Bhagya

Punjab State Board PSEB 7th Class English Book Solutions English Vocabulary Comparison using ‘as’ Exercise Questions and Answers, Notes.

PSEB 7th Class English Vocabulary Comparison using ‘as’

एक जैसे गुणों वाली दो वस्तुओं को तुलना के लिए as………. as का प्रयोग किया जाता है। किसी गुणधर्म को Highlight करने के लिए उसकी तुलना उसी गुणधर्म वाली किसी भी अन्य वस्तु से की जा सकती है।

1. as good as gold
2. as happy as a clown
3. as sharp as a razor
4. as big as elephant

5. as heavy as lead
6. as beautiful as moon
7. as gentle as lamb
8. as silent as dea$
9. as small as ant
10. as thin as paper
11. as white as snow
12. as tall as giraffe
13. as clever as fox
14. as costly as diamond.
15. as proud as peacock
16. as hot as sun
17. as pretty as flower
18. as sweet as honey
19. as red as rose
20. as strong as iron
21. as innocent as lamb
22. as clever as fox/monkey/owl/crow
23. as wild as lion
24. as soft as silk
25. as light as cotton
26. as black as night
27. as wise as crow
28. as brave as lion.
29. as sure as death
30. as deep as ocean
31. as cold as ice
32. as black as coal
33. as busy as a bee
34. as clear as bell
35. as common as dirt
36. as deaf as a post

37. as easy as A.B.C.
38. as large as life
39. as light as air
40. as straight as an arrow

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.
In quadrilateral ACBD. AC = AD and AB bisects ∠ A (see the given figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Answer:
In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠ BAC = ∠ BAD (AB bisects ∠ A)
AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (SAS rule)
∴ BC = BD (CPCT)
Thus, BC and BD are equal.

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see the given figure). Prove that (i) ∆ ABD ≅ ∆ BAC, (ii) BD = AC and (iii) ∠ ABD = ∠ BAC

Answer:
In ∆ ABD and ∆ BAC,
AD = BC (Given)
∠ DAB = ∠ CBA (Given)
AB = BA (Common)
∴ ∆ ABD ≅ ∆ BAC (SAS rule)
∴ BD = AC (CPCT)
∴ ∠ ABD = ∠ BAC (CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.

Answer:
AD and BC are equal perpendiculars to line segment AB.
∴ AD = BC and ∠ OAD = ∠ OBC = 90°.
Now, in ∆ ADO and ∆ BCO,
AD = BC
∠ OAD = ∠ OBC
∠ AOD = ∠ BOC (Vertically opposite angles)
∴ ∆ ADO ≅ ∆ BCO (AAS rule)
∴ OA = OB (CPCT)
CD intersects AB at O and OA = OB.
Hence, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that:
∆ ABC ≅ ∆ CDA.

Answer:
l || m and AC is their transversal.
∴ ∠ BCA = ∠ DAC (Alternate angles)
p l| q and AC is their transversal.
∴ ∠ BAC = ∠ DCA (Alternate angles)
Now, in ∆ ABC and ∆ CDA,
∠ BCA = ∠ DAC
∠ BAC = ∠ DCA
AC = CA (Common)
∴ ∆ ABC ≅ ∆ CDA (ASA rule)

Question 5.
Ray l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see the given figure). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Answer:
l is the bisector of ∠ PAQ and B is any point on l.
∴ ∠ PAB = ∠ QAB
BP and BQ are perpendiculars from B to AP and AQ.
∴ ∠ BPA = ∠ BQA = 90°.
Now, in ∆ APB and ∆ AQB,
∠ PAB = ∠ QAB
∠ BPA = ∠ BQA
AB = AB (Common)
∴ ∆ APB ≅ ∆ AQB (AAS rule)
∴ BP = BQ (CPCT)
BP and BQ are perpendiculars from B to arms AP and AQ of ∠ A.
∴ BP is the distance of B from AP and BQ is the distance of B from AQ.
Thus, B is equidistant from the arms of ∠ A.

Question 6.
In the given figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE

Answer:
∠ BAD = ∠ EAC
∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∴ ∠ BAC = ∠ DAE (Adjacent angles)
Now, in ∆ BAC and ∆ DAE,
AC = AE (Given)
AB = AD (Given)
∠ BAC = ∠ DAE
∴ ∆ BAC ≅ ∆ DAE (SAS rule)
∴ BC = DE (CPCT)

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see the given figure). Show that:
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE

Answer:
∠ BAD = ∠ ABE
∴ ∠ PAD = ∠PBE (∵ P lies on AB.)
∠ EPA = ∠ DPB
∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∴ ∠ APD = ∠ BPE (Adjacent angles)
P is the midpoint of AB.
∴ AP = BP
Now, in ∆ DAP and ∆ EBP
∠ PAD = ∠ PBE
∠ APD = ∠ BPE
AP = BP
∴ ∆ DAP ≅ ∆ EBP (ASA rule)
∴ AD = BE (CPCT)

