Bhagya

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Parts of Speech Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Parts of Speech

शब्दों को प्रयोग के अनुसार आठ भागों अथवा श्रेणियों में विभाजित किया जाता है, जिन्हें Parts of Speech (शब्द भेद) कहते हैं। ये आठ भाग हैं

1. Noun
2. Pronoun
3. Adjective
4. Verb
5. Adverb
6. Preposition
7. Conjunction
8. Interjection.

1. Noun- किसी व्यक्ति, स्थान, वस्तु अथवा गुण-दोष के नाम को Noun (संज्ञा) कहते हैं; जैसे Ram, Shyam, Mumbai, Pen, Pencil, Rose, Dog, Beauty, Honesty आदि।

2. Pronoun- जो शब्द किसी संज्ञा के स्थान पर प्रयोग होता है, उसे Pronoun (सर्वनाम) कहते हैं; जैसे – I, we, you, they, he, she, it आदि।

3. Adjective- जो शब्द किसी Noun अथवा Pronoun की विशेषता प्रकट करता है, उसे Adjective (विशेषण) कहते हैं; जैसे good, bad, thin, fat, tall, many, useful आदि। .

4. Verb- वह शब्द जिससे कार्य का होना अथवा करना प्रकट हो या जो शब्द किसी व्यक्ति, स्थान अथवा वस्तु के बारे में कुछ बतलाए, वह शब्द Verb (क्रिया) कहलाता है; जैसे-

• The girl reads a book.
• Ram is a good boy.
• Mohan goes to school daily.
• Seema feels sad.
• I helped the poor.

5. Adverb- जो शब्द किसी Verb, Adjective अथवा किसी अन्य Adverb की विशेषता प्रकट करे, तो उसे Adverb (क्रिया विशेषण) कहते हैं; जैसे-

• He ran fast. (Verb की विशेषता)
• Rose is very beautiful. (Adjective की विशेषता)
• He works too slowly. (Adverb की विशेषता)

6. Preposition- जो शब्द किसी Noun अथवा Pronoun से पहले प्रयोग किया जाता है और जो उनका सम्बन्ध वाक्य के अन्य शब्दों से जोड़ता है, उसे Preposition (सम्बन्धवाचक शब्द) कहते हैं; जैसे-

• Mohan is at the gate.
• Ram is proud of his teacher.
• He is fond of music.
• I am writing with a pen.

7. Conjunction-जो शब्द, शब्दों अथवा वाक्यों को जोड़ता है, उसे Conjunction (संयोजक) कहते हैं; जैसे-

• Ram and Shyam are friends.
• You will get through if you work hard.
• He was late because he missed the bus.
• Either he or his brother is at fault.

8. Interjection- अकस्मात् भावना को व्यक्त करने वाला शब्द Interjection (विस्मयादिबोधक अव्यय)
कहलाता है; जैसे-, Alas ! Hurrah! Oh! Lo ! आदि।

Exercises From Board’s Grammar (Solved)

I. Write in the space provided the name of the part of speech to which the underlined words belong in the following sentences:

1. Seema is a beautiful girl.
2. Alas! His dog is dead.
3. The sun sets in the west.
4. The lion is a ferocious animal.
5. Delhi is a very big city.
6. Rahim is poor but honest.
7. Honesty is the best policy.
8. The cat is under the table.
Hints:
1. Adjective
2. Interjection
3. Verb
4. Adjective
5. Adverb
6. Conjunction
7. Noun
8. Preposition.

II. Complete the following sentences with appropriate ‘Nouns’:

1. ………. is a good boy.
2. She goes to the ……….. everyday.
3. They go for a ……… daily.
4. ……… is the Capital of India.
5. ……….. is the best policy.
6. The ……….. rises in the …………….
7. Chandigarh is the ………. of Punjab and Haryana.
8. Shimla is a beautiful ………..
Hints:
1. Mohan
2. temple/Gurudwara
3. walk
4. Delhi
5. Honesty
6. sun, east
7. capital
8. place.

III. Fill in the blanks with suitable ‘Pronouns:

1. My son is playing with ………. toys.
2. ……….. father is working in Mumbai.
3. ………. has gone abroad for higher studies.
4. This is ……….. house.
5. The girls are doing ……….. homework.
6. ……… is the bread-winner of the family.
7. We should respect ……… parents.
8. He loves ……….. native place very much.
Hints:
1. his
2. My
3. He
4. our
5. their
6. She
7. our
8. his.

IV. Fill in the blanks with suitable ‘Adjectives:

1. She is my ……….. friend.
2. Pudding is my ……….. dish.
3. The scenery of Mussoorie is ………..
4. She likes to wear ……….. dresses.
5. The Bible is a ……….. book.
6. Cricket is a ……….. game.
7. John is an ……….. teacher.
8. Mango is a ……….. fruit.
Hints:
1. best
2. favourite
3. beautiful
4. white
5. holy
6. good
7. English
8 juicy.

V. Fill in the blanks with suitable ‘Verbs’:

1. She ……….. ice-cream.
2. They ……….. a lot.
3. My mother ……….. food.
4. John ………. in a factory.
5. They ………. football…
6. We should ………. a bath everyday.
7. The school peon ……….. the bell.
8. Seema ……… for a walk daily.
Hints:
1. is eating
2. worked
3. cooks
4. works
5. play
6. take
7. rings
8. goes.

VI. Fill up the blanks with suitable ‘Adverbs?:

1. Sohan walks ………….
2. Everyone should work ……..
3. She sings …………
4. His dad is a ……… respectable man.
5. Our teacher speaks ………. politely.
6. My sister sleeps ………
7. John is a ………. hardworking boy.
8. Girls sang …………..
Hints:
1. slowly
2. hard
3. sweetly
4. very
5. very
6. early
7. very
8. nicely.

VII. Complete the following sentences with suitable ‘Prepositions:

1. My grandfather is hard ………. hearing.
2. My mom is fond ……….. music.
3. A burglar broke ……….. our house last night.
4. The students should listen ……… their teachers attentively.
5. Yesterday we went ………. the Rose Garden.
6. Ayushi is playing ……….. the piano.
7. Children have been playing ……….. morning.
8. My uncle has been living ………. Canada ………. fifteen years.
9. I will play …….. finishing my homework.
10. I am standing ………. Anshu. Anshu is in front of me.
Hints:
1. of
2. of
3. into
4. to
5. to
6. on
7. since
8. in, for
9. after
10. behind.

