Class 9

Punjab State Board PSEB 9th Class English Book Solutions English Reading Comprehension  Picture/Poster Based Exercise Questions and Answers, Notes.

PSEB 9th Class English Reading Comprehension Picture/Poster Based

Answers have been given at the end of this set.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

Question 1.
Who did the fox invite to dinner ?
(a) The duck.
(b) The crane.
(c) The vixen.
(d) The deer.
Answer:
(b) The crane.

Question 2.
What did he serve his guest in the dinner ?
(a) Fruits.
(b) Meat.
(c) Soup.
(d) Eggs.
Answer:
(c) Soup.

Question 3.
The fox was very cunning. He placed a …….. before his guest.
(a) deep bowl
(b) flat dish
(c) narrow jar
(d) sound pitcher.
Answer:
(b) flat dish

Question 4.
What did the crane serve the fox when he invited the fox to dinner ?
(a) Cake.
(b) Milk.
(c) Rice.
(d) Chicken Curry
Answer:
(c) Rice.

Question 5.
The fox had to go hungry. Why ?
(a) Because the crane served the rice in a narrrow jar.
(b) Because the fox could not put his mouth in the narrow jar.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

Question 1.
An elephant and a ………. were very good friends.
(a) barber
(b) carpenter
(c) tailor
(d) cobbler
Answer:
(c) tailor

Question 2.
While going to the pond for water, the elephant would daily ………….
(a) stop at the tailor’s shop
(b) have a banana from the tailor
(c) both (a) and (b)
(d) neither (a) nor (b).
Answer:
(c) both (a) and (b)

Question 3.
One day when the elephant put his trunk into the shop,
(a) the tailor gave him a banana
(b) the tailor’s son gave him a banana
(c) the tailor pricked a needle into it
(d) the tailor’s son pricked a needle into it.
Answer:
(d) the tailor’s son pricked a needle into it.

Question 4.
The elephant had his revenge by ……….
(a) filling his trunk with muddy water
(b) throwing muddy water in the tailor’s shop
(c) both (a) and (b)
(d) neither (a) nor (b).
Answer:
(b) throwing muddy water in the tailor’s shop

Question 5.
The moral conveyed through these picture is ……….
(a) Might is right.
(b) Tit for tat.
(c) Do good have good.
(d) No pains no gains.
Answer:
(b) Tit for tat.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
The man is this picture is the famous cricketer
(a) Virat Kohli
(b) Irfan Pathan
(c) Sachin Tendulkar
(d) Mahendra Singh Dhoni.
Answer:
(c) Sachin Tendulkar

Question 2.
He is popularly known as ……. of cricket.
(a) Master Bowler
(b) Master Blaster
(c) Master Batsman
(d) Master Crickter.
Answer:
(b) Master Blaster

Question 3.
At the age of sixteen, he made his international debut against
(a) England
(b) Australia
(c) Sri Lanka
(d) Pakistan.
Answer:
(d) Pakistan.

Question 4.
Sachin took retirement from cricket in
(a) 2005
(b) 2018
(c) 2020
(d) 2013.
Answer:
(d) 2013.

Question 5.
Sachin Tendulkar was honoured with many prestigious awards like
(a) Arjuna Award and Rajiv Khel Ratna
(b) Padma Shri and Bharat Ratna
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer:
(c) Both (a) and (b)

Look at this poster and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
For what purpose is this poster designed ?
(a) To promote education for boys.
(b) To promote education for girls.
(c) To provide employment for boys.
(d) To provide employment for girls.
Answer:
(b) To promote education for girls.

Question 2.
Girls and boys have ……….. to education.
(a) no right
(b) equal right
(c) no interest
(d) equal interest.
Answer:
(b) equal right

Question 3.
How can the nation’s progress be accelerated ?
(a) By educating the boys.
(b) By educating the girls
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
We should not …….. the girls their rights.
(a) excuse
(b) give
(c) accept
(d) deny.
Answer:
(d) deny.

Question 5.
This poster teaches us to stop ……….
(a) the evil of dowry
(b) the evil of female foeticide
(c) the evil of bride burning
(d) the evil of gender discrimation.
Answer:
(b) the evil of female foeticide

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
Why did the fox jump into the well to drink water ?
(a) Because the water was very low.
(b) Because he wanted to eat the goat.
(c) Because he wanted to have a bath.
(d) None of these three.
Answer:
(a) Because the water was very low.

Question 2.
The fox drank water and ……….
(a) drenched his thirst
(b) quenched his thirst
(c) satisfied his thirst
(d) toasted his thirst.
Answer:
(b) quenched his thirst

Question 3.
How did the fox succeed in be fooling the goat who was passing that way ?
(a) He told the goat that it was very hot outside.
(b) He told the goat that it was very cold inside the well.
(c) He told that the water was very sweet.
(d) All of these three.
Answer:
(d) All of these three.

Question 4.
What happened when the foolished goat jumped into the well ?
(a) The fox at once climbed over her back.
(b) The fox jumped out of the well.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 5.
The moral of this story is
(a) As you sow so shall you reap
(b) Think before you speak.
(c) Look before you leap.
(d) All that glitters is not gold.
Answer:
(c) Look before you leap.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
Once upon a time a shepherd-boy ……… the sheep of the villagers.
(a) looked into
(b) looked after
(c) looked at
(d) looked for.
Answer:
(b) looked after

Question 2.
What mischief did he make one day ?
(a) He climbed up a tree.
(b) He started crying, “Wolf! Wolf!’’
(c) He shouted for help.
(d) All of these three.
Answer
(d) All of these three.

Question 3.
Why did the villagers who came to help the boy become cross with him ?
(a) Because they found no wolf there.
(b) Because the boy told them that he had shouted in fun only.
(c) Both (a) and (b).
(d) Neithor (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
What happened when a wolf really came there.
(a) The boy shouted for help but nobody came.
(b) The villagers didn’t believe the boy’s cries as he had be fooled them once.
(c) The wolf sprang upon the boy and. tore him to pieces.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
The moral of the story is —
(a) No pain no gain.
(b) Never give up hope in the hour of difficulty.
(c) Once a liar, always a liar
(d) Think before you speak.
Answer:
(c) Once a liar, always a liar

Look at these pictures and answer the questions given below :

Choose die correct option to answer each question:

Question 1.
An old farmer had three sons who were ………..
(a) very active
(b) very idle
(c) very naughty
(d) very hardworking.
Answer:
(b) very idle

Question 2.
What did the farmer said to sons before his death?
(a) He asked them to work hard in their uk.
(b) He asked them not to fight with each other after his death.
(c) He told them that there was a big treasure in his field.
(d) He asked them to live together happily.
Answer:
(c) He told them that there was a big treasure in his field.

Question 3.
What happened when the sons dig their field ?
(a) They found there a big treasure.
(b) They found there no treasure.
(c) They found there tools of farming.
(d) None of these three.
Answer:
(b) They found there no treasure.

Question 4.
What did the old man ask them to do?
(a) He asked them to sell that field to him.
(b) He asked them to sow seeds in their field.
(c) He asked them to dig their field more deeply.
(d) Any of these three.
Answer:
(b) He asked them to sow seeds in their field.

Question 5.
What happened when the Sons sow seeds in their field?
(a) There was a good crop that year.
(b) They got a lot of money for it.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 6.
The moral of this story is ……
(a) Do good find good.
(b) As you sow, so shall you reap.
(c) Hard work is the key to success.
(d) Hard work is man’s greatest treasure.
Answer:
(d) Hard work is man’s greatest treasure.

Look at these pictures and answer the questions given below:

Choose the correct option to answer each question :

Question 1.
What was the capseller doing in a forest ?
(a) He was passing through the forest to reach a village.
(b) He lay down under a tree to take some rest.
(c) He went there to sell his caps.
(d) Both (a) and (b).
Answer:
(d) Both (a) and (b).

Question 2.
What happened when the capseller fell asleep ?
(a) Some monkeys came there.
(b) The monkeys untied the bundle of caps.
(c) The monkeys took away all the caps.
(d) All of these three.
Answer:
(d) All of these three.

