Bhagya

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 7 Geometry Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 7 Geometry Ex 7.3

Question 1.
Circle the symmetrical figures of the following :

Solution:
Symmetrical figure : Aeroplane.

Question 2.
Draw symmetry line in the following :

Solution:

3. Draw line of symmetry of the following:

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 4.
Complete the figure, if dotted line is a line of symmetry :

Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 2 Fundamental Operations on Numbers MCQ Questions

Tick (✓) the right answer :

Question 1.
65432 + 34568
(a) 99999
(b) 100000
(c) 10000
(d) 99998.
Answer:
(b) 100000

Question 2.
35406 + 2580 + 43251 = 43251
+ ____ + 35406
(a) 35406
(b) 43251
(c) 2580
(d) 81237
Answer:
(c) 2580

Question 3.
99999 + 0
(a) 99990
(b) 99900
(c) 100000
(d) 99999
Answer:
(d) 99999

Question 4.
100000 – 1 = ___
(a) 10000
(b) 0
(c) 99999
(d) 100000.
Answer:
(c) 99999

Question 5.
Simar has ₹ 5832 and his sister Prabhjot has ₹ 3565. How much amount does Simar have more than his sister ?
(a) ₹ 2267
(b) ₹ 9397
(c) ₹ 2276
(d) ₹ 9973.
Answer:
(a) ₹ 2267

Question 6.
Surjeet has ₹ 50000 in her bank account and her hasband Charan Singh has ₹ 35682 in hisr account. What is total amount in both the accounts ?
(a) ₹ 14318
(b) ₹ 95682
(c) ₹ 85682
(d) ₹ 15318.
Answer:
(c) ₹ 85682

Question 7.
The population of a town is 12078. Out of that the number of men is 4872, women is 4729 and the rest are children. How many children are in the town ?
(a) 2477
(b) 20578
(c) 9601
(d) 8206.
Answer:
(a) 2477

Question 8.
98540 – ___ = 98539
(a) 0
(b) 1
(c) 98540
(d) 98539
Answer:
(b) 1

Question 9.
9999 + ___ = 100000
(a) 1
(b) 0
(c) 90001
(d) 9001.
Answer:
(c) 90001

Question 10.
1000 – __ = 999
(a) 1
(b) 0
(c) 90001
(d) 9001
Answer:
(a) 1

Question 11.
Find the difference between 5-digit smallest number and 4-digit largest number.
(a) 10000
(b) 9999
(c) 1
(d) 0
Answer:
(c) 1

Question 12.
Find the sum of the greatest and smallest 5-digit number using digits 2, 0, 4, 6, 7 ?
(a) 98687
(b) 96887
(c) 55953
(d) 76420.
Answer:
(b) 96887

Question 13.
1500 × 30 × 0 =
(a) 45000
(b) 30
(c) 0
(d) 450.
Answer:
(c) 0

Question 14.
7500 × 40 = 40 × ____
(a) 400
(b) 4000
(c) 750
(d) 7500.
Answer:
(d) 7500.

Question 15.
___ ÷ 100 = 1000
(a) 100
(b) 100000
(c) 10000
(d) 10.
Answer:
(b) 100000

Question 16.
The cost of a book ₹ 79. What is the cost of 12 books ?
(a) ₹ 948
(b) ₹ 938
(c) ₹ 790
(d) ₹ 793.
Answer:
(a) ₹ 948

Question 17.
Geeta has ₹ 175 with her. If she gives ₹ 25 to each child. How many children will get the money ?
(a) 6
(b) 9
(c) 1
(d) 8
Answer:
(c) 1

Question 18.
700 × ___ = 2800 × 1
(a) 5
(b) 6
(c) 4
(d) 3
Answer:
(c) 4

Question 19.
9999 ÷ 1 =
(a) 999
(b) 1
(c) 111
(d) 9999
Answer:
(d) 9999

Question 20.
8899 ÷ 8899 = ____
(a) 0
(b) 1
(c) 2
(d) 8899.
Answer:
(b) 1

Question 21.
99 × 99 = ____
(a) 99
(b) 9801
(c) 9901
(d) 1.
Answer:
(b) 9801

Question 22.
If price of 15 notebooks is ₹ 90. What is the price of one notebook ?
(a) ₹ 3
(b) ₹ 5
(c) ₹ 6
(d) ₹ 6.
Answer:
(c) ₹ 6

