Class 9

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.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

Question 1.
In the given figure, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).

Answer:

As AD is a median of ∆ ABC, it divides ∆ ABC into two triangles with equal areas.
∴ ar (ADB) = ar (ADC) …………… (1)
In ∆ ABC, AD is a median.
∴ BD = CD
Draw a line l through E and parallel to BC. Then, ∆ EDB and ∆ EDC are on the equal bases and between the same parallels l and BC.
∴ ar (EDB) = ar (EDC) ……………… (2)
Subtracting (2) from (1),
ar (ADB) – ar (EDB) = ar (ADC) – ar (EDC)
∴ ar (ABE) = ar (ACE)

Question 2.
In a triangle ABC, E is the midpoint of median AD. Show that ar (BED) = \(\frac{1}{4}\)ar (ABC).
Answer:

In ∆ ABC, AD is a median
∴ ar (ADB) = ar (ADC) = \(\frac{1}{2}\)ar (ABC)
Now, in ∆ ABD, E is the midpoint of AD.
∴ BE is a median of ∆ ABD.
∴ ar (BED) = ar (BEA) = \(\frac{1}{2}\)ar (ABD)
Thus, ar (BED) = \(\frac{1}{2}\)ar (ABD) and
ar (ABD) = \(\frac{1}{2}\)ar (ABC)
∴ ar (BED) = \(\frac{1}{2}\) × \(\frac{1}{2}\)ar (ABC)
∴ ar (BED) = \(\frac{1}{4}\)

Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer:

The diagonals of a parallelogram bisect each other.
Hence, in parallelogram ABCD, diagonals AC and BD bisect each other at M. Thus, M is the midpoint of AC as well as BD.
In ∆ ABC, BM is a median.
∴ ar (ABM) = ar (CBM) ……………. (1)
In ∆ BCD, CM is a median.
∴ ar (CBM) = ar (CDM) ……………. (2)
In ∆ ABD, AM is a median.
∴ ar (ABM) = ar (DAM) ……………. (3)
From (1), (2) and (3),
ar (ABM) = ar (CBM) = ar (CDM) = ar (DAM)
Thus, the diagonals of a parallelogram divide it into four triangles of equal areas.

Question 4.
In the given figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

Answer:
Drawing a line through A and parallel to CD, ∆ AOC and A AOD are on the equal bases and between the same parallels.
∴ ar (AOC) = ar (AOD) ……………… (1)
Similarly, drawing a line through B and parallel to CD, A BOC and A BOD are on the equal bases and between the same parallels.
∴ ar (BOC) = ar (BOD) ……………. (2)
Adding (1) and (2),
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
∴ ar (ABC) = ar (ABD)

Question 5.
D, E and F are respectively the midpoints of the sides BC, CA and AB of a A ABC. Show that:
(i) BDEF is a parallelogram.
(ii) ar (DEF) = \(\frac{1}{4}\)ar (ABC)
(iii) ar (BDEF) = \(\frac{1}{2}\)ar (ABC)
Answer:

In ∆ ABC, F and E are the midpoints of AB and AC respectively.
∴ FE || BC
∴ FE || BD
In ∆ ABC, E and D are the midpoints of AC and BC respectively.
∴ ED || AB
∴ ED || FB
In quadrilateral BDEF, FE || BD and ED || FB.
∴ Quadrilateral BDEF is a parallelogram. …… Result (i)
Similarly, quadrilaterals AFDE and FDCE are parallelograms.
In parallelogram BDEF, FD is a diagonal.
∴ ar (BDF) = ar (DEF) ……………….. (1)
In parallelogram AFDE, EF is a diagonal.
∴ ar (AFE) = ar (DEF) ……………….. (2)
In parallelogram FDCE, ED is a diagonal.
∴ ar (DCE) = ar (DEF) ………………. (3)
∆ ABC is made up of four non-overlapping triangles, ∆ BDF, ∆ AFE, ∆ DCE and ∆ DEF.
∴ ar (ABC)
= ar (BDF) + ar (AFE) + ar (DCE) + ar (DEF)
∴ ar (ABC)
= ar (DEF) + ar (DEF) + ar (DEF) + ar (DEF) [From (1), (2) and (3)]
∴ ar (ABC) = 4ar (DEF)
∴ ar (DEF) = \(\frac{1}{4}\)ar (ABC) …………. Result (ii)
Now, ar (BDEF) = ar (BDF) + ar (DEF)
∴ ar (BDEF) = ar (DEF) + ar (DEF)
∴ ar (BDEF) = 2ar (DEF)
∴ ar (BDEF) = 2 × \(\frac{1}{4}\)ar (ABC)
∴ ar (BDEF) = \(\frac{1}{2}\)ar (ABC) ……. Result (iii)