Question 8.
In right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = \(\frac{1}{2}\) AB

Answer:
In ∆ AMC and ∆ BMD,
AM = BM (∵ M is the midpoint of AB.)
CM = DM (Given)
∠ AMC = ∠ BMD (Vertically opposite angles)
∴ By SAS rule, ∆ AMC ≅ ∆ BMD [Result (i)]
∴ ∠ MCA = ∠ MDB (CPCT)
∠ MCA and ∠ MDB are alternate angles formed by transversal CD of lines AC and BD and they are equal.
∴ AC || BD
Now, ∠ DBC and ∠ ACB are interior angles on the same side of transversal BC of AC || BD.
∴ ∠ DBC + ∠ ACB = 180°
∴ ∠ DBC + 90° = 180° (Given : ∠ C = 90°)
∴ ∠ DBC = 90°
Thus, ∠ DBC is a right angle. [Result (ii)]
Now, ∆ AMC ≅ ∆ BMD
∴ AC = BD
In ∆ DBC and ∆ ACB,
BD = CA
∠ DBC = ∠ ACB (Right angles)
BC = CB (Common)
∴ ∆ DBC ≅ ∆ ACB [Result (iii)]
∴ DC = AB (CPCT)
DM = CM and M lies on line’ segment CD.
∴ DC = 2 CM
∴ AB = 2CM
∴ \(\frac{1}{2}\)AB = CM
∴ CM = \(\frac{1}{2}\)AB

Punjab State Board PSEB 7th Class English Book Solutions English Vocabulary Synonyms Exercise Questions and Answers, Notes.

PSEB 7th Class English Vocabulary Synonyms

A Synonyms is a word which is same in meaning (एक समान अर्थ) to another word; as-

Word – Synonym

auspicious – happy/lucky
alive – lively
along – side by side
aim – objective
beautiful – pretty

bow – to bend
blaze – fire
believe – trust
banks – shores
courtier – a member of a royal court
comer – end point of some place
crack – rift
complete – full
cultural – traditional
celebrate – to hold make merry
deep – far-down
deities – gods/goddesses
diamond – a precious stone
dawn – morning
expect – hope
energy – power
famous – well-known
false – untrue
fright – panic, threaten
fervous – zeal
glance – look
gentle – kind
grey – brown
handsome – smart
honest – truthful
immobile – stationary
mounds – heaps/piles
nice – fine
overcome – win over
peace – calmness
patience – toleration
prepared – ready
paddy – rice crop
real – genuine
strong – tough
slumber – sound sleep
sight – scene
slightly – a little
stout – well built/strong
silent – quiet, motionless

smooth – plain
stuck – got busy
to pass – to move onward
tiny – little
together – along with
tense – nervous
trade – business
tribal – related to a tribe
ugly – bad looking
woods – jungle

Punjab State Board PSEB 7th Class English Book Solutions English Vocabulary Antonyms Exercise Questions and Answers, Notes.

PSEB 7th Class English Vocabulary Antonyms

An Antonym is a word which is opposite in meaning to another word; as

Word – Antonym

departure – arrival
ugly – beautiful
after – before

white – black
big – small
break – make
begin – end
buy – sell
complete – incomplete
clean – dirty
come – go
cruel – kind
cool – warm
catch – throw
care – neglect
calm – noisy
dawn – dusk
day – night
earth – sky
east – west
friend – enemy
front – behind
first – last
found – lost
give – take
good – bad
hate – love
here – there
hide – seek
happy – sad
hot – cold
in – out
late – early
love – hate
long – short
land – water
morning – evening
many – few
now – then
near – far
new – old
open – close
obey – disobey
proud – humble
push – pull
pretty – ugly
rich – poor
right – wrong/left
some – many
sit – stand
soon – late

tall – short
top – bottom
up – down
virtue – vice
wild – pet/domestic
win – lose
weak – strong

Punjab State Board PSEB 7th Class English Book Solutions English Grammar Verb (Conjugation) Exercise Questions and Answers, Notes.

PSEB 7th Class English Grammar Verb (Conjugation)

वे शब्द जो किसी क्रिया (action) को व्यकत करते हैं, Verbs कहलाते हैं। Verbs दो प्रकार के होते हैं- Main Verbs (मुख्य क्रियाएं) तथा helping verbs (सहायक क्रियाएं)

नीचे लिखे वाक्यों को पढ़िए:

1. Dev is sleeping.
2. Geeta made a doll.
3. They laughed.
4. The girls are going out.
5. Our team will play a match.

ऊपर दिए गए वाक्यों में sleeping, made, laughed, going तथा play मुख्य क्रियाएं (Main verbs) है। परन्तु is, are, will सहायक क्रियाएं (Helping Verbs) हैं। वे मुख्य क्रियाओं की सहायता करती हैं।
सहायक क्रियाएं निम्नलिखित हैं:
Is, am, are, was, were, has, have, had, do, does, did, will, would, shall, should, can, could, may, might, must, ought to, need, dare, used to.