VIII. Fill in the blanks with suitable Conjunctions:

1. My younger brother is both intelligent ……….. hardworking.
2. He says ………. he is a doctor.
3. ……… Seema ………. Rita is at fault.
4. Our servant is poor ………. honest.
5. He ………. his nephew manage the shop.
6. Ram went on leave ………. he was injured.
7. You will get the tickets ……….. you reach there before 6 o’clock.
8. ……….. he was late, yet he was able to catch the bus.
Hints:
1. and
2. that
3. Either, or
4. but
5. and
6. because
7. if
8. Although

IX. Fill in the blanks with suitable ‘Interjections:

1. ……….. We have won the game
2. ………… The man is dead.
3. ………….. They have come.
4. ………….. Is this the place?
5. ……………. Well done.
6. ……………. He has lost all his money in gambling.
7. ……….. You are hurt.
8. ………….. He has failed again.

Hints:
1. Hurrah !
2. Alas!
3. Lo !
4. Oh !
5. Bravo!
6. Alas!
7. Oh !
8. Alas!

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Synonyms Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Synonyms

A Synonym is a word with the same meaning (एक सामान अये) as another word; as

Word – Synonym
abode – home
actual – real
allow – permit
add – plus
annual – yearly
arrive – reach
beauty – loveliness
beautiful – pretty
begin – start
big – large
brave – bold
briefa – short
calm – peaceful
centre – middle
clever – intelligent

close – shut
costly – expensive
corn – grain
damp – wet
daily – everyday
definite – certain
difficult – tough
enemy – foe
example – instance
excellent – wonderful
fear – terror/horror
foolish – stupid/silly
fun – enjoyment
grief – sorrow
happy – glad
hollow – empty
hot – warm
kind – generous
loving – affectiodate
perfect – ideal
quiet – silent
reply – answer
right – correct
small – tiny
smell – scent
soft – tender
taste – flavour
timeless – unending
unitet – cooperate
vacant – empty
wealthy – rich
wide – broad

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary One Word Substitution Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary One Word Substitution

Group of Words	Words
1. one who has become dependent on something	Addict
2. one who produces works of art	Artist
3. a number of sheep	Flock
4. one who writes books	Author
5. one who cuts hair	Barber
6. a building in which aircraft housed	Hanger
7. a building where an audience	auditorium
8. one who plays a note in a play or movie	sits
9. one who treats all teeth problems	Actor
10. one who grows crops	Dentist
11. one who works in gold	Farmers
12. a period of ten years	Goldsmith
13. a performance given by a number of musicians	decade
14. one who works in iron	Concert
15. one who sells meat	Blacksmith
16. one who works with wood	Butcher
17. a study of animals	Carpenter
18. one who lives at the same time	Zoology
19. one who sells medicines	Contemporary
20 one who mends shoes	Chemist
21. one who settles in another country	Cobbler
22. one who treats the sick	immigrant
23. person working in the same place	Doctor
24. one who sells flowers	Colleague
25. a doctor who treats mental illness	Florist

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Words as per Dictionary Order Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Words as per Dictionary Order

यदि English भाषा के शब्दों को अंग्रेज़ी Albhabet (A, B, C ….. Y, Z) के क्रम में लिखा जाए तो शब्दों का Dictionary order या Alphabetical Order कहा जाता है। Dictionary में कोई शब्द ढूंढ़ने के लिए हमारे लिए इस क्रम को जानना बहुत ज़रूरी है। वैसे भी शब्दों के किसी समूह को जब Dictionary Order में लिखा जाता है, तो उसमें से किसी भी शब्द को ढूंढ़ना आसान हो जाता है।

Dictionary Order में लिखने का तरीका-किसी दिए गए शब्द समूह को Dictionary Order में लिखने के लिए निम्नलिखित विधि अपनाएं:
1. सबसे पहले शब्दों के पहले letter (अक्षर) पर ध्यान दें। Alphabet में पहले आने वाले अक्षर से शुरू होने वाला शब्द सबसे पहले आयेगा। जैसे-पहले A, फिर B …..
2. यदि दो या दो से अधिक शब्दों का letter एक जैसा हो तो शब्द के दूसरे letter को follow करें।
3. हो सकता है किन्हीं दो शब्दों के पहले दो अक्षर एक जैसे हों। ऐसे में तीसरे अक्षर (letter) के अनुसार शब्दों को क्रम-बद्ध किया जाएगा।

इसी विधि का अनुसरण करते हुए आगे बढ़ते जाएं, जब तक कि पूरा शब्द-समूह क्रमबद्ध न हो जाए।

Worked Out Examples

Write the given words as per dictionary order:
Question 1.
Apple, Orange, Banana, Mango, Grapes.
Answer:
Apple, Banana, Grapes, Mango, Orange.

Question 2.
Tomato, Potato, Carrot, Radish, Onion.
Answer:
Carrot, Onion, Potato, Radish, Tomato.

Question 3.
Active, Absent, Addition, Aeroplane, Almond.
Answer: Absent, Active, Addition, Aeroplane, Almond.

Question 4.
January, June, July, April, August.
Answer:
April, August, January, July, June.

Question 5.
Black, Brown, Birthday, Blood, Beautiful.
Answer:
Beautiful, Birthday, Black, Blood, Brown.

Question 6.
Earth, Essay, Father, Farmer, Brother.
Answer:
Brother, Earth, Essay, Father, Farmer.

Question 7.
Watch, Which, Small, Smell, Bitch.
Answer:
Bitch, Small, Smell, Watch, Which.

Question 8.
School, Office, Post-office, Soldier, Knife.
Answer:
Knife, Office, Post-office, School, Soldier.

Question 9.
Barber, Author, Cobbler, Actor, Dancer.
Answer:
Actor, Author, Barber, Cobbler, Dancer.

Question 10.
Bear, Horse, Tiger, Lion, Lamb.
Answer:
Bear, Horse, Lamb, Lion, Tiger.

