Class 9

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 Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 9th Class English Reading Comprehension Unseen Passages

(A) Passages From Grammar Book Note :

• In all the passages, questions have been changed according to new Pattern of Testing.
• Answers have been given at the end of this set.

Passage 1

During the winter of 1945, I lived for several months in a house in Brooklyn. It was not a shabby place, but a pleasantly furnished one. It was well-kept by its owners – two elderly sisters. Mr. Jones lived in the room next to mine. My room was the smallest in the house, his the largest, a nice big sunshiny room, which Mr. Jones never left. All his needs meal, shopping, laundry? – were attended to by the middle-aged landladies.

Also, he was not without visitors; on an average, half-dozen various persons, men and women, young and old, in-between visited him from early morning till late in the evening. He was not a drug dealer or a fortune-teller“; no, they just came to talk to him and apparently they made him small gifts of money for his conversation and advice. If not, he had no obvious? means to support himself. I never had a conversation with him, because I was out most of the time. He was a handsome man about forty.

Slenders, black-haired and with distinctive face; a pale, lean face, high cheek bones, and with a birthmark on his left cheek. He wore gold-rimmed glasses with black lenses, for he was blind and cripple too. He was always dressed in pressed dark grey or blue three-piece suit and a light coloured tie — as though he was to set off for work.

Choose the correct option to answer each question :

Question 1.
Jones earned his living by …………
(a) selling drugs.
(b) telling future
(c) giving advice to people
(d) by working in farms.
Answer:
(c) giving advice to people

Question 2.
Mr. Jones was looked after by ……………
(a) the landladies
(b) the visitors
(c) the author
(d) the landlords.
Answer:
(a) the landladies

Question 3.
……….. came to visit Mr. Jones.
(a) Old people
(b) Young people
(c) Poor people
(d) People of all ages.
Answer:
(d) People of all ages.

Question 4.
What did the landladies do for Mr. Jones ?
(a) They prepared meals for him.
(b) They did shopping for him.
(c) They washed his clothes.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
Mr Jones had a birthmark on his ………
(a) left ear
(b) right ear
(c) left cheek
(d) right cheek.
Answer:
(d) right cheek.

Passage – 2

Yehudi Menuhin moved from Highgate into his early 19th century house in London’s Belgravia last July but has only lived in it for a couple of months. Born in 1917, the famous? violinist and conductor, who first began his public career at the age of seven in San Francisco, still spends nine months of the year on tour. His room is four storeys up on the top floor and a lift was waiting for us in the front hall. His wife greets us and we find the maestro waiting for us on the landing.

He leads the way up a further flight of polished wooden stairs to his studio. “This is my room and I absolutely love it.’ The idea is that the studio should look like a ship. Its walls are covered with pinewood and natural light comes in through the windows in the roof. On the floor, there are cotton rugs which were made in central Asia. The whole of one wall is covered with letters in frames, paintings and prints, mostly collected by his wife Diana. ‘Anything I have of beauty or value was given to me by my wife, including herself.’ He doesn’t like empty surfaces.

“I need many tables. The card table proves his point, with its neat rows of objects standing around a figure that was found in the Athens antique market. The grand piano belonged to Menuhin’s mother-in-law, who was a brilliant pianist. Rows of photographs are displayed on top. An Indian string instrument lying by the window contrasts with the record player and tape deck nearby:

Choose the correct option to answer each question :

Question 1.
Where does Yehudi Menuhin live ?
(a) London.
(b) Canada.
(c) Belmont
(d) New York
Answer:
(a) London.

Question 2.
Which instrument does he play?
(a) Piano.
(b) Guitar.
(c) Violin.
(d) Flute.
Answer:
(c) Violin.

Question 3.
When Menuhin was only seven years old …..
(a) he went to San Franciso.
(b) he performed for the public for the first time.
(c) both (a) and (b).
(d) neither (a) nor (b).
Answer:
(c) both (a) and (b).

Question 4.
The walls of Yehudi’s studio are covered with
(a) plaster.
(b) wall paper.
(c) pinewood.
(d) paintings.
Answer:
(c) pinewood.

Question 5.
Who does a brilliant pianist refer to ?
(a) Yehudi’s mother.
(b) Yehudi’s mother-in-law.
(c) Yehudi’s wife.
(d) Yehudi Menuhin.
Answer:
(b) Yehudi’s mother-in-law.