Question 23.
The Product of two numbers is 256. If one number is 256 then find the other number.
(a) 1
(b) 2
(c) 0
(d) 256.
Answer:
(a) 1

Question 24.
If 894 × 100 = 89400 then 894 × 10 = ____
(a) 894
(b) 89400
(c) 8940
(d) 8941.
Answer:
(c) 8940

Question 25.
26 ÷ 2 × 4 + 4 – 40 = ___
(a) 64
(b) 8
(c) 4
(d) 16.
Answer:
(d) 16.

Question 26.
A car has four wheels and auto has 3 wheels. Number of wheels of 2 cars and one auto is :

(a) 11
(b) 10
(c) 7
(d) 8.
Answer:
(a) 11

Question 27.
1500 × 30 × 0 = …….
(a) 1530
(b) 0
(c) 1
(d) 1230.
Answer:
(b) 0

Question.
Find the difference of the largest and the smallest number that can be formed by using the digits 5,1, 8, 6 and 7.
Solution:
The largest 5 digit number that can be formed by using the digits 5, 1, 8, 6 and 7 = 87651
The smallest 5 digit number that can be formed by using the digits 5, 1, 8, 6 and 7 = 15678
Their difference =

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 7 Geometry Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 7 Geometry Ex 7.2

1. Measure the following angles using protractor:

Question 1.

Solution:
70°

Question 2.

Solution:
105°

Question 3.

Solution:
90°

Question 4.

Solution:
130°

Question 5.

Solution:
90°

Question 6.

Solution:
115°

Question 7.

Solution:
20°

Question 8.

Solution:
50°

Question 9.

Solution:
35°

Question 10.

Solution:
50°

2. Draw the following angles by using a protractor:

Question 1.
15°
Solution:

Question 2.
40°
Solution:

Question 3.
42°
Solution:

Question 4.
53°
Solution:

Question 5.
65°
Solution:

Question 6.
75°
Solution:

Question 7.
90°
Solution:

Question 8.
110°
Solution:

Question 9.
117°
Solution:

Question 10.
135°
Solution:

Question 11.
157°
Solution:

Question 12.
180°
Solution:

3. Pick out the acute angle, obtuse angle and right angle from the following :

Question 1.
35°
Solution:
Acute angle

Question 2.
89°
Solution:
Acute angle

Question 3.
120°
Solution:
Obtuse angle

Question 4.
100°
Solution:
Obtuse angle

Question 5.
96°
Solution:
Obtuse angle

Question 6.
74°
Solution:
Acute angle

Question 7.
62°
Solution:
Acute angle

Question 8.
166°.
Solution:
Obtuse angle.

4. Fill in the blanks :

Question 1.
An angle between 0° and 90° is called ………………
Solution:
Acute angle

Question 2.
175° angle is ……………… angle.
Solution:
Obtuse

Question 3.
The hands of a clock make an angle of ……………….. at 3 a.m.
Solution:
right angle

Question 4.
Measurements of an angle between North and South direction is ………………….
Solution:
180°

Question 5.
An acute angle is ……………….. than right angle.
Sol.
smaller.

5. Tick the True and False :

Question 1.
Measurement of right angle is 90°.
Solution:
True

Question 2.
Right angle is greater than acute angle but smaller than obtuse angle.
Solution:
True

Question 3.
On the Internal and External scale of protractor, measurements are written up to 90°.
Solution:
False

Question 4.
85° is a right angle.
Solution:
False

Question 5.
115° is an obtuse angle.
Solution:
True

Question 6.
90° is an acute angle.
Solution:
False

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers InText Questions and Answers.

PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers InText Questions

Try These : (Textbook Page No.27)

Question 1.
Solve.