Exercise for Practice

Arrange the following group of words in Alphabetical/Dictionary Order:
1. Teacher, Artist, Student, Tailor, Gardener.
2. Bird, Book, Cycle, Circle, Circus.
3. Music, Monitor, May, Marker, March.
4. Agra, Delhi, Dehradun, Capital, Child.
5. Table, Toy, Chair, Friend, Tree.
6. Mouse, Magician, Rabbit, Manager, Prince.
7. King, Kite, Zebra, Zoo, Shirt
8. Grass, Green, Leaves, Pitcher, Picture.
9. Morning, Evening, Night, Afternoon, Dawn.
10. Coffee, Indian, Inkpot, Insect, Mountain.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3); (4, 1)
(ii)(-5, 7); (-1, 3)
(iii) (a, b); (-a, -b).
Solution:
(i) Given points are: (2, 3); (4, 1)
Required distance = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
\(\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}\)
= 2√2.

(ii) Given points are: (-5, 7); (-1, 3)
Required distance = \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
\(\sqrt{16+16}=\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= 4√2.

(iii) Given points are: (a, b); (-a, -b)
Required distance = \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{4 a^{2}+4 b^{2}}\)
= √4 \(\sqrt{a^{2}+b^{2}}\)
= \(2 \sqrt{a^{2}+b^{2}}\)

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B
discussed in section 7.2.
Solution:
Given points are: A (0, 0) and B (36, 15)
Distance, AB = \(\sqrt{(0-36)^{2}+(0-15)^{2}}\)
\(\sqrt{1296+225}=\sqrt{1521}\) = 39.
According to Section 7.2
Draw the distinct points A (0, 0) and B (36, 15) as shown in figure.

Draw BC ⊥ on X-axis.
Now, In rt. ∠d ∆ACB,
AB = \(\sqrt{\mathrm{AC}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(36)^{2}+(15)^{2}}\)
= \(\sqrt{1296+225}=\sqrt{1521}\)
= 39.
Hence, required distance between points is 39.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Given point are : A (1. 5); B (2.3) and C (- 2, – 11).
AB = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
= \(\sqrt{1+4}=\sqrt{5}\)

BC = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
= \(\sqrt{16+196}=\sqrt{212}\)

CA = \(\sqrt{(1+2)^{2}+(5+11)^{2}}\)
= \(\sqrt{9+256}=\sqrt{265}\)
From above distances, it is clear that sum of any two is not equal to third one.
Hence, given points are not collinear

Question 4.
Check whether (5, – 2); (6, 4) and (7, – 2) are the Vertices of an isosceles triangle.
Solution:
Given points be A (5, – 2); B (6, 4) and C (7, – 2).
AB = \(\sqrt{(5-6)^{2}+(-2-4)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

BC = \(\sqrt{(6-7)^{2}+(4+2)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

CA = \(\sqrt{(7-5)^{2}+(-2+2)^{2}}\)
= \(\sqrt{4+0}=2\)
From above discussion, it is clear that AB = BC = √37.
Given points are vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square”? Chameli disagrees. Using distance formula, find which of them is correct, and why?

Solution:
In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).
Now,
AB = \(\sqrt{(6-3)^{2}+(7-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

BC = \(\sqrt{(9-6)^{2}+(4-7)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

CD = \(\sqrt{(6-9)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

DA=\(\sqrt{(3-6)^{2}+(4-1)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

AC = \(\sqrt{(9-3)^{2}+(4-4)^{2}}\)
= \(\sqrt{36+0}=6\)

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\)
= \(\sqrt{0+36}\) = 6
From above discussion, it is clear that
AB = BC = CD = DA = √18 and AC = BD = 6.
ABCD formed a square and Champa is correct about her thinking.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1,- 2), (1, 0),(- 1, 2), (- 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2).
Solution:
(i) Given points be A (- 1, – 2); B(1, 0); C(- 1, 2) and D(- 3, 0).
AB = \(\sqrt{(1+1)^{2}+(0+2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

BC = \(\sqrt{(-1-1)^{2}+(2-0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

CD = \(\sqrt{(-3+1)^{2}+(0-2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

DA = \(\sqrt{(-1+3)^{2}+(-2+0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

AC = \(\sqrt{(-1+1)^{2}+(2+2)^{2}}\)
= \(\sqrt{0+16}=4\)

BD = \(\sqrt{(-3-1)^{2}+(0-0)^{2}}\)
= \(\sqrt{16+0}=4\)

From above discussion, it is clear that
AB = BC = CD = DA = √8 and AC = BD = 4.
Hence, given quadrilateral ABCD is a square.

(ii) Given points be A (- 3, 5); B (3, 1); C (0, 3) and D (- 1,- 4)
AB = \(\sqrt{(-3-3)^{2}+(5-1)^{2}}\)
= \(\sqrt{36+16}=\sqrt{52}=\sqrt{4 \times 13}\)
= 2√13

BC = \(\sqrt{(3-0)^{2}+(1-3)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)

CA = \(\sqrt{(0+3)^{2}+(3-5)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)
Now, BC + CA = \(\sqrt{13}+\sqrt{13}\) = 2√13 = AB
∴A, B and C are collinear then A, B, C and D do not form any quadrilateral.

(iii) Given points are A (4, 5); B (7, 6); C (4, 3) and D (1, 2)
AB = \(\sqrt{(7-4)^{2}+(6-5)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

BC = \(\sqrt{(4-7)^{2}+(3-6)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

CD = \(\sqrt{(1-4)^{2}+(2-3)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

DA = \(\sqrt{(4-1)^{2}+(5-2)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

AC = \(\sqrt{(4-4)^{2}+(3-5)^{2}}\)
= \(\sqrt{0+4}\) = 2

BD = \(\sqrt{(1-7)^{2}+(2-6)^{2}}\)
= \(\)

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.
i.e., opposite sides are equal but their diagonals are not equal.
Hence, given quadrilateral ABCD is a parallelogram.