Question 6.
What does the word ‘antique’ mean?
(a) old and valuable
(b) old and cheap
(c) new and valuable
(d) new and cheap.
Answer:
(a) old and valuable

Passage – 3

What kind of car will we be driving in 2030 ? Rather different from the type we know today, with the next 10 years bringing greater change than the past 50. The people who will be designing the models of tomorrow, believe that environmental problems may well accelerate the pace of the car’s development. Today, they are students of the transport design course at London’s Royal College of Art.

Their visions is of a machine with three wheels instead of four, electrically powered, environmentally clean, and able to drive itself along ‘intelligent’ roads with built-in power supplies. Future cars will pick up their fuel during long journeys from a power source built into the road, or store it in small quantities for travelling in the city.

Instead of today’s seating arrangement – two in front, two or three behind, all facing forward the 2010 car will have a different design with adults and children sitting in a family circle.

This view of the future car is based on a much more sophisticated” road system, with strips built into motorways to supply power to vehicles passing along them. Cars will not need drivers, because computers will provide safe driving control and route finding. All the driver will have to do is, say where to go and the computer will do the rest. It will become impossible for the cars to crash into one another. The technology already exists for the car to become a true automobile?.

Choose the correct option to answer each question :

Question 1.
Why will the new cars be developed ?
(a) Because of the fuel problems of today.
(b) Because of the traffic problems of today.
(c) Because of the environmental problems of today.
(d) None of these three.
Answer:
(c) Because of the environmental problems of today.

Question 2.
Who is going to develop them?
(a) The students of Belmont’s Royal College of Art.
(b) The students of London’s King College of Art.
(c) The students of Belmont’s King College of Art.
(d) The students of London’s Royal College of Art.
Answer:
(d) The students of London’s Royal College of Art.

Question 3.
How will the future cars be different from the present ones?
(a) They will cause no problem.
(b) They will be driven by computers.
(c) They will require no fuel.
(d) Any of the above.
Answer:
(b) They will be driven by computers.

Question 4.
How will the driving of cars become safer and easier in the 2030 ?
(a) Because the cars will be driven by computers.
(b) Because of the advanced road system.
(c) Because there will be less traffic on the roads.
(d) Because the people will obey the traffic rules.
Answer:
(a) Because the cars will be driven by computers.

Question 5.
The future cars will leave the environment clean. How ?
(a) They won’t cause any traffic problem.
(b) They won’t cause any noise pollution.
(c) They won’t give out any fumes.
(d) All of the above.
Answer:
(c) They won’t give out any fumes.

Question 6.
The word ‘supply’ in para 3 means ………
(a) provide
(b) allow
(c) support
(d) refuse.
Answer:
(a) provide

Passage 4

Tokyo is an ugly city. There are hardly any beautiful or even good buildings; there are very few parks; there are no mountains or even hills inside or outside the city; there is no green belt; there are few monuments worth looking at; the air pollution is terrifying; the perpetual noise deafening; the traffic murderous.

But not all is ugliness in Tokyo. There are a few good buildings and impressive? temples and shrines; there are a few parks worth visiting. And the overcrowding, the lack of space, has one advantage, pleasing at least to the eye. Everything has to be small in Tokyo; houses, rooms, shops – even, one feels, people, to fit into the small houses. Long side-streets consist of tiny houses only, and this often creates a toy-like, unreal quality, with small women tiptoeing along in their kimonos and equally small men sitting, motionless, inside their tiny shops.

Tokyo at night is a very different place from Tokyo in daytime. After the offices have closed and commuters have left the town, Tokyo puts on a new face. Millions of neon signs are switched on. At cafes, bars and nightclubs, sushi-places, yakitoriya, Chinese restaurants and theatres, cinemas, and many other places, this wild nightlife goes on and on and on – until 10.30 at night. Some nightclubs stay open till much later. By 11 p.m. (earlier on Sundays) all the gaiety is over, everyone is at home and in bed.

A town is not its buildings alone, it is an atmosphere, its ambience“, its feel, its pleasures, its sadness, its madness, its disappointments and above all its people. Tokyo may lack architectural beauty but it has character and excitement; it is alive. I found it a mysterious and lovable city.