Exercises

I. Pick out the helping verbs in the following sentences.

1. I have found a bag.
2. Where were you going yesterday ?
3. I am learning to swim.
4. He has lost his book.
5. You may go out now.
6. I will bring an orange for you.
7. They are cutting the hedge.
8. Where do you live?
9. I shall give you some money.
10. You can go now.
Hints:
1. have
2. were
3. am
4. has
5. may
6. will
7. are
8. do
9. shall
10. can.

II. Fill in the blanks with suitable helping verbs:

1. The teacher…………. not punish the whole class.
2. We………….go for a picnic tomorrow.
3. Where………….. the books kept ?
4. Who………….stolen my money?
5. Which games…………played in school ?
6. The winner…………..receive a prize from the chief guest.
7. …………you seen the Taj Mahal ?
8. When…………..your tutor come to teach you ?
9. Ravi…………go to London or America for his holidays.
10. ………….I help you with your problems ?

Hints:
1. did
2. shall
3. were
4. has
5. are
6. will
7. Have
8. does
9. will
10. May.

Punjab State Board PSEB 7th Class English Book Solutions English Grammar Translation Exercise Questions and Answers, Notes.

PSEB 7th Class English Grammar Translation

Present Indefinite Tense

1. मैं स्कूल जाता हूँ। – I go to school.
2. बच्चे खेलते हैं। – Children play.
3. माता जी खाना बनाते हैं। – Mother cooks food.
4. लड़के खेलते हैं। – Boys play.
5. मोहित दांत साफ़ करता है। – Mohit brushes his teeth.

Present Indefinite Tense (Interrogative)

1. क्या आप स्कूल जाते हैं ? – Do you go to school ?
2. आप क्या चाहते हो ? – What do you want ?
3. आप देर से क्यों आते हो ? – Why do you come late ?
4. आप कहाँ रहते हो। – Where do you live?
5. आपका स्कूल कब लगता है ? – When does your school start ?

Present Indefinite Tense (Negative)

1. रोहित झूठ नहीं बोलता है। – Rohit does not tell lies.
2. मैं खाना बेकार नहीं करता हूँ। – I don’t waste food.
3. आप गाना नहीं गाते हो। – You do not sing a song.
4. समय किसी के लिए भी नहीं रूकता है। – Time does not wait for anyone.
5. हम स्कूल लेट नहीं जाते। – We do not go to school late.

Past Indefinite Tense

1. मैं स्कूल गया। – I went to school.
2. आपने खाना खाया। – You took food.
3. मेरी माता जी ने भोजन बनाया। – My mother cooked food.
4. मैंने कल पाठ याद किया। – I learnt my lesson yesterday.
5. चपरासी ने घंटी बजाई। – The peon rang the bell.

Past Indefinite Tense (Interrogative)

1. क्या आप कल स्कूल गए थे ? – Did you go to school yesterday ?
2. आप कल कितने बजे बाज़ार गए थे ? – At what time did you go to the market yesterday ?
3. आप कल देर से क्यों आए ? – Why did you come late yesterday ?
4. उन्होंने मेरी बात क्यों नहीं सुनी ? – Why did they not listen to me ?
5. उषा ने खाना कब बनाया। – When did Usha cook food ?

Past Indefinite Tense (Negative)

1. मैंने झूठ नहीं बोला। – I did not tell a lie.
2. शरन ने दवाई नहीं ली। – Sharan did not take medicine.
3. रचना ने कसरत नहीं की। – Rachna did not take excercise.
4. भारत ने मन नहीं हारा। – Bharat did not lose heart.
5. चपरासी ने घंटी नहीं बजाई। – The poen did not ring the bell.

Future Indefinite Tense

1. कल बरसात होगी। – It will rain tomorrow.
2. मैं कल स्कूल आऊँगा। – I will come to school tomorrow.
3. वह कल दिल्ली जाएगा। – He will go to Delhi tomorrow.
4. आप कल रमेश से मिलोगे। – You will meet Ramesh tomorrow.
5. रमन कल मुझे ई-मेल करेगा। – Raman will send me an e-mail tomorrow.

Future Indefinite Tense (Negative)

1. बच्चे झूठ नहीं बोलेंगे। – The children will not tell lies.
2. पंछी गीत नहीं गाएंगे। – The birds will not sing songs.
3. नांव समुद्र में नहीं डूबेगी। – The boat will not sink in the sea.
4. आज बरसात नहीं होगी। – It will not rain today.
5. आप स्कूल से छुट्टी नहीं करोगे। – You will not take a leave from the school.

Future Indefinite Tense (Interrogative)

1. क्या आप कल मुझे मिलेगें ? – Will you meet me tomorrow ?
2. क्या चित्रकार चित्र बनाएगा ? – Will the painter paint a picture ?
3. क्या वह गाना गाएंगे ? – Will they sing a song ?
4. आप स्कूल कब आएंगे ? – When will you come to school ?
5. आप मेरा काम क्यों नहीं करोगे ? -Why will you not do my work ?