Question 6.
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB and ABCD is a parallelogram. [Hint: From D and B, draw perpendiculars to AC.]

Answer:
Draw DM ⊥ AC and BN ⊥ AC, where M and N are points on AC.
In ∆ DMO and ∆ BNO,
DO = BO (Given)
∠ DOM = ∠ BON (Vertically opposite angles)
∠ DMO = ∠ BNO (Right angles)
∴ ∆ DMO ≅ ∆ BNO (AAS rule)
∴ DM = BN (CPCT)
Now, in ∆ DMC and ∆ BNA,
DM = BN
hypotenuse DC = hypotenuse BA (Given)
∠ DMC = ∠ BNA (Right angles)
∴ ∆ DMC = ∆ BNA (RHS rule)
∴ ar (DMC) = ar (BNA) ……………… (1)
Moreover, ar (DMO) = ar (BNO) (∵ ∆ DMO ≅ ∆ BNO) ………………… (2)
Adding (1) and (2),
ar (DMC) + ar (DMO) = ar (BNA) + ar (BNO)
∴ ar (DOC) = ar (AOB) (Non-overlapping triangles) … Result (i)
Now, adding ar (COB) on both the sides,
ar (DOC) + ar (COB) = ar (AOB) + ar (COB)
∴ ar (DCB) = ar (ACB) ……… Result (ii)
∆ DCB and ∆ ACB are on the same base BC and their areas are equal.
∴ ∆ DCB and ∆ ACB are between the same parallels DA and CB.
∴ DA || CB ……………… (3)
∆ DMO = ∆ BNO gives OM = ON
∆ DMC = ∆ BNA gives CM = AN
∴ OM + CM = ON + AN
∴ OC = OA
Also, OD = OB (Given)
Thus, in quadrilateral ABCD, diagonals AC and BD bisect each other at O.
∴ ABCD is a parallelogram. ……………….. (4)
Taking (3) and (4) together,
DA || CB and ABCD is a parallelogram. ………. Result (iii)

Question 7.
D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
Answer:

ar (DBC) = ar (EBC)
∴ ∆ DBC and ∆ EBC have equal areas.
Here, ∆ DBC and ∆ EBC are on the same base BC and their areas are equal.
∴ ∆ DBC and ∆ EBC are between the same parallels DE and BC.
∴ DE || BC

Question 8.
XY is a line parallel to side BC of a ‘triangle ABC. If BE II AC and CF II AB meet ( XY at E and F respectively, show that ar (ABE) = ar (ACF).
Answer:

Suppose line XY drawn parallel to BC intersects AB and AC at M and N respectively.
In quadrilateral EBCN, EN || BC and BE || CN.
∴ EBCN is a parallelogram.
In quadrilateral MBCF, MF || BC and BM || CF.
∴ MBCF is a parallelogram.
Parallelograms EBCN and MBCF are on the same base BC and between the same parallels BC and EF
∴ ar (EBCN) = ar (MBCF)
∴ ar (BME) + ar (MBCN) = ar (MBCN) + ar (CFN)
∴ ar (BME) = ar (CFN) ……………….. (1)
Now, ∆ BME and ∆ CFN are between the same parallels EF and BC and their areas are equal.
∴ Their bases are equal.
∴ EM = NF
Through A, draw line PQ parallel to EF.
Then, ∆ AEM and ∆ ANF are on equal bases s and between the same parallels PQ and EF. !>
∴ ar (AEM) = ar (ANF) ……………… (2)
Adding (1) and (2),
ar (BME) + ar (AEM) = ar (CFN) + ar (ANF)
∴ ar (ABE) = ar (ACF)