Choose the correct option to answer each question :

Question 1.
Which characteristics goes against Tokyo ?
(a) There are very few parks.
(b) There are no green belts.
(c) There are no mountains or hills.
(d) All of these there.
Answer:
(d) All of these there.

Question 2.
Tokyo looks beautiful only ………..
(a) at night
(b) af dawn
(c) in the morning
(d) in the afternoon.
Answer:
(a) at night

Question 3.
What makes the city pleasant ?
(a) Its atomsphere
(b) Its character
(c) Its excitement.
(d) Any of these three.
Answer:
(d) Any of these three.

Question 4.
The author likes Tokyo and he calls it …………..
(a) a lovable city
(b) a mysterious city
(c) Both (a) and (b)
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b)

Question 5.
Which word in the passage means ‘atmosphere’?
(a) impressive
(b) ambience
(c) commuters
(d) architectural.
Answer:
(b) ambience

Passage 5

Even after three decades, the memory of that September afternoon is still fresh. It started and ended in a few seconds; but the disappointment haunts? me till the day. The toil, the tension, the torment, I’ve lived with them all. Today when I recall those moments, my heart bleeds. Isn’t it ironical that the best chapter of one’s life should end in pain ? For me, the pain is more than words can ever describe.

Missing an Olympic medal by a whisker caused me more disappointment than the happiness which I experienced after winning the medals in the Asian Games and from my winning sequence all over the Europe. Looking back, I would say it was a matter of luck.

I am sure Ron Clarke would agree with that. The great middle-distance runner set 17 world records but could not win an Olympic gold. Even to this day, I regreto not having entered the 200 metre race, where I could have figured among the medal winners. There is no question about it.

Choose the correct option to answer each question :

Question 1.
What disappointment does Milkha Singh talk about ?
(a) It is his failure to win the Asian gold.
(b) It is his failure to win the Olympic gold.
(c) both (a) and (b).
(d) neither (a) nor (b).
Answer:
(b) It is his failure to win the Olympic gold.

Question 2.
Why does his heart bleed ?
(a) He could not enter the 400 metre race.
(b) He could not win the Asian Gold medal.
(c) He had missed the Olympic medal by a whisker.
(d) All of these three.
Answer:
(c) He had missed the Olympic medal by a whisker.

Question 3.
Why does Milkha Singh mention Ron Clarke ?
(a) It was Ron Clarke who had challenged him.
(b) It was Ron Clarke who had won that mental.
(c) It was Ron Clarke who had won the Olympic medal.
(d) None of these three.
Answer:
(b) It was Ron Clarke who had won that mental.

Question 4.
“The great middle-distance runner set 17 world records ……… Who is the runner referred to in this line ?
(a) Ron Clarke
(b) P. T. Usha
(c) Milkha Singh
(d) Usain Bolt.
Answer:
(c) Milkha Singh

Question 5.
Which word in the passage means ‘a narrow margin’ ?
(a) haunts
(b) regret
(c) whisker
(d) decade.
Answer:
(c) whisker

Passage 6

An Irishman Foresees His Death
I know I should meet my fate?
Somewhere in the clouds above;
Those that I fight I do not hate,
Those I guard I do not love;

My country is Kiltartan’s poor,
No likely end could bring them loss
Or leave them happier than before. :
Nor law, nor duty bade? me fight,

Nor publicmen, nor cheering crowds?
A lonely impulse of delight
Drove this tumult in the clouds :

I balanced all, brought all to mind,
The years to come seemed waste of breath,
A waste of breath the years behind
In balance with this life, this death.

Choose the correct option to answer each question :

Question 1.
Which country does the airman belong to ?
(a) New Zealand
(b) Ireland
(c) Finland
(d) England
Answer:
(b) Ireland

Question 2.
Who does he not hate?
(a) Those he is fighting with.
(b) Those he is guarding.
(c) Those he is flying with.
(d) Those who are cheering him.
Answer:
(a) Those he is fighting with.

Question 3.
“Those I guard I do not love.’ This line means
(a) He likes the people he guards.
(b) He dislikes the people he guards.
(c) He does not know the people enough to love them.
(d) None of these three.
Answer:
(b) He dislikes the people he guards.