(a)

Solution:

(b)

Solution:

(c)

Solution:

(d)

Solution:

Question 2.
Fill in the blanks :
(a) 115 + 327 = 327 + ____
(b) 321 + 0 = ___
(c) 139 × 1 = ___
(d) 625 × 0 = ___
(e) 339 – 0 = ___
(f) 119 ÷ 119 = ___
(g) 128 ÷ 16 = ___
(h) 720 + 500 = ___
(i) 10000 ÷ 10 = ___
(j) 152 ÷ 19 = ___
Solution:
(a) 115 + 327 = 327 + 115
(b) 321 + 0 = 321
(c) 139 × 1 = 139
(d) 625 × 0 = 0
(e) 339 – 0 = 339
(f) 119 ÷ 119 = 1
(g) 128 ÷ 16 = 8
(h) 720 + 500 = 1220
(i) 10000 ÷ 10 = 100
(j) 152 ÷ 19 = 8

Question 3.
Let’s Do : 19 = 1

(a) In a school, there are 342 boys and 369 girls. How many total number of students are there in the school ?
Solution:

(b) In a godown, there are 459 bags of wheat and 813 bags of rice. How many bags are there in total ?
Solution:

(c) In a year, Harmanpreet Kaur scored 1790 runs and Mitali Raj scored 1299 runs. How many more runs were scored by Harmanpreet Kaur than Mitali Raj ?
Solution:
Harmanpreet Kaur scored runs = 1790
Mitali Raj scored runs = 1299
Number of more runs’ scored by Harmanpreet Kaur than Mitali Raj = 1790 – 1299 = 491.

(d) Harpreet took f 10000 from his father and bought a bicycle for ₹ 3540. How much amount is left with him ?
Solution:

(e) A shopkeeper has 625 packets of toffees. In each packet, there are 100 toffees. How many toffees in total the shopkeeper has ?
Solution:
Number of packets = 625
Number of toffees in each packet =100
Total number of toffees shopkeeper has = 625 × 100 = 62500.

(f) There is 250 litre diesel, in the diesel tank of a truck. It covers 9 km distance with one litre of diesel. How much distance can be covered with the diesel ?
Solution:
Quantity of diesel in the tank = 250 litres
Distance covered with one litre of diesel = 9 km

Total distance covered with the diesel = 9 × 250 km = 2250 km.

(g) In a school, there are 648 students. 18 students can sit in a school van to go for a picnic. How many vans are required to take all the students to picnic ?
Solution:
Total number of students in the school = 648
Number of students that can sit in a van = 18

Number of vans required = 648 ÷ 18
= 36.

(h) In a garden, there are 2568 guava trees. If there are 12 trees in a row then how many rows are there for 2568 guava trees ?
Solution:
Number of guava trees in the garden = 2568
Number of guava trees in a row = 12
Total number of TOWS = 2568 ÷ 12
= 214

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 7 Geometry Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 7 Geometry Ex 7.1

1. Identifythe acute angle, right angle and obtuse angle In the following: 

Question 1.

Solution:
Acute angle

Question 2.

Solution:
Obtuse angle

Question 3.

Solution:
Acute angle

Question 4.

Solution:
Right angle

Question 5.

Solution:
Obtuse Angle

Question 6.

Solution:
Acute angle

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.10 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.10

Solve the following:

Question 1.
42 ÷ 7 + 8
Solution:
42 ÷ 7 + 8 = 6 + 8 = 14

Question 2.
8 + 6 × 2
Solution:
8 + 6 × 2 = 8 + 12 = 20

Question 3.
7 × 8 ÷ 4 – 6
Solution:
7 × 8 ÷ 4 – 6
= 7 × 2 – 6
= 14 – 6 = 8

Question 4.
63 ÷ 9 × 4 + 28 – 15
Solution:
63 ÷ 9 × 4 + 28 – 15
= 7 × 4 + 28 – 15
= 28 + 28 – 15
= 56 – 15
= 41

Question 5.
25 × 3 + 42 ÷ 6 – 4
Solution:
25 × 3 + 42 ÷ 6 – 4
= 25 × 3 + 7 – 4
= 75 + 7 – 4
= 82 – 4
= 78

Question 6.
18 ÷ 6 × 21 + 17 – 18
Solution:
18 ÷ 6 × 21 + 17 – 18
= 3 × 21 + 17 – 18
= 63 + 17 – 18
= 80 – 18
= 62