Question 9.
The side AB of a parallelogram ABCD is ‘produced to any point P. A line through A and parallel to CP meets CB produced at and then parallelogram PBQR is completed (see the given figure). Show that: ar (ABCD) = ar (PBQR)
[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Answer:
Join AC and PQ.
∆ CAQ and ∆ PAQ are on the same base AQ and between the same parallels CP and AQ.
∴ ar (CAQ) = ar(PAQ)
∴ ar (ACB) + ar (ABQ) = ar (PBQ) + ar (ABQ) (Non-overlapping triangles)
∴ ar (ACB) = ar (PBQ) ……………. (1)
In parallelogram ABCD, AC is a diagonal.
∴ ar (ACB) = \(\frac{1}{2}\)ar (ABCD) ……………… (2)
In parallelogram PBQR, PQ is a diagonal.
∴ ar (PBQ) = \(\frac{1}{2}\)ar (PBQR) ………………. (3)
From (1), (2) and (3),
\(\frac{1}{2}\)ar (ABCD) = \(\frac{1}{2}\)ar (PBQR)
∴ ar (ABCD) = ar (PBQR)

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Answer:
∆ DAB and ∆ CBA are on the same base AB and between the same parallels AB and CD.
∴ ar (DAB) = ar (CBA)
∆ DAB and ∆ CBA are both formed by two non-overlapping triangles.
∴ ar (DAB) = ar (AOD) + ar (OAB) and
ar (CBA) = ar (BOC) + ar (OAB)
∴ ar (AOD) + ar (OAB) = ar (BOC) + ar (OAB)
∴ ar (AOD) = ar (BOC)

Question 11.
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)

Answer:
∆ ACB and ∆ ACF are on the same base AC and between the same parallels AC and BF.
∴ ar (ACB) = ar (ACF)
Adding ar (AEDC) on both the sides,
ar (ACF) + ar (AEDC) = ar (ACB) + ar (AEDC)
∴ ACF and quadrilateral AEDC are two non-overlapping figures and they form quadrilateral AEDF.
Similarly, ∆ ACB and quadrilateral AEDC are two non-overlapping figures and they form pentagon ABCDE.
∴ ar (AEDF) = ar (ABCDE)

Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the comers to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Answer:

Suppose, quadrilateral ABCD is the plot of Itwaari in the shape of a quadrilateral.
Draw diagonal AC. Through D draw a line parallel to AC to intersect BC produced at P
Join PA to intersect CD at Q.
Here, ∆ DAC and ∆ PAC are on the same base AC and between the same parallels DP and AC.
∴ ar (DAC) = ar (PAC)
∴ ar (DAQ) + ar (QAC) = ar (PQC) + ar (QAC)
∴ ar (DAQ) = ar (PQC)
Thus, we get two triangles – ∆ DAQ and ∆ PQC of equal areas.
So, the Gram Panchayat should take triangular region DAQ from the plot of Itwaari and in exchange give him triangular region PQC. Then, the plot of Itwaari will be of the triangular shape PAB and its area will be the same as the area of original plot ABCD.

Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.)
Answer:

Join CX.
∆ ADX and ∆ ACX are on the same base AX and between the same parallels AB and CD.
∴ ar (ADX) = ar (ACX) …………… (1)
A ACX and A ACY are on the same base AC and between the same parallels AC and XY.
∴ ar (ACX) = ar (ACY) ……………. (2)
From (1) and (2),
ar (ADX) = ar (ACY)

Question 14.
In the given figure, AP || BQ || OR. Prove that ar (AQC) = ar (PBR).

