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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Answer:

In ∆ ABC, ∠ B is a right angle.
∴ ∠ B = 90° and AC is the hypotenuse.
In ∆ ABC,
∠A + ∠B + ∠C = 180°
∴ ∠ A + 90° + ∠ C = 180°
∴ ∠ A + ∠ C = 90°
Now, ∠ A and ∠ C are both positive (in degrees) and their sum is 90°.
∴ ∠ A < 90° and ∠ C < 90°
∴ ∠ A < Z B and ∠ C < ∠ B
∴ BC < AC and AB < AC (Theorem 7.7)
Hence, hypotenuse AC is greater than each of the other two sides BC and AB. Thus, the hypotenuse is the longest side in a right angled triangle.

Question 2.
In the given figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠ QCB. Show that AC > AB.

Answer:
∠ PBC < ∠ QCB (Given)
∴ – ∠ PBC > – ∠ QCB (Multiplying an inequality by (-1), it reverses)
∴ 180° – ∠ PBC > 180° – ∠ QCB (Adding 180° on both the sides) …………. (1)
Now, ∠ ABC and ∠ PBC as well as ∠ ACB and ∠ QCB from a linear pair.
∴ ∠ ABC = 180° – ∠ PBC and
∠ ACB = 180° – ∠ QCB
Substituting these values in (1), we get
∠ ABC > ∠ ACB
Now, in ∆ ABC, ∠ ABC > ∠ ACB A
∴ AC > AB (Theorem 7.7)

Question 3.
In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Answer:
In ∆ OAB, ∠ B < ∠ A
∴ OA < OB (Theorem 7.7) …………….. (1)
In ∆ OCD, ∠ C < ∠ D
∴ OD < OC (Theorem 7.7) ……………… (2)
Adding (1) and (2),
OA + OD < OB + OC
∴ AD < BC (As O is the point of intersection of AD and BC, it lies on both the line segments.)

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD ,(see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

Answer:

Construction: In quadrilateral ABCD, draw diagonal AC.
AB is the smallest side of ABCD and CD is the longest side of ABCD.
∴ AB < BC and AD < CD.
In ∆ ABC, AB < BC
∴ ∠ ACB < ∠ BAC …… (1)
In ∆ CDA, AD < CD
∴ ∠ DCA < ∠ DAC ……… (2)
Adding (1) and (2),
∠ ACB + ∠ DCA < ∠ BAC + ∠ DAC
∴ ∠ BCD < ∠ BAD
∴ ∠ BAD > ∠ BCD
Thus, in quadrilateral ABCD, ∠ A > ∠ C. Similarly, after constructing diagonal BD and using the inequalities in A ABD and A CBD, it can be proved that ∠ B > ∠ D.

Question 5.
In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Answer:
PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS = \(\frac{1}{2}\) ∠QPR ……………. (1)
∠ PSR is an exterior angle of A PQS and ∠ PSQ is an exterior angle of A PRS.
∴ PSR = ∠ Q + ∠QPS and
∠PSQ = ∠R + ∠RPS …………….. (2)
Now, in A PQR, PR > PQ
∴ ∠ Q > ∠ R
∴ ∠Q + \(\frac{1}{2}\) ∠ QPR > ∠R + \(\frac{1}{2}\) ∠ QPR
∴ ∠Q + ∠QPS > ∠R + ∠RPS [from (1)]
∴ ∠PSR > ∠PSQ

Question 6.
Show that of all line segments drawn from a given point not on a given line, the perpendicular line segment is the shortest.
Answer:

AB is a line and P is a point not on AB.
PM is the perpendicular line segment drawn from P to line AB.
N is any point on AB, other than M.
In ∆ PMN, ∠ M = 90°
∴ ∠ N < 90°
Thus, in ∆ PMN, ZN < ZM.
∴ PM < PN
This is true for any location of point N.
Hence, of all the line segments drawn from a point not on a given line, the, perpendicular line segment is the shortest.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 7 Triangles MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
In ∆ ABC, ∠A = ∠C, AC = 5 and BC = 4. Then, the perimeter of ∆ ABC is ……………. .
A. 9
B. 14
C. 13
D. 15
Answer:
C. 13

Question 2.
In ∆ PQR, PQ = PR, QR is extended to S and ∠PRS = 110°. Then, ∠PQR = ……………… .
A. 30°
B. 50°
C. 80°
D. 70°
Answer:
D. 70°

Question 3.
In ∆ ABC and ∆ DEF, AB = DE, BC = EF and ∠B = ∠E. If the perimeter of ∆ ABC is 20, then the perimeter of ∆ DEF is ……………. .
A. 10
B. 20
C. 15
D. 40
Answer:
B. 20

Question 4.
In ∆ ABC and ∆ PQR, AB = PQ, ∠A = ∠P and ∠B = ∠Q. If ∠A + ∠C = 130°, then ∠Q = …………….. .
A. 65°
B. 130°
C. 50°
D. 100°
Answer:
C. 50°

Question 5.
In ∆ PQR, ∠P = ∠Q = ∠R. If PQ = 6, then the perimeter of ∆ PQR is ………………. .
A. 12
B. 9
C. 18
D. 24
Answer:
C. 18

Question 6.
In ∆ ABC, AB < AC. Then …………… holds good.
A. ∠A < ∠B
B. ∠B < ∠C
C. ∠C < ∠A
D. ∠C < ∠B
Answer:
D. ∠C < ∠B

Question 7.
In ∆ PQR, ∠R > ∠Q. Then, ………………. holds good.
A. PQ > PR
B. QR > PQ
C. PR > PQ
D. PQ > QR
Answer:
A. PQ > PR

Question 8.
In ∆ ABC, AB > BC and BC > AC. Then, the smallest angle of ∆ ABC is …………………. .
A. ∠A
B. ∠C
C. ∠B
D. ∠A or ∠C
Answer:
C. ∠B

Question 9
………………….. cannot be the measures of the sides of a triangle.
A. 10, 12, 14
B. 2, 3, 4
C. 8, 9, 10
D. 2, 4, 10
Answer:
D. 2, 4, 10

Question 10.
In ∆ PQR, PQ = 4, QR = 6 and PR = 5. Then, …………….. is the angle with greatest measure in ∆ PQR.
A. ∠P
B. ∠Q
C. ∠R
D. ∠QRP
Answer:
A. ∠P

Question 11.
In ∆ XYZ, ∠X = 45° and ∠Z = 60°. Then, …………….. is the longest side of ∆ XYZ.
A. XY
B. YZ
C. XZ
D. XY or YZ
Answer:
C. XZ