Question 7.
8 ÷ 8 + 8 × 8 – 8
Solution:
8 ÷ 8 + 8 × 8 – 8
= 1 + 8 × 8 – 8
= 1 + 64 – 8
= 65 – 8
= 57

Question 8.
72 + 48 × 36 ÷ 18 – 9
Solution:
72 + 48 × 36 ÷ 18 – 9
= 72 + 48 × 2 – 9
= 72 + 96 – 9
= 168 – 9
= 159

Question 9.
44 + 2 × 9 – 35 ÷ 5
Solution:
44 + 2 × 9 – 35 ÷ 5 .
= 44 + 2 × 9 – 7
= 44 + 18 – 7
= 62 – 7
= 55

Question 10.
18 + 126 ÷ 14 × 3 – 25
Solution:
18 + 126 ÷ 14 × 3 – 25
= 18 + 9 × 3 – 25
= 18 + 27 – 25
= 45 – 25
= 20

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.9

Question 1.
Find the estimated answers :
(a) 753 + 525
Solution:

(b) 11526 + 8748.
Solution:

(c) 980 – 489
Solution:

(d) 5897 – 2987
Solution:

(e) 440 × 28
Solution:

(f) 6198 × 13
Solution:

(g) 563 ÷ 34
Solution:
563 rounded off = 600
34 rounded off = 30
Estimated value = 600 ÷ 30 = 20

(h) 7541 ÷ 43
Answer:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.8

Question 1.
In a stadium, in the match of cricket there are 84000 people sitting in 24 rows. How many people are sitting in a row?
Solution:
Total number of people = 84000
Number of rows = 24
Number of people sitting in each row
= 84000 ÷ 24
= 3500

Question 2.
You have ₹ 99825 which is to be distributed equally among 33 friends. How much amount will each friend get ?
Solution:
Total amount = ₹ 99825
Number of friends = 33
Each friend will get = ₹ 99825 ÷ 33
= ₹ 3025

Question 3.
My grandfather divided ₹ 72000 equally among four brothers sisters. How much will each get?
Solution:
Total amount = ₹ 72000
Number of brothers and sisters = 4
Each will get
= 72000÷4
= 18000

Question 4.
What number must be multiplied with 26 to get 14508?
Solution:
Number to be obtained = 14508
The given number = 26
The required number = 14508 ÷ 26
= 558

Question 5.
The gardener has 23976 flowers to make garlands. One garland has 24 flowers in it. How many garland can be made from 23976 flowers ?
Solution:
Total number of flowers = 23976
Number of flowers in one garland = 24
Total number of garlands = 23976 ÷ 24
= 999

Question 6.
How many ₹ 2000 notes are there in forty thousand rupees ?
Solution:
Total amount = ₹ 40,000
Value of one note = ₹ 2000
Number of notes = ₹ 40,000 ÷ ₹ 2000
= 20

Question 7.
I need change of ₹ 25,000. How many following notes shall I get ?
(a) Number of notes of ₹ 1000 = ……….
(b) Number of notes of ₹ 500 = ………..
(c) Number of notes of ₹ 100 = ………..
Solution:
Total amount = ₹ 25,000
(a) Number of ₹ 1000 notes
= ₹ 25000 ÷ ₹ 1000

(b) Number of ₹ 500 notes
= ₹ 25000 ÷ ₹ 500

(c) Number of ₹ 100 notes
= ₹ 25000 ÷ ₹ 100

Question 8.
A J.C.B. machine picks 900 bricks in a round. How many rounds will it take to pick 99000 bricks ?
Solution:
Total number of bricks = 99,000
Number of bricks picked in one round = 900
Number of rounds
= 99,000 ÷ 900
= \(\frac{99000}{900}\) = 110

Question 9.
The cost of a railway ticket is ₹ 78. Palak gave ₹ 7722 for buying tickets. How many tickets will she get ?
Solution:
Cost of 1 ticket = ₹ 78
Total amount = ₹ 7722
Number of tickets she will get
= ₹ 7722 ÷ ₹ 78
= 99

Question 10.
A factory manufactures 45540 ice cream cones in the month of June. How many ice cream cones are manufactured in a day ?
Solution:
Total number of ice cream cones manufactured = 45540
Number of the days in the month of June = 30
Number of cones manufactured in one day = 45540 ÷ 30 = 1518