Question 3.
What did the capseller see when he woke up ?
(a) He found all his caps missing.
(b) He found the monkeys wearing his caps.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
What did he do to recover his caps ?
(a) He took off his caps and threw it down.
(b) The monkeys imitated him.
(c) The monkey threw down their caps.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
The moral of the story is ………
(a) Tit for tat.
(b) A stitch in time saves nine.
(c) God helps those who help themselves.
(d) Never give up hope in the hour of difficulty.
Answer:
(d) Never give up hope in the hour of difficulty.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

Question 1.
In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AGB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Answer:
∠AOC = ∠AOB + ∠BOC (Adjacent angles)
∴ ∠AQC = 60° + 30°
∴ ∠AOC = 90°
Now, 2 ∠ADC = ∠AOC (Theorem 10.8)
∴ ∠ADC = \(\frac{1}{2}\) ∠AOC
∴ ∠ADC = \(\frac{1}{2}\) × 90°
∴ ∠ADC = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Answer:

In the circle with centre O, chord AB is equal to radius PA.
∴ In ∆ PAB, PA = PB = AB
∆ PAB is an equilateral triangle.
∴ ∠ APB = 60°
Now, 2∠AYB = ∠APB (Theorem 10.8)
∴ ∠AYB = \(\frac{1}{2}\) ∠APB
= \(\frac{1}{2}\) × 60° = 30°
Quadrilateral AXBY is a cyclic quadrilateral.
∴ ∠X + ∠Y = 180° (Theorem 10.11)
∴ ∠X + 30°= 180°
∴ ∠X = 150°
Thus, the angle subtended by the chord at point X on the minor arc is 150° and the angle subtended by the chord at point Y on the major arc is 30°.

Question 3.
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Answer:
Here, reflex angle ∠POR = 2 × ∠PQR (Theorem 10.8)
∴ Reflex angle ∠POR = 2 × 100° = 200°
Now, ∠POR + Reflex angle ∠POR = 360°
∴ ∠POR + 200° = 360°
∴∠POR = 160°
In ∆ OPR. OP = OR (Radii)
∴ ∠OPR = ∠ORP
In ∆ OPR, ∠OPR + ∠ORP + ∠POR = 180°
∴ ∠OPR + ∠OPR + 160° = 180°
∴ 2∠OPR = 20°
∴ ∠OPR = 10°

Question 4.
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Answer:
In ∆ ABC, ∠ABC + ∠ACB + ∠BAC = 180°
∴ 69° + 31° + ∠BAC = 180°
∴ 100° + ∠BAC = 180°
∴ ∠BAC = 80°
Now, ∠BDC = ∠BAC (Theorem 10.9)
∴ ∠BDC = 80°

Question 5.
In the given figure, A, B, C and D are four s points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Answer:
In ∆ CDE, ∠BEC is an exterior angle.
∴ ∠BEC = ∠ECD + ∠EDC
∴ 130° = 20° + ∠BDC
∴ ∠BDC = 110°
Now, ∠BAC = ∠BDC (Theorem 10.9)
∴ ∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral Whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Answer:

∠DAC = ∠DBC (Theorem 10.9)
∴ ∠DAC = 70°
∠BAD = ∠BAC + ∠DAC (Adjacent angles)
∴ ∠BAD = 30° + 70°
∴ ∠BAD = 100°
In cyclic quadrilateral ABCD,
∠ BAD + ∠BCD = 180° (Theorem 10.11)
∴ 100° + ∠ BCD = 180°
∴ ∠BCD = 80°
In ∆ ABC, if AB = BC, then ∠ BAC = ∠ BCA
∴ 30° = ∠BCA
∴ ∠BCA = 30°
∠BCD = ∠BCA + ∠ACD (Adjacent angles)
∴ 80° = 30° + ∠ACD
∴ ∠ACD = 50°
∴ ∠ECD = 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Answer:

The vertices of cyclic quadrilateral ABCD lie on a circle with centre O and AC and BD are diameters of the circle.
As AC is a diameter, ∠ABC = ∠ADC = 90° (Angle in a semicircle)
As BD is a diameter, ∠BCD = ∠BAD = 90° (Angle in a semicircle)
Thus, all the four angles, ∠BAD, ∠ABC, ∠BCD and ∠ADC of quadrilateral ABCD are right angles.
Hence, quadrilateral ABCD is a rectangle.

Question 8.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer:

In trapezium ABCD, AB || CD and AD = BC.
Draw AM ⊥ CD and BN ⊥ CD, where M and N are points on CD.
In ∆ AMD and ∆ BNC,
∠AMD = ∠BNC (Right angles)
Hypotenuse AD = Hypotenuse BC (Given)
AM = BN (Distance between parallel lines)
∴ By RHS rule, ∆ AMD ≅ ∆ BNC
∴ ∠ADM = ∠BCN
∴ ∠ADC = ∠BCD
Now, AB || CD and AD is their transversal.
∴ ∠BAD + ∠ADC = 180° (Interior angles on the same side of transversal)
∴ ∠ BAD + ∠BCD = 180°
Thus, in quadrilateral ABCD, ∠A + ∠C = 180°.
Hence, ABCD is a cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
Answer:

∠ACP and ∠ABP are angles in the same segment.
∴ ∠ACP = ∠ABP (Theorem 10.9) …………… (1)
∠QCD and ∠QBD are angles in the same segment.
∴ ∠QCD = ∠QBD (Theorem 10.9) …………….. (2)
Now, ∠ABP and ∠QBD are vertically opposite angles.
∴ ∠ABP = ∠QBD ………………… (3)
From (1), (2) and (3),
∠ACP = ∠QCD

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer:
Circles are drawn taking sides AB and AC of ∆ ABC as diameters. These circles intersect each other at points A and P.
Draw common chord AP.
Since AB is a diameter, ∠APB is an angle in a semicircle.
∴ ∠APB = 90°
Since, AC is a diameter, ∠APC is an angle in a semicircle.
∴ ∠APC = 90°
Then, ∠APB + ∠APC = 90° + 90° = 180°
∠APB and ∠APC are adjacent angles with common arm AP and their sum is 180°.
∴ ∠APB and ∠APC form a linear pair.
Hence, the point of intersection of the circles with two sides of a triangle as diameters lies on the third side of the triangle.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC., Prove that ∠CAD = ∠CBD.
Answer:

In figure (1), line segment AC subtends equal angles at two points B and D lying on the same side of AC. Hence, by theorem 10.10, all the four points lie on the same circle.
Now, ∠CAD and ∠CBD are angles in the same segment.
∴ ∠CAD = ∠CBD (Theorem 10.9)
In figure (2), in quadrilateral ABCD,
∠B = ∠D = 90°.
∴ ∠B + ∠D = 180°
Hence, ABCD is a cyclic quadrilateral.
Again, ∠CAD and ∠CBD are angles in the same segment.
∴ ∠CAD = ∠CBD (Theorem 10.9)

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Answer:
Suppose ABCD is a cyclic parallelogram.
ABCD is a cyclic quadrilateral! .
∴ ∠A + ∠C = 180°
and ∠ B + ∠ D = 180° …….. (1)
ABCD is a parallelogram.
∴ ∠A = ∠C and ∠B = ∠D ……….. (2)
From (1) and (2),
∠A = ∠B = ∠C = ∠D = 90°
Thus, all the angles of quadrilateral ABCD are right angles.
Hence, ABCD is a rectangle.
Thus, a cyclic parallelogram is a rectangle.

Punjab State Board PSEB 9th Class English Book Solutions English Note-Making & Messages Exercise Questions and Answers, Notes.

PSEB 9th Class English Note-Making & Messages

Note-making

का अर्थ है किसी पैरे की मुख्य बातों को संक्षिप्त और साफ-सुथरे ढंग से प्रस्तुत करना। अच्छे Notes में निम्नलिखित विशेषताएं होती है –
1. वे संक्षिप्त होते हैं।

2. केवल प्रासंगिक बातें ही उनमें दी जाती हैं।

3. केवल शब्दों या वाक्यांशों का प्रयोग ही किया जाता है। पूरे वाक्यों की आमतौर पर आवश्यकता नहीं होती। अन्य शब्दों में हम कह सकते हैं कि Notes बनाते समय प्रयुक्त भाषा व्याकरण की दृष्टि से पूरी तरह सही नहीं भी हो सकती।

4. सूचना को सूचीबद्ध ढंग से प्रस्तुत किया जाता है। इसे विभाजित व उपविभाजित किया जाता है। विभाजन निम्न प्रकार से हो सकता है –
मुख्य खण्ड : 1, 2, 3, इत्यादि।
उपखण्ड : a, b, c, इत्यादि।

Passage 1:

If the young students in schools and colleges do not learn discipline, they will never be able to extract’ obedience from others in society. In fact, society will never accept them as persons fit for any responsible position in life. A school or college without discipline can never impart? suitable education to students.

