construct two lines perpendicular to AB, from P and Q using a ruler and compasses.
Solution:
1. Draw a line AB and Mark two points P and Q on either side of AB.

2. From point P draw an arc which intersect line AB at two points C and D.
3. Using any radius and taking C and D as centre draw two arcs that intersects at point say E on the other side as shown in figures.

4. Join PE to obtain perpendicular to AB.
5. From point Q draw an arc which intersects AB at two points X and Y.

6. Using any radius and taking X and Y as centre draw two arcs that intersects at point say R on the other side of line AB as shown in figures.
7. Join QR to obtain perpendicular to AB.
Thus, PE ⊥ AB and QR ⊥ AB

4. Draw a line segment of 7 cm and draw perpendicular bisector of this line segment.
Solution:
Steps of Construction:
1. Draw a line segment AB = 7 cm.
2. With A as centre and radius more than half of AB, draw an arc on both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

5. Draw a line segment PQ = 6.8 cm and draw its perpendicular bisector XY which bisect PQ at M. Find the length of PM and QM. Is PM = QM ?
Solution:
Steps of Construction:

1. Draw a line segment PQ = 6.8 cm
2. With P as centre and radius more than half of PQ draw arcs on both sides of PQ.

3. Now with Q as centre and the same radius as in step 2 draw arcs intersecting the previous drawn arcs at A and B respectively.
4. Join AB intersecting PQ at M. Then M bisects the line segment.
5. Measure the length of PM and QM
PM = 3.4 cm and QM = 3.4 cm
∴ PM = QM.

6. Draw perpendicular bisector of line segment AB = 5.4 cm. Mark point X anywhere on perpendicular bisector Join X with A and B. Is AX = BX ?
Solution:
Steps of construction.
1. Draw a line segment AB = 5.4 cm.
2. With A as centre and radius more than half of AB, draw an arc in both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O.
Then CD is the perpendicular bisector of AB.
Mark any point X on the perpendicular bisector CD. Drawn. Then join AX and BX.
On examination, we find that AX = BX.

7. Draw perpendicular bisectors of line segment of the following lengths.

Question (i)
8.2 cm
Solution:
Steps of Construction.
1. Draw a line regment AB = 8.2 cm

2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw an arcs intersecting the previous arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (ii)
7.8 cm
Solution:
Steps of Construction.

1. Draw a line segment AB = 7.8 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (iii)
6.5 cm.
Solution:
Steps of Construction.
1. Draw a line segment AB = 6.5 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2 draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

8. Draw a line segment of length 8 cm and divide it into four equal parts Using compasses. Measure each part.
Solution:
Steps of construction.

1. Draw a line segment AB of length 8 cm
2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at P and Q.
4. Join PQ intersecting AB at C then PQ is the perpendicular bisector of AB intersecting AB at C.
5. Similarly draw the perpendicular bisector of AC intersecting AC at D.
6. Draw the perpendicular bisector of CB intersecting CB at E.
By actual measurement, it can be verified that
AD = DC = CE = EB

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line it + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).
Solution:
Let line 2x + y – 4 = 0 divides the line segment joining the points A (2,- 2) and B(3, 7) at C (x, y) in the ratio k : 1

∴ Coordinates of C are x = \(\frac{3 k+2 \times 1}{k+1}=\frac{3 k+2}{k+1}\) and y = \(\frac{7 k+(-2) \times 1}{k+1}=\frac{7 k-2}{k+1}\)
∴ C \(\left[\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right]\). must lie on the line 2x + y – 4 = 0

i.e., 2\(\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)\) – 4 = 0
or \(\frac{6 k+4+7 k-2-4 k-4}{k+1}\) = 0
or 9k – 2 = 0
or 9k = 2
or k = \(\frac{2}{9}\).
∴ ratio k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9.
Hence required ratio is 2 : 9.

Question 2.
Find a relation between x and y if (x, y) ; (1, 2) and (7, 0) are collinear.
Solution:
Let given points are A (x, y); B (1, 2) and C (7, 0).
Here x₁ = x, x₂ = 1, x₃ = 7
y₁ = y, y₂ = 2, y₃ = 0
∵ Three points are collinear
iff \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) x (2 – 0) + 1 (0 – y) + 7 (y – 2)] = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0
or x + 3y – 7 = 0 is the required relation.