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 8 Perimeter and Area MCQ Questions

Multiple Choice Questions

Tick (✓) the right answer :

Question 1.
Which type of figure is notebook’s page ?
(a) Square
(b) Rectangle
(c) Triangle
(d) Pentagon.
Answer:
(b) Rectangle

Question 2.
What is the perimeter of the square if its side is 6 cm ?
(a) 36 cm
(b) 18 cm
(c) 24 cm
(d) 21 cm.
Answer:
(c) 24 cm

Question 3.
The four sides of square are
(a) different
(b) equal
(c) two equal pairs
(d) none.
Answer:
(b) equal

Question 4.
The length and breadth of rectangle is 6 m and 4 m. Find its perimeter.
(a) 36 m
(b) 16 m
(c) 20 m
(d) 10 m.
Answer:
(c) 20 m

Question 5.
A rectangular park is 65 m long and 35 m wide. Mukesh takes 4 rounds of it. How much distance is covered by him ?
(a) 100 m
(b) 200 m
(c) 400 m
(d) 800 m.
Answer:
(d) 800 m

Question 6.
What will be the area of a square whose side is 13 cm ?
(a) 169 cm
(b) 169 sq. cm
(c) 52 sq. cm
(c) 26 sq. cm.
Answer:
(a) 169 cm

Question 7.
A chart is 125 cm long and 8 cm wide. Its area = ………………..
(a) 100 sq. cm
(b) 1000 sq. cm
(c) 1250 sq. cm
(d) 1100 sq. cm.
Answer:
(b) 1000 sq. cm

Question 8.
If length and breadth of the rectangle is equal then it is called ………………
(a) Rectangle
(b) Length
(c) Square
(d) Perimeter.
Answer:
(c) Square

Question 9.
Side × Side is the area of a ………………
(a) Square
(b) Rectangle
(c) Breadth
(d) Circle.
Answer:
(a) Square

Question 10.
Area of a rectangle is 96 sq. cm. If its length is 12 cm then its breadth is :
(a) 8 cm
(b) 9 cm
(c) 10 cm
(d) 108 cm.
Ans.
(a) 8 cm

Question 11.
Find the area of given rectangle.

(a) 10 sq. cm
(b) 10 cm
(c) 8 sq. cm
(d) 12 sq. cm.
Answer:
(a) 10 sq. cm

Question 12.
Look at the following two figure carefully and select the correct option :

(a) Area of fig. 1 is more than area of fig. 2.
(b) Area of fig. 1 is less than area of fig. 2.
(c) Area of fig. 1 is equal to area of fig. 2.
(d) Area of both the figures is equal.
Answer:
(b) Area of fig. 1 is less than area of fig. 2.

Question 13.
Below is given a picture of a field. Find the area of field.

Answer:
Length of field = 96 m
Breadth of field = 64 m
Area of field = Length × Breadth
= 96 × 64 sq. m
= 6144 sq. m

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 8 Perimeter and Area Ex 8.2

1. Find the area of following rectangles whose length and breadth are as follows :

Question 1.
9 m and 7 m
Solution:
Length of rectangle = 9 m
Breadth of rectangle = 7 m
Area of rectangle = Length × Breadth
= 9 m × 7 m
= 63 m²

Question 2.
85 cm and 76 cm
Solution:
Length of rectangle = 85 cm
Breadth of rectangle = 76 cm
Area of rectangle = Length × Breadth
= 85 cm × 76 cm
= 6460 cm²

Question 3.
23 mm and 18 mm
Solution:
Length of rectangle = 23 mm
Breadth of rectangle = 18 mm
Area of rectangle = Length × Breadth
= 23 mm × 18 mm
= 414 mm²

Question 4.
5 m and 85 cm
Solution:
Length of rectangle
= 5 m
= 5 × 100 cm
= 500 cm
Breadth of rectangle = 85 cm
Area of rectangle = Length × Breadth
= 500 cm × 85 cm
= 42500 cm²