Such a school or college is no better than a factory that turns out imperfect’ men and women. Sense of discipline plays a very important part in the playground and the battlefield. A disciplined team is likely to win the match in spite of its weakness but a very good team may not fare well for want of discipline. The rule of discipline equally applies to soldiers in the battlefield.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Need for discipline in schools and colleges for good education.
2. Indisciplined students fail’ to win any respect or position later on in their life.
3. Importance of discipline, for players on the playground
4. Importance of discipline, for soldiers in the battlefield.

Passage 2

Early rising leads to health and happiness. The man who rises late, can have little rest in the course of the day. Anyone who lies in bed late is compelled to work till a late hour in the evening. He has to go without the morning exercise which is so necessary for his health. In spite of all efforts’, his work will not produce as good results as that of the early riser. The reason for this is that he cannot take advantage of the refreshing hours in the morning.

Some people say that the quiet hour of midnight is the best time for working. Several great thinkers say that they can write best only when they burn the midnight oil. Yet it is true to say that few men have a clear brain at midnight when the body needs rest and sleep. Those who work at that time soon ruin their health. Bad health must, in the long run, have a bad effect on the quality of their work.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Advantages of early rising :
(i) health
(ii) Disadvantages

2. Disadvantages of late rising
(i) work till late in the evening
(ii) go without morning exercise
(iii) work not done properly.

(3) (i) Burning midnight oil bad for health.
(ii) Bad health, poor quality of our work.

Passage 3

Games, though essential, should not become the be-all and end-all of student life. Generally, the sportsmen waste too much time on them, and fail in their examinations. One must never devote more than an hour to sports and after that, one should not even think about them. Again, if a player plays a game rashly’, there is every danger of breaking bones.

If it is played without the spirit of sportsmanship, it can lead to bad blood and quarrels. In some of the colleges, there is a tradition that if the visiting team is winning a match, the home team plays foul, picks a quarrel and breaks the bones of the visitors. But in spite of these minor defects, sports are very useful in keeping the students busy and in developing their personalities.

India expects its citizens to have the qualities of true sportsmen. If we all acquire these qualities, there will be no narrow-mindedness, no corruption, and no injustice. There will be independence in the real sense of the word.

Read the given passage carefully and make notes on it in points only, using headings and sub-headings.

1. Sports essential for students, but not the be-all and end-all.
2. Wasting too much time → failure in examinations.
3. Playing rashly → breaking of bones.
4. Lack of sportsmanship → quarrels between teams.
5. True sportsmanship can end narrow – mindedness, corruption and injustice.

Passage 4

Of all amusements which can possibly be imagined for a hard-working man after his daily toil, there is nothing like reading an entertaining book. It calls for no bodily exertion of which he has had enough. It relieves his home of its dullness. It transports him to a livelier and more interesting scene, and while he enjoys himself there, he may forget the evils of the present moment.

It accompanies him to his next day’s work and if the book he has been reading be anything above the very idlest and the dullest, it gives him something to think about besides the drudgery of his everyday occupation. If I were to pray for a taste which should stand me in good stead under every variety of circumstances and be. a source of happiness and cheerfulness through life, it would be a taste for reading.

Give a man this taste,.and the means of gratifying it, and you can hardly fail to make him happy unless indeed you put into his hand a most perverse selection of books.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Reading of an interesting book a good diversion after the day’s hard work.
2. (i) Removes dullness of home,
(ii) Transports one into a livelier world.
3. (i) A good book food for thought.
(ii) Stands in good srcad under every circumstance.
4. Taste for reading, a source of great happiness.

Passage -5

English is important not because a number of people know it in India, although it is a factor to be remembered. It is not important because it is the language of Milton and Shakespeare, although that has to be considered. English is important because it is the major window for us on the modern world. And we dare not close that window. If we close it, we imperil our future. We think of Industrialisation, scientific development, research and technology.

But every door of modern knowledge will be closed if we do not have one or more foreign languages. We need not have English : we can have Russian, French or German, if you like, but obviously it is infinitely simpler for us to deal with a language which we know than to shift over to Russian, French or German which will be a tremendous job.
Certainly, we want to learn foreign languages, because we deal with the people of those languages in business, trade and science. So in the present stage of our development, we cannot go ahead without English and other foreign languages.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Mistakes do little harm

• when admitted.
• set tight before they a do any carnage

2. Delay in admitting mistakes

• harmful for the task in hand.
• harmful for the reputation.

3. Person who admits his mistakes

• is liked by everybody
• wins the confidence and respect of others.

4. Person who hides his mistakes

• is considered a fool.
• nobody likes him

Passage – 7

Teachers have a great responsibility at this time when our society is undergoing transformation. The future of the teaching profession in India will depend on the decision which the teachers take on vital questions relating to social change. In normal times, when society is comparatively more stable, the teachers’ primary task is transmitting culture. But in a period of transition, like the one through which we are passing, they have sometimes to set aside the culture in which they live, make a proper appraisal9 of it, pick out its salient features and reinterpret them for the new generation. The oncoming generations can rise to a high level of wisdom and cultivation only when teachers guide them carefully during this period of change.

Read the given passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
A. Teachers’ role in normal times :

• transmitting culture.

B. Modern rimes :

• not normal
• a period of transition.

C. teacher roles:

• proper appraisal of old culture
• pick out its salient features
•  re-interpreting them for future generations.

Passage – 8
Each one of us must realize that the only future for India and her people is one of tolerance and co-operation, which have been the basis of our culture for ages past. We have laid down in our constitution that Indians a Secular State. This does not mean irreligion. It means equal respect for all faiths and equal opportunities for those who profess different faiths. We have, therefore, always to keep in mind this vital aspect of our culture which is also of the highest importance in India today. Those who put up barriers between one Indian and another and who promote disruptive tendencies do not serve the cause of India and her culture. They weaken us at home and discredit us abroad.
Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer
(A) Tolerance and co-operation

• basis of our past culture
• pillars of our future

(B) Secularism

• equal respect for all faiths
• equal opportunities
• of highest importance in present-day India.

(C) Disruptive tendencies

• serve no cause
• discredit the country

Passage – 9

As a result of a long series of discoveries, mans life has been altered more radically and more rapidly during the last one hundred and fifty years than during the whole of the preceding two thousand years. In what ways does this alteration chiefly show itself ? In the first place, most of the external enemies to which our species in the past had been exposed are either overcome or are in a fair way to being overcome.

Look back over mans life in the past and you cannot but realize what sordid, meagre, frightened affair it must have been. His crops and, therefore, his livelihood have been at the mercy of forces which he could neither understand nor control; forces of fire and flood, of earthquake and drought; his communities were swept by pestilence and famine; and with the sweat of his brow, he wrung meager sustenance from nature. Today, thanks to science, all these enemies to man’s well-being have either disappeared or have been reduced to comparative impotence.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer
1. Discoveries of science during the
2. Man’s life completely changed.
3. External enemies overpowered :

• floods and fires
• earthquakes
• famines.
• pestilence

4. Now getting one’s livelihood not so difficult as it used to bo’

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Answer:

The circle with centre O and radius 5 cm intersects the circle with centre P and radius 3 cm at points A and B.
Hence, AB is their common chord.
Then, OP = 4 cm (Given),
OA = 5 cm and PA = 3 cm.
In ∆ OAP, OA² = 5² = 25 and
OP² + AP² = 4² + 3² = 16 + 9 = 25
Thus, in ∆ OAP, OA² = OP² + AP²
∴ ∆ OAP is a right triangle in which ∠OPA is a right angle and OA is the hypotenuse.
Thus, in the circle with centre O, OP is perpendicular from centre O to chord AB.
∴ OP bisects AB.
AB = 2PA = 2 × 3 = 6 cm
Thus, the length of the common chord is 6 cm.