Present Continuous Tense

1. बच्चे पढ़ रहे हैं। – Children are reading.
2. अंजू कार चला रही है। – Anju is driving a car.
3. आपको पसीना आ रहा है। – You are perspiring/sweating.
4. पानी बहुत तेज़ बह रहा है। – Water is running very fast.
5. अध्यापक पढ़ रहा है। – The teacher is teaching.

Present Continuous Tense (Interrogative)

1. क्या पानी तेज़ी से बह रहा है ? – Is the water running fast ?
2. लोग वृक्षों की छाया क्यों ढूंढ़ रहे हैं ? – Why are the people looking for the shade of the trees. ?
3. क्या रानी निशी की मदद कर रही है ? – Is Rani helping Nishi ?
4. मोहन घर कब जा रहा है ? – When is Mohan going home ?
5. उसकी माता कौन-सा खाना बना रही है ? – Which food is her mother cooking ?

Present Continuous Tense (Negative)

1. छोटे बच्चे खाना नहीं खा रहे हैं। – Little children are not eating food.
2. रोहित गाड़ी तेज़ नहीं चला रहा है। – Rohit is not driving fast.
3. सीता अपना काम ठीक नहीं कर रही है। – Sita is not doing her work.
4. बरसात नहीं हो रही है। – It is not raining.
5. मकैनिक कार ठीक नहीं कर रहा है। – Mechanic is not repairing the car.

Past Continuous Tense

1. वह सफलता के लिए प्रार्थना कर रहा था। – He was praying for success.
2. मेरा भाई स्कूल जा रहा था। – My brother was going to school.
3. मां पैसे बचा रही थी। – Mother was saving money.
4. सिमर नई किताब लिख रही थी। – Simar was writing a new book.
5. बरसात हो रही थी। – It was raining.

Past Continuous Tense (Negative)

1. साइकिल तेज़ नहीं चल रही थी। – The bicycle was not running fast.
2. पंछी उड़ नहीं रहे थे। – The birds were not flying.
3. मेरी सहेली गाना नहीं गा रही थी। – My friend was not singing a song.
4. मेरी माता जी गुस्सा नहीं कर रही थी। – My mother was not angry.
5. बच्चे शोर नहीं मचा रहे थे। – The Children were not making a noise.

Past Continuous Tense (Interrogative)

1. क्या सभी अपना पाठ याद कर रहे थे ? – Were all learning their lesson ?
2. पारस अपना जन्मदिन कैसे मना रहा था ? – How was Paras celebrating his birthday ?
3. लेखक कौन सी किताब लिख रहा था ? – Which book was the writer writing ?
4. राकेश घर कब जा रहा होगा ? – When was Rakesh going home ?
5. बरसात कब हो रही थी ? – When was it raining ?

Future Continuous Tense

1. मैं कल बस से दिल्ली जा रहा होगा। – I will be going to Delhi by bus tomorrow.
2. मेरे माता जी इस वक्त खाना बना रहे होंगे। – My mother will be cooking food at this time.
3. प्रीति गाना गा रही होगी। – Preeti will be singing a song.
4. रचित फिल्म देख रहा होगा। – Rachit will be watching a movie.
5. कल मनजीत पेपर दे रहा होगा। – Manjeet will be taking a test tomorrow.

Future Continuous Tense (Negative)

1. अगले सप्ताह किसान फ़सल नहीं काट रहा होगा। – The farmer will not be reaping the crop next week.
2. परीक्षा के दौरान हम टी०वी० नहीं देख रहे होंगे। – We were not watching T.V. during the examinations.
3. मुस्कान शाम के समय चाय नहीं पी रही होगी। – Muskan will not be taking tea in the evening.
4. वह घर नहीं जा रही होगी। – She will not be going home.
5. बच्चा इस समय सो नहीं रहा होगा। – The child will not be sleeping by this time.

Future Continuous Tense (Interrogative)

1. गुरमुख शाम का क्या कर रहा होगा ? – What will he be doing in the evening ?
2. क्या कल सुबह सोनू पढ़ रहा होगा ? – Will Sonu be reading in the morning tomorrow ?
3. क्या नरेश नई किताब खरीद रहा होगा ? – Will Naresh be buying new book ?
4. क्या सोनू क्रिकेट खेल रहा होगा ? – Will Sonu playing cricket ?
5. निशी शाम को घर कैसे जा रही होगी ? – How will Nishi going home in the evening ?