Answer:
∆ CBQ and ∆ RBQ are on the same base BQ and between the same parallels BQ and CR.
∴ ar (CBQ) = ar (RBQ) ……. (1)
∆ ABQ and ∆ PBQ are on the same base BQ and between the same parallels BQ and AE
∴ ar (ABQ) = ar (PBQ) ……… (2)
Adding (1) and (2),
ar (CBQ) + ar (ABQ) = ar (RBQ) + ar (PBQ)
∴ ar (AQC) = ar(PBR) (Non-overlapping triangles)

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a , trapezium.
Answer:

ar (AOD) = ar (BOC)
Adding ar (OAB) on both the sides,
ar (AOD) + ar (OAB) = ar (BOC) + ar (OAB)
∴ ar (DAB) = ar (CAB)
Thus, ∆ DAB and ∆ CAB are on the same base AB and their areas are equal.
Hence, by theorem 9.3, ∆ DAB and ∆ CAB are between the same parallel.
∴ DC || AB
In quadrilateral ABCD, DC || AB.
Hence, ABCD is a trapezium.

Question 16.
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer:
ar (DRC) = ar (DPC)
Thus, ∆ DRC and ∆ DPC are on the same base DC and their areas are equal.
Hence, by theorem 9.3, they are between the same parallels.
∴ DC || RP
In quadrilateral DCPR, DC || RE .
∴ Quadrilateral DCPR is a trapezium.
Now, ar (BDP) = ar (ARC)
∴ ar (DPC) + ar (BDC) = ar (DRC) + ar (ADC) (Non-overlapping triangles)
But, ar (DPC) = ar (DRC)
∴ ar (BDC) = ar (ADC)
Thus, ∆ BDC and ∆ ADC are on the same base DC and their areas are equal.
Hence, by theorem 9.3, they are between the same parallels.
∴ AB || DC
In quadrilateral ABCD, AB || DC.
∴ Quadrilateral ABCD is a trapezium.

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

PSEB 9th Class English Vocabulary Antonyms

Match the Antonyms correctly :

A — B
admit — destroy
public — straight
junior — conclude
bent — stale
defend — senior
dark — private
fresh — well-lit
create — descend
ascend — dangerous
safe — attack
start — modern
ancient — deny
Answer:
admit → deny; public → private ; junior → senior; bent → straight; defend → attack; dark → well-lit; fresh → stale ; create → destroy; ascend → descend; safe → dangerous; start → conclude ; ancient → modem.

Complete the following sentences with the Antonyms given in the box. The italicized words will help you to choose the right words:

modern;  spend; withdraw; vacant; remember; success; punctual; expensive; plus; import

1. Ramesh, you are always ………… but why are you late today?
2. We will not ………… all the money now. We will save it for the rainy day.
3. Seven seats are still ………… They were all full by now last year.
4. I will deposit the whole amount. I can ………… it at any time I want to.
5. The Goyals ………… wool from Australia, make garments and export them to the U.K.
6. Don’t forget your sweater here ………….. to wear it when it is cold in Srinagar.
7. ……….. and failure go hand in hand in life.
8. This umbrella is very ……… I need a cheap one.
9. Rome has both ancient and ……….. buildings.
10. Can you tell me the minus and ……….. points of this proposal?
Answer:
1. punctual
2. spend
3. vacant
4. withdraw
5. import
6. Remember
7. Success
8. expensive
9. modem
10. plus.

Antonyms

जिन दो शब्दों के अर्थ एक-दूसरे के विपरीत हों, उन्हें Antonyms कहते हैं; जैसे
clean → dirty
deep → shallow
sorrow → joy

List Of Antonyms

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

PSEB 9th Class English Vocabulary Synonyms

Match the words in column A with their synonyms in column B :

A — B
just — reach
leave — wrath
permit — refuge
brave — fearless
certain — vanish
disappear — sure
anger — prevent
forbid — fair
shelter — inactive
attempt — depart
lazy — allow
arrive — try
Answer:
just = fair ; leave = depart ; permit = allow ; brave = fearless ; certain = sure ; disappear = vanish; anger = wrath ; forbid = prevent; shelter = refuge; attempt = try; lazy = inactive; arrive = reach.