Question 3.
Find the centre of a cirçle passing through the points (6, —6); (3, —7) and (3,3).
Solution:
Let O (x, y) be the required centre of the circle which passes through points P(6, – 6); Q(3, – 7) and R (3, 3).
∴ radii of circle are equal.

∴ OP = OQ = OR
or (OP)² = (OQ)² = (OR)²
Now, (OP)² = (OQ)²
(x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
or x² + 36 – 12x + y² + 36 + 12y = x² + 9 – 6x + y² + 49 + 14y
or – 12x + 12y + 72 = – 6x + 14y + 58
or – 6x – 2y + 14 = 0
or 3x + y – 7 = 0 ………………(1)
Also, (OQ)² = (OR)²
or (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
or (y + 7)² = (y – 3)²
or y² + 49 + 14y = y² + 9 – 6y
or 20y = – 40
y = \(\frac{-40}{20}\) = – 2
Substitute this value of)’ in (1), we get
3x – 2 – 7 = 0
or 3x – 9 = 0
or 3x = 9
or x = \(\frac{9}{3}\) = 3
∴ Required centre is (3, – 2).

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.
Solution:
Let two opposite vertices of a square ACBD are A (- 1, 2) and B (3, 2) and coordinates of C are (x, y)
∵ Length of each sides of square are equal.
∴ AC = BC
or (AC)² = (BC)²
or (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²

or (x + 1)² = (x – 3)²
or x² + 1 + 2x = x² + 9 – 6x
or 8x = 8
or x = \(\frac{8}{8}\) = 1
Now, in rt ∠d ∆ACB,
Using Pythagoras Theorem,
(AC)² + (BC)² = (AB)²
(x + 1)² + (y – 2)² + (x – 3)² + (y – 2)² = (3 + 1)² + (2 – 2)²
or x² + 1 + 2x + y² + 4 – 4y + x² + 9 – 6x + y² + 4 – 4y = 16
or 2x² + 2y² – 4x – 8y + 2 = 0
or x² + y² – 2x – 4y + 1 = 0
Putting the value of x = 1 in (1), we get
(1)² + y² – 2 (1) – 4y + 1 = 0
or y² – 4y = 0
or y (y – 4) = 0
Either y = 0 or y – 4 = 0
Either y = 0 or y = 4
∴ y = 0, 4
∴ Required points are (1. 0) and (1.4).

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There ¡s a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of A PQR if C is the origin? Also calculate the areas of the triangles In these cases. What do you observe?
Solution:
Case I:
When taking A as origin then AD is X-axis and AB is Y-axis.
∴ Coordinates of triangular grassy Lawn
PQR are P (4, 6); Q (3, 2) and R(6, 5).
Here x₁ = 4, x₂ = 3, x₃ = 6
y₁ = 6, y₂ = 2, y₃ = 50
Now, area of ∆PQR = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
= \(\frac{1}{2}\) [- 12 – 3 + 24] = \(\frac{9}{2}\)
= 4.5 sq. units.

Case II: When taking C as origin then CB is X – axis and CD is Y – axis.
∴ Coordinates of triangular grassy lawn PQR
are P(12, 2); Q (13,6) and R (10, 3)
Here x₁ = 12, x₂ = 13, x₃ = 10
y₁ = 2, y₂ = 6, y₃ = 3
Now, area of ∆PQR = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [12 (6 – 3) + 13 (3 – 2) + 10 (2 – 6)]
= \(\frac{1}{2}\) [36 + 13 – 40]
= \(\frac{9}{2}\) = 4.5 sq. units.
From above two cases, it is clear that area of triangular grassy lawn is same.

Question 6.
The vertices of a ∆ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. (Recall Theorem 6.2 and Theorem 6.6).
Solution:
The vertices of ∆ABC are A (4, 6); B (1, 5) and C (7, 2)

A line is drawn to intersect sides AB and AC at D (x₁, y₁) and E (x₂, y₂) respectively such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\).