Question 12.
In ∆ ABC, ∠B = 30° and BC is extended to D. If ∠ACD = 110°, then the longest side of ∆ ABC is ………………. .
A. AB
B. BC
C. CA
D. AB or AC
Answer:
B. BC

Question 13.
In ∆ ABC, AB = 4 and BC = 7, Then, ……………… holds good.
A. AC < 7
B. AC > 4
C. 4 < AC < 7
D. 3 < AC < 11
Answer:
D. 3 < AC < 11

Question 14.
In ∆ PQR, PQ = 3 and QR = 7. Then, …………….. holds good.
A. PR = 4
B. PR = 10
C. 10 > PR > 4
D. 7 > PR > 3
Answer:
C. 10 > PR > 4

Question 15.
In ∆ ABC, the bisectors of ∠B and ∠C intersect at I. If ∠A = 70°, then ∠BIC = ………………… .
A. 35°
B. 75°
C. 100°
D. 125°
Answer:
D. 125°

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

1. Find the perimeter and the area of a rectangle having :
(i) Length = 28 cm, Breadth = 15 cm
(ii) Length = 9.4 cm Breadth = 2.5 cm
Solution:
(i) Given length of rectangle = 28 cm
Breadth of rectangle = 15 cm
Perimeter of rectangle = 2 [length + Breadth]
= 2 [28 + 15]
= 2 × 43
= 86 cm

Area of rectangle = length × Breadth
= 28 × 15
= 420 cm²

(ii) Perimeter of rectangle = 2 [9.4 + 2.5]
= 2 × 11.9
= 23.8 cm
Area of rectangle = 9.4 × 2.5
= 23.5 cm²

2. Find the perimeter and the area of a square whose side measures
(i) 29 cm
(ii) 8.3 cm
Solution:
(i) Given side of square = 29 cm
Perimeter of square = 4 × side
= 4 × 29
= 116 cm
Area of square = (side)²
= (29)²
= 841 cm²

(ii) Perimeter of square = 4 × 8.3
= 33.2 cm
Area of square = 8.3 × 8.3
= 68.89 cm²

3. The perimeter of a square park is 148 m. Find its area.
Solution:
Given the perimeter of square park = 148 m
Side of the square park = \(\frac{perimeter}{4}\)
= \(\frac {148}{4}\)
Area of the square park = (side)²
= (37)²
= 1369 m²

4. The area of a rectangle is 580 cm². Its length is 29 cm. Find its breadth and also, the perimeter.
Solution:
Given area of rectangle = 580 cm²
Length of the rectangle = 29 cm
Let breadth of the rectangle = b cm
Area of the rectangle = length × breadth
580 = 29 × b
\(\frac {580}{29}\) = b
b = 20 cm
Perimeter of rectangle = 2 [length + breadth]
= 2 [29 + 20]
= 2 × 49
= 98 cm

5. A wire is in the shape of a rectangle. Its length is 48 cm and breadth is 32 cm. If the same wire is rebent into the shape of a square, what will be the measure of each side. Also, find which shape encloses more area and by how much ?
Solution:
Given length of the rectangle = 48 cm
Breadth of the rectangle = 32 cm
Perimeter of the rectangle = 2 [length + breadth]
= 2 [48 + 32]
= 2 × 80
= 160 cm
Let side of square = a cm
Perimeter of the square = 4 × a
Since wire is rebent into the shape of a square
Perimeter of square = Perimeter of rectangle
4 a = 160
Therefore, a = \(\frac {160}{4}\)
= 40 cm
Area of square = (side)²
= 40 × 40
= 1600 cm²
Area of rectangle = length × breadth
= 48 × 32
= 1536 cm²
∴ Square encloses more area by 64 cm²

6. The area of a square park is the same as that of a rectangular park. If the side of the square park is 75 m and the length of the rectangular park is 125 m, find the breadth of the rectangular park. Also, find the perimeter of rectangular park.
Solution:
Given side of square park = 75 m
Area of square park = (75)²
= 75 × 75
= 5625 m²
Length of rectangular park = 125 m
Let breadth of rectangular park = b m
Area of rectangular park = length × breadth
= 125 × b m²
Given that
Area of rectangular park = Area of square park
125 × b = 5625
= 45 m
Perimeter of rectangular park = 2 [length + breadth]
= 2 [125 + 45]
= 2 × 170
= 340 m

7. A door of length 2.5 m and breadth 1.5 m is fitted in a wall. The length of wall, is 9 m and breadth is 6 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 30 per m².

Solution:
Length of door = 2.5 m
Breadth of door = 1.5 m
Area of door = length × breadth
= 2.5 × 1.5
= 3.75 m²
Area of wall = 9 × 6
= 54 m²
Area of wall painting = Area of wall including door – Area of door
= 54 – 3.75
= 50.25 m²
Cost of painting 1 m² of wall = ₹ 30
Cost of painting 50.25 m² of wall = ₹ 50.25 × 30
= ₹ 1507.50

8. A door of dimensions 3 m × 2 m and a window of dimensions 2.5 m × 1.5 m is fitted in a wall. The length of the wall is 7.8 m and breadth is 3.9 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 25 per m².
Solution:
Area of door = 3 × 2 = 6 m²
Area of window = 2.5 m × 1.5 m
= 3.75 m²
Area of wall = 7.8 m × 3.9 m
= 30.42 m²
Area of painting the wall = Area of wall – Area of door – Area of window
= 30.42 – 6 – 3.75
= 20.67 m²
Cost of painting the wall = ₹ 25 × 20.67
= ₹ 516.75

9. Find the area and the perimeter of the following figures.

Solution:
(i) Perimeter of the given figure
= AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 3.5 + 3 + 2 + 5 + 3.5 + 10 + 9
= 38 cm²
Area of the figure = Area of rectangle ABCJ + Area of rectangle JDEI + Area of rectangle IFGH
= 2 × 3.5 + 5 × 2 + 10 × 3.5
= 7 + 10 + 35
= 52 cm²

(ii) Perimeter of the given figure
= 8cm + 5 cm + 1.5 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1.5 cm + 1.5 cm + 2.5 cm + 1.5 cm
= 29 cm
Area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III
= 8 cm × 1.5 cm + 3.5 cm × 1.5 cm + 1.5 cm × 1.5 cm
= 12 cm² + 5.25 cm² + 2.25 cm
= 19.5 cm²