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answer:

In the circle with centre O, equal chords AB and CD intersect at E.
Draw OM ⊥ AB and ON ⊥ CD.
∴ AM = BM = \(\frac{1}{2}\)AB and CN = DN = \(\frac{1}{2}\)CD.
But, AB = CD
∴AM = BM = CN = DN …………….. (1)
Chords AB and CD, being equal, are equidistant from the centre.
∴ OM = ON
In ∆ OME and ∆ ONE,
∠OME = ∠ONE (Right angles)
OE = OE (Common)
OM = ON
By RHS rule, ∆ OME ≅ ∆ ONE
∴ME = EN (CPCT) ……………… (2)
From (1) and (2),
AM + ME = CN + NE
∴ AE = CE
Similarly, BM – ME = DN – NE
∴ BE = DE
Thus, if two equal chords of a circle intersect within the circle, the segments of one chord are equal to corresponding segments of the other chord.

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer:
As the data of example 2 and example 3 are same, we use the proof of example 2 up to the required stage and do not repeat it here.
In example 2, we proved that,
∆ OME ≅ ∆ ONE ,
∴ ∠ OEM = ∠ OEN
∴ ∠ OEA = ∠ OEC
Thus, the line joining the point of intersection of two equal chords of a circle to the centre makes equal angles with the chords.

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see the given figure).

Answer:

From centre O, draw perpendicular OM to line AD.
In the outer circle, OM is the perpendicular drawn from centre O to chord AD.
Hence, M is the midpoint of AD.
∴ MA = MD …………… (1)
In the inner circle, OM is the perpendicular drawn from centre O to chord BC.
Hence, M is the midpoint of BC.
∴ MB = MC ………….. (2)
Subtracting (2) from (1),
MA – MB = MD – MC
∴ AB = CD

Question 5.
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Answer:

Here, OR = OM = OS = 5 m (Radius of the circle) and RS = SM = 6 m.
In quadrilateral ORSM, OR = OM = 5 m and RS = SM = 6 m.
∴ Quadrilateral ORSM is a kite.
∴ It diagonal OS bisects the diagonal RM at right angles.
∴ ∠RKO = 90° ………………. (1)
OK is perpendicular from centre O to chord RM.
Hence, K is the midpoint of RM.
∴ RM = 2RK ………………… (2)
From centre O, draw perpendicular OL to chord RS.
∴ RL = \(\frac{1}{2}\)RS = \(\frac{1}{2}\) × 6 = 3 m
In ∆ RLO, ∠ L = 90°
∴ RO² = OL² + RL²
∴ 5² = OL² + 3²
∴ 25 = OL² + 9
∴ OL² = 16
∴ OL = 4 m
Now, area of ∆ ROS = \(\frac{1}{2}\) × RS × OL
= \(\frac{1}{2}\) × OS × RK [by (1)]
∴RS × OL = OS × RK
∴ 6 × 4 = 5 × RK
∴ 24 = 5 × RK
∴ RK = \(\frac{24}{5}\) = 4.8 m
Then, RM = 2RK [by (2)]
∴ RM = 2 × 4.8
∴ RM = 9.6 m
Thus, the distance between Reshma and Mandip is 9.6 m.

Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk with each other. Find the length of the string of each phone.
Answer:

Here, the circle with centre O represents the park and the points A, S and D represent the positions of Ankur, Syed and David respectively. Since Ankur, Syed and David are sitting at equal distances from the others, ∆ ASD is an equilateral triangle.

Then, drawing the perpendicular bisector of SD from its midpoint M, it will pass through O as well as A.
Suppose, SM = x m
∴ SD = 2SM = 2xm
Area of equilateral ∆ ASD = \(\frac{\sqrt{3}}{4}\) (side)²
∴ Area of equilateral ∆ ASD = \(\frac{\sqrt{3}}{4}\) × (2x)²
∴ Area of equilateral ∆ ASD = √3x² …………. (1)
In ∆ OMS, ∠M = 90°
∴ OM² = OS² – SM² = (20)² – (x)² = 400 – x²
∴ OM = \(\sqrt{400-x^{2}}\)
Now, area of ∆ OSD = \(\frac{1}{2}\) × SD × OM
∴ Area of ∆ OSD = \(\frac{1}{2}\) × 2x × \(\sqrt{400-x^{2}}\)
∴ Area of ∆ OSD = x\(\sqrt{400-x^{2}}\) …………….. (2)
Here, ∆ OAS, ∆ OSD and ∆ ODA are congruent triangles.
Area of ∆ ASD = Area of ∆ OAS + Area of ∆ OSD + Area of ∆ ODA
∴ Area of ∆ ASD = 3 × Area of ∆ OSD
∴ √3 ∙ x<sup2 = 3 × x\(\sqrt{400-x^{2}}\)
∴x = √3 ∙ \(\sqrt{400-x^{2}}\)
∴ x² = 3(400 – x²2)
∴ x²= 1200 – 3x²
∴ 4x² = 1200
∴x² = 300
∴x= 10 √3
SD = 2x = 2 × 10 √3 = 20 √3 m
Thus, the length of the string of each phone is 20 √3m.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Answer:

Thus, given a pair of circles, the maximum number of common points they have is 2.

Question 2.
Suppose you are given a circle. Give a construction to find its centre.
Answer:

• In the given circle, draw chords AB and BC with one endpoint B in common.
• Draw l-the perpendicular bisector of AB and m-the perpendicular bisector of BC.
• Let l and m intersect at O.
• Then, O is the centre of the given circle.

Note: Here, any two chords without an end-point in common can be drawn.

Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Answer:

Here, two circles with centre O and P intersect each other at points A and B.
AB and OP intersect at M.
In ∆ OAP and ∆ OBR
OA = OB (Radii of the circle with centre O)
PA = PB (Radii of the circle with centre P).
OP = OP (Common)
∴ ∆ OAP ≅ ∆ OBP (SSS rule)
∴ ∠ AOP = ∠ BOP (CPCT)
∴ ∠ AOM = ∠BOM
Now, in ∆ AOM and ∆ BOM,
AO = BO (Radii of the circle)
∠ AOM = ∠ BOM
OM = OM (Common)
∴ ∆ AOM = ∆ BOM (SAS rule)
∴ AM = BM and ∠ AMO = ∠ BMO (CPCT)
But, ∠AMO + ∠BMO = 180° (Linear pair)
∴ ∠ AMO = ∠ BMO = \(\frac{180^{\circ}}{2}\) = 90°
Thus, line OM is the perpendicular bisector of AB.
Hence, line OP is the perpendicular bisector of AB.
Thus, the centres O and P of the circle intersecting in points A and B lie on the perpendicular bisector of common chord AB.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.2

Question 1.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Answer:

Two circles with centres O and P are congruent. Moreover, chord AB of the circle with centre
O and chord CD of the circle with centre P are congruent.
In ∆ OAB and ∆ PCD,
OA = PC and OB = PD (Radii of congruent circles)
And, AB = CD (Given)
∴ ∆ OAB ≅ ∆ PCD (SSS rule)
∴ ∠AOB = ∠ CPD
Thus, equal chords of congruent circles subtend equal angles at their centres.

Question 2.
Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Answer:

Two circles with centres O and P are congruent. Moreover, ∠ AOB subtended by chord AB of the circle with centre O and ∠CPD subtended by chord CD of the circle with centre P at their respective centres are equal.
In ∆ OAB and ∆ PCD,
OA = PC and OB = PD (Radii of congruent circles)
And, ∠AOB = ∠CPD (Given)
∴ ∆ OAB ≅ ∆ PCD (SAS rule)
∴ AB = CD
Thus, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.1

Question 1.
Fill in the blanks :
(i) The centre of a circle lies in ……………………….. of the circle, (exterior/interior)
Answer:
interior

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ………………….. of the circle, (exterior/interior)
Answer:
exterior

(iii) The longest chord of a circle is a ………………………. of the circle.
Answer:
diameter

(iv) An arc is a ……………….. when its ends are the ends of a diameter.
Answer:
semicircle

(v) Segment of a circle is the region between an arc and …………………………… of the circle.
Answer:
a chord

(vi) A circle divides the plane, on which it lies, in ………………………….. parts.
Answer:
three

Question 2.
Write True or False. Give reasons for your answers.
(i ) Line segment joining the centre to any point on the circle is a radius of the circle.
Answer:
The given statement is true, because according to the definition of a radius, a line segment joining the centre to any point on the circle is a radius of the circle.

(ii) A circle has only finite number of equal chords.
Answer:
The given statement is false, because a circle has infinitely many equal chords, e.g., all the diameters of a circle are chords and they are all equal and uncountable.

