Class 9

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m² costs ₹ 20.
Answer:
The plastic box to be made is open at the top. Hence, the plastic sheet is required for the lateral surfaces and the base.
Here, for the box to be made,
length l = 1.5 m;
breadth b = 1.25 m and
height h = 65 cm = 0.65 m.
Area of the plastic sheet required for open box = Lateral surface area + Area of base
= 2 h(l + b) + l × b
= 2 × 0.65 (1.5 + 1.25) + 1.5 × 1.25 m²
= 1.3 × 2.75 + 1.875 m²
= 3.575 + 1.875 m²
= 5.45 m²
Cost of 1 m² sheet = ₹ 20
∴ Cost of 5.45 m² sheet = ₹ (5.45 × ₹ 20)
= ₹ 109

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².
Answer:
For the given room, length 1 = 5 m; breadth b = 4 m and height h = 3 m.
Area of the region to be white washed
= Area of four walls + Area of ceiling
= 2 h(l + b) + l × b
= 2 × 3 (5 + 4) + 5 × 4 m²
= 54 + 20 m²
= 74 m²
Cost of white washing 1 m² region = ₹ 7.5
∴ Cost of white washing 74 m² region
= ₹ (74 × 7.5)
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Answer:
Area painted at the cost of ₹ 10 = 1 m²
∴ Area painted at the cost of ₹ 15,000
= \(\frac{15000}{10}\)
= 1500 m²
∴ Area of the four walls = 1500m²
∴ Lateral surface area = 1500 m²
∴ Perimeter Of the floor × Height = 1500 m²
∴ 250 m × Height = 1500 m²
∴ Height = \(\frac{15000}{250}\)
∴ Height = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm× 7.5 cm can be painted out of this container?
Answer:
For each brick, length l = 22.5 cm; breadth b = 10 cm and height h = 7.5 cm.
Total surface area of one brick
= 2 (lb + bh + hl)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2 (225 + 75 + 168.75) cm²
= 2 (468.75) cm²
= 937.5 cm²
= \(\frac{937.5}{10000}\) m² = 0.09375 m²
No. of bricks that can be painted with paint sufficient to paint 0.09375 m² area = 1
∴ No. of bricks that can be painted with paint sufficient to paint 9.375 m² area 9.375
= \(\frac{9.375}{0.09375}\) = 100

Question 5.
A cubical box has each edge 10 cm and |> another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much ?
Answer:
For the cubical box, edge a = 10 cm and for the cuboidal box, length l = 12.5 cm; breadth b = 10 cm and height h = 8 cm
(i) Lateral surface area of cubical box
= 4a²
= 4 (10)² cm²
= 400 cm
Lateral surface area of cuboidal box
= 2h(l + b)
= 2 × 8(12.5 + 10) cm²
= 16 × 22.5 cm²
= 360 cm²
Thus, the lateral surface area of cubical box is greater by 40 cm² (400 – 360).

(ii) Total surface area of cubical box = 6a2
= 6 (10)² cm²
= 600 cm²
Total surface area of cuboidal box
= 2 (lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm²
= 2 (125 + 80 + 100) cm²
= 2 (305) cm²
= 610 cm²
Thus, the total surface area of cubical box is smaller by 10 cm² (610 – 600).

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass ?
(ii) How much of tape is needed for all the 12 edges ?
Answer:
(i) For the cuboidal greenhouse, length l = 30 cm; breadth fa = 25 cm and height h = 25 cm.
Area of glass used
= Total surface area of cuboid
= 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 25 + 25 × 30) cm²
= 2 (750 + 625 + 750) cm²
= 2 (2125) cm²
= 4250 cm²