10. Multiple Choice Questions :

Question (i).
What is the area of a rectangle of dimensions 12 cm × 10 cm ?
(a) 44 cm²
(b) 120 cm²
(c) 1200 cm²
(d) 1440 cm²
Answer:
(b) 120 cm²

Question (ii).
Find the breadth of a rectangle whose length is 12 cm and perimeter is 36 cm.
(a) 6 cm
(b) 3 cm
(c) 9 cm
(d) 12 cm
Answer:
(a) 6 cm

Question (iii).
If each side of a square is 1 m then its area is ?
(a) 10 cm²
(b) 100 cm²
(c) 1000 cm²
(d) 10000 cm²
Answer:
(d) 10000 cm²

Question (iv).
Find the area of a square whose perimeter is 96 cm.
(a) 576 cm²
(b) 626 cm²
(c) 726 cm²
(d) 748 cm².
Answer:
(a) 576 cm²

Question (v).
The area of a rectangular sheet is 500 cm². If the length of the sheet is 25 cm, what is its breadth ?
(a) 30 cm
(b) 40 cm
(c) 20 cm
(d) 25 cm.
Answer:
(c) 20 cm

Question (vi).
What happens to the area of a square, if its side is doubled ?
(a) The area becomes 4 times, the area of original square.
(b) The area becomes \(\frac {1}{4}\) times, the area of original square.
(c) The area becomes 16 times, the area of original square.
(d) The area becomes \(\frac {1}{6}\) times, 6 the area of original square.
Answer:
(a) The area becomes 4 times, the area of original square.

Punjab State Board PSEB 7th Class English Book Solutions English Reading Comprehension Conversation / Dialogue Based Exercise Questions and Answers, Notes.

PSEB 7th Class English Reading Comprehension Conversation / Dialogue Based

Read the following conversation carefully and answer the questions that follow:

1. Akbar : Birbal make me a painting. Use your imagination in it.
Birbal : I am not an artist, how can I paint ?
Akbar : If I don’t get a good painting, you shall be punished.
Birbal : (Next day) I am here with the painting.
Akbar : I am happy to see that you obeyed me. Please show me the painting.
Birbal : Huzoor, have a look. I am opening the covered frame.
Akbar : This painting is nothing but only ground and sky. There is a little grass on the ground. What is this?
Birbal : A cow eating grass, Huzoor!
Akbar : Where is the cow and grass ?
Birbal : I used my imagination. The cow ate the grass and returned to its shed.

Question 1.
The given conversation is between ……………
(a) Birbal and Begum
(b) Birbal and his Son
(c) Akbar and his Son
(d) Akbar and Birbal.
Answer:
(d) Akbar and Birbal.

Question 2.
What did Akbar want Birbal to make for him?
(a) a cake
(b) a painting
(c) a dish
(d) all of the above.
Answer:
(b) a painting.

Question 3.
Who ate the grass ?
(a) cow
(b) cat
(c) camel
(d) goat.
Answer:
(a) cow.

Question 4.
What was in the painting ?
(a) ground and sky
(b) boy and sky
(c) girl and kite
(d) cat and dog.
Answer:
(a) ground and sky.

Question 5.
Birbal carried a ……………. with him.
(a) covered glass
(b) covered frame
(c) covered bowl
(d) covered box.
Answer:
(b) covered frame.

2. Sadab : Good afternoon.
Ritu : Good afternoon.
Sadab : So, where are you going now?
Ritu : I am going to meet my friend Garima in Kolkata Market.
Sadab : Oh, so nice, I have to buy a new school uniform.
Ritu : Will you accompany me ?
Sadab : Yes, it will be great fun for me.
Ritu : So, can we move now so that we can reach on time.
Sadab : Sure.

Question 1.
This conversation is between ………..
(a) Sadab and Garima
(b) Ritu and Garima
(c) Ritu and Sadab
(d) Garima and Kolkata Market.
Answer:
(c) Ritu and Sadab.

Question 2.
Where is Ritu going ?
(a) to meet her sister Garima
(b) to meet her friend Garima
(c) to meet her teacher
(d) to buy a school uniform.
Answer:
(b) to meet her friend Garima.

Question 3.
Garima will be found ……………
(a) in Kolkata market
(b) in Bombay market
(c) in Kolkata street
(d) in Sadab market.
Answer:
(a) in Kolkata market.

Question 4.
What does Sadab want to buy ?
(a) some new books
(b) a new mobile phone
(c) a new school uniform
(d) new shoes.
Answer:
(c) a new school uniform.

Question 5.
What was the time of conversation ?
(a) early morning
(b) afternoon
(c) late evening
(d) night.
Answer:
(b) afternoon.

3. Headman Ant asked, “What happened ? Why are you running ?” Rabbit said, “We’re running for our lives. I heard the news of Tsunami, very high sea waves.”

Headman Ant asked everyone to stay calm and explained that they were just testing the warning system. It didn’t happen for real. Let us all be prepared to deal with disasters.

Next day at the village meeting, everyone gave their ideas.

Elephant : “We should always follow news carefully.”
Butterfly : “I have already put my valuable things away in a safe place.”
Rats : “I have prepared survival bag which will always be close to me. I have kept a bottle of water, medicines, dried food, some beans, the radio for news and a family photo.”
Mother Frog : “My kids will not go out to play by themselves in hazardous places.”
Mother Goat : “We must run to higher grounds or to the safe areas. I shall make a map of the village.
Headman Ant : “Why didn’t you run like asked Grandpa others ?
OX : ……..
Grandpa OX : “We are too old. We don’t want to be a burden to anyone.”
Headman Ant : (said politely) “No one is a burden. What would your grandchildren do without you ??

Question 1.
Rabbit was running for life because he had heard the news of ……………
(a) Earthquake
(b) Tsunami
(c) Flash flood
(d) Land slide.
Answer:
(b) Tsunami.

Question 2.
What did the Headman Ant tell the animals about the news ?
(a) It was real
(b) It was to create panic only
(c) It was not real
(d) It was to entertain the old people.
Answer:
(c) It was not real.

Question 3.
Who is most careful about news ?
(a) Grandpa Ox
(6) Rabbit
(c) Elephant
(d) Frog.
Answer:
(c) Elephant.

Question 4.
Which one of the following is not true according to the conversation ?
(a) The elephant was not present at the meeting
(6) Grandpa Ox does not want to be a burden to anyone
(c) Mother Goat wants to run to the higher places
(d) Rats have prepared survival bags.
Answer:
(a) The elephant was not present at the meeting.