(iii) If a circle is divided into three equal arcs, each is a major arc.
Answer:
The given statement is false, because if a circle is divided into three equal parts, each part is a minor arc.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
Answer:
The given statement is true, because a chord of a circle which is twice as long as its radius passes through the centre of the circle and a chord passing through the centre is called a diameter of the circle.

(v) Sector is the region between the chord and its corresponding arc.
Answer:
The given statement is false, because the region between a chord an corresponding arc is called a segment, not a sector.

(vi) A circle is a plane figure.
Answer:
The given statement is true, because circle is a collection of all the points in a plane which are at a fixed distance from a fixed point in the plane.

Punjab State Board PSEB 9th Class English Book Solutions English Vocabulary Formation of Words Exercise Questions and Answers, Notes.

PSEB 9th Class English Vocabulary Formation of Words

A. Compound Words

किन्हीं दो या दो से अधिक शब्दों के मेल से बने नये शब्द को Compound word कहते हैं; जैसे book + shop = bookshop grand + father = grandfather कुछ Compound words किन्हीं दो या दो से अधिक शब्दों में योजक चिन्ह (-) लगाकर जोड़ने से भी ; the Sister-in-law; ready-to-serve; air-conditioned.

Match the words in column A with the words in column B to make Compound words :

A —- B
basket —- wife
grand —- life
sun —- yard
milk —- ball
houses —- book
vine —- post
lamp —- glasses
wild —- maid
over —- worked
world —- father
text —- grocer
green —- wide
Answer:
basketball; grandfather; sunglasses; milkmaid; housewife; vineyard; lamp post; wildlife; overworked; worldwide; textbook; greengrocer.

Rewrite the word by inserting a hyphen (-), if required :

fiftynine — headache
easygoing — welloiled
preschool — uptodate
mothertobe — selfstudy
busybody — inlaws
highway — incometax
snowbound — waterbased
snowstorm — household
Answer:
fifty-nine; easy-going; pre-school; mother-to-be, busybody; highway; snowbound; snowstorm; headache; well-oiled; up-to-date; self-study, in-laws; income tax; water-based; household.

Choose suitable compound words from the given list to complete the sentences :

world-famous ; oil-based ; handmade ; bulletproof ; air-conditioned ; absent-minded ; eyesight ; downtown.

1. In summer, many people like to travel by ……………. buses.
2. Vikram Seth is a ……………. writer.
3. Chaman Lal got his house painted with ………….. paints.
4. Where did you buy this ……………. paper ?
5. He goes ……………. every week to buy his grocery.
6. Get your ……………. checked; I think you need glasses.
7. The policeman was saved because he was wearing a …………… jacket.
8. My father is becoming ……………; he never pays his bills on time these days.
Answer:
1. air-conditioned
2. world-famous
3. oil-based
4. handmade
5. downtown
6. eyesight
7. bulletproof
8. absent-minded.

Formation Of Words

शब्दों को उनके प्रयोग और रूप के अनुसार भिन्न-भिन्न वर्गों में बांटा जा सकता है; जैसे

1. Noun
2. Adjective
3. Verb
4. Adverb
अनेक शब्द ऐसे होते हैं जिनके एक रूप को हम दूसरे रूप में बदल सकते हैं; जैसे

राब्द के एक रूप को दूसरे रूप में बदलने (अर्थात् शब्द-रचना करने) का कोई विशेष नियम नहीं होता है। अपने शब्द-ज्ञान को बढ़ाने के लिए हमें शब्दों के विभिन्न रूपों को कण्ठस्थ ही करना पड़ता है।

B. Prefixes & Suffixes

Prefix (उपसर्ग) उस शब्दांश को कहा जाता है जिसे किसी शब्द के आरम्भ में जोड़ने से एक नया शब्द बन जाता है। Prefix का अपना कोई अर्थ नहीं होता, किन्तु वह दूसरे शब्द से जुड़ कर उसके अर्थ में परिवर्तन कर देता है; जैसे happy के आरम्भ में un उपसर्ग लगाने से एक नया शब्द unhappy बन जाता है। take के आरम्भ में mis उपसर्ग लगाने से एक नया शब्द mistake बन जाता है।
Suffix (प्रत्यय) उस शब्दांश को कहा जाता है जिसे किसी शब्द के अन्त में जोड़ने से एक नया शब्द बन जाता है।
Suffix का अपना कोई अर्थ नहीं होता है, किन्तु वह दूसरे शब्द में जुड़ कर उसके अर्थ में परिवर्तन कर देता है; जैसे
king के अन्त में dom प्रत्यय लगाने से एक नया शब्द kingdom बन जाता है।
quarrel के अन्त में some प्रत्यय लगाने से एक नया शब्द quarrelsome बन जाता है।

Complete the sentences using the correct form of the words given in the brackets :

1. There were a lot of games for …………… at my cousin’s party. (amuse)
2. After the ………….. of the bridge, the labourers will be sent to some other place. (complete)
3. She is learning French in …………… to English and Punjabi. (add)
4. He was asked to show his passport for ………… . (verify)
5. Due to the ………….., the wall of the house collapsed. (seep)
6. I am going to write a letter to the …………….. of that newspaper. (edit)
7. Many children receive awards for their …………… on Republic Day every year. (brave)
8. ………… classes are held in Adarsh Colony to train the needy women. (sew)
9. Some people kill animals and birds for …..
10. The main …………….. of some tribals in Rajasthan is camel breeding. (occupy)
Answer:
1. amusement
2. completion
3. addition
4. verification
5. seepage
6. editor
7. bravery
8. Sewing
9. pleasure
10. occupation.

Match the Verbs under column A with their Nouns under column B :

A —- B
vibrate —- burial
permit —- preference
prosper —- actor
prefer—- permission
act —- settlement
employ —- relation
relate —- vibration
settle —- authority
bury —- employee
authorize —- prosperity
Answer:
vibrate—vibration; permit—permission; prosper—prosperity; prefer—preference; act—actor; employ—employee; relate—relation; settle—settlement; bury—burial; authorize—authority.

Form Nouns from the following Verbs and use them in sentences of your own :

preach, create, appear, arrive, enjoy, apologize, develop, meet, deliver, memorize.
Answer
1. Preacher — Satguru Ram Singh Ji was a great preacher.
2. Creation — The cake was a delicious creation of sponge, cream and fruit.
3. Appearance — Appearances are often deceptive.
4. Arrival — Are you sure about the late arrival of the train ?
5. Enjoyment — Gardening is one of my chief enjoyments.
6. Apology — I owe you an apology.
7. Development — He bought the land for development.
8. Meeting — What was decided at Friday meeting ?
9. Delivery — Your order is ready for delivery.
10. Memory — He has a good memory for dates.

Fill in the correct words in the blanks with the help of words given in the brackets :

1. We will …………… our house by growing flowering plants. (beauty)
2. Don’t …………… your life by going near the fire. (danger)
3. In a few years, the government is likely to …………….. several villages. (electricity)
4. She couldn’t …………… her stay abroad for so many months. (justice)
5. You can’t ………….. me with your lies any more. (fool)
6. My friends ………….. playing in the sun even in the summer. (joy)
7. Can you ……………. the bad points of smoking ? (list)
8. I won’t ……………. you by talking again about that accident. (terror)
Answer:
1. beautify
2. endanger
3. electrify
4. justify
5. befool
6. enjoy
7. enlist
8. terrify.

Match the Nouns in column A with the Adjectives from column B :

Α —- B
expense — yearly
year — intelligent
economy — defensive
edit — exemplary
flower — needful
example — floral
defence — editorial
intelligence — economical
need — expensive
Answer:
expense — expensive; year — yearly; economy — economical; edit — editorial; flower — floral; example — exemplary; defence — defensive; intelligence — intelligent; need — needful.

Use a prefix / suffix with the words given in the brackets. Make necessary changes in the words, if required :

1. There are many …………… hotels in Mumbai. (luxury)
2. A ………….. function was held on the eve of Diwali. (colour)
3. Is it ………….. to travel by. air ? (economy)
4. The stay in Singapore was very …………….. (expense)
5. Sunil acts quite …………….. at times. (child)
6. The discussion took place in a …………….. atmosphere. (friend)
7. I am going to make my ……………. trip to Varanasi in June. (year)
8. It turned very …………….. in the evening. (dust)
9. The money will be given to some …………….. persons. (need)
10. Abdul is a very …………. person; he works 14 hours a day. (industry)
Answer:
1. luxurious
2. colourful
3. economical
4. expensive
5. childishly
6. friendly
7. yearly
8. dusty
9. needy
10. industrious.