(ii) 12 edges of the cuboidal greenhouse is made-up of 4 lengths, 4 breadths and 4 heights.
∴ Length of tape needed for 12 edges
= 4l + 4b + 4h
= 4 (l + b + h)
= 4 (30 + 25 + 25) cm
= 4 (80) cm
= 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
For bigger cuboidal boxes, length l = 25 cm;
breadth b = 20 cm and height h = 5 cm.
Total surface area of a bigger box
= 2 (lb + bh + hl)
= 2 (25 × 20 + 20 × 5 + 5 × 25) cm²
= 2 (500 + 100 + 125) cm²
= 1450 cm²
Area of cardboard required for overlap
= 5 % of 1450 cm²
= 72.5 cm²
Thus, the total area of cardboard required for 1 bigger box = 1450 + 72.5 cm²
= 1522.5 cm²
∴ The total area of cardboard required for 250 bigger boxes = (1522.5 × 250) cm²
For smaller cuboidal boxes, length l = 15 cm; breadth b = 12 cm and height h = 5 cm.
Total surface area of a smaller box
= 2 (lb + bh + hl)
= 2 (15 × 12 + 12 × 5 + 5 × 15) cm²
= 2(180 +60 + 75) cm²
= 2 (315) cm²
= 630 cm²
Area of cardboard required for overlap
= 5% of 630 cm²
= 31.5 cm²
Thus, the total area of cardboard required for 1 smaller box = 630 + 31.5 cm² = 661.5 cm²
∴ The total area of cardboard required for 250 smaller boxes = (661.5 × 250) cm²
Now, the total area of cardboard required for all the boxes
= (1522.5 × 250) + (661.5 × 250) cm²
= 250(1522.5 + 661.5) cm²
= 250 × 2184 cm²
Cost of 1000 cm² cardboard = ₹ 4
∴ Cost of 250 × 2184 cm² cardboard
= ₹ \(\left(\frac{4 \times 250 \times 2184}{1000}\right)\)
= ₹ 2184

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m ?
Answer:
For the box-like structure without base, length
l = 4m; breadth b = 3m and height h = 2.5m.
Area of tarpaulin required
= Area of lateral surfaces + Area of top
= 2 h(l + b) + l × b
= 2 × 2.5 (4 + 3) + 4 × 3 m²
= 35 + 12 m²
= 47 m²

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 12 Heron’s Formula 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.
The sides of a triangle measure 8cm, 12cm and 6 cm. Then, the semiperimeter of the triangle is ……………………… cm.
A. 26
B. 52
C. 13
D. 6.5
Answer:
C. 13

Question 2.
Each side of an equilateral triangle measures 8 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 4
B. 24
C. 12
D.36
Answer:
C. 12

Question 3.
In a right angled triangle, the length of the hypotenuse is 15 cm and one of the sides forming right angle is 9 cm. Then, the semiperimeter of the triangle is ……………………….. cm.
A. 36
B. 18
C. 12
D. 15
Answer:
B. 18

Question 4.
The ratio of the measures of the sides of a triangle is 3:4:5. If the semiperimeter of the < triangle is 36 cm, the measure of the longest side of the triangle is ……………………. cm.
A. 12
B. 15
C. 20
D. 30
Answer:
D. 30

Question 5.
The area of a triangle is 48 cm2 and one of its sides measures 12 cm. Then, the length of the altitude corresponding to this side is …………………. cm.
A. 4
B. 8
C. 16
D. 6
Answer:
B. 8

Question 6.
The sides of a triangle measure 12 cm, 17 cm and 25 cm. Then, the area of the triangle is ……………………….. cm².
A. 54
B. 90
C. 180
D. 135
Answer:
B. 90

Question 7.
Two sides of a triangle measure 9 cm and 10 cm. If the perimeter of the triangle is 36cm, then its area is …………………. cm².
A. 17
B. 36
C. 72
D. 18
Answer:
B. 36

Question 8.
The area of an equilateral triangle with each side measuring 10 cm is ………………….. cm².
A. \(\frac{5 \sqrt{3}}{2}\)
B. 25√3
C. 5√3
D. 3√5
Answer:
B. 25√3

Question 9.
∆ ABC is an isosceles triangle in which BC = 8 cm and AB = AC = 5 cm. Then, area of ∆ ABC = ……………………….. cm².
A. 6
B. 12
C. 18
D. 24
Answer:
B. 12

Question 10.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar(ABCD) = …………………. cm².
A. 18
B. 9
C. 36
D. 27
Answer:
C. 36