Question 5.
Tsunami is a ……………..
(a) safe place from high sea wave
(b) very high sea wave
(c) cool breeze coming from sea
(d) dry wind blowing towards sea.
Answer:
(b) very high sea wave.

4. Ali : Hello ! I’m Ali. What’s your name?
Ravi : Hello ! My name’s Ravi. I’m ten years old.
Ali . : I’m ten, too. Do you play cricket ?
Ravi : Not very well. I’m fond of football.
Ali : Oh, good ! Will you help me with my homework?
Ravi : Of course, I will. Come to me during the break.
Ali : Thanks. I’ll help you play cricket.
Ravi : That will be great. Thank you !
Ali : : Where do you live ?
Ravi : I live near the school and Ali, where do you live?
Ali : I live in the school hostel.

Question 1.
What is common between Ali and Ravi ?
(a) both are ten years old
(b) both play cricket very well
(c) both live in the school hostel
(d) both are fond of playing football.
Answer:
(a) both are ten years old.

Question 2.
Where does Ali live ?
(a) near the school
(b) in the school hostel
(c) near the football club
(d) near the cricket club.
Answer:
(b) in the school hostel.

Question 3.
Ali will help Ravi ……………
(a) play football
(b) do his homework
(c) play cricket
(d) all the above.
Answer:
(c) play cricket.

Question 4.
Ravi will help Ali ………
(a) play cricket
(b) play football
(c) do his homework
(d) none of these.
Answer:
(c) do his homework.

Question 5.
When will Ravi help Ali with his homework?
(a) after playing cricket
(b) after playing football
(c) when the school starts
(d) during the break.
Answer:
(d) during the break.

5. Ravi : Do you know what time it is now, please ?
Aman : No, I’m really very sorry, I don’t have a watch.
Rama : Could you tell me the time, please ?
Radha : It’s about eleven-thirty.
Rama : Thanks a lot.
Radha : At what time do you come to school ?
Rama : At 8:30 in the morning.
Radha : At what time does the school get over ?
Rama : At 3 o’clock in the afternoon.
Ajay : What time is the next bus, please ?
Anil : It leaves at 5:30 p.m.
Ajay : Is there any bus after 10 p.m. ?
Anil : The last bus leaves at 10:30 p.m.
Ajay : Thanks a lot for the information.

Question 1.
The main point in this conversation is …………… .
(a) bus
(b) time
(c) watch
(d) school.
Answer:
(b) time.

Question 2.
Aman is unable to tell Ravi time because …………….
(a) he is in a hurry
(b) he is very sorry
(c) he has no watch
(d) he has no clock.
Answer:
(c) he has no watch.

Question 3.
When does Rama go to school ?
(a) at 8:30
(b) at 3 o’clock
(c) at 11:30
(d) at 10 o’clock.
Answer:
(a) at 8:30.

Question 4.
What does Ajay want to know ?
(a) times about school
(b) times about buses
(c) time by Aman’s watch
(d) none of these.
Answer:
(b) times about buses.

Question 5.
Who tells Ajay time for last bus ?
(a) Rama
(b) Anil
(c) Radha
(d) Aman.
Ans.
(b) Anil.

6. Raj : Could you help me, please ?
Shopkeeper : Certainly.
Raj : Thank you.
Shopkeeper : What can I get you ?
Raj : A black shoe polish.
Rani : Excuse me please, where are the pens ?
Shopkeeper : They are in the second row. Let me help you.
Rani : How much do the two pens cost ?
Shopkeeper : The price is written on them. They cost Rs. 10 each.
Rani : Thank you.

Question 1.
Whose help does Raj ask for ?
(a) Rani’s
(b) shopkeeper’s
(c) nobody’s
(d) his own.
Answer:
(b) shopkeeper’s.

Question 2.
Raj wants to buy a ……………..
(a) brown shoe polish
(b) black shoe polish
(c) some pens and a polish
(d) none of these.
Answer:
(b) black shoe polish.

Question 3.
The pens are lying
(a) in a box
(b) in the first row
(c) in the second row
(d) on the upmost shelf.
Answer:
(c) in the second row.

Question 4.
How many pens does Rani want to buy ?
(a) two
(b) three
(c) one.
(d) four.
Answer:
(a) two.

Question 5.
How much do the two pens cost ?
(a) Rs. 20
(6) Rs. 10
(c) Rs. 30
(d) Rs. 15.
Answer:
(a) Rs. 20.

7. Aryan : Why are you wearing this funny dress, Praveen ?
Praveen : Aryan, what is funny about it?
Aryan : Look at your clothes ! They look so tight, and your shoes have wheels. And what’s that you have on your head ?
Praveen : (laughs) Oh, are you talking about this dress ? I’m going for my skating class and this is my helmet to save me from injuries.
Aryan : Skating ?
Praveen : Yes, skating. That’s why I’m wearing these roller skating shoes.
Aryan : Where are you going for your skating class ?
Praveen : I’m going to the ‘Roller Mall’. It has a skating rink.
Aryan : Could I come with you to see what you do there?
Praveen : Sure, come along. (They reach the skating rink.) Aryan (amazed) Oh, so many people moving on the wheels !
Praveen : They’re all skaters moving on roller skates.
Aryan : Look ! How they all are gliding on the floor. I wish I could do the same. Will you teach me how to skate ?
Praveen : Oh yes, whenever you are ready.
Aryan : Thanks.

Question 1.
Praveen’s dress looks funny to Aryan. Why?
(a) It is very tight.
(b) His shoes have wheels.
(c) There is a, helmet on his head.
(d) all these.
Answer:
(d) all these.

Question 2.
Praveen is going to …………
(a) his school to take part in a drama
(b) a park for a picnic
(c) perform a magic show
(d) his skating class.
Answer:
(d) his skating class.

Question 3.
A helmet saves us from …………
(a) foot injuries
(b) severe cold on ice
(c) head injuries
(d) all the above.
Answer:
(c) head injuries.

Question 4.
What does Aryan want to learn ?
(a) skating
(b) use of wheeled shoes
(c) using a helmet
(d) wearing a tight dress.
Answer:
(a) skating.

Question 5.
Who will teach Aryan how to skate ?
(a) skating teacher
(b) his elder brother
(c) Praveen
(d) class teacher.
Answer:
(c) Praveen.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

1. Construct ΔABC in which AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
Solution:
Given : Three sides of a triangle as AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
To construct : A triangle with these three sides.
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment BC = 5 cm.