Question 5.
840 cm and 7 m
Solution:
Length of rectangle = 840 cm
Breadth of rectangle = 7 m
= 7 × 100 cm
= 700 cm
Area of rectangle = Length × Breadth
= 840 cm × 700 cm
= 588000 cm²

2. Find the area of a square whose side is :

Question 1.
25 cm
Sol.
Side of square = 25 cm
Area of square = side × side
= 25 cm × 25 cm
= 625 cm²

Question 2.
48 cm
Solution:
Side of the square = 48 cm
Area of the square = side × side
= 48 cm × 48 cm
= 2304 cm²

Question 3.
27 mm
Solution:
Side of the square = 27 mm
Area of the square = side × side
= 27 mm × 27 mm
= 729 mm²

Question 4.
87 m
Solution:
Side of the square = 87 m
Area of the square = side × side
= 87 m × 87 m
= 7569 m²

Question 3.
Find the area of rectangular park whose length is 62 m and breadth is 38 m.
Solution:
Length of the rectangular park = 62 m
Breadth of the rectangular park = 38 m
Area of the rectangular park = Length × Breadth
= 62 m × 38 m
= 2356 m²

Question 4.
The side of a carrom-board is 60 cm. Find its area.
Solution:
Side of the carrom-board = 60 cm
Area of the carrom-board = side × side
= 60 cm × 60 cm
= 3600 cm²

Question 5.
The length and breadth of a rectangular field is 100 m and 45 m. What is the cost of levelling its floor at the rate of ₹ 8 per sq. m?
Solution:
Length of the rectangular field = 100 m
Breadth of the rectangular field = 45 m
Area of the rectangular field = Length × Breadth
= 100 m × 45 m
= 4500 m²
The rate of levelling = ₹ 8 per m²
The cost of levelling = ₹ 8 × 4500
= ₹ 36000

Question 6.
A carpet has a length 8 m and breadth 5 m. In an auditorium, 125 such carpets are being set on the floor. Find the area of the floor of the auditorium.
Solution:
Length of the carpet = 8 m
Breadth of the carpet = 5 m
Area of each carpet = Length × Breadth
= 8 m × 5 m
= 40 m²
Area of 125 carpets = 125 × 40 m²
= 5000 m²
Therefore, area of the floor of the auditorium = 5000 m².

Question 7.
The verandah of Gurpreet’s home is 52 m long and 32 m wide and the verandah of Pankaj’s home is of square shape with side 41 m. which person’s home has a roof of verandah bigger and by how much ?
Solution:
Gurpreet :
Length of verandah = 52 m
Breadth of verandah = 32 m
Area of verandah = Length × Breadth
= 52 m × 32 m
=1664 m²

Pankaj :
Side of square verandah = 41 m
Area of square verandah = Side × side
= 41 m × 41 m
= 1681 m²
Area of verandah of Pankaj’s home is bigger by
= 1681 m² – 1664 m²
= 17 m²

Question 8.
Roof of Amarjeet’s home is of length 9 m and breadth 6 m. There is a leakage of water from the roof. He wants to fix tiles of size 30 cm long and 20 cm wide for plugging the leakage. How many tiles does he need ?
Solution:
Length of the roof = 9 m
= 9 ² 100 cm
= 900 cm
Breadth of the roof = 6 m
= 6 ² 100 cm
= 600 cm
Area of the roof = Length × Breadth
= 900 cm × 600 cm
= 540000 cm²
Length of each tile = 30 cm
Breadth of each tile = 20 cm
Area of each tile = Length × Breadth
= 30 cm × 20 cm
= 600 cm²
Area of the roof
Number of tiles = \(\frac{\text { Area of the roof }}{\text { Area of each tile }}\)
= \(\frac{540000}{600}\) = \(\frac{900 \times 600}{600}\)
= 900.

9. Fill in the blanks :

Question 1.
Area of rectangle = ……………… × ………………
Solution:
Length × Breadth

Question 2.
Area of square = ……………… × ………………
Solution:
side × side

Question 3.
1 sq. m. = ……………… sq. cm.
Solution:
10000

Question 4.
The space covered by a closed figure is called its ………………
Solution:
Area.

Question 10.
Complete the table :

Solution:
(a) 56 m²
(b) 2 cm
(c) 6 mm
(d) 700 cm²