∴ D and E divides AB and AC in the ratio 1 : 3.
∴ Coordinates of D are
x₁ = \(\frac{1(1)+3(4)}{1+3}=\frac{1+12}{4}=\frac{13}{4}\) and y₁ = \(\frac{1(5)+3(6)}{1+3}=\frac{5+18}{4}=\frac{23}{4}\)

∴ Coordinates of D are (\(\frac{13}{4}\), \(\frac{23}{4}\))
Now, coordinates of E are
x₂ = \(\frac{1(7)+3(4)}{1+3}=\frac{7+12}{4}=\frac{19}{4}\) and y₂ = \(\frac{1(2)+3(6)}{1+3}=\frac{2+18}{4}=\frac{20}{4}=5\)

∴ Coordinates of E are (\(\frac{19}{4}\), 5).

In ∆ADE
x₁ = 4, x₂ = \(\frac{13}{4}\), x₃ = \(\frac{19}{4}\)
y₂ = 6, y₂ = \(\frac{23}{4}\), y₃ = 5
area of ∆ADE = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= \(\frac{1}{2}\left[4\left(\frac{23}{4}-5\right)+\frac{13}{4}(5-6)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]\)

= \(\frac{1}{2}\left[4\left(\frac{23-20}{4}\right)+\frac{13}{4}(-1)+\frac{19}{4}\left(\frac{24-23}{4}\right)\right]\)

= \(\frac{1}{2}\left[3-\frac{13}{4}+\frac{19}{16}\right]\)

= \(\frac{1}{2}\left[\frac{48-52+19}{16}=\frac{15}{16}\right]\)
= \(\frac{15}{32}\) sq. units.

In ∆ABC
x₁ = 4, x₂ = 1, x₃ = 7
y₂ = 6, y₂ = 5, y₃ = 2
Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= \(\frac{1}{2}\) [4 (5 – 2) + 1 (2 – 6) + 7 (6 – 5)]
= \(\frac{1}{2}\) [12 – 4 + 7] = \(\frac{15}{2}\) sq.units.

Now, \(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\frac{15}{32}}{\frac{15}{2}}=\frac{15}{32} \times \frac{2^{1}}{16_{1}}\)

= \(\frac{1}{16}=\left(\frac{1}{4}\right)^{2}\)

= \(\left(\frac{A D}{A B}\right)^{2} \text { or }\left(\frac{A E}{A C}\right)^{2}\).

Question 7.
Let (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the potnt P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note : The point which is common to all the three medians ¡s called centroid and this point divides each median in the ratio 2: 1]
(v) if A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
Given that vertices of ∆ABC are A (4, 2); B (6, 5) and C (1, 4).
(i) AD is the median from the vertex A.
∴ D is the mid point of BC.

then x = \(\frac{6+1}{2}=\frac{7}{2}\) and y = \(\frac{5+4}{2}=\frac{9}{2}\)
Hence, coordinates of D is (\(\frac{7}{2}\), \(\frac{9}{2}\)).

(ii) Let P(x, y) be point on AD such that AP : PD = 2 : 1

then x = \(\frac{2\left(\frac{7}{2}\right)+1(4)}{2+1}\)
= \(\frac{7+4}{3}=\frac{11}{3}\)

and y = \(\frac{2\left(\frac{9}{2}\right)+1(2)}{2+1}\)
= \(\frac{9+2}{3}=\frac{11}{3}\)

Hence, Coordinates of P is (\(\frac{11}{3}\), \(\frac{11}{3}\)).

(iii) Le BE and CF are the medians of ∆ABC to AC and AB respectively.
∴ E and F are mid points of AC and AB respectively.