Step 3 : From B; point A is at a distance 3.5 cm. So, with B as centre, draw an arc of radius
3.5 cm (Now point A will be some where on this arc. Our job is to find where exactly A is.)

Step 4. From C, point A is at a distance of 7 cm. So, with C as centre; draw an arc of radius 7 cm. (A will be some where on this arc. We have to fix it).

Step 5. A has to be on both the arcs drawn. So it is the point of intersection of arcs.
Mark the point of intersection of arcs as A. Join AB and AC.
Thus we obtain ΔABC.

2. Construct a triangle ABC in which AB = BC = 6.5 cm and CA = 4 cm. Also name the kind of triangle drawn.
Solution:
Given : Three sides of triangle as AB = BC = 6.5 cm. and CA = 4 cm.
To construct : A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment AC = 4 cm.

Step 3. From A; point B is at a distance of 6.5 cm. So, with A as centre, draw an arc of radius 6.5 cm. (Now point B will be somewhere on this arc. Our job is to find where exactly A is.)

Step 4. From C; point B is at a distance of 6.5 cm. So; with C as centre; draw an arc of radius 6.5 cm. (B will be some where on this arc. We have to fix it.)

Step 5. B has to be on both the arcs drawn. So it is the point of intersection of arcs. Mark the point of intersection of arcs as B. Join AB and BC.

We observe that AB = BC = 6.5 cm.
Since two sides are of equal length. Thus we obtain an isosceles ΔABC.

3. Construct a triangle XYZ such that length of each side is 5 cm. Also name the kind of triangle drawn.
Solution:
Given : A triangle XYZ in which XY = YZ = ZX = 5 cm..
To Construct. A triangle XYZ with each side 5 cm.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle XYZ and indicate the lengths of the three sides.

Step 2. Draw a line segment YZ = 5 cm.

Step 3. From Y; point X is at a distance of 5 cm. So, with Y as centre, draw an arc of radius 5 cm. (Now point X will be somewhere on this arc. Our job is to find where exactly X is.)

Step 4. From Z, point X is at a distance of 5 cm. So, with Z as centre, draw an arc of radius 5 cm. (X will be somewhere oh this arc. We have to fix it.)

Step 5. Point X has to be on both the arcs drawn. So, it is the point of intersection of arcs.
Mark the point of intersection of arcs as X. Join XY and XZ.

Thus we obtain an equilateral ΔXYZ each of whose side is 5 cm.

4. Construct a triangle PQR such that PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm. Measure ∠PQR and also name the kind of triangle drawn.
Solution:
Given. Three sides of triangle as PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm.
To construct. A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the lengths of the three sides.

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

Step 3. From Q; point P is at a distance of 2.5 cm. So, with Q as centre, draw an arc of radius 2.5 cm. (Now point P will be some where on this arc. Our job is to find where exactly P is.)

Step 4. From R; point P is at a distance of 6.5 cm. So; with R as centre; draw an arc of radius 6.5 cm. (P will be somewhere on this arc. We have to fix it.)

Step 5. Point P has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark a point of intersection of arcs as P. Join PQ and PR.

Thus we obtain ΔPQR; on measuring ∠PQR, we observe that ∠PQR = 90° so it is a right angled triangle.

5. Construct a triangle ABC, in which AB = 6 cm, BC = 2 cm, CA = 3 cm. (If possible). If not possible give the reason.
Solution:
Since AB = 6 cm, BC = 2 cm, CA = 3 cm
Here BC + CA = 2 cm + 3 cm
= 5 cm < 6 cm < AB
Which is not possible because the sum of two sides of a triangle is always greater than third side of the triangle.

6. Question (i).
Which of the following can be used to construct a triangle ?
(a) The lengths of the three sides
(b) The perimeter of the triangle
(c) The measures of three angles
(d) The name of three vertices
Answer:
(a) The lengths of the three sides

Question (ii).
A triangle can be constructed by taking its sides as :
(a) 1.8 cm, 2.6 cm, 4.4 cm
(b) 3 cm, 4 cm, 8 cm
(c) 4 cm, 7 cm, 2 cm
(d) 5 cm, 4 cm, 4 cm.
Answer:
(d) 5 cm, 4 cm, 4 cm.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 3 Playing with Numbers MCQ Questions

Multiple Choice Questions

Question 1.
Which number is a factor of every, number?
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 2.
How many even numbers are prime?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 3.
The smallest composite number is:
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(d) 4.

Question 4.
Which of the following number is a perfect number?
(a) 8
(b) 6
(c) 12
(d) 18.
Answer:
(b) 6

Question 5.
Which of the following is not a multiple of 7?
(a) 35
(b) 48
(c) 56
(d) 91.
Answer:
(b) 48

Question 6.
Which of the following is not a factor of 36?
(a) 12
(b) 6
(c) 9
(d) 8.
Answer:
(d) 8.

Question 7.
The number of prime numbers upto 25 are:
(a) 9
(b) 10
(c) 8
(d) 12.
Answer:
(a) 9

Question 8.
Which mathematician gave the method to find prime and composite numbers?
(a) Aryabhatta
(b) Ramayan
(c) Eratosthenes
(d) Goldbach.
Answer:
(c) Eratosthenes

Question 9.
The statement “Every even number greater than 4 can be expressed as the sum of two odd prime numbers” is given by:
(a) Goldbach
(b) Eratosthenes
(c) Aryabhatta
(d) Ramanujan.
Answer:
(a) Goldbach

Question 10.
Which of the following is a prime number?
(a) 221
(b) 195
(c) 97
(d) 111.
Answer:
(c) 97

Question 11.
Which of the following number is divisible by 4?
(a) 52369
(b) 25746
(c) 21564
(d) 83426.
Answer:
(c) 21564

Question 12.
Which of the following is not true?
(a) If a number is factor of two numbers then it is also factor of their sum.
(b) If a number is factor of two numbers then it is also factor of their difference.
(c) 15 and 24 are co-prime to each other.
(d) 1 is neither prime nor composite.
Answer:
(c) 15 and 24 are co-prime to each other.