Question 7.
Find the points on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).
According to question,
PA = PB
(PA)² = (PB)²
or (2 – x)² + (- 5- 0)² = (- 2 – x)² + (9 – 0)²
or 4 + x² – 4x + 25 = 4 + x²+ 4x + 81
-8x = 56
x = \(\frac{4}{4}\) = – 7
Hence, required point be (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
Given points are P (2, – 3) and Q (10, y)
PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)
= \(\sqrt{64+y^{2}+9+6 y}\)
= \(\sqrt{y^{2}+6 y+73}\)
According to question,
PQ = 10
or \(\sqrt{y^{2}+6 y+73}\) = 10
Squaring
or y² + 6y + 73 = 100
or y² + 6y – 27 = 0
or y² + 9y – 3y – 27 = 0
S = 6 P = – 27
or y (y + 9) – 3 (y + 9) = 0
or (y + 9) (y – 3) = 0
Either y + 9 = 0 or y – 3 = 0
y = – 9 or y = 3
Hence, y = – 9 and 3.

Question 9.
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Given points Q (0, 1); P (5, – 3) and R (x, 6)
QP = \(\sqrt{(5-0)^{2}+(-3-1)^{2}}\)
= \(\sqrt{25+16}=\sqrt{41}\)

and QR = \(\sqrt{(x-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{x^{2}+25}\)

According to question,
QP = QR
or \(\sqrt{41}=\sqrt{x^{2}+25}\)
Squaring
or 41 = x² + 25
or x² = 16
or x = ± √16 = ± √4.

When x = 4 then R (4, 6).
QR = \(\sqrt{(4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{1+81}=\sqrt{82}\)

When x = – 4 then R (- 4, 6).
QR = \(\sqrt{(-4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(-4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{81+81}=\sqrt{162}\).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let required points be P (x, y) and given points are A (3, 6) and B (- 3, 4)
PA = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
= \(\sqrt{9+x^{2}-6 x+36+y^{2}-12 y}\)
= \(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\)

and PB = \(\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
= \(\sqrt{9+x^{2}+6 x+16+y^{2}-8 y}\)
= \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)

According to question,
PA = PB
\(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\) = \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)
sq,. both sides, we have,
or x² + y² – 6x – 12y + 45 = x² + y² + 6x – 8y – 25
or -12x – 4y + 20 = 0
or 3x + y – 5 = 0 is the required relation.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Answer:
For the cuboidal matchbox, length l = 4 cm; breadth b = 2.5 cm and height h = 1.5 cm.
Volume of a cuboidal matchbox
= l × b × h
= 4 × 2.5 × 1.5 cm³
= 15 cm³
Then, volume of 12 matchboxes = 12 × 15 cm³ = 180 cm³
Thus, the volume of a packet containing 12 matchboxes is 180 cm³.

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 l)
Answer:
For the cuboidal water tank, length l = 6m; breadth b = 5 m and height h = 4.5 m.
Capacity of the cuboidal tank = l × b × h
= 6 × 5 × 4.5 m³
= 135 m³
1 m³ = 1000 litres
∴ 135 m³ = 135000 litres
Thus, the given cuboidal water tank can hold 1,35,000 litres of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer:
For the cuboidal vessel, length l = 10 m;
breadth b = 8 m and capacity = 380 m³.
Capacity of a cuboidal vessel = l × b × h
∴ 380 m³ = 10 m × 8 m × h m
∴ h = \(\frac{380}{10 \times 8}\) m
∴ h = 4.75 m
The height of the cuboidal vessel must be made 4.75 m.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹ 30 per m³.
Answer:
For the cuboidal pit, length l = 8m; breadth b = 6 m and height (depth) h = 3 m.
Volume of the earth to be dugout to make the cuboidal pit = Volume of a cuboid
= l × b × h
= 8 × 6 × 3 m³
= 144 m³
Cost of digging out 1 m³ of earth = ₹ 30
∴ Cost of digging out 144 m³ of earth
= ₹ (30 × 144)
= ₹ 4320
Thus, the cost of digging the cuboidal pit is ₹ 4320.

Question 5.
The capacity of a cuboidal tank is 50,000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Answer:
For the cuboidal tank, length l = 2.5 m;
height (depth) h = 10 m and
capacity = 50,000 litres.
1000 litres = 1 m³
∴ 50,000 litres = \(\frac{50,000}{1000}\) m³ = 50 m³
Capacity of cuboidal tank = l × b × h
∴ 50 m³ = 2.5 m × b m × 10 m
∴ b = \(\frac{50}{2.5 \times 10}\) m
∴ b = 2 m
Thus, the breadth of the cuboidal tank is 2 m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Answer:
Total requirement of water per day
= No. of people × daily requirement per person
= 4000 × 150 litres
= 6,00,000 litres
= \(\frac{6,00,000}{1000}\) m³
= 600 m³
For the cuboidal tank, length l = 20 m;
breadth b = 15 m and height h = 6 m
Capacity of the cuboidal tank = l × b × h
= 20 × 15 × 6 m³
= 1800 m³
600 m³ of water can last for 1 day in the village.
∴ 1800 m³ of water can last for \(\frac{1800}{600}\) = 3 days in the village.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Answer:
For the cuboidal godown, length l = 40 m;
breadth b = 25 m and height h = 15 m.
Capacity of cuboidal godown = l × b × h
= 40 × 25 × 15 m³
For the wooden cuboidal crate, length l = 1.5 m; breadth b = 1.25 m and height h = 0,5 m.
Volume of 1 cuboidal crate
= l × b × h
= 1.5 × 1.25 × 0.5 m³
∴The no. of crates that can be stored in the godown = \(\)
= \(\left(\frac{40}{1.25}\right) \times\left(\frac{25}{0.5}\right) \times\left(\frac{15}{1.5}\right)\)
= 32 × 50 × 10
= 16,000

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube ? Also, find the ratio between their surface areas.
Answer:
For the original cube, edge a =12 cm.
Volume of original cube = a³ = 123 cm³
= 1728 cm³
8 cubes of equal volume are made from the original cube.
∴ Volume of each new cube = \(\frac{1728}{8}\) cm³
= 216 cm³
Let the edge of new cube be A cm.
Volume of new cube = A³
∴ 216 cm³ = A³
∴ A = \(\sqrt[3]{216}\) cm = 6 cm
Thus, the side of each new cube is 6 cm.
Total surface area of original cube
= 6a²
= 6 (12)² cm²
Total surface area of a new cube = 6A²
= 6 (6)² cm²
\(\frac{\text { Total surface area of original cube }}{\text { Total surface area of a new cube }}\) = \(\frac{6(12)^{2} \mathrm{~cm}^{2}}{6(6)^{2} \mathrm{~cm}^{2}}\)
= \(\left(\frac{12}{6}\right)^{2}\)
= 4
= 4:1
Thus, the required ratio of the total surface area of the original cube and the total surface area of a new cube is 4:1.
Note: If the ratio of TSA of the original ‘ cube and TSA of all the new cubes is required, then it will be 1 : 2.