Question 11.
ABCD is a parallelogram. If ar (ABC) = 18 cm², then ar (ABCD) = …………………. cm².
A. 3.6
B. 7.2
C. 7.5
D. 6
Answer:
B. 7.2

Question 12.
In quadrilateral ABCD, AC = 10 cm. BM and DN are altitudes on AC from B and D respectively. If BM = 12cm and DN = 4 cm, then ar (ABCD) = …………………. cm².
A. 160
B. 80
C. 320
D. 480
Answer:
B. 80

Question 13.
The perimeter of rhombus ABCD is 40 cm and BD =16 cm. Then, ar (ABCD) = ……………………. cm².
A. 96
B. 48
C. 24
D. 72
Answer:
A. 96

Question 14.
The area of a rhombus is 72 cm² and one of its diagonals measures 16 cm. Then, the length of the other diagonal is ………………… cm.
A. 12
B. 9
C. 18
D. 15
Answer:
B. 9

Question 15.
PQRS is a square. If PQ = 10 cm, then PR = ……………………….. cm.
A. 10
B. 20
C. 10√2
D. 2√10
Answer:
C. 10√2

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

PSEB 9th Class English Reading Comprehension Picture/Poster Based

Answers have been given at the end of this set.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

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

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

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

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

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

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

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

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

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

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

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

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

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

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

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

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

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

Look at this poster and answer the questions given below :

Choose the correct option to answer each question :

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

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

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

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

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

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

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

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

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

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

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

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

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

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

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

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

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

Look at these pictures and answer the questions given below :

Choose die correct option to answer each question:

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

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

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

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

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

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

Look at these pictures and answer the questions given below:

Choose the correct option to answer each question :

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

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

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

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

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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer:

Steps of construction :

1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect at P on one side of line MAB.
4. Draw ray AC passing through E
   Thus, ∠CAB is the required angle of 90°.

Justification:
Draw PX and PY.
In ∆ PAX and ∆ PAY,
AX = AY (Radii of same arc)
PX = PY (Radii of congruent arcs)
PA = PA (Common)
∴ By SSS rule, ∆ PAX ≅ ∆ PAY
∴ ∠PAX = ∠PAY (CPCT)
But, ∠PAX + ∠PAY = 180° (Linear pair)
∴ ∠PAY = \(\frac{180^{\circ}}{2}\) = 90°
∴ ∠CAB = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:

Steps of construction:

1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect at P on one side of line MAB.
4. Draw ray AC passing through E Thus, ∠CAB of 90° is received.
5. Name the point of intersection of the arc with centre A and ray AC as Z.
6. Taking Y and Z as centres and radius more than \(\frac{1}{2}\)YZ, draw arcs to intersect each other at Q.
7. Draw ray AQ.
   Thus, ∠QAB is the required angle of 45°.

Justification:
In example 1, we have already justified that ∠CAB = 90°. So, we do not repeat that part’ here.
Draw QZ and QY.
In ∆ AYQ and ∆ AZQ,
AY = AZ (Radii of same arc)
YQ = ZQ (Radii of congruent arcs)
AQ = AQ (Common)
∴ By SSS rule, ∆ AYQ ≅ ∆ AZQ
∴ ∠QAY = ∠QAZ (CPCT)
But, ∠QAY + ∠QAZ = ∠ZAY = ∠CAB = 90°
∴ ∠QAY = \(\frac{90^{\circ}}{2}\) = 45°
∴ ∠QAB = 45°

Question 3.
Construct the angles of the following measurements:
(i) 30°
Answer:

Steps of construction:

1. Draw any ray AB. With centre A and any radius, draw an arc to intersect AB at X.
2. With centre X and the same radius [as in step (1)], draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
3. Draw ray AT, the bisector of ∠YAB.
   Thus, ∠TAB is the required angle of 30°.