Question 13.
Which of the following pair is co-prime?
(a) (12, 25)
(b) (18, 27)
(c) (25, 35)
(d) (21, 56).
Answer:
(a) (12, 25)

Question 14.
Which of the following number is divisible by 8?
(a) 123568
(b) 412580
(c) 258124
(d) 453230.
Answer:
(a) 123568

Question 15.
Prime factorisation of 84:
(a) 2 × 2 × 3 × 2 × 7
(b) 7 × 2 × 3 × 3
(c) 2 × 3 × 7 × 2
(d) 3 × 2 × 3 × 2 × 7.
Answer:
(c) 2 × 3 × 7 × 2

Question 16.
H.C.F. of 25 and 45 is:
(a) 15
(b) 5
(c) 225
(d) 135.
Answer:
(b) 5

Question 17.
If L.C.M. of two numbers is 36 then which of the following can not be their H.C.F.?
(a) 9
(b) 12
(c) 8
(d) 18.
Answer:
(c) 8

Question 18.
The L.C.M. of two co-prime numbers is 143. If one number is 11 then find other number.
(a) 132
(b) 154
(c) 18
(d) 13.
Answer:
(d) 13.

Question 19.
Find the greatest number which divides 145 and 235 leaving the remainder 1 in each case.
(a) 24
(b) 18
(c) 19
(d) 17.
Answer:
(b) 18

Question 20.
The greatest 4 digit number which is divisible by 12,15 and 20.
(a) 9990
(b) 9000
(c) 9960
(d) 9999.
Answer:
(c) 9960

Question 21.
Which of the following is a prime number?
(a) 23
(b) 51
(c) 39
(d) 26.
Answer:
(a) 23

Question 22.
Which of die following is a prime number?
(a) 32
(b) 30
(c) 31.
(d) 33.
Answer:
(c) 31.

Question 23.
Which of the following is a composite number?
(a) 12
(b) 19
(c) 29
(d) 31.
Answer:
(a) 12

Question 24.
Which of the following is an even number?
(a) 13
(b) 15
(c) 16
(d) 19.
Answer:
(c) 16

Question 25.
Which of the following is an odd number?
(a) 12
(b) 13
(c) 14
(d) 20.
Answer:
(b) 13

Fill in the blanks:

Question 1.
…………… is an even prime number?
Answer:
2

Question 2.
…………… is the greatest prime number between 1 and 10.
Answer:
7

Question 3.
……………. is neither prime nor composite number.
Answer:
1

Question 4.
A number which has only two factors is called a …………….. number.
Answer:
prime number

Question 5.
A number which has more than two factors is called a ……………… number.
Answer:
composite number

Write True/False:

Question 1.
The sum of three odd number is even. (True/False)
Answer:
False

Question 2.
All prime numbers are odd. (True/False)
Answer:
False

Question 3.
All even numbers are composite numbers. (True/False)
Answer:
False

Question 4.
1 neither prime nor composite. (True/False)
Answer:
True

Question 5.
If a number is factor of two numbers then it is also factor of their sum. (True/False)
Answer:
True

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 6 Lines and Angles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 6 Lines and Angles 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 measure of the complementary angle of an angle with measure 40° is ………………….. .
A. 40°
B. 20°
C. 140°
D. 50°
Answer:
D. 50°

Question 2.
The measure of the supplementary angle of an angle with measure 70° is ………………… .
A. 20°
B. 35°
C. 70°
D. 110°
Answer:
D. 110°

Question 3.
∠ ABC and ∠ ABD form a linear pair. If ∠ ABC = 30°, then ∠ ABD = ………………. .
A. 30°
B. 60°
C. 150°
D. 15°
Answer:
C. 150°

Question 4.
∠P and ∠Q are supplementary angles such that ∠P = 2x – 5 and ∠Q = 3x + 10. Then ∠Q = …………….. .
A. 35°
B. 65°
C. 105°
D. 115°
Answer:
D. 115°

Question 5.
The measure of an angle is four times the measure of its complementary angle. Then, the measure of that angle is ………………… .
A. 18°
B. 72°
C. 40°
D. 10°
Answer:
B. 72°

Question 6.
The measures of two supplementary angles differ by 20°. Then, the measure of the acute angle among them is …………….. .
A. 50°
B. 80°
C. 100°
D. 20°
Answer:
B. 80°

Question 7.
The measure of an angle is twice the measure of its supplementary angle. Then, the measure of that angle is ………………. .
A. 60°
B. 120°
C. 50°
D. 100°
Answer:
B. 120°

Question 8.
∠ ACD is an exterior angle of ∆ ABC. If ∠ ACD = 110° and ∠ A = 60°, then ∠ B = ………………….. .
A. 50°
B. 60°
C. 70°
D. 55°
Answer:
A. 50°

Question 9.
In ∆ ABC, ∠ A = 70° and ∠ B = 60°. Then, the measure of an exterior angle of ∆ ABC can be ……………….. .
A. 50°
B. 110°
C. 100°
D. 70°
Answer:
B. 110°

Question 10.
In ∆ ABC, ∠ B = 55° and ∠ C = 65°. Then the measure of an exterior angle of ∆ ABC cannot be ………………. .
A. 125°
B. 120°
C. 115°
D. 110°
Answer:
D. 110°

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

PSEB 7th Class English Notice Writing

Notice लोगों को किसी घटना की जानकारी देने का संदेश अथवा समाचार होता है। उदाहरण के लिए किसी स्कूल के Notice-board पर विद्यार्थियों को स्कूल की विभिन्न गतिविधियों की जानकारी दी जा सकती है। Notice लिखते समय अग्रलिखित बातों का उल्लेख अवश्य करें:

1. शीर्षक-इसमें स्पष्ट किया जाना चाहिए कि Notice किसके लिए है।
2. Notice लिखे जाने की तिथि
3. संबंधित घटना का दिन, समय तथा स्थान
4. Notice के नीचे लिखने वाले का नाम, पद, पता आदि।
5. शब्द-सीमा लगभग 30 शब्दों तक।

1. You are the PTI of your school. Write a notice asking the students to enrol for free yoga classes.

Free Yoga Classes

Attention!

20 March 20 ……

All students interested in attending free yoga classes from 10th April every. morning from 6 a.m. to 7 a.m. should contact the undersigned before 7th April.

Sd/
B.S. Bedi
PTI

2. You are the librarian of your school. Write a notice asking the students to return borrowed books before the school closes down.

Attention!

10 March 20…….

The school is going to be closed down for the summer vacation next week. All students who have borrowed any book from the library must return it before the school closes down.

Raman Kumar
Librarian

3. You are Anupam, the editor of the school magazine, and want to hold an interclass competiton to collect poems and cartoons for the magazine before Sept. 9. Draft a notice for the students ‘Notice-board inviting entires.

Interesting Contest, Amazing Prizes

We are going to hold an inter-class compition to collet poems and cartoons for the school magazine. Entries for the same are invited to reach the under signed befors 9th September. The result of the contest will be declared on 15th. The best poem and cartoon shall win a free school blazer. There will be some consolation prizes to.