Complete the sentences with the words given in the box. The italicized words will help you to choose the right words 

inactive ; pouring ; bright ; finish ; annoyed ; enormous ; collect ; wear ; try

1. Is the new student intelligent ? Yes, he is very ……………..
2. It was raining heavily. In fact, it was ………………
3. The show stopped at 10 p.m. because the policemen asked us to ………….. early.
4. She was very angry with my brother. I don’t know why she was so …………….. with him.
5. It was a very huge animal. We had never seen such an ……………. animal before.
6. He likes to assemble photographs. He has decided to …………….. about one thousand rare ones.
7. Don’t put on the red cap, …………….. the black one.
8. She is quite lazy. I don’t know why she is so ……………..
Answer:
1. bright
2. pouring
3. finish
4. annoyed
5. enormous
6. collect
7. wear
8. inactive.

Synonyms

जिन शब्दों का अर्थ एक ही हो अथवा मिलता-जुलता हो, उन्हें Synonyms कहते हैं; जैसे
hard = difficult
pray = request
fact = truth

List Of Synonyms

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

PSEB 9th Class English Vocabulary Words as different Parts of Speech

Write n’ for noun, ‘adj.’ for adjective, ‘adv.’ for adverb and ‘ for verb for the italicized word given in the sentences :

1. One must travel light while travelling by air.
2. This clock is ten minutes fast.
3. The chief guest’s speech was brief and to the point.
4. The officer was briefed about the facts of the case.
5. This bicycle cost him twelve hundred rupees.
6. Manvinder ran fast enough to reach the school in time.
7. What is your examiner’s schedule for tomorrow ?
8. You can buy many fancy items from Meena Bazar.
9. That tastes real good. Where did you get it from ?
10. What is the cost of this pair of socks ?
Answer:
1. adv.
2. adj.
3. adj.
4. v
5. v
6. adv.
7. n
8. adj.
9. adv.
10. n.

Use the given words in sentences as directed :

1. Fare (noun) – He has not paid the bus fare.
Fare (verb) – Ram has fared well in the examination.

2. Bear (noun) – There are many types of bear in the zoo.
Bear (verb) – He has borne much suffering in his life.

3. Wound (noun) – My wound has healed.
Wound (verb) – He was wounded seriously.

4. Round (adjective) – The earth is round.
Round (preposition)- The earth goes round the sun.
Round (adverb) – Everybody joins hands and dances round.

5. Fast (adjective) – He is a fast friend of mine.
Fast (adverb) – She is running very fast.
Fast (noun) – I observe fast on every Monday.

6. Stand (noun) – I saw him near the cycle-stand.
Stand (verb) – Will you please stand up?

7. Produce (noun) – This shop sells only fresh local produce.
Produce (verb) – She produced a hot meal for us within 20 minutes.

एक ही शब्द को भाषा के विभिन्न रूपों में इस्तेमाल किया जा सकता है। उदाहरण के रूप में निम्नलिखित वाक्यों को देखिए

They fought to the last. (Noun)
The war lasted one month. (Verb)
He was the last man to come. (Adjective)
He spoke last at the meeting. (Adverb)

यहां शब्द ‘last’ को पहले वाक्य में Noun के रूप में, दूसरे वाक्य में एक Verb के रूप में, तीसरे वाक्य में एक Adjective तथा चौथे वाक्य में एक Adverb के रूप में इस्तेमाल किया गया है।
Examples :

Act
Verb : He acted upon my advice.
Noun : It was an act of kindness.

All
Noun : I lost my all in this business.
Adjective : All the boys reached the playground.

Bail
Verb : He will be bailed out today.
Noun : He was set free on bail.

Back
Verb : He becked my proposal.
Noun: He has pain in his back.
Adjective : He came through the back door.

Better
Adjective : This book is better than that.
Adverb: He fared better in the test than he had hoped.
Verb : We hope to better the conditions of our workers.
Noun (pl.) : Follow your betters.