(ii) 22\(\frac{1}{2}\)°
Answer:

Steps of construction:

1. Draw any ray AB. Produce AB on the side of A to get line CAB.
2. Taking A as centre and any radius, draw an arc of a circle to intersect line CAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect each other at L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 45°.
5. Draw ray AN, the bisector of ∠MAB. Then, ∠NAB = 22\(\frac{1}{2}\)°.
   Thus, ∠NAB is the required angle of 22\(\frac{1}{2}\)°.

(iii) 15°
Answer:

Steps of construction:

1. Draw any ray AB. Taking A as centre and any radius, draw an arc of a circle to intersect AB at X.
2. Taking X as centre and the same radius as before, draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
3. Draw ray AL, the bisector of ∠YAB. Then, ∠LAB = 30°.
4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 15°.
   Thus, ∠MAB is the required angle of 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
(i) 75° and (ii) 105°
Answer:

Steps of construction:

1. Draw any ray AB and produce it on the side of A to get line CAB. Taking A as centre and any radius draw an arc of a circle to intersect line CAB at X and Y.
2. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect each other at point L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
3. Taking X as centre and radius AX, draw an arc of a circle to Intersect the first arc (arc XY) with centre A at Z.
4. Draw ray AZ. Then, ∠ZAB = 60°.
5. Now, draw ray AM, the bisector of ∠LAZ. Then, ∠MAB = 75° and ∠MAC = 105°.
   Thus, ∠MAB and ∠MAC are the required angles of measure 75° and 105° respectively.

(iii) 135°
Answer:

Steps of construction:

1. Draw line CAB. Taking A as centre and any radius, draw an arc of a circle to Intersect line CAB at X and Y.
2. Taking X and Y as centres and radius more them \(\frac{1}{2}\)XY, draw arcs to intersect each other at P on one side of line CAB.
3. Draw ray AP Then, ∠PAB = ∠PAC = 90°.
4. Draw ray AQ, the bisector of ∠PAC.
5. Then, ∠QAB = 135°.
   Thus, ∠QAB is the required angle of 135°.

Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Answer:
Line segment XY is given. We have to construct an equilateral triangle with each side being equal to XY.

Steps of construction:

1. Draw any ray BM.
2. With centre B and radius XY, draw an arc of a circle to intersect BM at C.
3. Taking B and C as centres and rhdius XY, draw arcs to intersect each other at A on one side of line AC.
4. Draw AB and AC.
   Thus, ∆ ABC is the required equilateral triangle with each side being equal to XY.

Justification:
The arc drawn with centre B and radius XY intersects ray BM at C. ∴ BC = XY. The arcs drawn with centres B and C and radius XY intersect at A.
∴ AB = XY and AC = XY.
Thus, in ∆ ABC, AB = BC = AC = XY.
Hence, ∆ ABC is an equilateral triangle in which all sides are equal to XY.
Note: If the measure of sides are given numerically, e.g., 4 cm, 5 cm, etc., then we have to use graduated scale instead of straight edge.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 10 Circles 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.
In a circle with centre P, AB and CD are congruent chords. If ∠PAB = 40°, then ∠CPD = ………………..
A. 40°
B. 80°
C. 100°
D. 50°
Answer:
C. 100°

Question 2.
In a circle with radius 5 cm, the length of a chord lying at distance 4 cm from the centre is …………………. cm.
A. 3
B. 6
C. 12
D. 15
Answer:
B. 6

Question 3.
In a circle with radius 13 cm, the length of a chord is 24 cm. Then, the distance of the chord from the centre is ……………….. cm.
A. 10
B. 5
C. 12
D. 6.5
Answer:
B. 5

Question 4.
In a circle with radius 7 cm, the length of a minor arc is always less than ………………… cm.
A. 11
B. 22
C. 15
D. π
Answer:
B. 22

Question 5.
In a circle with centre P, AB is a minor arc. Point R is a point other than A and B on major arc AB. If ∠APB = 150°, then ∠ARB = …………… .
A. 150°
B. 75°
C. 50°
D. 100°
Answer:
B. 75°

Question 6.
In a circle with centre P, AB is a minor arc. Point R is a point other than A and B on major arc AB. If ∠ARB = 80°, then ∠APB = ……………. .
A. 40°
B. 80°
C. 160°
D. 60°
Answer:
C. 160°