Question 4.
What did he feel about the life he had lived so far?
(a) He felt it had been a mere waste of breath.
(b) He felt it had been very joyful.
(c) He felt it had been very successful.
(d) He felt it had been lived in the service of the nation.
Answer:
(a) He felt it had been a mere waste of breath.

Question 5.
Why does the Irishman show no fear of death?
(a) Because he is very brave.
(b) Because he has no hope for the future.
(c) Because he is dying for his country.
(d) Because he is enjoying his flight in the clouds.
Answer:
(b) Because he has no hope for the future.

Question 6.
What is the tone of the poem ?
(a) Cheerful.
(b) Sad.
(c) Encouraging.
(d) Sarcastic.
Answer:
(b) Sad.

Passage 7

The Road Not Taken
Two roads diverged in a yellow wood?
And sorry I could not travel both;
And be one traveller, long I stood

And looked down one as far as I could
To where it bent in the undergrowth?
Then took the other, just as fair,
And having perhaps the better claim
Because it was grassy and wanted wear:
Though as far that the passing there

Had worn them really about the same.
And both that morning equally lay,
In leaves no step had trodden’ black;
Oh, I kept the first for another day!
Yet knowing how way leads on to way,
I doubted if I should ever come back.
I shall be telling this with a sigh,

Somewhere ages and ages hence :
Two roads diverged in a wood, and I –
I took the one less travelled by’,
And that has made all the difference. – Robert Frost

Choose the correct option to answer each question :

Question 1.
Where did the two roads diverge ?
(a) In a green wood.
(b) In a red wood.
(c) In a yellow wood.
(d) Nowhere.
Answer:
(c) In a yellow wood.

Question 2.
Why did the poet choose the grassy road ?
(a) Because it was much travelled by.
(b) Because it was less travelled by.
(c) Because it looked pleasant and beautiful.
(d) Because it led to a green wood.
Answer:
(b) Because it was less travelled by.

Question 3.
The phrase ‘wanted wear’ means
(a) the road needed repair.
(b) the road was too difficult to travel on.
(c) not many people travelled on that road.
(d) any of these three.
Answer:
(c) not many people travelled on that road.

Question 4.
What was it that the poet doubted ? .
(a) He doubted if way leads on to way.
(b) He doubted if he would ever come back.
(c) He doubted if he could reach home safely.
(d) He doubted if the two roads diverged in a wood.
Answer:
(b) He doubted if he would ever come back.

Question 5.
The poet uses the word ‘road to talk about ……………. in life.
(a) meeting failures
(b) taking decisions
(c) facing distractions
(d) none of these three.
Answer:
(b) taking decisions

(B) Some Other Passages

Passage 1

The pressured of vehicles on roads in Delhi has gone almost to a breaking point. The government has undertaken’ the task of building the flyovers. This has been done to ease the traffic on roads. But it would not solve the traffic problem unless a modern and an efficient public transport system is developed. In cities like Mumbai or Kolkata, the suburban train system is the lifeline. In Berlin, it is possible to go up to a distance of almost seventy-five kilometres in thirty to forty minutes. In Delhi, public transport system is not in a good shape”. Taxis and autos are expensive. The buses don’t run on time and are overcrowded most of the time. Measures should be taken to discourage the use of private transport.’

Choose the correct option to answer each question :

Question 1.
What is the purpose of building flyovers ?.
(a) To solve the parking problem.
(b) To develop efficient public transport system.
(c) To ease the pressure of traffic on roads.
(d) None of these three.
Answer:
(c) To ease the pressure of traffic on roads.

Question 2.
What is needed for an effective solution of the traffic problem?
(a) Public transport system should be discouraged.
(b) Private transport system should be encouraged.
(c) An efficient public transport system should be developed.
(d) An efficient private transport system should be developed.
Answer:
(c) An efficient public transport system should be developed.

Question 3.
What is the main source of transport in Mumbai and Kolkata ?
(a) The town tram way system.
(b) Automated urban metro system.
(c) The suburban train system.
(d) Any of these three.
Answer:
(c) The suburban train system.