Coordinate of E are
x₁ = \(\frac{4+1}{2}=\frac{5}{2}\)
and y₁ = \(\frac{4+2}{2}=\frac{6}{2}\) = 3
Coordinate of E are (\(\frac{5}{2}\), 3)
Coordinate of F are
x₂ = \(\frac{4+6}{2}=\frac{10}{2}\) = 5
and y₂ = \(\frac{5+2}{2}=\frac{7}{2}\)
∴ Coordinate of F are (5, \(\frac{7}{2}\))
Now, Q divides BE such that BQ : QE = 2: 1

∴ Coordinate of Q are \(\left(\frac{2\left(\frac{5}{2}\right)+6(1)}{2+1}, \frac{2(3)+1(5)}{2+1}\right)\)

= \(\left(\frac{5+6}{3}, \frac{6+5}{3}\right)\) = \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

Also, R divides CF such that CR : RF = 2 : 1

∴ Coordinate of R are = \(\left(\frac{2(5)+1(1)}{2+1}, \frac{2\left(\frac{7}{2}\right)+(4)}{2+1}\right)\)

= \(\left(\frac{10+1}{3}, \frac{7+4}{3}\right)\)

= \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iv) From above discussion, it is clear that coordinates of P, Q and R are same and coincide at a point, is known as centroid of triangle, which divides each median in the ratio 2: 1.

(v) The vertices of given ∆ABC are
A (x₁, y₁); B (x₂, y₂) and C (x₃, y₃).

Let AD is median of E, ∆ABC.
∴ D is the mid point of BC then coordinates of D are \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

Now, G be the centroid of ABC, which divides the median AD in the ratio 2: 1
∴ Coordinates of G are [using (iv)]

= \(\left[\frac{2\left(\frac{x_{2}+x_{3}}{2}\right)+1\left(x_{1}\right)}{2+1}, \frac{2\left(\frac{y_{2}+y_{3}}{2}\right)+1\left(y_{1}\right)}{2+1}\right]\)

= \(\left[\frac{x_{2}+x_{3}+x_{1}}{3}, \frac{y_{2}+y_{3}+y_{1}}{3}\right]\)

= \(\left[\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right]\).

Question 8.
ABCD is a rectangle formed by the points A (- 1, – 1), B (- 1, 4), C (5, 4) and D (5, – 1). P, Q R and S are the mid points
of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle? or a rhombus ? Justify your answer.
Solution:
Given: The vertices ot’ given rectangle ABCD are
A(- 1, – 1); B(- 1, 4); C(5, 4) and D (5, – 1).

∵ P is the mid point of AB.
∴ Coordinates of P are \(\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right)=\left(-1, \frac{3}{2}\right)\)
∵ Q is the mid point of BC.
∴ Co-ordinates of Q are \(\left(\frac{-5+5}{2}, \frac{4+4}{2}\right)\) = (2, 4)
∵ R is the mid point of CD.
∴ Coordinates of R are \(\left(\frac{5+5}{2}, \frac{4+1}{2}\right)=\left(5, \frac{3}{2}\right)\)

∵ S is the mid point of AD.
∴ Co-ordinates of S are \(\left(\frac{5-1}{2}, \frac{-1-1}{2}\right)\) = (2, -1)

PQ = \(\sqrt{(2+1)^{2}+\left(4-\frac{3}{2}\right)^{2}}\)

= \(\sqrt{9 \times \frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

PQ = \(\sqrt{\frac{61}{4}}\)

QR = \(\sqrt{(5-2)^{2}+\left(\frac{3}{2}-4\right)^{2}}\)

= \(\sqrt{(3)^{2}+\left(\frac{3-8}{2}\right)^{2}}\)

= \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

QR = \(\sqrt{\frac{61}{4}}\)

RS = \(\sqrt{(2-5)^{2}+\left(-1-\frac{3}{2}\right)^{2}}\)

= \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

RS = \(\sqrt{\frac{61}{4}}\)

and SP = \(\sqrt{(-1-2)^{2}+\left(\frac{3}{2}+1\right)^{2}}\)

SP = \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\)

Also PR = \(\sqrt{(5+1)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}}\)
PR = \(\sqrt{36+0}=\sqrt{36}\) = 6
QS = \(\sqrt{(2-2)^{2}+(4+1)^{2}}\)
= \(\sqrt{0+25}=\sqrt{25}\) = 5.

Form above discussion it is clear that PQ = QR = RS = SP.
Also, PR ≠ QS.
⇒ All sides of quad. PQRS are equal but their diagonals are not equal.
Quad. PQRS is a rhombus.