Question 7.
In cyclic quadrilateral ABCD, ∠A – ∠C = 20°.
Then, ∠A = ………………. .
A. 20°
B. 80°
C. 100°
D. 50°
Answer:
C. 100°

Question 8.
In cyclic quadrilateral PQRS, 7∠P = 2∠R.
Then, ∠P = ………………….. .
A. 20°
B. 40°
C. 140°
D. 100°
Answer:
B. 40°

Question 9.
The measures of two angles of a cyclic quadrilateral are 40° and HOP. Then, the measures of other two angles of the quadrilateral are ……………….. .
A. 40° and 110°
B. 50° and 100°
C. 140° and 70°
D. 20° and 120°
Answer:
C. 140° and 70°

Question 10.
In cyclic quadrilateral PQRS, ∠SQR = 60° and ∠QPR = 20°. Then, ∠QRS = ……………… .
A. 40°
B. 60°
C. 80°
D. 100°
Answer:
D. 100°

Question 11.
In cyclic quadrilateral ABCD, ∠CAB = 30° and ∠ABC = 100°. Then, ∠ADB =
A. 50°
B. 100°
C. 75°
D. 60°
Answer:
A. 50°

Question 12.
Equilateral ∆ ABC is inscribed in a circle with centre P. Then, ∠BPC = ……………. .
A. 60°
B. 90°
C. 120°
D. 75°
Answer:
C. 120°

Question 13.
∆ ABC is inscribed in a circle with centre O and radius 5 cm and AC is a diameter of the circle. If AB = 8 cm, then BC = ………………… cm.
A. 10
B. 8
C. 6
D. 15
Answer:
C. 6

Question 14.
In cyclic quadrilateral ABCD, ∠A = 70° and ∠B + ∠C = 160°. Then, ∠B = ………………. .
A. 35°
B. 25°
C. 50°
D. 130°
Answer:
C. 50°

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Question 1.
Prove that the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:

Circles with centres O and P intersect each other at points A and B.
In ∆ OAP and ∆ OBR
OA = OB (Radii of circle with centre O)
PA = PB (Radii of circle with centre P)
OP = OP (Common)
∴ By SSS rule, ∆ OAP = ∆ OBP
∴ ∠OAP = ∠OBP (CPCT)
Thus, OP subtends equal angles at A and B. Hence, the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Answer:

Draw the perpendicular bisector of AB to intersect AB at M and draw the perpendicular bisector of CD to intersect CD at N.
Both these perpendicular bisectors pass through centre O and since AB || CD; M, O and N are collinear points.
Now, MB = \(\frac{1}{2}\)AB = \(\frac{5}{2}\) = 2.5 cm,
CN = \(\frac{1}{2}\)CD = \(\frac{11}{2}\) = 5.5 cm and MN = 6 cm.
Let ON = x cm s
∴ OM = MN – ON = (6 – x) cm
Suppose the radius of the circle is r cm.
∴ OB = OC = r cm
In ∆ OMB, ∠M = 90°
∴OB² = OM² + MB²
∴ r² = (6 – x)² + (2.5)²
∴ r² = 36 – 12x + x² + 6.25 ………….. (1)
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + CN²
∴ r² = (x)² + (5.5)²
∴ r² = x² + 30.25 ………………. (2)
From (1) and (2),
36 – 12x + x² + 6.25 = x² + 30.25
∴ – 12x = 30.25 – 6.25 – 36
∴- 12x = – 12
∴x = 1
Now, r² = x² + 30.25
∴ r² = (1)2 + 30.25
∴ r² = 31.25
∴ r = √31.25 (Approximately 5.6)
Thus, the radius of the circle is √31.25 (approximately 5.6) cm.
Note: If the calculations are carried out in simple fractions, then MB = \(\frac{5}{2}\) cm, CN = \(\frac{11}{2}\) cm and radius is \(\frac{5 \sqrt{5}}{2}\) (approximately 5.6) cm.

Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chords is at distance 4 cm from the cehtre, what is the distance of the other chord from the centre?
Answer:

In a circle with centre O, chord AB is parallel to chord CD, AB = 8 cm and CD = 6 cm.
Draw OM ⊥ AB, ON ⊥ CD, radius OB and radius OC.
Then, MB = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 8 = 4 cm,
NC = \(\frac{1}{2}\)CD = \(\frac{1}{2}\) × 6 = 3cm and ON = 4cm.
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + NC² = 4² + 3² = 16 + 9 = 25
∴ OC = 5 cm
∴ OB = 5 cm (OB = OC = Radius)
In ∆ OMB, ∠M = 90°
∴ OB² = OM² + MB²
∴ 5² = OM² + 4²
∴ 25 = OM² + 16
∴ OM² = 9
∴ OM = 3 cm
Thus, the distance of the other chord from the centre is 3 cm.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer:

In ∆ ABE, ∠AEC is an exterior angle.
∴ ∠AEC = ∠ABE + ∠BAE
∴ ∠ABE = ∠AEC – ∠BAE
∴ ∠ABC = ∠AEC – ∠DAE ……………. (1)
Now, ∠AEC = \(\frac{1}{2}\) ∠AOC (Theorem 10.8)
and ∠ DAE = \(\frac{1}{2}\) ∠DOE (Theorem 10.8)
Substituting above values in (1),
∠ABC = \(\frac{1}{2}\) ∠AOC – \(\frac{1}{2}\)∠DOE
∴ ∠ABC = \(\frac{1}{2}\) (∠AOC – ∠DOE)
Here, ∠AOC is the angle subtended by chord AC at the centre and ∠DOE is the angle subtended by chord DE at the centre.
Thus, ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Note: There is no need for chords AD and CE to be equal.

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:

ABCD is a rhombus and its diagonals intersect at M.
∴ ∠BMC is a right angle.
A circle is drawn with diameter BC.
There are three possibilities for point M:
(1) M lies in the interior of the circle,
(2) M lies in the exterior of the circle.
(3) M lies on the circle.
According to (1), if M lies in the interior of the circle, then BM produced will intersect the circle at E. Then, ∠BEC is an angle in a semicircle and hence a right angle, i.e.,
∠MEC = 90°.
In ∆ MEC, ∠ BMC is an exterior angle.
∴ ∠ BMC > ∠ MEC, i.e., ∠ BMC > 90°. In this situation, ∠ BMC is an obtuse angle which contradicts that ∠ BMC = 90°.
Similarly, according to (2), if M lies in the exterior of the circle, then ∠BMC is an acute angle which contradicts that ∠BMC 90°. Thus, possibilities (1) and (2) cannot be true.
Hence, only possibility (3) is true, i.e., M lies on the circle.
Thus, the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary at) E. Prove that AE = AD.
Answer:

Here, the circle through A, B and C intersects CD at E.
∴ Quadrilateral ABCE is cyclic.
ABCD is a parallelogram.
∴ ∠ABC = ∠ADC
∴ ∠ABC = ∠ADE
In cyclic quadrilateral ABCE,
∠ABC + ∠AEC = 180°
∴ ∠ADE + ∠AEC = 180° ……………… (1)
Moreover, ∠AEC and ∠AED form a linear pair.
∴ ∠AED + ∠AEC = 180° ………………. (2)
From (1) and (2),
∠ADE + ∠AEC = ∠AED + ∠AEC
∴ ∠ ADE = ∠ AED
Thus, in ∆ AED, ∠ADE = ∠AED.
∴ AE = AD (Sides opposite to equal angles)
Note: If the circle intersect CD produced, l then also the result can be proved in similar way.

Question 7.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Answer:

Chords AC and BD of a circle bisect each other at point O.
Hence, the diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD Is a parallelogram.
∴ ∠BAC = ∠ACD (Alternate angles formed by transversal AC of AB || CD)
Moreover, ∠ACD = ∠ABD (Angles in same segment)
∴ ∠BAC = ∠ABD
∴ ∠BAO = ∠ABO
∴ In A OAB, OA = OB.
But, OA = OC and OB = OD
∴ OA = OB = OC = OD
∴ OA + OC = OB + OD
∴ AC = BD
Thus, the diagonals of parallelogram ABCD are equal.
∴ ABCD is a rectangle.
∴ ∠ABC = 90°
Hence, ∠ABC is an angle in a semicircle and AC is a diameter.
Similarly, ∠BAD = 90°.
Hence, ∠BAD is an angle in a semicircle and BD is a diameter.