Question 4.
How can you say that the public transport system in Delhi is not in a good shape ?
(a) Taxis are expensive.
(b) Buses are over crowded.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 5.
In Berlin, we can cover the distance of seventy five kilometers in 30-40 minutes by …………
(a) bus
(b) train
(c) aeroplane
(d) ship.
Answer:
(b) train

Passage 2

Language is a wonderfull gift given to man. No animal possesses this gift, but they have their own way of expressing themselves. When a rabbit sees an enemy it runs away into its hole. Its tail, which is white, bobs up and down as it runs. The other rabbits see it and they run too. They know that there is a danger. When a cobra is angry, it raises its hoods and makes itself look fierce. This warns? other animals.

When a bee has found some food, it goes back to the have. It cannot tell the other bees where the food is by speaking to them, but it does a kind of dance in the air. Some animals say things by making sounds. A dog barks when a stranger comes near.

Choose the correct option to answer each question :

Question 1.
How is man different from other animals ?
(a) Man has the gift of knowledge.
(b) Man has the gift of a sharp mind.
(c) Man has the gift of speech.
(d) Man has the gift of science.
Answer:
(c) Man has the gift of speech.

Question 2.
How does a rabbit react when it sees an enemy?
(a) It starts crying in fear. .
(b) It calls the other rabbits for help.
(c) It runs up a tree.
(d) It runs away into its hole.
Answer:
(d) It runs away into its hole.

Question 3.
How does the rabbit give a signal of danger to other rabbits ?
(a) It holds up its ears.
(b) It makes loud sharp cries.
(c) It bobs its tail up and down as it runs.
(d) It speaks to them in its own language.
Answer:
(c) It bobs its tail up and down as it runs.

Question 4.
How does a bee, tell the other bees about where the food is ?
(a) It does a kind of dance in the air.
(b) It makes a sweet humming sound.
(c) It keeps flying round the food it has found.
(d) It gives a kind of light from its body.
Answer:
(a) It does a kind of dance in the air.

Question 5.
The word ‘possesses’ means ………
(a) owns
(b) passes
(c) gives
(d) lives.
Answer:
(a) owns

Passage 3.

There was once an engine driver who was a cheerfull person. He always looked on the bright side of things, and was fond of telling people that there was sure to be some good in their misfortune?, whether they could see it or not. One day, his train ran into another and he was terribly injured.

When he was taken to hospital, it was found necessary to amputate one of his legs. A few days later, a party of friends visited him and one of his friends said, “I am afraid the poor fellow will have some difficulty in seeing the bright side of this affair.” Hearing this, the engine driver smiled and said, “Not at all. I shall have only one boot to buy and clean in future. Cheerfulness is better than grumblingo.”

Choose the correct option to answer each question :

Question 1.
What did the engine driver feel about misfortune ?’
(a) He felt that misfortune came when God wanted to punish us.
(b) He felt that there was some good in every misfortune.
(c) He felt that misfortune kills many people.
(d) He felt that misfortune is a punishment for our sins.
Answer:
(b) He felt that there was some good in every misfortune.

Question 2.
How was the engine driver injured ?
(a) He was injured in a bus accident.
(b) He was injured in a fight.
(c) He was injured when he fell down from his engine.
(d) He was injured when his train ran into another.
Answer:
(d) He was injured when his train ran into another.

Question 3.
What was done to him in the hospital ?
(a) His wound was dressed.
(b) His leg was amputated.
(c) He was given an injection.
(d) He was given a new leg.
Answer:
(b) His leg was amputated.

Question 4.
Who came to visit the engine-driver ?
(a) A doctor from the hospital.
(b) A clerk from the railway office.
(c) A party of friends.
(d) An officer from the police station.
Answer:
(c) A party of friends.

Question 5.
What is meant by the word ‘misfortune’?
(a) bad luck
(b) bad action
(c) bad company
(d) bad news.
Answer:
(a) bad luck

Passage 4 :

Once a smart-looking young man visited the office of a business firm to ask for a job. The manager, though pleased with his behaviour”, said, “There is no vacancy here for a clerk.” The young man was very sad and turned to go. As he was passing out of the doorway, he found a pin lying near it. He at once picked it up and placed it on the table.

The manager was-greatly “impressed’. He thought that the young man was good enough to be employed in his office. So he called him back and appointed him as a clerk in his office. The young man, in due course of time, became the head of that firm. It was his love of order and economy that brought him success in life.