Book
Verb : Get your luggage booked.
Noun : I have read this book.

Close
Noun : The meeting came to a close in time.
Verb : The school is closed for summer vacation.

Dawn
Verb : The truth dawned upon him yesterday.
Noun : I always get up before dawn.

Drive
Verb : Suresh is driving the car at top speed.
Noun : He had a good drive in the morning.

Effect
Verb : The prisoner effected his escape.
Noun : My advice had the desired effect.

Fare
Verb : I have fared badly in my test.
Noun : We paid the bus fare.

Fix
Noun : We are in a fix now.
Verb : Let us fix the programme.

Face :
Verb: He has faced many difficulties in his life.
Noun Wash your face.

Fast
Noun: They keep a fast every Friday.
Adjective: This is a fast train.
Verb: I fast once every week.

Hand
Verb: The thief was handed over to the police.
Noun: Wash your hands.

Iron
Verb: Iron your clothes.
Noun: Iron is a useful metal.
Adjective: This knife has an iron handle.

Idle :
Verb: Do not idle away your precious time.
Noun: The idle fail in their life.
Adjective: Sohan is an idle boy.

Like
Verb: I like her ways.
Noun: Everybody has his likes and dislikes.
Adjective: Like poles repel each other.

Light
Noun: There was no light in the room.
Adjective: Always take a light diet to keep fit.
Verb: Light the lamp.
Adverb: Travel light if you must.

Less
Adjective: He is paying less attention to studies these days.
Verb: He is less intelligent than his brother.
Noun: He won’t be satisfied with less.

Near :
Adjective: He is a near relation of the headmaster.
Verb: He is nearing his end.
Adverb: Come near

Right
Noun: We must fight for our rights.
Adjective: This is my right hand.
Verb: Every wrong should be righted.

Round
Noun: Let us have a round of cards.
Adjective: The earth is round.
Verb: He rounded his lips in anger.
Adjective: He turned round.
Preposition: She wore a necklace round her neck.

Second :
Verb: I will second your proposal.
Noun: I will come back in a second.
Adjective: February is the second month of the year.

Still
Noun: In the still of the night, a thief entered our house.
Adjective: The night was still.

Stone
Adjective: It is a stone wall.
Verb: The boys stoned the dog to death.
Noun: The boys threw stones at the beggar.

Time
Noun: What is the time by your watch ?
Verb: The train is timed to reach here at 5 p.m.

Well
Noun: We have a well in our garden.
Adjective: The patient is now well.
Adverb: He has fared well in the examination.
Verb: Tears welled up in her eyes.

While
Noun: Rest a little while.
Verb: He is whiling away his time.

Will
Noun: Where there is a will, there is a way.
Verb: Have what you will.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Question 1.
In the given figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Answer:
The area of a parallelogram is the product of its base and the altitude corresponding to that base.
Here, in parallelogram ABCD, the altitude corresponding to base DC is AE and the altitude corresponding to base AD is CF.
∴ ar(ABCD) = DC × AE = AD × CF
∴ DC × AE = AD × CF
∴ AB × AE = AD × CF (∵ AB = DC In parallelogram ABCD)
∴ 16 × 8 = AD × 1O
∴ AD = \(\frac{16 \times 8}{10}\)
∴ AD = \(\frac{128}{10}\)
∴ AD = 12.8 cm

Question 2.
If E, F, G and H are respectively the midpoints of the sides of a parallelogram ABCD, show that ar(EFGH) = \(\frac{1}{2}\)ar (ABCD).
Answer:
In parallelogram ABCD. E, F, G and H are the midpoints of AB, BC. CD and DA respectively.
Draw GE.