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – \(\frac{1}{2}\)A, 90° – \(\frac{1}{2}\)B and 90° – \(\frac{1}{2}\)C.
Answer:

The bisectors of ∠A, ∠B and ∠ C of ∆ ABC intersect the circumcircle of ∆ ABC at D, E and F respectively. .
∠FDE = ∠FDA + ∠EDA (Adjacent angles)
= ∠ FCA + ∠ EBA (Angles in same segment)
= \(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠B (Bisector of angles in ∆ ABC)
= \(\frac{1}{2}\)(∠ B + ∠ C)
= \(\frac{1}{2}\)(180° – ∠A) [∠A + ∠B + ∠C = 180°)
= 90° – \(\frac{1}{2}\) ∠A
Thus, ∠FDE = 90° – \(\frac{1}{2}\) ∠A.
Similarly, ∠ DEF = 90° – \(\frac{1}{2}\) ∠B and
∠ EFD = 90° – \(\frac{1}{2}\) ∠C.

Question 9.
Two congruent circles intersect each Other at points A and B. Through A any line segment PAQ is drawn so that P 9 lie-on. , the two circles. Prove that BP = BQ.
Answer:

Two congruent circles with centres X and Y intersect at A and B.
Hence, AB is their common chord.
In congruent circles, equal chords subtend equal angles at the centres.
∴ ∠AXB = ∠AYB
In the circle with centre X, ∠AXB = 2∠APB and in the circle with centre Y, ∠AYB = 2∠AQB.
∴ 2∠ APB = 2∠ AQB
∴ ∠APB = ∠AQB
∴ ∠QPB = ∠PQB
Thus, in ∆ BPQ, ∠QPB = ∠PQB
∴ QB = PB (Sides opposite to equal angles)
Hence, BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:

In ∆ ABC, the bisector of ∠A intersects the circumcircle of ∆ ABC at D.
∴∠BAD = ∠CAD
Aso, ∠BAD = ∠BCD and ∠CAD = ∠CBD (Angles in same segment)
∴ ∠BCD = ∠CBD
Thus, in ∆ BCD, ∠BCD = ∠CBD
∴BD = CD (Sides opposite to equal angles)
Thus, point D is equidistant from B and C.
Hence, D is a point on the perpendicular bisector of BC.
Thus, the bisector of ∠ A and the perpendicular bisector of side BC intersect at D and D is a point on the circumcircle of ∆ ABC.
Thus, in ∆ ABC, if the angle bisector of ∠A and the perpendicular bisector of side BC intersect, they intersect on the circumcircle of ∆ ABC.
Note: In ∆ ABC, if AB = AC, then the bisector of ∠A and the perpendicular bisector of side BC will coincide , and would not intersect in a single point.

Punjab State Board PSEB 9th Class English Book Solutions English E-mail Message Writing Exercise Questions and Answers, Notes.

PSEB 9th Class English E-mail Message Writing

Inviting friend to Watch a Play

Suppose you are Surjit Singh. Write an e-mail to your friend, Vipin Goyal, inviting him to watch a play.

Hi Vipin.
I am going to Government College for Women, Amritsar, to watch a play on 6 July, 20 – -.
Would you like to come? Let me know by Tuesday so that I can buy your ticket too.
Love
Surjit Singh.

House for Rent

Suppose you are Ramneek. Write an e-mail to your cousin, Darshan Pal, to put up a notice on his college notice-board to rent out your house.

Dear Pal
My father wants to rent our the Second floor of our house. There are two rooms, a kitchen and two attached bathrooms. He should like to have ₹ 2000 as rent. He will rake two months’ rent in advance. He warns to rent out the house to students. Please put up a notice on your college notice-board.
Regards
Ramneek.