Choose the correct option to answer each question :

Question 1.
What was the young man in need of ?
(a) He was in need of money.
(b) He was in need of books.
(c) He was in need of a job.
(d) He was in need of new clothes.
Answer:
(c) He was in need of a job.

Question 2.
What was it that pleased the manager ?
(a) The young man’s looks.
(b) The young man’s dress.
(c) The young man’s knowledge.
(d) The young man’s behaviour.
Answer:
(d) The young man’s behaviour.

Question 3.
What made the young man sad ?
(a) He was told that there was no vacancy.
(b) He was told to get out at once.
(c) He was told that he was not fit for the job.
(d) He was told that he had failed.
Answer:
(a) He was told that there was no vacancy.

Question 4.
What job was the young man given ?
(a) The job of a clerk.
(b) The job of a manager
(c) The job of a head cashier.
(d) The job of a peon.
Answer:
(a) The job of a clerk.

Question 5.
What is meant by the word ‘vacancy’ ?
(a) unfilled post
(b) empty room
(c) decent behaviour
(d) honest labour.
Answer:
(a) unfilled post

Passage 5

Late in the afternoon, Swami Vivekananda spoke on Hinduism in the great meeting. He was dressed in the yellow robes? of a Sanyasi. When he came and stood before the people, they were charmed by his appearance. He was silent for some time and then he felt a divine power in him and began his speech. He addressed the gathering as ‘Sisters and Brothers of America’.

People clapped their hands and gave him hearty+ cheers”. When the clapping ceased“, Swamiji spoke on Hinduism. He said that all the religions of the world were the same. They were all true. Only the paths leading to the goal were different. He also said that Hinduism regards every man, woman and child as a part of God. To a Hindu, the service of man is the true service of God.

Choose the correct option to answer each question :

Question 1.
What did Swami Vivekananda say about the religions of the world ?
(a) He said that all the religions are the same.
(b) He said that all the religions are different.
(c) He said that there should be only one religion.
(d) He said that the religions of the world are useless.
Answer:
(a) He said that all the religions are the same.

Question 2.
What did he feel in him before his speech ?
(a) A great sense of fear.
(b) A great sense of pride.
(c) A feeling of divine power in him.
(d) A deep love for all the people of his country.
Answer:
(c) A feeling of divine power in him.

Question 3.
How did he address the people at the meeting ?
(a) He called them “Great Men of America”.
(b) He called them ‘Men and Women of his country’.
(c) He called them ‘Sisters and Brothers of America’.
(d) He called them ‘Children of the same God’.
Answer:
(c) He called them ‘Sisters and Brothers of America’.

Question 4.
What subject did Swami Vivekananda speak on ?
(a) Education.
(b) Politics.
(c) Hinduism.
(d) Astrology
Answer:
(c) Hinduism.

Question 5.
What is the noun form of the word ‘true’ ?
(a) truly
(b) truthful
(c) truth
(d) truthfully
Answer:
(c) truth

Passage 6

A few days later, Prem Chand resigned his job of Inspector of Schools after having worked in the department for twenty years. He was a free man after all. Now he could write novels and stories about his country and its people. In his books, he dealt with the lives of the peasants and workers. He revealed the greed and meanness of the moneylenders, landlords and priests. He attacked the social evils like dowry and early marriage. He held society responsible for the sins of women. The heroes of Prem Chand’s stories and novels fight against cruelty and injustice“. Prem Chand valued love and tolerance, particularly Hindu-Muslim unity.

Choose the correct option to answer each question :

Question 1.
What was Prem Chand ?
(a) The Headmaster..
(b) The Police Inspector.
(c) The Inspector of schools.
(d) The Deputy Commissioner.
Answer:
(c) The Inspector of schools.

Question 2.
Why did he resign his job ?
(a) He had worked in that department for twenty years.
(b) He wanted to write novels and stories.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 3.
What had made the life of women hard ?
(a) Dowry.
(b) Early marriage.
(c) Gender discrimination.
(d) Both (a) and (b).
Answer:
(d) Both (a) and (b).