Other Sentences

1. मेरा बात सुनो। – Listen to me.
2. विनम्रता से बात करो। – Speak gently.
3. अपनी किताब मुझे दो। – Give me your book.
4. एक-दूसरे को धक्का ना मारो। – Don’t push one another.
5. आपस में मत झगड़ो। – Don’t quarrel among yourselves.
6. वह आदमी अमीर है। – That man is rich.
7. चाय गर्म है। – Tea is hot.
8. वह घर सुन्दर है। – That house is beautiful.
9. वह अच्छा अध्यापक है। – He is a good teacher.
10. धरती सूर्य के इर्द-गिर्द घूमती है। – The earth moves around the Sun.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

1. Construct ΔABC such that AB = 4 cm, ∠B = 30°, BC = 4 cm. Also name the type of triangle on the basis of sides.
Solution:
Given : Two sides of ΔABC as AB = 4 cm, BC = 4 cm and ∠B = 30°.
To construct: A triangle with these two sides and included angle.
Step of Construction :
Step 1. We first draw a rough sketch of the ΔABC and indicate the measure of these two sides and included angle.

Step 2. Draw a line segment BC of length 4 cm.

Step 3. At B draw BX making an angle of 30° with BC (The point A must be somewhere on this ray of the angle).

Step 4. (To fix A, the distance AB has been given) With B as centre, draw an arc of radius 3 cm. It cuts BX at the point A.

Step 5. Join AC.
ΔDEF is now obtained.

Since two sides of triangle are equal.
Therefore ΔABC is an isosceles triangle.

2. Construct ΔABC with AB = 7.5 cm, BC = 5 cm and ∠B = 30°.
Solution:
Given. Two sides of ΔABC as AB = 7.5 cm,
BC = 5 cm
and ∠B = 30°
To construct A triangle with these two sides and included angle.
Steps of Construction.
Step 1. We first draw a rough sketch of the ΔABC and indicate the measures of these two sides and included angle.

Step 2. Draw a line segment BC of length 5 cm.

Step 3. At B draw BX making an angle of 30° with BC. (The point A must be somewhere on this ray of the angle)

Step 4. (To fix A; the distance BC has been given) With B as centre draw an arc of radius 7.5 cm. It cuts CX at the point A.

Step 5 : Join AC.
ΔABC is now obtained.

3. Construct a triangle XYZ, such that XY = 6 cm, YZ = 6 cm and ∠Y = 60°. Also name the type of this triangle.
Solution:
Step 1. Draw a rough sketch of XYZ with given measures.

Step 2. Draw a line segment XY of length 6 cm.

Step 3. With the help of compass, at Y, draw a ray YA making an angle 60°

Step 4. With Y as centre and radius 6 cm. draw an arc intersecting the ray YX at point Z.

Step 5. Join XZ.ΔXYZ is required triangle, Measure the third side. We see that ZX = 6 cm
∴ In Δ XYZ
XY = YZ = ZX = 6 cm
Therefore XYZ is an equilateral triangle.

4. Which of the following triangle can be constructed using SAS criterion.
(a) AB = 5 cm, BC = 5 cm, CA = 6 cm
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°
(c) ∠A = 60°, ∠B = 60°, ∠C = 60°
(d) BC = 5 cm, ∠B = ∠C = 45°
Answer:
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3

1. Carry out the following divisions:

Question (i)
28x⁴ ÷ 56x
Solution:

Question (ii)
– 36y³ ÷ 9y²
Solution:

Question (iii)
66pq²r³ ÷ 11qr²
Solution:

Question (iv)
34x³y³z³ ÷ 51 xy²z³
Solution:

Question (v)
12a⁸b⁸ ÷ (- 6a⁶b⁴)
Solution:

2. Divide the given polynomial by the given monomial:

Question (i)
(5x² – 6x) ÷ 3x
Solution:

Question (ii)
(3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
Solution:

Question (iii)
8 (x³y²z² + x²y³z² ÷ 4 x²y²z²)
Solution:

Question (iv)
(x³ + 2x² + 3x) ÷ 2x
Solution:

Question (v)
(P³ q⁶ – p⁶ q³) ÷ p³ q³
Solution:

3. Work out the following divisions:

Question (i)
(10x – 25) ÷ 5
Solution:
= \(\frac{10 x-25}{5}\)
= \(\frac{5(2 x-5)}{5}\)
= 2x – 5

Question (ii)
(10x-25) ÷ (2x – 5)
Solution:
= \(\frac{10 x-25}{2 x-5}\)
= \(\frac{5(2 x-5)}{(2 x-5)}\)
= 5

Question (iii)
10y (6y + 21) ÷ 5 (2y + 7)
Solution:
= \(\frac{10 y(6 y+21)}{5(2 y+7)}\)
= \(\frac{2 \times 5 \times y \times 3 \times(2 y+7)}{5(2 y+7)}\)
= 2 × y × 3
= 6y

Question (iv)
9x²y²(3z – 24) ÷ 27xy (z – 8)
Solution:

Question (v)
96 abc (3a – 12) (5b – 30) ÷ 144 (a – 4)(b – 6)
Solution:
= \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times a \times b \times c \times 3 \times(a-4) \times 5 \times(b-6)}{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4) \times(b-6)}\)
= 2 × 5 × a × b × c
= 10 abc