Congratulation on Engagement

Suppose you are Shvinder Gill. Write an e-mail to your friend, Alok Wasn, congratulating him on his engagement.

Hi Alok
I have learnt that you are engaged. Congratulating! who is the lucky girl? who does she live and what does she do? Let me know when are you getting married? Is the dare fixed.
Love
Shivinder Gill.

Trip to the South

Suppose you are Varsha Gill. Write an e-mail to your friend, Asha Lakhpal, describing your visit to the south.

Hello Asha
Sorry, I couldn’t write to you earlier. I visited the south with my friend last month. We spent eight days there. We liked the Meenakshi Temple ar Madurai very much. The sunset at Kanyakumari was fascinating. We also went to Aurbindo Ashratn at Puducherry. It was very powerful there.
Love
Varsha.

E-mail (electronic mail) is the medium of communication that sends and receives messages through a specially designed computer network. With the revolution in information technology along with the rapid growth of the Internet, e-mail has become the most popular medium of communication. More and more people are using e-mail to send their messages. Due to its high speed, efficiency and low cost, e-mail has become one of the most important channels of communication. As e-mails are faster than letters, they are used for a quick transmission of all sorts of information.

Specimen of an Informal E-mail

Hi Paul
Sorry to say I’ll he a bit late for tonight’s rehearsal as something’s come up at home and I won’t be able to get away on time. I hope to make it by 7.15.
D. Paul.

Specimen of an Formal E-mail

Dear Ms. Maya
The books you ordered last week are now in stock and awaiting collection. I attach a list of the course books currently in stock at the bookshop.
Julie
Assistant Manage!

HFI Bookshop
Tel : 01123318301
Fax : 01123317931

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

PSEB 9th Class English Note-Making & Messages

Note-making

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

2. केवल प्रासंगिक बातें ही उनमें दी जाती हैं।

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

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

Passage 1:

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

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

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

Passage 2

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

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

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

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

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

Passage 3

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

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

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

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

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

Passage 4

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

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

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

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

Passage -5

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

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

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

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

2. Delay in admitting mistakes

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

3. Person who admits his mistakes

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

4. Person who hides his mistakes

• is considered a fool.
• nobody likes him

Passage – 7

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

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

• transmitting culture.

B. Modern rimes :

• not normal
• a period of transition.

C. teacher roles:

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

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

• basis of our past culture
• pillars of our future

(B) Secularism

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

(C) Disruptive tendencies

• serve no cause
• discredit the country

Passage – 9

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

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

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

• floods and fires
• earthquakes
• famines.
• pestilence

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

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

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Punjab State Board PSEB 9th Class English Book Solutions English Reading Comprehension Unseen Passages Exercise Questions and Answers, Notes.

PSEB 9th Class English Reading Comprehension Unseen Passages

(A) Passages From Grammar Book Note :

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

Passage 1

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage – 2

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

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

Passage – 3

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

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

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

Passage 4

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

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 5

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 6

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

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

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

Passage 7

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

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

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

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

Choose the correct option to answer each question :

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

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

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

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

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

(B) Some Other Passages

Passage 1

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 2

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 3.

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 4 :

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 5

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 6

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 7

His first “Satyagraha’ in India was in Champaran, in Bihar. The peasants of that district were being cruelly treated by the British indigo planters. Gandhiji left for Champaran to find out the truth. The news that a Mahatma had arrived to inquire into their suffering attracted thousands of peasants who flocked to have his darshan. The government got alarmed?, and Gandhiji was asked to leave the district. He refused and was asked to appear before the magistrate. Later, the case was withdrawn?. Gandhiji lived with the peasants for some time in order to learn about their hard lot. But he also taught them to be free and to stand on their feet. At last, he succeeded in securing justice for the poor peasants.

Choose the correct option to answer each question :

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

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

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

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

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

Passage 8

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 9

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

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

Choose the correct option to answer each question :

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

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

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

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

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

Passage 10

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

Choose the correct option to answer each question :

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

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

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

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

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