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer:
2 km = 2000 m and 1 hour = 60 minutes
Rate of flow of water in the river
= 2 km/hour
= \(\frac{2000}{60}\) m/min
Thus, during 1 minute, water of length will flow in the sea.
Then, the water falling in sea per minute takes cuboidal shape with length l = \(\frac{2000}{60}\) m,
breadth b = 40 m and height (depth) h = 3 m.
Volume of water falling in sea per minute
= l × b × h
= \(\frac{2000}{60}\) × 40 × 3 m³
= 4000 m³
Thus, 4000 m³ of water will fall into the sea in a minute.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm
Answer:
For the given sphere,
radius r = 10.5 cm = \(\frac{21}{2}\) cm.
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm² = 1386 cm²

(ii) 5.6 cm
Answer:
For the given sphere, radius r = 5.6cm.
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × 5.6 × 5.6 cm²
= 394.24 cm²

(iii) 14 cm
Answer:
For the given sphere, radius r= 14 cm.
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × 14 × 14 cm²
= 2464 cm²

Question 2.
Find the surface area of a sphere of diameter:
(i) 14cm
Answer:
For the given sphere, diameter d = 14 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{14}{2}\) cm = 7 cm
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × 7 × 7 cm²
= 616 cm²

(ii) 21cm
Answer:
For the given sphere, diameter d = 21 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{21}{2}\) cm
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm²
= 1386 cm²

(iii) 3.5 m
Answer:
For the given sphere, diameter d = 3.5 cm.
Then, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) m
= \(\frac{35}{20}\) m
Surface area of a sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{35}{20}\) × \(\frac{35}{20}\) m²
= 38.5 m²

Note: We can also use the formula “Surface area of a sphere = πd²” as
4πr² = π × 4r² = π × (2r)² = πd², where r and d are radius and diameter of the sphere respectively.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π =3.14)
Answer:
For the given hemisphere, radius r = 10 cm.
Total surface area of a hemisphere
= 3πr²
= 3 × 3.14 × 10 × 10 cm²
= 942 cm²

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Answer:
For the first case, radius r₁ of the spherical balloon is 7 cm.
Surface area of the spherical balloon in the
first case = 4πr₁²
= 4 × \(\frac{22}{7}\) × 7 × 7cm²
For the second case, radius r₂ of the spherical balloon is 14 cm.
Surface area of the spherical balloon in the second case = 4πr₂²
Then, the required ratio of surface areas in two cases
= \(\frac{4 \times \frac{22}{7} \times 7 \times 7}{4 \times \frac{22}{7} \times 14 \times 14}\)
= \(\frac{1}{4}\) = 1 : 4
Thus, the required ratio is 1 : 4.
Note: Here, the ratio of radii = 7 : 14 = 1 : 2
Hence, the ratio of surface areas = \(\left(\frac{1}{2}\right)^{2}\) = \(\frac{1}{4}\) = 1 : 4, because in the formula of surface area of sphere, the degree of r is 2.

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm².
For the given hemispherical bowl, diameter = 10.5 cm.
Then, the radius r of the bowl = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= \(\frac{\frac{21}{2}}{2}\) cm
= \(\frac{21}{4}\) cm
Inner curved surface area of the hemispherical bowl
= 2 πr²
= 2 × \(\frac{22}{7}\) × \(\frac{21}{4}\) × \(\frac{21}{4}\) cm²
= \(\) cm²
= 173.25 cm²
Cost of tin-plating 100 cm² region = ₹ 16
∴ Cost of tin-plating 173.25 cm² region
= ₹ \(\left(\frac{16 \times 173.25}{100}\right)\)
= ₹ 27.72
Thus, the cost of tin-plating on the inner surface of the bowl is ₹ 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Answer:
For the given sphere, surface area = 154 cm².
Surface area of a sphere = 4 πr²
∴ 154 cm² = 4 × \(\frac{22}{7}\) × r² cm²
∴ r² = \(\frac{154 \times 7}{4 \times 22}\) cm²
∴ r² = \(\frac{49}{4}\) cm²
∴ r = \(\frac{7}{2}\) cm
∴ r = 3.5 cm
Thus, the radius of the given sphere is 3.5 cm.

Question 7.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Answer:
Suppose, the diameter of the moon = d₁
The diameter of the earth = 4 × d₁ = 4d₁
Then, the radius of the moon r₁ = \(\frac{d_{1}}{2}\) and
the radius of the earth r₂ = \(\frac{4 d_{1}}{2}\) = 2d₁.
Now, \(\frac{\text { The surface area of the moon }}{\text { The surface area of the earth }}\) = \(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}\)
= \(\frac{r_{1}^{2}}{r_{2}^{2}}\)
= \(\frac{\left(\frac{d_{1}}{2}\right)^{2}}{\left(2 d_{1}\right)^{2}}\)
= \(\frac{d_{1}^{2}}{4} \times \frac{1}{4 d_{1}^{2}}\)
= \(\frac{1}{16}\)
= 1 : 16
Thus, the ratio of the surface area of the moon and the surface area of the earth is 1 : 16.

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Answer:
For the given hemispherical bowl, the inner radius is 5 cm and the thickness of steel is 0.25 cm.
∴ Outer radius r of the given hemispherical bowl = 5 + 0.25 cm = 5.25 cm.
Curved surface area of a hemisphere
= 2πr²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25 cm²
= 2 × \(\frac{22}{7}\) × \(\frac{525}{100}\) × \(\frac{525}{100}\) cm²
= \(\frac{693}{4}\) cm²
= 173.25 cm²
Thus, the outer curved surface area of the given hemispherical bowl is 173.25 cm².