4. You are Sangeets, the secretary of the school quiz club. You want to hold an inter-class competition to decide on entires for an inter-school competition 2 weeks from now. Draft a notice for the students notice-board inviting participants.

Prizes Through Quizzes

An inter-school quiz-competition is going to be held two weeks from now. In order to decide entreis for the same, we have decided to hold on inter-class competition on Monday, the 15th. All those who desire to participets schould give their name to the undersigned by tomorrow. The three best participants shall be awarded free school blazers Sangeeta.

(Secretary, School Quiz Club.)

5. You are Kulbir Singh of Class VII. You have lost your new water bottle. Write the notice that you would like to put up on the school notice-board.

Lost ! Lost ! Lost !

12 March 20……..

I have lost my new water bottle somewhere in the school garden. The bottle is of Eagle make with blue colour. The finder is requested to return it to me or deposit it with the school office.

Kulbir Singh
Roll No. 2
VII B.

6. You are the Sarpanch of your village. Write a notice inviting adults to donate blood at the blood donation camp to be held at the community centre.

Blood Donation Camp

8 March 20 ………

The village Panchayat is going to organise a blood donation camp in the community centre on 8th April from 9 a.m. to 11 a.m. All the adults of the village should come forward and donate blood to save lives of many.

Balwan Singh
Village Sarpanch.

7. You have misplaced a library book ‘Panchtantra Tales’. Write a notice that you would like to put up in the classroom.

Book Misplaced

10 March 20 …….

I have misplaced my book “Panchtantra Tales’ somewhere in the classroom. I borrowed it from the library yesterday. The finder is requested to return it to me or deposit it with the class teacher.

Rajni
Roll No. 5
VII A.

8. You are the sports captain of your school. Write a notice to all participants to submit their names and event in which they are taking part.

Attention ! Sports Participants

02 May 20 …….

All the participants in the school sports are requested to submit their names with the undersigned before Sunday, the 6th May. They must mention the event they are taking part in.

Geeta Sharma
Sports Captain
Khalsa Girls High School, Moga.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.2

1. Using number line write the integer which is:

Question (a)
5 less than -1
Solution:
5 less than -1 = ?
We need to find the integer which is 5 less than -1.
So, we shall start with ‘-1 ’ and proceed 5 steps to the left of -1 as shown below:

Therefore 5 less than -1 is -6

Question (b)
5 more than -5
Solution:
5 more than -5 = ?
We need to find the integer which is 5 more than -5.
So, we shall start with -5 and proceed 5 steps to the right of -5 as shown below:

Therefore 5 more than -5 is zero.

Question (c)
2 less than 5
Solution:
2 less than 5 = ?
We need to find the integer which is 2 less than 5.
So, we shall start with 5 and proceed
2 steps to the left of 5 as shown below:

Therefore 2 less than 5 is 3.

Question (d)
3 less than -2.
Solution:
3 less than – 2 = ?
We need to find the integer which is 3 less than -2.
So, we shall start with -2 and proceed 3 steps to the left of -2 as shown below:

Therefore 3 less than -2 is -5.

2. Using number line, add the following integers:

Question (a)
9 + (-3)
Solution:
On number line we shall start from 0 and move 9 steps to the right of zero. Then we shall move three steps to the left of ‘+9’. We finally reach at +6. Thus, +6 is the answer.

Hence, 9 + (-3) = +6

Question (b)
5 + (-11)
Solution:
On number line we shall start from 0 and move 5 steps to the right of zero. Then we shall move eleven steps to the left of ‘+5’. We finally reach at ‘-6’. Thus, -6 is the answer.

Hence, 5 + (-11) = -6

Question (c)
(-1) + (-4)
Solution:
On number line we first move 1 step to the left of zero. Then we moved ahead 4 steps to the left of ‘-1\ We finally reaches at -5. Thus -5 is the answer.

Hence, (-1) + (-4) = -5

Question (d)
(-5) + 12
Solution:
On number line we shall move 5 steps to the left of zero. Then we shall move 12 steps to the right of ‘-5’ and finally reach at ‘+7’. Thus +7 is the answer.

Hence, (-5) + 12 = +7

Question (e)
(-1) + (-2) + (-4)
Solution:
Step I: On number line we shall move one step to the left of zero suggested by minus sign of ‘-1’.
Step II: Then we shall move 2 steps to the left of -1 suggested by minus sign of -2 and we reach at ‘-3’.
Step III: Then again we shall move 4 steps to the left of ‘-3’ suggested by minus sign of -4 and finally we reach at-7.

Hence, (-1) + (-2) + (-4) = -7

Question (f)
(-2) + 4 + (-5)
Solution:
Step I: On number line we shall move 2 steps to the left of zero suggested by minus sign of ‘-2’.
Step II: Then we shall move 4 steps to the right of ‘-2’ suggested by plus sign of ‘44’ and we reach at +2.
Step III: Then we shall move 5 steps to the left of ‘+2’ as suggested by minus sign of ‘-5’ and finally we reach at -3. Thus answer is ‘-3’.

Hence, (-2) + 4 + (-5) = -3

Question (g)
(-3) + (5) + (-4).
Solution:
Step I: On number line we shall move 3 steps to the left of zero as suggested by minus sign of ‘-3’.
Step II: Then we shall move 5 steps to the right of ‘-3’ as suggested by plus sign of ‘+5’ and we reach at ‘+2’.
Step III: Then we shall move 4 steps to the left of ‘+2’ as suggested by minus sign of ‘4’ and finally we reach at ‘-2’. Thus ‘-2’ is the answer.