In parallelogram ABCD, AB || CD and AB = CD.
∴ BE || CG and BE (\(\frac{1}{2}\) AB) = CG (\(\frac{1}{2}\) CD)
∴ Quadrilateral EBCG is a parallelogram.
∴ GE || BC
Now, ∆ EFG and parallelogram EBCG are on the same base GE and between the same parallels GE and BC.
∴ ar(EFG)= \(\frac{1}{2}\)ar (EBCG) ………………. (1)
Similarly, ∆ EHG and parallelogram AEGD are on the same base GE and between the same parallels GE and DA.
∴ ar(EHG) = \(\frac{1}{2}\)ar (AEGD) ……………….. (2)
Adding (1) and (2),
ar (EFG) + ar (EHG) = \(\frac{1}{2}\) ar (EBCG) + \(\frac{1}{2}\) ar (AEGD)
∴ ar (EFGH) = \(\frac{1}{2}\) [ar (EBCG) + ar (AEGD)]
∴ ar (EFGH) = \(\frac{1}{2}\) ar (ABCD)

Question 3.
p and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

Answer:
Here, ∆ APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
∴ ar (APB) = \(\frac{1}{2}\)ar (ABCD) …………… (1)
Similarly, ∆ BQC and parallelogram ABCD are on the same base BC and between the same parallels BC and DA.
∴ ar (BQC) = \(\frac{1}{2}\)ar (ABCD) ……………… (2)
From (1) and (2),
ar(APB) = ar(BQC)

Question 4.
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that,
(i) ar (APB) + ar (PCD) = \(\frac{1}{2}\)ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint: Through P, draw a line parallel to AB.]

Answer:
Through P, draw a line parallel to AB which intersects BC at Q and AD at R.
Now, in quadrilateral ABQR,
AB || QR (By construction)
BQ || AR (In parallelogram ABCD, BC || AD)
∴ Quadrilateral ABQR is a parallelogram.
Similarly, DCQR is a parallelogram.
∆ APB and parallelogram ABQR are on the same base AB and between the same parallels AB and QR.
∴ ar(APB) = \(\frac{1}{2}\)ar (ABQR) ……………….. (1)
Similarly, ∆ PCD and parallelogram DCQR are on the same base DC and between the same parallels DC and QR.
∴ ar(PCD) = \(\frac{1}{2}\)ar (DCQR) ………………… (2)
Adding (1) and (2),
= ar(APB) + ar (PCD)
= \(\frac{1}{2}\)ar (ABQR) + \(\frac{1}{2}\)ar (DCQR)
∴ ar (APB) + ar (PCD)
= \(\frac{1}{2}\) [ar (ABQR) + ar (DCQR)]
∴ ar (APB) + ar (PCD) = \(\frac{1}{2}\)ar (ABCD) …………………. (3)
Now, through P, draw a line parallel to AD which intersects AB at S and CD at T.
Then, as above, it can be proved that
ar(APD) + ar(PBC) = \(\frac{1}{2}\)ar (ABCD) ……………………. (4)
From (3) and (4),
ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Question 5.
In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar (ABRS)
(ii) ar(AXS) = \(\frac{1}{2}\)ar (PQRS)

Answer:
Parallelograms PQRS and ABRS are on the same base RS and between the same parallels PB and SR.
∴ ar (PQRS) = ar (ABRS) …………….. (1)
∆ AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.
∴ ar (AXS) = \(\frac{1}{2}\)ar (ABRS) ………………… (2)
From (1) and (2),
ar (AXS) = \(\frac{1}{2}\)ar (PQRS)

Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field Is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Answer:

By taking any point A on RS and joining it to points P and Q, the field is divided into three triangular parts as ∆ PSA, ∆ APQ and ∆ QRA.
Here, ∆ APQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and SR.
∴ ar (APQ) = \(\frac{1}{2}\)ar (PQRS)
∴ ar (PSA) + ar(QRA) = \(\frac{1}{2}\)ar(PQRS)
Thus, ar (APQ) = ar (PSA) + ar (QRA)
Now, as the farmer wants to sow wheat and pulses in equal portions of the field separately.
she has two options as below to do so:
(1) She should sow wheat in ∆ APQ and pulses in ∆ PSA as well as ∆ QRA.
(2) She should sow pulses in ∆ APQ and wheat in ∆ PSA as well as ∆ QRA.