Question 9.
A right circular cylinder just encloses a sphere of radius r (see the given figure). Find :
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Answer:
Here,
radius of the cylinder = radius of the sphere = r and height of the cylinder h
= 2 × radius of the sphere = 2r
(i) Surface area of the sphere = 4πr²

(ii) Curved surface area of the cylinder
= 2 πrh
= 2 × 1 × r × 2r
= 4 πr²

(iii) Ratio of areas obtained in ( i ) and (ii)
= \(\frac{4 \pi r^{2}}{4 \pi r^{2}}\)
= \(\frac{1}{1}\)
= 1 : 1

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.3

1. Identify the shape having perpendicular lines:

Solution:
(ii), (iii), (v).

2. Identify the examples having perpendicular lines:

Question (i)
(i) Lines of railway track.
(ii) Adjacent edges of a table.
(iii) Line segment forming letter ‘L’.
Solution:
(ii) Adjacent edges of a table.
(iii) Line segment forming letter ‘L’.

3. Let \(\overrightarrow{\mathbf{AB}}\) be perpendicular to \(\overrightarrow{\mathbf{PQ}}\) and they intersect at O. What is the measure of \(\angle \mathbf{AOP}\)?
Solution:

\(\angle \mathbf{AOP}\) = 90°, because AB ⊥ PQ.

4. Line m is perpendicular to line l in the given figure. Each point on the line l is marked at equal intervals. Study the diagram and state true or false.

Question (i)
Line m is ⊥ bisector of line segment AI.
Solution:
True

Question (ii)
CE = EG
Solution:
True

Question (iii)
DF = 2DE.
Solution:
True.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.2

1. Classify the angles as acute, obtuse, right, straight or reflex angles:

Solution:
(i) Acute angle
(ii) Obtuse angle
(iii) Reflex angle
(iv) Straight angle
(v) Acute angle
(vi) Right angle
(vii) Obtuse angle
(viii) Right angle
(ix) Reflex angle
(x) Acute angle.

2. Classify the angles:

Question (i)
80°
Solution:
80° is between 0° and 90°.
∴ It is an acute angle.

Question (ii)
172°
Solution:
172° is between 90° and 180°
∴ It is an obtuse angle.

Question (iii)
90°
Solution:
90° is a right angle.

Question (iv)
0°
Solution:
0° is a zero angle.

Question (v)
179°
Solution:
179° is between 90° and 180°.
∴ It is an obtuse angle.

Question (vi)
215°
Solution:
215° is between 180° and 360°.
∴ It is an reflex angle.

Question (vii)
360°
Solution:
360° is a complete angle.

Question (viii)
350°
Solution:
350° is between 180° and 360°.
∴ It is a reflex angle.

Question (ix)
15°
Solution:
15° is between 0° and 90°.
∴ It is an acute angle.

Question (x)
180°
Solution:
180° is a straight angle.

3. Measure the following angles with protractor and write their measurement:

Solution:
(i) 60°
(ii) 125°
(iii) 110°
(iv) 80°
(v) 120°
(vi) 105°
(vii) 80°
(viii) 135°
(ix) 88°
(x) 90°.

4. How many degrees are there in

Question (i)
Two right angles
Solution:
1 right angle = 90°
∴ Two right angles = 2 × 90°
= 180°

Question (ii)
\(\frac {2}{3}\) right angles
Solution:
1 right angle = 90°
∴ \(\frac {2}{3}\) right angles = \(\frac {2}{3}\) × 90°
= 2 × 30°
= 60°

Question (iii)
Four right angles?
Solution:
1 right angle = 90°
∴ Four right angles = 4 × 90°
= 360°

5. What fraction of a clockwise revolution does the hour hand of a clock turn through when it goes from:

Question (i)
3 to 9
Solution:
3 to 9 : Half or \(\frac {1}{2}\)

Question (ii)
5 to 8
Solution:
5 to 8 : Quarter or \(\frac {1}{4}\)

Question (iii)
10 to 4
Solution:
10 to 4 : Half or \(\frac {1}{2}\)

Question (iv)
2 to 11
Solution:
2 to 11 : 3 Quarters or \(\frac {3}{4}\)

Question (v)
6 to 3
Solution:
6 to 3 : 3 Quarters or \(\frac {3}{4}\)

Question (vi)
2 to 7.
Solution:
2 to 7 : \(\frac {5}{12}\)

6. Find the number of right angles turned through by the hour hand of a dock when it goes from

Question (i)
5 to 8
Solution:
5 to 8 : 1 right angle

Question (ii)
1 to 7
Solution:
1 to 7 : 2 right angles

Question (iii)
4 to 10
Solution:
4 to 10 : 2 right angles

Question (iv)
9 to 12
Solution:
9 to 12 : 1 right angles

Question (v)
11 to 2
Solution:
11 to 2 : 1 right angles

Question (vi)
9 to 6
Solution:
9 to 6 : 3 right angles

Question (vii)
2 to 11
Solution:
2 to 11 : 3 right angles

Question (viii)
10 to 1
Solution:
10 to 1 : 1 right angles

Question (ix)
12 to 6
Solution:
12 to 6 : 2 right angles

Question (x)
5 to 2.
Solution:
5 to 2 : 3 right angles.

7. Where will be the hand of a clock stop if it starts at:

Question (i)
12 and make \(\frac {1}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{4}\) revolution, the hour hand takes \(\frac {1}{4}\) × 12 hours = 3 hours.
If hour hand starts at 12 and make \(\frac {1}{4}\) revolution clockwise it will stop at 3.

Question (ii)
2 and make \(\frac {1}{2}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{2}\) revolution, the hour hand takes \(\frac {1}{2}\) × 12 hours = 6 hours.
If hour hand starts at 2 and make \(\frac {1}{2}\) revolution clockwise it will stop at 8.

Question (iii)
5 and make \(\frac {1}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours.
For \(\frac {1}{4}\) revolution, the hour hand takes \(\frac {1}{4}\) × 12 hours = 3 hours
If hour hand starts at 5 and make \(\frac {1}{4}\) revolution clockwise it will stop at 8.

Question (iv)
5 and make \(\frac {3}{4}\) revolution clock-wise.
Solution:
For 1 revolution, the hour hand takes 12 hours .
For \(\frac {3}{4}\) revolution, the hour hand takes \(\frac {3}{4}\) × 12 hours = 9 hours.
If hour hand starts at 5 and make \(\frac {3}{4}\) revolution clockwise it will stop at 2.