Hence, (-3) + (5) + (-4) = -2

3. Add without using number line:

Question (a)
18 + 13
Solution:
18 + 13 = 31

Question (b)
18 + (- 13)
Solution:
18 + (- 13) = + (18 – 13) = +5

Question (c)
(-18) + 13
Solution:
(-18) + 13 = – (18 – 13) = -5

Question (d)
(-18) + (-13)
Solution:
(-18) + (-13) = -(18 + 13) = -31

Question (e)
180 + (-200)
Solution:
180 + (-200) = -(200 – 180) = -20

Question (f)
111 + (-67)
Solution:
111 + (-67) = (777 – 67) = 710

Question (g)
1262 + (-366) + (-962)
Solution:
1262 + (-366) + (-962)
= 1262 – (366 + 962)
= 1262 – 1328
= -(1328 – 1262) = -66

Question (h)
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
Solution:
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
= 30 + 21 + 11 + (-27) + (-19) + (-3) + (-9)
= 62 + (-58) = 62 – 58 = 4

Question (i)
(-7) + (-9) + 4 + 16
Solution:
(-7) + (-9) + 4+16 = (-16) + 20 = 4

Question (j)
37 + (-2) + (-65) + (-8).
Solution:
37 + (-2) + (-65) + (-8)
= 37 – (2 + 65 + 8)
= 37 – 75 = -38

4. Write the successor and predecessor of the following:

Question (a)
-15
Solution:
Successor of -15 = -15 + 1 = -14
Predecessor of -15 = -15 – 1 = -16

Question (b)
27
Solution:
Successor of 27 = 27 + 1 = 28
Predecessor of 27 = 27 – 1 = 26

Question (c)
-79
Solution:
Successor of -79 = -79 + 1 = -78
Predecessor of -79 = -79- 1 = -80

Question (d)
0
Solution:
Successor of 0 = 0+ 1 = 1
Predecessor of 0 = 0 – 1 = -1

Question (e)
29
Solution:
Successor of 29 = 29 + 1 = 30
Predecessor of 29 = 29 – 1 = 28

Question (f)
-18
Solution:
Successor of -18 = -18 + 1 = -17
Predecessor of -18 = -18 – 1 = -19

Question (g)
-21
Solution:
Successor of -21 = -21 + 1 = -200
Predecessor of -21 = -21 – 1 = -22

Question (h)
99
Solution:
Successor of 99 = 99 + 1 = 100
Predecessor of 99 = 99 – 1 = 98

Question (i)
-1
Solution:
Successor of-1 = -1 + 1 = 0
Predecessor of -1 = -1 – 1 = -2

Question (j)
-13.
Solution:
Successor of -13 = -13 + 1 = -12
Predecessor of -13 = -13 – 1 = -14

5. Complete the following addition table:

+	-3	-4	-2	+1	+2	+3
-2
-3
0
+1
+2

Solution:

+	-3	-4	-2	+ 1	+2	+3
-2	-5	-6	-4	-1	0	+ 1
-3	-6	-7	-5	-2	-1	0
0	-3	-4	-2	+ 1	+2	+3
+ 1	-2	-3	-1	+2	+3	+4
+2	-1	-2	0	+3	+4	+5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

Question 1.
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at D show that,

(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP ≅ ∆ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Answer:
∆ ABC and ∆ DBC are isosceles triangles on the same base BC.
∴ In ∆ ABC, AB = AC and in ∆ DBC, DB = DC.
In ∆ ABD and ∆ ACD,
AB = AC
DB = DC
and AD = AD (Common)
∴ ∆ ABD s ∆ ACD (SSS rule) [Result (i)]
∴ ∠ BAD = ∠ CAD (CPCT)
In ∆ ABP and ∆ ACP
AB = AC
∠ BAP = ∠ CAP (∵ ∠ BAD = ∠ CAD)
and AP = AP (Common)
∴ ∆ ABP ≅ ∆ ACP (SAS rule) [Result (ii)]
∴ BP = CP (CPCT)
In ∆ DBP and ∆ DCR
DB = DC
BP = CP
and DP = DP (Common)
∴ ∆ DBP ≅ ∆ DCP (SSS rule)
From ∆ ABP ≅ ∆ ACR ∠ BAP = ∠ CAP (CPCT)
∴ AP bisects ∠A.
From ∆ DBP ≅ ∆ DCR ∠BDP = ∠ CDP (CPCT)
∴ DP bisects ∠D.
Thus, AP bisects ∠A as well as ∠D. [Result (iii)]
∆ A ABP ≅ ∆ ACP
∴ BP = CP and ∠ APB = ∠ APC (CPCT)
But, ∠ APB + ∠ APC = 180° (Linear pair)
∴ ∠ APB = ∠ APC = \(\) = 90°
Thus, BP = CP and AP ⊥ BC.
∴ AP is the perpendicular bisector of BC. [Result (iv)]

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠ A

Answer:
AD is an altitude of A ABC.
∴ ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
hypotenuse AB = hypotenuse AC (Given)
∠ ADB = ∠ ADC (Right angles)
AD = AD (Common)
∴ ∆ ADB ≅ ∆ ADC (RHS rule)
∴ BD = CD and ∠ BAD = ∠ CAD (CPCT)
Now, BD = CD means D is the midpoint of BC.
Hence, AD bisects BC. [Result (i)]
Moreover, ∠ BAD = ∠ CAD and
∠ BAD + ∠ CAD = ∠ BAC.
Hence, AD bisects ∠A. [Result (ii)]

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see the given figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR

Answer:
In ∆ ABC, AM is a median.
∴ BM = CM = \(\frac{1}{2}\) BC
In ∆ PQR, PN is a median.
∴ QN = RN = \(\frac{1}{2}\) QR
Now, BC = QR (Given)
∴ \(\frac{1}{2}\) BC = \(\frac{1}{2}\) QR
∴ BM = QN
In ∆ ABM and ∆ PQN,
AB = PQ (Given)
AM = PN (Given)
BM = QN (Proved)
∴ ∆ ABM ≅ ∆ PQN (SSS rule) [Result (i)]
∴ ∠ ABM = ∠ PQN (CPCT)
∴ ∠ ABC = ∠ PQR
Now, in ∆ ABC and ∆ PQR,
AB = PQ
∠ ABC = ∠ PQR
BC = QR .
∴ ∆ ABC ≅ ∆ PQR (SAS rule) [Result (ii)]

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:
In ∆ FBC and ∆ ECB,
CF = BE (Given)
∠ CFB = ∠ BEC = 90° (Given)
BC = CB (Common)
∴ A FBC ≅ A ECB (RHS rule)
∴ ∠ FBC = ∠ ECB (CPCT)
∴ ∠ ABC = ∠ ACB
Now, in ∆ ABC, ∠ ABC = ∠ ACB
∴ AC = AB (Theorem 7.3)
Hence, ∆ ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
Answer:

In ∆ ABC, AP is an altitude.
∴ ∠ APB = ∠ APC = 90°
In ∆ APB and ∆ APC,
∠ APB = ∠ APC = 90°
AB = AC (Given)
AP = AP (Common)
∴ ∆ APB ≅ ∆ APC (RHS rule)
∴ ∠ ABP = ∠ AGP (CPCT)
∴ ∠ ABC = ∠ ACB
Thus, in ∆ ABC, ∠ B = ∠ C.