4. Divide as directed:

Question (i)
5 (2x + 1) (3x + 5) ÷ (2x + 1)
Solution:
= \(\frac{5(2 x+1)(3 x+5)}{(2 x+1)}\)
= \(\frac{5 \times(3 x+5)}{1}\)
= 5(3x + 5)

Question (ii)
26xy (x + 5) (y – 4) ÷ 13x (y – 4)
Solution:
= \(\frac{26 x y(x+5)(y-4)}{13 x(y-4)}\)
= \(\frac{2 \times 13 \times x \times y(x+5)(y-4)}{13 x(y-4)}\)
= 2y(x + 5)

Question (iii)
52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
Solution:
= \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)}\)
= \(\frac{52 \times p \times q \times r \times(p+q)(q+r)(r+p)}{2 \times 52 \times p \times q \times(q+r)(r+p)}\)
= \(\frac{r \times(p+q)}{2}\)
= \(\frac {1}{2}\)r (p + q)

Question (iv)
20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
Solution:
= \(\frac{20(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= \(\frac{2 \times 2 \times 5 \times(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= 2 × 2 × (y² + 5y + 3)
= 4(y² + 5y + 3)

Question (v)
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
Solution:
= \(\frac{x(x+1)(x+2)(x+3)}{x(x+1)}\)
= \(\frac{(x+2)(x+3)}{1}\)
= (x + 2)(x + 3)

5. Factorise the expressions and divide them as directed:

Question (i)
(y² + 7y + 10) ÷ (y + 5)
Solution:
First we factorise
y² + 7y + 10
= y² + 5y + 2y + 10
= y (y + 5) + 2 (y + 5)
= (y + 5) (y + 2)
∴ (y² + 7y + 10) ÷ (y + 5)
= \(\frac{(y+5)(y+2)}{(y+5)}\)
= y + 2

Question (ii)
(m² – 14m – 32) ÷ (m + 2)
Solution:
First we factorise
m² – 14m – 32
= m² – 16m + 2m – 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)
∴ (m² – 14m – 32) ÷ (m + 2)
= \(\frac{(m-16)(m+2)}{m+2}\)
= m – 16

Question (iii)
(5p² – 25p + 20) ÷ (p – 1)
Solution:
First we factorise
5p² – 25p + 20
= 5 (p² – 5p + 4)
= 5 (p² – 4p – p + 4)
= 5 [p (p – 4) – 1 (p – 4)]
= 5 (p – 4) (p – 1)
∴ (5p² – 25p + 20) ÷ (p – 1)
= \(\frac{5(p-4)(p-1)}{p-1}\)
= 5 (p – 4)

Question (iv)
4yz (z² + 6z – 16) ÷ 2y (z + 8)
Solution:
= \(\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}\)
= \(\frac{2 z\left(z^{2}+6 z-16\right)}{z+8}\)
Now, factorise
z² + 6z – 16
= z² + 8z – 2z – 16
= z (z + 8) – 2 (z + 8)
= (z + 8) (z – 2)
∴ 4yz (z² + 6z – 16) ÷ 2y (z + 8)
= \(\frac{2 z(z+8)(z-2)}{z+8}\)
= 2z (z – 2)

Question (v)
5pq (p² – q²) ÷ 2p(p + q)
Solution:

Question (vi)
12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
Solution:
= \(\frac{12 x y\left(9 x^{2}-16 y^{2}\right)}{4 x y(3 x+4 y)}\)
= \(\frac{12 x y(3 x-4 y)(3 x+4 y)}{4 x y(3 x+4 y)}\)
= 3 (3x – 4y)

Question (vii)
39y³ (50y² – 98) ÷ 26y² (5y + 7)
Solution:

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

1. Factorise the following expressions:

Question (i)
a² + 8a + 16
Solution:
= (a)² + 2 (a)(4) + (4)²
= (a + 4)²

Question (ii)
p² – 10p + 25
Solution:
= (p)² – 2 (p)(5) + (5)²
= (P – 5)²

Question (iii)
25m² + 30m + 9
Solution:
= (5m)² + 2 (5m) (3) + (3)²
= (5m + 3)²

Question (iv)
49y² + 84yz + 36z²
Solution:
= (7y)² + 2 (7y)(6z) + (6z)²
= (7y + 6z)²

Question (v)
4x² – 8x + 4
Solution:
= 4(x² – 2x + 1)
= 4 [(x)² – 2 (x)(1) + (1)²]
= 4 (x – 1)²

Question (vi)
121b² – 88bc + 16c²
Solution:
= (11b)² – 2 (11b)(4c) + (4c)²
= (11b – 4c)²

Question (vii)
(l + m)² – 4lm [Hint: Expand (1 + m)² first]
Solution:
= l² + 2lm + m² – 4lm
= l² + 2lm – 4lm + m²
= l² – 2lm + m²
= (l)² – 2 (l) (m) + (m)²
= (l – m)²

Question (viii)
a⁴ + 2a²b² + b⁴
Solution:
= (a²)² + 2 (a²)(b²) + (b²)²
= (a² + b²)²