8. What part of revolution have you turned through if you stand facing:

Question (i)
East and turn clockwise to North
Solution:
I turned through \(\frac {3}{4}\) part of a revolution.

Question (ii)
South and turn clockwise to North
Solution:
I turned through \(\frac {1}{2}\) part of a revolution.

Question (iii)
South and turn clockwise to East
Solution:
I turned through \(\frac {3}{4}\) part of a revolution.

Question (iv)
West and turn clockwise to East
Solution:
I turned through \(\frac {1}{2}\) part of at revolution.

9. Find the angle measure between the hands of the clock in each figure:

Solution:
(i) Angle measure between the hands of the clock at 3.00 a.m.
= \(\frac {3}{12}\) × 360° = 90°
(ii) Angle measure between the hands of the clock at 6.00 a.m.
= \(\frac {6}{12}\) × 360° = 180°
(iii) Angle measure between the hands of the clock at 2.00 a.m.
= \(\frac {2}{12}\) × 360° = 60°

10. Draw the following angles by protractor:

Question (i)
(i) 40°
(ii) 75°
(iii) 105°
(iv) 90°
(v) 130°
Solution:

11. State true or false:

Question (i)
The sum of two right angles is always a straight angle.
Solution:
True

Question (ii)
The sum of two acute angles is always a reflex angle.
Solution:
False

Question (iiii)
The obtuse angle has measurement between 90° to 180°.
Solution:
True

Question (iv)
A complete revolution has four right angles.
Solution:
True

12. Fill in the blanks:

Question (i)
The angle which is greater than 0° and less than 90° is called ………….. .
Solution:
acute angle

Question (ii)
The angle whose measurement equal to two right angle is …………….. .
Solution:
straight angle

Question (iii)
The angle between 90° and 180° is ……………. .
Solution:
obtuse angle.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

Question 1.
In figure, PS is bisector of ∠QPR of ∆PQR. Prove that = \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\).

Solution:
Given: ∆PQR. PS is bisector of ∠QPR
i.e., ∠1 = ∠2
To prove: \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\)
Construction : Through R draw a line parallel to PS to meet QP produced at T.
Proof: In ∆QRT, PS || TR

∠1 = ∠4 (Corresponding angle)
but ∠1 = ∠2 (given)
∴ ∠3 = ∠4
In ∆PRT,
∠3 = ∠4 (Proved)
PT = PR
[Equal side have equal angle opposite to it]
In ∆QRT,
PS || TR
∴ \(\frac{\mathrm{QP}}{\mathrm{PT}}=\frac{\mathrm{QS}}{\mathrm{SR}}\)
[By Basic Proportionality Theorem]
\(\frac{\mathrm{QP}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\) (PT = PR)
\(\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\)
Which is the required result.

Question 2.
In the given fig., D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC, DN ⊥ AB, prove that:
(i) DM² = DN.MC
(ii) DN² = OMAN.

Solution:
Given: ∆ABC, DM ⊥ BC, DN ⊥ AB
To prove: DM² = DN . AC
DN² = DM . AN.
Proof: BD ⊥ AC (Given)
⇒ ∠BDC = 90°
⇒ ∠BDM + ∠MDC = 90°
In ∠DMC, ∠DMC = 90°
[∵ DM ⊥ BC (Given)]
⇒ ∠C + ∠MDC = 90°
From (1) and (2),
∠BDM + ∠MDC = ∠C + ∠MDC
∠BDM =∠C
[Cancelling ∠MDC from both sides]
Now in ∆BMD and ∆MDC,
∠BDM = ∠C [Proved)
∠BMD = ∠DMC [Each 90°]
∆BMD ~ ∆MDC [By AA criterion of similarity]
⇒ \(\frac{\mathrm{DM}}{\mathrm{BM}}=\frac{\mathrm{MC}}{\mathrm{DM}}\)
[∵ Corresponding sides of similar triangles are proportional]
⇒ DM² = BM × MC
⇒ DM² = DN × MC [∵ BM = DN]
Similarly, ∆NDA ~ ∆NBD
⇒ \(\frac{\mathrm{DN}}{\mathrm{BN}}=\frac{\mathrm{AN}}{\mathrm{DN}}\)
[∵ Corresponding sides of similar triangles are próportional]
⇒ DN² = BN × AN
⇒ DN² = DM × AN .
Hence proved.

Question 3.
In fig., ABC is triangle in which ∠ABC > 90° and AD ⊥ BC produced, prove that AC² = AB² + BC² + 2BC.BD.

Solution:
Given: ∠ABC, AD ⊥ BC when produced, ∠ABC > 90°.
To prove : AC² = AB² + BC² + 2BC. BD.
Proof: Let BC = a,
CA = b,
AB = c,
AD = h
and BD = x.
In right-angled ∆ADB,
Using Pythagoras Theorem.
AB² = BD² + AD²
i.e., c² = x² + h²
Again, in right-angled AADC,
AC² = CD² + AD²
i.e.. b² = (a + x)² + h²
= a² + 2ax + x² + h²
= a² + 2ax + c²; [Using (1)]
b² = a² + b² + 2w.
Hence, AB² = BC² + AC² + 2BC × CD.

Question 4.
In fig., ABC is a triangle in which ∠ABC < 90°, AD ⊥ BC, prove that AC² = AB² + BC² – 2BC.BD.

Solution:
Given: ∆ ABC, ∠ABC < 90°, AD ⊥ BC.
To prove : AC² = AB² + BC² – 2BC BD.
Proof: ADC is right-angled z at D
AC² = CD² + DA² (Pythagora’s Theorem) ……………..(1)
Also, ADB is right angled ∆ at D
AB² = AD² + DB² ……………….(2)
From (1), we get:
AC² = AD² + (CB – BD)²
= AD² + CB² + BD² – 2CB × BD
or AC² = (BD² + AD²) + CB² – 2CB × BD
AC² = AB² + BC² – 2BC × BD. [Using (2)]

Question 5.
In fig., AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) AC² = AD² + BC. DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB² = AD² ± BC.DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC² + AB² = 2 AD² + \(\frac{1}{2}\) BC²