Form Adjectives from the following Nouns :

accident; adventure; abuse; east; fault; hand; guilt; might; difference; example
Answer:
accident – accidental; adventure – adventurous; abuse – abusive; east – eastern; fault – faulty; hand – handy; guilt – guilty; might – mighty; difference – different; example – exemplary.

Form Nouns by adding the suffixes -ity, -th, -dom, -ness, -ence to the words given in the brackets :

1. Many areas of Bihar are known for their …………….. (backward)
2. I felt very uncomfortable in Chennai because of the (humid)
3. “What’s the …………… of your turban ?” the foreigner asked. (long)
4. Because of her ……………. she could not go there. (ill)
5. Nelson Mandela went to jail for the ……………. of his people. (free)
6. Is there any …………….. of the train coming late ? (possible)
7. There is ………….. in her behaviour. (warm)
8. Ramanand Jewellers are known for the …………….. of their gold. (pure)
9. No one spoke in the …………….. of the police. (present)
10. His ………….. was felt by all. (absent)
Answer:
1. backwardness
2. humidity
3. length
4. illness
5. freedom
6. possibility.
7. warmth
8. purity
9. presence
10. absence.

Form Verbs from the following Adjectives :

able, broad, black, deep, false, popularsad, sick, glorious, minimum, deep
Answer:
able — enable; broad — broaden; black — blacken; deep — deepen; false — falsify; popular — popularize; sad — sadden; sick — sicken; glorious — glorify; minimum — minimize.

Add suffixes / prefixes to the words given in the brackets and write them in the space provided :

1. Go to the Rose Garden. The roses will …………….. (glad) you.
2. You can …………… (rich) your knowledge by reading good books.
3. Some children cannot ……………. (different) between p and b.
4. I think the mystery will further …………….. (deep) in the novel I am reading.
5. Buy a cycle; it will …………. (able) you to reach your school in time.
6. I am trying to ….. (minimum) my expenses.
7. The computer will …………… (right) the error if you give it the correct command.
8. Sukhbir will like to …………….. (special) in medicine.
Answer:
1. gladden
2. enrich
3. differentiate
4. deepen
5. enable
6. minimize
7. rectify
8. specialize.

Match the Verbs in column A with the Adjectives in Column B :

A — B

agree — admirable
admire — selective
select — collective
doubt — helpful
collect — removable
change — agreeable
remove — changeable
help — doubtful
Answer:
agree – agreeable; admire – admirable; select – selective; doubt – doubtful; collect – collective; change – changeable; remove – removable; help – helpful.

Form Adverbs from the following Adjectives and use them in sentences of your own :

brief, broad, bitte, calm, easy, frequent, generous occasional, peaceful
Answer:
1. Briefly—Answer these questions briefly.
2. Broadly — Broadly speaking, I agree with you.
3. Bitterly — He was talking bitterly to his wife.
4. Calmly — He listened to my whole problem calmly.
5. Easily — I could solve all the questions easily.
6. Frequently — Buses run frequently from the station to the airport.
7. Generously — He helps the poor generously.
8. Occasionally — He comes here only occasionally.
9. Peacefully — Men should learn to live peacefully with each other.

Match the words in column A with their Abstract Nouns in column B :

A — B
beggar — brotherliness
brother — earldom
chemist — membership
earl — begging
friend — inspection
inspector — patriotism
member — friendship
patron — chemistry
patriot — widowhood
widow — patronage
Answer:
beggar — begging; brother—brotherliness; chemist — chemistry; earl — earldom; friend — friendship; inspector — inspection; member — membership; patron — patronage; patriot — patriotism; widow — widowhood.

Form Abstract Nouns from the following words and use them in sentences of your own :

act, agent, child, infant, mother, hero, partner, recruit, move

Answer:
1. Action – What is your next plan of action ?
2. Agency – There is an advertisement agency near our house.
3. Childhood – We spent our childhood in great joy.
4. Infancy – Many poor children die in their infancy.
5. Motherhood – There is no joy like the joy of motherhood.
6. Heroism – Our soldiers showed great heroism during the war.
7. Partnership – I have a partnership in this firm.
8. Recruitment -The recruitment of new workers will take place next month.
9. Movement – The movement of goods from one place to the other is a big problem.

Write the opposite of the statements given below. Use the prefixes ir-, un-, in-, im-, il-, diswith the italicized words :

1. Mr. Reddy is known for making logical statements.
2. The speaker made several relevant points in his speech.
3. The fire-fighters were able to rescue the child trapped inside the house.
4. Savita is a very mature person.
5. Is it legal to have two wives ?
6. Some students are regular in attending classes.
7. Your handwriting is quite legible.
8. My father likes boys who have long hair.
9. Quite a lot of people are literate in any colony.
10. The foreigners were very polite to me.
Answer:
1. Mr. Reddy is known for making illogical statements.
2. The’ speaker made several irrelevant points in his speech.
3. The fire-fighters were unable to rescue the child trapped inside the house.
4. Savita is a very immature person.
5. Is it illegal to have two wives?
6. Some students are irregular in attending classes.
7. Your handwriting is quite illegible.
8. My father dislikes boys who have long hair.
9. Quite a lot of people are illiterate in any colony.
10. The foreigners were very impolite to me.

Use fore-, pre-, mono-, anti-, post-, out-, ex-, under- with the words given in the brackets and use them to complete the sentences :

1. It is proved that our …………….. (father) were monkeys.
2. To avoid illness, take ………………. (malaria) tablets in the rainy season.
3. Soon ………………. (rail) will be introduced in many big, crowded cities in India.
4. Mrs. Kapoor is so ……………. (spoken) that a few people like to talk to her.
5. The …………….. (independence) progress is quite remarkable in our country.
6. The …………….. (headmaster) of our school was the Chief Guest at the Annual Function.
7. The pilot was …………….. (warned) about the bad weather. ………….. (age) children are not allowed to see the A movies in cinema halls.
9. My three-year-old nephew is studying in a ………… …… (nursery) class.
10. ……………… (aircraft) guns are commonly used in wars.
Answer:
1. forefathers
2. anti-malaria
3. mono-rail
4. outspoken
5. post-independence
6. ex-headmaster
7. fore-warned
8. Underage
9. pre-nursery
10. Anti-aircraft.

Formation of Word By The Use Of Phefixes & Suffixes

(i) Forming Nouns From Verb :

(ix) Negative Prefixes:

in – : inactive, incomplete, inanimate, inhuman
dis – : disappear, dislike, disrespect
un – : unable, unkind, untie
im – : impossible. impolite, immature
ir – : irregular, irresponsible, irrelevant
il – : illegal, illegible, illiterate
mis – : misplaced, misfortune, mislead
mal – : malfunction, maladjustment, malpractice

(x) Prefixes That Denote Degree :

extra – : extracurricular, extraordinary
mini – : mini-skirt, mini-track
out – : outshine, outspoken, outshoot
over – : overdose, overdraw, overage
semi – : semi-darkness, semi-commercial, semi-liquid
sub – : sub-region, sub-depot,
super – : supernatural, superman
under – : underage, underhand, undergraduate

(xi) Prefixes That Express Time of Sequence:

ex -: ex-principal, ex-inspector
fore – : forewarn, forecast, forefather
post – : post-independence, post-haste
pre – : preoccupy, pre-eminent
re – : recast, remarry recall

(xii) Prefixes That Express Number:

bi – : bicycle, hi-yearly
mono – : mono-drama, mono-type, mono-rail
tri -: tripod, tri-partition, tricycle

(xiii) Prefixes That Express Attitude:

anti – : antiseptic, anti-tank
co – : co-accused, co-education
counter – : counrâpan, counterbalance
pro – : pro-establishment

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 9 Areas of Parallelograms and Triangles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 9 Areas of Parallelograms and Triangles 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.
Area of a parallelogram = ………………….
A. \(\frac{1}{2}\) × base × corresponding altitude
B. \(\frac{1}{2}\) × the product of diagonals
C. base × corresponding altitude
D. \(\frac{1}{2}\) × the product of adjacent sides.
Answer:
C. base × corresponding altitude