2. Factorise:

Question (i)
4p² – 9q²
Solution:
= (2p)² – (3q)²
= (2p – 3q) (2p + 3q)

Question (ii)
63a² – 112b²
Solution:
= 7 (9a² – 16b²)
= 7 [(3a)² -(4b)²]
= 7 (3a – 4b) (3a + 4b)

Question (iii)
49x² – 36
Solution:
= (7x)² – (6)²
= (7x – 6) (7x + 6)

Question (iv)
16x⁵ – 144x³
Solution:
= 16x³(x² – 9)
= 16x³ [(x)² – (3)²]
= 16x³ (x-3) (x + 3)

Question (v)
(l + m)² – (l – m)²
Solution:
=[(l + m) + (l – m)] [(l + m) – (l – m)]
= [l + m + l – m] [l + m – l + m]
= (2l) (2m)
= 4lm

Question (vi)
9x²y² – 16
Solution:
= (3xy)² – (4)²
= (3xy – 4) (3xy + 4)

Question (vii)
(x² – 2xy + y²) – z²
Solution:
= (x – y)² – (z)²
= [(x – y) – z] [(x – y) + z]
= (x – y – z) (x – y + z)

Question (viii)
25a² – 4b² + 28bc – 49c²
Solution:
= (25a²) – (4b² – 28bc + 49c²)
= (5a)² – (2b – 7c)²
= [(5a) – (2b – 7c)] [(5a) + (2b – 7c)]
= (5a – 2b + 7c) (5a + 2b – 7c)

3. Factorise the expressions:

Question (i)
ax² + bx
Solution:
= x (ax + b)

Question (ii)
7p² + 21q²
Solution:
= 7 (p² + 3q²)

Question (iii)
2x³ + 2xy² + 2xz²
Solution:
= 2x(x² + y² + z²)

Question (iv)
am² + bm² + bn² + an²
Solution:
= am² + bm² + an² + bn²
= m² (a + b) + n²(a + b)
= (a + b) (m² + n²)

Question (v)
(lm + l) + m + 1
Solution:
= l (m + 1) + 1 (m + 1)
= (m + 1) (l + 1)

Question (vi)
y(y + z) + 9(y + z)
Solution:
= (y + z)(y + 9)

Question (vii)
5y² – 20y – 8z + 2yz
Solution:
= 5y² – 20y + 2yz – 8z
= 5y (y – 4) + 2z (y – 4)
= (y- 4) (5y + 2z)

Question (viii)
10ab + 4a + 5b + 2
Solution:
= 2a (5b + 2) + 1 (5b + 2)
= (5b + 2) (2a + 1)

Question (ix)
6xy – 4y + 6 – 9x
Solution:
= 6xy – 4y – 9x + 6
= 2y (3x-2)-3(3x-2)
= (3x-2) (2y – 3)

4. Factorise:

Question (i)
a⁴ – b⁴
Solution:
= (a²)² – (b²)²
= (a² – b²) (a² + b²)
= ((a)² – (b²)] (a² + b²)
= (a – b) (a + b) (a² + b²)

Question (ii)
p⁴ – 81
Solution:
= (p²)² – (9)²
= (p² – 9) (p² + 9)
= ((p)² – (3)²] (p² + 9)
= (p – 3)(p + 3)(p² + 9)

Question (iii)
x⁴ – (y + z)⁴
Solution:
= (x²)² – (a²)² (∵ y + z = a)
= (x² – a²) (x² + a²)
= (x – a) (x + a) (x² + a²)
= [x – (y + z)] [x + (y + z)] [x² + (y + z)²] (∵ a = y + z)
= (x – y – z) (x + y + z) [x² + (y + z)²]

Question (iv)
x⁴ – (x – z)⁴
Solution:
= (x²)² – [(x – z)²]²
= [x² – (x – z)²] [x² + (x – z)²]
= [x² – (x² – 2xz + z²)] [x² + (x² – 2xz + z²)]
= (x² – x² + 2xz – z²) (x² + x² – 2xz + z²)
= (2xz – z²) (2x² – 2xz + z²)
= z (2x – z) (2x² – 2xz + z²)

Question (v)
a⁴ – 2a²b² + b⁴
Solution:
= (a²)² – 2(a²)(b²) + (b²)²
= (a² – b²)²
= (a² – b²) (a² – b²)
= (a – b) (a + b) (a – b) (a + b)

5. Factorise the following expressions:

Question (i)
p² + 6p + 8
Solution:
= p² + 6p + 9 – 1
= (p² + 6p + 9) – (1)
= (p + 3)² – (1)²
= (p + 3 + 1) (p + 3 – 1)
= (P + 4) (p + 2)
Here, last term is 8.
∴ 9 – 1 = 8.