Solution:
Given: ∆ABC, AM ⊥ BC,
AD is median of ¿ABC.
To prove:
(i) AC² = AD² + BC. DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB² = AD² ± BC.DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC² + AB² = 2 AD² + \(\frac{1}{2}\) BC²
Proof: In ∆AMC.
AC² = AM² + MC²
= AM² + (MD + DC)²
AC² = AM² + MD² + DC² + 2MD × DC
AC² = (AM² + MD²) + \(\left(\frac{\mathrm{BC}}{2}\right)^{2}\) + 2 . MD \(\left(\frac{\mathrm{BC}}{2}\right)\)
AC² = AD² + BC . MD + \(\frac{\mathrm{BC}^{2}}{4}\) …………(1)

(ii) In right angled triangle AME,
AB² = AM² + BM²
= AM² + (BD – MD)²
=AM² + BD² + MD² – 2BD × MD
= (AM² + MD²) + BD² – 2(\(\frac{1}{2}\) BC) MD
= AD² + (\(\frac{1}{2}\) BC)² – BC . MD
[∵ In ∆ AMD; AD² = MA² + MD²]
AB² + AD² (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) – BC . MD ………….(2)

(iii) Adding (1) and (2),
AB² + AC² = AD² + BC.MD + (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) + AD² + (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) – BC . MD
= 2 AD² + \(\frac{\mathrm{BC}^{2}}{4}+\frac{\mathrm{BC}^{2}}{4}\)
= 2AD² + 2 \(\frac{\mathrm{BC}^{2}}{4}\)
AB² + AC² = 2AD² + \(\frac{\mathrm{BC}^{2}}{2}\)
Which is the required result.

Question 6.
Prove that sum of squares of the diagonals of a parallelogram is equal to sum of squares of its sides.
Solution:

Given:
Let ABCD be a parallelogram in which diagonalš AC and BD intersect at point M.
To prove: AB² + BC² + CD² + DA² = AC² + BD²
Solution:
Proof: Diagonals of a parallelogram bisect each other.
∴ In || gm ABCD,
Diagonal BD and AC bisect each other.
Or MB and MD are medians of ∆ABC and ∆ADC respectively.
We know that, if AD is a medians of ¿ABC,
then AB² + AC² = 2AD² + BC²
Using this result, we get:
AB² + BC² = 2 BM² + \(\frac{1}{2}\) AC² ………..(1)
and AD² + CD² = 2 DM² + \(\frac{1}{2}\) AC² ………….(2)
Adding (1) and (2), we get:
AB² + BC² + AD² + CD² = 2 (BM² + DM²) + (AC² + AC²)
AB² + BC² + AD² + CD² = 2 (\(\frac{1}{2}\) BD² + \(\frac{1}{4}\) BD²) + AC²
AB² + BC² + AD² + CD² = BD² + AC²
Hence, sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Question 7.
In fig., two chords AB and CD intersect each other at the point P prove that:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.

Solution:
Given: Circle, AB and CD are two chords intersects each other at P.

To prove:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.
Proof:
(i) In ∆APC and ∆DPB,
∠1 = ∠2 (Vertically opposite angle)
∠3 = ∠4 (angle on same segment)
∴ ∆APC ~ ∆DPB [AA similarity criterion]

(ii) ∆APC ~ ∆DPB (Proved above)
\(\frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{PC}}{\mathrm{PB}}\)
(If two triangles are sitnilar corresponding sides are proportional)
AP.PB = PC.DP
Which is the required result.

Question 8.
In fig., two chords AB and CD of a circle intersect each other at point P (when produced) outside the circle prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.

Solution:
Given: AB and CD are two chord of circle intersects each other at P (when produced)
To prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.
Proof:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common)
∠PAC = ∠PDB.
(Exterior angle of cyclic quadrilqteral is equal to interior opposite angle)
∴ ∆PAC ~ ∆PDB [AA similarity criterion]

(ii) ∆PAB ~ ∆WDB
∴ \(\frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}\)
[If two triangles are similar corresponding sides are proportional]
PA × PB = PC × PD.

Question 9.
In fig., D is a point on side BC of BD AB ∆ABC such that \(\frac{\mathbf{B D}}{\mathbf{D C}}=\frac{\mathbf{A B}}{\mathbf{A C}}\). Prove that: AD is bisector of ∠BAC.

Solution:
Given: A ∆ABC, D is a point on BC such that \(\frac{\mathbf{B D}}{\mathbf{D C}}=\frac{\mathbf{A B}}{\mathbf{A C}}\)
To prove: AD bisects ∠BAC
i.e., ∠1 = ∠2
Construction: Through C draw CE || DA meeting BA produced at E.

Proof:
In ∆BCE, we have:
AD || CE ………(const.)
So, by Basic Proportionality Theorem,
But \(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AE}}\)
\(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ \(\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AE = AC

In ∆ACE, we have:
AE = AC
⇒ ∠3 = ∠4 ………. (∠s opp. to equal sides)
Since CE || DA and AC cuts them, then:
∠2 = ∠4 ……….(alt ∠s)
Also CE || DA and BAE cuts them, then:
∠1 = ∠3 …………(Corr. ∠s)
Thus we have:
∠3 = ∠4
⇒ ∠3 = ∠1
But ∠4 = ∠2
⇒ ∠1 = ∠2.
HenCe AD bisects ∠BAC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds?

Solution:
A right angled triangle, ABC, in which,
AB = 1.8 cm,
BC = 2.4 cm.
∠B = 90°
By Pythagoras Theorem,
AC² = AB² + BC²
AC² = (1.8)² + (2.4)²
AC² = 3.24 + 5.76 =9
AC² = (3)²
AC = 3 cm
Now, when Nazima pulls in the string at the rate of 5 cm/sec ; then the length of the string decrease = 5 × 12 = 60 cm
= 0.6 m in 12 seconds.
Let after 12 seconds, position of the fly will be at D.
∴ AD = AC – distance covered in 12 seconds
AD = (3 – 0.6) m
AD = 2.4 m
Also, in right angled ∆ABD,
Using Pythagoras Theorem,
AD² = AB² + BD²
(2.4)² = (1.8)² + BD²
BD² = 5.76 – 3.24
BD² = 2.52 m
BD = 1.587 m.
∴ Horizontal distance of the fly from Nazima = BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m
Hence, original length of string and horizontal distance of the fly from Nazima is 3 m and 2.79 m.