Question 2.
Area of a triangle = ……………………
A. base × corresponding altitude
B. base + corresponding altitude
C. \(\frac{1}{2}\) × base × corresponding altitude
D. 2 × base × corresponding altitude
Answer:
C. \(\frac{1}{2}\) × base × corresponding altitude

Question 3.
ABCD is a rectangle. If AB = 10 cm and ar (ABCD) = 150 cm², then BC = ………………….. cm.
A. 7.5
B. 15
C. 30
D. 12
Answer:
B. 15

Question 4.
ABCD is a square. If ar (ABCD) = 36 cm², then AB = ………………… cm.
A. 18
B. 9
C. 6
D. 12
Answer:
C. 6

Question 5.
In ∆ ABC, BC = 10 cm and the length of altitude AD is 5 cm. Then, ar (ABC) = …………………. cm².
A. 50
B. 100
C. 25
D. 15
Answer:
C. 25

Question 6.
In ∆ ABC, AD is an altitude. If BC = 8 cm and ar (ABC) = 40 cm², then AD = …………………. cm.
A. 5
B. 10
C. 15
D. 20
Answer:
B. 10

Question 7.
In ∆ PQR, QM is an altitude and PR is the hypotenuse. If PR = 12 cm and QM = 6 cm, then ar (PQR) = ……………………. cm².
A. 18
B. 72
C. 36
D. 24
Answer:
C. 36

Question 8.
In ∆ XYZ, XZ is the hypotenuse. If XY = 8cm and YZ = 12 cm, then ar (XYZ) = ……………….. cm².
A. 20
B. 40
C. 96
D. 48
Answer:
D. 48

Question 9.
In parallelogram ABCD, AM is an altitude corresponding to base BC. If BC = 8 cm and AM = 6 cm, then ar (ABCD) = …………………. cm².
A. 48
B. 24
C. 12
D. 96
Answer:
A. 48

Question 10.
In parallelogram PQRS, QR = 10 cm and ar (PQRS) = 120 cm². Then, the length of altitude PM corresponding to base QR is ……………………… cm.
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12

Question 11.
For parallelogram ABCD, ar (ABCD) = 48 cm².
Then, ar (ABC) = …………………….. cm².
A. 96
B. 48
C. 24
D. 12
Answer:
C. 24

Question 12.
ABCD is a rhombus. If AC = 6 cm and BD = 9 cm, then ar (ABCD) = ………………….. cm².
A. 15
B. 7.5
C. 54
D. 27
Answer:
D. 27

Question 13.
PQRS is a rhombus. If ar (PQRS) = 40 cm² and PR = 8 cm, then QS = ………………….. cm.
A. 20
B. 10
C. 25
D. 40
Answer:
B. 10

Question 14.
In ∆ PQR, ∠Q = 90°, PQ = 5 cm and PR = 13 cm.
Then, ar (PQR) = …………………….. cm².
A. 15
B. 30
C. 45
D. 60
Answer:
B. 30

Question 15.
In ∆ ABC, P Q and R are the midpoints of AB, BC and CA respectively. If ar (ABC) = 32 cm²,
then ar (PQR) = ………………………. cm².
A. 128
B. 16
C. 8
D. 64
Answer:
C. 8

Question 16.
In ∆ ABC, P, Q and R are the midpoints of AB, BC and CA respectively. If ar (ABC) = 32 cm², then ar (PBCR) = ………………….. cm².
A. 10
B. 20
C. 30
D. 40
Answer:
C. 30

Question 17.
In ∆ ABC, P, Q and R are the midpoints of AB, BC and CA respectively. If ar (PBQR) = 36 cm², then ar (ABC) = ……………………….. cm².
A. 18
B. 36
C. 54
D. 72
Answer:
D. 72

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

Question 1.
Parallelogram ABCD and rectangle ABEF are on the, same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Rectangle ABEF is a parallelogram too.
Now, parallelograms ABCD and ABEF are on the same base AB and they have equal areas. Hence, they are between the same parallels FC and AB.
In ∆ AFD, ∠F, being an angle of rectangle ABEF, is a right angle and so, AD is the hypotenuse.
∴ AD > AF
∴ AD + AB > AF + AB
∴ 2 (AD + AB) > 2 (AF + AB)
∴ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF

Question 2.
In the given figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark : Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide AABC into n triangles of equal areas.]

Answer:
Here, in ∆ ABE, D is a point on BE such that BD = DE.
So, in ∆ ABE, D is the midpoint of BE and AD is a median.
∴ ar (ABD) = ar (ADE) ……………… (1)
Similarly, in A ADC, E is the midpoint of DC and AE is a median.
∴ ar (ADE) = ar (AEC) ……………. (2)
From (1) and (2),
ar (ABD) = ar (ADE) = ar (AEC)
Thus, in ∆ ABC, by joining the points of trisection of BC, i.e., D and E to vertex A, the triangle is divided into ∆ ABD, ∆ ADE and ∆ AEC which have the same area.

Now, the answer to the question which was left unanswered in the ‘Introduction’ is ‘Yes’. The manner in which Budhia divided her field, the area of all the three parts are equal.

Question 3.
In the given figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Answer:
Opposite sides of a parallelogram are equal.
∴ In parallelogram ABCD, AD = BC, in parallelogram DCFE, DE = CF and in parallelogram ABFE, AE = BF.
Now, in ∆ ADE and ∆ BCE
AD = BC, DE = CF and AE = BE
∴ By SSS rule, ∆ ADE = ∆ BCF
∴ ar (ADE) = ar (BCF)

Question 4.
In the given figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at E show that ar (BPC) = ar (DPQ).
[Hint: Join AC.]

Answer:
Join AC.
In parallelogram ABCD, BC || AD and BC = AD.
BC is produced to point Q such that AD = CQ.
Thus, AD = CQ and AD || CQ.
∴ Quadrilateral ACQD is a parallelogram.
Diagonals of a parallelogram divide it into four triangles of equal areas.
∴ ar (DPQ) = ar (DPA) = ar (APC) = ar (CPQ)
∴ ar (DPQ) = ar (APC) ……………. (1)
Now, ∆ APC and ∆ BPC are on the same base PC and between the same parallels PC and AB.
∴ ar (APC) = ar (BPC) ………….. (2)
From (1) and (2),
ar (BPC) = ar (DPQ)

Question 5.
In the given figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, show that
(i) ar (BDE) = \(\frac{1}{4}\)ar (ABC)
(ii) ar (BDE) = \(\frac{1}{2}\)ar (BAE)
(iii) ar (ABC) = 2ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v ) ar (BFE) = 2ar (FED)
(vi) ar (FED) = \(\frac{1}{8}\)ar (AFC)
[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]