Question 4.
Which things did Prem Chand value ?
(a) Love.
(b) Tolerance
(c) Hindu-Muslim unity.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
Which people did Prem Chand attack for their greed and meanness?
(a) Money lenders
(b) Land lords
(c) Priests.
(d) Any of these three.
Answer:
(d) Any of these three.

Passage 7

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 suffering attracted thousands of peasants who flocked to have his darshan. The government got alarmed?, and Gandhiji was asked to leave the district. He refused and was asked to appear before the 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.

Choose the correct option to answer each question :

Question 1.
What do you mean by Satyagraha ?
(a) It means civil disobedience.
(b) It means zeal for truth.
(c) It means non-violent protest having a political aim.
(d) All of these three.
Answer:
(d) All of these three.

Question 2.
What was the object of Champaran Satyagraha ?
(a) To have justice for the poor peasants.
(b) To save the peasants from the cruel British planters.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 3.
Why did peasants flock at Champaran ?
(a) To leave their district.
(b) To protest against the British Indigo planters.
(c) To see Mahatma Gandhi.
(d) None of these three.
Answer:
(c) To see Mahatma Gandhi.

Question 4.
Why was Gandhiji asked to appear before the magistrate ?
(a) He was asked to leave Champaran.
(b) He refused to leave Champaran.
(c) Both (a) and (b)
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b)

Question 5.
Why did he live among the peasants for some time?
(a) To teach them to be fire.
(b) To learn about their hard lot.
(c) To teach them to stand on their feet.
(d) None of these three.
Answer:
(b) To learn about their hard lot.

Passage 8

Milk is the best food. It has in it water, sugar, fat, vitamins and proteins. People drink milk from different animals. In England, New Zealand and many other cool lands, there are cows. In hot, dry countries like Arabia and the middle of Asia there are camels. In India, there are buffaloes as well as cows. In many places there are goats. The Eskimos have herds of reindeers. They live in the very cold countries of North America. If people keep cows or these animals, they get a lot of milk. From milk they can make butter and cheese. It is essential that the milk we use should be pure and germ-free. Impure milk does more harm than good to the human body…

Choose the correct option to answer each question :

Question 1.
Why is milk called the best food ?
(a) Because it has water and sugar.
(b) Because it has in it sugar and fat.
(c) Because it has in it vitamins and proteins.
(d) All of these three.
Answer:
(d) All of these three.

Question 2.
We get milk from ………….
(a) Cows and buffaloes
(b) Goats and camels
(c) Camels and reindeers
(d) All of these three.
Answer:
(d) All of these three.

Question 3.
Which animals are kept for milk in hot, dry countries ?
(a) Camels.
(b) Reindeers.
(c) Buffaloes.
(d) Goats.
Answer:
(a) Camels.

Question 4.
Why should we use pure milk ?
(a) Because impure milk harms the human body.
(b) Because it is not costly.
(c) Because it is germ-free.
(d) Because it contains vitamins and proteins.
Answer:
(c) Because it is germ-free.

Question 5.
Which word in the passage means ‘necessary’ ?
(a) impure
(b) essential
(c) herd
(d) different.
Answer:
(b) essential

Passage 9

Games, though essentiall, 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. 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 all these minor defects, sports are very useful in keeping the students busy and in developing their personalities.

Choose the correct option to answer each question :

Question 1.
What harm do games do to some students ?
(a) They waste too much time on games.
(b) They fail in exams.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 2.
How much time should one devote to games ?
(a) Half an hour.
(b) An hour
(c) Two hours.
(d) Three hours.
Answer:
(b) An hour

Question 3.
What may happen if a game is played rashly ?
(a) There can be danger of quarrels.
(b) There can be danger of breaking bones.
(c) There can be danger of losing the game.
(d) There can be danger of playing foul.
Answer:
(b) There can be danger of breaking bones.

Question 4.
How do games help the students in building up their personalities?
(a) It keeps them busy.
(b) It develops in them team spirit.
(c) It teaches them discipline.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
Which word in the passage means ‘custom” ?
(a) tradition
(b) rashly
(c) devote
(d) spirit.
Answer:
(a) tradition

Passage 10

Life is not a bed of roses, but a bed of thorns. It is full of dangers and difficulties. In the race of life, we should not be afraid of the risk which is but natural. Success in any work in life goes to those persons who welcome risk. Science would not have made such wonderfull achievements if our scientists had not risked their lives and comforts. The more difficult a work is, the harder should be our efforts to perform it. Life is not a smooth sailing. Petty difficulties frighten a weak heart who is not prepared to take a risk. But brave hearts achieve fame and honour, because they enjoy taking risks. In short, risk brings success and works miracles.