OR
p² + 6p + 8
Here, ab = 8 and a + b = 6
On solving equations, a = 4, b = 2
Now, p² + 6p + 8
= p² + 4p + 2p + 8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question (ii)
q² – 10q + 21
Solution:
= q² – 10q + 25 – 4
= (q² – 10q + 25) – (4)
= (q – 5)² – (2)²
= (q – 5 + 2) (q – 5 – 2)
= (q – 3) (q – 7)
Here, last term is 21.
∴ 25 – 4 = 21.

OR
q² – 10q + 21
Here, ab = 21 and a + b = (- 10)
Possible values of a = 7 or (-7)
b = 3 or (- 3)
Let us check, 7 + 3 = 10 ≠ (- 10)
∴ a = – 7, b = – 3
Now, q² – 10q + 21
= q² – 7q – 3q + 21
= q (q – 7) – 3 (q – 7)
= (q – 7) (q – 3)

Question (iii)
p² + 6p – 16
Solution:
= p² + 6p + 9 – 25
= (P² + 6p + 9) – (25)
= (p + 3)² – (5)²
= (p + 3 – 5) (p + 3 + 5)
= (p – 2) (p + 8)
Here, last term is (-16).
∴ (-25) + 9 = (-16)

OR

p² + 6p – 16
Here, ab = – 16 and a + b = 6
Possible values of a = 8 or (-8) b = 2 or (-2)
Let us check, 8 + 2 = 10 ≠ 6
(- 8) + 2 = (-6) ≠ 6
8 + (-2) = 8 – 2 = 6
Now, p² + 6p – 16
= p² + 8p – 2p – 16
= P (P + 8) – 2 (p + 8)
= (p + 8) (p – 2)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1

1. Find the common factors of the given terms:

Question (i)
12x, 36
Solution:
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Common factors of 12x and 36 = 2 × 2 × 3
= 12

Question (ii)
2y, 22xy
Solution:
2y = 2 xy
22xy = 2 × 11 × x × y
∴ Common factors of 2y and 22xy
= 2 × y = 2y

Question (iii)
14pq, 28p²q²
Solution:
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
∴ Common factors of 14pq and 28p²q²
= 2 × 7 × p × q = 14pq

Question (iv)
2x, 3x², 4
Solution:
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
∴ Common factors of 2x, 3x² and 4 = 1 [Note: 1 is a factor of each term.]

Question (v)
6abc, 24ab², 12a²b
Solution:
6abc = 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
∴ Common factors of 6abc, 24ab² and 12a²b
= 2 × 3 × a × b
= 6ab

Question (vi)
16x³, – 4x², 32x
Solution:
16x³ = 2 × 2 × 2 × 2 × x × x × x
– 4x² = (-1) × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ Common factors of 16x³, – 4x² and 32x = 2 × 2 × x = 4x

Question (vii)
10pq, 20qr, 30rp
Solution:
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ Common factors of 10pq, 20qr and 30rp = 2 × 5 = 10

Question (viii)
3x²y³, 10x³y², 6x²y²z
Solution:
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ Common factors of 3x²y³, 10x³y² and 6x²y²z
= x × x × y × y = x²y²

2. Factorise the following expressions:

Question (i)
7x – 42
Solution:
7x = 7 × x
42 = 2 × 3 × 7
∴ 7 is common in both terms.
7x – 42 = 7 (x – 6)

Question (ii)
6p – 12q
Solution:
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
∴ 2 × 3 = 6 is common in both terms.
6p – 12q = 6 (p – 2q)

Question (iii)
7a² + 14a
Solution:
7a² = 7 × a × a
14a = 2 × 7 × a
∴ 7 × a = 7a is common in both terms.
∴ 7a² + 14a = 7a (a + 2)

Question (iv)
– 16z + 20z³
Solution:
– 16z = -2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
∴ 2 × 2 × z = 4z is common in both terms.
∴ – 16z + 20z³ = 4z (- 4 + 5z²)

Question (v)
20l²m + 30alm
Solution:
= 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m
= 10lm (2l + 3a)

Question (vi)
5x²y – 15xy²
Solution:
= 5xy (x – 3y)

Question (vii)
10a² – 15b² + 20c²
Solution:
= 5 (2a² – 3b² + 4c²)

Question (viii)
– 4a² + 4ab – 4ca
Solution:
= 4a (- a + b – c)

Question (ix)
x²yz + xy²z + xyz²
Solution:
= xyz (x + y + z)

Question (x)
ax²y + bxy² + cxyz
Solution:
= xy (ax + by + cz)

3. Factorise:

Question (i)
x² + xy + 8x + 8y
Solution:
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)

Question (ii)
15xy – 6x + 5y – 2
Solution:
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

Question (iii)
ax + bx- ay – by
Solution:
= x (a + b) – y (a + b)
= (a+ b) (x- y)

Question (iv)
15pq + 15 + 9q + 25p
Solution:
= 15pq + 9q + 25p + 15
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

Question (v)
z – 7 + 7xy – xyz
Solution:
= z – 7 – xyz + 7xy
= 1 (z – 7) – xy (z – 7)
= (z – 7) (1 – xy)