Answer:
Join EC and AD.
In equilateral ∆ ABC, ∠ ACB = 60°
In equilateral ∆ BDE, ∠ DBE = 60°
∴ ∠ CBE = 60°
Thus, ∠ ACB = ∠ CBE
But, ∠ ACB and ∠ CBE are alternate angles formed by transversal BC of AC and BE and they are equal.
∴BE || AC .
Similarly, ∠ ABD = ∠ BDE = 60°
∴ DE || AB
Now, in ∆ ABC, D is the midpoint of BC.
Hence, AD is a median of ∆ ABC.
∴ ar (ADB) = ar (ADC) = \(\frac{1}{2}\)ar (ABC)
∆ ABC and AAEC are on the same base AC and between the same parallels AC and BE.
∴ ar (ABC) = ar (AEC)
∴ ar (ABC) = ar (ADC) + ar (EDC) + ar (AED) …………….. (1)
In ∆ EBC, ED is a median.
∴ ar (EDC) = ar (BDE) = \(\frac{1}{2}\)ar (EBC) ………………… (2)
∆ AED and ∆ BDE are on the same base DE and between the same parallels AB and DE.
∴ ar (AED) = ar (BDE) …………… (3)
From (1), (2) and (3),
ar (ABC) = \(\frac{1}{2}\)ar (ABC) + ar (BDE) + ar (BDE)
∴ ar (ABC) – \(\frac{1}{2}\) ar (ABC) = 2ar (BDE)
∴\(\frac{1}{2}\)ar (ABC) = 2ar (BDE)
∴ ar (BDE) = \(\frac{1}{4}\)ar (ABC) ….. Result (i)
∆ BAE and ∆ BCE are on the same base BE and between the same parallels BE and AC.
∴ ar (BAE) = ar (BCE) ……………. (4)
In ∆ BEC, ED is a median.
∴ ar (BDE) = \(\frac{1}{2}\)ar (BCE)
∴ ar (BDE) = \(\frac{1}{2}\)ar (BAE) [by (4)] ……. Result (ii)
The diagonals of trapezium ABED intersect at F.
∴ ar (AFD) = ar (BFE) ……………… (5)
The diagonals of trapezium ABEC intersect at F.
∴ ar (ABF) = ar (EFC) ………………….. (6)
In ∆ ABC, AD is a median. s
∴ ar (ABC) = 2ar (ADB) S
∴ ar (ABC) = 2[ar (ABF) + ar (AFD)l
∴ ar (ABC) = 2[ar (EFC) + ar (BFE)] [by (5) and (6)]
∴ ar (ABC) = 2ar (BEC) … Result (iii)
In trapezium ABED, AB || ED and diagonals intersect at F.
∴ ar (BFE) = ar (AFD) …….. Result (iv)
By result (i),
ar (BDE) = \(\frac{1}{4}\)ar (ABC)
∴ ar (BDE) = \(\frac{1}{4}\) 2ar (ABD)
∴ ar (BDE) = \(\frac{1}{2}\)ar (ABD)
∆ BDE and ∆ ABD have the common base s BD.
∴ Altitude on BD in ∆ BDE = \(\frac{1}{2}\) × altitude on BD in ∆ ABD.
Now, the altitude on base BD in ∆ BDE is the same as the altitude on base BF in ∆ BEF and the altitude on base BD in ∆ ABD is the same as the altitude on base FD in ∆ AFD.
∴ Altitude on base BF in ∆ BEF
= \(\frac{1}{2}\) × altitude on base FD in ∆ AFD.
But, ar (BFE) = ar (AFD) …Result (iv)]
∴ BF = 2 × FD
Now, in ∆ BFE and ∆ FED, the altitudes corresponding to base BF and FD respectively are the same.
∴ ar (BFE) = 2ar (FED) … Result (v)
Suppose, in ∆ ABD, the altitude on base BD = h.
∴ In ∆ AFC, the altitude on base FC = h.
Also, in ∆ BDE, the altitude on base BD = \(\frac{h}{2}\)
∴ In A FED, the altitude on base FD = \(\frac{h}{2}\).
Now, ar (FED) = \(\frac{1}{2}\) × FD × \(\frac{h}{2}\) = \(\frac{h \times \mathrm{FD}}{4}\).
∴ FD = \(\frac{4 {ar}(\mathrm{FED})}{h}\) ………….. (7)
and ar (AFC) = \(\frac{1}{2}\) × FC × h = \(\frac{h}{2}\) × FC
= \(\frac{h}{2}\) (CD + FD)
= \(\frac{h}{2}\) (BD + FD) [∵ BD = CD]
= \(\frac{h}{2}\) (BF + FD + FD)
= \(\frac{h}{2}\) (2FD + FD + FD) [∵ BF = 2FD]
= \(\frac{h}{2}\) × 4FD
∴ ar (AFC) = 2 × h × FD
= 2 × h × \(\frac{4 {ar}(\mathrm{FED})}{h}\) [by (7)]
∴ ar (AFC) = 8 ar (FED)
∴ ar (FED) = \(\frac{1}{8}\) ar (AFC) … Result (vi)

Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at E Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[(Hint: From A and C, draw perpendiculars to BD.]
Answer:

Draw AM ⊥ BD and CN ⊥ BD, where M and N are points on BD.
∴ ar (APB) × ar (CPD)
= (\(\frac{1}{2}\) × PB × AM) × (\(\frac{1}{2}\) × PD × CN)
= (\(\frac{1}{2}\) × PB × CN) × (\(\frac{1}{2}\) × PD × AM)
Thus, ar (APB) × ar (CPD) = ar (APD) × ar (BPC)

Question 7.
P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the midpoint of AR show that
(i) ar (PRQ) = \(\frac{1}{2}\) ar (ARC)
(ii) ar (RQC) = \(\frac{3}{8}\) ar (ABC)
(iii) ar (PBQ) = ar (ARC)
Answer:

In ∆ ABC, AQ and CP are medians. In ∆ APC, CR is a median, In ∆ APQ, QR is a median. In ∆ PBC, PQ is a median, In ∆ RBC, RQ is a median.

(i) ar (PRQ) = ar(ARQ) }
= \(\frac{1}{2}\)ar (APQ)
= \(\frac{1}{2}\)ar (BPQ)
= \(\frac{1}{2}\)ar(PQC)
= \(\frac{1}{2}\) ∙ \(\frac{1}{2}\)ar (PBC)
= \(\frac{1}{4}\)ar (PBC)
= \(\frac{1}{4}\) ∙ \(\frac{1}{2}\)ar (ABC)
= \(\frac{1}{8}\)ar (ABC)
\(\frac{1}{2}\)ar (ARC) = \(\frac{1}{2}\) ∙ \(\frac{1}{2}\)ar(APC)
= \(\frac{1}{4}\)ar (APC)
= \(\frac{1}{4}\) ∙ \(\frac{1}{2}\)ar (ABC)
= \(\frac{1}{8}\)ar (ABC)
∴ar (PRQ) = \(\frac{1}{2}\)ar (ARC)

(ii) ar(RQC) = ar(RBQ)
= ar (PBQ) + ar (PRQ)
= \(\frac{1}{2}\)ar (PBC) + \(\frac{1}{8}\)ar (ABC)
= \(\frac{1}{2}\) ∙ \(\frac{1}{2}\)ar (ABC) + \(\frac{1}{8}\)ar (ABC)
= \(\frac{1}{4}\)ar (ABC) + \(\frac{1}{8}\)ar (ABC)
= \(\frac{3}{8}\)ar (ABC)

(iii) ar (PBQ) = \(\frac{1}{2}\)ar (PBC) = \(\frac{1}{2}\) ∙ \(\frac{1}{2}\)ar (ABC)
= \(\frac{1}{4}\)ar (ABC)
ar (ARC) = \(\frac{1}{2}\)ar (APC) = \(\frac{1}{2}\) ∙ \(\frac{1}{2}\)ar (ABC)
= \(\frac{1}{4}\)ar (ABC)
∴ ar (PBQ) = ar (ARC)

Question 8.
In the given figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN „ are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:
(i) ∆ MBC S ∆ ABD
(ii) ar (BYXD) = 2ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) ∆ FCB ≅ ∆ ACE
( v ) ar (CYXE) = 2ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler j! proof of this theorem in Class X.

Answer:
(i) ∠ ABM = ∠ CBD = 90°
∴ ∠ABM + ∠ABC = ∠CBD + ∠ABC
∴ ∠ MBC = ∠ ABD
In ∆ MBC and ∆ ABD,
MB = AB, ∠ MBC = ∠ ABD and BC = BD
∴ By SAS rule, ∆ MBC ≅ ∆ ABD

(ii) ar (BYXD) = 2ar (ABD)
∴ ar (BYXD) = 2ar (MBC) [∆ MBC ≅ ∆ ABD]

(iii) ar (BYXD) = 2ar (ABD)
ar (ABMN) = 2ar (MBC)
But, ar (MBC) = ar (ABD)
∴ ar (BYXD) = ar (ABMN)

(iv) ∠ FCA = ∠ ECB = 90°
∴ ∠FCA + ∠ACB = ∠ECB + ∠ACB
∴ ∠FCB = ∠ACE
In ∆ FCB and ∆ ACE,
FC = AC, ∠ FCB = ∠ACE and CB = CE
∴By SAS rule, ∆ FCB ≅ ∆ ACE

(v) ar (CYXE) = 2ar (ACE)
∴ar (CYXE) = 2ar (FCB) [∵ ∆ FCB ≅ ∆ ACE]

(vi) ar (CYXE) = 2ar (FCB)
and ar (ACFG) = 2ar (FCB)
∴ ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (CYXE) + ar (BYXD)
∴ ar (BCED) = ar (ACFG) + ar (ABMN) [By result (iii) and (vi)]
∴ ar (BCED) = ar (ABMN) + ar (ACFG)