Choose the correct option to answer each question :

Question 1.
Why is life a bed of thorns ?
(a) Because it is full of dangers.
(b) Because it is full of difficulties.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 2.
Why should we take risk in life?
(a) Without taking risks we can have experience.
(b) Without taking risks we can’t succeed in life.
(c) Without taking risk we can’t work hard.
(d) None of these three.
Answer:
(b) Without taking risks we can’t succeed in life.

Question 3.
What helped science make wonderful achievements ?
(a) For this our scientist had risked their lives.
(b) For this our scientist had forgotten their comforts.
(c) Both (a) and (b).
(d) None of these three.
Answer:
(c) Both (a) and (b).

Question 4.
What lesson do you learn from the passage ?
(a) Life is not a bed of roses.
(b) Hard work is the key to success.
(c) Risks bring success.
(d) All of these three.
Answer:
(c) Risks bring success.

Question 5.
Which word in the passage means ‘wonders’?
(a) comforts
(b) miracles
(c) achievements
(d) efforts.
Answer:
(b) miracles

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)

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

PSEB 9th Class English Vocabulary Homonyms

Homonyms

जिन शब्दों का उच्चारण एक जैसा हो परन्तु उनके अर्थ तथा हिज्जे (spellings) अलग-अलग हों, उन्हें Homonyms कहते है, जैसे
break, brake; write, right; sight, site; weight, wait; etc.

List Of Homonyms

1. Berth – Please get a berth reserved for me in the Flying Mail.
Birth – She gave birth to a male child.

2. Brake – The driver applied brakes to save the child.
Break – Glass breaks easily.

3. Cell – This remote control works on two pencil cells.
Sell – We want to sell our old furniture.

4. Died – His father died at the age of eighty.
Dyed – She dyed her hair dark brown.

5. Dose – This bottle contains six doses.
Doze – He was dozing in the class.

6. Hair – She was combing her hair.
Hare – The hare can run very fast.

7. Heal – The wound healed slowly.
Heel – The thief took to heels.

8. Pain – I have pain in my stomach.
Pane – Who has broken the window pane ?

9. Pair – I have bought a pair of shoes.
Pare – Pare your nails.

10. Peace – Who does not want peace ?
Piece – I gave him a piece of bread.

11. Pray – I pray to God for your health and happiness.
Prey – The tiger jumped on its prey.

12. Principal – My mother went to the school to meet the principal.
Principle – He is a man of high principles.

13. Root – This tree has very deep roots.
Route – We took the shortest route.

14. Stair – The man slipped while climbing the stairs.
Stare – It is a bad habit to stare at anyone.

15. Storey – This house has three storeys.
Story – My grandmother told me a very interesting story.

16. Their – Their house is small but comfortable.
There – We went there in a group.

17. Wait – I had to wait for a long time.
Weight – My weight is fifty kilograms.

18. Waist – The water in the river soon rose above his waist.
Waste – Don’t waste your time.

19. Weather – The weather has suddenly turned cold.
Whether – I want to know whether this answer is correct.

20. Heir – Kanwar Mahendra Singh is the next heir to the throne.
Air – Go out for a walk in the fresh air.

Choose the word from the pairs of words given and complete the sentences. You may have to change the form of the word in some cases.

fair, fare; groan, grown; practise, practice; principle, principal; feet, feat; vain, vein; stationery, stationary; wait, weight.

1. It is my ………… not to lend money to anyone.
2. The player was badly hurt and was ………… with pain.
3. Can you ………… for some time? The officer is very busy at the moment.
4. A passenger train hit a ………… goods train near Pune.
5. The ………… of buses may go up by 10% next month.
6. Have you done enough ………… to win the match ?
7. Mamta tried in ………… to climb to the top of the building.
8. The Lotus Temple in Delhi is a great ………… of engineering.
Answers
1. principle
2. groaning
3. wait
4. stationary
5. fare
6. practice
7. vain
8. feat.